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If I send two rockets from the Earth in opposite directions, at, say, 60% of the speed of light relative to the Earth, then relative to each other they are travelling at 120% of the speed of light. What is my problem in reasoning?

Is it to do with the fact that due to SR their relative velocities are not 120%, in either of their reference frames? If this is the case, what is the maximum relative velocity that two bodies can have from any reference frame?

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marked as duplicate by ACuriousMind, yuggib, Kyle Kanos, Qmechanic May 28 '15 at 22:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ possible duplicate of Relative Speed vs speed of light $\endgroup$ – Gonenc Mogol May 28 '15 at 9:50
  • $\begingroup$ when you say 'their relative speeds', you set yourself in the rockets frames. In your frame with local datas, you don't need special relativity transformations $\endgroup$ – user46925 May 28 '15 at 10:05
  • $\begingroup$ @gonenc : but the answer isn't complete $\endgroup$ – user46925 May 28 '15 at 10:12
  • $\begingroup$ @igael what do you mean? $\endgroup$ – Gonenc Mogol May 28 '15 at 10:19
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/7446/2451 and links therein. $\endgroup$ – Qmechanic May 28 '15 at 10:50
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You've assumed the law to add velocities is the same in special relativity as it is in Gallilean kinematics. This is wrong (as the paradox you've reached shows). Its possible to derive a way to "add" velocities which is compatible with relativity.

Your version is:

\begin{equation} s=v_1 - v_2 \end{equation}

The correct version is:

\begin{equation} s= \frac{v_1 - v_2}{1- \frac{v_1v_2}{c^2}} \end{equation}

With your velocities ($v_1=0.6c$, $v_2=-0.6c$) we get:

\begin{align} s &= \frac{1.2c}{1+ 0.6^2}\\ &\approx0.88c \end{align}

This is how things actually work and is significantly different to the version you used when the velocities are large enough. In particular if you use it you'll always find that two observers will see each other move at less than the speed of light if any third observer sees them both moving less than the speed of light.

On the other hand when the velocities in question are small then this version is very similar to the Gallilean one which is why our instinct works for everyday speeds. For example with $v_1=-v_2 = 0.01c$:

\begin{align} s&= \frac{(0.01+0.01)c}{1+0.01^2}\\ &\approx 0.019998c \end{align}

Very close to the Gallilean answer of $0.02c$.

Edit: I should point out that these formulae are only correct for motion in the same line. More general motion has a more general (and messier) formulae.

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  • $\begingroup$ Ok, but as they are travelling in opposite directions why are the velocities both positive? $\endgroup$ – bnosnehpets May 28 '15 at 9:15
  • $\begingroup$ I was a bit sloppy with my minus signs. The formula I gave is correct for this situation with the two velocities in opposite directions. For general motion you have a minus sign between the two terms in the numerator and in the denominator but these minus signs cancel with the one you get from these velocities being in opposite directions. I'll edit my answer to save further confusion. $\endgroup$ – or1426 May 28 '15 at 9:22
  • $\begingroup$ you might compute the speeds in each frame. For you , the distance between the rockets grows without lorentz transformation , at 0.6 c - - 0.6 c = 1.2 c since all the datas are from the SamE frame. Indeed, an observer in his rocket frame will need to transform the speeds from you frame and will find a final sum of 0.88 c. $\endgroup$ – user46925 May 28 '15 at 10:01
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Its been a while since I studied relativity, so I wont attempt the math, but I remember covering a similar situation at university. From what I remember, the issue is the reference frame you are considering it from.

When you say they are moving in opposite directions, that is from a specific reference frame, the earth. In that frame, the separation between the two may be increasing at greater than the speed of light, but neither of the individual objects is moving faster than the speed of light relative to you, the observer on earth, therefore not violating any laws of relativity.

From the perspective of one of the rockets, this would be different, the other rocket would in fact be moving away from you at the 0.88c as calculated in or1426's answer.

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    $\begingroup$ Quopting from: en.wikipedia.org/wiki/… "Special relativity does not prohibit this. It tells us that it is wrong to use Galilean relativity to compute the velocity of one of the particles, as would be measured by an observer traveling alongside the other particle. That is, special relativity gives the right formula for computing such relative velocity." $\endgroup$ – RyanO May 28 '15 at 11:54
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    $\begingroup$ @gonenc : that isn't right I'm afraid. We use the velocity addition formula to say how each observer gauges the other observer's speed. But an observer on Earth in the middle would say they're moving apart at 1.2c. See the Wiki page RyanO referred to. It's the flip side of a closing speed. $\endgroup$ – John Duffield May 28 '15 at 12:08

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