# Follow up on Velocity 4-Vectors

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1. Dec 10, 2016

### MikeLizzi

I did not use the template for this posting.

My understanding for the relativistic transformation of a velocity u to u' is given by

$$\begin{bmatrix} \gamma_{u'} \\ \gamma_{u'} u'_x \\ \gamma_{u'} u'_y \\ \gamma_{u'} u'_z \end{bmatrix} = \begin{bmatrix} \gamma & -\gamma\beta_x & -\gamma\beta_y & -\gamma\beta_z \\ -\gamma\beta_x & 1+\frac{(\gamma -1)\beta_x^2}{\beta^2} & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & \frac{(\gamma -1)\beta_x \beta_z}{\beta^2} \\ -\gamma\beta_y & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & 1+\frac{(\gamma -1)\beta_y^2}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} \\ -\gamma\beta_z & \frac{(\gamma -1)\beta_x\beta_z}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} & 1+\frac{(\gamma -1)\beta_z^2}{\beta^2} \end{bmatrix} \begin{bmatrix} \gamma_{u} \\ \gamma_{u} u_x \\ \gamma_{u} u_y \\ \gamma_{u} u_z \end{bmatrix}$$
Where v is the velocity of reference frame S' with respect to reference frame S and
$$\beta_x = v_x/c \\ \beta_y = v_y/c \\ \beta_y = v_y/c \\ \beta^2 = \beta_x^2 + \beta_y^2 + \beta_z^2 \\ \gamma = \frac{1}{\sqrt{1-\beta^2}} \\ \gamma_u = \frac{1}{\sqrt{1-\beta_u^2}} \\ \gamma_{u'} = \frac{1}{\sqrt{1-\beta_{u'}^2}}$$
This seems to work for any object whose velocity is less that c. But, I got the impression from a previous post that I could transform the velocity of light rays the same way I transformed the velocity of massive objects.

If I try to do that using the definitions I have above it means calculating $$\gamma_u$$ for a light ray. That's undefined. I have a 3-vector version of the velocity addition formula that doesn't require such a calculation and it works fine. Did I get the 4-vector structure wrong?

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