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Obviously, $\beta=v/c$. But in this case, I'm not too sure what $v$ represents. I've mostly done 1-D special relativity and therefore it is pretty clear in those cases. However say there are two reference frames, one moving at some velocity $u$ with respect to the other. Let's say that there is some particle moving at an angle $\theta$ and speed $v$ from the first reference frame.

We can ignore the transformation in the vertical direction as the particle's vertical velocity is the same for both reference frames, therefore we can just take the velocity in the horizontal direction, let's call it $v_x$.

Is the $\beta$ factor, then, $\beta=v/c$ or $\beta=v_x/c$?

Qmechanic
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agaminon
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1 Answers1

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It depends on the context. I have seen authors define a 3-vector $\vec\beta=\left(\beta_x, \beta_y, \beta_z\right) = \left(v_x/c, v_y/c, v_z/c\right)$. However in the general Lorentz transformation formulas, when you see the factor $\gamma\equiv (1-\beta^2)^{-1/2}$ appear, $\beta$ is the magnitude of the velocity divided by $c$, $\beta=v/c$.

In general, the author of a source you are reading can define whatever notation they want. The key task for you as the reader is to look for where the notation is defined in the source you are reading. Don't assume that different authors will use the same conventions for all notation.

Andrew
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