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Lorentz Velocity Transformation: $$v_x = \frac{v_x' + u}{1+\frac{uv_x'}{c^2}}\tag1$$

$$v_x' = \frac{v_x - u}{1-\frac{uv_x}{c^2}}\tag2$$

The speed of spaceship and scoutship are given relative to Earth whereas the speed of the Robot Space Probe is given relative to the spaceship.

Assume we want to find the speed of Robot Space Probe relative to Earth I understand that the formula must use formula (1) where $v_x$ is the speed required. This is clearly obtainable from the fact that the Earth is at rest relative to the moving spaceship and hence this reduces to a situation similar to the usual "Observer - Train" situation. This is also recognizable from the derivation where we assumed that the reference frame S' moves at speed u relative to the reference frame S.

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However, assume that we have to find the speed of the Scoutship relative to the Spaceship what Lorentz Transform Equation should we use? Here the case is kind of confusing because the two spaceships are moving what should we consider as our S frame and which one should we consider as S' frame?

I know that the math shows that taking the wrong reference frame gives us a value of magnitude greater than the speed of light and hence we would have to take the other Lorentz Equation. However, I need to learn the concept.

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Images are taken from: University Physics with Modern Physics Sears and Zemansky 13th Edition.

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HINT: The easiest way to solve this problem is to let the spaceship be in the "rest frame", and let Earth be the "moving frame". What is the velocity of the Earth relative to the "rest frame"?


As an aside, you say that:

I know that the math shows that taking the wrong reference frame gives us a value of magnitude greater than the speed of light and hence we would have to take the other Lorentz Equation.

Taking the wrong reference frame in the velocity transformation equations may give you the wrong answer, but it should never give an answer greater than $c$ unless one of the "input" velocities ($u$ or $v_x$) is itself greater than $c$. If you got an answer greater than $c$ with both $|u|$ and $|v_x|$ less than $c$, you probably applied the equation incorrectly.

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  • $\begingroup$ Understood, however for the second part can I solve it assuming: Frame of Reference (S) : Earth Frame of Reference (S'): Spaceship Event : Scoutship Then use equation (2) with " u " being that of spaceship , $v_x $ being speed of Scoutship ? $\endgroup$ – Hasan Hammoud Jun 10 '18 at 17:54
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First, I recommend using the symbol $v$ to be the relative velocity of the two reference frames, i.e., the primed inertial reference frame (IRF) has velocity $v$ relative to the unprimed IRF. Note that $|v| < c$ (there are no IRFs with relative speed $c$)

Second, use $u$ ($u'$) for the velocity of the object in the unprimed (primed) IRF.

Then, by the relativistic velocity addition formula,

$$u = \frac{v + u'}{1 + \frac{vu'}{c^2}}$$

Let's see if this makes sense. If the object's velocity in the primed IRF is $u'$ and and if the primed IRF has velocity $v$ in the unprimed IRF, then it seems intuitive that the object's velocity in the unprimed IRF involves the sum of $v$ and $u'$.

Now, solve for $u'$:

$$u' = \left(1 + \frac{vu'}{c^2}\right)u - v \rightarrow u'\left(1 - \frac{vu}{c^2}\right) = u-v \Rightarrow u' = \frac{u - v}{1 - \frac{vu}{c^2}} $$

so your 2nd equation has a sign error is correct.

The speed of spaceship and scoutship are given relative to Earth whereas the speed of the Robot Space Probe is given relative to the spaceship.

Since the velocity $u'$ of the Robot Space Probe (RSP) is given relative to the spacecraft and since we know the velocity $v$ of the spacecraft relative to the Earth, the velocity $u$ of the RSP relative to the Earth is given by the 1st equation above.

$$u = \frac{0.9 + 0.7}{1 + (0.9)(0.7)}c = 0.981c $$

However, assume that we have to find the speed of the Scoutship relative to the Spaceship what Lorentz Transform Equation should we use?

You can use either as long as properly identify the velocities based on what I wrote above. But we know the spaceship has velocity $v$ relative to the Earth and we know the velocity $\nu$ of the scoutship relative to the Earth (I use $\nu$ here to distinguish the velocity of the scoutship from the velocity of the RSP).

Thus, we can use the 2nd equation above to find the velocity $\nu'$ of the scoutship relative to the spaceship:

$$\nu' = \frac{\nu - v}{1 - \frac{v\nu}{c^2}} = \frac{0.95 - 0.9}{1 - (0.95)(0.9)}c = 0.345c$$

Let's double check:

$$\nu = \frac{0.9 + 0.345}{1 + (0.9)(0.345)}c = 0.95c$$

However, you could have used the first equation by choosing the spaceship to be the unprimed IRF. I will leave that as an exercise for the interested reader.

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  • $\begingroup$ The second equation doesnt have a sign error you just used a different symbolic convention . $\endgroup$ – Hasan Hammoud Jun 10 '18 at 18:52
  • $\begingroup$ @HasanHammoud, oops, not sure why I thought there was a sign error. $\endgroup$ – Alfred Centauri Jun 10 '18 at 19:08

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