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A spaceship is heading towards earth at 0.99c, when it is 1*10^9 metres away from the earth as measured by an earthly observer, the spaceship ejects a life pod. Is the distance between the earth and the spaceship at the instance the life pod is ejected the same for both the spaceship pilot and earthly observer?

Intuitively I think that as the situation is identical from both observer's frame of reference with the other heading toward them at 0.99c, the measured distance would also be the same. However, when I tried to solve this with Lorentz transforms, I failed miserably.

I tried to first align the frames of reference at: time(earth) = time(spaceship) = 0s
from this, I can find the time at which the life pod was ejected in earth's frame of reference by solving: timeEarth(ejection) = -(1*10^9)/(0.99c) = -3.38 seconds, which means the spaceship ejects the life pod 3.38 seconds before it passes the earth, from the earth's frame of reference.

I then assign spacetime coordinates for this event in the earth's frame of reference and let everything occur 1 dimensionally along an x-axis, so the event occurs at (110^9 m, -3.38 s) in the earth's frame of reference where the position of the space ship is 110^9 metres to the right. If I then apply Lorentz transforms to find the spacetime coordinates of the event from the spaceship's frame of reference:

Lorentz factor (γ) = 1 / sqrt( 1 - (0.99c / c)^2 ) = 7.089
timeSpaceship(ejection) = γ * ( -3.38 - (0.99c / c^2) * 1* 10^9) = -47.45 seconds
PositionSpaceship(ejection) = γ * (1*10^9 - 0.99c * -3.38) = 1.42 * 10^10 metres

This result clearly looks wrong, because imagining the frame of reference of the spaceship which remains still, the life pod must be ejected at the spaceship's position, and I would expect PositionSpaceship(ejection) = 0, since time was synchronised to zero when the spaceship passes the earth and so the spaceship would remain at x=0 in its frame of reference, while the earth moved toward the spaceship from the left at 0.99c.

I'm sure I've just made a mistake in using Lorentz transforms, but any help would be greatly appreciated.

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  • $\begingroup$ I would recommend choosing $ v: \gamma(v) = 2$ for convenience, when numerically checking thought experiments, just for ease of multiplication. $\endgroup$
    – g s
    Oct 27, 2022 at 18:27
  • $\begingroup$ Since the two reference frames do not coincide at t=t'=0, it is necessary to use the Poincaré transformation. physics.stackexchange.com/questions/499760/… $\endgroup$
    – The Tiler
    Oct 27, 2022 at 18:44

2 Answers 2

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The time should be $-3.37\text{ s}$, not $-3.38\text{ s}$, which actually makes a big difference in the calculation below, because the terms almost cancel.

That aside, it's just a sign error. You wrote (reformatting a bit)

$$t^{\text{Spaceship}}_\text{ejection} = γ ( -3.38\text{ s} - (0.99c / c^2) \cdot 10^9\text{ m}) = -47.45\text{ s}$$

but it should be

$$t^{\text{Spaceship}}_\text{ejection} = γ ( -3.37\text{ s} + (0.99c / c^2) \cdot 10^9\text{ m}) = -0.48\text{ s}$$

Also, the position should be zero, not $1.42 \times 10^{10}\text{ m}$, since the spaceship is at rest at the origin in the spaceship's rest frame. You can use that as a sanity check that you got the sign right.

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  • $\begingroup$ Yes! thank you, I see how I made that mistake, I've got it sorted out to have the event occur at X2=0, I just found it odd that the two times are different as the situation appears symmetrical? As from either reference frame the other is coming toward them at the same super high speed, I imagine it's due to the event being observed exactly at the spaceship's position for the spaceship frame of reference, compared to the event observed from earth? $\endgroup$
    – Lucas Williams
    Oct 27, 2022 at 19:43
  • $\begingroup$ @LucasWilliams Your final question (in the above comment) is exactly to the point. The situation is not symmetric for exactly this reason. $\endgroup$
    – WillO
    Oct 27, 2022 at 22:35
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Space is not a perfect vacuum.... particle density is low but at .99c it's like driving thru a sand blaster. Your escape pod would have to be ejected faster than the ship or dropped behind it. In any case it's mush.....

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  • $\begingroup$ This is at most a comment and not an answer. On the other hand, it is wrong. at 0.99c the particles that you meet in space are not sand at any rate. They are a powerful penetrating radiation instead. $\endgroup$
    – fraxinus
    Oct 27, 2022 at 18:09

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