# I Why do people keep saying photons are timeless?

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1. Sep 12, 2017

### weezy

If you search for "does a photon experience time", almost every other link says that they travel at the speed of light and so STR tells us that its clock doesn't tick at all. However why do they use the arguments for special relativity which was developed for massive particles moving close to the speed of light and extrapolate them to at speed of light? As far as I remember the time dilation formula was devised by Einstein on analyzing the light clock which always moved less than the speed of light. If the clock now runs at velocity 'c' then the pulse of light bouncing off one mirror completely misses out the second mirror during the instantaneous acceleration of the light clock apparatus from subluminal to luminal speed and our clock thus 'breaks'. Instead if we have a light clock which always moved at the speed of light, then a massless observer moving along the light clock, were he to shoot a pulse of light towards the bottom mirror would just rebound fine and he would experience normal time if he chooses to ignore the rest of the universe outside his frame. This last bit is given light still holds the same meaning as it does at subluminal observer speeds.

I don't know if my reasoning is sound but I will say that claiming the photons don't experience time is just speculation and not a fact. We don't know about if relativity applies at speeds of light

2. Sep 12, 2017

### Orodruin

Staff Emeritus
There is no clock that can move at the speed of light. Photons do not have a rest frame.

This colloquial statement (generally in popular science) stems from the length of the world-line (which is the proper time for time-like world lines) is zero for light-like world lines.

The light-clock is an argument for popular science and introductory relativity. The time dilation formula can be devised without reference to an actual constructed clock from the first principles behind SR.

Some very common misconceptions. Photons do not have rest frames and there is no such thing as "outside a frame". In SR, everything exists in all frames. Different frames are just different tools for describing the same physical situation. If something happens in one frame, it must happen in all frames.

3. Sep 12, 2017

### Ibix

Special relativity applies to everything, in the absence of gravity.

However, that the speed of light is constant in all inertial reference frames is one of the principles of relativity. So a rest frame for light - where it is stationary - is a contradiction in terms. So trying to describe what a photon does or does not experience is impossible - there is no way to "put yourself in its shoes".

That doesn't seem to stop people trying.

4. Sep 12, 2017

### I like Serena

Hmm... actually photons would only travel at the speed of light in vacuum - and true vacuum does not exist - so they can't.
So in practice photons would always go a bit slower, since they're always in a medium.
I wonder... doesn't that mean that the clock does tick for them after all?

Last edited: Sep 12, 2017
5. Sep 12, 2017

### Biker

I dont think that is true. Light is always moving at c but the light wave is slower in mediums. There are a lot of explanations such as absorption and emitting but that is wrong.

The only one that is true is something about group velocity.

6. Sep 12, 2017

### SiennaTheGr8

I don't think it's true that a photon ever moves slower than $c$.

I'm no expert(!!!)
, but my understanding is that the QM explanation of light propagating through a medium involves not slower-than-$c$ photons but rather "quasiparticles" (polaritons?) that are something like a superposition of photons and the excited particles in the medium (excited because they're interacting with the photons). It's these quasiparticles that you might say travel slower than $c$.

Perhaps someone who knows better could correct me or improve my phrasing.

(I believe that a common incorrect explanation is the delay between an atom's absorption and emission of a photon—i.e., that the photons travel at $c$ from atom to atom but are delayed every time they're absorbed.)

7. Sep 12, 2017

### DrStupid

Practical photons are irrelevant for clocks. In the second postulate Einstein refers to plane light waves in vacuum. And it is not even neccecary to refer to light at all. It is sufficient to say that there is an invariant speed. This speed can be determined experimentally and it appears to be equal to the speed of light in vacuum.

8. Sep 12, 2017

### stoomart

I imagine one needs to seriously study relativity before claiming it is speculation. My understanding of photons being timeless comes from posts like the one from @Orodruin, and researching the issue myself. Per Wikipedia's Spacetime article:

Last edited: Sep 12, 2017
9. Sep 12, 2017

### GrayGhost

Hello weezy, nice to make your acquaintance.

The reason they do that, is because there is no other option to predict a speed c POV. The photon cannot be held in a state of rest, and thus has no inertial POV (even though it travels at a steady speed). We only know what material POVs experience. We cannot know what a photon's experience is. So we ask the "what if" ... "what if" a material POV could attain a relative c, what would its experience be? Or, "what if" a particle of rest mass does exist that always travels at c (never seen to date)? All that we can do, is to explore what the mathematics suggests said experience should be. We consider an SR thought experiment, for higher and higher speeds, to determine the trend in the relativistic effects, then we extrapolate kinematically what the experience should be in the limit as v attains c.

In the initial instant an inertial wonder-POV attains a relative c (call it a wonder clock), it hands then reach a zero tick-rate per normal inertial cosmic POVs. Let's say the clock's hands then read 2:30. Observers record the clock to move at c thru 3-space over duration, the clock's hands never moving always reading 2:30 as it travels. So per normal inertial observers, the speed-c clock should not experience the passage of time as it goes, since its hands never move while at c. But wrt any finite traversal distance considered, and per the wonder clock itself, it is everywhere AT ONCE along its propagation path for the portion of its path while its hands read 2:30. Existing at multiple contiguous locations AT ONCE is to traverse a finite distance in zero duration, ie an infinite velocity. AT ONCE, requires no passage of time. This is why it is said that the hands of a clock carried by a photon never move, or that a photon does not experience the passage of time.

Yet, photons do not possess a material POV since they cannot be held in a state of rest. As such, it is said that the extrapolated experience of a speed c wonder clock cannot be known to correctly represent the actual experience of a photon, assuming it could have such a thing. There are other obstacles as well, physically speaking. For example, to attain a relative speed c wrt a single cosmic body, you must also attain a relative c wrt every other cosmic body. It must all happen AT ONCE per you. You do not attain a speed c wrt this guy, then later wrt that guy, and so on. As such, and logically speaking (and besides the fact that all heavenly material entities would be infinite in mass per you), it seems quite impossible to attain a relative speed c. A relative speed c plays the role of an infinite velocity (per SR) ... and it's most difficult to imagine infinities in nature.

Best regards,
GrayGhost

Last edited: Sep 13, 2017
10. Sep 12, 2017

### Orodruin

Staff Emeritus
I would certainly hope not, because that was not my intended meaning at all. It was meant as an explanation of why people might say things like that. Also, Wikipedia is generally not considered a reliable source.

11. Sep 13, 2017

### Ibix

Something with mass travelling at c is another contradiction in terms (it must have a null energy-momentum vector because that's what travelling at c means - but mass is the modulus of the energy-momentum vector, and the modulus of a null vector is zero). So what this quote boils down to is "we try to imagine the logical consequences of contradicting ourselves". That's why the outcome is inevitably nonsense.

Last edited: Sep 13, 2017
12. Sep 13, 2017

### puzzled fish

This is gonna be my last post in PhysicsForums as I have decided to go.

One must be very careful when dealing with infinities. Does infinity minus infinity when both x and t are infinite in spacetime interval, tend to zero? Or does 0 multiple infinite times ( for our little trivial measurements of our theories) tend to zero?

13. Sep 13, 2017

### weezy

I can say for sure that an infinite sum of 0s will always be 0. Not sure about the rest.

14. Sep 13, 2017

### weezy

Why is the absorption emitting explanation wrong? How do you explain refraction at photon level then?

15. Sep 13, 2017

### Biker

Lets assume that the situation is true, We know the atoms absorb specific wavelengths, right? so the refractive index of these wavelengths should change drastically because they will get absorbed a lot which means delayed but that doesn't happen. Yes the refractive index depend on the wavelength but not as discrete but smoothly. Also specific wavelength shouldn't be absorbed so I guess no change for the speed of light?

2nd thing, Absorbing and re-emitting is just a random process. It takes different amount of time. You could say that it kinda averages out but I don't think that happens( I need more insight on that). Whenever you measure the time it takes light to pass it is always constant.

16. Sep 13, 2017

### DrStupid

As Biker already mentioned, repeatedly absorption and emission would result in a random walk and that doesn't fit to experimental observations.

By describing the photon as an elecromagnetic wave. The speed of electromagnetic waves depends on the permittivity and permeability of the environment. And these properties change in presence of matter.

17. Sep 13, 2017

### SiennaTheGr8

18. Sep 13, 2017

### I like Serena

Wiki explains that an electromagnetic pulse has a front speed, a phase speed, and a group speed (link).
And the front speed (under certain assumptions) is always equal to c, the speed of light in vacuum.
The phase speed and group speed deviate because the electromagnetic wave interacts with the particles in the medium.

To be honest, I still don't quite get this, since it seems to suggest that whether we shoot a laser through a vacuum or through a medium, that the first effect of the waves arrive on the other side at the same time.

Last edited: Sep 13, 2017
19. Sep 13, 2017

### vanhees71

Well, don't read only Wikipedia but also some good physics book ;-). For the classical dispersion theory, have a look in

A. Sommerfeld, Lectures on theoretical physics, vol. 4 (Optics)

Sommerfeld and Brillouin wrote two very famous papers to clarify the apparent problem with the special theory of relativity and dispersion in the realm of anomalous dispersion, where the phase as well as the group velocity formally get $>c$. The true answer is that, in this realm the group velocity looses its usual physical meaning, because the corresponding approximate solution of the involved Fourier integrals, using the saddle-point method, is not applicable in this region. Of course, the full solution is always causal in accordance with special relativity. Last but not least it's one of the most beautiful applications of complex-function theory (theorem of residues) ever . You find it all in Sommerfeld's textbook.

I'm not aware, whether the original articles by Sommerfeld and Brillouin have ever been translated to English (in any case they should have been!). Here are the German references:

A. Sommerfeld, Über die Fortpflanzung des Lichtes in dispergierenden Medien, Ann. Phys. (Leipzig), 349 (1914), p. 177–202.
http://dx.doi.org/10.1002/andp.19143491002

L. Brillouin, Über die Fortpflanzung des Lichtes in dispergierenden Medien, Ann. Phys. (Leipzig), 349 (1914), p. 203.
http://dx.doi.org/10.1002/andp.19143491003

20. Sep 13, 2017

### haushofer

Ask for an operational definition of how much time elapsed for a photon. That will shut them up. :P