# Photon existence question

1. Mar 26, 2015

### debeselis

I have a question about photon speed and time releation. When photon is speeding at the speed of light, the time for him stops as i understand. So that means that from photon perspective to travel 1 light year takes no time and that means, that from it's perspective at the same time it is in all possible positions in the spacetime and what is more confusing for me, that universe will come to the end and photon will "die" and basically it should mean that photon traveling at the speed of light is existing and not exisiting in the same time from his time reference.
Please explain me where am I wrong with my perception

2. Mar 26, 2015

### bhobba

You understand incorrectly.

Here are the facts.

First you need to see a correct derivation of the Lorentz transformations:
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

Its got nothing to do with light etc - rather its a theory of space-time symmetry.

Now in those equations put c as the velocity - you get 1/0 which is undefined. Its physical non-sense to talk about time at all for an object travelling at c.

Thanks
Bill

Last edited by a moderator: Mar 26, 2015
3. Mar 26, 2015

### Ibix

The basic problem is that it isn't possible to talk about "the photon's reference frame" or "time experienced by a photon". Neither concept really makes sense, which is why you are coming up with odd consequences when you try to think about it.

It is true that the so-called "interval" along the path of a massive object corresponds to the time measured by a clock attached to that object. Brian Greene (from whom I presume you go the idea of photons being timeless) extends the argument to the case of photons, where the interval along their path is zero, and argues that therefore photons experience zero time.

One can approach the case in a different way. In Special Relativity light travels in straight lines, and it is rather easy (using the Lorentz transforms that Bhobba mentioned) to consider a couple of events on those lines and show that, from the "photon's point of view", they both occur at infinite time with infinite time passing between the two. Which contradicts my previous paragraph.

The point is that time is what clocks measure, and you cannot make a clock that travels at the speed of light. So it is not immediately clear that time is defined for anything travelling at light speed. The fact that the maths gives contradictory answers when you try to work out what a hypothetical speed-of-light clock would read tells you that there's something illegitimate about the concept.

Unfortunately, in this case, Greene has simplified what's going on to the point that he's not really making sense - as you observed in your first post.

Last edited: Mar 26, 2015
4. Mar 26, 2015

### bhobba

Exactly - SR is a theory about transformations between inertial frames. No inertial frame can travel at c - hence the question of time for an object travelling at c is meaningless.

I have to say writing about physics for a populist audience is very very difficult. I read such books occasionally and scientists of the highest calibre, which Brian Greene is, often say things that are, well quite dubious. I remember Brian Cox, who I like a lot, got into similar problems with a comment he made about the energy levels of a diamond if you warm it up - he said every other energy level in the universe would have to shift imperceptibly to accommodate it. It was wrong - there are an infinite number of free spaces available and all that happens is they slot into it. But it did generate a lot of interesting discussion.

Thanks
Bill

Last edited: Mar 26, 2015
5. Mar 26, 2015

### harrylin

"A photon experiences no time" merely means that if a clock could go as fast as a photon (which is impossible), it would stop ticking. And of course, a photon cannot experience anything anyway. Transposed from QM to SR it's just a wave packet that is not at rest in any reference system. So, "experiences no time" should be understood in the sense of a clock that doesn't tick: it cannot be used as a reference for measuring time.

6. Mar 27, 2015

### GeorgeBaxter

I am not sure that I agree with the statements about neither making sense. Einstein hit upon the idea that time would appear to slow down in classical physics, when the observer of a clock sees that the hands would appear to be moving more slowly and appear to stop if they were also moving at the speed of light. Nothing new there but SR did come out of the thought experiments. So velocities of c are not possible for non-lightlike events.

And yet the speed of light is there, and invariant, to any inertial frame and it translations. If you consider that the photon does have a frame of reference, then invariance of c becomes explainable in a holographic universe. It is an event with zero time on the holographic boundary. The projection of the event in our holographic universe is not as a point, but some event with a duration and a velocity of c That would be the observation of the first observer of the zero time event. A second moving observer of the same holographic, zero time event, must also generate the same velocity c even though it is moving relative to the first observer. Thus they both obtain the photon was travelling at c and so photons appear to travel at c irrespective of the motion of the observer. So photons do have a frame of reference, but do not experience time.

7. Mar 27, 2015

### Staff: Mentor

Do you have a reference for this? It looks like a personal theory to me. Personal theories are off topic in this forum.

8. Mar 27, 2015

### bhobba

You mean assuming something that contradicts the assumptions the theory is based on?

I would like to see it as well. It doesn't really make sense to me - but I am all ears.

Thanks
Bill

9. Mar 27, 2015

### Staff: Mentor

The term "the frame of reference of <something>" is an imprecise but convenient shorthand for the more precise "a frame in which <something> is at rest". The speed of a light signal is $c$ in all frames so there are no frames in which the light signal is at rest, and hence nothing that we can call its frame of reference.

It's tempting to set $v=c$ in the Lorentz transformations, just to see what happens if you do assume that there is a frame in which the light signal is at rest. But the starting point for the derivation of these formulas is that the speed of light is $c$ in all frames, so they cannot be applied in the $v=c$ case... we're just putting garbage in and getting garbage out.

10. Mar 27, 2015

### Khashishi

Brian Greene isn't really wrong here. We can't define a frame going the speed of light, but we can take the limit as speed approaches the speed of light, and then time dilation goes to infinity--proper time stops. It's not so bad to think this way. It makes it clear why neutrino oscillation implies that neutrinos have mass. If they had no mass, they would travel at the speed of light, and they couldn't oscillate, since they have no proper time.

11. Mar 27, 2015

### bhobba

I can take limits of all sorts things - it being physically relevant is another matter.

As Wheeler says - forward is always forward so you cant attach an internal frame to objects travelling at C.

Since the Lorentz transformations are between inertial frames, then since an inertial frame cant be attached to something moving at c, it doesn't apply.

Thanks
Bill

Last edited: Mar 27, 2015
12. Mar 27, 2015

### Staff: Mentor

You can't just say "take the limit"; you have to specify what you are taking the limit of.

In the case of proper time, you are taking the limit of arc length between two distinct points on a curve. But changing the "speed" means changing the curve; and making the speed approach the speed of light means changing the curve from a timelike curve into a null curve. Physically, this doesn't make sense. So while you might be able to mathematically formalize such a limit, that doesn't guarantee that it means anything physically.

Physically, the key difference between timelike curves and null curves is that, for timelike curves, arc length (proper time) is an affine parameter, whereas for null curves, it isn't (since the arc length is zero between any two distinct points). In other words, we can use proper time (arc length) to distinguish points on a timelike curve, and our entire concept of "proper time" is based on being able to do this. But we can't use arc length to distinguish points on a null curve, so our whole conceptual scheme for proper time breaks down for null curves.

So when we say that the concept of "proper time" doesn't make sense for photons, we're saying something stronger than just "proper time is zero for null curves, but we can still take limits and draw analogies with timelike curves"; we're saying "applying the concept of proper time at all to photons is not valid, because null curves are fundamentally different from timelike curves".

Given the umpteen number of threads here on PF based on misconceptions arising from this way of thinking, I strongly disagree.

This is not a valid argument; if it were true, it would prove that photons can't oscillate either. But they can. You can have oscillatory variations of fields along null curves just as you can along timelike curves (or spacelike curves, for that matter). You just can't parameterize the oscillations along null curves by arc length.

Just for the record, what neutrino oscillations imply is not that all three neutrino species must have mass, but that the three species must have different masses. This means that at least two of them must have nonzero mass; but it still allows one to have zero mass. (The current belief is that all three have nonzero mass, but that's based on much more detailed experiments that attempt to measure matrix elements for various reactions involving the three neutrino species; it's not based on the simple fact of neutrino oscillations.)

13. Mar 28, 2015

### GeorgeBaxter

The link below is a discussion of diverse alternatives to the origins of space and time, and reality

http://www.nature.com/news/theoretical-physics-the-origins-of-space-and-time-1.13613

Holographic universe is one postulated interpretation of reality, where all the real events occur on the boundary of the universe. The "real" photons, and "real" particles exist there. That is their true frame of reference, and not the universe that we experience. Therefore they do have a frame of reference.

Both SR and GR have breaking points. They are both known to be fundamentally incomplete and incompatible with quantum mechanics. They both end with infinities ( SR at v=c , GR with singularities in black holes ). To say that SR does not allow clocks to travel at c, therefore photons cannot have a frame of reference is poor logic. In the same link, it considers that gravity can be determined as a result of a deeper level of thermodynamics, and yet GR predicts singularities. But the thermodynamic theory side-steps the singularities.

Brian Greene has already been cited as a considering zero interval ( both in space and time ). To me, that is simply a point, and that point exists in the holographic boundary. I do not think that constitutes “a personal theory”. The counter that is not an inertial frame of reference at rest seems to me to be the personal theory. Who says that photons have to exist “at rest” in a frame of reference? I am saying that if photons exist then they have a frame of reference. It does not need to be an inertial from. GR is built on non-inertial frames. GR does not assume the constancy of c either.

14. Mar 28, 2015

### bhobba

I am not an expert on the Holographic universe but my understanding is that is not what it says - it says the description of what occurs inside the boundary is encoded on the boundary - not that the real events occur on the boundary - which considering the definition of space-time event doesn't really make sense. But I am all ears for what someone who is expert on it has to say.

That's incorrect. Re-normalisation resolves the infinities in QFT (that's SR combined with QM). It turns out to simply be the choice of a lousy thing to perturb about - what they perturbed about was secretly dependant on a cut-off and went to infinity when the cut-off went to infinity - the solution was to use something without that weird behaviour:
http://arxiv.org/pdf/hep-th/0212049.pdf

QM with GR is in fact resolved in the same way. It too has exactly the same issue - but isn't resolved by the re-normalisation method. However Wilson suggested a new way of looking at renormalisation (he got a Nobel prize for it) - that theories are only valid to some cut-off - below that cut-off we have what are called effective theories:
http://arxiv.org/abs/1209.3511

What the issue is, is we don't have theories valid past about the Plank scale - but that is different to saying they are incompatible with QM - incomplete yes - but gee all our theories are incomplete. Even classical theories are incomplete eg in EM we have issues with the so called Loretz Dirac equation:
http://arxiv.org/abs/gr-qc/9912045

For some reason people don't get excited as much about it as combining QM with GR - yet its the same thing - push it to far - beyond its range of applicability and it blows up.

Did you read the link I gave. By DEFINITION SR is a theory about transformations between inertial frames which, by DEFINITION, are defined as a frame where space and time are homogeneous and space isotropic. That has very strong heuristic support as well as experimental evidence such exist to a high degree of accuracy in interstellar space.

Thanks
Bill

Last edited: Mar 28, 2015
15. Mar 28, 2015

### PWiz

Special relativity is very compatible with quantum mechanics (100% actually). That branch of physics is called relativistic quantum mechanics, and you can read more about it on the Internet.

The reason why GR can't be used with quantum mechanics is because the resulting theory is not renormalizable, not because of fundamental incompatiblity. The theory of relativity is complete. Only quantum gravitational effects at the Planck's scale are unexplained.
EDIT: I guess I was 15 minutes late :P

16. Mar 28, 2015

### Staff: Mentor

I read that article and never found any mention of a connection between the holographic universe and the validity of a photon's frame of reference. The reference does not support the point that is being objected to, specifically the part about a photon having a frame of reference.

17. Mar 30, 2015

### Staff: Mentor

18. Mar 31, 2015

### mac_alleb

Is it the only way for photon to move at c? It's just max possible or?

19. Mar 31, 2015

### Staff: Mentor

Yes. Any object with zero invariant mass can only move at $c$.