# What is the mass of Photon or Light Particle?

We know that any particle moving at speed of light will have infinite mass.

Since even Light has particles. Then what is the size of the Particle? How to imagine this with the mass of light particles.

• Recommended reading: The shape and size of a photon – user3408085 Jun 25 '18 at 18:18
• We know that any particle moving at the speed of light has ZERO mass. – Andrei Geanta Jun 25 '18 at 18:26
• In that article I read: "But an electromagnetic wave is not like a ‘probability wave’. So… Do they have a de Broglie wavelength as well?" Yes it is and of course they do. This is where De Broglie got the idea. – my2cts Jun 25 '18 at 19:09
• We know that any particle moving at speed of light will have infinite mass. Would it not hurt a bit, when you looked at any light source, if this were correct? – user198207 Jun 25 '18 at 19:20
• "We know that any particle moving at speed of light will have infinite mass." It is worth thinking about the (possibly unstated) assumptions of whatever "proof" of this "fact" you've read. Do you see the loophole? – dmckee --- ex-moderator kitten Jun 26 '18 at 1:21

The mass of a photon is anywhere between zero and $10^{-18} ~ eV/c^2$.

https://en.wikipedia.org/wiki/Photon

No need to write photon or light with capitals, by the way.

• This is a funny way to state the experimental upper limit on the photon's mass, because it suggests that the probability distribution is uniform. The most likely value for the photon mass is zero. The limit $10^{-18}\,\mathrm{eV}/c^2$ is where a massive photon would affect the way the sun's magnetic field interacts with the solar wind. Making the same sort of argument on the scale of galactic magnetic fields apparently reduces the upper limit by eight or ten orders or magnitude! For details, see the Particle Data Group review. – rob Jun 25 '18 at 22:26
• @rob I do not understand. 1) Stating a range does not suggest a uniform probability distribution per your own reference. 2) The Particle Group Review states the same limit that I quoted from wikipedia. – my2cts Jun 25 '18 at 22:34
• Fair enough (1) The "anywhere" is what suggested to me the uniform probability distribution. Compare to my prediction that your height is anywhere between zero and one thousand meters. Technically correct, but the upper limit sets the wrong scale: I could be talking about a person or a mountain. (2) My link is the source for the Wikipedia article. – rob Jun 25 '18 at 22:53
• @rob This is misleading. You know that my length is in much smaller range. You don't know that the photon mass is in a smaller range. You may like the idea that the photon has zero mass but you don't know that it is not somewhere else in the range specified by the particle group. No one does. Thus is why they specify this range. – my2cts Sep 15 '18 at 21:53
• I would agree with you if there weren't reasonable arguments for a much smaller limit, but the reasons those arguments don't set the PDG limit are too complex for a comment. I would be okay if you had said something like "most likely zero, but could be as large as $10^{-26}\rm\,eV$ without destroying the galactic magnetic field. Changes in our assumptions about just quite what a photon is could make the mass as large as $10^{-18}\rm\,eV$." – rob Sep 15 '18 at 23:34

Massive objects simply can not move at the speed of light. Light, or photons, are mass-less particles. They have 0 mass - which is why they can (and have to) move at the speed of light.

From the special relativity theory you got the usefull relationship

$$E^2 = \Big(m\cdot c^2 \Big)^2 + \Big(p \cdot c \Big)^2 \quad .$$

All experimental results up to now show that $m=0$ for a photon. Therefore

$$E^2 = 0 + (p\cdot c)^2 \\ E = p\cdot c = \hbar \cdot {k} \cdot c = \hbar\cdot \omega$$

BTW: In my opinion it is false to talk about relativistic masses as you did in your post. The mass is invariant!

The equation can be simple derived by using the equation for a relativistic moment

$$p = \cfrac{m\cdot v}{\sqrt{1-\frac{v^2}{c^2}}} \qquad ,$$

and the relativistic energy

$$E = \cfrac{m\cdot c^2}{\sqrt{1-\frac{v^2}{c^2}}} \qquad .$$