B Is Time Experienced by Photons at the Speed of Light?

[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.

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

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.

 

[.Scott]

OK. Let's do this one more time.
It doesn't make sense to say "approaching the speed of light" without offering a frame of reference.
In this case, I was trying to describe a speed relative to Earth that was so close to the speed of light that it would be at least 0.99999999c relative to anything with mass: pot bellies, LHC particles, cosmic rays, anything.

And just to be clear, it would not matter if it turned out that there was something else in the universe that turned out to be going the same speed as my reference frame. It would invalidate my description of the speed as being "more than 0.9999999c relative to anything", but it would not invalidate its use in context.

It was not my intention to use terms that could be interpreted in such a variety of ways. I thought there was sufficient context to make my intent clear.

 

[YoungPhysicist]

cosmic rays close to the speed of light

Sorry for the unrelated content, but is that the OMG particle?

https://en.m.wikipedia.org/wiki/Oh-My-God_particle

 

[.Scott]

Sorry for the unrelated content, but is that the OMG particle?

https://en.m.wikipedia.org/wiki/Oh-My-God_particle

Yes. And it's the reason I picked it in response to the LHC example.

 

[m4r35n357]

In this case, I was trying to describe a speed relative to Earth that was so close to the speed of light that it would be at least 0.99999999c relative to anything with mass: pot bellies, LHC particles, cosmic rays, anything.

Your insistence here indicates that you do not fully understand what relative motion means and how to specify it. "Relative to anything with mass" means nothing whatsoever (for example, is the mass in question moving WRT you or not?). Did you understand my previous comment?

 

[.Scott]

Your insistence here indicates that you do not fully understand what relative motion means and how to specify it. "Relative to anything with mass" means nothing whatsoever (for example, is the mass in question moving WRT you or not?). Did you understand my previous comment?

Yes I do understand. In the original context, I was referring to speeds relative to the observer that were so close to c that they would be close to c no matter what massive reference frame was selected.
My point was not to discount the relativity of velocities but to emphasize the extremeness of the reference frame I was describing.

 

[m4r35n357]

Yes I do understand. In the original context, I was referring to speeds relative to the observer that were so close to c that they would be close to c no matter what massive reference frame was selected.
My point was not to discount the relativity of velocities but to emphasize the extremeness of the reference frame I was describing.

At this point I must give up. There is no such thing as an "extreme" (inertial) reference frame, they are all equivalent. Your point is consistently obscured by your reluctance to adopt the language we are all trying to get you to use.

 

[.Scott]

At this point I must give up. There is no such thing as an "extreme" (inertial) reference frame, they are all equivalent. Your point is consistently obscured by your reluctance to adopt the language we are all trying to get you to use.

It can only be extreme relative to another reference frame. I never said otherwise.
And in the original context, that type of extremeness was important because we wanted to consider how the rest of the universe would look from such an extreme reference frame.

I also get that just because you are traveling at 0.99999999999999c relative to Earth, you are still no closer to the speed of light than you were before you left Earth.

That doesn't stop me from considering reference frames where the speed of our galaxy (relative to that frame) appears to be very close to the speed of light.

I think the issue here is that I am willing to use that as a possible window into questions related to "how a photon sees the universe" - admitting that the photon, by my calculation or anyone else's, has zero time to "see" anything.

 

[SiennaTheGr8]

Tangentially related is the GZK limit.

 

[timmdeeg]

That doesn't stop me from considering reference frames where the speed of our galaxy (relative to that frame) appears to be very close to the speed of light.

And you can always imagine an observer who’s speed is even closer to the speed of light. But you can never imagine an obsrever who measures the speed of light less than c in his frame. And that’s the reason that the question “how a photon sees the universe” doesn’t make any sense.

 

[Ibix]

That doesn't stop me from considering reference frames where the speed of our galaxy (relative to that frame) appears to be very close to the speed of light.

Of course you can. But in such frames, just like the rest frame of the galaxy, the speed of light is 3×10⁸m/s. So you aren't really any closer to knowing what the perspective of light is. Fundamentally, that's because the concept doesn't make sense, certainly not as an extension of standard inertial frames.

 

[.Scott]

And you can always imagine an observer who’s speed is even closer to the speed of light. But you can never imagine an observer who measures the speed of light less than c in his frame. And that’s the reason that the question “how a photon sees the universe” doesn’t make any sense.

That's interesting. In the process of refuting that the photon reference frame exists, without intending to, you just presumed something about that frame. Specifically, you have presumed that in such a frame, the speed of light would appear to be less than c. I know you didn't mean to do that - but since you've brought it up, I would have presumed that even in a photon's reference frame, the speed of light would remain to be c. Since there's no stop-watch in that frame, who's to say otherwise.

 

[Ibix]

I would have presumed that even in a photon's reference frame, the speed of light would remain to be c.

Exactly. And it would have to be zero also - or otherwise what do you mean by "photon's reference frame" other than its rest frame?

 

[.Scott]

Exactly. And it would have to be zero also - or otherwise what do you mean by "photon's reference frame" other than its rest frame?

What doesn't bother me about that is that from a photon's reference frame, there is no time and therefore, there is no velocity. Velocity is simply something that cannot be observed from that reference frame.

 

[timmdeeg]

What doesn't bother me about that is that from a photon's reference frame, there is no time and therefore, there is no velocity. Velocity is simply something that cannot be observed from that reference frame.

If doesn’t bother you if at the end of the day you agree that a photon’s reference frame does not exist.

 

[Nugatory]

I would have presumed that even in a photon's reference frame, the speed of light would remain to be c.

What doesn't bother me about that is that from a photon's reference frame, there is no time and therefore, there is no velocity. Velocity is simply something that cannot be observed from that reference frame.

OK, this thread has gone off the deep end... it is closed.