# Position of the front of an asteroid using special relativity

## Homework Statement

I'm stuck on part (d) but I've included the previous subquestions in case they're useful.

The length of an asteroid is exactly 300 m = 1 µls (micro light-second) when at rest. Draw a carefully labelled space-time diagram to illustrate the following:
(a) Depict the rest frame F′ of the asteroid, when it is moving along the x-axis at speed $v = \sqrt{3/2}c$ in the diagram reference frame F. The two reference frames can be taken in standard configuration
(b) In the diagram of part (a), depict the asteroid in both reference frames, when the rear end of the asteroid lies at the origin. Find the length of the asteroid in F.
(c) For a given position of the back end, the front ends of the asteroid in F and F′ are different events. Find, in F, the spatial locations of these events (i.e. of the front ends of the asteroid in both frames) when the rear end of the asteroid lies at the origin (i.e. when t = t′ = 0).
(d) Depict the asteroid in both reference frames at a later time, when the rear end lies at x = 4 µls in frame F. Find the locations in F of the front ends of the asteroid in both frames. What is the location in F′ of the front end at this time?

## Homework Equations

Lorentz transformation: $x'=\gamma (x-vt)$ and $x=\gamma (x'+vt')$

## The Attempt at a Solution

I'm confused about the wording of the question: find the location of the front end of the asteroid in both frames in F, and finally in F'? I don't quite get it.

Anyway, $\gamma=2$ so I think the length in F should be (where L is the length at rest) $l=\frac{L}{\gamma} = 0.5\mu ls$. This implies that the position of the front end in F is at $x=4.5\mu ls$. I'm not sure how to do anything else, since I don't know the time in either frame.

Any help would be greatly appreciated!

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## Homework Statement

I'm stuck on part (d) but I've included the previous subquestions in case they're useful.

The length of an asteroid is exactly 300 m = 1 µls (micro light-second) when at rest. Draw a carefully labelled space-time diagram to illustrate the following:
(a) Depict the rest frame F′ of the asteroid, when it is moving along the x-axis at speed $v = \sqrt{3/2}c$ in the diagram reference frame F. The two reference frames can be taken in standard configuration
(b) In the diagram of part (a), depict the asteroid in both reference frames, when the rear end of the asteroid lies at the origin. Find the length of the asteroid in F.
(c) For a given position of the back end, the front ends of the asteroid in F and F′ are different events. Find, in F, the spatial locations of these events (i.e. of the front ends of the asteroid in both frames) when the rear end of the asteroid lies at the origin (i.e. when t = t′ = 0).
(d) Depict the asteroid in both reference frames at a later time, when the rear end lies at x = 4 µls in frame F. Find the locations in F of the front ends of the asteroid in both frames. What is the location in F′ of the front end at this time?

## Homework Equations

Lorentz transformation: $x'=\gamma (x-vt)$ and $x=\gamma (x'+vt')$

## The Attempt at a Solution

I'm confused about the wording of the question: find the location of the front end of the asteroid in both frames in F, and finally in F'? I don't quite get it.

Anyway, $\gamma=2$ so I think the length in F should be (where L is the length at rest) $l=\frac{L}{\gamma} = 0.5\mu ls$. This implies that the position of the front end in F is at $x=4.5\mu ls$. I'm not sure how to do anything else, since I don't know the time in either frame.

Any help would be greatly appreciated!
The problem statement gives you the distance traveled in F and the speed in F.

The problem statement gives you the distance traveled in F and the speed in F.
(Correction for original post: the speed is supposed to be $\frac{\sqrt{3}}{2}c$, which then gives $\gamma=2$.)
Thanks, I can't believe how long I spent without noticing that. So, the time in $F$ is: $$t=\frac{D}{v}=\frac{8}{3}\sqrt{3}\mu s$$ Therefore the time in $F'$ is: $$t'=\frac{t}{\gamma}=\frac{4}{3}\sqrt{3}\mu s$$
The front of the asteroid in $F'$ is at $x'=1\mu ls$ because that's the rest frame of the asteroid, and the front of the asteroid in $F$ is at $x_1=4.5\mu ls$ because the length in $F$ is $0.5\mu ls$. Lastly, the position in $F$ corresponding to the front end of the asteroid in $F'$ is: $$x_2=\gamma (x'+vt')=6 \mu ls$$
I hope I haven't made any massive mistakes? Thanks very much for your help :-)