# B 4-momenta question

1. Feb 24, 2017

### AishaGirl

The 4-momenta before a photon collides with a stationary election $P_m=(mc^2,0,0,0)$ or is it $P_m=(mc,0,0,0)$

Which is the difference between the two?

Thanks.

Last edited: Feb 24, 2017
2. Feb 24, 2017

### Staff: Mentor

The four momentum is usually written in units of momentum, so (mc,0,0,0). However, it is even more common to use units where c=1, so that energy momentum and mass are all the same unit.

3. Feb 24, 2017

### Staff: Mentor

Units of energy versus units of momentum. I was trained in experimental particle physics, so I prefer units of energy $(E, p_x c, p_y c, p_z c)$ but I recognize that others prefer units of momentum $(E/c, p_x, p_y, p_z)$. Whichever units you prefer, be consistent within a calculation!

4. Feb 25, 2017

### Staff: Mentor

Interesting, does the particle physics community still call it the four momentum if they commonly use energy units?

5. Feb 25, 2017

### Orodruin

Staff Emeritus
We do not differentiate between the two since we use units where c=1.

6. Feb 25, 2017

### Staff: Mentor

When I was a grad student it was general practice to talk of momentum in units of MeV and GeV.

But we nevertheless called it (3-)momentum or 4-momentum as appropriate.

7. Feb 25, 2017

### AishaGirl

Thanks for the replies. So using (mc,0,0,0) is just as valid as using (mc^2,0,0,0) ?

8. Feb 25, 2017

### vanhees71

Sure, you only must be consistent with your convention or, most elegantly, set $c=1$, as the particle physicists do (see the postings above). If I keep $c \neq 1$, then I usually prefer to use the convention that the four-vector quantities have the dimension of the spatial parts, i.e., for momentum
$$(p^{\mu})=\begin{pmatrix} E/c \\ \vec{p} \end{pmatrix},$$
where $E=c \sqrt{m^2 c^2+\vec{p}^2}$ is the relativistic energy (i.e., kinetic + rest energy) of the particle. Then the energy-momentum relation (often called "on-shell condition") can be written in manifestly covariant form as
$$p_{\mu} p^{\mu}=m^2 c^2.$$

For the photon of momentum $\vec{k}$, following thie convention you have
$$(k^{\mu}) = \begin{pmatrix} |\vec{k}|,\vec{k} \end{pmatrix}.$$
Because it's massless the energy of the photon is $E_{\gamma}(\vec{k})=c |\vec{k}|$.