salvestrom
- 226
- 0
John232 said:This is like saying that S'=S, but they don't. The whole reason of assigning the prime symbol is to indicate that it is a different frame of reference. Then any value in frame S' would follow to have a prime symbol and any value in S frame would not have a prime symbol. You would then only need to use one of the equations, because it gives the relation to the other frame. An equation should work the same forwards and backwards, if you have 1 sec for a value of time you will only get one answer, if you put that same answer into the other value you should get 1 sec for the first value. Neither frame has to be at rest, they can both be moving but regardless of that the prime frame should never exchange values to the unprimed frame. You can say that the unprimed frame is also in constant motion it doesn't have to read 1 sec for time, then you would only use the equation τ=t√(1-v^2/c^2), where τ is the proper time in the primed frame, then to convert to the unprimed frame you would only need to put in the same value for τ.
The primed frame can't assume values that is unprimed because then it would no longer be a valid frame. So then you can't assume that t<t', you would have to know that t>t' when you assigned the two frames.
t>t' and t'>t. This is the whole point of SR and Lorentz symmetry: t>t' for the unprimed observer and t'>t for the primed observer. Each sees the other's time as passing slower. The symmetry only disappears when clocks get reunited, and even then it depends on how they are brought back together.
The ' is for distinguishing frames. You may find it useful to consider that the frame which is assigned the ' is arbitrary and any given observer will usually, quite biasedly, assign themselves as the unprimed frame.
EDit: Also, using your statement "it's like saying S'=S", what I'm actually saying is: "the difference in time as seen by S' is the same as the difference in time seen by S".
http://img513.imageshack.us/img513/43/spacetimediagram066c.png
I apologise for the scruffyness of the image. The blue lines are constant time for t' and the black lines constant time for t. Note where these lines transect the x=0 line (thick black upright bar) and x'=0 line (thick green, leaning bar). Ignore the mirad red lines that I failed to edit out.
Last edited by a moderator: