# Does light have a non-zero mass? [duplicate]

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I heard that the light has zero mass. But I searched in internet, some people say light has mass and some people said doesn't. I am not sure what is the right answer.

And also I heard that the reason we can't travel in light speed is because we have mass but light doesn't. Please tell me the right answer.

## marked as duplicate by Emilio Pisanty, David Z♦Sep 19 '18 at 22:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Not enough prior research. – my2cts Sep 19 '18 at 22:32
• @my2cts Well, Kevin did mention that he searched the internet first, which definitely counts for something. On the other hand, it's true that there are several other questions about this on the site. – David Z Sep 19 '18 at 22:42
• @David Z If he searched the internet he should have found something. – my2cts Sep 20 '18 at 19:59
• @my2cts Sure, and it sounds like he found a whole lot, which is why this question was posted. – David Z Sep 20 '18 at 20:10
• @David Z I would not call that prior research. Also I would require links in place of "some people say". Some people say just about anything. No problem, I am quite willing to answer this one. See physics.stackexchange.com/questions/413654/… – my2cts Sep 20 '18 at 20:54

## 2 Answers

A photon (elementary particle that makes up light) does not have a (rest) mass, but it does have energy $E$ and momentum $p$, given by the relation $$E=pc$$ where $c$ is the speed of light.

(The full form of this equation for a general particle is $E^2=m^2c^4+p^2c^4$ - if you consider a stationary particle, which is not the case for a photon, then you end up with the famous $E=mc^2$ equation).

To accelerate towards the speed of light, a particle with mass would need infinite energy - this falls out of special relativity formulae. In addition to the above general formula I gave, another way of expressing the energy of a particle with mass is $$E=\gamma mc^2$$ where the "Lorentz factor" features here, which is: $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ which explodes to infinity as $v\rightarrow c$ and so if you want to get a massive particle with $v=c$, you would need infinite energy.

According to energy-momentum relation, energy possessed by a particle of mass $$m_0$$(rest mass) and having a momenta $$p$$ is $$E^2=m_0^2c^4+p^2c^2$$. Please note that the mass that goes into this equation is rest mass. Now the relativistic mass is given as $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$. According to Einstein, rest mass of photon is $$m_0=0$$ and as it is moving at speed $$c$$, it's relativistic mass is indeterminant. The energy possessed by a photon is simply $$E=pc$$

• "As Einstein said energy possessed by a particle..." This is incorrect. The Energy–momentum relation was first established by Paul Dirac in 1928. "According to Einstein, rest mass of photon is $m_0=0$" - This is incorrect. The photon does not have a defined rest mass. Also, the zero invariant mass of any quantization of light follows from the Coulomb's law and Maxwell's equations. – safesphere Sep 19 '18 at 23:48
• I have never quite understood the equation E=Pc. It seems redundant to me. P and C both have velocities. We know what C is, what is the velocity in P? – Bill Alsept Sep 20 '18 at 0:03
• @safesphere I think Einstein defined the zero rest mass of photon in his famous paper of photoelectric effect. There he used $E=h\nu$. – Jitendra Sep 20 '18 at 19:47
• @BillAlsept You should post this question on the website. One way to understand $E=pc$ is by using de-broglie's relation $\lambda=\frac{h}{p}$. Now equation becomes $E=\frac{hc}{\lambda}$. The momentum $p$ that we are taking here is not just the mechanical momentum but it also have electromagnetic momentum in it. $p$ should really be called canonical momentum. – Jitendra Sep 20 '18 at 19:55
• @safesphere 1) Your assertion on the EM relation is worth a reference. 2) One can define the photon rest mass to be zero without difficulty, except for the infrared catastrophe, which is solved by assuming an arbitrarily small non-zero photon mass. 3) no it does not. It would follow from the unattainable experimental proof that Maxwell's equations and the Coulomb's law are exact to any level of precision. – my2cts Sep 20 '18 at 21:51