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Say two spaceships start at the same point and from the vantage point of your inertial reference frame $S$, Spaceship A travels at $.75c$ and Spaceship B travels at $.25c$, travelling in the same direction. The goal is to compute the velocity of spaceship A relative to spaceship B.

In the reference frame $S$, the distance $x$ between the spaceships after a particular time $t$ is given as follows.

$$x=0.75ct-0.25ct=0.5ct$$

Relative to the reference frame of spaceship $B$, there is a Lorentz factor equal to the following value.

$$\gamma = \frac{1}{\sqrt{1-0.25^2}} = \sqrt{\frac{16}{15}}$$

Thus, this same time interval $t$ is observed in reference frame $B$ as $t_B = t\gamma$ and the distance interval between the spaceships is $x_B=x/\gamma$. Thus, the velocity of spaceship $A$ in the reference frame of spaceship $B$ should be

$$v = \frac{x_B}{t_B} = \frac{x/\gamma}{t\gamma} = 0.5 \cdot \frac{16}{15} \cdot c \approx 0.46875 c.$$

However, this value is LESS than the difference in velocities of the spaceships as observed in $S$, when it should instead be more. What is incorrect about this reasoning?

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2 Answers 2

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As is often the case, the "culprit" is the relativity of simultaneity. In the $S$ frame, the events of A being at location $.75ct$ and B being at $.25ct$ are simultaneous, but that's not the case in the $A$ frame or the $B$ frame.

You can't just use the $\gamma$ separately on distance & time and combine them, you need to use the full Lorentz transformation to transform $(t, x)$ coordinates between frames.

Fortunately, for this simple case, we can use the relativistic formula for addition and subtraction of velocities, which can be derived from the Lorentz transformation. Using geometrical units, where $c=1$, the formula is:

$$w = \frac{u \pm v}{1 \pm uv}$$

Plugging in your velocities, we get:

$$w = \frac{\frac34 - \frac14}{1 - \frac34\cdot\frac14}$$ $$w = \frac{\frac12}{1 - \frac3{16}} = \frac8{13}$$

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  • $\begingroup$ Got it, thank you. I had a hunch I was making a false assumption about simultaneity, but couldn't quite nail down where. Thanks. $\endgroup$ Commented Jun 14, 2019 at 17:21
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Both the distance traveled and the time elapsed are multiplied by $\gamma$, so if you were to just divided the distance between the two ships by the time elapsed, you would get .5c. However, Lorentz boosts don't simply scale space and time, they convert some time to space and some space to time.

Imagine at time $t_0$ you were to measure the positions of $A$ and $B$ in S. Call those $A_0, B_0$ respectively. Then you measure them again at $t_1$. Call those $A_1, B_1$. You will find that $((A_1-B_1)-(A_0-B_0))/(t_1-t_0)=.5c$, i.e. the distance between them is increasing at .5c. Now suppose you measure them in the reference frame of $B$. Let's call those $A_0', B_0', A_1', B_1'$. If someone on Spaceship B calculates $(A_1'-B_1')-(A_0'-B_0')$ and $t_1-t_0$, they will find that both have been multiplied by $\gamma$, and so they will agree with S that the ratio is .5c.

But, they will not agree that it follows that $A$ is travelling at .5c relative to them. The important thing to remember in relativity is that everything is four dimensional. $A_1$ and $B_1$ are events in spacetime. Someone in S says that they are the position of $A$ and $B$ at time $t_1$, and will then say that we can find the distance between $A$ and $B$ by taking $A_1-B_1$. However, in relativity there is no universal concept of simultaneity. Someone in S will say that $A_1$ and $B_1$ are the positions that $A$ and $B$ have at the same time. Someone in Spaceship B, however, will say "No, those are positions of $A$ and $B$ at different times. If we want the distance between the two spaceships, we need to take their positions at the same time." When they look at where Spaceship B is at the "same" time, they will see a time that S says is later. B will therefore say that the distance between the spaceships is larger, and thus will say that $A$ is traveling faster than .5c.

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  • $\begingroup$ Both the distance traveled and the time elapsed are multiplied by $\gamma$ Good point, which I forgot to address in my answer. $\endgroup$
    – PM 2Ring
    Commented Jun 14, 2019 at 20:16

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