Is Time Experienced by Photons at the Speed of Light?

[sqljunkey]

What field of physics deals with what a photon experiences as far as it comes to time and such. Like two photons traveling side by side, or having some kind of interaction. Is that the realm of quantum mechanics?

 

[Nugatory]

What field of physics deals with what a photon experiences as far as it comes to time and such. Like two photons traveling side by side, or having some kind of interaction. Is that the realm of quantum mechanics?

No field of physics deals with what a photon experiences or with two photons traveling side by side, for the same reason that no field of biochemistry deals with the metabolism of pink unicorns - these concepts make no sense.

Photon-photon interactions do happen, and they are described by quantum electrodynamics.

 

[weirdoguy]

What field of physics deals with what a photon experiences as far as it comes to time and such.

Did you even read what was said in this thread?

 

[Chris Miller]

The Lorentz equations cannot be applied to photons. Their time dilation is not zero, it is either undefined or meaningless.

 

[vanhees71]

What do you mean by "Lorentz equations"? A photon is described by the relativistic QFT named QED, which is completely compatible with special relativity (as are the classical Maxwell equations).

 

[NoTe]

In the Netherlands, on the Wikipedia page of the lemma 'foton', part of the physics section, it says: "Volgens de speciale relativiteitstheorie staat de lokale tijd van een lichtdeeltje stil." Translation: "According to Special Relativity the local time of a photon stands still." It's been there for fifteen years or more and apparently no one seems to care. I have always thought (i.e. known) it couldn't be right, but I'm not going to edit Wikipedia pages... but it shows as a nice example of the reliability of Wikipedia and such!

 

[vanhees71]

Well in Wikipedia there are some people allowed to write and they quite often don't write in completely correct terms. I've no clue what a local time might be. If they mean "proper time" it's of course nonsense, because there's no way to define proper time of a photon.

If they mean the eikonal approximation for free electromagnetic fields it doesn't make sense either, because light-like curves do not admit the definition of a uniquely defined proper time. All you can do is to use some parameter to parametrize the light-like ray. You can even use an affine parameter, but that doesn't define a uniquely defined proper time either. The proper time is well-defined for worldlines of massive particles, which are time-like. Then it's the affine parameter $\tau$ which is defined uniquely by the equation
$$\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} \tau}=c^2.$$

In short, the Wikipedia statement is (a) inaccurate using undefined non-standard terms ("local time") and if with the best will you try to understand what they might mean, it doesn't make sense either.

 

[Alfredo Tifi]

Well, fundamental particles don’t really experience anything. However, the time that human beings experience (and clocks and other animals and ...) is called “proper time”, and photons do not have any proper time.

The reference system of the “fast” object Measures a velocity c respect to the photon. It’s time is relented only if watched by another reference system, but it’s own time “proper time” is unchanged and its “perceptions” are unchanged. Given that a photon don’t have a reference system, we can’t say anything about the time flow in the life of a photon.
Nevertheless the photon has a lifespan and a frequency if viewed from any inertial observer. And a frequency is a vibration of something in the time of observer. The faster the observer flies towards the photons’ source, the higher will be the observed frequency. The faster the observer would run away from the photon source, the lower will be the observed frequency. In this case we can go to the limit: if the object goes away from the source at c-ɛ, as ɛ → 0 it will observe a frequency f → 0. But this is still the time of a reference system, not the photon’s time. So, we can’t claim that nothing oscillates from the point of view of the photon.

 

[PeterDonis]

The Lorentz equations cannot be applied to photons.

It depends on what you mean by "the Lorentz equations". You can describe the motion of a photon in any inertial frame just fine, and the description transforms between inertial frames according to the Lorentz transformation just fine. What you can't do is define an inertial frame in which the photon is at rest.

 

[Tom Atkinson]

So if I was traveling at the speed of light my clock would not be moving relative to me? I would be frozen in time? I wouldn't notice it but to others that's how it looks like?

If you were traveling near the speed of light time would be just fine, your clocks running, to an observer you'd be literally glowing and no way to read out the clock at that speed!

 

[Nugatory]

If you were traveling near the speed of light time would be just fine, your clocks running...

Yes

to an observer you'd be literally glowing

If you are approaching the observer, yes. But not if you were moving away, then Doppler would red-shift the light coming from you.

 

[.Scott]

As far as I know, a object will experience time slower when its speed is close to the speed of light.
But photons themselves moves at the speed of light, does that mean that they experience no time?

Notwithstanding the very valid issues already posted, it is possible to answer this question in the spirit in which it was asked.

Although you cannot be accelerated to the speed of light, you can, in theory, be accelerated to speeds which approach the speed of light.
So we can talk about what would happen as you reach speeds of about 0.999999999c (relative to Earth). From a Earth-bound observer, your clock will have essentially stopped and your relativistic mass would have increased enormously. If your destination was 1000 light-years away, you will be observed to arrive there in roughly 1000 years, but your clock will have advanced by only a couple of weeks.

From your perspective, you will have completed the journey in only weeks. But wouldn't that make it seem as though, from your point of view, you were traveling at a speed much faster than light? It won't because from your point of view, the entire universe would appear to be foreshortened. By your measurement, your distance traveled would be on the order of the diameter of the solar system.

So what would a photon see?
First, it doesn't have time to see anything. Even if it had a clock, it would never tick. This is why neutrinos, that seem to be able to change while travelling, are determined to be traveling at something less than the speed of light.
Second, from the photons perspective, the starting and ending points are coincident. It travels a distance of zero in zero time.
Third, photons don't really travel like that. They follow quantum mechanical rules that defy the notion of a straight path from a source point to a destination point.

 

[kurt101]

The proper time is well-defined for worldlines of massive particles, which are time-like. Then it's the affine parameter $\tau$ which is defined uniquely by the equation
$$\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} \tau}=c^2.$$

Is there any online reference that explains this equation in more detail? What is $\tau$, x_$\mu$, and x^$\mu$? Thanks.

 

[weirdoguy]

and your mass would have increased enormously.

Relativistic mass which is not what is meant nowadays when physicists say 'mass' (as been pointed out zilion times, even in this thread...).

Second, from the photons perspective

This whole thread is about the fact that there is no such thing as "photons perspective", so why you write things like this?

 

[.Scott]

Relativistic mass which is not what is meant nowadays when physicists say 'mass' (as been pointed out zilion times, even in this thread...).

The sentence started "From an Earth-bound observer". I don't need to explicitly say "relativistic".

This whole thread is about the fact that there is no such thing as "photons perspective", so why you write things like this?

And my point is that we can still look at the situation as the limit is approached. I think I did a very good job in explaining some of the problems with the notion of the "photon's perspective".

 

[jbriggs444]

The sentence started "From an Earth-bound observer". I don't need to explicitly say "relativistic".

If you say "mass" and mean "relativistic mass" then yes, you do need to say "relativistic mass". Or, better yet, rephrase what you are saying in terms of invariant mass, momentum or energy.

 

[vanhees71]

Is there any online reference that explains this equation in more detail? What is $\tau$, x_$\mu$, and x^$\mu$? Thanks.

You should find this in any textbook on relativity. It's very basic when you treat relativistic dynamics. My attempt to explain relativity can be found here (Chpt. 2):

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[weirdoguy]

The sentence started "From an Earth-bound observer". I don't need to explicitly say "relativistic".

Mass as is understood in relativity nowadays is an invariant and does not change with speed. What changes with speed is relativistic mass, so you have to say explicity what you mean.

 

[.Scott]

Relativistic mass which is not what is meant nowadays when physicists say 'mass'

If you say "mass" and mean "relativistic mass" then yes, you do need to say "relativistic mass".

I stand corrected.

 

[PeterDonis]

from the photons perspective, the starting and ending points are coincident

It has already been pointed out that a photon does not have a "perspective" in the sense of an inertial frame. However, even if we choose a coordinate chart in which the photon has three out of four coordinates constant (which is possible, although such a chart will not correspond to any inertial frame), it is not the case that "the starting and ending points are coincident". A photon's worldline still consists of distinct points. In more technical language, a null worldline can still be affinely parameterized; you just can't use arc length along the worldline as the affine parameter (whereas you can with timelike and spacelike curves).

 

[.Scott]

It has already been pointed out that a photon does not have a "perspective" in the sense of an inertial frame. However, even if we choose a coordinate chart in which the photon has three out of four coordinates constant (which is possible, although such a chart will not correspond to any inertial frame), it is not the case that "the starting and ending points are coincident".

Certainly as one approaches c (relative to everything else), the universe will appear to contract in the direction of travel. So wouldn't it be correct to say that as v approaches c, the apparent distance between your starting and ending points approaches zero? Then from there, as a limit, the distance would be zero?

Certainly the inertial reference frame becomes degenerate as the velocity approaches c. And if you want to define it as being completely non-existent at c (as opposed to just degenerate), that is fine.
It's like saying that your don't want to call zero a number because someone might want to put it in a denominator. That's a fine thing to do, but you should give it some other name.
Responding to a question about the photon's perspective by saying is has none becomes an exercise in semantics. It is okay to have the question - badly worded or otherwise.

 

[Ibix]

Then from there, as a limit, the distance would be zero?

There isn't a limit here. At any time you will see light passing you at c, and still have 300,000km/s to make up to catch up to it.

It is okay to have the question - badly worded or otherwise.

Of course it's ok to have the question. Just like it's ok to ask which way is north from the north pole. The answer, though, is going to be that the question doesn't make sense because it's based on assumptions about arc length that don't apply to null paths.

 

[PeterDonis]

wouldn't it be correct to say that as v approaches c, the apparent distance between your starting and ending points approaches zero?

The apparent distance in an appropriately defined inertial frame, yes.

Then from there, as a limit, the distance would be zero?

No, because the limit you are trying to take is not well-defined.

 

[.Scott]

There isn't a limit here. At any time you will see light passing you at c, and still have 300,000km/s to make up to catch up to it.

That's right. The limit in velocity is only seen be the Earth-bound people. The limits seen be the traveler is with the velocity of the universe rushing by (approaching c), the limit of the distance between the start and end point (approaching zero), and the travel time (also approaching zero).

 

[.Scott]

No, because the limit you are trying to take is not well-defined.

Mathematically, being poorly defined at the limit value doesn't invalidate the process of taking limit of a function. If anything, it could suggest that a different answer could be derived if the limit was taken differently.

 

[Nugatory]

Mathematically, being poorly defined at the limit value doesn't invalidate the process of taking limit of a function.

No, but it does invalidate any attempt to equate the value as we approach the point at which the function is poorly defined with "the value at that point" (scare-quotes because of course that value doesn't exist - if it did it wouldn't be poorly defined). There is no mathematically sound way of getting from "the separation in an inertial frame approaches zero as the relative velocity approaches $c$" to "the separation is zero in the inertial frame in which the relative velocity is $c$".

It is somewhat unfortunate that setting $v=c$ in the time dilation, length contraction, and relativistic mass formulas yields such a convincing hand-waving heuristic argument for the speed of light limit, because taking that argument seriously leads to some misunderstandings later. That's why you're getting so much resistance when you offer this answer "in the spirit in which [the original question] was asked".

 

[l0st]

Sorry for a philosophical take on it, but I just wanted to remind, that you, as in human mind, are not just made of matter. There are also electrical fields propagating inside at the speed of light (but not in vacuum, if you look from a high level), interacting with matter. So when answering the original question, it might make sense to understand/define what "experience" means for them.

 

[SiennaTheGr8]

It is somewhat unfortunate that setting $v=c$ in the time dilation, length contraction, and relativistic mass formulas yields such a convincing hand-waving heuristic argument for the speed of light limit, because taking that argument seriously leads to some misunderstandings later.

Indeed.

It's become a pet peeve of mine to see statements like "It would take infinite energy to accelerate an object to the speed of light, and that's why it can't be done." Such explanations have it backwards and emphasize the wrong thing, IMO (though they aren't "wrong"). Energy isn't the barrier; the postulates are (or geometry is).

But much worse is the completely illegal insertion of $c$ for both speeds in the velocity-addition formula:

$\dfrac{c + c}{1 + cc/c^2}$,

sometimes used to "show" that $c$ is invariant. For example, at 6:11 here (from the wonderful Don Lincoln no less!):



The error is immediately apparent when you do the same thing for the inverse formula (velocity-subtraction):

$\dfrac{c - c}{1 - cc/c^2}$,

which is undefined. Again, the issue is that the postulates simply forbid two frames from have relative speed $c$, period!

 

[vanhees71]

Is this the same Don Lincoln from Fermi Lab who writes great articles in The Physics Teacher?

 

[SiennaTheGr8]

Is this the same Don Lincoln from Fermi Lab who writes great articles in The Physics Teacher?

Must be.

 

[Mister T]

Certainly as one approaches c (relative to everything else), the universe will appear to contract in the direction of travel.

But it makes no sense to say your speed approaches $c$ relative to everything else. Protons in the Large Hadron Collider have speeds that approach $c$ relative to the laboratory, but relative to each other they have speeds that approach zero.

 

[.Scott]

But it makes no sense to say your speed approaches $c$ relative to everything else. Protons in the Large Hadron Collider have speeds that approach $c$ relative to the laboratory, but relative to each other they have speeds that approach zero.

I needed to say relative to something. Since I was looking to approach the limit, saying "relative to everything else" works.
Alternatively, I could have said, "relative to even the fastest cosmic rays".

 

[m4r35n357]

I needed to say relative to something. Since I was looking to approach the limit, saying "relative to everything else" works.
Alternatively, I could have said, "relative to even the fastest cosmic rays".

The only thing that makes any sense here is along the lines of "relative to something that was stationary before I went off, and stayed stationary".

 

[TeethWhitener]

Alternatively, I could have said, "relative to even the fastest cosmic rays".

I'm pretty sure my pot belly sitting in my office chair is traveling close to $c$ relative to the fastest cosmic rays.

 

[Mister T]

I needed to say relative to something. Since I was looking to approach the limit, saying "relative to everything else" works. Alternatively, I could have said, "relative to even the fastest cosmic rays".

But the comment I made about LHC protons applies just as well to cosmic ray particles. Perhaps what you say about "needing" to make your claim is true. But the fact that you need to make a nonsense claim to make a point ought to tell you something about that point.