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 energy. Enter E=mc^{2}. 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 m^{2}/s^{2}]. 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 matter? 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 mass, we can plug the values into the equation and see just what we get:

E=mc^{2}

E= (86.18kg)(3.00 × 108m/s)^{2}

E= 7.76 × 10^{18} 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 = 4.18×10^{12} 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 10^{18} J

7.76 x 1018 J x 1 kT TNT/ 4.18 x 10^{12} 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….

**Explore further:**
Superman's solar-powered feats break a fundamental law of physics

## antialias_physorg

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.

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.

## thefurlong

## john_p_tarver

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

## Milou

## thefurlong

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)

## thefurlong

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

## john_p_tarver

Light has no mass and is never a padticle.

## thefurlong

What?

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.

## julianpenrod

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.

## john_p_tarver

## Milou

## thefurlong

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.

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.

## john_p_tarver

## antialias_physorg

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.

## john_p_tarver

## john_p_tarver

## thefurlong

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.

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?

## thefurlong

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

Why?

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.

## Code_Warrior

## john_p_tarver

## Code_Warrior

## john_p_tarver

## Jeppe

Sep 25, 2014## john_p_tarver

## Code_Warrior

## antialias_physorg

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.

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.

## johanfprins

Sep 26, 2014## johanfprins

Sep 26, 2014## thefurlong

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.

## johanfprins

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.

## johanfprins

This is what happes when a photon wave has moved through two slits This must obviously be so since the absorber is not the size of the screen but the size of a electronic atomic orbital. Correct! The expanded wavefront resonates with the atomic electrons and collapses in size. A SPOT is formed!!

## thefurlong

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?

## johanfprins

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".

## johanfprins

## thefurlong

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.

## johanfprins

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?

## Anda

## johanfprins

Sep 26, 2014## Uncle Ira

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.

## bluehigh

It's the energy required for containment.

- 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.

## whitewalking

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.

## thefurlong

If you think I said something intelligent, I must be doing something wrong.

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.

## johanfprins

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!!

## thefurlong

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?

## Code_Warrior

## johanfprins

Sep 26, 2014## thefurlong

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,

And

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.

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

## johanfprins

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!!

## thefurlong

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.

## thefurlong

No, Johan, I have solved Maxwell's equations a few times, now. I could probably show you a thing or two concerning them.

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?

## johanfprins

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!!

## johanfprins

Sep 26, 2014## johanfprins

## Uncle Ira

## thefurlong

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.

Nobody cares.

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.

## thefurlong

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.

## thefurlong

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)

## thefurlong

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.

## thefurlong

Which would then have to be peer reviewed.

Which would then have to be peer reviewed.

Which would then have to be peer reviewed.

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.

## thefurlong

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.

## johanfprins

Sep 27, 2014## johanfprins

Sep 27, 2014## Code_Warrior

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.

## thefurlong

Oh please. Stop being such a predictable crackpot. Put some oomph into that misplaced condescension.

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.

## thefurlong

Neeeigggh

(he's wrong)

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.

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.

## thefurlong

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.

## gwrede

## Goika

Sep 27, 2014## johanfprins

Sep 28, 2014## johanfprins

You are beneath contempt!

## johanfprins

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

## johanfprins

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.

## johanfprins

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.

## johanfprins

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.

## Code_Warrior

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.

## Uncle Ira

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 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.

## johanfprins

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!!

## johanfprins

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"!

## Code_Warrior

## Code_Warrior

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.

## johanfprins

Sep 28, 2014## johanfprins

Sep 28, 2014## johanfprins

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.

## johanfprins

Sep 28, 2014## thefurlong

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?

## thefurlong

Uh, no, Johan. He claimed that the speed of light, as measured in any IRF is c. Stop being dense.

Yes, because I understand how coordinate systems work.

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?

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!

## thefurlong

Oh? And here I thought it was developed by rabidly yelling at people who correct you! How silly of me!

Uh, Johan, you don't know what paper I published.

I haven't started it yet. Oh, Johan, you and Clever Hans are so cute when you assume things.

Because I don't yell with spittle flecks at people who disagree with me?

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?

## thefurlong

What is English, and how does it work?

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.

## thefurlong

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?

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.

## thefurlong

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:

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)

## johanfprins

Sep 28, 2014## thefurlong

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.

## thefurlong

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.

## thefurlong

Yes, you do. You believe that uncertainty, time dilation, and length contraction are VOODOO.

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.

## johanfprins

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.

## johanfprins

Sep 29, 2014## johanfprins

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".

## johanfprins

Sep 29, 2014## johanfprins

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.

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!

## johanfprins

Sep 29, 2014## johanfprins

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

## johanfprins

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.

## johanfprins

## johanfprins

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!

## johanfprins

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?

## thefurlong

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.

## johanfprins

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?

## thefurlong

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)

## thefurlong

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.

## johanfprins

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

## thefurlong

Yep. An observer in K' would measure light as moving relative with the mirror with speed 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?

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)

## thefurlong

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)

## thefurlong

"change in distance"/"change in time" = (L'*G(1-v/c))/(L'*G/c) = (c - v)

## thefurlong

when I meant to write

## johanfprins

This is irrelevant POEPHOL!

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.

## thefurlong

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)

## johanfprins

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.

## johanfprins

## thefurlong

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)

## thefurlong

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.

## johanfprins

Sep 30, 2014## johanfprins

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) .

## thefurlong

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

with

Read my argument again, and stop jumping to conclusions.

## thefurlong

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.

## Code_Warrior

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.

## thefurlong

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.

## Gawad

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

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.

## johanfprins

## thefurlong

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

## Code_Warrior

NEEDS MOAR CAPS! BWAHAHAHAHA!

## johanfprins

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'.

## thefurlong

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. :)

## Code_Warrior

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!

## johanfprins

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.

## johanfprins

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, ......

## johanfprins

Sep 30, 2014## Code_Warrior

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....

## Gawad

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

:^)

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.)

## thefurlong

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

## Captain Stumpy

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

## Code_Warrior

I see what you did there. Nicely done!

## Accounts

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