A fun way of understanding Einstein's General Theory of Relativity

A fun way of understanding Einstein's General Theory of Relativity
Einstein’s famous equation. Credit: Pixabay.

Many people fail to realize just how much energy there is locked up in matter. The nucleus of any atom is an oven of intense radiation, and when you open the oven door, that energy spills out; oftentimes violently. However, there is something even more intrinsic to this aspect of matter that escaped scientists for years.

It wasn't until the brilliance of Albert Einstein that we were able to fully grasp this correlation between mass and . Enter E=mc2. This seemingly simple algebraic formula represents the correlation of energy to matter (energy equivalence of any given amount of mass). Many have heard of it, but not very many understand what it implies. Many people are unaware of just how much energy is contained within matter. So, for the next few minutes, I will attempt to convey to you the magnitude of your own personal potential energy equivalence.

First, we must break down this equation. What do each of the letters mean? What are their values? Let's break it down from left to right:

E represents the energy, which we measure in Joules. Joules is an SI measurement for energy and is measured as kilograms x meters squared per seconds squared [kg x m2/s2]. All this essentially means is that a Joule of energy is equal to the force used to move a specific object 1 meter in the same direction as the force.

m represents the mass of the specified object. For this equation, we measure mass in Kilograms (or 1000 grams).

c represents the speed of light. In a vacuum, light moves at 186,282 miles per second. However in science we utilize the SI (International System of Units), therefore we use measurements of meters and kilometers as opposed to feet and miles. So whenever we do our calculations for light, we use 3.00 × 108m/s, or rather 300,000,000 meters per second.

So essentially what the equation is saying is that for a specific amount of mass (in kilograms), if you multiply it by the speed of light squared (3.00×108)2, you get its energy equivalence (Joules). So, what does this mean? How can I relate to this, and how much energy is in ? Well, here comes the fun part. We are about to conduct an experiment.

This isn't one that we need fancy equipment for, nor is it one that we need a large laboratory for. All we need is simple math and our imagination. Now before I go on, I would like to point out that I am utilizing this equation in its most basic form. There are many more complex derivatives of this equation that are used for many different applications. It is also worth mentioning that when two atoms fuse (such as Hydrogen fusing into Helium in the core of our star) only about 0.7% of the mass is converted into total energy. For our purposes we needn't worry about this, as I am simply illustrating the incredible amounts of energy that constitutes your equivalence in mass, not illustrating the fusion of all of your mass turning into energy.

Let's begin by collecting the data so that we can input it into our equation. I weigh roughly 190 pounds. Again, as we use SI units in science, we need to convert this over from pounds to grams. Here is how we do this:

1 Josh = 190lbs
1 lbs = 453.6g
So 190lbs × 453.6g/1 lbs = 86,184g
So 1 Josh = 86,184g

Since our measurement for E is in Joules, and Joule units of measurement are kilograms x meters squared per seconds squared, I need to convert my mass in grams to my mass in kilograms. We do that this way:

86,184g × 1kg/1000g = 86.18kg.

So 1 Josh = 86.18kg.

Now that I'm in the right unit of measure for , we can plug the values into the and see just what we get:

E=mc2
E= (86.18kg)(3.00 × 108m/s)2
E= 7.76 × 1018 J

That looks like this: 7,760,000,000,000,000,000 or roughly 7.8 septillion Joules of energy.

This is an incredibly large amount of energy. However, it still seems very vague. What does that number mean? How much energy is that really? Well, let's continue this experiment and find something that we can measure this against, to help put this amount of energy into perspective for us.

First, let's convert our energy into an equivalent measurement. Something we can relate to. How does TNT sound? First, we must identify a common unit of measurement for TNT. The kiloton. Now we find out just how many kilotons of TNT are in 1 Joule. After doing a little searching I found a conversion ratio that will let us do just this:

1 Joule = 2.39 × 10-13 kilotons of explosives. Meaning that 1 Joule of energy is equal to .000000000000239 kilotons of TNT. That is a very small number. At better way to understand this relationship is to flip that ratio around to see how many Joules of energy is in 1 kiloton of TNT. 1 kiloton of TNT = 4.18×1012 Joules or rather 4,184,000,000,000 Joules.

Now that we have our conversion ratio, let's do the math.

1 Josh (E) = 7.76 x 1018 J
7.76 x 1018 J x 1 kT TNT/ 4.18 x 1012 J = 1,856,459 kilotons of TNT.

Thus, concluding our little mind experiment we find that just one human being is roughly the equivalence of 1.86 MILLION kilotons of TNT worth of energy. Let's now put that into perspective, just to illuminate the massive amount of power that this equivalence really is.

The bomb that destroyed Nagasaki in Japan during World War II was devastating. It leveled a city in seconds and brought the War in the Pacific to a close. That bomb was approximately 21 kilotons of explosives. So that means that I, 1 human being, have 88,403 times more explosive energy in me than a bomb that destroyed an entire city… and that goes for every human being.

So when you hear someone tell you that you've got real potential, just reply that they have no idea….

A fun way of understanding Einstein's General Theory of Relativity
Hydrogen Bomb Blast. Credit: Pixabay.

Explore further

Superman's solar-powered feats break a fundamental law of physics

Source: Universe Today
Citation: A fun way of understanding Einstein's General Theory of Relativity (2014, September 25) retrieved 25 May 2019 from https://phys.org/news/2014-09-fun-einstein-theory-relativity.html
This document is subject to copyright. Apart from any fair dealing for the purpose of private study or research, no part may be reproduced without the written permission. The content is provided for information purposes only.
1 shares

Feedback to editors

User comments

Sep 25, 2014
E represents...
m represents...
c represents...

The really vexing thing is (and I suspect it'd be an instant Nobel prize if someone figured it out): what is the physical meaning of c^2 ?
Energy and mass are easily visualized, but the speed of light squared (and why squared in the first place but not cubed or linear?) - that's the real poser.

So that means that I, 1 human being, have 88,403 times more explosive energy in me than a bomb that destroyed an entire city

To get it out you'd need an equal amount of antimatter, though (which would give you twice as large a bang.). So as long as no one chucks around copious amounts of antimatter we're pretty good/safe storage for that amount of energy.

Sep 25, 2014
This is not an article about understanding the General Theory of Relativity. E=mc^2 is the domain of Special Relativity, not General. This article is not about understanding a theory. It's about understanding the meaning of an equation.

Sep 25, 2014
The strong atomic force is also relativity and this equation can be demonstrated using simple combustion.

It is c**2 due to units, as any freshman physics student knows.

Sep 25, 2014
I still do not understand why we are more TNT than a bomb?

Sep 25, 2014
what is the physical meaning of c^2 ?
Energy and mass are easily visualized, but the speed of light squared (and why squared in the first place but not cubed or linear?) - that's the real poser.

Well, I am not sure what kind of explanation you are looking for, but one short (and slightly unsatisfactory, though important) comes from dimensional analysis. If Energy is proportional to mass, then we must find a factor that gives the RHS of the equation the correct units.

E has units kg*m^2/s^2. So, since mass has units pf kg, then we must multiply mass by some constant with unites m^2/s^2.

A longer answer is that relativity predicts that momentum is of the form m*v/sqrt(1-v^2/c^2). If we use this to evaluate the work integral, we get E^2 = m^2c^4 + p^2c^2. So, when the object is at rest, p=0, and E = mc^2. (to be continued)

Sep 25, 2014
(continued)
Perhaps the best answer (to me), though, comes from Fritz Rohrlich, who gave the thought experiment of an object that, in the rest frame, radiates photons to the right and left with equal frequency. In the moving frame, these photons would be respectively red shifted, and blue shifted. However, photons carry momentum that depends on their frequency. Their momenta would thus not be equal in magnitude. Since the object never changed velocity, the only way for momentum to be preserved is if there were an equivalent amount of mass lost with the radiation of this energy. Hence, E = mc^2

Sep 25, 2014
The photon is a violation of General Relativity, as the proof of a photon is identical to the proof of General Relativity; a photograph of a 1919 solar eclipse. Insight into further knowledge based on the photon are as imaginary as the photon itself.

Light has no mass and is never a padticle.

Sep 25, 2014
The photon is a violation of General Relativity, as the proof of a photon is identical to the proof of General Relativity; a photograph of a 1919 solar eclipse. Insight into further knowledge based on the photon are as imaginary as the photon itself.

What?

Light has no mass and is never a padticle.

There are different types of energy, such as kinetic, heat, electrical, and so on, and one type of energy can be converted into another by various physical processes. Mass stores a certain amount of energy--mass energy. An analogous classical concept is the energy stored in a spinning flywheel. In this case, the energy is associated with the moment of inertia. That energy can be converted to kinetic energy, or to EM energy. That does not imply that either the kinetic energy or EM energy have moments of inertia.

As for light not being a particle, that's not true. We have overwhelmingly demonstrated the existence photons in various experiments.

Sep 25, 2014
In fact, there is a great deal of misrepresentation in this.
Among other things, the only case in which, purportedly, energy is turned to matter and vice versa, the decay of radioactive nuclei. But, remember, among other things, the mass in the equation is rest mass. A nucleus is not a solid particle, it is subatomic particles moving about each other at high speeds. And, when they decay, the particles are not at rest, again, they are moving at high rates of speed. There is no real derivation of rest mass.
In fact, if you look at the kinematic derivation of the formula, the formula acts as the zero point for "relativistic" mechanical energy at velocities, the "relativistic" kinetic energy is an expression added to the equation. To say that that necessarily has meaning is to "conclude", from the formula of the parabolic path of a tossed object, that, when it reaches the ground, it must travel underground.

Sep 25, 2014
Light is completely described by Maxwell and is always a wave. If light had mass it would be described by General Relativity.

Sep 25, 2014
nkalanaga: Thanks for trying to explain but, I need a new brain. I am now more confused than ever. As long as things work the way they do, I am fine with that.

Sep 25, 2014
Light is completely described by Maxwell and is always a wave.

No, it isn't. You can't solve the Ultraviolet catastoprophe by assuming that light is completely described by Maxwell's equations. You also can't explain, using Maxwell's equations, the photoelectric effect, nor how the smallest amounts of light always appears in localized packets against photographic plates.
If light had mass it would be described by General Relativity.

Nobody but you is saying that light has mass. Light is one type of energy that can result from getting rid of mass. This is because mass has mass-energy, just like a spinning object has rotational energy.

Sep 25, 2014
The energy to move an electron from one shell to another is quantized, not light. You are confusing fascist politics for science.

Sep 25, 2014
t is c**2 due to units, as any freshman physics student knows.

Units is not a physical explanation. That's just 'making it fit so it looks nice'.
I am aware of the way the formula is derived. Still, despite its unarguable success, it is somewhat irksome that there's no fundamental understanding (yet) of how to tie that factor into some physical reality.

That it's a square suggests a topological explanation (curvature?). But what of is unfortunately unclear. Probably every student of physics has tried to crack that nut at one time or another. I guess we need a new Einstein before it comes clear.

Sep 25, 2014
Actually, a dimensional analysis is basic physics. It is a requirement that the units for energy are on both sides of the equal sign. It is derived from Maxwell as Maxwell is the basis of Relatvity.

Sep 25, 2014
Sorry to ignore your curvature comment, if we add some trig to GR to account for the spral nature of our galaxy the dark matter disappears. A little more physical in the curvatufe.

Sep 25, 2014
The energy to move an electron from one shell to another is quantized, not light.

Well, yes, that energy required is quantized, but that doesn't change the fact that small units of light are absorbed in packets. In other words, photons don't disperse. They are always absorbed as discrete packets. Again, if light could purely be described by Maxwell's equations, then photons would disperse. They don't.
You are confusing fascist politics for science.

Sheesh! What is it about crackpots and linking scientific discourse with fascism?

You are so close to satisfying Godwin's law. I might as well get it out of the way, so that we can move on. HITLER! HITLER! SS! NAZIS! MUSSOLINI! HITLER!

There, now can we please stop talking about silliness like fascism being the arbiter of peer review, and talk instead about actual physics arguments? Or are you going to keep screaming "fascism!!1!" in place of anything meaningful?

Sep 25, 2014
it is somewhat irksome that there's no fundamental understanding (yet) of how to tie that factor into some physical reality.

By fundamental, do you mean geometric? That's what the next thing you write seems to imply.
That it's a square suggests a topological explanation (curvature?).

Why?
But what of is unfortunately unclear. Probably every student of physics has tried to crack that nut at one time or another. I guess we need a new Einstein before it comes clear

I confess that this has never bothered me. I would think it has to do with how EM waves impart energy. I'd have to think about it more, though. What bothers me much more is why the first postulate of relativity is true in the first place. People just seem to accept is as a postulate, and nobody seems to be concerned with why it should be that way. It seems like if we could understand that, maybe we could find a way to subvert it.

Sep 25, 2014
it is somewhat irksome that there's no fundamental understanding (yet) of how to tie that factor into some physical reality.

That it's a square suggests a topological explanation (curvature?). But what of is unfortunately unclear. Probably every student of physics has tried to crack that nut at one time or another. I guess we need a new Einstein before it comes clear.
In my head I've always pictured fundamental particles with mass being like spacetime that has curled up into a tiny little sphere with photons racing around trapped inside of it, but not like a black hole or anything. I'm quite sure that this won't stand up to even the most cursory scrutiny, but it gives me a warm and fuzzy feeling of something moving inside at speed c that kinda-sorta makes it feel like the mass is related to the kinetic energy of the trapped photons except for that pesky 1/2 thingy in (1/2)mv^2 with v=c. It gets me through life while I wait for someone a lot smarter than me to figure it out.

Sep 25, 2014
The energy well for moving an electron from one shell to another is a discrete energy packet, not EM. The photon is imaginary and is a political deal to pay for scuence.

Sep 25, 2014
The energy well for moving an electron from one shell to another is a discrete energy packet, not EM. The photon is imaginary and is a political deal to pay for scuence.
What are the characteristics of this energy packet? Where did it come from? Does it contain an electric field? What happens to the discrete energy packet after it's done moving the electron? Does another discrete energy packet move the electron to a lower energy shell? Where do used discrete energy packets go? Does they move away? If so, how fast do they move? If they can move, what is the mechanism?

Sep 25, 2014
Once the energy well is filled that energy is consumed by the transition to the next shell. That same quantity of energy is released should the electron transition back to the lower shell. The EM returns to it's wave form.

Sep 25, 2014
This comment has been removed by a moderator.

Sep 25, 2014
There is a bunch of fake science from fascist theology. Pastor Nye evangelizes it.

Sep 26, 2014
Once the energy well is filled that energy is consumed by the transition to the next shell. That same quantity of energy is released should the electron transition back to the lower shell. The EM returns to it's wave form.
What fills the energy well? Why doesn't the electron move part way toward the next shell when the energy well is only partially full? Since photons don't exist, an EM wave requires an oscillating electric dipole to propagate it. Does the energy well operate as an oscillating electric dipole? If so, why doesn't the well constantly dissipate EM energy? Does the frequency of the propagated EM wave get lower as the energy well loses it's energy to the EM wave? If not, then, assuming that the energy well releases an EM wave of a specific frequency, and from the fact that you cannot emit a fraction of an EM wave cycle, how many integral EM wave cycles does the energy well hold and what determines the frequency?

Sep 26, 2014
Actually, a dimensional analysis is basic physics.

Yes it is: but it doesn't MEAN anything. You can do all kinds of absurd things with dimensional analysis and come out with the right units (but completley nonsensical results). That doesn't mean that any times the units fit you have a good formula on your hands. At some point you have to be able to say WHY you use the particular components you do.

By fundamental, do you mean geometric? That's what the next thing you write seems to imply.

Yes. That it's topological in nature is just a hunch. It has the 'smell' of a (spherical?) surface to me. I had a (very naive) go at tying it in with the holographic theory of spacetime at one point, but no luck.

Sep 26, 2014
This comment has been removed by a moderator.

Sep 26, 2014
This comment has been removed by a moderator.

Sep 26, 2014
It is refreshing to hear sense and not the usual garbage that the furlong is posting. A continuous light wave need not consist of "photons" to eject electrons when it irradiates a metal.

Sigh. And yet, whenever we fire light waves with energy h*c/lambda, where h is planck's constant, and lambda is the wavelength, despite your protestations, it never disperses, so it is not an EM wave. It is a quantization of the EM field, which is different.

But, suppose you were correct, and it were an EM wave. You would have problems. First, it would disperse. So, as the wavefront expands, the energy gets distributed uniformly across the wave front. However, from experiment, we always ever see the photon get absorbed locally. Hence, due to conservation of energy, the particle that absorbs the photon must somehow use up all the energy carried by this wave, meaning that it must somehow sap up the entire expanding wave front. This is not what happens in classical EM.

Sep 26, 2014
Light has no mass and is never a padticle.
A photon is a coherent electromagnetic wave with a single frequency nu, and its energy is the lowest energy that a source can emit. This energy, its shape and size, like that of any EM-wave, is determined by the source and it will not spread like a wave-packet.

However, when it encounters new boundary conditions, its shape and size morphs to fit the new boundary condictions. For this reason it can form two lobes each of which moves through a slit when a photon-wave encounters two slits. This is what waves do and have always done. Why claim that a photon-wave s a "particle"? It is insane!

Obiously according to Maxwell's equations a photon-wave MUST consist of distribured EM energy which when integrated gives E=h*nu. But energy is m*c^2. Thus the distributed EM energy is alo distributed mass energy: And distributed mass ALWAYS has a centre-of-mass. (COM) Thus a photon has mass m=(h*nu)/c^2 and momentum p=m*c acting at its COM.

Sep 26, 2014
But, suppose you were correct, and it were an EM wave. You would have problems. First, it would disperse.
Why? A light-pulse with a single frequency emitted by laser source does not disperse. So why must a light pulse with a single frequency nu, butwith less energy h*nu disperse?

So, as the wavefront expands, the energy gets distributed uniformly across the wave front.
This is what happes when a photon wave has moved through two slits
However, from experiment, we always ever see the photon get absorbed locally.
This must obviously be so since the absorber is not the size of the screen but the size of a electronic atomic orbital.
Hence, due to conservation of energy, the particle that absorbs the photon must somehow use up all the energy carried by this wave, meaning that it must somehow sap up the entire expanding wave front.
Correct! The expanded wavefront resonates with the atomic electrons and collapses in size. A SPOT is formed!!

Sep 26, 2014
Correct! The expanded wavefront resonates with the atomic electrons and collapses in size. A SPOT is formed!!

No, Johan, that's not how EM works. I mean, what does that even mean? Have you ever seen a wave behave like that in reality? When a tremor radiates through the ground during an earthquake, and a building resonates with it, does the building absorb the entire tremor? God, your brain is broken.

...and what happens when you have two electrons positioned equidistant from the wave, then?
The wave would resonate with them, would they not? Which electron absorbs the wave?

Sep 26, 2014
This is not what happens in classical EM.
Oh yes an absorber of radio-waves also collapses EM-energy to appear within the antenna of the radio, no matter how spread out the wavefronts are.

Simlarly in the case of electron-"antennas" within metal, the electron (radio-antenna) determines how much energy it must disentangle and absorb. In the latter case an electron-antenna cannot disentangle more energy than h*nu. So even when sending in a homogeneous laser beam, an electron in the metal will slice off (disentangle) light of energy h*nu. This does NOT require that the impinging light consists of separate photon-waves. A single light-wave is thus a uniform Maxwell-field; NOT a "quantized-field".

Sep 26, 2014
Correct! The expanded wavefront resonates with the atomic electrons and collapses in size. A SPOT is formed!!


No, Johan, that's not how EM works. I mean, what does that even mean? Have you ever seen a wave behave like that in reality?
Yes every morning when I switch on my radio and tune it to resonate with a carrier wave, the resonance collapses EM energy into the antenna of my radio-wave. But ONLY when thre i resonace!
When a tremor radiates through the ground during an earthquake, and a building resonates with it, does the building absorb the entire tremor?
It depends on the building and the energy of the tremor. In a metal the electron is the "building" and it also does NOT absorb the wholeSINGLE laser wave if you illuminate the metal with a SINGLE coherent wave with frequency nu. It only absorbs what it can absorb h*nu: Just like the building!!!! I nearly called you a knucklehead again". But I will refrain.

Sep 26, 2014
Oh yes an absorber of radio-waves also collapses EM-energy to appear within the antenna of the radio, no matter how spread out the wavefronts are.

Haha. What?

No, Johan, that's not how reality works. So, what happens when you turn on a radio, Johan? By your logic, no other radio would work, because it would absorb all of the waves. Gee, I wonder why I don't believe you when you claim to be an accredited physics researcher.

Sep 26, 2014
...and what happens when you have two electrons positioned equidistant from the wave, then? The wave would resonate with them, would they not? Which electron absorbs the wave?
Congratulations! For a change you asked an intelligent question. Yes it can resonate with either one of them and will collapse within the one it resonates first. A single photon-wave thus has a 50/50 probability to collpase into the one or the other.

That is why when you have two detectors behind the slits of a Young's apparatus, the photon-wave THAT MOVED THROUGH BOTH SLITS, only collapses into one of the two detectors. If you are MORON, you will conclude that the photon is a "particle" which ONLY moved through the slit at which the detector in which it collapsed is pointed.

In fact he experiment proves that you collapsed the diffracted wavefronft before it reached the screen since the diffraction pattern disappears. Why invoke Voodoo and claim that the photon wave is a probability distribution?


Sep 26, 2014
This is weird... ignorants arguing with PhD's...

Sep 26, 2014
This comment has been removed by a moderator.

Sep 26, 2014
]Because you are too stupid to understand even the most simple physics like that of EM-waves.


Skippy why you got to call him stupid?

Maybe he just understands it different than the way you misunderstand it, eh Cher. You ever think about that? He does a good job of explaining things that doesn't sound stupid to me.

Now if I was to try to explain it, you could call me stupid and maybe I would call me that too if I tried to explain it. But the furlong-Skippy does a lot better than most peoples here.

Sep 26, 2014
So that means that I, 1 human being, have 88,403 times more explosive energy ..
- article

It's the energy required for containment.

To get it out you'd need an equal amount of antimatter,
- AA

The antimatter for complete release of the contained energy would need to be constituted from a full 'mirror' of antimatter. Eg: A lump of anti-hydrogen of the equivalent mass to the human subject would certainly liberate energy but not completely. Atomic weights.


Sep 26, 2014
I have to point out a numerical prefix mis-use, and I hope the author fixes it because it drives me bananas to see it.

Here's the offending section:

E= 7.76 × 1018 J
That looks like this: 7,760,000,000,000,000,000 or roughly 7.8 septillion Joules of energy.

In American usage (as opposed to the British usage, which would refer to an even larger number) septillion refers to a number with 24 digits after the first comma (a number with 25-27 digits before the decimal point.)

Here's the usage progression of these prefixes
billions 7,760,000,000
trillions 7,760,000,000,000
quadrillions 7,760,000,000,000,000
quintillions 7,760,000,000,000,000,000 what the author was trying to reference
sextillions 7,760,000,000,000,000,000,000
septillions 7,760,000,000,000,000,000,000,000 what he ended up referencing.

Of course he could have just stuck with the digits themselves.

Sep 26, 2014
Congratulations! For a change you asked an intelligent question.

If you think I said something intelligent, I must be doing something wrong.
Yes it can resonate with either one of them and will collapse within the one it resonates first. A single photon-wave thus has a 50/50 probability to collpase into the one or the other.

Haha, uh, no. The electrons are placed equidistant from the wave source. By symmetry, both electrons will interact with the wave in the same exact manner. EM is a deterministic theory. There is no chance involved. Either both will be excited by the wave or neither. You can't just choose one or the other.

Sep 26, 2014
Yes it can resonate with either one of them and will collapse within the one it resonates first. A single photon-wave thus has a 50/50 probability to collpase into the one or the other.

The electrons are placed equidistant from the wave source. By symmetry, both electrons will interact with the wave in the same exact manner. EM is a deterministic theory.
Stop being such a fool. Light-waves and resonance with matter is totally modelled by Maxwell's equations; also in this case!

Even if these electrons resonate simultaneously with the photon-wave, they cannot each absorb 1/2 of h^nu since each cannot absorb less than h*nu and also not more than h*nu. Thus, if there is only energy h*nu, it has to end up in either the one or the other. If the wave has energy that is much larger than h*nu, like a continuous laser pule wave, each can simultaneously absorb energy h*nu; and even more electrons can each slice off energy h*.nu. This is how EM wave-energy is absorbed!!

Sep 26, 2014
Even if these electrons resonate simultaneously with the photon-wave, they cannot each absorb 1/2 of h^nu since each cannot absorb less than h*nu and also not more than h*nu. Thus, if there is only energy h*nu, it has to end up in either the one or the other.

No. Your reasoning is bad, and you should feel bad. If neither electron can absorb 1/2 of the energy, and the situation is symmetric, then NEITHER WILL absorb the energy. As I said, EM is DETERMINISTIC. You can't just roll dice and have one electron absorb the energy half the time, and have the other absorb energy half the time--not when the situation is symmetric...unless, of course, you have a model that is actual probabilistic. Probabilistic...probabilistic...where have I heard of that before?

Oh! Right! Quantum Mechanics!

I mean, do you even understand the concept of determinism? That if you have the same initial conditions, you must get the same outcome every single time? Does that ring a bell?

Sep 26, 2014
In my head I've always pictured fundamental particles with mass being like spacetime that has curled up into a tiny little sphere with photons racing around trapped inside of it, but not like a black hole or anything. I'm quite sure that this won't stand up to even the most cursory scrutiny, but it gives me a warm and fuzzy feeling of something moving inside at speed c that kinda-sorta makes it feel like the mass is related to the kinetic energy of the trapped photons except for that pesky 1/2 thingy in (1/2)mv^2 with v=c. It gets me through life while I wait for someone a lot smarter than me to figure it out.
LOL! Someone actually wasted the time to give this obvious cheeky response a 1! What a loser! Please, by all means 1 the crap out of it and this one too! I should turn it into the next crackpot theory and use it to troll the site - compactedspacetimephotodynamics! Hell, this article's comment section is replete with crackpots! If you can't beat em' join em! LOL!

Sep 26, 2014
This comment has been removed by a moderator.

Sep 26, 2014
LOL! Someone actually wasted the time to give this obvious cheeky response a 1! What a loser!

I gave your comment a 5 to offset the 1. I don't think it merited a 1/5. I mean, there's nothing wrong with speculation. It's not like you didn't also write,
I'm quite sure that this won't stand up to even the most cursory scrutiny

And
It gets me through life while I wait for someone a lot smarter than me to figure it out.

I wish all the crackpots on this site would take a look at the difference between what you wrote and what they write. You didn't pretend to be an authority, or smarter than everyone else, and most of all, you didn't act like scientists are incompetent buffoons. Keep doing what you're doing. I, for one, don't mind.
Hell, this article's comment section is replete with crackpots! If you can't beat em' join em! LOL!

Haha. Yeah. In reading these comments, you'd think physorg was a misnomer.

Sep 26, 2014
I mean, do you even understand the concept of determinism? That if you have the same initial condition


This does not apply to waves and resonance interactions; as known since 1850. At least those of us who have solved Maxwell's equations for electromagnetic wave-fields subject to different boundary conditions. It is twerps like you who have NEVER done any Engineering Science who spout the excrement that you are spouting!

Maxwell's equations already contain ALL the physics required to model the interactions of electrons with light-waves on the atomic scale; provided you understand how to apply and solve these equations.At the beginning of the 20th century the leading scientists like Lorentz an Poincare failed to do this simple exercise!

It is insane to argue that there is a Voodoo-beak in wave mechanics from Maxwell's equations to Bohr's "wave-particle" duality. Everything can be modelled in terms of Maxwell's equations; unless you want to stay within the looney bin!!

Sep 26, 2014
Of course all physics is deterministic, but when you roll dice or shoot a marble at different possibilities of which ONLY one can manifest one has to use statistics: But the latter is REAL statistics NOT VOODOO statistics as Born in his utter stupidity has postulated!!

Sigh. No, Johan, that's not how dice or marbles work, either. You see, when you have the same, exact initial conditions, you will always have the same, exact outcome. Now, of course where those are concerned, there is sensitivity to initial conditions, so a slight deviation from the initial setup will lead to drastically different outcomes--but it is still deterministic. If you roll a die exactly the same way both times, assuming classical mechanics, you will get the same, exact, outcome every time. The same is true for charged objects that resonate with EM waves. You are just wrong. Just. Stop. Seriously, I am tired of correcting every single thing you say.

Sep 26, 2014
This does not apply to waves and resonance interactions; as known since 1850. At least those of us who have solved Maxwell's equations for electromagnetic wave-fields subject to different boundary conditions.

No, Johan, I have solved Maxwell's equations a few times, now. I could probably show you a thing or two concerning them.
Maxwell's equations already contain ALL the physics required to model the interactions of electrons with light-waves on the atomic scale;

Ohhh, except for the ultra-violet catastrophe, the photoelectric effect, the Aharonov–Bohm effect, the Stark effect, the Zeeman effect, etc, but why consider all discoveries up to the present, when you are so comfortable focusing on one narrow window of physics results from 100+ years ago?

Sep 26, 2014
Sigh. No, Johan, ... You see, when you have the same, exact initial conditions, you will always have the same, exact outcome.
Sigh. If you have EVER solved Maxwell's wave equations, you will realise that a change in boundary conditions changes the wave Heisenberrg stupidly called these changes during resonance "uncertainties" since resonance changes the motion, shape and size of a Maxwell lightwave.

The concept of "uncertainties" are Voodoo concepts, which you believe in as if they are built into the laws of nature. If you have a wave of a certain shape and size emitted with the same boundary conditions from a source, it will move deterministically until it encounters a change in boundary conditions. If this change only allows a SINGLE outcome, YOU WILL HAVE A SINGLE OUTCOME. But if the boundary conditions allow different outcomes FOR IDENTICAL WAVES, you will have different outcomes. It happens for purely classical radio-waves every day! SIGH!!

Sep 26, 2014
This comment has been removed by a moderator.

Sep 26, 2014
Maxwell's equations already contain ALL the physics required to model the interactions of electrons with light-waves on the atomic scale;

Ohhh, except for the ultra-violet catastrophe,
It is fully explained in my publications
the photoelectric effect,
It is fully explained in my publications
the Aharonov–Bohm effect,
It is fully explained in my publications
the Stark effect, the Zeeman effect, etc,
I have not yet written up the latter effects but will soon do so: They are quite simple.
but why consider all discoveries up to the present, when you are so comfortable focusing on one narrow window of physics results from 100+ years ago?
Why postulating Voodoo like "time-dilation", "wave-particle-duality" etc. when they are not required to model all these effects in terms of Maxwell's waves?

Sep 26, 2014
@ johypringle-Skippy why you got to use all the bad words all the time? If you are losing your control and becoming wild and weird, maybe it's time to sit this one out. Peoples with mental conditions that make them that mad should stay away from the interweb because it makes it worser for them.

Sep 26, 2014
Sigh. If you have EVER solved Maxwell's wave equations, you will realise that a change in boundary conditions changes the wave

This has nothing to do with my argument. EM is deterministic. If you have electrons in the same state that are equidistant from the wave source, then they MUST, BY SYMMETRY, AND DETERMINISM, respond in the exact same manner. The whole point of this is that the boundary conditions are the same FOR BOTH ELECTRONS. Thus, you cannot have a situation where one electron is excited, and the other isn't.
The concept of "uncertainties" are Voodoo concepts,

Nobody cares.
which you believe in as if they are built into the laws of nature.

Like I said to Clever Hans, it doesn't matter what I believe. What matters is what experiment shows, and what we can conclude from demonstrated postulates. Your religious and philosophical protestations mean squat in the eyes of experiment.

Sep 26, 2014
But if the boundary conditions allow different outcomes FOR IDENTICAL WAVES, you will have different outcomes.

You are talking about degenerate states or modes. But the thing about those, is they result from missing information. An example would be a drum membrane. Every resonant frequency has different modes. That does not mean that if you hit the drum membrane in the exact same way, in the exact same place, that it will be a random mode, every time. This should not be a difficult concept. But it is, for you, and it really wouldn't be a problem with me if you didn't then visit click bait articles about Relativity and QM and spew your nonsense. You wouldn't even hear from me if it weren't for all the mouth-foamers like you who come on here and have the gall to sit in their armchairs, pretend that high school math is cutting edge physics, and act as if their misunderstanding of basic concepts gives them license to belittle decades of experiment and research.

Sep 26, 2014
So how the Hell can a MORON like you "correct me"?

I did it at least twice, Johan, my boy. You know, I wouldn't even harp on it, except for the fact that you feel you need to hold your pretend CV over our heads as if it has anything to do physics arguments. Obviously, you feel inadequate. Otherwise, you wouldn't mention it. I have never, once, heard Strassler, Baez, or Woit, all avid physics commentators, mention their CV in relation to arguments. Hell, even Lubos Motl, no stranger to provocation, doesn't do this. Why? Because they know their stuff. You may not agree with them, but they don't need to mention their accolades. Their arguments testify to that quite handily.
(to be continued)

Sep 26, 2014
(continued)

I have co-written and published one paper. I am not a grad student. I was working toward a second bachelor's degree in physics, until I was forced to quit because the company I worked for tanked. Until things start going better for me, I am stuck in the limbo of being a professional for 8 hours a day, and sneaking in physics studies here, and there, in between chores. And if I, a lowly undergrad with 3 years of physics training (and 2 years of personal study) can correct you like that, imagine what Strassler would do to you.

Sorry, Johan, but if you want to be taken seriously, perhaps you should actually demonstrate it by not making ridiculous basic errors, that an undergrad can spot. That would convince me of your accolades more than any silly self-published internet site ever could.

Sep 26, 2014
It is fully explained in my publications

Which would then have to be peer reviewed.
It is fully explained in my publications

Which would then have to be peer reviewed.
It is fully explained in my publications

Which would then have to be peer reviewed.
I have not yet written up the latter effects but will soon do so: They are quite simple.

And when you do, those, too, will have to be peer reviewed. If somebody recognizes your work as more than insane rantings, then the next step is for a whole bunch of experts to argue about it until a consensus is reached.

Of course, one thing that might help you reach that next step is not to say inane things like equating x^2 - (ct)^2 to x'^2 -(ct')^2 is equivalent to dividing by 0. I mean, yes, unfortunately, there is probably too much nepotism in academia, but you could at least put yourself on the map by not sounding so mouth-foamy.

Sep 26, 2014
Why postulating Voodoo like "time-dilation", "wave-particle-duality" etc. when they are not required to model all these effects in terms of Maxwell's waves?

You know, I have to say that one positive thing about arguing with your broken-ass mind, is that I continuously come up with better ways of understanding relativity. The more I try to break SR, the more intuitive it becomes to me. Since I first started arguing with you, I have gone from an outsider's cursory perspective, to finding new, and clever ways (at least to me) of deriving the Lorentz transform, dealing with simultaneity, and analyzing the Twin Paradox, without using the typical methods they use, which crackpots like you latch on to.
Case-in-point: I have found a way to derive the Lorentz transform without having to bounce signals back and forth at all. I'll have to post it to my blog one of these days. So, thanks for that, at least.

Sep 27, 2014
This comment has been removed by a moderator.

Sep 27, 2014
This comment has been removed by a moderator.

Sep 27, 2014
@johanfprins,
I've read some of your excerpts on your web site and you spend more time ranting and calling other scientists stupid than anything else. If an idea is truly revolutionary, then the idea will stand on its own merit. It may be widely criticized at first, and the person who put forth the idea may not be recognized in their lifetime, especially if they are an insufferable douche like you, but if their idea is truly revolutionary, it will eventually be recognized and it's author recognized accordingly. The problem with insufferable douches like you is that the need to be recognized as great is their primary motivation. Frankly, you deserve the treatment you get on this site and you likely deserve the treatment you have received from your peers. If your ideas are truly revolutionary, then you are unlikely to be recognized for them until after you have passed. Such is the inevitable fate of the insufferable douche.

Sep 27, 2014
You have not yet learned to think for yourself

Oh please. Stop being such a predictable crackpot. Put some oomph into that misplaced condescension.
You have not yet learned to think for yourself and thus consider it your duty to defend mediocre physics that passed the censorship of anonymous "peers" with a zealot's zeal.

Nope! I consider it my duty to correct things that are undeniably wrong. We're not having a discussion about String Theory or SUSY, here. We're not even discussing something like Heim theory, which is generally considered crackpot science, but too obscure to be readily understood.

We're talking about rookie-level misconceptions that you have about physics and math. I don't need to know your conclusions, because the mistakes you make are fundamental ones. It's not so much about upholding the "dogma" of science as it is recognizing that you don't even understand basic concepts, like what a coordinate system is, and how to use it.

Sep 27, 2014
Take for example Reg Mundy:

Neeeigggh
I do not know whether he is right or wrong

(he's wrong)
but my scientific ethics...will not allow me to attack his ideas before I have studied his book

Sure you can. If Reg Mundy told you we live on Mars, you wouldn't need to read his book to correct him. Reg Mundy claims, among other silly things, that everything returns under gravity. This is demonstrably false. I don't need to read his book in order to recognize that he's wrong.
YOU on the other hand decide a priori that he MUST be wrong since what he claims is "heretical"

Haha, nope! It has nothing to do with his ideas being unorthodox. It has everything to do with his ideas being demonstrably wrong. It would be one thing for him to claim we live on a 4D brane. It's crazy, yes, but not demonstrably wrong. It's entirely another thing for him to claim that there is no such thing as escape velocity.

Sep 27, 2014
@johanfprins
Likewise, I am not claiming you are incorrect because your ideas are heretical. I am claiming you are incorrect because you are making demonstrably false claims. For example, you claim that it is impossible to have a vector space where nonzero vectors are assigned a magnitude of 0. This is demonstrably false. You also claim that equating two identically zero quantities to each other is equivalent to dividing by 0. This is demonstrably wrong. You also claim that A cannot measure light to be approaching B at relative speed, c-v. This, also is demonstrably wrong. You claim that a person on the ground would measure the impulse on a ball shot vertically out of a moving cannon as having a horizontal component. This is demonstrably wrong.

The mistakes you make are rookie mistakes, but you come in here, and act like an arrogant, know-it-all ass who deserves deference because of a pretend CV, and yell at anybody who suggests you might be wrong. This is why I mock you.

Sep 27, 2014
If somebody were to create a piece of software that can filter out three kinds of people (brainless, schizophrenics, sadists) then I could start paying for discussion forums. And that person would die rich and famous.

Sep 27, 2014
This comment has been removed by a moderator.

Sep 28, 2014
This comment has been removed by a moderator.

Sep 28, 2014
. It has everything to do with his ideas being demonstrably wrong. It would be one thing for him to claim we live on a 4D brane. It's crazy, yes, but not demonstrably wrong. It's entirely another thing for him to claim that there is no such thing as escape velocity.
Einstein claimed that the relative speed of light is c: YOU claim it can be c+v or c-v: Thus either YOU or Einstein must be wrong. Since I have studied Einstein in depth, I can claim that YOU are wrong. Ideas in physics are usually developed by reasoning and you are an arrogant prick who thinks that after having published a single paper (obviously mediocre paper: Since only such papers nowadays pass "peer review" in "established fields" of physics), not finishing your PhD (for which you do not seem to have the required objectivity) you can argue that a person is wrong without reading the full text of what the person has published!
You are beneath contempt!

Sep 28, 2014
@johanfprins
I am claiming you are incorrect because you are making demonstrably false claims. For example, you claim that it is impossible to have a vector space where nonzero vectors are assigned a magnitude of 0.
I have not claimed that you cannot have a zero vector. I claim that a space can only have the same distances when using different coordinates when there is ONLY ONE such a zero vector, which gives the position of the origin from which all coordinate points are measured. This is simple high school high school graph plotting.

You also claim that equating two identically zero quantities to each other is equivalent to dividing by 0.
Not quantities, but two mathematical expressions using different coordinates on the same manifold.: Like f(x,y,z,w)=0=F(x',y',z'.w'). This means that you can leave x',y',z',w' the same and choose different sets of values for x,y,z,w, so that f(x1,y1,z1,w1)=0= f( x2,z2,y2,w2). Thus, there is not.... continued

Sep 28, 2014
a one-to-one correspondence when trahsforming the preselected coordinates x',y',z',w' : They can be tranformed into either x1,y1,z1,w1 or into x2,y2,z2,w2. In other words the relationship between the coordiantes x',y',z',w' and the other coordinates x,y,z,w, cannot be defined. This is the same when you divide by zero. Which is effectivley what you are doing when you post that f(x,y,z,w)=0=F(x',y',z',w'). This is wha Einstein stupidly did and what is still being done in textbooks.It is thus fortuitous that Einstein got the corrct equations. In my book in section 6.6,
it is shown that one must write that rho^2=x^2+y^2=(ct)^2-x^2, and that rho'^2=x'^2+y'^2=(ct')^2-x'^2: When one now chooses rho=rho' one can equate WITHOUT dividing by zero: to obtain that:

(ct)^2-x^2=(ct')-x'^2.

I do this in section 6.6 of my book and then obtain for a spherical light wavefront emitted by a source at 0' within K' that the coordinates x',y',z',w' transform into K by means of the Lorentz equations.

Sep 28, 2014
I also find from this correct derivation that although t is larger than t', these two times are not simultaneously present on the clocks at 0 and 0'. Ths is not surprising since "simultaneous" means THE SAME TIME. The derivation proves that when tgtime on the clock in K' is t', the simultaneous time on the cllock in K IS ALSO t'. And when the time on the clock in K is t, the simultaneous time on the clock in K' IS ALSO t. But of course you are "so clever" that you just know that these results must be wrong without studying the derivation.

You also claim that A cannot measure light to be approaching B at relative speed, c-v.
THe Lorentz equations demand that the speed cannot be c-v: As I have already derivedthis on another thread. If the speed can be c-v, the Lorentz equations must be wrong.


Sep 28, 2014
You claim that a person on the ground would measure the impulse on a ball shot vertically out of a moving cannon as having a horizontal component.
One of your usual lies! I claim that if another cannon is stationary it can only reproduce the motion of the cannon-ball shot vertically from a moving cannon, by adding a horizontal component. You are not a scientist's arsehole, my friend!

The mistakes you make are rookie mistakes,
You are not even a rookie yet: You stupidly believe that established physics is sacrosant, and therefore you do not have to read any arguments that it might be wrong. The most important principle when you are a physicist is to accept that everything we at present believe is correct physics can at any time be proved to have been wrong all along. Only in this way can physics be self-correcting. Physicists are not following this rule anymore; and therefore physics has stopped to be self-correcting during the 20th century.

Sep 28, 2014
@johanfprins,
LOL! I had Fritz Zwicky in mind when I was talking about ideas that receive posthumous recognition despite their author being an insufferable douche while alive. You're no Fritz Zwicky. Like every insufferable douche (I.D.), you won't live to see your ideas recognized. That's the inevitable fate of the I.D. - THE GUARANTEE THAT THEY WON'T SEE THEIR IDEAS RECOGNIZED IN THEIR LIFETIME! Also, I stated that REVOLUTIONARY ideas would eventually be recognized and attributed, not plain old run of the mill ideas.

If I had a revolutionary idea, I would be testing it against existing observations and looking to find currently unexplained problems that it could solve. I would probably write to some of the recognized people, get their input, and consider their critiques with the understanding that they may just ignore me. I would be looking to be proven wrong so that I could fix or abandon the idea. I wouldn't be seeking glory unlike you, you insufferable douche.

Sep 28, 2014
@johanfprins,
LOL! I had Fritz Zwicky in mind when I was talking about ideas that receive posthumous recognition despite their author being an insufferable douche while alive. You're no Fritz Zwicky.


I read the book about him last month. He was a funny man. I wish I could met him for no reason but to get some ideas for saying some funny stuffs to peoples. I especially like the story about him making his telescope-assistant-Skippy shoot the gun out of the telescope building hole to get the the air to smooth out so he could see better, he could be the Honorary Coonass for that one.

I wouldn't be seeking glory unlike you, you insufferable douche.


I bet you would not be trying to get them to agree with you by calling them stupid-idiot-bad word-morons and some other things I forget right now. When I calls people that I am not so stupid that I get mad when they don't agree with me.

Sep 28, 2014
I had Fritz Zwicky in mind when I was talking about ideas that receive posthumous recognition despite their author being an insufferable douche while alive. You're no Fritz Zwicky.
And if you can wipe the SNOT from our eyes you will see that I was talking about Herbert Dingle who was far more competent than Zwickey when it came to Astronony and Theoretical Physics.

If I had a revolutionary idea, I would be testing it against existing observations and looking to find currently unexplained problems that it could solve. I would probably write to some of the recognized people, get their input, and consider their critiques with the understanding that they may just ignore me. I would be looking to be proven wrong so that I could fix or abandon the idea
This is EXACTLY what I did after I demonstrated superconduction at room temperature in 1999. But, like Dingle i was censored since it also proves that the accepted mechanism for SC (the BCS-model) IS WRONG!!

Sep 28, 2014
I wouldn't be seeking glory unlike you,
Only an obnoxious moron like you will make this assertion. Just like "furbrain" you jump to stupid conclusions before knowing the whole history of what has happened.

Why do you think physicists publish? My first reason is to advance this field of science; and when I do succeeed I expect, LIKE ALL SCIENTISTS DO, to get recognition for this. So what is wrong with the latter? If this is "seeking glory", then I do not know of any research physicist who should NOT be accused of this?

What is, however, even worse, is a physicist who has received "glory" for wrong physics (like Brian Josephson has for his wrong model of superconducting tunnelling) and who then censor new ideas to protect his "glory"!

Sep 28, 2014
Not quantities, but two mathematical expressions using different coordinates on the same manifold.: Like f(x,y,z,w)=0=F(x',y',z'.w'). This means that you can leave x',y',z',w' the same and choose different sets of values for x,y,z,w, so that f(x1,y1,z1,w1)=0= f( x2,z2,y2,w2). Thus, there is not a one-to-one correspondence when trahsforming the preselected coordinates x',y',z',w' : They can be tranformed into either x1,y1,z1,w1 or into x2,y2,z2,w2. In other words the relationship between the coordiantes x',y',z',w' and the other coordinates x,y,z,w, cannot be defined.
You are trying to claim that each element of the set of coordinates where f = 0 can be transformed into every element in the set of coordinates where F = 0. Coordinate transformations depend only on the coordinate systems themselves, not on some arbitrary functions that may be defined on them. Your logic is flat wrong.

Sep 28, 2014
@johanfprins,
If you were just a humble seeker of knowledge then your approach would be different. You would willingly accept challenges to your ideas and counter them without feeling like the challenges were insults against you. You wouldn't run around exclaiming that the entire physics community is wrong and you are right and they are stupid. You elevate yourself by belittling others and boasting about your credentials. All of it is intended to glorify yourself at the expense of others. You are an obvious ego maniac who is utterly unable to accept any challenge to his ideas - especially if the challenge comes from someone who lacks credentials you deem worthy.

Frankly, after reading your mathematical nonsense and commenting on it, this will be my last post. I'm quite sure you will want to get the last word, but it won't help you to have it. Everyone reading this comment thread has already made their assessment of you by this point and I'm quite sure I needn't say any more.

Sep 28, 2014
This comment has been removed by a moderator.

Sep 28, 2014
This comment has been removed by a moderator.

Sep 28, 2014
You are an obvious ego maniac who is utterly unable to accept any challenge to his ideas - especially if the challenge comes from someone who lacks credentials you deem worthy.
The rule in physics is that you must be judged by "your peers". That means by persons who are as capable as you are. Since the latter persons have refused to engage me for 14 years, I decided to try and see if I could pick up a "peer" on discussion forumsl So far I have I only encountered trolls like YOU and furbrain.

Last year the editor of Phys. Essays forced the reviewers to engage me with logic. After giving me all the same nonsense that furbrain is vomiting here, they had to concede that they cannot prove that I am wrong when it comes to STR. Two papers have thus appeared in the December issue and the March issue. You will not find this honesty and integrity in publications like Nature, PRL, Royal Soc publicaions etc. This is the reason why the latter publications hate Phys. Essays.

Sep 28, 2014
This comment has been removed by a moderator.

Sep 28, 2014
It has everything to do with his ideas being demonstrably wrong. It would be one thing for him to claim we live on a 4D brane. It's crazy, yes, but not demonstrably wrong. It's entirely another thing for him to claim that there is no such thing as escape velocity.

Einstein claimed that the relative speed of light is c: YOU claim it can be c+v or c-v...

This has nothing to do with the point I made, Johan. Do you, at least, agree that if Reg Mundy says something that is undeniably false, that there is no compelling reason to read his book? Or, should I take your silence on the matter to mean that you, too, believe that there is no such thing as escape velocity?

Sep 28, 2014
Einstein claimed that the relative speed of light is c

Uh, no, Johan. He claimed that the speed of light, as measured in any IRF is c. Stop being dense.
YOU claim it can be c+v or c-v

Yes, because I understand how coordinate systems work.
Thus either YOU or Einstein must be wrong.

Nope! There's another possibility, Johan, my boy. And that is, that you don't understand what Einstein said. You are creating a false dichotomy. What is logic, and how does it work?
Since I have studied Einstein in depth, I can claim that YOU are wrong.

And since I have derived all Special Relativity results I know about on my own without consulting textbooks and research papers, and in some cases, deriving results that I hadn't read about, I can claim you are wrong. Read that again. I derived known results WITHOUT KNOWING THEM. What, now?
Oh, right. Your pretend CV has the magical power to automatically make everything I say wrong. How silly of me!

Sep 28, 2014
Ideas in physics are usually developed by reasoning

Oh? And here I thought it was developed by rabidly yelling at people who correct you! How silly of me!
and you are an arrogant prick who thinks that after having published a single paper (obviously mediocre paper: Since only such papers nowadays pass "peer review" in "established fields" of physics),

Uh, Johan, you don't know what paper I published.
not finishing your PhD

I haven't started it yet. Oh, Johan, you and Clever Hans are so cute when you assume things.
(for which you do not seem to have the required objectivity)

Because I don't yell with spittle flecks at people who disagree with me?
you can argue that a person is wrong without reading the full text of what the person has published!

Read what I said above. The reason we have this ongoing "debate" is because I attempted to read your book the very first time I met you, and immediately found major errors. Shall I list them for you?

Sep 28, 2014
@johanfprins
I am claiming you are incorrect because you are making demonstrably false claims. For example, you claim that it is impossible to have a vector space where nonzero vectors are assigned a magnitude of 0.

I have not claimed that you cannot have a zero vector.

What is English, and how does it work?
I claim that a space can only have the same distances when using different coordinates when there is ONLY ONE such a zero vector, which gives the position of the origin from which all coordinate points are measured.

If we were working with space equipped with a positive definite scalar product, then yes, that would be true. But once you choose a different scalar product, then that's no longer true. And the reason we equip it with this scalar product is because it's convenient--it is a quantity that is preserved after a Lorentz transformation.

Sep 28, 2014
This means that you can leave x',y',z',w' the same and choose different sets of values for x,y,z,w, so that f(x1,y1,z1,w1)=0= f( x2,z2,y2,w2).

But you don't have a point. First of all, 0, has no special status here. I could just as easily replace it with 2, and what you just said would still be true.
Here, I'll show you.
Suppose x^2-(ct)^2 = 2
L*(x'^2 - (ct')^2) = 2, where L is constant.
So,
x^2-(ct)^2 = 2 = L*(x'^2 - (ct')^2)

But wait, as long as I choose x' and t' such that L*(x'^2 - (ct')^2) = 2 always holds, this equation will always hold.

Wait! What? OMG! I can leave x and t the same and choose different x' and t'? It's UNPOSSIBLE! What is this impossible thing known as grade-school algebra, and how does it work?
This is the same when you divide by zero.

And this is why I call you are a rookie. Because that BASIC result is not true. You aren't a heretic. You just suck horribly at basic math.

Sep 28, 2014
This is wha Einstein stupidly did and what is still being done in textbooks.It is thus fortuitous that Einstein got the corrct equations. In my book in section 6.6,

Haha, no, Johan. As I keep saying, you don't actually understand what Einstein was saying. Look VERY CAREFULLY at http://www.bartle...a1.html, which contains his derivation.

Specifically, pay close attention to this:
Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation
(x'-ct') = lambda*(x-ct)
is fulfilled in general, where lambda indicates a constant;

All he is doing is guessing that the two expressions are proportional to each other by a constant, lambda, based on the fact that both equate to 0 for the space-time coordinates of a light signal. In other words, HE IS GUESSING THIS RELATION HOLDS IN GENERAL, not just for a light signal traveling along the x axis. You are making stuff up to suit your bias. (to be continued)

Sep 28, 2014
This comment has been removed by a moderator.

Sep 28, 2014
(continued)
Furthermore, since he GUESSES that the relation holds IN GENERAL, then we can no longer say that x' - ct' = 0 and x - ct = 0. In particular, x-ct might be, say, 2. Then, by his guess, x'-ct' = lambda*(x-ct) = lambda*2.

Then, he follows a similar chain of logic along the negative x axis. So, now, we have two unknowns, lambda, and mu. THESE are the variables he then solves for, using relativistic arguments.

I mean, all you need to do is shed your preconceived notions and ask yourself what would make more sense, that Einstein divided by 0, or that you just don't understand what he was doing? I know, I know.

You have studied it for years, so YOU MUST BE RIGHT. But, see, that logic doesn't hold up. You know who else studied this stuff for years? Most professional physicists, and the majority agree that you are not correct. So, sorry, that particular argument won't fly either.

Sep 28, 2014
I do not know the reasons why he claims what he is claiming; and as a responsible physicist, not an uneducated mentally-disturbed abortion like you, I will first read his book before bombarding him hysterically as is your infantile approach. ARSEHOLE!!!!

Oh, ok, Johan. Well, then, I suggest you mosey on over to http://theflatear...rg/cms/. You see, they have been studying the phenomenon of earth flatness FOR YEARS, and compiled a lot of literature on the subject. I mean, personally, I think they're making claims that are demonstrably wrong, but I haven't read their literature, so what do I know? But, as a responsible physicist, you have the obligation to read everything they've written before deciding they're not correct. You'd better get cracking.

Sep 28, 2014
I do not "believe" anything in physics,

Yes, you do. You believe that uncertainty, time dilation, and length contraction are VOODOO.
since ALL that we believe is correct today can be proved to be wrong tomorrow. Therefore I keep an open mind

But not open enough to consider the likelihood that you just misunderstood what Einstein wrote. I mean, clearly, it is far more believable that he, a man who changed the face of modern physics, was wrong and divided by 0.

Sep 29, 2014
Uh, no, Johan. He claimed that the speed of light, as measured in any IRF is c. Stop being dense.
YOU are the one that is dense: Are you claiming that the speed c is NOT a relative speed? STR is in a gravity-free space: There are NO gravitational forces: EACH and EVERY body with mass moves with velocities relative to all other bodies with mass. THUS each one can be used to define an IRF: In fact ecah one IS an IRF. Thus to state that the speed of light must be c relative to all IRF's IS THE SAME AS STATING that the speed of light MUST ALWAYS be c relative ALL moving bodies with mass: Therefore c+v and c-v are NEVER possible!In

Yes, because I understand how coordinate systems work.
You do not have a clue about RELATIVISTIC COORDINATE TRANSFORMATIONS since you stupidly believe that they are the same as coordinate transformations within a single IRF which leave the physics equations covariant. The relativistic transformations do not do so! cont.

Sep 29, 2014
This comment has been removed by a moderator.

Sep 29, 2014
I claim that a space can only have the same distances when using different coordinates when there is ONLY ONE such a zero vector, which gives the position of the origin from which all coordinate points are measured.

If we were working with space equipped with a positive definite scalar product, then yes, that would be true. But once you choose a different scalar product, then that's no longer true. And the reason we equip it with this scalar product is because it's convenient--it is a quantity that is preserved after a Lorentz transformation.
This is of course BULLSHIT. If you claim a 4D manifold in which the mapping of 4D distances are the same for different coordinates (4D isometric mappings) like Minkowski is claming, this is ONLY possible when the scalar product is positive-definite. Choosing another "convenient" scalar product, which violates the fundamental rules of mathematics, is insane. You cannot willy-nilly violate mathematics since it is "convenient".

Sep 29, 2014
This comment has been removed by a moderator.

Sep 29, 2014
I claim that a space can only have the same distances when using different coordinates when there is ONLY ONE such a zero vector, which gives the position of the origin from which all coordinate points are measured.

If we were working with space equipped with a positive definite scalar product, then yes, that would be true. But once you choose a different scalar product, then that's no longer true.
You cannot claim that the Lorentz transformation defines isometric 4D distances UNLESS the scalar product is positive definite. Clearly this is not the case for Minkowski's "space-time" and therefore this concept is mathematically impossible.

And the reason we equip it with this scalar product is because it's convenient--it is a quantity that is preserved after a Lorentz transformation.
No it is NOT! It only SEEMS to be preserved when you are stupid enough to argue that dfferent times can be simultaneous! SIMULTANEOUS=THE SAME TIME!

Sep 29, 2014
This comment has been removed by a moderator.

Sep 29, 2014
Specifically, pay close attention to this:
Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation
(x'-ct') = lambda*(x-ct)
is fulfilled in general, where lambda indicates a constant;
Oh you do read some books! I thought you can do it all on your own: Crackpot! This derivation is wrong since it ignores he coordinates y and z. And if you go further it assumes that the Lorentz transformation has an inverse; which it does not have since it tranforms physics, NOT JUST COORDINATES.
All he is doing is guessing that the two expressions are proportional to each other by a constant, lambda,
He should motivate this in terms of the rules of mathematics. Sometimes it is usefull to "guess" when doing research, but if a guess seems to give an acceptable result a REAL physicist will afterwards determine why the guess seems to work. If it works for the wrong reasons, as in this case, the guess is wrong

Sep 29, 2014
Furthermore, since he GUESSES that the relation holds IN GENERAL, then we can no longer say that x' - ct' = 0 and x - ct = 0. In particular, x-ct might be, say, 2.
So you think a guess proves physics? How absurdly ridiculous you are;
Then, by his guess, x'-ct' = lambda*(x-ct) = lambda*2.
A guess is not scientific proof Mr. Crackpot.

I mean, all you need to do is shed your preconceived notions and ask yourself what would make more sense, that Einstein divided by 0, or that you just don't understand what he was doing? I know, I know.
Very few physicistss teach this proof to their students since it is clearly a division by zero, which fortuitously leads to the Lorentz equations.

For this reason most textooks use the emission of a spherical wavefront from a point source when 0' in K' and 0 in K coincide. But they do not specify in which IRF the point source is and still end up dividing by zero. To avoid the latter one must use the derivation in 6.6 of my book.

Sep 29, 2014
I do not know the reasons why he claims what he is claiming; and as a responsible physicist, not an uneducated mentally-disturbed abortion like you, I will first read his book before bombarding him hysterically as is your infantile approach. ARSEHOLE!!!!

Oh, ok, Johan. Well, then, I suggest you mosey on over to http://theflatear...rg/cms/.
To jump to the conclusion that Reg Mundy is propagating the ideas of the flat earth society, you must first read Mundy's book and prove that he does. That is what a responsible physiciat will do. But of course you do not understand the ethics and responsibilty of being a physicist: You just shoot from the hip before doing your homwework: You are a real crackpot!

Sep 29, 2014
Yes, you do. You believe that uncertainty, time dilation, and length contraction are VOODOO.
I believed that these concepts are correct when they were first taught to me, but has in the meantime found that they lead to absurd conclusions, and threfore I HAD to conclude by reductio ad absurdum that they are VODOO!
since ALL that we believe is correct today can be proved to be wrong tomorrow. Therefore I keep an open mind

But not open enough to consider the likelihood that you just misunderstood what Einstein wrote.
I do keep such an open mind, but I have had many experts on STR who had to agree in the end that they cannot fault me: They all raised your pathetic dogmatic arguments. I know these arguments well and have argued these points well enough to know that you are wrong! If you are able to read my book with the same open mind that I read anybody elses publications, you might learn something: But I despair: You can only post mantras not any logic!

Sep 29, 2014
Arm of MM experiment from x'=0 to x'=L'. Wavefront leaves x'=0 at t'=0 and arrives at x'=L'at

t'=L'/c-----------------Eq. 1:

The latter is so since the speed of light is c relative to the end-mirror within K'. Do you agree furbrain, or are you, as usually too stupid?

Now what hapens within K. Do you agree that this can only be obtained from the Lorentz transformation from K' to K? If you do not agree then you MUST reject the Lorentz transformation as being incorrect.

Thus, the position within K when the light reaches the mirror is x=G*(x'+vt') and the time is t=G*(t+(v/c)*(x'/c)). where x'=L' and t' is given by Eq. 1. Thus, the transformed coordinates become:

x=G*(L'+v*(L'/c))= L'*G*(1+v/c)-------------Eq. 2

And

t=G*(L'/c+(v/c)*(L'/c))=(L'/c)*G*(1+v/c)------Eq. 3.

THE SPEED c(K) within K with which the light approaches the mirror is given by c(K)=x/t=(Eq.2)/(Eq.3)

AND THIS IS: c(K)=L'/(L'/c)=c!!!! QED! Where is your c-v crackpot?

Sep 29, 2014
Are you claiming that the speed c is NOT a relative speed?

Here is exactly what I am saying: Let's say that you want to measure how quickly light is moving WRT YOURSELF. When you do this, you will always get a value of c. This says NOTHING about how quickly you would measure light to be moving WRT someone else.

And again, this isn't complicated. Let's say you are at rest with the origin of K. If you see someone moving with trajectory vt, and you fire a light signal at them at time t0, this light signal will travel with trajectory c(t-t0). So, you will measure the light to have gotten to this person by solving
vt = c(t-t0),
which can be rewritten as
(c-v)t - ct0 = 0.

The LHS of last equation is the relative position of light to B WRT K.
It moves with velocity (c-v) WRT K.

We can say this another way, but I have to leave now. I will be back later.

Sep 29, 2014
Are you claiming that the speed c is NOT a relative speed?

Here is exactly what I am saying: Let's say that you want to measure how quickly light is moving WRT YOURSELF. When you do this, you will always get a value of c. This says NOTHING about how quickly you would measure light to be moving WRT someone else.
What this speed is, is determined by the Lorentz transformation and according to the Lorentz transformation the light approaches this object moving relative to you with the speed c within YOUR reference frame, AND NOT with the speed c-v. See above! Thus, you can ONLY be correct in your reasoning when the Lorentz transformation DOES NOT apply.

My reasoning is based on th assumption that the Lorentz tranformation is correct: And then the speed is c and not c-v. It is ONLY c-v when the Galilean tranformation applies. You are hopelessly addled in your reasoning. If you want to claim c-v, you must reject the validity of the Lorentz Transformation. Do you?

Sep 29, 2014
What this speed is, is determined by the Lorentz transformation

No, it isn't. It's determined by you actually measuring how quickly light approaches B.

Let's go over this again, Johan. Assume the usual setup with B passing A, as their origins coincidem and also assume that A, looong before this thought experiment, placed a detector with a clock synchronized to his own at position, c((c*(1 s)/(c-v)) - 1 s) WRT K. Then, at time 1 s on A's clock, A fires a light signal.
B moves with trajectory vt WRT K. The light signal moves with trajectory c(t-1 s) WRT K.

Now, at time, t = c*(1 s)/(c-v) on the detector's clock, both the light signal and B will reach the detector.

Proof:
vt = v*c*(1 s)/(c-v)
c(t - 1s) = c*(c*1s.(c-v) - 1s) = c*((c - c + v)*1 s/(c-v)) = c*v*1 s/(c - v)

Therefore, vt = c(t - 1 s).
(to be continued)

Sep 29, 2014
(continued)
So, the detector and A will agree that it took c*1 s/(c - v) - 1 s time for the light signal to get to B.
This amount of time is equivalent to v*1 s/(c-v).

Now, at time t = 1s WRT K, B was a distance v*1 s away from the light signal. At time t = c*1 s/(c-v), B was a distance of 0 m way from the light signal.

So, in time v*1 s/(c-v) WRT A and the detector, the light signal went from being v*1s away from B, to being 0 m away from B.

Now, divide v*1s, the distance, by v*1s/(c-v), the time. You get c-v. So, A measures the light signal to approach B with speed (c - v).

Now, if we were asking how quickly B was measuring the light signal to approach him, that would be a different story. By the 1st postulate, B would measure the signal to approach him with speed c.

Sep 29, 2014
What this speed is, is determined by the Lorentz transformation

No, it isn't. It's determined by you actually measuring how quickly light approaches B.
Lets us go over it again: The light starts off at time t'=t at position x'=x=0. Wthin K' it reaches x'=L' at the time t'=L'/c. In other words x'/t' gives the speed with which the light approaches the mirror.

The corrsponding coordinates within K when the light meets the mirror are, AS HAVE ALREADY BEEN TWICE DERIVED FROM THE LORENTZ TRANSFORMATION, given by.

x=L'*G*(1+v/c) and t=(L'/c)*G*(1+v/c). Since the light has started to move at x=0 at time t=0 within K and reached the mirror at time t=(L'/c)*G*(1+v/c) within K when the mirror is at x=L'*G*(1=v/c) within K, it MUST have moved towards the mirrior with the speed c(K) relative to the mirror, where c(K)=x/t: AND THIS GIVES c(K)=c. Thus, it is NOT and CANNOT be moving with the speed c-v towards the mirror within K if the Lorentz tranformation is valid! QED

Sep 29, 2014
Lets us go over it again: The light starts off at time t'=t at position x'=x=0. Wthin K' it reaches x'=L' at the time t'=L'/c. In other words x'/t' gives the speed with which the light approaches the mirror.

Yep. An observer in K' would measure light as moving relative with the mirror with speed c.
The corrsponding coordinates within K when the light meets the mirror are, AS HAVE ALREADY BEEN TWICE DERIVED FROM THE LORENTZ TRANSFORMATION, given by.

x=L'*G*(1+v/c) and t=(L'/c)*G*(1+v/c).

....aaand, what about the corresponding coordinates of the mirror in K? You are forgetting about those. WRT K, where did the mirror start out at time t=0?
Since the light has started to move at x=0 at time t=0 within K...

But you are forgetting where the mirror started out WRT K. Where did it "start out" WRT K?
Let's examine that.
(to be continued)

Sep 29, 2014
(continued)
WRT K', at t'=0, the mirror was at L'.
Using the Lorentz transformation,
at time t = (t' + x'v/c^2)*G = (L'*v/c^2)*G, the mirror was at position
x = (x'+vt')*G = L'*G
all WRT K.
Where was the light signal with respect to K at this time?

Well, the light signal has trajectory ct.
So, the light signal was at position v*L'*G/c WRT K at this same time.
So, the distance between the light signal and the mirror at this time WRT K was L'*G(1 - v/c)

Then, as you calculated, WRT, the light signal "met" the mirror at time t=(L'/c)*G*(1+v/c), at position x = L'*G*(1+v/c), meaning that the distance between them at this time was 0 m.
So, in (L'/c)*G*(1+v/c) - (L'*v/c^2)*G = L'*G/c seconds WRT K, the light signal and mirror went from being L'*G(1 - v/c) away from each other to being 0 m away from each other.

So, "difference in positions"/"difference in time" = (L'*G(1-v/c))/(L'*G/c) = (c - v)

Sep 29, 2014
sorry, I meant
"change in distance"/"change in time" = (L'*G(1-v/c))/(L'*G/c) = (c - v)

Sep 29, 2014
Also, I wrote
Then, as you calculated, WRT, the light signal "met" the mirror at time t=(L'/c)*G*(1+v/c), at position x = L'*G*(1+v/c),

when I meant to write
Then, as you calculated, WRT K, the light signal "met" the mirror at time t=(L'/c)*G*(1+v/c), at position x = L'*G*(1+v/c),

Sep 29, 2014
Yep. An observer in K' would measure light as moving relative with the mirror with speed c.
Good!
The corrsponding coordinates within K when the light meets the mirror are, AS HAVE ALREADY BEEN TWICE DERIVED FROM THE LORENTZ TRANSFORMATION, given by.

x=L'*G*(1+v/c) and t=(L'/c)*G*(1+v/c).

....aaand, what about the corresponding coordinates of the mirror in K? You are forgetting about those. WRT K, where did the mirror start out at time t=0?
This is irrelevant POEPHOL!
But you are forgetting where the mirror started out WRT K. Where did it "start out" WRT K?

This is not required, since the LT gives the position and time WITHIN K when the light-pulse leaves the source and the position and time WITHIN K when it reaches the mirror within K. As anybody with a modicum of brains should know, any other information is superfluous and irrelevant when calculating the speed.

Sep 29, 2014
This is not required, since the LT gives the position and time WITHIN K when the light-pulse leaves the source and the position and time WITHIN K when it reaches the mirror within K. As anybody with a modicum of brains should know, any other information is superfluous and irrelevant when calculating the speed.

But you want to know the distance BETWEEN the light signal and the mirror, not just the position of the light signal, because you are measuring the relative distance between the light signal and the mirror WRT K. You can't do that unless you know how far the mirror is from the light signal in K.

Here, let me use your strategy to show you how wrong you are.
(to be continued)

Sep 29, 2014
Using the Lorentz transformation,
at time t = (t' + x'v/c^2)*G = (L'*v/c^2)*G, the mirror was at position x = (x'+vt')*G = L'*G all WRT K.
This is where the mirror will be at the time t=(L'*v/c^2)*G. The mirror is not at this position at time t=0: It is at x=L' at t=t'=0. Only when the mirror acts as a moving source that sends out a wavefront at time t=0 and position x=L', will it be concluded at 0 within K that this wavefront was emitted at time t=(L'*v/c^2)*G and at position x = (x'+vt')*G = L'*G owing to the Doppler effect.
But even so, YOU only need to worry where the souce is within K and the time at the source when it emits the wavefront, and what the position of the mirror is within K and the time at the position of the mirror when the wavefront reaches the mirror, in order to calculate the speed of light: And the LT gives this information impeccably (see above): And this information gives the speed of light as c NOT c-v.


Sep 29, 2014
But you want to know the distance BETWEEN the light signal and the mirror, not just the position of the light signal, because you are measuring the relative distance between the light signal and the mirror WRT K.
This is what the LT gives me POEPHOL! It gives me the distance between the point x=0 at which the light signal started at time t=0 within K and the point at which the light signal reaches the mirror within K; namely x=L'*G*(1+v/c): AND it gives me the time that the light signal moved to complete this distance namely t=(L'/c)*G*(1+v/c). Therefore the speed of light which is approaching the mirror MUST be x/t=c: NOT c-v!

Sep 29, 2014
(continued)
Ok, Johan, so I am going to use your strategy to show you how wrong you are.
Ready?
So, first consider when the mirror's clock reads
t' = -L'*v/c^2. Since the mirror is at rest with K', then, x'=L'

We apply the LT to determine where and when the mirror existed in this state WRT K.
x = (x'+vt')*G = (L' - L'*v^2/c^2)*G = L'/G
t = (t' +x'v/c^2)*G = 0s

So, I just used the LT exactly how you used it to determine that, WRT K, the mirror was at position L'/G, at time t = 0 s

Now, consider when x' = 0m, t' = 0s.

Applying the LT exactly as you have been applying it, we get,
x=0, t=0
This is to be expected. At 0s, the light signal was at position 0 m WRT K.

So, according to someone at rest with K, at time t = 0s, the mirror was L'/G away from the light signal.

(to be continued)

Sep 29, 2014
(continued)
Now, you calculated that when the distance between the mirror and the light signal is 0 m, the time WRT K is t = (L'/c)*G*(1+v/c)

So, what has happened WRT K? Well, at time t = 0, the mirror and signal were L'/G apart. At time t = (L'/c)*G*(1+v/c), they were 0 m apart.

Now, taking "change in distance"/"change in time", we get

(L'/G)/((L'/c)*G*(1+v/c)) = c/(G^2 * (1+v/c)) = c(1-v^2/c^2)/(1+v/c) = c(1-v/c)(1+v/c)/(1+v/c) = c-v

So, I just used the LT exactly how you have been using it to find the corresponding positions and times of the mirror and the light signal WRT K. And, surprise, surprise, I, once again, got c-v.

Now, you can't tell me that this is invalid because I just used the LT EXACTLY how you have been using it to map events from K' to K.

You've been using the wrong approach to compute the relative speed between the mirror and the signal WRT K because you aren't applying the LT to the position of the mirror. You've just been applying it to the signal.

Sep 30, 2014
This comment has been removed by a moderator.

Sep 30, 2014
Thus the time within K at the instantaneous oposition of the clock within K, MUST always be the same as the time on the clock at 0 within K.

Any person who have passed grade 2 will know that in order to determine the speed within a coordinate system you must know within this coordinate system at which point the motion starts and at which time; Within K it is at x=0 and t=0. Next you must know at which point within the coordinate system the motion stops and at which time the motion stops. Within K this time is t=L'*G*(1+v/c) and the position is x=(L'/c)*G*(1+v/c). Since the starting time of the motion is t=0 and the position x=0, The total time of motion is t=L'*G*(1+v/c) and the total distance is x=(L'/c)*G*(1+v/c).

Thus, according to grade 2 competency (which YOU obviously do not have), the speed of the motion c(K) must be given by the total distance x divided by the total time t, AND THEREFORE:

c(K)={(L'/c)*G)(1+v/c)}/{L'*G*(1+v/c)}={L'/c}/{L'}=c.

There is NO (c-v) .

Sep 30, 2014
The clock is at 0' within K' ARSEHOLE, and the other clock is at 0 within K ARSEHOLE! And botth clocks show t'=t=0. Einstein spent a lot of time to argue in his 1905 manuscript that the time at ALL positions in an IRF is the SAME as at the origin of this IRF: ARSEHOLE!

This has nothing to do with my argument, Johan. Read it more carefully. I used the time on the mirror's clock, but I could have just as easily used the time on the clock at 0'. If it bothers you, then replace
So, first consider when the mirror's clock reads
t' = -L'*v/c^2. Since the mirror is at rest with K', then, x'=L'

with
So, first consider when clock at 0' reads
t' = -L'*v/c^2. Since the mirror is at rest with K', then, x'=L'


Read my argument again, and stop jumping to conclusions.

Sep 30, 2014
Any person who have passed grade 2 will know that in order to determine the speed within a coordinate system you must know within this coordinate system at which point the motion starts and at which time;

But Johan, we're not calculating light's speed WRT K. Where are calculating how quickly somebody at rest with K measures the light signal to have approached the mirror--at least that's what I'm talking about. Who on earth knows what you're talking about?

Now, I have used the LT exactly as you have been using it to take events in K' and map them to K. And I have STILL calculated the relative velocity WRT K as c-v.

I can't believe you can sit there, and in serious tell me that taking the change in distance over the change in time does not give relative speed as measured from K. Again, why even have coordinate systems if you're just going to ignore what they mean Now, read my argument again, please.

Sep 30, 2014
@thefurlong
I wasn't going to post anymore in this thread, but this is getting ridiculous. You won this argument like 20 posts ago or something (acutally you won it with your first post...).

There is a reason that johan can claim that his peer reveiwers can't find any msitakes in his math - he has imaginary peer reviewers! Johan may have had the intellectual chops to get his PHD back in the day, but those days have long since passed him by and now he's like Russel Crowe in a Beautiful Mind only he's at the stage where he thinks the peer reviewers are real and he has a shed full of physics papers and yarn lines that are proof of the physics conspiracy against the world. He's probably making secret reports and stuffing them in the mailbox of an abandoned home somewhere.

Save yourself. Let the man live in his fantasy land. He's already destroyed whatever professional reputation he may have enjoyed. His fantasies are all he has left.

Sep 30, 2014
Save yourself. Let the man live in his fantasy land. He's already destroyed whatever professional reputation he may have enjoyed. His fantasies are all he has left.

You're probably right. I know he's crazy. I suppose I just have this delusional optimism that he'll see the error of his thinking. Mind you, I do have have evidence of this. He has admitted he was wrong twice before. You'd think that that would have indicated to him that he should double check the rest of his reasoning, but it seems to have only reinforced his delusions.

I find it astonishing how somebody could function with such a broken mind. I mean, can he tie his own shoes? Can he cross the street by himself, or does someone have to constantly pull him out of the way because he thinks cars approaching him aren't actually there?

Anyway, you are right. I should stop.

Sep 30, 2014
Save yourself. Let the man live in his fantasy land. He's already destroyed whatever professional reputation he may have enjoyed.


Lol! Lex Luther to Furlong's Clark Kent :)

I find it astonishing how somebody could function with such a broken mind. I mean, can he tie his own shoes? Can he cross the street by himself, or does someone have to constantly pull him out of the way because he thinks cars approaching him aren't actually there?


In all seriousness, tying his shoes is probably the least of his problems (besides, I'm willing to bet he has loafers), but his live interactions with other human beings, if he has any at all anymore, must be quite painful to watch. There's just no way he doesn't carry on like this about mundane things as well. It should be clear by now that he can't be brought to his senses, only left to indulge in his delusions...hopefully harmlessly. Anyway, he's got what, ten years left, tops? His blood pressure must be through the roof.

Sep 30, 2014
This has nothing to do with my argument, Johan. Read it more carefully. I used the time on the mirror's clock, but I could have just as easily used the time on the clock at 0'.
At the time t'=t=0, the time at ALL positions within BOTH K' and K are THE SAME TIME namely ZERO! Thus, at the stationary position of the mirror in K' the time is t'=0 and at the instantaneous position of the mirror in K, THE CLOCK WITHIN K AT THIS POSITION IS ALSO AT t=t'=0. And when the light-front reaches the mirror within K' at time t' (not equal to zero) the time on a clock at the instantaneous position of the mirror within K is ALSO at t=t', SINCE THE TIME AT ANY INSTANT CANNOT BE SIMULTANEOUSLY DIFFERENT AT THE SAME INSTANT IN TIME. ONLY AN INSANE PERSON WHO EMBRACES THE ABSURD AS REAL WILL ARGUE THAT THE SAME INSTANT IN TIME CAN OCCUR AT DIFFERENT TIMES. The same instant in time HAS ALWAYS MEANT THE SAME TIME: IT CANNOT MEAN ANYTHING ELSE EVER!


Sep 30, 2014
At the time t'=t=0, the time at ALL positions within BOTH K' and K are THE SAME TIME namely ZERO! Thus, at the stationary position of the mirror in K' the time is t'=0 and at the instantaneous position of the mirror in K, THE CLOCK WITHIN K AT THIS POSITION IS ALSO AT t=t'=0

Well, it's very simple. If observers at rest with K' measure two events as simultaneous, observers at rest with K, will, in general not measure them as simultaneous. In this case, the observer at rest with K' measures the event of the origins coinciding and the event of the mirror being at L' WRT K' as simultaneous. This is not the case with K.

Also, Johan, I just used the Lorentz transformation to map t'= -L'*v/c^2,x'=L'

to

t=0, x=L'/G

You are the one who keeps demanding that I use the LT from K' to K to find out where and when things actually occurred WRT K.

eppur si muove

Sep 30, 2014
At the time t'=t=0, the time at ALL positions within BOTH K' and K are THE SAME TIME namely ZERO! Thus, at the stationary position of the mirror in K' the time is t'=0 and at the instantaneous position of the mirror in K, THE CLOCK WITHIN K AT THIS POSITION IS ALSO AT t=t'=0. And when the light-front reaches the mirror within K' at time t' (not equal to zero) the time on a clock at the instantaneous position of the mirror within K is ALSO at t=t', SINCE THE TIME AT ANY INSTANT CANNOT BE SIMULTANEOUSLY DIFFERENT AT THE SAME INSTANT IN TIME. ONLY AN INSANE PERSON WHO EMBRACES THE ABSURD AS REAL WILL ARGUE THAT THE SAME INSTANT IN TIME CAN OCCUR AT DIFFERENT TIMES. The same instant in time HAS ALWAYS MEANT THE SAME TIME: IT CANNOT MEAN ANYTHING ELSE EVER!

NEEDS MOAR CAPS! BWAHAHAHAHA!

Sep 30, 2014
The wavefront actually reaches the mirror at the time t=t' at the postion within K given by x=L'+vt. At the SAME time in K' and K and at the concident positions x'=L' and x=L'+vt'.

But the Lorentz transformation demands that this event must be recorded in terms of the speed of light and another time t(L) and another distance x(L) relative to 0 within K. Since such a reference of the event is the same as sending out a wavefront at the mirror towards 0, the event is not recorded within K at the actual position x=x'+vt and t=t' but at the distance x(L)=L'*G*(1+v/c) and time t(L)=(L'/c)*G*(1+v/c). This is required since the frequency of the light signal of this wavefront is different within K and K'.

As any idiot should know, the factor G*(1+v/c) is the Doppler factor for a light wave that comes from a source (in this case the mirror) which is moving away from 0. Within K' the mirror is not moving way from 0' and the recorded time is thus the time t' at the position x'=L'.

Sep 30, 2014
@Johanfprins
Your mind is a hydra of failure. It is clear that continuing this conversation is pointless, and possibly even sadistic on my part. I am obviously antagonizing a delusional, old, man. As ornery as you may be, it would be cruel for you to learn that you have devoted the last few years of your precious life to tilting at windmills, instead of applying your intellect to more useful pursuits.

As Code_Warrior said, I won this argument about 20 posts ago, and anyone who reads this, who isn't insane, will be able to follow my methodical reasoning, and justifiably reject yours. They'll see that, despite your protestations that I am upholding "dogma" and "not thinking for myself", that I have very clearly noted the basic errors in your reasoning, never once invoking my accomplishments, or the name of some luminary in place of actual argument, as you have.

I leave you with the last word. Goodbye, Johan. :)

Sep 30, 2014
@johanfprins, I gave you a 5 by mistake. Don't get excited. Think of it as charity.

Also, take a lesson from Russel Crowe's character in the movie "A Beautiful Mind" and notice that as you age your imaginary peer reviewers never get any older - especially the little girl physics genious who agrees with you.

Either that, or you have real peer reviewers who have just decided to give up and tell you that you're right because they're sick of your shit and hope you will leave them alone.

Whatevs. Carry on. Fight the good fight and all that. And USE MOAR CAPS IN YOUR POSTS FERGODSAKES!

Sep 30, 2014
The Lorentz transformation is used to transform the time and position coordinates of an event in K' into the corresponding coordintaes in K. The time is determined by the relationship of the event relative to origins of K' and K. When an event occurs at the SAME instant in time within BOTH K' and K, and the event in K' occurs at a stationary body within K', like the mirror at x'=L', this event must be recorded relative to both 0' and 0 by means of a light signal from the coincident event at time t'=t moving to both 0' and 0.
Within K' the light signal comes from a stationary mirror, while in K the mirror is moving. Thus, in the latter case there is a Doppler shift. And since c=(omega)/k, and since c must be constant, a change in (omega) must be proportional to a change in k. (omega) is inversely proportional to time and k inversely proportional to length. They must both change from the actual coordinates at which the event occurs within K: i.e. t=t' a x=x'+v, by the Doppler factor.

Sep 30, 2014
And as I have found above this is true since the Doppler shift for a body moving away from 0 is D=G*(1+v/c): So that x=D*L' and t=D*t'.

Consider now a MM arm where the source is at x'=0 and the mirror at x'=(minus)L'. The wavefront will now move along the negative direction of x' and reach x'=(minus)L' after a time t'=L'/c. The LT coordinates in K is thus x(L)=G((minus)L'+vt')= (minus)L'*G*(1-v/c) AND the time is t(L)_=G*(t'-(v/c)*(L'/c))=(L'/c)*G*(1-v/c)) so that the speed of light is given by c(K)=x(L)/t(L)=(minus)c: as it should be according to postulate that the relative speed of light MUST ALWAYS have the magnitude of c.

Furthermore the factor which has to be multiplied with L'/c and L' is G*(1-v/c), which s the Doppler factor for he mirror whuich is now approaching the origin 0 of K. In other words: Now the time and the distance is shortened.

You all are a bunch of moronic arseholes for not understanding what relativity is all about! When the event moves away from 0, ......

Sep 30, 2014
This comment has been removed by a moderator.

Sep 30, 2014
Can you not see what a bunch of arseholes you all have been all along? You are the SCUM of the earth!

Good riddance!!! If you have any integrity you will all go and commit suicide. Unfortunately you will not do this to rid this earth of rubbish like you!

You know, the real irony here is that the title of the article is:
"A fun way of understanding Einstein's General Theory Of Relativity"

I don't think johanfprins had very much fun. He seems upset by the whole thing. I guess some people just don't know how to have fun. Johan, dude, you seriously need to get laid....

Sep 30, 2014
Good riddance!!! If you have any integrity you will all go and commit suicide. Unfortunately you will not do this to rid this earth of rubbish like you!


That's pretty rich for someone who accuses everyone one else of being a criminal without reason. *Counseling suicide* is actually a bona fide criminal act in a number of countries.

Sep 30, 2014
You know, the real irony here is that the title of the article is:
"A fun way of understanding Einstein's General Theory Of Relativity"


:^)

Maybe PhysOrg staff can ask Johan to contribute the article "A Sad Way of Misunderstanding Einstein's Special Theory Of Relativity". (He's not even at the level of misunderstanding GR yet.)


Sep 30, 2014
Maybe PhysOrg staff can ask Johan to contribute the article "A Sad Way of Misunderstanding Einstein's Special Theory Of Relativity". (He's not even at the level of misunderstanding GR yet.)

Congratulatons. You almost made me spit Dr. Pepper on my keyboard.

Sep 30, 2014
Good riddance!!! If you have any integrity you will all go and commit suicide. Unfortunately you will not do this to rid this earth of rubbish like you!

That's pretty rich for someone who accuses everyone one else of being a criminal without reason. *Counseling suicide* is actually a bona fide criminal act in a number of countries
@Gawad
you know, after reading this link: https://en.wikipe...h_Africa
and looking at some South African laws, it may well be that you can report Johan to the internet crimes department for his threats and counseling above!
It IS a crime according to what I have read so far (on law sites for SA)... as well as being Hate Speech and a threat, which can be called Battery it might be terrorist threatening as well..


SA has a big Web-crime ctr: http://cybercrime...2013.pdf

according to the PDF, you can report johannie for FRAUD as well, given his non-peer reviewed book
:-D

Sep 30, 2014

Maybe PhysOrg staff can ask Johan to contribute the article "A Sad Way of Misunderstanding Einstein's Special Theory Of Relativity". (He's not even at the level of misunderstanding GR yet.)

I see what you did there. Nicely done!

Oct 04, 2014
Just a couple of words. Terms like "large" and "small" have no physical significance in this oft repeated science fact. It's common, for example, that the reason nuclear bombs are so "big" is that the speed of light is so big. Sheesh, how poorly we teach our childern.

The speed of light in not descibed through adjectives like "big". Physics doesn't have a meanig for "big". It's all just a matter of numan scale. It's big to US but not physics. It's trivial, for example, to define light as one (1), just 1. It's just a matter of how you choose your units.

And simple E = M is a lot more meaningful than the usual. Mwss and energy are the same thing. There'e not a "lot" of energy in mass. They are exactly the same size.

Why care? Because we constantly infect our young scientists with these improper, common sense, view of physics even though real science has moved far beyond this.

Same thing every time we say quantuum mechanics. It's not. It's classical mechanics that

Please sign in to add a comment. Registration is free, and takes less than a minute. Read more