Quick Check for Relativity HW, rocket passing by space station

[PhDeezNutz]

Homework Statement

Part (C), The book contends $\gamma$ should be in the denominator where as I think it should be in the numerator$$

Relevant Equations

The Lorentz Transform (see below)

 

[TSny]

Check this.

 

[PhDeezNutz]

Check this.

Alright, I think I got it and it doesn't make the least bit of sense but here we go. I want to practice using Lorentz Transforms instead of time dilation and length contraction formulas.

Event 1: is when $A'$ crosses $A$. The coordinates for these in the station frame and ship frame are $S_1 = \left( 0,0\right)$ and $S'_1 = \left(0,0\right)$ respectively.

Event 2: is when $B'$ crosses $A$. The coordinates for these in the station frame and ship frame are $S_2 = \left( 0,t_2\right)$ and $S'_2 = \left(-\gamma v t_2,\gamma t_2\right)$ respectively where $S'_2$ is found by Lorentz Transforming $S_2$ according to

$x' = \gamma x - \gamma v t$

$t' = - \frac{\gamma v}{c^2} + \gamma t$

But $\gamma t_2 = \frac{l_0}{v}$. So, therefore $t_2 = \frac{l_0}{\gamma v}$.

I actually think this is wrong since I made the same mistake that you warned me about but I somehow got the right answer. I can't physically reconcile the last step, all I know is that it gets me the right answer.

 

[TSny]

It should be easy to write down the space and time coordinates for the second event in the primed frame. They are easily expressed in terms of $l_0$ and $v$. In your first post, you have the correct expression for the time coordinate of this event in the primed frame. But, you mistakenly wrote the space coordinate as $0$. What is the space coordinate of the tail of the ship in the primed frame (at any time)?

 

[PhDeezNutz]

It should be easy to write down the space and time coordinates for the second event in the primed frame. They are easily expressed in terms of $l_0$ and $v$. In your first post, you have the correct expression for the time coordinate of this event in the primed frame. But, you mistakenly wrote the space coordinate as $0$. What is the space coordinate of the tail of the ship in the primed frame (at any time)?

$- l_0$? I say minus because in my mind it's to the left of the origin of $S'$.

 

[TSny]

$- l_0$? I say minus because in my mind it's to the left of the origin of $S'$.

Yes.

 

[PhDeezNutz]

Yes.

I'm afraid I don't follow when I take the event $S'_{B'\:crosses\:A} = \left( -l_0, \frac{l_0}{v}\right)$ and transform it to $S$ I get $S_{B'\:crosses\:A} = \left( -\gamma l_0 - \gamma l_0, -\frac{\gamma v}{c^2}\left( \frac{l_0}{v}\right) + \frac{\gamma l_0}{v}\right) = \left( 0, \frac{\gamma v}{c^2}\left( \frac{l_0}{v}\right) + \frac{\gamma l_0}{v} \right)$, but the time coordinate of this event in $S$ isn't $\frac{l_0}{\gamma v}$ (The answer that the book gets).

I'm going to continue to work on this problem throughout the day. I need to think more about what is actually happening instead of just plugging and chugging imo anyway.

 

[PhDeezNutz]

Alright I think I got it by really slowing down to think about it. There's still one part I haven't reconciled with intuition. Namely $t' = \frac{l_0}{v}$ when $B'$ crosses $A$ because it almost looks classical; $v$ is the speed of $B'$ in $S$ and $l_0$ is proper length. Don't quite get it. I'll think about it some more.Now for the solution handled as generally as possible.

Start with equations.

(x′t′)=(γ−γv−γvc2γ)(xt)(x′t′)=(γ−γv−γvc2γ)(xt)

(xt)=(γγvγvc2γ)(x′t′)(xt)=(γγvγvc2γ)(x′t′)​

1) Observer at $A'$ is adamant that the position of $B'$ is $-l_0$ so we have $S'_{B'} = \left( - l_0, t' \right)$

2) Observer at $A$ is adamant that the velocity of $B'$ is $(+v)$ but its position isn't $-l_0$. Let's call the position of $B'$ in $S$; $l_s$ so it has coordinates $S_{B'} = \left( -l_s + vt, t \right)$.

(ls+vtt)=(γγvγvc2γ)(−l0t)(ls+vtt)=(γγvγvc2γ)(−l0t)​

$B'$ crosses $A$ when $-l_s + vt = 0 \Rightarrow t = \frac{l_s}{v}$

(0lsv)=(γγvγvc2γ)(−l0t)(0lsv)=(γγvγvc2γ)(−l0t)​

We need to solve for $l_s$

−γl0+γvt′=0⇒t′=l0v−γl0+γvt′=0⇒t′=l0v​

The above part is kind of surprising for aforementioned reasons

lsv=−γvc2l0+γl0vlsv=−γvc2l0+γl0v

lsv=γl0(−vc2+1v)lsv=γl0(−vc2+1v)

ls=γl0(−vc2+1)ls=γl0(−vc2+1)​

$$l_s = \gamma l_0 \left( 1 - \frac{v^2}{c^2} \right) = \gamma \left( \frac{1}{\gamma^2 \right) l_0 = \frac{l_0}{\gamma}$$

As mentioned earlier $t = \frac{ls}{v}$ so we have

t2=l0γvt2=l0γv​

Well apparently my LaTeX got messed up.

 

[PhDeezNutz]

 

[TSny]

There's still one part I haven't reconciled with intuition. Namely $t' = \frac{l_0}{v}$ when $B'$ crosses $A$ because it almost looks classical; $v$ is the speed of $B'$ in $S$ and $l_0$ is proper length. Don't quite get it.

From S' viewpoint, the point A (in S) moves to the left with speed v and the distance A must travel to get from the front to the back of the ship is the length of the ship in S'.