Relativistic rod moving in two directions.

[center o bass]

Homework Statement

A rigid rod has rest length L0 and moves relative to a system S' for which it's coordinates
is x'=0, y' = b-ut', z' = 0. In this reference frame the rod is at all times parallel with the x'-axis.
a) A point (A) on the rod is measured to be a distance= a, away from x' = 0 in S'. What are
the time dependent coordinates of this point in the same reference frame?
b) The inertial frame S moves with velocity v along the x-axis of another inertial frame S. (The axes of the two frames are parallel.) Find the time dependent coordinates (x, y, z) of the point A in this reference frame.
c) What is the orientation of the rod relative to the coordinate axes of S, and what is the length of the rod measured in this frame?

Homework Equations

Lortenz transformations

The Attempt at a Solution

a) In the S' frame the coordinates of this point should obviously be

x'= a, y' = b-ut', z'=0.
b)
Since the x-axis of S' moves with velocity v along the x-axis of S the x coordinate is given by the lorentz transformation as
x = G(a + vt'),
where G = G(v) is the lortenz factor.

I am not completely sure about the other coordinates, but I think that since there are no relative motion between the two frames in the y-directions

y = y' = b- ut',

but the time coordinates still disagree by
t' = G(t -(v/c)a),

and z = z' = 0.
c) Intuitively it would make sense that the rod has the same orientation in both of the frames. Since the y coordinates of both the point (A) and the middle of the rod should agree this means that the rod is also parallel in the frame S. However the rod is length contracted in the x-direction as measured in S, by L0/G.


- Are these arguments correct? What I worry about is the additional motion of the rod along
the y-axis. I first thought about introducing a third system moving along with the rod in that direction, but that would introduce an additional time coordinate. So is the argument that since there is no relative motion between S and S' in the y-direction their respective y-coordinates will agree?

 

[tiny-tim]

hi center o bass!

y = y'

and z = z'
…
So is the argument that since there is no relative motion between S and S' in the y-direction their respective y-coordinates will agree?

yes, y = y' and z = z' …

don't start imagining that relativity is even more difficult than it is! ​

 

[center o bass]

Hehe, okay! Thanks!

Follow up question:

When there is a relative velocity in two of the directions, how is the dime dilation formula
modified? Can one use

t' = G(t' - (v/c) x) where v² = vx² + vy² or would one then also have to rotate the coordinate system such that the relative motion is again along the x-axis?

 

[tiny-tim]

hi center o bass!

When there is a relative velocity in two of the directions, how is the dime dilation formula modified?

making your money last longer?

Can one use

t' = G(t' - (v/c) x) where v² = vx² + vy² or would one then also have to rotate the coordinate system such that the relative motion is again along the x-axis?

i'm not sure i understand the set-up …

if it's only time dilation, why would the direction matter?

but in most cases, yes, it's best to rotate the coordinate system ​

 

[center o bass]

If there were only relative motion along the x-axis we would just have
t' = G(t - (v/c) x).

But suppose no that the motion were directed at some angle u relative to the x-axis. So that
v = v(cos(u)i + sin(u)j). I guess my question is weather one would have to use something like

t' = G(t - (v/c)sqrt(x² + y²))

where s= sqrt(x² + y²) is the distance along the direction of motion.

 

[tiny-tim]

But suppose no that the motion were directed at some angle u relative to the x-axis. So that
v = v(cos(u)i + sin(u)j). I guess my question is weather one would have to use something like

t' = G(t - (v/c)sqrt(x² + y²))

where s= sqrt(x² + y²) is the distance along the direction of motion.

yes, except it would be x/√(x² + y²)

(but changing the coordinates is probably safer)