# RELATIVITY: Co-ordinates of events

PeterPeter

## Homework Statement

A meter ruler moves at velocity u to the right past a stationary observer. The observer is at (0,0) in his rest frame. Give the (x,t) co-ordinates of the following events.
1. The right end of the ruler passes the observer (in the observer's frame)
2. The left end of the ruler passes the observer (in the observer's frame)
3. The observer passes the right end of the ruler (in the ruler's frame)
4. The observer passes the left end of the ruler (in the ruler's frame)

## Homework Equations

t2 = $gamma$*(t2 dash + u * x2 dash / $c^2$)
x2 = $gamma$*(x2 dash + u * t2 dash)

Where the dashed co-ordinates are in the ruler's rest frame.

## The Attempt at a Solution

1. The right end of the ruler passes the observer (in the observer's frame) (0,0)
2. The left end of the ruler passes the observer (in the observer's frame) (0,t2)
3. The observer passes the right end of the ruler (in the ruler's frame) (0,0)
4. The observer passes the left end of the ruler (in the ruler's frame) (-1,t2 dash)

which when I use the above equations I get t2 dash = 1/u

Where have I gone wrong?

PS Not sure how you get subscripts and dashes in LaTex

Staff Emeritus
Homework Helper
Gold Member
It would be easier to tell you what goes wrong if you show us how you arrive at your answer.

Regarding the LaTeX part, I have fixed your equations for you:
$$t_2 = \gamma(u) \left(t'_2 + \frac{u x'_2}{c^2}\right)$$
$$x_2 = \gamma(u) (x'_2 + u t'_2)$$
if you want to know how it was done - quoting this message will show you the code.

Mentor

## Homework Statement

A meter ruler moves at velocity u to the right past a stationary observer. The observer is at (0,0) in his rest frame. Give the (x,t) co-ordinates of the following events.
1. The right end of the ruler passes the observer (in the observer's frame)
2. The left end of the ruler passes the observer (in the observer's frame)
3. The observer passes the right end of the ruler (in the ruler's frame)
4. The observer passes the left end of the ruler (in the ruler's frame)

## Homework Equations

t2 = $gamma$*(t2 dash + u * x2 dash / $c^2$)
x2 = $gamma$*(x2 dash + u * t2 dash)

Where the dashed co-ordinates are in the ruler's rest frame.

## The Attempt at a Solution

1. The right end of the ruler passes the observer (in the observer's frame) (0,0)
2. The left end of the ruler passes the observer (in the observer's frame) (0,t2)
3. The observer passes the right end of the ruler (in the ruler's frame) (0,0)
4. The observer passes the left end of the ruler (in the ruler's frame) (-1,t2 dash)

which when I use the above equations I get t2 dash = 1/u

Where have I gone wrong?

PS Not sure how you get subscripts and dashes in LaTex
Your results for parts 3&4 are correct. You can use the LT to get the results for parts 1 & 2 from the results of parts 3 & 4.

Chet

PeterPeter
Your results for parts 3&4 are correct. You can use the LT to get the results for parts 1 & 2 from the results of parts 3 & 4.

Chet

Thanks.

I was expecting to get something like length/gamma ie length contraction!

Staff Emeritus
Homework Helper
Gold Member
You will, if you do not remove t2' from your equations and instead use the relation t2' = 1/u to break out a factor t2' from your first equation.

I think it would be more instructive to give the ruler an arbitrary rest length L to be honest. That way units in your expression will make sense and you do not fall into the trap of removing t2'.

Mentor
Thanks.

I was expecting to get something like length/gamma ie length contraction!
You would only get that if both ends of the stick are observed at the same time in the unprimed frame.

Chet

Last edited:
PeterPeter
I have reworked using a ruler length of L instead of 1m. I get:

1. The right end of the ruler passes the observer (in the observer's frame) (0,0)
2. The left end of the ruler passes the observer (in the observer's frame) (0,L/(u*gamma))
3. The observer passes the right end of the ruler (in the ruler's frame) (0,0)
4. The observer passes the left end of the ruler (in the ruler's frame) (-L,L/u)

So the stationary observer would see that the ruler is moving with velocity u and that it takes L/(u*gamma) seconds to pass him.

So I guess he would conclude that the ruler is u*L/(u*gamma) = L/gamma metres long?

PS How does one get the greek letter gamma? I clicked the "Quick symbols" by nothing happened.

Staff Emeritus
Homework Helper
Gold Member
Yes, but finding L this way assumes that u is known or measured by the observer. The problem is more well suited to motivate time dilation.

There is a gamma in the quick symbols: γ (between β and δ). If you want the gamma in LaTeX mode: \gamma

You would only get that if both ends of the stick are observed at the same time in the unprimed frame.

As seen in OPs solution, this is not strictly necessary if the velocity u of the object is known. If an object with constant velocity u passes you in time t, you would probably deduce its length as ut and this would be equivalent to using the positions of the end points at the same time in your system.

• 1 person
Mentor
Yes, but finding L this way assumes that u is known or measured by the observer. The problem is more well suited to motivate time dilation.

There is a gamma in the quick symbols: γ (between β and δ). If you want the gamma in LaTeX mode: \gamma

As seen in OPs solution, this is not strictly necessary if the velocity u of the object is known. If an object with constant velocity u passes you in time t, you would probably deduce its length as ut and this would be equivalent to using the positions of the end points at the same time in your system.
Thanks. I guess I had a mental lapse, because i had known that previously. "Senior moment."

Chet

dauto
Unfortunately, the font used by the quick symbols is awfully inadequate. The symbol for gamma "γ" looks too much like an "y". You can always use the latex "##\gamma##" instead.

Staff Emeritus