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Just doesn't make sense to me that a mass-less particle (photon) which should not "experience" time does in fact require time to travel from point A to point B if the distance is long enough.

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- Thread starter MattAndMatthe
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- #1

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Just doesn't make sense to me that a mass-less particle (photon) which should not "experience" time does in fact require time to travel from point A to point B if the distance is long enough.

- #2

Filip Larsen

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While it may make little sense when it is proposed just like that, you can consider what happens if we give progressively lighter particles a fixed amount of kinetic energy before they are to transit from point A to B (which are at rest relative to you). As the the particles become lighter (i.e. having less rest mass) they will have to move faster relative to you (but still less than c) to have the same fixed amount of kinetic energy. From the perspective of each of these particles (ie. from a reference frame at rest relative to each particle) the distance A B has been shortened due to length contraction. In the limit where the mass go to zero, the distance each particle travel also goes to zero, meaning they would travel any finite distance however big in zero time. Now change the particles with ones that are born at A and absorbed at B and you have, more or less, "the life of a photon".

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Filip Larsen

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OK, I am trying to reason through your thorough reply with my limited understanding of relativity theory. So, let's see if I understand correctly: A photon traveling in a linear direction from Sirius to Earth traverses the entire distance in one moment relative to itself, but relative to us the photon's travel is not instantaneous because time dilates differently from our point of observation.

This is almost correct. You should not try understand this by starting with a photon that moves nowhere in no time and then apply transformations from the special relativity to get from the photons perspective back to ours. This will not make sense. You should start with the photon as we (and everyone else) observe it to move with speed c and then, possibly via a limit argument as I mentioned earlier, work towards what "happens" as seen from the photon.

measured in a unit of time equal to the distance of the line divided by c?

I'm not exactly sure what you mean here. Again, for photons it does not make much sense to say that it sees the remaining universe move at c since this both space and time is in some sense zero for it.

the photon's path is contracted to us?.

No. If anything it should be the other way around. Any finite distance is infinitely contracted from the photons perspective.

Let me once again stress, that it does not necessarily make sense to try understand what "happens" to time and space for a photon. The photon is kinda "born" in a different "world" than ours and to "translate" our concept of time and space (and speed) to a photon gives no physical meaning. The only physical meaning I can recognize is, as mentioned, to see what happens to moving particles in the limit when their rest mass goes to zero.

Maybe others can offer a better explanation.

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So, let's see if I understand correctly: A photon traveling in a linear direction from Sirius to Earth traverses the entire distance in one moment relative to itself (measured in a unit of time equal to the distance of the line divided by c?)

There is a relativistic formula that calculates the time dilation factor, among other things;

[TEX]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/TEX]

And thus travel at the speed of light results in [TEX]\frac{1}{0}[/TEX] in a vacuum.

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rcgldr

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diazona

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No it isn't. You have to be careful with perspective here:Photons have a frequency, which is evidence that photons "experience" time.

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rcgldr

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You could consider the oscillations to be a photon's clock. It's a cyclic change of state with repect to time, one that gets affected by gravity intensity in the same manner that sub-light speed clocks do.To a photon, the difference between crossing your room and crossing the galaxy is in the number of oscillations they undergo, not in the time it takes them (which is non-existent to them.)

- #10

Cleonis

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[...]

You should not try understand this by starting with a photon that moves nowhere in no time and then apply transformations from the special relativity to get from the photons perspective back to ours. This will not make sense. You should start with the photon as we (and everyone else) observe it to move with speed c and then, possibly via a limit argument as I mentioned earlier, work towards what "happens" as seen from the photon.

[...]

It seems to me that describing the state of photons is beyond the means of relativistic physics.

What I mean is that the only reasoning that we can bring to bear is to compare with the case of moving at a velocity that is very near to the speed of light, and then "take a limit".

As someone pointed out earlier, electromagnetic radiation has a frequency. Between emission and subsequent absorption the electromagnetic wave goes through a certain number of cycles of its frequency. While distance travelled and elapsed time are frame dependent, the

It follows logically that such a thing as "the photon frame" cannot exist in the same sense as any other frame of reference. Because if it would then there is a self-contradiction. The photon has to go through a certain number of cycles of its frequency, but as mapped "in the photon frame" there is no time for that.

I find the status of photons in relativistic physics just as paradoxical as particle/wave duality. That doesn't stop me from relying on relativistic physics, but it does illustrate, I think, that relativistic physics is just as much a non-classical theory as quantum physics is.

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Hope that helps.

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It seems to me that describing the state of photons is beyond the means of relativistic physics.

What I mean is that the only reasoning that we can bring to bear is to compare with the case of moving at a velocity that is very near to the speed of light, and then "take a limit".

As someone pointed out earlier, electromagnetic radiation has a frequency. Between emission and subsequent absorption the electromagnetic wave goes through a certain number of cycles of its frequency. While distance travelled and elapsed time are frame dependent, thenumber of cycles is not frame dependent.

It follows logically that such a thing as "the photon frame" cannot exist in the same sense as any other frame of reference. Because if it would then there is a self-contradiction. The photon has to go through a certain number of cycles of its frequency, but as mapped "in the photon frame" there is no time for that.

I find the status of photons in relativistic physics just as paradoxical as particle/wave duality. That doesn't stop me from relying on relativistic physics, but it does illustrate, I think, that relativistic physics is just as much a non-classical theory as quantum physics is.

I agree with you.As for the frequency thing,one has to notice that the frequency is frame-dependent.It is NOT Lorentz-invariant.

But the photons in relativity isn't paradoxical in my opinion.When you apply Lorentz transformation to photon frame,you will get a invertible transform matrix which is against the principle of relativity.But it is no paradox because there is NO photon frame.

Since there is no way to observe in a photon's view,it is totally meaningless to discuss what it looks like in a photon's eye.You cannot prove or falsify it with experiments,so why bother to think?

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This is almost correct. You should not try understand this by starting with a photon that moves nowhere in no time and then apply transformations from the special relativity to get from the photons perspective back to ours. This will not make sense. You should start with the photon as we (and everyone else) observe it to move with speed c and then, possibly via a limit argument as I mentioned earlier, work towards what "happens" as seen from the photon.

I'm not exactly sure what you mean here. Again, for photons it does not make much sense to say that it sees the remaining universe move at c since this both space and time is in some sense zero for it.

No. If anything it should be the other way around. Any finite distance is infinitely contracted from the photons perspective.

Let me once again stress, that it does not necessarily make sense to try understand what "happens" to time and space for a photon. The photon is kinda "born" in a different "world" than ours and to "translate" our concept of time and space (and speed) to a photon gives no physical meaning. The only physical meaning I can recognize is, as mentioned, to see what happens to moving particles in the limit when their rest mass goes to zero.

Maybe others can offer a better explanation.

So, what you're saying is that for a photon emitted during the Big Bang, no time has elapsed and no distance travelled (based on the photons frame of reference) - is that correct?

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Photons are ripples in the electromagnetic field, which in turn appears to be somehow part of space/time itself. Does time experience time?

It feels like you're asking if rain ever gets wet.

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DaveC426913

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So, what you're saying is that for a photon emitted during the Big Bang, no time has elapsed and no distance travelled (based on the photons frame of reference) - is that correct?

Correct. A photon is a fixed, static straight line in 4-dimensional spacetime that joins point A at x

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DaveC426913

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True.first, the light does not require any type of a medium to propagate in,

False.and the light experience time, but according to the theory of special relativity, the time it does experience is less than the time you (the observers on planet earth) think it.

Light (photons) does not experience time

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