Question from the FAQ on Rest Frame of a Photon

[harrylin]

[..]

To adapt from your "riding the lightwave" illustration, I guess what I'm wanting to do is mind-meld with the brain of the photon and grasp exactly what it "sees" as it travels. What is a photon's inherent age after traveling from the star to me? (I know -- theoretically undefined..., but it's still got to have an age, right?) Or is a photo ageless since time is zero?

These questions are not without some redeeming merit -- there is a practical interest that is a portion of the motivation. Those who study the stars tell us that it takes so many light-years for light to reach us from the stars. But the discussion on this thread has indicated to me that such estimates are (seemingly) based upon measurements of time and distance on our side. It would, however, seem that relativity would suggest that the more accurate determination of the time can not be determined on this side, but on light's side (theoretically undefined, though it may be).

That's a different one, if you want to have not a practical answer (which I and others gave) but a more philosophical one. Einstein had thought a lot about "riding the lightwave" and he also gave the answer for the hypothetical (but unreachable) v=c .
Here, in http://www.fourmilab.ch/etexts/einstein/specrel/www/ , section 4:

"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."

Indeed, if you could go almost at the speed of light, then you would hardly age at all on your travel from Sirius to Earth. In that sense the same practical effect can be obtained as with a nearly infinite velocity in classical mechanics.

Did that help?


Harald

 

[rede96]

Ah, a fellow newbie -- a welcome sight. <ha>

(I know exactly how you feel.)

I also found this link useful.

https://www.physicsforums.com/showpost.php?p=1684890&postcount=20"

 

[DonB]

Quite so; you need to construct (or define) a reference system for doing measurements - and a system in which rulers have zero length and infinite mass is not an option.

... and where bathroom scales always read 175 lbs.

I'm afraid that you did not really read what I wrote (and apparently, nor did you read what others wrote): in the real world there is nothing "undefined" about light traveling from Star X to your eye. As I already explained, you can choose for example the solar "frame" (that is, the system in which the Sun is in rest at its centre), and describe the trajectory with respect to that system. This is not different at all from how it's done in classical physics.

Is it finally clear now?

I picked up the word "undefined" from the DH article. And as the discussion here has progressed I've come to see that relativity holds that a ref. frame from light's perspective (incl. in any way attempting to measure time or distance) is "undefined", as DH explained. Sure, other ref. frames are possible, as you pointed out, but that really wasn't addressing the question on my mind. I wanted to grasp time and distance from the photon's perspective..., and it seems that gives me a "divide by zero" error. <ha>

"Is it clear now?" I'm sitting here trying to emphatically experience a photon's travel from a star to Earth in some beyond-science-fiction environment where there is no such thing as time and distance... So, ahhh..., no, no where near it. <ha> But I am making some progress as we volley this stuff back and forth. So, it has been productive.

 

[DonB]

(I know exactly how you feel.)

I also found this link useful.

https://www.physicsforums.com/showpost.php?p=1684890&postcount=20"

Interesting read. I'll have to study on that a while.

Thanks!

 

[ghwellsjr]

Hi DrDon.

Like you I am a relative newbie. I’ve tried the same as you have to understand the nature of light too and asked a lot of the questions you mentioned here. (And still do!)

What I’ve found is that in order to understand some if the implications that you mentioned in your first post, I had to think about how to ask my questions using the proper convention.

So I thought of an experiment that might help address some of the implications you mentioned. (Please ignore if this doesn't relate to your question.)

By the way, this is a genuine question as I am not sure what the correct answer is.

Imagine a large ‘T’ in space.

Person A is at rest with respect to the point in space that we could describe as the intersection of where the horizontal path would cross the vertical path of the ‘T’. A is also at a distance of 10 light minutes away from this point. (So at the bottom of the T)

Person B is in his spaceship moving in the horizontal (left to right) part of the T with some small velocity of 100 mph wrt to the point of intersection. (I wanted to use a small velocity to approximate Euclidean space so we don’t have to complicate things with time dilation/length contraction.)

Person A has a laser gun with a powerful telescopic lens and he sets his sights in the 12 o’clock direction waiting for person B to pass the intersection of the ‘T’. B has a target in the middle of his ship that A is trying to hit.

However, person A assumes that if he waits for B to be directly in his line of sight before firing, then B would have actually passed this point by 10 minutes by the time the laser reaches B

So A does some calculations and re-aligns his sights so it is directed at a point in space before the intersection of the T and in such a way that when he sees B’s target, he knows B is the right time away from the intersection. So he fires the laser and thus assumes should hit B at the point when B arrives at the intersection.

So the question is:

a) Does B detect the laser hitting his target as he passes the intersection of the ‘T’

b) Does B actually intersect the laser beam, as it was already there when he passes the intersection? (This analogous to time standing still for light.)

c) The laser misses B’s target all together.

Since it will take ten minutes for the image of B to reach A and another ten minutes for the laser pulse to travel from A to the intersection, A needs to pull the trigger twenty minutes before it appears that B would arrive at the intersection, in other words when B is 33.333 miles before the intersection.

So when A sees B at that point and fires his laser gun, if he could watch the laser pulse, it would appear to him to travel at 1/2 the speed of light and arrive at the intersection just as B arrives there.

 

[DonB]

...Indeed, if you could go almost at the speed of light, then you would hardly age at all on your travel from Sirius to Earth. In that sense the same practical effect can be obtained as with a nearly infinite velocity in classical mechanics.

Did that help?


Harald

What you say here fits with my earlier quandary -- the faster you go, the shorter the time to cover the distance. Then when one crosses over the "light barrier", and v=c, time and distance have reduced down to zero. Thus the light beam leaves the star and is instantaneously here at Earth for me to see.

But others have shown me that that is not exactly what happens. Rather, once the light barrier has been crossed, it's more like the whole thing has reached critical mass, and the disruption leaves one without any system of time or space. It's not that one now reads zero units of measure for time/distance, but there simply is no units of measure for such things in this realm. At least that's how I'm understanding the explanation.

 

[DonB]

I thought you said that you do not reject answers. I have told you several times now that DH's article does not state "that at c time equals zero and distance equals zero". Why do you persist in repeating this false claim?

Man, I've tried to avoid an argument over all this -- it's not helpful to anyone. But since you insist, here is what DH said...

"Time and length cease to have meaning in the limit v→c. In that limit, all time and length intervals shrink to zero."

 

[ghwellsjr]

What you say here fits with my earlier quandary -- the faster you go, the shorter the time to cover the distance. Then when one crosses over the "light barrier", and v=c, time and distance have reduced down to zero. Thus the light beam leaves the star and is instantaneously here at Earth for me to see.

But others have shown me that that is not exactly what happens. Rather, once the light barrier has been crossed, it's more like the whole thing has reached critical mass, and the disruption leaves one without any system of time or space. It's not that one now reads zero units of measure for time/distance, but there simply is no units of measure for such things in this realm. At least that's how I'm understanding the explanation.

The faster you go, the shorter the time it appears to you to cover the distance, but also the distance is shortened for you. But not for those you left behind, it still takes a long time and it is a long distance.

But you can never cross over the "light barrier", let alone reach it, so you never have to worry about the system reaching critical mass and leaving you without a system of time and space. In fact, let's say you have fired your powerful rockets for a long enough time that you have accelerated to 99% the speed of light. Then you measure the speed of light by the technique I described earlier and you get c. So now you are at rest with respect to the speed of light and repeat the process again so that you have accelerated to 99%c. You turn off your thrusters and measure the speed of light to be c again. You can repeat this any number of times and you will never be any closer to reaching the light barrier as you were before you started.

Do you understand that?

Do you believe that?

Will you stop talking about reaching the "light barrier"?

Will you stop talking about crossing over the "light barrier"?

Please?

 

[DonB]

The faster you go, the shorter the time it appears to you to cover the distance, but also the distance is shortened for you. But not for those you left behind, it still takes a long time and it is a long distance.

But you can never cross over the "light barrier", let alone reach it...

If you will note the full context of my comment, including the earlier posts to which it refers, you will note that it ultimately refers back to a photon traveling from a star to earth. As such, it does in fact reach the speed of light.

Will you stop talking about reaching the "light barrier"?

Will you stop talking about crossing over the "light barrier"?

Please?

If my one (?) prior casual description of "light barrier" evokes this level of response..., well, I think it speaks for itself. And sorry, but no I find no reason to discontinue its use.

 

[ghwellsjr]

Man, I've tried to avoid an argument over all this -- it's not helpful to anyone. But since you insist, here is what DH said...

"Time and length cease to have meaning in the limit v→c. In that limit, all time and length intervals shrink to zero."

As I explained in post #10, but you seem to have ignored:

You don't understand this nomenclature which you quoted from the FAQ:

Time and length cease to have meaning in the limit v→c. In that limit, all time and length intervals shrink to zero.​

What that means is as a velocity approaches the speed of light, but never getting there, the time and length intervals approach zero, but they never get there. The whole purpose of the FAQ is to address the meaninglessness of the concept of a rest frame for a photon.

Not only are you misunderstanding the nomenclature, you are taking the quote out of context. Here is the entire paragraph:

Time and length cease to have meaning in the limit v→c. In that limit, all time and length intervals shrink to zero. In the rest frame of a photon, the coordinates of any point in the universe at any time in the past, any time in the future is identically zero. That just doesn't make a bit of sense.​

Now does that look like DH is promoting the idea of the time and distance for a photon being zero? No, he says it "doesn't make a bit of sense."

This isn't like an argument where we both have different valid opinions and it doesn't help that you are trying to avoid confrontation. You need help to learn the truth. That's why I keep asking you if you understand and agree with the statements I'm making. It doesn't help when you don't answer because you want to avoid an argument, you'll never make progress in understanding the nature of the real world or the Theory of Special Relativity unless you try hard to understand what we are saying. It's OK to be a slow learner, we would love nothing more than for you to finally grasp these ideas but you have to make an effort.

 

[nitsuj]

What does this even mean? To me, this sounds like you're saying photons can be described by something of the form $k^{\mu} = (0,0,0,T)$, but photons must be describable by null vectors as far as I know.

Saying it differently, EM travels distances in the time dimension, not in definitive xyz coordinates like mass things do.

I have no idea what "vectors are, but wiki says this of Null Vectors

Null vectors fall into three classes:

1.)the zero vector, whose components in any basis are (0,0,0,0),
2.)future directed null vectors whose first component is positive, and
3.)past directed null vectors whose first component is negative.

I'd guess this is special treatment for photons being described in coordinates xyz.

 

[DonB]

This isn't like an argument where we both have different valid opinions and it doesn't help that you are trying to avoid confrontation. You need help to learn the truth. That's why I keep asking you if you understand and agree with the statements I'm making. It doesn't help when you don't answer because you want to avoid an argument, you'll never make progress in understanding the nature of the real world or the Theory of Special Relativity unless you try hard to understand what we are saying. It's OK to be a slow learner, we would love nothing more than for you to finally grasp these ideas but you have to make an effort.

Okay, my friend, here it is. I'm not avoiding a confrontation. I am avoiding your pushiness. And I am avoiding what seems to me to be a condescending attitude. And that is it in a nutshell.

I don't need to be told I need help learning -- I've not only said as much, but the fact of my being here in the first place shows I know that. Others have stepped up to the plate and allowed me to work through this at a pace I can process. Maybe I'm dumb; maybe I'm retarded (even though an earned doctorate in a different field suggests otherwise). And when it comes down to responding to two (or more) posts -- one without the attitude, the pushiness, and that puts stuff down on my level; and the other that isn't that way -- I'll work with what I find workable.

I don't say this to criticize you; and wouldn't typically say it at all. But when you insist that I defend my stance..., then brother, that is my defense.

'Nouf said.

 

[Dale]

Yes, I understand the nomenclature. But what I don't understand is why you're addressing these issues with me -- they are not my words.

They may not be your words, but they are the substance of your question.

The folks here at PF have set this article (with its specific terminology) as authoritative, not me; and if you believe the wording to be self-contradictory I don't think I'm the one that can do anything about it.

I have no complaint about the FAQ, it is correct and I agree with it. Since you didn't understand the FAQ I am trying to explain to you in a different way why the concept of a photon's rest frame is meaningless.

Do you understand how the concept of an inertial frame where light is at rest is logically self-contradictory?

 

[WannabeNewton]

What does this even mean? To me, this sounds like you're saying photons can be described by something of the form $k^{\mu} = (0,0,0,T)$, but photons must be describable by null vectors as far as I know.

You are right that $k^{\mu }k_{\mu } = 0$ along a photon's world line so I don't understand what he means either.

 

[WannabeNewton]

I'd guess this is special treatment for photons being described in coordinates xyz.

It is a statement of how the tangent vector along a null geodesic is orthogonal to itself so it is independent of the coordinate system. I don't understand why you keep saying the photon travels only in time; it seems to go against the nullity of the tangent vector or, in Pengwuino's example, the wave 4 - vector along the photon's path.

 

[nitsuj]

You are right that $k^{\mu }k_{\mu } = 0$ along a photon's world line so I don't understand what he means either.

[STRIKE]What does $k^{\mu }k_{\mu } = 0$ mean?[/STRIKE] nvm, i won't understand your reply

 

[DonB]

Do you understand how the concept of an inertial frame where light is at rest is logically self-contradictory?

I am familiar with the arguments, yes.

 

[nitsuj]

What does this even mean? To me, this sounds like you're saying photons can be described by something of the form $k^{\mu} = (0,0,0,T)$, but photons must be describable by null vectors as far as I know.

A fair amount of reading and a third attempt at rewording.

A photon is an elementary particle and is apparently (wiki) "treated" mathematically as a point particle which has no [physical] dimension.

 

[Pengwuino]

A fair amount of reading and a third attempt at rewording.

A photon is an elementary particle and is apparently (wiki) "treated" mathematically as a point particle which has no dimension.

Ok, with no disrespect intended, I suggest you retract your argument. Please don't try to piece together an argument based off things you don't understand on a wikipedia page. We're trying to explain something to someone and contributing statements that are nonsensical and vague will do far more harm than good.

 

[nitsuj]

Ok, with no disrespect intended, I suggest you retract your argument. Please don't try to piece together an argument based off things you don't understand on a wikipedia page. We're trying to explain something to someone and contributing statements that are nonsensical and vague will do far more harm than good.

sorry for the harm i caused

 

[Dale]

I am familiar with the arguments, yes.

Good. Then is there anything remaining to discuss?

 

[DonB]

Good. Then is there anything remaining to discuss?

Maybe in due time. Plenty to just chew on for the time being.

 

[harrylin]

What you say here fits with my earlier quandary -- the faster you go, the shorter the time to cover the distance. Then when one crosses over the "light barrier", and v=c, time and distance have reduced down to zero.

Right -except that you meant "reaching the light barrier" (which cannot be reached...)

Thus the light beam leaves the star and is instantaneously here at Earth for me to see.

I think that what you mean is wrong - and others already pointed this out to you.
The frozen clock that the photon uses (although that idea is nonsense) to observe the Earth, is *not* the clock that you use to observe the photon. Your "time" experience cannot be affected by a stuck clock that you don't use.

But others have shown me that that is not exactly what happens. Rather, once the light barrier has been crossed, [...]

Nothing to do with "crossing a light barrier".

In order to understand it, stick with nearly the speed of light as I suggested, and reflect on that. After you understand what goes on for that case, then you can extrapolate to the unattainable limit for a clock with v=c, as Einstein did.

Harald

 

[DonB]

Right -except that you meant "reaching the light barrier" (which cannot be reached...)

No, actually I meant crossing the light barrier, but I made a shift (ref. to light that travels at c) that wasn't clear. (See my original posting of this statement.)

I think that what you mean is wrong - and others already pointed this out to you.

Yes, and that is what I said ("But others have shown me that that is not exactly what happens.")

The frozen clock that the photon uses (although that idea is nonsense) to observe the Earth...

Or non-clock, I think some would say.

Nothing to do with "crossing a light barrier".

Ah, but it does, within the discussion that I was pursuing. If we consider traveling in general, as we amp up the speed, we get closer and closer to c, and then as we near that peak we find that one thing (light -- the immediate context of my earlier post) has crossed that line -- has exceeded that barrier. That was my point.

 

[harrylin]

[..] If we consider traveling in general, as we amp up the speed, we get closer and closer to c, and then as we near that peak we find that one thing (light -- the immediate context of my earlier post) has crossed that line -- has exceeded that barrier. That was my point.

That point was either wrong or poorly formulated - and now it's clearer what that point was, so we're still making progress.

According to SR:
- light in vacuum always propagates at c; it doesn't "cross a line" or "exceed a barrier".
- Material objects (such as you) cannot reach the speed of light.

You could say that light always propagates at a speed that is the limit speed for material objects.

Also, you did not say that your "time" experience cannot be affected by a stuck clock that you don't use. You seemed to say the opposite...Harald

 

[rede96]

Since it will take ten minutes for the image of B to reach A and another ten minutes for the laser pulse to travel from A to the intersection, A needs to pull the trigger twenty minutes before it appears that B would arrive at the intersection, in other words when B is 33.333 miles before the intersection.

Thanks ghwellsjr.

So as Harold pointed out here:

In order to understand it, stick with nearly the speed of light as I suggested, and reflect on that. After you understand what goes on for that case, then you can extrapolate to the unattainable limit for a clock with v=c, as Einstein did.

Harald

I though I would try and do just that.

Imagine that instead of A firing a laser, A just decides to set off in his super fast ship and fires himself at B.

So ignoring acceleration for now, if A was capable of traveling at 0.9999986111095911 c, then A would feel that he covered the 10 light minutes in just 1 second.

A must also see B travel the 33.3333 miles B was away from the intersection in just one second too, as A will still hit B, but will just be a little off target.

However B would still feel like it took him 10 minutes.

So if we imagine just for a moment what a photon might experience from A's laser in the original experiment, once the photon was created, it would be instantaneously at B's target where it would be absorbed.

B, which was 33.333 miles from the intersection would also have to be instantaneously at the intersection for the photon to hit the target.

Although A and B still feel the passing of time as normal, for the photon, as everything happens instantaneously, there is nothing for the photon to experience. Which I guess is why we can't look at this problem from the frame of a photon. There is nothing to look at!


I know that the above is not strictly in keeping with the principles of SR, but I thought it might help address DrDon's questions.