# Four-momentum of a photon

I've searched all over the web to get an answer for this, to no avail. So I decided to ask this question here and I'm hoping someone can.

What is the four-momentum of a photon?

I saw online that the 4-momentum is $$P = m_{0}U$$ but then this leaves me with another question and that is, what is the four-velocity $$U$$ of a photon. I don't believe it exists because a photon is never at rest.

Can anyone help me with this? Many thanks beforehand for any advice and assistance.

Dale
Mentor
2020 Award
Hi protonchain, welcome to PF

That definition of the four-momentum is nice, because it is a nice direct analogy of the usual Newtonian momentum. However, as you note, the four-velocity of a photon is not defined, so that particular definition of the four-momentum is not useful for a photon. However, even for a photon the following definition of energy applies:
E² = m²c^4 + c²(p.p)

As does the following definition of the four-momentum:
P = (E/c,p)

Since m=0, combining those two gives the four-momentum of a photon as:
(|p|,p)
or equivalently for a photon travelling in the x direction:
(E/c,E/c,0,0)

Dear Dale,

Thank you very much for the welcome, and of course for the beautiful answer to my question.

May I just clarify one small detail? I'm sorry to trouble you.

So the Energy of the photon that you had given was

1. $$E^{2} = m^{2}c^{4} + c^{2}(p \cdot p)$$

I've seen in my textbook and in references that

2. $$E = h \nu = U_{\alpha}p^{\alpha}$$

where $$U_{\alpha}$$ is the 4-velocity and $$p^{\alpha}$$ is the 4-momentum.

So from my understanding $$U_{\alpha}$$ is obviously not the photon velocity but of an observer of some kind, and then $$p^{\alpha}$$ is the one you just described.

Does that mean I can equate the square root of equation 1 with 2, aka

$$E = \sqrt{c^{2}(p \cdot p)} = U_{\alpha}p^{\alpha}$$

I thought that $$U_{\alpha}$$ was related to time in some form, so when I glanced at my reference book, and at your equation for energy, I was a little perplexed as to how the 2 equated.

Again sorry to trouble you with even more questions, I promise this is the last

DrGreg
Gold Member
$U^{\alpha}$ is the 4-velocity of the observer who is measuring the energy, and so in the observer's coordinates,

$$U^0 = c$$

$$U^1 = U^2 = U^3 = 0$$​

Dear Dr. Greg,

Thank you for clarifying that note. The advice from both you as well as Dale are much appreciated! :)

Dale
Mentor
2020 Award
You are quite welcome!

i heard that speed of light was the upper limit of the velocity but in Cerenkov Radiation the particle is going in speed more than that of light and if charged particles do so in a medium doesn't it show cooling effect? can any one answer this questions please...

does particles having more than the speed of light as tachyons only have four momentum or every thing has four momentum??

how come photons have four momentum in what conditions does it have four momentum? c

DrGreg
Gold Member
i heard that speed of light was the upper limit of the velocity but in Cerenkov Radiation the particle is going in speed more than that of light and if charged particles do so in a medium doesn't it show cooling effect? can any one answer this questions please...
Welcome to PF!

The fastest speed is c, the speed of light in vacuum. In water, for example, it is possible for particles to travel faster than the speed of light in water, which is about ¾c, but still slower than c.
does particles having more than the speed of light as tachyons only have four momentum or every thing has four momentum??
how come photons have four momentum in what conditions does it have four momentum? c
All particles, including photons and hypothetical tachyons, have 4-momentum. You take the particle's energy and its three components of momentum and use those to form a 4-dimensional vector, as indicated by previous posts in this thread.

thanks a lot dr greg that explains a lot.