How does time expansion/contraction work for opposite moving observers?

[randyu]

If A travels at .99c in his negative coordinate direction then his length expands and time increases observed by B moving at v=.6, the opposite of normally discussed situations. How is this explained?

 

[ghwellsjr]

If A travels at .99c in his negative coordinate direction then his length expands and time increases observed by B moving at v=.6, the opposite of normally discussed situations. How is this explained?

In an Inertial Reference Frame in which A is traveling at -0.99c and B is traveling at 0.6c, gamma for A is just over 7 and gamma for B is 1.25. These are the factors that the time on their respective clocks is dilated and the reciprocal of those factors are their amount of length contraction along the direction of motion. That's as good an explanation as I can provide but it doesn't seem to coincide with what you are suggesting since it is not the opposite of normally discussed situations.

 

[randyu]

it doesn't seem to coincide with what you are suggesting since it is not the opposite of normally discussed situations.

Thanks George,

I was thinking something like standard setup, t'=x=x'=0 @ t=0
then let x=-1 @ t=1

t'=Y(t-vx/c2)

t' = 1.25 (1 - 0.6 (-1)/1²) = 2

therefore, Δt'=2
a doubling of time intervals

 

[ghwellsjr]

Thanks George,

I was thinking something like standard setup, t'=x=x'=0 @ t=0
then let x=-1 @ t=1

t'=Y(t-vx/c2)

t' = 1.25 (1 - 0.6 (-1)/1²) = 2

therefore, Δt'=2
a doubling of time intervals

Let me start by giving you a helpful hint: instead of using the letter "Y" to represent gamma, if you hit the "Go Advanced" button below the place where you compose your response, it will take you to a new page that has some "Quick Symbols" off to the right where you can click on "γ" to insert it into your response. Just make sure you unhighlight it or it will disappear. It's not a very good looking gamma but at least it has its own symbol.

Ok, now to move on to the main point: When you want to use the Lorentz Transform to show the Time Dilation of a moving clock in one Inertial Reference Frame (IRF), you have to pick two events that are at the same location in the IRF that is moving at the same speed as the clock was moving in the first frame. You can also use the origin as one of those events so that you only have to do one transform. So if you have a clock that is moving at -0.6c in one frame and you want to see what time it has on it when the coordinate time is 1, you would figure out where it was at that time. Since its speed is -0.6c, its location will be -0.6. You would transform the event x=-0.6 @ t=1 and get t'=0.8. (Don't forget to make the speed equal to -0.6c.)

Alternately, you could ask a different question: What is the time coordinate of the event in the original IRF when the Proper Time on the moving clock equals 1? Now we are working the problem backwards so we use a speed with the opposite sign (0.6c) and we transform x=0 @ t=1 which gives us t'=1.25.

Both of these answers are correct, the first one shows us the Proper Time on the moving clock in the original IRF when the coordinate time =1 and the second one shows us the coordinate time in the original IRF when the Proper Time on the moving clock equals 1.

 

[randyu]

Thanks George, humm... I'll need to think about those.

To clarify, in my setup frame x' is moving at +.6c, v=+.6c
the clock in x is moving to the left at -.99c

then @ t=1, x=-.99

Δt'= γ(Δt-vx/c2)

Δt' = 1.25 (1 - 0.6 (-.99)/1²) = 1.9925 ≈ 2

still wrong I guess but why?

 

[JesseM]

Thanks George, humm... I'll need to think about those.

To clarify, in my setup frame x' is moving at +.6c, v=+.6c
the clock in x is moving to the left at -.99c

then @ t=1, x=-.99

Δt'= γ(Δt-vx/c2)

Δt' = 1.25 (1 - 0.6 (-.99)/1²) = 1.9925 ≈ 2

still wrong I guess but why?

That's correct, but it's not dealing with the time dilation of a physical clock at rest in either the x,t frame or the x',t' frame, since Δx is not zero and neither is Δx'. The time dilation equation Δt' = γΔt is intended to answer the question "if you have a clock that elapses a time Δt between two events on its worldline (so Δt is both the clock's proper time and the coordinate time in the frame where the clock is at rest, meaning Δx=0 in that frame), then what is the time Δt' between those same two events in the frame where the clock is moving at speed v?"

Similarly, the length contraction equation L' = L/γ is meant to answer the question "if you have an object that has a length L in its rest frame (so that if you pick two events on either end of it, Δx for those events is L; "length" normally implies that you also measure the position of both ends simultaneously, but in this case the two ends are at rest so it doesn't matter), then what is the length L' in a frame where it's moving at v? (meaning that if the two events on either end are simultaneous in this frame, so Δt'=0, then Δx'=L/γ)."

If you meet these conditions, you find that Δt' = γΔt and L' = L/γ apply regardless of the direction of the objects. If you don't meet the conditions (i.e. you pick pair of events where there is a change in position in both frames and find the time between them in both frames, or you pick a pair of events which are non-simultaneous in both frames and find the distance between them in both frames) then you are not really calculating "time dilation" or "length contraction" for any physical clock or extended object.

 

[randyu]

thanks guys, I got it now