# Light -- Zero Rest Mass and Momentum **URL:** \ **Category:** Factual Questions\ **Created:** [June 18, 2003, 8:01pm UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778 "2003-06-18T20:01:12Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![Major\_Kong](https://avatars.discourse-cdn.com/v4/letter/m/b9bd4f/32.png) [@Major\_Kong](https://boards.straightdope.com/u/Major_Kong)\ **Post date:** [June 18, 2003, 8:01pm UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778/1 "2003-06-18T20:01:12Z")
How can light have both zero rest mass and momentum? Isn’t rest mass required for momentum? Does light have mass when moving at c? Can light only move at c? ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [June 18, 2003, 8:31pm UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778/2 "2003-06-18T20:31:18Z")
Maybe the link below will help. I’ll post a bit of it so you can see if it would be interesting to you but I don’t want to offend the copyright gods either by posting too much: > [@](#): > > **Question** > > My science data book says that the mass of a photon is zero. But how can photon exert radiation pressure, as it occurs in the sun? > > Asked by: Suraj > > **Answer** > > Actually what your science book says or is trying to say anyway,is that the photon’s rest mass is zero. The concept of the rest mass derives from special relativity. The rest mass is the mass of a particle (in our case the photon) as measured by an observer who sees the particle still and with zero speed. In other words, the particle is at rest as far as this observer is concerned. Thus comes the term REST mass. But according to special relativity, light ALWAYS travels with the light speed c, and is NEVER at rest. And so it has zero REST mass. > > SOURCE: [http://www.physlink.com/Education/AskExperts/ae180.cfm](http://www.physlink.com/Education/AskExperts/ae180.cfm) ---
**Author:** ![Exapno\_Mapcase](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/exapno_mapcase/32/1051_2.png) [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)\ **Post date:** [June 18, 2003, 9:36pm UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778/3 "2003-06-18T21:36:42Z")
The equation to use is not the classic E = mc[sup]2[/sup], which goes to zero when mass is zero, but E[sup]2[/sup] = p[sup]2[/sup]c[sup]2[/sup] + m[sup]2[/sup]c[sup]4[/sup], where p is momentum. With photons having zero rest mass, the equation goes to a simple E = pc. This is what shows the energy of a zero rest mass particle. [http://www.colorado.edu/physics/phys2170/phys2170\_spring96/notes/2170\_notes1\_4.html](http://www.colorado.edu/physics/phys2170/phys2170_spring96/notes/2170_notes1_4.html) ---
**Author:** ![Francis\_E\_Dec\_Esq](https://avatars.discourse-cdn.com/v4/letter/f/48db29/32.png) [@Francis\_E\_Dec\_Esq](https://boards.straightdope.com/u/Francis_E_Dec_Esq)\ **Post date:** [June 18, 2003, 10:45pm UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778/4 "2003-06-18T22:45:28Z")
> [@](#): > > \*Originally posted by Exapno Mapcase \* > \*\*With photons having zero rest mass, the equation goes to a simple E = pc. > \*\* Right. No momentum, then no energy. Here is another way to think of it. Photons are just modes of vibration of the electromagnetic field. And we know waves of the EM field _have_ to be able to carry momentum. Jiggle an electron at X, an EM wave travels away from it to Y causing electron at Y to jiggle. The momentum of electron X was transferred to Y via the field. The wave doesn’t have mass, but it does transfer momentum. A photon has zero _rest_ mass, but that shouldn’t bother you because **photons are never at rest**. But in one sense they could be considered to have mass. The relativistic expression for a particle’s momentum is\* p = m[sub]R[/sub]v = mv/(1-v[sup]2[/sup]/c[sup]2[/sup])[sup]1/2[/sup]\* where _m_ is the rest mass and _m[sub]R[/sub] = m/(1-v[sup]2[/sup]/c[sup]2[/sup])[sup]1/2[/sup]_ is the relativistic mass. As you can see, the relativistic mass is also equal to _p/v_, just as it is in Newtonian physics. Although the rest mass of a photon vanishes, the ratio _p/v = p/c_ does not and you can define that as the “mass” of the photon. This is perfectly valid and quite in accordance with the intuitive notion of inertial mass. The only problem is that the relativistic mass varies with the speed and most people are used to thinking of the rest mass as **the** mass, but it is really just a matter of definition. In fact, this definition of mass is more in accordance with the OP’s intuition than rest mass is. ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [June 19, 2003, 12:47am UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778/5 "2003-06-19T00:47:05Z")
> [@](#): > > \*Originally posted by Francis E Dec, Esq \* > \*\* > This is perfectly valid and quite in accordance with the intuitive notion of inertial mass. The only problem is that the relativistic mass varies with the speed and most people are used to thinking of the rest mass as **the** mass, but it is really just a matter of definition. In fact, this definition of mass is more in accordance with the OP’s intuition than rest mass is. \*\* I wouldn’t say that relativistic mass is completely wrong but it’s certainly considered verboten by a very large majority of physicists. **Francis E Dec, Esq** you might want to try this thought experiment. Let’s say that you apply a longitudinal force to a relativistic particle of rest mass m, how would you calculate the relativistic mass so that F = ma (in terms of the rest mass)? ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [June 19, 2003, 1:13am UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778/6 "2003-06-19T01:13:30Z")
I should probably give the answer to my proposed question. Longitudinal direction: \*_F = Gamma[sup]3[/sup]m[sub]0[/sub]_ And for a force applied in the transverse direction: **F = Gamma\*m[sub]0[/sub]** Gamma = 1 / (1 - v[sup]2[/sup] / c[sup]2[/sup]) [sup]1/2[/sup] Instead of a nice invariant scalar quantity we now have a matrix quantity. ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [June 19, 2003, 1:16am UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778/7 "2003-06-19T01:16:37Z")
Damn I meant to hit preview not submit. I should probably give the answer to my proposed question. Longitudinal direction: \*_m[sub]r[/sub] = Gamma[sup]3[/sup]m[sub]0[/sub]_ And for a force applied in the transverse direction: **m[sub]r[/sub] = Gamma\*m[sub]0[/sub]** Gamma = 1 / (1 - v[sup]2[/sup]/c[sup]2[/sup])[sup]1/2[/sup] Instead of a nice invariant scalar quantity we now have a matrix quantity. ---
**Author:** ![Francis\_E\_Dec\_Esq](https://avatars.discourse-cdn.com/v4/letter/f/48db29/32.png) [@Francis\_E\_Dec\_Esq](https://boards.straightdope.com/u/Francis_E_Dec_Esq)\ **Post date:** [June 19, 2003, 4:08pm UTC](https://boards.straightdope.com/t/light-zero-rest-mass-and-momentum/182778/8 "2003-06-19T16:08:21Z")
Thanks, **Ring**. Very interesting. I’ll have to play with that some more.