# Photons, Mass, Gravity, Light, Rest Mass, Invariant Mass, Energy, Momentum

*Most recent answer: 01/09/2011*

- Stian Dahl (age 21)

Vestfold Norway

Much of the confusion arises because the word "mass" has been used in two different ways in physics. The "m" in E=mc

^{2}is (as the equation makes clear) just another symbol for energy, expressed in different units. This same "m" also appears in the equation for momentum p=mv, where v is velocity. Light has energy and momentum, so it has "m" in this sense. This m is the same thing that appears in General Relativity (or even Newtonian gravity) as the source of gravitational effects. So light is definitely affected by gravity. Since light has energy, it is also a source of gravitational effects on other objects, although not a very strong one under ordinary circumstances.

Now when people are describing the "mass" of different objects, including particles, it's much more convenient to talk about the rest mass, also called the "invariant mass", the mass a particle has in a frame in which its momentum is zero. That way you don't have to ask what reference frame you're using, and can just give a specific mass for each object. In that sense, the mass of a photon is zero. However, that's not the term that enters into the gravitational equations.

Although the "invariant mass" is indeed invariant under choice of reference frame, it is not invariant under choice of how to group things into objects. For example, take two similar blips of light traveling opposite directions. Each one has energy E, momentum |p|=E/c, and an invariant mass of zero. Since the momenta are opposite, we are already using the reference frame in which the momentum of the two-blip object is zero. The invariant mass of the two-blip object is then 2E/c

^{2}, not zero. Even when things have no interaction, the invariant mass of the sum is not the sum of the invariant masses. A big box of photons has energy, zero average momentum, and thus has some invariant mass. It acts gravitationally just like anything else with the same energy and no momentum.

Mike W.

*(published on 01/09/2011)*

## Follow-Up #1: light, gravity, again

- Philip (age 21)

England

etc.

*(published on 02/23/2011)*

## Follow-Up #2: Higgs: inertial, gravitational, and rest masses

- Mark Cliffod (age 18)

Santa Fe

gives some particles rest mass. The other particles, lacking rest mass, still have inertial mass, which is identical to gravitational mass.

On another point, I'm curious why you say that "the scientific world was shocked at the discovery, or rather proof of the Higgs bosons existence". This particle has been expected for decades. It would have been shocking if it weren't there. The particle seen has not yet been pinned down to be the simplest sort of Higgs particle, but it is some sort of rest-mass-generating boson.

Mike W.

*(published on 07/27/2012)*

## Follow-Up #3: light, mass, momentum

- petrus snel (age 55)

france

See also:

Mike W.

*(published on 07/28/2012)*

## Follow-Up #4: can gravity make light accelerate?

- Andrew (age 21)

New Zealand

We've only dealt briefly with the issue of light "accelerating". Why not have a look at and then follow-up for more clarification?

Mike W.

*(published on 10/11/2012)*

## Follow-Up #5: Do photons have mass?

- Hamuda (age 16)

Maldives

Mike W.

*(published on 05/14/2013)*

## Follow-Up #6: momentum of a massless particle

- Anushka (age 16)

India

I've marked this as a follow-up to ones that may answer it. The key is to pay attention to the *different *uses of the symbol "m", either as rest-mass or as the factor in **p**=m**v**.

"How do we know the momentum of a massless particle?" There are several ways. One is to directly measure the momentum by measuring, for example, the force exerted on a mirror by a stream of photons. Here one uses that **p** is conserved and also that the particle number can be determined using universal quantum relation E=hf, where E is energy, h is Planck's constant, and f is frequency. Another way is to use an argument from Maxwell's equations that requires E=pc if momentum is to be conserved. That gives a momentum density in terms of the electric and magnetic fields. It can be converted to a momentum per particle again using E=hf. Another way is to look for the missing momentum in events involving a few massive particles and a photon or two. Other ways include using the universal quantum relation |**p|**=h**/**λ, where λ is the wavelength. λ can be measured with diffraction gratings or other methods.

Mike W.

*(published on 06/04/2013)*

## Follow-Up #7: mass, light, and matter

- muhammad uzair ahmad (age 15)

pakistan

Inertial mass is a well defined quantity that appears in physical effects, such as momentum conservation and gravity. Those effects show the inertial mass (not rest mass) of light. There's more discussion in the earlier part of the thread in which I've put your question.

I'm not sure what the word "matter" means.

Mike W.

*(published on 12/21/2013)*

## Follow-Up #8: gravity and light

- Suroj Dey (age 14)

India, ASSAM

same answer!

Mike W.

As we have said many times the full equation for energy is E^{2} = p^{2} + mc^{2} where p is the momentum. LeeH

*(published on 04/18/2014)*

## Follow-Up #9: photon energy

- Travis (age 27)

Richland, wa

Actually, we do happen to use E=hf in one answer. Most of these questions, however, center on the relation between momentum and energy and gravity, not on the size of the quantum packets.

Mike W.

*(published on 03/24/2016)*

## Follow-Up #10: gravity and mass

- Sam L (age 33)

Phoenix, AZ

Just on your core question:

They *all* interact via gravity.

Mike W.

*(published on 01/19/2018)*

## Follow-Up #11: light, mass, and energy again

- Austin (age 18)

South Carolina

We think that this thread should answer your question.

The quantum aspect doesn't have much to do with the issue, since light's energy and momentum were already described by Maxwell's classical electromagnetism.

Mike W.

*(published on 04/27/2020)*