# I Velocity 4-Vector for Light Ray

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1. Dec 10, 2016

### MikeLizzi

I hope I'm not violating Forum protocol, again.
I tried posting this question in the Homework section but it got locked for violating homework protocol.

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?

2. Dec 10, 2016

### vanhees71

For a "light ray" there is no four-velocity in the usual sense, because it's light-like. You can take a photon (which is a quantum of the quantized electromagnetic field) momentum as a substitute. The corresponding four-momentum is
$$(p^{\mu})=\begin{pmatrix} |\vec{p}| \\ \vec{p} \end{pmatrix}.$$
It transforms as a four vector. The mass of a photon is $p_{\mu} p^{\mu}=0$, i.e., it's a massless quantum. Of course this doesn't change under Lorentz transformations, thus a photon has a lightlike momentum in any frame, and the three-velocity is
$$\vec{v}=\frac{c\vec{p}}{|\vec{p}|} \; \Rightarrow \; |\vec{v}|=c,$$
i.e., the photon travels with the speed of light in any frame.