Word Problem about Special Relativity

[philcsar]

hi. i am getting really confused about which should be the primed and unprimed values. id just like to check if what i did was right. thank you. :D

Problem:
A meterstick is glued to the wall with its 100cm end farther to the right in the positive direction. It has a clock at its center and one on each end. You walk by the meterstick in the possitive direction at speed .
(a) when you each the center clock, it reads 0. what do the other two read at this instant in your frame.
(b) you instantly reverse direction. The clock at the center is still reading 0 and so is yours. what do the others read?
(c) How does this relate to the twin paradox?

My solution:
(a) since the person is walking with velocity v, he composes the frame S'
since the meterstick is stationary, it composes the frame S

using Lorentz Transfromation:
t' = γ(-xv/c^2 +t)

x => x/2 since you reached the center clock
t = 0 since the reading is zero

thus, t' = -0.5γxv/c^2 (left clock) and it will be +0.5γxv/c^2 (right clock)


(b) right clock will be -0.5γxv/c^2 and left clock will be +0.5γxv/c^2 (by symmetry)

(c) i don't know

 

[Simon Bridge]

hi. i am getting really confused about which should be the primed and unprimed values.

Welcome to PF;
That is common - the secret is to concentrate on who is doing the observing and remember the physical result rather than the equations.

Generally the primed value is the one measured in the unprimed frame, of something that is moving in that frame.

Problem:
A meterstick is glued to the wall with its 100cm end farther to the right in the positive direction. It has a clock at its center and one on each end. You walk by the meterstick in the possitive direction at speed .
(a) when you each the center clock, it reads 0. what do the other two read at this instant in your frame.
(b) you instantly reverse direction. The clock at the center is still reading 0 and so is yours. what do the others read?
(c) How does this relate to the twin paradox?

My solution:
(a) since the person is walking with velocity v, he composes the frame S'
since the meterstick is stationary, it composes the frame S

There is no absolute motion - every observer is stationary in their own frame.
- you need to relate the "moving" and "stationary" objects to who's taking the measurements.

In this case "you" are making the observations.
Therefore "you" are stationary. The ruler is doing the moving.

This means that "you go at some speed v" means that the meter ruler is traveling backwards at some speed v.

The rest-length (or "proper" length) of the meter stick is the length measured in the W for "wall" reference frame. It's length in the Y for "you" reference frame is L=L₀/γ

... notice I am using a subscript-zero to refer to the rest-length?
There are only two frames, so I don't have to give the you-length a special subscript.
If you like, you can compare that to the primed/unprimed notation.

All the laws of physics take their normal form inside the same reference frame.
So when the 50cm mark on the ruler is next to you, and your clock reads 0s, then the 0cm mark of the ruler was next to you when your clock reads Δt=-L/2v (because the ruler traveled a distance L/2 at speed v). In terms of the ruler rest-length, that is Δt=-L₀/2γv

When you look at a clock, the time you see on it's display will be the time elapsed, in its rest frame, since it last read 0, minus the time it takes the light from the clock to reach you.