# Symmetry Argument

1. Oct 14, 2009

### Gear300

This is pretty much like the twin paradox question: if person A and person B were moving with a velocity v relative to each other and away from each other, then person B would observe a time dilation in person A's reference frame while person A would observe a time dilation in person B's reference frame...which seems contradictory. What would be the resolution to this apparent contradiction?

Last edited: Oct 14, 2009
2. Oct 14, 2009

### Staff: Mentor

Where do you see a contradiction? (To compare measurements between frames, one must consider length contraction and clock synchronization as well as time dilation.)

3. Oct 14, 2009

### JesseM

4. Oct 14, 2009

### Gear300

Hmmm...it took me a while to get what you're saying...but to make sure of things...what you're saying is that because the notion of space-time is not universal as assumed in the equations, my assumption of being able to connect the two observations as if they were at the same time or in the same reference frame was not valid?

5. Oct 14, 2009

### JesseM

I don't know what you mean by "the notion of space-time is not universal". It's the notion of simultaneity that's not universal--two events which are assigned the same t-coordinate in one frame can be assigned different t-coordinates in another. In Newtonian physics it's already true that two events could have the same x-coordinate (or same y or z) in one coordinate system but different x-coordinates in another; relativity extends this to time coordinates as well.

In Newtonian physics, if you know the coordinates x,y,z,t of an event in one inertial frame, and you want to know the coordinates x',y',z',t' of the same event from the perspective of a different inertial frame whose origin is moving at velocity v along the x-axis of the first frame, then you'd use the Galilei transformation:

x' = x - vt
y' = y
z' = z
t' = t

But in relativity, you'd instead use the Lorentz transformation:

x' = gamma*(x - vt)
y' = y
z' = z
t' = gamma*(t - vx/c^2)

where gamma = 1/sqrt(1 - v^2/c^2). You can see that with the Galilei transformation the x coordinate of the event in the first frame can be different than the x' coordinate of the event in the second frame, but the t coordinate is always equal to the t' coordinate. On the other hand, with the Lorentz transformation, the t coordinate of an event in frame #1 can be different from the t' coordinate of the event in frame #2.

6. Oct 14, 2009

### Gear300

So then what you're saying is that it only seems like a contradiction because we're used to thinking of these sorts of things as contradictions (and it would be a contradiction in a Newtonian system but not in a Relativistic system)?

7. Oct 14, 2009

### JesseM

Yes, we're used to thinking there should be some truth about who is "really" older, but if simultaneity is relative than the answer will depend on your choice of reference frame. Keep in mind, though, that disagreements about simultaneity only happen for events at different locations in space, if the two of them got back together to compare ages at a single location, every frame would agree about who was older (but in order for them to get back together one of them would have to accelerate to turn around, so then you have a typical twin paradox type situation where the one that accelerates is always going to be the one who has aged less when they reunite.

8. Oct 14, 2009

### Gear300

So then lets say that both person A and B were holding mirrors when they set their time to 0s, and that they held the mirrors so that one person could see his/her clock in the other person's mirror. If person B's mirror held the reflection of person A's clock, and if we had to choose between whether the reflection showed the time dilation of A or followed the same time intervals as B's clock, then would we choose the latter because the reflection is occurring right next to person B?

9. Oct 14, 2009

### JesseM

If they're right next to each other then the delay between the light leaving A's clock and being reflected off the mirror and reaching B's eyes will be negligible, so the delay between B seeing his own clock reading 0 and seeing A's clock reading 0 will be negligible. But time dilation is about the rate each clock ticks over time, not about the time they show at a particular instant (like the instant when they are right next to each other)--if they watch the other guy's clock for any extended length of time, whether by looking in a mirror or by looking directly at the other guy's clock (I don't actually understand the point of the mirror in this thought-experiment), they will see it running slower than their own, due to a combination of time dilation and the Doppler effect (remember my point that what you 'observe' in your rest frame is different from what you see visually--the Doppler effect doesn't have anything to do with how fast a clock ticks in your frame, but it does effect how fast you see it ticking visually when you're watching it with your eyes).

10. Oct 14, 2009

### Gear300

The reason I included the mirrors was because I was assuming the situation would be different if person B could view the behavior of person A's clock from a different point in space (if the source was a mirror stationary with respect to B). I was also thinking that person B could view his/her own clock from a moving source (person A's mirror). So what you're saying is that it doesn't matter whether the mirrors are there or not?

11. Oct 14, 2009

### JesseM

If you have mirrors at different points in space far away from A or B that would be a different issue because the light from a given event would take some noticeable extra time to reach the mirror and then return to the eyes of A or B. But you said the mirrors were just being held by A and B, which makes them seem totally pointless--the light from a given tick of A's clock will reach B's mirror at almost the exact same time it reaches B's eyes since the mirror is right next to B's eyes (and the time for the light to bounce off the mirror into B's eyes is negligible since they're right next to each other), so why not just ask what B sees when he looks directly at A's clock, instead of asking what he sees when he looks in his own mirror at the reflection of A's clock? Unless you're asking what A will see when he looks at his own clock in B's mirror or something like that.

12. Oct 14, 2009

### Gear300

That was actually the question I was originally intending to ask when I started this mirror question. Would A see a time dilation in his/her clock when looking at B's mirror?

13. Oct 14, 2009

### JesseM

He will see his clock slowed down, but this will be purely due to the signal delays and not due to time dilation--he would see his own clock slowed down by the same amount in a Newtonian universe too (the signal delays have to do with the fact that the light from each successive tick of his clock has a greater distance to travel to reach the mirror and return to him than the light from the previous tick, since the mirror is constantly moving away from him, so there'll be a greater time between reflected ticks than there would be if he was watching his own clock in a mirror which was stationary relative to himself). On the other hand, if he watches B's own clock which is being held next to the mirror that holds the reflection of his clock, he'll see it running even slower than the reflected image of his clock, due to time dilation--this would not be the case in a Newtonian universe.

14. Oct 14, 2009

### Gear300

I think I'm starting to get the bigger picture (starting, of course). Thanks for the replies.

One last question (don't worry...this is the last...hopefully). Let us say that we were both familiar with a particular movie and its run-time. Let us also say that there is a direct wireless network between my computer and your computer and that one of us hacked into the other's computer (we'll say that you hacked into my computer...whatever hits your preference). The frame I'm in is actually moving at 0.6c relative to your frame (but we're not observing this because we are looking at our computers). I'm actually watching a movie on my computer (the movie we're both familiar with), and because you've hacked into my computer, you're also watching it. Taking into account that my frame is moving at 0.6c relative to your frame, would you notice a slower run-time of the movie?

Last edited: Oct 14, 2009
15. Oct 14, 2009

### JesseM

Whoever's computer is running the movie will see it playing at normal speed, while the one who hacked into the other one's computer will see it playing at an altered rate, due to a combination of signal delays due to motion (i.e. the Doppler effect) and genuine time dilation.

16. Oct 14, 2009

### Gear300

Alright then (I was just checking the objectivity of relativistic phenomenon...in this case, if the two didn't observe the relative velocity; I was suspecting the time dilation but forgot to - or simply didn't - take into account the Doppler effect). Thanks again for all the replies.

17. Oct 15, 2009

### Rasalhague

Wouldn't B's clock always show a later time than A's reflected clock, albeit not later by as much as in a Newtonian universe?

The time shown by the reflected image of A's clock is the time shown by A's clock at the instant when the light left A's clock. Call the event of this light being emitted from A's clock A1. Call the event of A receiving the reflected image of A's clock showing this time A3.

Unless I'm mistaken, the time observed on B's clock should be equal to (1-(u/c)2)1/2 times the time shown by A's clock at an event A2 half way along A's worldline between events A1 and A3. (Half way because light travels at a constant speed, so its journey to the mirror will take as long as its journey back to A.) This time value will be less than that shown by A's clock at event A2 because, although in A's rest frame A2 is simultaneous with the event of the image being reflected from B's mirror, this event of reflection, call it B2, will happen before A2 in B's rest frame. This is the slowing down related to "time dilation".

The reason why I don't think A could ever observe B's clock showing an earlier time than the reflected image of A's own clock is because, for this to happen, B2 would have to be simultaneous in B's rest frame with an event on A's worldline separated from the mutal origin by a smaller spacetime interval than A1 is separated from the origin. But this couldn't happen unless perhaps B was traveling faster than light...

18. Oct 15, 2009

### JesseM

Hmm, when I actually do the calculations below it seems you're right. But since both B's clock and the reflected image of A's clock read 0 when B was right next to A, this must mean that A sees B's clock continually ticking faster than the reflected image of his own clock, not slower as I said. But A will definitely see B's clock ticking slower than his own (non-reflected) clock by the amount predicted by the relativistic Doppler effect, so he must see the reflected image of his own clock slowed down (relative to his own non-reflected clock) by an amount even greater than would be given by either the relativistic Doppler effect or the classical Doppler effect. When I think about it, this makes sense though--just thinking in purely Newtonian terms, it makes sense that if you are watching your own clock in a mirror which is moving away from you, it should be slowed down by a factor greater than the ordinary classical Doppler effect, since light from successive ticks is not being emitted by the mirror at one tick per second, as there is also a Doppler-type delay for the light from these ticks reaching the mirror as a consequence of the fact that the mirror is continually getting farther and farther from your clock and so the light from each tick has farther to travel to reach it. So I wasn't wrong when I said that the amount that A sees his own reflected image slowed down would be the same in relativity as it is in Newtonian physics, I was just wrong when I imagined that the rate it would be slowed down in Newtonian physics would be given by the classical Doppler equation.
Yes, that's correct. When A's own clock reads A3, he will be seeing the reflected image of his own clock reading A1, and he will be seeing B's clock reading A2*sqrt(1 - v2/c2), where A2 = (A3 - A1)/2. So if B is moving away from A at constant velocity v, then at time t=A1, B will be at a distance of v*A1 (in A's frame). Then since B continues to move away at velocity v while the light from A at A1 moves toward him at velocity c, the time needed for the light to catch up with B will be (v*A1)/(c - v). That will be the time interval between A1 and A2, which means the actual time of A2 will be given by A1 + (v*A1)/(c - v) = (c*A1 - v*A1)/(c - v) + (v*A1)/(c - v) = c*A1/(c - v). And since we know B's clock reads A2*sqrt(1 - v2/c2) at the moment the light from A1 reaches it, that means B's clock reads A1*c*sqrt(1 - v2/c2)/(c - v) = A1*sqrt(c2 - 2)/(c - v) = A1 * sqrt((c + v)*(c - v))/(c - v) = A1 * sqrt(c + v)/sqrt(c - v). This number will always be greater than A1, so A will always see B's time as ahead of the time on the reflected image on his own clock.

Just to pick a numerical example, if B is moving away from A at 0.6c, and A1 is the event of A's clock reading 10 seconds, then at this moment B is 6 light-seconds away, so it takes another 15 seconds for the light to catch up with B, at which time B's clock reads 25*sqrt(1 - 0.62) = 25*0.8 = 20 seconds, which is in fact equal to 10*sqrt(c + v)/sqrt(c - v) = 10*sqrt(1.6)/sqrt(0.4) = 20. So in this case B appears to be ticking twice as fast as the reflected image of A.

Last edited: Oct 15, 2009
19. Oct 15, 2009

### Rasalhague

I suppose the closer B's speed (in A's rest frame) to c, the smaller the difference between the time A sees on clock B and the time A sees on the reflected clock A, and the slower they'll both be compared to A's clock as A views it directly, approaching the limit at c of infinite slowness, i.e. not ticking at all.

20. Oct 15, 2009

### JesseM

It's true that both clocks will approach infinite slowness in their visual ticking rate in the limit as B's velocity in A's frame approaches c, but it's not right that the difference in times between B and the reflected image of A gets smaller as you approach c...according to the formula I found, when A sees the reflected image of his own clock showing time A1, then at the same moment the image of B's clock will show time A1*sqrt(c + v)/sqrt(c - v), and that factor of sqrt(c + v)/sqrt(c - v) approaches infinity in the limit as v approaches c. So although the images of the two clocks are approaching infinite slowness, they are approaching it at different rates, and the ratio between their two rates of apparent ticking is getting larger and larger. It's in the opposite limit as v approaches 0 that they approach ticking at the same apparent rate, unless there's a mistake in my derivation.