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The problem came up when I was trying to solve the classical problem of a wire with current due to which a charged particle experiences a force. This force should depend only on the lorentz factor $\gamma$ between different reference frames as $F'=\gamma F$. I did the calculation ones using two boosts (first into rest frame of the current then to the particle) and ones using a single boost directly into the rest frame of the particle. The results did not match so I started looking into combination of Lorentz boosts.

Generally results should not depend on how you transformed from one frame to another as long as you use valid Lorentz transformations (in this case in only consider Lorentz boosts in one dimension). So the following should be true for three systems 0, 1 and 2 with $\gamma_{ij}$ being the boost from system i to j:

$$ F_1 = \gamma_{01} F_0 \qquad F_2 = \gamma_{02} F_0 = \gamma_{12} F_1 = \gamma_{12} \gamma_{01} F_0 $$ $$ \Rightarrow \gamma_{02} = \gamma_{01}\gamma_{12} $$

With the relativistic velocity addition: $$ \beta_{02} = \frac{\beta_{01} + \beta_{12}}{1 + \beta_{01}\beta_{12}} $$

But applying this result to the Lorentz factor composition from above (and squaring and inverting for simplicity) I get

$$ (1-\beta_{01}^2)(1-\beta_{12}^2) \ne 1 - \left(\frac{\beta_{01}^2 + \beta_{12}^2}{1 + \beta_{01}\beta_{12}}\right)^2. $$

So how do you properly combine Lorentz boosts regarding Lorentz factors?

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  • $\begingroup$ This is my first post on this SE community, I tried to follow question guidelines, but please inform me about my (in-)conformity with guidelines here. $\endgroup$
    – imsodin
    Jun 4 '15 at 16:00
  • $\begingroup$ First impression, I think you did pretty well :-) Certainly far above the average first post we get here. Just to clarify: you're only talking about Lorentz boosts in one dimension, right? So the noncommutativity of rotations is not a factor? $\endgroup$
    – David Z
    Jun 4 '15 at 16:11
  • $\begingroup$ Great :) Yes I am, no rotations involved, just boosts in one direction. $\endgroup$
    – imsodin
    Jun 4 '15 at 16:16
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You cannot just combine the Lorentz boosts in that way - the boosts do not form a subgroup of the Lorentz group, the successive application of two boosts is, in general, not a boost, but a boost followed by a rotation, as you may see by explicitly writing down the $4\times 4$-matrices corresponding to the boosts and computing their product, which simply isn't the matrix of a pure boost. This phenomenon is known as Thomas precession.

Nevertheless, one can derive a formula for the composition of two boosts in terms of the velocities, but the formula for the rotation induced by the two boosts is... either ugly or requires going on about gyrovectors, so I will not reproduce it here.

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  • $\begingroup$ Thanks, thats exactly the (huge) misconception I had for boosts. My intuition still does not really want to accept it, but then intuition is your foe in SRT :P $\endgroup$
    – imsodin
    Jun 4 '15 at 16:32

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