Does time stop for a photon? Why is that a nonsensical question?

In summary, the answer to this question is that photons have zero mass and always travel the speed of light. It is impossible to imagine what it would be like for a photon to be at rest, as this would lead to paradoxes.
  • #36
Getting back to the OP's question

JamieForum said:
From this link http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/headlights.html I don't understand the following..."Does time stop for a photon?. . . It is really not possible to make sense of such questions and any attempt to do so is bound to lead to paradoxes. There are no inertial reference frames in which the photon is at rest so it is hopeless to try to imagine what it would be like in one."

In particular this statement "There are no inertial reference frames in which the photon is at rest". Can anyone explain that to me?

I offer an old post of mine:
www.physicsforums.com/showthread.php?p=899778#post899778
which tries to directly address this question
by first trying to
DEFINE what one might mean by a reference frame.
 
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  • #37
Yes Passion flower, it makes sense :)

It will be our measurements, relative our clocks, that defines a SpaceTime path for each one of us measuring some other. But, for the one moving 'faster' relative the other, the Lorentz contraction he measures will make his path 'shorter' than the 'slower ones' measure of it.

And it is true in one way more. You won't ever find your own 'local clock' to differ depending on motion or mass, only others. That's the most interesting one to me too.
 
  • #38
robphy said:
I offer an old post of mine:
www.physicsforums.com/showthread.php?p=899778#post899778
which tries to directly address this question
by first trying to
DEFINE what one might mean by a reference frame.

How about saying that time doesn't pass for light because for two light waves in the same direction, they always maintain the same phase, so if you are situated at the peak of one light wave, the peaks and troughs of the other light wave never pass you, so time is frozen for you if you use the other wave as a clock. (This is my "physical interpretation" of the null wave vector.)
 
  • #39
atyy said:
Can't I build a "clock" that travels on a null geodesic and returns a null reading? Say the clock is two beams of light of different frequencies. The "elapsed time" is the change in phase between them from the initial phase.

Well, then the "elapsed time" by this definition wouldn't return a null result, would it? If the beams started in phase at the emitter, then in general they would not be in phase at the detector (unless the detector were located very precisely relative to the emitter, so that an exactly integral number of wave fronts of both waves spanned the distance between them).

EDIT: I see I misunderstood what you were visualizing. What I said just above is still true, but it's not really a response to your question. See post #41 below for a better response.
 
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  • #40
A photon doesn't exist in its 'propagation' Jamie. It only has a 'source' (laser), and its 'sink' (eye). But it isn't 'there' for you in the same way a moving ball is. Physics enjoy different interpretations about that 'propagation'. Some want to use the fact that all photons are 'identical', to define a path from measuring those identical photons at different positions and time, and so create a 'average path'.

It all depends on where you stand and look at that fact.

Most of what a photon does, as its 'recoil' is explained through conservation laws, not 'classically' as we explain that ball in form of its acceleration. What we know is that it has a constant invariant 'c' though.
 
  • #41
atyy said:
How about saying that time doesn't pass for light because for two light waves in the same direction, they always maintain the same phase, so if you are situated at the peak of one light wave, the peaks and troughs of the other light wave never pass you, so time is frozen for you if you use the other wave as a clock. (This is my "physical interpretation" of the null wave vector.)

Einstein tried this thought experiment, and he realized that a "standing" electromagnetic wave like this violates Maxwell's Equations. This was one of the lines of reasoning that led him to SR. However, that still does leave the question of what exactly is wrong with the plausible-seeming picture you present. The resolution is that the wave fronts of each light beam, the surfaces of constant phase that you are calling "peaks and troughs", are not spacelike surfaces; they are null surfaces (because they are orthogonal to the null wave vectors, and a surface orthogonal to a null vector must itself be null). But your reasoning in the quote above (and the reasoning that shows that a "standing electromagnetic wave" violates Maxwell's Equations) depends implicitly on the wave fronts being spacelike; otherwise there is no physical meaning to "being situated at the peak of one light wave", because the "peak" is not a spatial location.
 
  • #42
PeterDonis said:
Einstein tried this thought experiment, and he realized that a "standing" electromagnetic wave like this violates Maxwell's Equations. This was one of the lines of reasoning that led him to SR. However, that still does leave the question of what exactly is wrong with the plausible-seeming picture you present. The resolution is that the wave fronts of each light beam, the surfaces of constant phase that you are calling "peaks and troughs", are not spacelike surfaces; they are null surfaces (because they are orthogonal to the null wave vectors, and a surface orthogonal to a null vector must itself be null). But your reasoning in the quote above (and the reasoning that shows that a "standing electromagnetic wave" violates Maxwell's Equations) depends implicitly on the wave fronts being spacelike; otherwise there is no physical meaning to "being situated at the peak of one light wave", because the "peak" is not a spatial location.

How about just considering a normal plane wave solution of Maxwell's equations. Coordinatize it in an inertial frame. The relative phase of two plane waves does not change.

I don't think I want to use light front coordinates since it those coordinates, "light cone time" passes for a photon.

The basic kludge I'd like to keep is that neutrino oscillations indicate they have mass, since they couldn't oscillate if time did not pass for them. Something like http://physics.stackexchange.com/qu...no-oscillations-imply-nonzero-neutrino-masses or http://m-Newton.ex.ac.uk/teaching/resources/eh/phy3135/lecture11.pdf [Broken]. (I swear this was taught to me twice as an undergrad - doesn't mean it's right, but that's how I was brainwashed - the same guys loved relativistic mass!)
 
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  • #43
atyy said:
How about just considering a normal plane wave solution of Maxwell's equations. Coordinatize it in an inertial frame. The relative phase of two plane waves does not change.

Careful. As I said in an earlier post, if you consider two plane waves with different frequencies/wavenumbers, between a source and a detector, the "relative phase" does change in the sense that if the waves are in phase at the source, they will be out of phase at the detector unless the distance from the source to the detector is chosen very carefully (to include an integral number of wave fronts of both waves). So you can't just say categorically that the relative phase does not change; you have to be more specific about what you mean by "relative phase", and being more specific makes it clear that the surfaces of constant phase are null, not spacelike.

atyy said:
I don't think I want to use light front coordinates since it those coordinates, "light cone time" passes for a photon.

It's not a question of coordinates. The wave vectors are null vectors, therefore the wave front surfaces orthogonal to them are also null. That's a coordinate-independent geometric statement; it's true even if you use an ordinary timelike-spacelike coordinate system to describe the vectors and wave fronts.

atyy said:
The basic kludge I'd like to keep is that neutrino oscillations indicate they have mass, since they couldn't oscillate if time did not pass for them. Something like http://physics.stackexchange.com/qu...no-oscillations-imply-nonzero-neutrino-masses .

I don't think anything I've said is inconsistent with this, and I agree that neutrino oscillations imply a non-zero neutrino rest mass. However, as Frank H's response on the stackexchange page makes clear, it's not really because "time must pass for neutrinos in order for them to oscillate". The key is that oscillations require that the mass eigenstates must be unequal (so at least two of the three mass eigenstates must be nonzero--normally all three are taken to be nonzero), and also that the mass eigenstates must be different than the flavor eigenstates.
 
  • #44
PeterDonis said:
Careful. As I said in an earlier post, if you consider two plane waves with different frequencies/wavenumbers, between a source and a detector, the "relative phase" does change in the sense that if the waves are in phase at the source, they will be out of phase at the detector unless the distance from the source to the detector is chosen very carefully (to include an integral number of wave fronts of both waves). So you can't just say categorically that the relative phase does not change; you have to be more specific about what you mean by "relative phase", and being more specific makes it clear that the surfaces of constant phase are null, not spacelike.

Yes, that's right.

PeterDonis said:
It's not a question of coordinates. The wave vectors are null vectors, therefore the wave front surfaces orthogonal to them are also null. That's a coordinate-independent geometric statement; it's true even if you use an ordinary timelike-spacelike coordinate system to describe the vectors and wave fronts.

So that seems to leave this possibility. What's wrong with it if we already allow that here the "time" direction is null? It seems ok to me, except that now this seems to prove that time does pass for photons.

Let me sketch out my initial thoughts on this. Evidently some of it doesn't work, but let's see if it can be corrected:

Time does not pass for photons -> photons travel on null lines -> photons have null wave vectors -> null wave vectors mean no dispersion -> the shape of a wave composed of different wavelengths does not change as it travels -> Time does not pass for photons.

PeterDonis said:
I don't think anything I've said is inconsistent with this, and I agree that neutrino oscillations imply a non-zero neutrino rest mass. However, as Frank H's response on the stackexchange page makes clear, it's not really because "time must pass for neutrinos in order for them to oscillate". The key is that oscillations require that the mass eigenstates must be unequal (so at least two of the three mass eigenstates must be nonzero--normally all three are taken to be nonzero), and also that the mass eigenstates must be different than the flavor eigenstates.

Yes to the mass eigenstate explanation, but what about the handwavy explanation? Do we chuck that out completely? A quick google shows it's not an uncommon handwave.
 
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  • #45
JamieForum said:
From this link http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/headlights.html I don't understand the following..."Does time stop for a photon?. . . It is really not possible to make sense of such questions and any attempt to do so is bound to lead to paradoxes. There are no inertial reference frames in which the photon is at rest so it is hopeless to try to imagine what it would be like in one."

In particular this statement "There are no inertial reference frames in which the photon is at rest". Can anyone explain that to me?

Hi Jamie welcome at physicsforums. :smile:

I'm afraid that the replies may have become overly technical. What that statement means in practice, is that a clock that would go at the speed of light would not only stand still, but also have zero(!) length, and infinite inertia(mass). Such a clock cannot exist.

Apart of that, the authors there differ in opinion from other authors such as Einstein who (in 1905) nevertheless thought it useful to discuss this limit speed as follows:

"For v=c all moving objects—viewed from the “stationary” system—shrivel up into plane figures. For velocities greater than that of light our deliberations become meaningless; we shall, however, find in what follows, that the velocity of light in our theory plays the part, physically, of an infinitely great velocity."

I agree with his way of saying it, as long as it is well understood that objects can never fully reach that speed.

Harald
 
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  • #46
harrylin said:
I'm afraid that the replies may have become overly technical. What that statement means in practice, is that a clock that would go at the speed of light would not only stand still, but also have zero(!) length, and infinite inertia(mass). Such a clock cannot exist.
By suggesting "what would happen", you open the door to people coming up with flawed counter-suggestions about how to mitigate those results. It happens all the time.

It is more than just such a clock cannot exist; it is that the universe won't let it.

What you need to do is talk about what happens to the clock as it approaches the speed of light, and then it becomes obvious to the OP why it can never get there. An easy one is the approaches infinite mass property. As it approaches infinite mass, it also requires an amount of energy approaching infinity to accelerate it further. There simply isn't that amount of energy.
 
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  • #47
atyy said:
So that seems to leave this possibility. What's wrong with it if we already allow that here the "time" direction is null?

Because a null direction is not a timelike direction. They're fundamentally different physically, and calling a null direction "the time direction of a photon" does not make them the same, nor does it mean that "time passes for a photon".

atyy said:
Time does not pass for photons -> photons travel on null lines -> photons have null wave vectors -> null wave vectors mean no dispersion -> the shape of a wave composed of different wavelengths does not change as it travels -> Time does not pass for photons.

Except for the first and last sentences, this looks OK. :wink: The problem with saying that "time does not pass for photons" is that it invites further inferences which are not justified. Saying that "the concept of time passing does not apply to photons" is better because it makes it clear that there is not just a binary choice, "does time pass or not?"; rather, more complex physics is involved. See below.

atyy said:
Yes to the mass eigenstate explanation, but what about the handwavy explanation? Do we chuck that out completely? A quick google shows it's not an uncommon handwave.

The problem I have with the handwave as it stands (i.e., if it is not supplemented with the more accurate explanation) is that it invites the view that a photon is "frozen", which invites the view that "everything is simultaneous to a photon", or "to a photon, the entire universe looks like a point", and so on. I realize these are popular "handwaves" too, but that doesn't make them right. IMO it's better to state right up front that objects moving on null worldlines are fundamentally different, physically, from objects moving on timelike worldlines, and some concepts, like "proper time" and "simultaneity", that apply to the latter simply don't apply to the former.
 
  • #48
DaveC426913 said:
[..] As it approaches infinite mass, it also requires an amount of energy approaching infinity to accelerate it further. There simply isn't that amount of energy.

Yes, exactly - thanks for the elaboration!
 
  • #49
PeterDonis said:
Because a null direction is not a timelike direction. They're fundamentally different physically, and calling a null direction "the time direction of a photon" does not make them the same, nor does it mean that "time passes for a photon".

I see. Your view is that we have some notion of time already, and it doesn't apply. I agree with that, except that I then say, well, anyway, time is just what we define it to be, and we can have many definitions of time. So can we find another definition in which the sentence is true? So I have no problems with calling the null direction the "light front time". I think it's nice because it makes it clear that with this definition, time does pass for a photon. Anyway, no physics disagreement.

PeterDonis said:
Except for the first and last sentences, this looks OK. :wink: The problem with saying that "time does not pass for photons" is that it invites further inferences which are not justified. Saying that "the concept of time passing does not apply to photons" is better because it makes it clear that there is not just a binary choice, "does time pass or not?"; rather, more complex physics is involved. See below.

OK, again, no physics disagreement.

PeterDonis said:
The problem I have with the handwave as it stands (i.e., if it is not supplemented with the more accurate explanation) is that it invites the view that a photon is "frozen", which invites the view that "everything is simultaneous to a photon", or "to a photon, the entire universe looks like a point", and so on. I realize these are popular "handwaves" too, but that doesn't make them right. IMO it's better to state right up front that objects moving on null worldlines are fundamentally different, physically, from objects moving on timelike worldlines, and some concepts, like "proper time" and "simultaneity", that apply to the latter simply don't apply to the former.

What I was trying to do with the null wave vector is to bring the idea of dispersion forward (ie. photons are dispersionless), since the neutrino oscillation explanation uses the massive dispersion relation to get different masses to travel at different speeds - and in this way link the handwavy "timeless" idea with the dispersion idea which is sound.
 
  • #52
PeterDonis said:
I guess I'll have to add Schutz to my list of physicists I want to hit over the head, right after Brian Greene. :grumpy:

And I guess that I agree with Schutz and Einstein on this one - with the caveat that I gave in post #45. :wink:
 
<h2>1. Does time actually stop for a photon?</h2><p>No, time does not actually stop for a photon. This is a common misconception based on the idea that as an object approaches the speed of light, time slows down. However, this only applies to observers outside of the photon's frame of reference. From the photon's perspective, time appears to pass normally.</p><h2>2. Why is it a nonsensical question to ask if time stops for a photon?</h2><p>As mentioned before, the concept of time stopping for a photon is based on a misunderstanding of the theory of relativity. Time dilation, or the slowing of time, only applies to objects in motion relative to an observer. Since a photon is always traveling at the speed of light, there is no frame of reference from which to observe time dilation.</p><h2>3. What happens to time for a photon?</h2><p>For a photon, time appears to pass normally. It is only when observed from a different frame of reference that time dilation becomes apparent. This is because the speed of light is constant in all frames of reference, and time dilation is a result of the relationship between an object's speed and the speed of light.</p><h2>4. Can anything travel at the speed of light?</h2><p>No, according to the theory of relativity, nothing can travel at the speed of light except for light itself. As an object approaches the speed of light, its mass and energy increase infinitely, making it impossible to reach the speed of light. Therefore, it is not possible for anything to experience time stopping, as it would require traveling at the speed of light.</p><h2>5. What is the significance of understanding that time does not stop for a photon?</h2><p>Understanding that time does not stop for a photon is important for accurately understanding and applying the theory of relativity. It also helps to dispel common misconceptions about the nature of time and light. Additionally, it highlights the importance of considering different frames of reference when studying the behavior of objects in motion.</p>

1. Does time actually stop for a photon?

No, time does not actually stop for a photon. This is a common misconception based on the idea that as an object approaches the speed of light, time slows down. However, this only applies to observers outside of the photon's frame of reference. From the photon's perspective, time appears to pass normally.

2. Why is it a nonsensical question to ask if time stops for a photon?

As mentioned before, the concept of time stopping for a photon is based on a misunderstanding of the theory of relativity. Time dilation, or the slowing of time, only applies to objects in motion relative to an observer. Since a photon is always traveling at the speed of light, there is no frame of reference from which to observe time dilation.

3. What happens to time for a photon?

For a photon, time appears to pass normally. It is only when observed from a different frame of reference that time dilation becomes apparent. This is because the speed of light is constant in all frames of reference, and time dilation is a result of the relationship between an object's speed and the speed of light.

4. Can anything travel at the speed of light?

No, according to the theory of relativity, nothing can travel at the speed of light except for light itself. As an object approaches the speed of light, its mass and energy increase infinitely, making it impossible to reach the speed of light. Therefore, it is not possible for anything to experience time stopping, as it would require traveling at the speed of light.

5. What is the significance of understanding that time does not stop for a photon?

Understanding that time does not stop for a photon is important for accurately understanding and applying the theory of relativity. It also helps to dispel common misconceptions about the nature of time and light. Additionally, it highlights the importance of considering different frames of reference when studying the behavior of objects in motion.

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