Extending Einstein: Researchers demonstrate a new kind of quantum entanglement

Dec 14, 2012

Physicists at the University of Calgary and at the Institute for Quantum Computing in Waterloo have published new research in Nature Physics which builds on the original ideas of Einstein and adds a new ingredient: a third entangled particle.

is one of the central principles of , which is the science of sub-atomic particles. Multiple particles, such as photons, are connected with each other even when they are very far apart and what happens to one particle can have an effect on the other one at the same moment, even though these effects can not be used to send information faster than light.

The new form of three-particle entanglement demonstrated in this experiment, which is based on the position and momentum properties of photons, may prove to be a valuable part of future communications networks that operate on the rules of , and could lead to new fundamental tests of that deepen our understanding of the world around us.

"This work opens up a rich area of exploration that combines fundamental questions in quantum mechanics and quantum technologies," says Christoph Simon, paper co-author and researcher at the University of Calgary.

This research extends the theories of Einstein, seventy-seven years later. In 1935, Albert Einstein, Boris Podolsky, and Nathan Rosen, commonly referred to as EPR, published a thought experiment designed to show that quantum mechanics, by itself, is not sufficient to describe reality.

Using two EPR tried to demonstrate that there must be some hidden parameters that quantum mechanics does not account for. Later John Bell and others showed that the kind of hidden parameters EPR had in mind are incompatible with our observations. The mystery at the heart of quantum mechanics thus remains intact. But the entanglement first proposed by EPR is now a valuable resource in emerging like , , and quantum precision measurements.

"It is exciting, after all this time, to be able to finally create, control, and entangle, quantum particles in this new way. Using these new states of light it may be possible to interact with and entangle distant quantum computer memories based on exotic atomic gases, " says Thomas Jennewein, whose group at the University of Waterloo carried out the experiment.

The next step for the researchers is to try to combine the position and momentum entanglement between their three photons with more traditional types of entanglement based on angular momentum. This will allow the creation of hybrid quantum systems that combine multiple unique properties of light at the same time.

Explore further: Physicists design quantum switches which can be activated by single photons

More information: Three-photon energy–time entanglement: www.nature.com/nphys/journal/vaop/ncurrent/abs/nphys2492.html

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Maggnus
2.1 / 5 (7) Dec 14, 2012
and what happens to one particle can have an effect on the other one at the same moment, even though these effects can not be used to send information faster than light.


I'm confused, if what happens to one particle happens to another instantaneously, regardless of distance, why can't this be used to send information faster than light?
Silentsam
1 / 5 (4) Dec 14, 2012
and what happens to one particle can have an effect on the other one at the same moment, even though these effects can not be used to send information faster than light.


I'm confused, if what happens to one particle happens to another instantaneously, regardless of distance, why can't this be used to send information faster than light?

The information does not need to be transmitted faster than light because the particles where once near each other and left each other at the speed of light or less.
Lurker2358
1.7 / 5 (6) Dec 14, 2012
The information does not need to be transmitted faster than light because the particles where once near each other and left each other at the speed of light or less.


Wrong.

They left one another at the speed of light or less, but once they are a decent distance away from one another, they can be much farther away than a light second, silly.

The Moon is farther than one light second. Therefore if had a moon base with a pair entangled with the Earth, and somebody makes a change in the first of an entangled pair, the change will be observed in the other member instantly.

It doesn't matter how fast or how far the ship flies, because the "information" is entangled.

Further, you need not know what type of information is being entangled, because you can use the mere fact of "any change" occurring as a signal.

Example in macroscopic world would be like if you see a smoke signal, that is a 1 no matter what it looks like, and if you see a pause, that is a 0, no matter what.
Lurker2358
2 / 5 (8) Dec 14, 2012
To be sure you are sent a "0" when you are supposed to be sent a zero, and a "1" when you are supposed to be sent a one, you can use two pairs of entangled particles to represent one "bit".

If there is a change in particle A1, you interpret that as a "1".

If there is a change in particle A2, you interpret that as a "0".

You can use a third pair of entangled particles to send a "change" as a control, A3, any time you are intentionally sending a signal to either of the other two particles (A1 or A2). This way you reduce the number of errors.

So if you see a change in either particle A1 or A2 simultaneously with a Change in particle A3, then you know an intentional message has been sent. If the change was in A1 it represents a "1". If it was in A2, it represents a "0".

You do not need to know the size or type of change ahead of time to simply recognize that "a change" has happened.

Now measuring a change may require additional tricks or apparatus...
Lurker2358
2 / 5 (8) Dec 14, 2012
Anyway, the point is:

1, Sending no information is itself information.
2, Sending a zero/0 can be accomplished by sending "any change" to a second entangled particle. This avoids the notion of "waiting twice as long" to represent a zero.
3, Sending a one/1 is very easy.

So you need at least 3 particles to send 1 formatted bit without error.

One is used as a time keeper to prove that the people on the other end are in fact intending to send a signal.
One is used as a "half-bit" to represent binary 1.
One is used as a "half-bit" to represent binar 0.

They've been trying so hard to abuse quantum behavior that they've forgotten to USE normal behavior and concepts...
clay_ferguson
3.6 / 5 (17) Dec 15, 2012
Lurker, when particles are entangled you don't know WHICH state EITHER particle is in. So you can't send information, because you cannot assign any states when this "both unknown" (i.e. entanglement) situation exists. Entanglement means you know NOTHING. However, once you 'observe' one particle then you can SEE its state, and you KNOW for sure the other particle must have the opposite state. However that's NOT new information to you. You can't send info this way. It's mathematically provable you cannot.
Anonimo_Napoletano
2 / 5 (4) Dec 15, 2012
I did not understand if these entangled particles are just twins
(created equal, so if one knows a particle, he knows that the other is the same)
or are twins and symbiotic (born equal, but if one has a change the other also has the same change)
in the second case if one produces two particles from a coherent source, inducing a change in any one particle, see the change in the second and then can send information.
Although it will never exceeded the speed of light.
However, if these particles were numerous and formed an atom having two entangled atoms and separating them at a great distance, if I supply power to one (for example, causing an emission of light), I might have the same effect on the other and instantly.
Noumenon
2.7 / 5 (37) Dec 15, 2012
and what happens to one particle can have an effect on the other one at the same moment, even though these effects can not be used to send information faster than light.


I'm confused, if what happens to one particle happens to another instantaneously, regardless of distance, why can't this be used to send information faster than light?


Because each observation is intrinsically indeterminate. One cannot control the result of the measurement so one cannot control the 'conversation', thus one cannot send information.

So in order to observe entanglement, one must bring both sets of measurements together at <=c and compare. Only at this time does one see correlations that are stronger than could be explained by 'classical physics'.

Noumenon
2.8 / 5 (38) Dec 15, 2012
So if you see a change in either particle A1 or A2 simultaneously with a Change in particle [...] You do not need to know the size or type of change ahead of time to simply recognize that "a change" has happened.


The fact that you think the matter can be resolved with such triviality, should be an indication to you that you don't understand qm properly.

For example, it does not make any sense to say 'when a change occurs', or to 'wait for a change to occcur'. You don't know the state of each particle to begin with, so how do you know if it changed?

The change that occurs to the state is the result of measurement. The 'actual state' before measurement is a linear superposition of all possible states, so it is not in 'one specific state' where you can see if it changed.
antialias_physorg
3.7 / 5 (6) Dec 15, 2012
I'm confused, if what happens to one particle happens to another instantaneously, regardless of distance, why can't this be used to send information faster than light?

Because information transmission is defined thus:
1) You have 'a priori' knowledge (knowledge of what you encode)
2) After sending and receiving the carrying entity, you have 'a posteriori' knowledege (knowledge of what you received).

The correlation (preferrably fully correlated) of these two states is what makes up information transmission.

But with entanglement you don't have 1)

You don't know what state your particle is in a priori - because as soon as you look (or act on it to force it into a state) you break the entanglement: You can't encode anything on it.

You only ever have a posteriori knowledge - and that is useless for information transmission because you can't correlate to the unknown a priori state (but useful enough for encryption - which is not information transmission).
antialias_physorg
3.7 / 5 (6) Dec 15, 2012
1, Sending no information is itself information.

No it is not. If you mean by just establishing a ticking clock at each end and saying that "if at tick X no signal arrives then interpret that as a zero" then that still doesn't allow FTL information transmission because the guy at the sending end must decide when not to send something (encode the information). And the observation of him not encoding stuff is limited by the speed of light.
Sending a zero/0 can be accomplished by sending "any change" to a second entangled particle.

But since you don't know what state it was in initially you can't know if there was a change at all. And anyhow. Doing something to the other entangled entity at the source doesn't change the particle at the destination at all. Only observing it will tell you what you will observe in your particle. You can't FORCE it into a state and have the other one comply. That's not what entanglement means.
Maggnus
1.2 / 5 (5) Dec 15, 2012
Thank you to Noumenon and Antialias for your explanations. That makes perfect sense to me.
Pressure2
2.3 / 5 (6) Dec 15, 2012
If you don't know the state either is in what proof do you have that they are in both states or superposition? They could just as easily be in the state they were created in when you latere detect one or the other.

I see no difference between this and splitting a coin in half, mixing them up and tossing them into the air. When checking one you instantly know what the other is.
Lurker2358
1.4 / 5 (9) Dec 15, 2012
However that's NOT new information to you. You can't send info this way. It's mathematically provable you cannot.


Silly.

How would you prove entanglement happened at all if there is no "information" being sent? An "entanglement" theory or experiment would be non-falsifiable if that were actually the case.

Doing something to the other entangled entity at the source doesn't change the particle at the destination at all. Only observing it will tell you what you will observe in your particle.


If that's true then entanglement is non-falsifiable, as you can't even prove it ever actually happened.

Being non-falsifiable doesn't prove something isn't true, it just proves you can't prove it.

However, I certainly hope the entanglement buzz has not been about "non-falsifiable" measurements, because their are entirely too many people working on this for that to be acceptable.
Lurker2358
1.5 / 5 (8) Dec 15, 2012
No it is not.

"No news is good news".
The simultaneous absence of any change in any of the 3 particles, especially the third, means no signal has been sent. If no signal has been sent then you know you don't need to do anything.

But since you don't know what state it was in initially you can't know if there was a change at all. And anyhow. Doing something to the other entangled entity at the source doesn't change the particle at the destination at all. Only observing it will tell you what you will observe in your particle. You can't FORCE it into a state and have the other one comply. That's not what entanglement means.


If that is happening, then how do you prove entanglement exists at all, except as a consequence of mutually related classical causes?

I can "entangle" two points on a disk just by rotating it, since they must obey the formula of a circle and angular momentum. A change in one particle, which is controllable, changes the other automatically.
Lurker2358
1.4 / 5 (9) Dec 15, 2012
Now that's a classical example, but what you have described is actually no different, except that you're saying you can't tell a change actually happened at all.

Something is absurd about that.

What you are describing is nothing more than a conservation law trick, where the observe is blind to the initial conditions.

Which is to say, if you split a photon, you'd obviously expect the two photons to sum to the initial one in energy and properties, due to conservation. So if you randomly split a photon, daughter B is predicted by the characteristics of Daughter A, because of conservation laws. There's nothing special about that just because you have an observer who is "blind" at the moment the split happens.

I could infer, or a rocket scientist could anyway, about how much fuel was "probably spent" by a rocket flying by just based on it's relative velocity. Now assumptions about it's origin may change probabilities, but overall, the properties are knowable..
Noumenon
2.7 / 5 (37) Dec 15, 2012
I see no difference between this and splitting a coin in half, mixing them up and tossing them into the air. When checking one you instantly know what the other is.


That analogy assumes that the states are determined before you 'toss them into the air', as in a hidden variable theory. That has been proven not to represent quantum entanglement, which is to say, not account for the statistical correlations that actually occur.

There is no intuitive analogy to it. It appears as though the two particles or the two photons, exist as One object, an entangled system, until it is measured.
Noumenon
2.7 / 5 (37) Dec 15, 2012
If that is happening, then how do you prove entanglement exists at all, except as a consequence of mutually related classical causes?


You can make two independent catalogs of measurements for each entangled particle then bring the two together and compare the results. It is found that the resulting statistical correlation is stronger than what can be intuitively explained, or what can be explained by assuming 'hidden variables'.
Pressure2
2.6 / 5 (5) Dec 15, 2012
If that is happening, then how do you prove entanglement exists at all, except as a consequence of mutually related classical causes?


You can make two independent catalogs of measurements for each entangled particle then bring the two together and compare the results. It is found that the resulting statistical correlation is stronger than what can be intuitively explained, or what can be explained by assuming 'hidden variables'.

Well what does that prove. If you bring the two coin halves back together again there is a 100% chance of a correlation.
Noumenon
2.8 / 5 (40) Dec 15, 2012
If that is happening, then how do you prove entanglement exists at all, except as a consequence of mutually related classical causes?


You can make two independent catalogs of measurements for each entangled particle then bring the two together and compare the results. It is found that the resulting statistical correlation is stronger than what can be intuitively explained, or what can be explained by assuming 'hidden variables'. -Noumenon

Well what does that prove. If you bring the two coin halves back together again there is a 100% chance of a correlation.


I just told you in effect that the coin analogy (hidden variables) fails.
Pressure2
2.6 / 5 (5) Dec 15, 2012
If that is happening, then how do you prove entanglement exists at all, except as a consequence of mutually related classical causes?


You can make two independent catalogs of measurements for each entangled particle then bring the two together and compare the results. It is found that the resulting statistical correlation is stronger than what can be intuitively explained, or what can be explained by assuming 'hidden variables'. -Noumenon

Well what does that prove. If you bring the two coin halves back together again there is a 100% chance of a correlation.


I just told you in effect that the coin analogy (hidden variables) fails.

Tell me why it would fail with coins but not with two polarized photons?
Pkunk_
1.8 / 5 (5) Dec 15, 2012
Lurker, when particles are entangled you don't know WHICH state EITHER particle is in. So you can't send information, because you cannot assign any states when this "both unknown" (i.e. entanglement) situation exists. Entanglement means you know NOTHING. However, once you 'observe' one particle then you can SEE its state, and you KNOW for sure the other particle must have the opposite state. However that's NOT new information to you. You can't send info this way. It's mathematically provable you cannot.

That "limitation" can be easily sidestepped by taking a page from the way computers work in transmitting serial/parallel data. You just need two sets of entangled quantum objects. One used exclusively for sending and one for receiving in one end and vice-versa on the other. Since one side is only observing it cannot "confuse" the other end which is only transmitting.
ValeriaT
1 / 5 (8) Dec 15, 2012
There is no intuitive analogy to it. It appears as though the two particles or the two photons, exist as one object
At the water surface the density fluctuations propagate with higher speed, than the speed o surface waves, so they can mediate and "instantaneous" phenomena. This is why so many quantum phenomena are modeled with surface waves so easily (1,2,3,4,..) Of course, these analogies don't fit the anti-aether religion as promoted with many mainstream physics trolls - but who cares... The future is not for these people.
kochevnik
1 / 5 (7) Dec 15, 2012
Nature is an unbroken whole which can only be perceived as an infinitely deep chain of nested paradoxes.
Noumenon
3.5 / 5 (22) Dec 15, 2012
If that is happening, then how do you prove entanglement exists at all, except as a consequence of mutually related classical causes?


You can make two independent catalogs of measurements for each entangled particle then bring the two together and compare the results. It is found that the resulting statistical correlation is stronger than what can be intuitively explained, or what can be explained by assuming 'hidden variables'. -Noumenon

Well what does that prove. If you bring the two coin halves back together again there is a 100% chance of a correlation.


I just told you in effect that the coin analogy (hidden variables) fails.

Tell me why it would fail with coins but not with two polarized photons?


For one thing, each experimenter can decide independently which direction to check the polarization, while there is not such multiple concepts of heads and tails.
Pressure2
1 / 5 (3) Dec 15, 2012
The coin heads and tails can also be sent into any unknown direction, I see no difference.
Q-Star
3.3 / 5 (7) Dec 15, 2012
At the water surface the density fluctuations propagate with higher speed, than the speed o surface waves, so they can mediate and "instantaneous" phenomena. This is why so many quantum phenomena are modeled with surface waves so easily


But when the waves of the flat surface ripple beneath the surface all you are left with is the subcosmological variables.

The future is not for these people.


The future belongs to Zephyr.
ValeriaT
1 / 5 (4) Dec 15, 2012
The future belongs to no one.
antialias_physorg
5 / 5 (4) Dec 15, 2012
If you don't know the state either is in what proof do you have that they are in both states or superposition? They could just as easily be in the state they were created in when you latere detect one or the other.

Because you can observe that the entangled properties are always perfectly correlated at the source and the destination over as many test runs as you'd like to make. If entanglement weren't real then there would be no correlation. (Because you can also make this test with non-entangled entities and you will find no correlation whatsoever.)
Lurker2358
1 / 5 (5) Dec 15, 2012
For one thing, each experimenter can decide independently which direction to check the polarization, while there is not such multiple concepts of heads and tails.


Actually, there is, nobody thinks about it though, because in conventional cases nobody is checking the "bottom" of the coin first.

We could easily do that with a video camera.

Because you can observe that the entangled properties are always perfectly correlated at the source and the destination over as many test runs as you'd like to make. If entanglement weren't real then there would be no correlation. (Because you can also make this test with non-entangled entities and you will find no correlation whatsoever.)


Then you're doing soimething wrong.

If events have a common cause, such as in classical physics, then they are in fact correlated, even if the observer is too lazy or ignorant to realize it.
Pressure2
1 / 5 (3) Dec 15, 2012
If you don't know the state either is in what proof do you have that they are in both states or superposition? They could just as easily be in the state they were created in when you latere detect one or the other.

Because you can observe that the entangled properties are always perfectly correlated at the source and the destination over as many test runs as you'd like to make. If entanglement weren't real then there would be no correlation. (Because you can also make this test with non-entangled entities and you will find no correlation whatsoever.)

Perfectly correlated at the source? Well that does not prove one iota that the two particles are entangled or occupy both states (superposition) until one is detected. They could be in the same state when created and detected just like the coin halves.
Where is your proof that they occupied both states in transit?
antialias_physorg
4.4 / 5 (7) Dec 15, 2012
If events have a common cause, such as in classical physics, then they are in fact correlated, even if the observer is too lazy or ignorant to realize it.

The standard setup goes like this: Create a pair of (non-entangled) photons of which you know that one is vertically and one horizontally polarized. Then send them to different locations where they each encounter a 45degree polarization filter. Photons are probabilistic (as can be seen with the interference experiments) so each has a 50/50 chance of passing or not passing that filter.
Whether photon A passes is in this case independent of whether photon B passes, since the LOCAL chance of passing the 45 degree filter is 50%. So far, so good.

But that's not what happens with entangled photons. Either both pass or don't pass DESPITE the LOCAL chance for each passing is 50% (in real experiments its not quite like that - the pass/don't pass scenarios aren't 100% linked but 'merely' highly correlated)
antialias_physorg
3.7 / 5 (3) Dec 15, 2012
They could be in the same state when created and detected just like the coin halves. Where is your proof that they occupied both states in transit?

If they're merely in the same state by chance then they behave as expected in the above detailed experiment: The pass/don't pass events are independent of one another.
Lurker2358
1 / 5 (5) Dec 15, 2012
But that's not what happens with entangled photons. Either both pass or don't pass DESPITE the LOCAL chance for each passing is 50% (in real experiments its not quite like that - the pass/don't pass scenarios aren't 100% linked but 'merely' highly correlated)


Are the filters the exact same distance from one another?

If you visualize a wave form (whether or not the reality is exactly the same way,) You can see that if the photons are created with related wave patters, i.e. rotated by 90 degrees, and you pass them through a 45 degree filter, the mathematical model would suggest that both waves will in fact either both strike or both miss.

There is nothing special about that.

If you change the distance of just one of the slits by a half-wavelength of light, or perhaps a quarter wavelength, and change nothing else about the experiment, then it should be possible to have one photon go through the slit while the other fails to go through.
Lurker2358
1 / 5 (5) Dec 15, 2012
In other words, the problem is a matter of composition.

If two photons have a related cause (which is actually hard to avoid in an experiment because your very decision to participate in an experiment created a related cause,) anyway, if they have a related cause they will be expected to have related properties (wavelength, polarization, etc.)

The standard setup goes like this: Create a pair of (non-entangled) photons of which you know that one is vertically and one horizontally polarized.


The decision to perform the experiment causes a classical causality relationship.

If you start an engine, one piston goes up and another goes down, and there's nothing spectacular about that.

If a random viewer stops the engine at a random time and looks at the state of the engine, they don't know it ahead of time, but that didn't change anything about the relationship. The state appears random, but that's only due to lack of information.
antialias_physorg
5 / 5 (3) Dec 15, 2012
Are the filters the exact same distance from one another?

Not necessarily.

If you change the distance of just one of the slits by a half-wavelength of light, or perhaps a quarter wavelength, and change nothing else about the experiment, then it should be possible to have one photon go through the slit while the other fails to go through.

I think you still don't understand. You cannot 'force' one to go through a 45degree filter (or not) by moving a filter. It's a 45degree filter!
The chance of passing is ALWAYS 50% for such a filter given that your incident photon is either horizontally or vertically polarized. If you were to observe just one of these filters then 50% of all photons that arrive (entangled or not) pass through and 50% don't.
But if you look at the results from BOTH filters you see that the pass/block events for entangled photons are correlated and those for non-entangled photons are not.
Lurker2358
1 / 5 (5) Dec 15, 2012
You see, when a transverse wave has reached the point in it's cycle so that it's centered on the X axis (as viewed from the side in it's propagation,) it has minimum "height" at that instant.

At that moment, the wave could pass through a filter of nearly any orientation. If two photons are travelling the same distance, and they are related by common cause, then they would both be expected to reach the filter at the same time, and therefore the "probabilities" are exactly the same for each photon. But if they are related, they would both be at the same phase in their wave cycle when they reach the slits. This is just a classical relationship.
Lurker2358
1 / 5 (5) Dec 15, 2012
But if you look at the results from BOTH filters you see that the pass/block events for entangled photons are correlated and those for non-entangled photons are not.


In what way are you "non-entangling"?

If you are sending photons with different source or type of production, they will have different classical properties anyway, or the phase of their wavelengths will be unrelated, or the polarization, therefore they would not be expected to have linked behavior...

It is a matter of classical causality, whether you see that or not.

I'm not sure why you're having trouble seeing that.

Remember Newton "For every action there is an equal and opposite reaction".

If you split photons the act of splitting automatically produces like characteristics in the daughter photons due to classical conservation laws, therefore they would be expected to have linked behavior...
antialias_physorg
4 / 5 (4) Dec 15, 2012
You see, when a transverse wave has reached the point

It doesn't matter where you position the second analyzer or whether you test the second photon billions of years after the first. You can even decide on entanglement AFTER measurement - which is completely unexplainable by local variables.

Here's a simple experiment, very well explained:
http://scienceblo...xperime/
Lurker2358
1 / 5 (5) Dec 15, 2012
It doesn't matter where you position the second analyzer or whether you test the second photon billions of years after the first. You can even decide on entanglement AFTER measurement - which is completely unexplainable by local variables.

Here's a simple experiment, very well explained:
http://scienceblo...xperime/


Unfortunately, that doesn't work either.

The common cause is still in the decision to do an experiment at all, as the production of all of the photons is in fact entangled classically in that moment.

"I'd like to do an entanglement experiment. Let's get to work!!"

Boom. You have already classically entangled the photons, as YOU are the common cause...

the decision to entangle photons was made ahead of time. You just entangled "random" photons, and moreover the "randomness" of the choice, is actually an introduction of classical causal relationship...
ValeriaT
1 / 5 (3) Dec 15, 2012
Mechanical analogy of 2-photon and 3-photon entanglement measured in the above study.
Lurker2358
1 / 5 (5) Dec 15, 2012
The opposite sides of a standard 6 sided die always add up to 7.

If you have 6 detectors and you rolled a 6 sided die, the outcome on one face determines the outcome on the opposite face.

Better to use colors to represent numbers so you don't contaminate the test.

It's true that you won't know the exact orientation of the other 4 based on only a color(number), but there is nothing special about that because it's a matter of partial knowledge.

If you knew enough about the physics of the dice toss you could predict the orientation of all 6 sides.

Knowing any one other face beyond top and bottom gives you the state of the entire system. In other words, if you know any 3 faces you can predict the other 3 exactly, including an entire pair you have not seen...
antialias_physorg
4.2 / 5 (5) Dec 15, 2012
You have already classically entangled the photons, as YOU are the common cause...

That makes no sense whatsoever - since it is perfectly possible to create unentangled photons (and much more easily than entangled ones).

From that statement alone it seems you don't have the faintest clue what entanglement even means. Please look it up before commenting.

And these experiments have been done. Reality trumps your "can't be" any time.
Q-Star
3 / 5 (6) Dec 15, 2012
2-photon and 3-photon entanglement measured in the above study.


It looked more like a dozen or so photons riding on that transverse wave as they meet a longitudinal disconnect. The red ones are the photons, right?
Lurker2358
1 / 5 (5) Dec 15, 2012
And these experiments have been done. Reality trumps your "can't be" any time.


The experiment is not the problem.

The problem is interpretation of the experiment.

I never said it hadn't been done, and you don't get it.

"Splitting" a photon into a pair is classical, causal entanglement. It can't help being so any more than rolling a die and comparing opposite faces.

Sure you can create non-entangled events.

The results of rolling two die, for example, one die result is not necessarily dependent on the other, because the two die are not entangled.

But the opposite faces of the same die are entangled.

Two daughter photons are "entangled" but it is nothing more than opposite faces of the "die" which was the original photon.
frajo
4 / 5 (4) Dec 15, 2012
How would you prove entanglement happened at all if there is no "information" being sent? An "entanglement" theory or experiment would be non-falsifiable if that were actually the case.
Falsifiability is a property of (scientific) statements and theories, not a property experiments may have. Experiments may be reproducible or not.

If that's true then entanglement is non-falsifiable, as you can't even prove it ever actually happened.

As entanglement is neither a statement nor a theory it cannot have the property "falsifiability".

Being non-falsifiable doesn't prove something isn't true, it just proves you can't prove it.

While entanglement is not a statement, the statement "there exist entangled photons" is perfectly falsifiable (by measurements).

However, I certainly hope the entanglement buzz has not been about "non-falsifiable" measurements ...
I'm afraid you got the concept of falsifiability wrong.
Lurker2358
1 / 5 (5) Dec 15, 2012
By the way...

The subtractive causal relationship of a die roll:

7 - top = bottom
7 - bottom = top
7 - left = right
7 - right = left
7 - front = back
7 - back = front

Is always correct, 100% of the time. So it is not possible for an alleged "quantum entanglement" experiment to have a higher correlation than that.

You can pick any 3 sides and prove the other 3.

You can pick any 2 non-opposite sides and actually prove the other 4 sides, if you have the die memorized.
Lurker2358
1 / 5 (5) Dec 15, 2012
On a standard die, which uses 1 through 6 and has dots, it's possible to infer several things about the 4 unknowns in some cases.

If you have a "2" for your top number, you know the "5" automatically.

However, since the dots are on a diagonal, and the diagonal has 180 rotation symmetry, not 90 degrees, you can infer the orientation of the 3-4 axis and the orientation of the 6-1 axis. So that by making one observation, seeing the "2," you will know the opposite face is a 5, and you'll know the orientation of the axis of all the other 4 faces.

The same can be done with the "3" and the "6" because they too have only 180 degree rotation symmetry, not 90 degrees.

This is why I suggested using a colored die to reduce the giveaways, since the standard die is actually even more entangled that color alone (representing numbers,) would be.

The standard die gives you "information" not only about the opposite face, but in the case of three faces (2, 3, and 6,) they give you almost everything
Lurker2358
1 / 5 (5) Dec 15, 2012
Now finally, if you used all digits instead of dots: "1,2,3,4,5,6," then everything about the die would be entangled.

If you rolled a "1" you would be able to predict the value and orientation of all five of the other sides, without looking at any of them, thus it would be impossible to roll the die and not get all the information about all 6 faces simultaneously.
Lurker2358
1 / 5 (4) Dec 15, 2012
Trolls: 1 me if you like. You know nothing.

Try this:

Look into the vertex at the top of a corner in a room. Us the handle of a flashlight or something like a book to break up the outline of the corner. Stare until your eyes deceive you and you see the walls and ceilings reverse their apparent spacial orientation, so that they appear to be a surface, popping out towards you instead of receding away.

Well, we know they are not popping out. How do we know this?

Context clues in the environment. We know the surrounding reality well enough to realize the illusion is not actually happening.

If you disregard the information of reality that you already know, ti can be open to interpretations which may be absurd.

However, if you interpret the observation through your knowledge of the surroundings, you will know the illusion is certainly not real.
Pressure2
1 / 5 (3) Dec 15, 2012
Big deal AP, if the polarized photons are correlated they either go through a 45 degree offset or not. Why wouldn't they both go through the filter after all both are offset by the same amount, 45 degrees. Naturely if you have two polarized photons that are not 90 degrees out of phase and you use a 45 degree offset on both one might go through while the other is filter out.

This absolutely proof nothing, it is simple classical physics!

Now provide me proof that the photons are in both states, superposition, at the same time until one is read.
Tausch
1 / 5 (3) Dec 15, 2012
If all the elements are probabilistic in nature for a system, then words like "instantaneous change" are unavoidable and necessary (and insufficient) when describing the information of experiments whose outcomes demand such wording.
We are not at a loss for lack of interpretations.
Our non mathematical vocabulary sucks.

"For exact understanding exact language is necessary."

Welcome back, frajo.
Lurker2358
1 / 5 (2) Dec 15, 2012
Ok, AP, try this:

Two entangled particles.

Hit one of the pair with it's own anti-particle before you measure the other member, this collision will be a "measurement" by default, but it will also change the values of the measurement, possibly, since the "output" of the measurement will be the energy and particles released in the annihilation.

What do you expect to happen to the other member from the entangled pair and why?
daywalk3r
2.7 / 5 (7) Dec 15, 2012
People declaring others as being wrong because they "don't understand qm properly".. How ironic.
Photons are probabilistic

Only the result of their interaction, and only in the scope of a statistical model/interpretation..

What happens "inbetween" is not handled by such models whatsoever. And claiming that there is no "inbetween" just because it is not required to predict the outcome, is borderline delusional at best..

Ergo, I don't need to know what exactly happens when a nuclear warhead goes off in close vicinity to a group of people, to accurately predict that they will be all dead after..

That's the core difference between statistics and reality..

I see no problems with statistical models of reality, as long as they are not proclaimed to be reality itself - instead of just being BASED on it(!).

There are people in the field who actually understand this (at least to a certain extent), but sadly the "quantum sensationalist" crowd seems to be alot more prominent nowadays..
clay_ferguson
1.9 / 5 (7) Dec 15, 2012
@noumenon, once one particle is observed in state A you know the other entangled partner has state B. You never have to bring them back together to compare. You KNOW for sure the second one is B.

@pressure2, both particles 'can' be either A or B, but it's not known until you measure one. That's what they mean by superposition.

@lurker @valeriaT, you're an idiot as always, completely ruining any thread you comment on.
ValeriaT
1 / 5 (5) Dec 15, 2012
@valeriaT, you're an idiot as always
A)I'm not lurker and B) I made just two brief posts in this 60 posts long thread. If you believe, I can ruin such a thread just with two posts, then my words have apparently much higher power, than it seems...;-)
daywalk3r
2 / 5 (4) Dec 15, 2012
if the polarized photons are correlated they either go through a 45 degree offset or not. Why wouldn't they both go through the filter after all both are offset by the same amount,45 degrees.

The thing about the polarizer is, that when two photons are "entangled" they are EXACTLY 180 degrees phase shifted, and so have naturally a higher chance to produce the same result when passing through the same polarizer..

What was "proven" by the experiments is, that they more often produce a similar result than not (ergo not 50/50) - which is a clear sign of correlation. Rest is a matter of interpretation, really..

I would like to emphasize that the exact 180deg shift really is the key component here. If non-entangled pairs were used, the Bell curve phase shift distribution (179.999, 180.001, etc.) would make the results converge at the expected 50% chance.

But not in the case of "entangled" pairs, where the shift is allways exactly 180.0000(0) deg., with a distribution width of ZERO.
clay_ferguson
1 / 5 (6) Dec 16, 2012
daywalk, Dirac agrees with you! lol. But personally I believe that higher dimensional beings than us would see that zero distribution width as non-zero, just like "for them", they can see entangled pairs as 'still connected' in some way by seeing things we cannot directly observe.

ValeriaT, I lumped you in because you are a well known troll on this site.
Noumenon
3.4 / 5 (22) Dec 16, 2012
@noumenon, once one particle is observed in state A you know the other entangled partner has state B. You never have to bring them back together to compare. You KNOW for sure the second one is B.


Yes, I know this. Please reread my post. I said one must bring together the two catalogs of measurements (the list of results), in order to see a correlation,... not the photons.
clay_ferguson
1.6 / 5 (7) Dec 16, 2012
Noumenon, you did say bring "measurements" together didn't you ? My bad. But you said bring them together at "< c" it seemed like you meant particles. But you were just emphasizing that even the measurement data could not be transmitted faster than 'c'. Yes you appear to know what you are talking about! :)
Pressure2
1 / 5 (3) Dec 16, 2012
Daywalk3r, I don't think they are 180 degrees out of phase, if they were you would not be able to tell one from the other. I would think they are 90 degrees out of phase with 45 degrees being 1/2 way between both.
clay_ferguson
1 / 5 (5) Dec 16, 2012
@Pressure2, saying 180 degrees phase shifted simply means in opposite phase (antiphase). Saying something is "Polarized at 45 degrees" (or 135) is different because it is not RELATIVE to anything except for "unpolarized"...or zero.
clay_ferguson
1 / 5 (5) Dec 16, 2012
For example, if one photon is at precisely "vertical polarization" then you know that once you verify that by observing it at time 't', you can be sure any entangled partner photon will be at precisely "horizontal polarization" also at time 't' with 100% certainty.
antialias_physorg
4.2 / 5 (5) Dec 16, 2012
What do you expect to happen to the other member from the entangled pair and why?

Nothing whatsoever. CHANGING the value of one entangled pair has no effect on the other in any way (if it did you could transmit information via entanglement - which you can't). It only breaks entanglement.
Only OBSERVATION will fix the value of the other partner. But this is useless for information transmission because you don't know (a priori) which state you will see when observing the particle local to you.
As explained before: To transmit information you need a priori AND a posteriori knowledge. Entanglement only gives you a posteriori knowledge.

Ergo, I don't need to know what exactly happens when a nuclear warhead goes off in close vicinity to a group of people, to accurately predict that they will be all dead after..

Only because that is MANY qm events that average out. For singular events definite predictions are impossible. THAT is reality.
clay_ferguson
1 / 5 (5) Dec 16, 2012
@antialias, changing one particle has a probabilistic effect on any entangled particle right ? I say this because the sum of the probabilities of the two still have to remain 1. Bras and Kets right ?
Pressure2
1.2 / 5 (5) Dec 16, 2012
I agree with you A&P on the information issue, what I don't get is what are they trying to prove with these experiments? It would seem to me to be an exercise in futility.

I also think there is a way to prove the two polarized photons are never in a superposition. Have the sender know which phase he is sending in which direction. I think he could predict with 100% accuracy which receiver would receive which phase proving they were never in a superposition.
antialias_physorg
5 / 5 (3) Dec 16, 2012
changing one particle has a probabilistic effect on any entangled particle right ?

No. Why should it?
I say this because the sum of the probabilities of the two still have to remain 1

Not after you changed one of them. Then you have broken entanglement and the two are just like any other two (non-entangled) entities.

what I don't get is what are they trying to prove with these experiments?

There are uses for entangled entities in encryption and quantum computing.
For quantum computing you need to be able to multiply entangled entities (because occasionally stuff does interact and breaks entanglement - so it's good to have more than one entangled pair to retain at least one pair of entangled entities)
Pressure2
1 / 5 (4) Dec 16, 2012
A&P quote: "There are uses for entangled entities in encryption and quantum computing."

I agree entangled particles could be used for those purposes. But there are two problems that would appear to make this impossible.

1. This would be sending information which even you say is not possible.
2. It has never been proven that the photons in these experiments are in fact even entangled.

I submitted a test above to find out once and for all if they are entangled. I say they are not, they are no different than two coin halves tossed into the air. They are what they are from the moment they were created until they are received.
antialias_physorg
5 / 5 (3) Dec 16, 2012
This would be sending information which even you say is not possible.

No. Encryption does not add information. That's why entanglement works here. (Look up how information is defined and you will see why).
It has never been proven that the photons in these experiments are in fact even entangled.

Go to the links posted throughout the thread. Entanglement has been shown many times. If these entities weren't entangled the test results would be entirely different.

they are no different than two coin halves tossed into the air.

That analogy is wrong on so many levels 1000 characters don't even suffice to begin to explain. Just go take a look at what quantum mechanics even is (and then look up what entaglement is) before posting such nonsense.
Pressure2
1 / 5 (4) Dec 16, 2012
Quote A&P: "Just go take a look at what quantum mechanics even is (and then look up what entaglement is) before posting such nonsense."

Do the test to see if these photons are even entangled, I say they are not. The burden of proof is on the one making the claim. We would soon find out who is posting nonsense then.

When you encrypt something that is placing information in IT so if you send it some place else you ARE sending informtion. You are contradicting yourself here.

Just because QM believes something is entangled doesn't make it entangled, you need PROOF, and the burden of proof is on the one making the claim. Extraordinary claims require extraordimary, I have yet see or read anywhere there is one iota a proof these photon are entangled any more than the two coin halves.
clay_ferguson
1 / 5 (5) Dec 16, 2012
A&P, I agree with you on the broken entanglement thing. I thought we were talking about persistent entanglement rather than broken entanglement, which is why I mentioned the probabilistic relationship. Seems to me like breaking entanglement could only happen by observing the entangled observable, but I've never researched entanglement breaking. :(
Widdekind
1.8 / 5 (5) Dec 16, 2012
When two particles are entangled, they are not separate. One particle is not "here" and the other "there". Instead, both particles have "schizophrenic" wave-functions. Both particles are "partially here" AND "partially there", in cross-correlated pairwise ways. What seems to be the particle "there" (say) is the composite overlap, of parts, of both particles. If that entangled pseudo-particle is Observed; then the ensuing wave-function collapse simultaneously affects both particles, since each particle was (partially) involved in the Observation, and ensuing collapse. Simplistically, one particle becomes wholly present "there", and the other wholly absent "there" (i.e. wholly present "here"). When two particles are entangled, the pseudo-particles that are then separated, are not the original, previously distinct, particles. Rather, you wind up with two "half of one, half of the other" pseudo-particles (oil & water --> 2x vinaigrettes, which revert to oil & water on Observation).
Widdekind
2.1 / 5 (7) Dec 16, 2012
Quantum entanglement cannot transmit meaningful information, without also controlling the wave-function collapse process. If "oil & water" are entangled, into two "vinaigrettes", then to transmit meaningful messages from "here" to "there", i would have to be able to willfully choose "oil" (to send you "water"), or choose "water" (to send you "oil"). But, simply Observing a quantum object, allows it to "decide" (randomly) how its wave-function collapses. i could force my "vinaigrette" pseudo-particle to collapse, forcing it to "decide" to become either oil or water. Then, you would instantly wind up with water or oil. But, i didn't choose which outcome occurred, so i wasn't "speaking" intelligently or meaningfully. Other than that i "spoke", i.e. caused a collapse, no other information would be transmitted. It would be like i walked up to you, and pointed at a deck of cards, drawing one card randomly. Other than "some guy caused a card to be drawn", no other info would be sent.
Widdekind
1 / 5 (4) Dec 17, 2012
Perhaps an oscillating wave-function, "sloshing" side to side (as in some quantum oscillators), could have its halves separated? Then the transmitter "here" would have a time-varying wave-function, oscillating from fully present, to fully absent, to fully present, etc. Meanwhile, the receiver "there" would receive the complimentary wave-function, oscillating from fully absent, to present, to absent, etc. By judicious timing of Measurements, the transmitter might be able to nearly guarantee themselves winding up with "all" or "nothing", so transmitting "nothing" or "all", i.e. 0 or 1 ?
antialias_physorg
5 / 5 (3) Dec 17, 2012
Do the test to see if these photons are even entangled, I say they are not.

Just google 'demonstrating entanglement' and you will get thousands of links (preferrably go to google scholar for links to hard science papers). There is plenty of proof there.

Without entanglement stuff like quantum encryption wouldn't work at all - and that has been done since 1991.
http://www.econom...tography

COMMERCIAL applications (like in the swiss elections) are becoming more prevalent. This is already way beyond the 'mere theory' stage.

When you encrypt something that is placing information in IT

No you do not. Encyrption itself does not convey information. Again: please look up what information is. You may want to start with Claude Shannon who established the field of information theory around the time of WWII. His definitions are still very much the basis for everything in the field.
Pressure2
1 / 5 (4) Dec 17, 2012
All words with no proof of anything in the link you provided A&P. The only alluded to proof uses the word "similar" in their description. Similar is not proof. If and when they call a computer a quantum computer if will operate by laws of classical physics. There is no entanglement action at a distance. The split coin anology suffices.

Quote from link:

"In 2007 the Swiss authorities used a secure fibre-optic quantum network constructed along similar principles to Dr Ekert's to transmit voting data during a general election."

http://www.econom...tography
antialias_physorg
5 / 5 (3) Dec 17, 2012
If you don't see it then no one can help you. But from what you write I'm fairly sure you don't even know what entanglement (or information) even means 8in a scientific manner. Not in an "I read the word once"-manner).
So it's a moot point trying to convince you of its reality.

You think you can comment on something in a meaningful way that is not trivial/everyday knowledge? Keep on deluding yourself. You may not notice it, but to those who have even a cursory knowledge of the subject that looks beyond ridiculous.

It's like explaining colors to a blind man.
Sure you can deny that color exists. But that doesn't stop people from making actual products that use that property. Sticking your fingers in your ears and saying "It ain't real" isn't going to change that.

Do some work. Get an education. Then come back and comment.
Pressure2
1 / 5 (4) Dec 17, 2012
Here is some education for you also. I have done some research on quantum cryptography and it does not use quantumly entangled photons. Read the Wiki link below.

By the way quantum cryptography has nothing to do with entanglement. It a good thing because quantum entanglement hasn't been proven to exist yet and cannot be use send information.

http://en.wikiped...tography
clay_ferguson
1.6 / 5 (7) Dec 17, 2012
Note to everyone: Pressure2 ValeriaT (there are more) is the same person and only comes here to troll. He writes stuff he knows is false, just to see people respond.
Tausch
1 / 5 (3) Dec 17, 2012
Entanglement enthusiasts let go of what's real - there's no way to ignore the entity and/or events called 'you', and there's no 'local' notion.

In a way today's entanglement enthusiasts remind me of the first proponents of a round world or an absolute time.
The food for thought is evolution - diets change.
antialias_physorg
5 / 5 (5) Dec 17, 2012
I have done some research on quantum cryptography

Erm. No. You have done no such research (or even a google search) on quantum cryptography. I call BS on that statememnt.
ValeriaT
1 / 5 (3) Dec 17, 2012
Note to everyone: Pressure2 ValeriaT (there are more) is the same person
It isn't. Refused experimentally...
Pressure2
1 / 5 (4) Dec 17, 2012
Note to everyone: Pressure2 ValeriaT (there are more) is the same person
It isn't. Refused experimentally...

Agreed, we are not the same person.
I wounder though if Lite, Q-star and A&P might not be the same person? ? Lite seems to alway give a 1 to someone giving A&P a problem he cannot answer while Q-star continually gives 5 to A&P.
clay_ferguson
1 / 5 (5) Dec 17, 2012
ValeriaT is DEFINITELY a 100% confirmed troll on this site, spewing as much false claims as possible, and usually contradicting confirmed science left and right, with every sentence he writes being nonsensical.

Pressure2's trollness is still only a 75% probability.
ValeriaT
1 / 5 (5) Dec 17, 2012
spewing as much false claims as possible, and usually contradicting confirmed science left and right
This is not so surprising, because I'm describing the reality from perspective of human observer at the distance scale around one meter. So I must contradict with both quantum mechanics, both general relativity. The general relativity predicts, all massive objects will collapse into pin-point singularities with no mercy. Whereas the quantum mechanics predicts instead, all objects will expand into infinity like quantum wave packets. So when I'm describing the Universe from perspective of quasistable organism, which neither expands, neither collapses, then I must contradict both theories from left and right.

The trick is, the most general description of Universe must be distance scale agnostic, so I consider both classical physics perspective, both quantum physics, both general relativity perspectives at the same moment. Nothing very much remains from mainstream physics after then.
ValeriaT
1 / 5 (5) Dec 17, 2012
Why the mainstream physics adheres on combination of general relativity and quantum mechanics, although these theories contradict mutually and they actually don't work at all at the human observer scale? Because in hyperdimensional universe the hyperdimensional reality can be observed both from inside, both from outside at the same moment and at the certain distance scale such a view becomes low-dimensional and as such deterministic sufficiently. We can actually recognize these dimensional scales rather easily, because just at these scales the observable reality becomes as simple and deterministic, as possible - the observable objects appear like symmetrical spheres at just these scales. The validity domain of quantum mechanics is the scale of atoms and the validity domain of general relativity are stars composed mostly of atoms. Outside of these two distance/energy density scales the mainstream physics gets into troubles - actually the more, the more it gets distant from them.
ValeriaT
1 / 5 (5) Dec 17, 2012
In AWT the appearance of Universe is given with geometry, in which huge random observer (Boltzman brain) interacts with the rest of random environment via transverse waves. The key in understanding of dense aether model is therefore the understanding of the mutual deterministic interaction of hyperdimensional geometries. In AWT the hyperdimensional objects can be modeled like the fuzzy random blobs composed of another isolated smaller blobs, recursively. So we can imagine the observable reality like the sorta moire effect, when two hyperdimensional objects (Universe and its observer) overlap mutually along time dimension. In two concentric zones the distance scale of universe and its observer will fit mutually and we will observe the deterministic reality from both inside both outside like the spherically symmetrical objects floating inside of emptiness. Outside of these two zones the shape of objects will not be symmetrical anymore and the space between them will not be empty and flat
clay_ferguson
1 / 5 (6) Dec 17, 2012
Actually ValeriaT, you made coherent sense this time! (on last three posts). I mean you didn't go off the deep end, and make statements like there is no evidence for QM or whatnot. Please keep that up. You probably just did it to prove me wrong however. lol.

back to business: I'm going to do blog soon that describes how I think both General and Special Relativity are showing us that actually particles can completely "loose a dimension". I'll hope I can get you to read it. It's like this, 1) everything exists only on an event horizon. 2) black holes are a 2-dimensional thing (its event horizon), because when some particles fall on to the E. Horizon they become part of a 2-dimensional (subuniverse) that is only the mathematical surface of the EH. (that's the Gravity part-general relativity) 3) And also when you reach the speed of light (if you did) you would exist on a 2-Dimensional shape also, which is the Plane perpendicular to your motion vector, and everything flattens out.
clay_ferguson
1 / 5 (6) Dec 17, 2012
So what we see as a 4D universe we are in, is really just an event horizon in a higher-dimensional universe. That explains also why galaxies have black holes in them, and mainly only one BH per galaxy. So it is a recursive definition of what spacetime is.

Also, back to black holes for a moment, that event horizon would still have clumps on it where there is mass converging together in that 2-D sphere, and eventually there wold be a sort of 2D blackhole where there is a mathematical impossibility (singularity type thing) where the 2D space would bend itselfso much that light could no longer circle it, and it would THEN create a One Dimensional line from that spot into the center of the BH. At the center of the BH the same thing can STILL happen again where that line looses the final dimension and the particles become 0-dimensional (sort of a dark energy of sorts), and if -1 dimensions exists then that may explain why the number 'i' exists or something wierd like that. That's it.
clay_ferguson
1 / 5 (6) Dec 17, 2012
And to finally clarify, what I mean is that the surface of black holes (event horizons) have spots on them where there is a 2-D line that formed, and went to the center. Think of a spherical porcupine! lol. That's what the inside of a black hole would look like! It's all mathematical modelling and this is all my pure speculation. In my crazy mind we must live on the 'surface' of a 4D (arguably 3) event horizon ourselves...AND there would be possibly 2D 'patterns' of particles on black holes that can recreate themselves and would look maybe like "The Game of Life" type appearance or possibly be 'intelligences ?' Ok I'm finally done.
ValeriaT
1 / 5 (5) Dec 17, 2012
@clay_ferguson I'm aware of your line of thoughts. Here you can for example read, what the hypothetical observer should see if he could pass the event horizon - his experience would be pretty similar to the observation of sky, which is penetrated with many tiny black holes inside of galaxies. But it's geometric, relativistic perspective, so it's limited in some aspects due the black hole complementarity. Therefore the interior of black hole wouldn't appear like the "porcupine" - if we could live inside of it, it would rather appear quite similarly to our Universe - just with substantially simpler and more pronounced geometry (the clouds of dark matter would appear like the clouds on the sky, for example). But because the space-time inside of black holes is topologically inverted, we will probably evaporate into gravitons during it.
frajo
3 / 5 (2) Dec 18, 2012
Note to everyone: Pressure2 ValeriaT (there are more) is the same person ...

No. While it is true that ValeriaT is only one of several dozens of nicks of the same person (long time users call him Zephyr) Pressure2 definitely is a different person.

Zephyr is easily recognizable because he isn't a native English speaker (seems to be Czech) and therefore produces idiosyncratic linguistic glitches he is unable to avoid.
Q-Star
3 / 5 (4) Dec 18, 2012
while Q-star continually gives 5 to A&P.


I give 5's to comments based on the quality of the science and the facility with which it is presented.

I give 5's also to comments that reflect that the person is thinking even if I might thing they are not quite on target.

I give 1's to comments based on the lunacy, convolution, weirdness, foolishness and plain falsity of it.

I don't rate comments made to me whether they support or dispute me. Sometimes, I do when they contain a large amount of cursing or rise to a level of stupid that is just beyond the pale.

I for one don't think you are ValeriaT. But everyone knows that he is in fact Zephyr, why he quite using that name is perplexing, the stupid level is not changed.
clay_ferguson
1 / 5 (5) Dec 18, 2012
@ValeriaT, you pointed us to an article that discusses classical relativity with respect to black holes. What I'm talking about is a mathematical loss of a dimension, with lightspeed creating a mathematical plane, while gravity creates a mathematical sphere (shell only, not interior). That actually shows that you were NOT aware of my line of thoughts. You just googled "what happens if I fall in a black hole" probably. It's ok tho. We all enjoy the discussion, which is why we come back here!
clay_ferguson
1 / 5 (5) Dec 18, 2012
So we know from relativity that "time is local" (to each particle), so I'm extending that to say we need to model "dimensions" as local also. So that it's possible to have particles that can only move in a certain number of dimensions, while being "Fixed in position" relative to whatever dimension it lost. Kind of like coming to rest in that dimension. So light speed may be just the rate of expansion of the 4D event horizon we live in, and Gravity strength (universal gravitational constant) may be caused by the "Radius" of the 4D event horizon we are in. Think of the radius of a 10D hypersphere. That would also explain the lack of us finding dark matter to account for our universe changing rate of expansion. Think of two flatland creatures living on the surface of a normal-type classical event horizon, as the radius expands on that 2D sphere, the points on it all begin to spread apart, but the CAUSE of expansion being simply "more stuff fell in", so the radius of the EH increased.
antialias_physorg
5 / 5 (4) Dec 19, 2012
I wounder though if Lite, Q-star and A&P might not be the same person?

Since lite hands me the occasional 1, I'm pretty sure I'm not him (as I'm not anyone else, though I did have another account by the name of "antialias" when I first got here because I wanted separate accounts from work and home at the time. But that isn't exactly duping anyone into thinking I was someone else, is it?)

I've never had the need (nor the leisure) to go for sockpuppets. Ratings/titles/etc. don't mean anything: It's what you think and say that counts. Sockpuppets are strictly for losers who play by the Calvin* rules: "If you can't win by reason - go for volume"

...I have no respect for adolsecents (or even adults) who act like small children using sockpuppets. Not even on something as ephemeral as an internet comment section.

*from Calvin&Hobbes
clay_ferguson
1 / 5 (5) Dec 19, 2012
I think most of us are genuinely interested in science, and not "going for volume" whatever that means. However I *have* posted too much, and do need to lay off the coffee!
ValeriaT
1 / 5 (3) Dec 19, 2012
What I'm talking about is a mathematical loss of a dimension, with lightspeed creating a mathematical plane
I know about these models: the same effects are assumed at the particle horizon of observable Universe, where the space-time becomes time-like and it's losing its spatial dimensions. The same effects occur at the water surface, where 2D light waves are scattering into underwater. After all, the galaxies appear flat as well, or not? But for topologically inverted space-time inside of black hole the same effect occurs reciprocally and both perspectives are separated with many extra dimensions each other (these extra-dimensions appear like the particle cloud forming the galaxy for us). Which is why I don't consider these simplified low-dimensional holographic models seriously.
ValeriaT
1 / 5 (4) Dec 19, 2012
There is an interesting question, whether we can have very massive black hole separated from its parent galaxy: IMO there is a dynamic equilibrium and such a black hole would start to evaporate, until it would get into new equilibrium. In this way the radiation of black holes at the center of large galaxies is often a consequence of radiative evaporation of the whole galaxy.
Q-Star
2 / 5 (4) Dec 19, 2012
There is an interesting question, whether we can have very massive black hole separated from its parent galaxy: IMO there is a dynamic equilibrium and such a black hole would start to evaporate, until it would get into new equilibrium. In this way the radiation of black holes at the center of large galaxies is often a consequence of radiative evaporation of the whole galaxy.


Would apply to flat water waves that lay in the intervortex medium? Or only along the contraradiative edges?
clay_ferguson
1 / 5 (5) Dec 19, 2012
@Valeria, water is not really a good analogy IMO. You've been watching too many cartoon explanations of the slit experiment. :) Why would light under water appear 2D ? Please explain. Is it the transmogenated vectorized hyperdoodle quadramagic luminescent water transverse giggernating gravitational super-ocillator mechanics, or just the jiggerwakky motornator configuraor vector field matrix matriculator phenomena ?
Widdekind
1 / 5 (4) Dec 19, 2012
Hypothetical exploitation, of the Quantum Instantaneity of Quantum Entanglement, for FTL communication, seemingly would require (1) stretching an entangled wave-function between the two communicators; (2) CONTROL-ing the wave-function collapse. (1) is conceptually simply, and has already been repeatedly accomplished, over short distances, in labs. (2) is therefore the impediment to FTL. In the absence of a mechanism to control wave-function collapse, upon probing & Measurement, only random, and hence meaningless, gibberish could be transmitted. Speculatively, if a "conscious observer" could become entangled with the signals sent, then the "decision" by the experimenter to do one thing, or another, would cause the joint experimenter-signal wave-function to collapse, as the former willed. Inexpertly, that would require an intelligent "decider" to enter into, and remain in, the Quantum State, as long as required to propagate the signal; and then "decide" from within the Quantum State
ValeriaT
1 / 5 (4) Dec 19, 2012
water is not really a good analogy IMO. Why would light under water appear 2D
Light is the analogy of surface waves. When the initial wavelength of ripples is lower than the size of density fluctuations of the underwater, then they do scatter into longitudinal ones, which are much faster and the wavelength of surface waves increases. When the original wavelength is smaller than this limit, their wavelength decreases instead. This is rather nontrivial behavior, but thelight does the very same, when it's spreading through CMBR noise. The similar effect happens, when the light spreads through system of particles, the wavelength of whose is comparable with the wavelength of light.
You've been watching too many cartoon explanations of the slit experiment.
I know, most of these cartoon explanations are mine...;

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