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Having trouble understanding SR

So say you have an observer in a reference frame S and he observes two light beams moving side by side away from him at the speed of light. Then lets say we move to the reference frame of one of the beams of light and view the other beam of light, presumably it sees that beam of light moving at the speed of light.
Now from my (limited) understanding of Relativity i know that if one entity sees a second entity moving away from it at a velocity u, then the second entity will see the first moving in the opposite direction with a velocity -u.
So my question is which direction will the light beam see the other light beam moving if they were moving side by side?
(edited 11 years ago)
Reply 1
If you somehow manage to travel at the speed of light there would be no perception of motion because time stands still, at the same time all distance is also 0.

*I'm not completely familiar with the mathematics, this is inferred from the fact that as you travel faster (this is of course, relative to another inertial frame), time becomes increasingly dilated while your measurements of distance would decrease.
Reply 2
Original post by XiaoXiao1
If you somehow manage to travel at the speed of light there would be no perception of motion because time stands still, at the same time all distance is also 0.

*I'm not completely familiar with the mathematics, this is inferred from the fact that as you travel faster (this is of course, relative to another inertial frame), time becomes increasingly dilated while your measurements of distance would decrease.


I thought the speed of light and all other laws of physics were the same in any inertial reference frames, a frame moving at the speed of light is surely still inertial. So i thought a beam of light still sees another beam of light moving at the speed of light, after all it doesnt know its moving, because there is no absolute space or time - thats the principle of relativity.
speed = distance/time
I thought to maintain a constant speed of light time dilated (as you say) so this measurement gets smaller and the distance actually measured gets larger, because your measure of length contracts, your rulers contract and therefore you measure the spatial distance as being larger. I could be wrong though. Overall the equation linking relative speeds is v` = u +v / 1+ (uv/c^2). So if you let u and v = c as is the case here then v` is still c. What im struggling with is in what direction?
Reply 3
Original post by Einy
I thought the speed of light and all other laws of physics were the same in any inertial reference frames, a frame moving at the speed of light is surely still inertial. So i thought a beam of light still sees another beam of light moving at the speed of light, after all it doesnt know its moving, because there is no absolute space or time - thats the principle of relativity.
speed = distance/time



In addition to it being unattainable, there must be something "special"/contradictory about a reference frame of a photon. If you move at c, then from the principle of relativity you would be travelling at the speed of light relative to any observer, you can thus see that the lorentz transforms for time and distance break down as
Unparseable latex formula:

1 - \frac{v^}{c^2}

becomes 0. You would not be able to do any experiments to verify this as time comes to a halt.

I thought to maintain a constant speed of light time dilated (as you say) so this measurement gets smaller and the distance actually measured gets larger, because your measure of length contracts, your rulers contract and therefore you measure the spatial distance as being larger. I could be wrong though. Overall the equation linking relative speeds is v` = u +v / 1+ (uv/c^2). So if you let u and v = c as is the case here then v` is still c. What im struggling with is in what direction?


The ruler contraction is how I think about it too, the equation V' = u + v/(1+(uv/c^2) is how you "add" two relativistic speeds, and shows you that even if someone shines a beam of light inside a flying spaceship, you would still measure the speed of the beam as c. However in the situation you describe, something moving at the speed of light would not be able to measure any speed because theories point towards distance and time become somewhat meaningless.

If you let both u and v = c, then the situation would be somewhat analogous to a light beam being emitted inside something moving at the speed of light, this isn't physically possible.

Feel free to tear down what I said; I haven't done relativity rigorously.
Reply 4
Original post by XiaoXiao1
In addition to it being unattainable, there must be something "special"/contradictory about a reference frame of a photon. If you move at c, then from the principle of relativity you would be travelling at the speed of light relative to any observer, you can thus see that the lorentz transforms for time and distance break down as
Unparseable latex formula:

1 - \frac{v^}{c^2}

becomes 0. You would not be able to do any experiments to verify this as time comes to a halt.

To the best of our knowledge i think the speed of light is attainable for photons only because they are massless and not because there is something special about their frame. From my understanding of SR time is flowing perfectly normally for the photon as relative to it its at rest and the rest of the universe is moving in the opposite direction at the speed of light and thus its the rest of the universe whos time has come to a hault.
Yes the lorentz transforms do break down when u equals c. However because you divide one by the other to get speed - the gamma function cancels and obviously to make it mathematically correct youd probably have to stick a limit in there, but i think generally it is accepted that the gammas cancel even when they equal 1/0.


Original post by XiaoXiao1

The ruler contraction is how I think about it too, the equation V' = u + v/(1+(uv/c^2) is how you "add" two relativistic speeds, and shows you that even if someone shines a beam of light inside a flying spaceship, you would still measure the speed of the beam as c. However in the situation you describe, something moving at the speed of light would not be able to measure any speed because theories point towards distance and time become somewhat meaningless.

As I said above I think the space and time co-ordinates of a light beam are only bizarre relative to a frame observing it, but the symmetry of sr means that relative to the photon its at rest and the rest of the universe is moving at the speed of light and suffers time dilation and length contraction relative to it. After all there is no absolute motion and therefore no absolute time dilations or length contractions.

Original post by XiaoXiao1

If you let both u and v = c, then the situation would be somewhat analogous to a light beam being emitted inside something moving at the speed of light, this isn't physically possible.

Agreed that wouldnt be physically possible unless the something had a rest mass of 0, which it does in the case i described. :smile:

Original post by XiaoXiao1
Feel free to tear down what I said; I haven't done relativity rigorously.

Feel free to reciprocate - im enjoying the debate :smile:
(edited 11 years ago)
The proper time on null like world lines is 0 thus you cannot define what it means for something to be inertial "in the frame of reference of the photon".
This isn't so strange if you consider the newtonian analogue where light is propagated instantly. What would it mean to travel infinitely quickly?
(edited 11 years ago)
Reply 6
To the best of our knowledge i think the speed of light is attainable for photons only because they are massless and not because there is something special about their frame. From my understanding of SR time is flowing perfectly normally for the photon as relative to it its at rest and the rest of the universe is moving in the opposite direction at the speed of light and thus its the rest of the universe whos time has come to a hault.
Yes the lorentz transforms do break down when u equals c. However because you divide one by the other to get speed - the gamma function cancels and obviously to make it mathematically correct youd probably have to stick a limit in there, but i think generally it is accepted that the gammas cancel even when they equal 1/0.


If you work from the equation E2=p2c2+m02c4E^2 = p^2c^2 + m_0^2c^4 , then any object with a rest mass above 0 would have infinite energy relative to any observer, this is impossible and shows that there is indeed something special about the speed of light.

As I said above I think the space and time co-ordinates of a light beam are only bizarre relative to a frame observing it, but the symmetry of sr means that relative to the photon its at rest and the rest of the universe is moving at the speed of light and suffers time dilation and length contraction relative to it. After all there is no absolute motion and therefore no absolute time dilations or length contractions.



This is how I'm approaching time dilation: You're moving at speed u relative to system S', for every TT that passes in your frame, a time Tγ T\gamma passes in S', at light speed, it would take infinite time to pass in S' for any progression in time in your frame.

Similarly, a distance LL measured by you would measure LγL\gamma in S', so any object with non-zero length would correspond to something infinite in s hence time "coming to a stop" and distances reducing to 0: the issue is not dividing distance by time, but because detecting speed requires a change in these quantities, which cannot occur when you're travelling at the the speed of light.
(edited 11 years ago)
Reply 7
Original post by XiaoXiao1
If you work from the equation E2=p2c2+m02c4E^2 = p^2c^2 + m_0^2c^4 , then any object with a rest mass above 0 would have infinite energy relative to any observer, this is impossible and shows that there is indeed something special about the speed of light.




This is how I'm approaching time dilation: You're moving at speed u relative to system S', for every TT that passes in your frame, a time Tγ T\gamma passes in S', at light speed, it would take infinite time to pass in S' for any progression in time in your frame.

Similarly, a distance LL measured by you would measure LγL\gamma in S', so any object with non-zero length would correspond to something infinite in s hence time "coming to a stop" and distances reducing to 0: the issue is not dividing distance by time, but because detecting speed requires a change in these quantities, which cannot occur when you're travelling at the the speed of light.


But didnt einstein himself believe that if you looked in a mirror whilst travelling at the speed of light you would still see your reflection? How is this consistant with your explanations?
Original post by Einy
But didnt einstein himself believe that if you looked in a mirror whilst travelling arbitrarily close to the speed of light you would still see your reflection? How is this consistant with your explanations?


fixed

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