Relativistic Kinematics: Distance of B When A Coincides with L

[Amrator]

Homework Statement

A and B both start at the origin and simultaneously head off in opposite directions, each with speed $\frac{3c}{5}$ with respect to the ground. A moves to the right and B moves to the left. Consider a mark on the ground at $x = L$. As viewed in the ground frame, A and B are a distance $2L$ apart when A passes this mark. As viewed by A, how far away is B when A coincides with the mark?

Relevant Equations

$$t_{observed} = \gamma t_{proper}$$
$$L_{observed} = \frac{L_{proper}} {\gamma}$$
$$t_R - t_L = \frac{Lv}{c^2}$$
$$V = \frac{u+v}{1+\frac{uv}{c^2}}$$

I don't know why I'm so puzzled by this problem; it's only one star. So first, I drew a picture of A and B in the ground frame. Then I drew B and the ground in A's frame. I then used the velocity addition formula to obtain the velocities of both B and the ground relative to A.
$$\frac{v_{B}^{A} + v_{gound}^{A}}{1 + \frac{v_{B}^{A} v_{gound}^{A}}{c^{2}}} = \frac{-\frac{3c}{5} - \frac{3c}{5}}{1 + \frac{9}{25}} = -\frac{15c}{17}$$
And I don't where to go from here.

 

[anuttarasammyak]

What is time reading of A's clock when A is on the mark ? Then multiply it with the velocity you calculated.

 

[Amrator]

What is time reading of A's clock when A is on the mark ? Then multiply it with the velocity you calculated.

I got $V_{B}^{A}t_{A}^{ground} = \frac{25L}{17}$. So wait, that's the answer? We don't have to take into account time dilation?

 

[PeroK]

I got $V_{B}^{A}t_{A}^{ground} = \frac{25L}{17}$. So wait, that's the answer? We don't have to take into account time dilation?

That can't be right. You must use time from A's frame.

 

[Amrator]

That can't be right. You must use time from A's frame.

Ah! Now I see! The mark $L$ is part of the ground frame, so in A's frame, the distance between the origin and $L$ will be contracted! So the time it takes for A to coincide with $L$ in A's frame is $$\Delta t_{A at L} = \frac{\frac{L}{\sqrt{1-\frac{9}{25}}}} {-\frac{3c}{5}}$$. But this gives me a negative time.

 

[PeroK]

Ah! Now I see! The mark $L$ is part of the ground frame, so in A's frame, the distance between the origin and $L$ will be contracted! So the time it takes for A to coincide with $L$ in A's frame is $$\Delta t_{A at L} = \frac{\frac{L}{\sqrt{1-\frac{9}{25}}}} {-\frac{3c}{5}}$$. But this gives me a negative time.

I don't understand that calculation.

Why not stick with your original plan to calculate $v_B\Delta t_A$, where $v_B$ is the speed of $B$ and $\Delta t_A$ is the elapsed time both in $A$'s frame?

 

[Amrator]

I don't understand that calculation.

Why not stick with your original plan to calculate $v_B\Delta t_A$, where $v_B$ is the speed of $B$ and $\Delta t_A$ is the elapsed time both in $A$'s frame?

That is what I'm going to do. I'm just trying to get $\Delta t_A$ first. I'm treating the ground in A's frame as almost like a train with proper length $L$. So then I can calculate the proper time of A $\Delta t_A$ when it coincides with the back of the "train" (i.e $L$). Is what I'm trying to do nonsense? Please let me know.

 

[PeroK]

That is what I'm going to do. I'm just trying to get $\Delta t_A$ first. I'm treating the ground in A's frame as almost like a train with proper length $L$. So then I can calculate the proper time of A $\Delta t_A$ when it coincides with the back of the "train" (i.e $L$). Is what I'm trying to do nonsense? Please let me know.

That's fine. In A's frame the origin and the point at $L$ (in the ground frame) are length contracted. Imagine a marker at the origin and another at $L$ in the ground frame. That marker is moving in $A's$ frame and takes $t_A$ to reach $A$.

 

[Amrator]

That's fine. In A's frame the origin and the point at $L$ (in the ground frame) are length contracted. Imagine a marker at the origin and another at $L$ in the ground frame. That marker is moving in $A's$ frame and takes $t_A$ to reach $A$.

Yup, that's exactly what I'm trying to do. But I'm still getting a negative time from the velocity of the markers in A's frame.

 

[PeroK]

Yup, that's exactly what I'm trying to do. But I'm still getting a negative time from the velocity of the markers in A's frame.

That's silly! You have a distance divided by a speed. There should be no negatives involved.

 

[Amrator]

I'm getting -25L/12c.

 

[PeroK]

I'm getting -25L/12c.

What would you do if this was a non-relativistic problem? Would you still get negative time?

 

[Amrator]

What would you do if this was a non-relativistic problem? Would you still get negative time?

No, because then I wouldn't be taking into account sign of the velocity. I would just be dividing by the speed. The markers are moving to the left at 3c/5, thus, -3c/5. I then divide $L/\lambda$ by the aforementioned velocity which gives a negative value.

 

[PeroK]

No, because then I wouldn't be taking into account sign of the velocity. I would just be dividing by the speed. The markers are moving to the left at 3c/5, thus, -3c/5. I then divide $L/\lambda$ by the aforementioned velocity which gives a negative value.

Why do you only take into account the sign of the velocity because it's SR? You're saying that if the speed is $10m/s$, then you use $10m/s$ and if it's $1km/s$ you use $1km/s$, but at a speed of some significant fraction of $c$, you switch to a negative speed?

At what speed do you switch from from positive to negative speeds?

 

[PeroK]

If you want to use velocity, which may be negative, you must use displacement, which can also be negative.

 

[Amrator]

If you want to use velocity, which may be negative, you must use displacement, which can also be negative.

Yes. Sorry, I confused the frames. $\Delta t_A$ is therefore $25L/12c$. $v_{B}^{A} \Delta t_{A} = (-\frac{15c}{17})(\frac{25L}{12c}) = 375L/204$.

 

[PeroK]

Yes. Sorry, I confused the frames. $\Delta t_A$ is therefore $25L/12c$. $v_{B}^{A} \Delta t_{A} = (-\frac{15c}{17})(\frac{25L}{12c}) = 375L/204$.

That doesn't look right.

What did you get for $\gamma$?

 

[Amrator]

That doesn't look right.

What did you get for $\gamma$?

I got 5/4.

 

[PeroK]

I got 4/5.

You mean $5/4$?

 

[Amrator]

You mean $5/4$?

Fixed it.

 

[PeroK]

Fixed it.

I'll show how I would approach this problem. One option, of course, is to apply the Lorentz Transformation. Alternatively (I'll use $v$ as a fraction of $c$, sometimes called $\beta$):

First, in A's frame at $t_A = 0$ the point at $L$ is a distance $L/\gamma$ away. And, moving towards $A$ at speed $v$. It reaches $A$ after time $\Delta t_A = \frac{L}{\gamma v}$.

Meanwhile, B has been moving away from $A$ at speed $v_B = \frac{2v}{1 + v^2}$. Hence is a distance:
$$d = v_B \Delta t_A = \frac{2L}{\gamma(1 + v^2)}$$
Now you can plug in whatever $v$ you've been given.

 

[Amrator]

I'll show how I would approach this problem. One option, of course, is to apply the Lorentz Transformation. Alternatively (I'll use $v$ as a fraction of $c$, sometimes called $\beta$):

First, in A's frame at $t_A = 0$ the point at $L$ is a distance $L/\gamma$ away. And, moving towards $A$ at speed $v$. It reaches $A$ after time $\Delta t_A = \frac{L}{\gamma v}$.

Meanwhile, B has been moving away from $A$ at speed $v_B = \frac{2v}{1 + v^2}$. Hence is a distance:
$$d = v_B \Delta t_A = \frac{2L}{\gamma(1 + v^2)}$$
Now you can plug in whatever $v$ you've been given.

Is that gamma associated with the speed of A, B, or ground, and in which frame? That's what I'm confused about right now.

 

[PeroK]

Is that gamma associated with the speed of A, B, or ground, and in which frame? That's what I'm confused about right now.

$\gamma$ is associated with speed $v$.

 

[Amrator]

I'm getting 125L/68. That doesn't seem right.

 

[PeroK]

I'm getting 125L/68. That doesn't seem right.

I agree it doesn't look right.

Let me show you one advantage of my method. (I'm very anti plug-and-chug for this reason). I got:
$$d = \frac{2L}{\gamma(1 + v^2)}$$
And I can sanity check this formula. If $v$ is non-relativistic, i.e. $v \approx 0$ and $\gamma \approx 1$, then we get $d = 2L$, which is the correct non-relativistic limit.

Also, as $v \rightarrow c$, we get $d \rightarrow 0$. Which, if you think about it makes sense.

That doesn't mean my formula is right. But, all you have are numbers that tell you nothing about the physics.

All you have is $125L/68$, which tells you nothing about the physics of the problem.

In any case, how did you get that number?

 

[Amrator]

I agree it doesn't look right.

Let me show you one advantage of my method. (I'm very anti plug-and-chug for this reason). I got:
$$d = \frac{2L}{\gamma(1 + v^2)}$$
And I can sanity check this formula. If $v$ is non-relativistic, i.e. $v \approx 0$ and $\gamma \approx 1$, then we get $d = 2L$, which is the correct non-relativistic limit.

Also, as $v \rightarrow c$, we get $d \rightarrow 0$. Which, if you think about it makes sense.

That doesn't mean my formula is right. But, all you have are numbers that tell you nothing about the physics.

All you have is $125L/68$, which tells you nothing about the physics of the problem.

In any case, how did you get that number?

I see.

I actually did the gamma wrong. Now I'm getting 4L/17. I did the following: $$v_{B}\Delta t_{A} = \frac{2L\sqrt{1-\frac{9}{25}}}{1+\frac{9}{25}}$$.

 

[PeroK]

I see.

I actually did the gamma wrong. Now I'm getting 4L/17. I did the following: $$v_{B}\Delta t_{A} = \frac{2L\sqrt{1-\frac{9}{25}}}{1+\frac{9}{25}}$$.

$$\frac{2L\sqrt{1-\frac{9}{25}}}{1+\frac{9}{25}} \ne \frac{4L}{17}$$.

 

[Amrator]

$$\frac{2L\sqrt{1-\frac{9}{25}}}{1+\frac{9}{25}} \ne \frac{4L}{17}$$.

Right, arithmetic mistake. Now, I'm getting 20L/17.

 

[Amrator]

Thank you for the help, @PeroK . I'm going to go back over this to make sure I understand the physics.