# B Mass of a photon

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1. Apr 17, 2016

### Stephanus

Dear PF Forum,
Yesterday (on Sunday) my friend texted me, "the mass of a photon is 10-18 Ev/c2$If my number is correct. And I also I check it in Wiki, and it's true. Now what I want to ask is this first. I Understand, my wording is confusing. Now, I'll rearrange my question for a new question. A: If something travels at 0.99999c wrt (A), than it can travel 0.1c wrt (B). It CANNOT travels at c execpt wrt light. B: If something ELSE travels at c wrt object with mass, it will always travel at c wrt anything, even against light. Is A and B true? I think so. I just want to rearrange my questions and get confirmations. Now about this photon, it seems that it's not massless, It's 10-18eV/c2 Or 1.6 * 10-37 joules/c2. Or around 0.18 * 10-53 kg. C. Does proton ALWAYS travels at c against something? It has mass no matter how tiny does it, right? D. If photon represents light, than light doesn't always travel at c? I think C and D can contradict A and B E. I read in wiki that light doesn't always travel at c. c is the universal constant where maximum speed is allowed in this universe. Or perhaps 10-53 kg is such too small number that we can ignore its mass? I check the mass of photon/proton is around 10-30, so anything below 10-30 mass of a proton can be considered massless? 2. Apr 17, 2016 ### PeterDonis ### Staff: Mentor No, it isn't. What is true is that our best current experimental tests show that the mass of a photon cannot be larger than$10^{-18}$Ev/c2. (That appears to be the current upper bound given by the Particle Data Group.) But that doesn't mean the photon has nonzero mass; it just means that there are unavoidable error bars around any measurement, so we can't say we've explicitly measured the photon's mass to be exactly zero; we can only say that we've measured it to be within some small error range around zero. That is perfectly consistent with it being exactly zero. 3. Apr 18, 2016 ### Ibix With A, I'm a little uncomfortable with your "wrt to light" since that implies light has a perspective on this. That there are no reference frames travelling at the speed of light is part of this issue. Otherwise, A and B are fine. As Peter already pointed out you are mis-interpreting the photon mass issue. We believe its mass is zero. We have confirmed by experiment that its mass is less than some upper bound. I'm sure the bound will continue to tighten but, even assuming the photon genuinely is massless, it will never actually reach zero. Assuming the photon is massless it always travels at c. If photons did have mass they would not be able to travel at c. However, the bound on any mass they could have is so low that the slightest touch would make them race off at near c, which would explain why we haven't noticed yet. That doesn't affect relativity (although particle physics would be in uproar, I gather). Relativity requires that there is an invariant speed. It does not require that anything actually travel at it. 4. Apr 18, 2016 ### Stephanus 5. Apr 18, 2016 ### vanhees71 It's also important to remark that the energy unit used here is called electron volts which is written eV (NOT Ev). It's the energy a particle with the charge of an electron (the socalled elementary charge) gains when running through a electrostatic potential difference of 1 V. 6. Apr 18, 2016 ### Stephanus Thanks vanhees71. It's a typo. I know about electron volt as you can see in my next paragraph. Oh no. Did I write proton? I guess you all know that what I meant is actually photon. Thanks, I'll read eV again. But as far as I know, eV is a unit of energy not voltage. Thanks again for the correction. I still have to read @Ibix and @PeterDonis answer above. 7. Apr 18, 2016 ### vanhees71 As I said, it's the ENERGY a particle with the charge of an electron gains when running through a electrostatic potential difference of 1 V. So you really ask about the photon not the proton. First of all nowadays we use exclusively the invariant mass if we say simply mass. A photon is described in the Standard Model by a massless vector field, and the empirical evidence for this assumption is quite good. The upper limit of the invariant mass of the photon is about$10^{-18} \text{eV}$. 8. Apr 20, 2016 ### Stephanus Sure, it's definetely a typo. I was thinking of how much lighter is a photon (in light) compared to a proton (in the atomic nucleus) and my fingers just typing out of my brain.. Massless vector field? I'll study it later. Shouldn't energi per speed square to become mass?$10^{-18} \frac{eV}{c^2}$Dear advisors, If I may summarize your answers... A. Photon has mass B. The mass of a photon is very tiny that it can travel AT the speed of light? C. The mass of a photon is very tiny that it can travel NEAR the speed of light?. Which is right? Which is wrong? According to this$V_{r} = \frac{V_{g}-V_{p}}{1 - V_{p}V{g}}$Vg is the velocity of a galaxy very far away that travel near c. (about 2 gly from us?) Vp is the velocity of a photon. And if I shine a light torch in that galaxy, assuming that my light is strong enough to reach it, so this photon can hover near a pedestrian in that galaxy? If Vr becomes say... 0.001 km / sec. I can tamper those numbers, but I think you already know the answer. The mass of a photon as my calculation goes... is about 10-53 kg. So either this photon can travel at C or Near C, what about it's relativistic mass? If it has mass, it can't ever travel at C, right? Or this universe simply discards the mass and consider photon is massless?. What is this, my extended floating point variable is 80 bits. Our universe just runs out of double precision that it "thinks" the mass of the photon is zero? Thanks 9. Apr 20, 2016 ### PeterDonis ### Staff: Mentor A, B, and C are all wrong according to our best current knowledge. Photons have zero invariant mass and travel exactly at the speed of light. Photons have energy, but energy is not the same as invariant mass. Other meanings of the word "mass", such as "relativistic mass", are out of date and should not be used; they are basically equivalent to "energy", and we already have a perfectly good word for that, namely "energy". 10. Apr 20, 2016 ### Nugatory ### Staff: Mentor None of the above. When we do an experiment to measure the rest mass of a photon, our instruments report that it is zero. However, our best instruments of this type are only accurate to$\pm{10}^{-18}$eV so we can't say "the rest mass of a photon is zero". Instead, we say "the rest mass of a photon is zero according to our best instruments but they could be wrong by as much as$\pm{10}^{-18}$eV so all the experiment proves is that the rest mass is less than that". And indeed it is less than that - it is zero, which is less than that. 11. Apr 20, 2016 ### Stephanus Oh, so there IS really an experiment. And this number 10-18 eV is the limitation of the instrument. So the mass of this photon can't be bigger then 10-18eV (now I follow suit, telling mass in a unit of energy). Okay... thank you very much. Now, this might be off topic, because I know that the mass of a photon is known. So why 10-18eV? Perhaps in the next generation if the scientist have improved the instrument, the mass of a photon changes to 10-19eV? What is so siginificant about this number? 12. Apr 20, 2016 ### PeterDonis ### Staff: Mentor Why not? Every instrument has some finite accuracy. No, the mass of the photon will still be zero. But the accuracy of the instrument will be$10^{-19}$eV instead of the$10^{-18}## eV that it is now.

Why does it have to be "significant"? It's just the finite accuracy of a particular device.

13. Apr 20, 2016

### Staff: Mentor

We would not say that the mass of the photon of the photon has changed from 10-18eV to 10-19eV. We'd say that the mass of the photon is zero just like it was last year and the instrument we used to measure it this year is accurate to 10-19eV while last year's instrument was only accurate to 10-18eV.
Nothing. It just happens to be more or less how good the measuring technology is today.

14. Apr 21, 2016

### haushofer

I like the reasoning "I checked it on Wiki, and it's true".