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if so, what is that and how is resting mass different than just mass?

- Thread starter psuedoben
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- #1

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if so, what is that and how is resting mass different than just mass?

- #2

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- #3

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Ok, thank you. I was just confused I had because I had always been told they had no mass at all (because they don't interact with the Higgs fields right?) so it didn't make sense to me as to why they would have resting mass

- #4

ChrisVer

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In almost all the cases, the photons are massless.. Only under certain conditions, you can give them an **effective** mass.

They are massless (rest mass =0 ) because they don't interact with Higgs, but ... we knew before Higgs that they should be massless, for many reasons- one of which is that they travel at [itex]c[/itex], or that Coulomb's law make the force have infinite range, etc...

They are massless (rest mass =0 ) because they don't interact with Higgs, but ... we knew before Higgs that they should be massless, for many reasons- one of which is that they travel at [itex]c[/itex], or that Coulomb's law make the force have infinite range, etc...

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- #5

jtbell

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Please, in English, we say "rest mass", not "resting mass."

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mfb

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Ah. Thanks.

- #9

strangerep

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Heh,... and I really wish we would always say "invariant mass" instead of "rest mass".Please, in English, we say "rest mass", not "resting mass."

- #10

mfb

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If the system is not a single particle (or decay products of a single particle), "invariant mass" is clearer, but not necessary.

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I am guessing the 'mass' you are referring to is possibly the relativistic mass. It is related to energy, which is not an invariant quantity. General relativistic gravitation is related to the energy-momentum density; in that sense, for a photon gas, there would be frame involving energy density and pressure and this energy density might have an effective gravitating mass density (you still have to take the gravitational effect of pressure relativistically though).

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The invariant mass of a free photon is 0. In a medium you can define the retarded photon propagator and a spectral function from it (in thermal equilibrium it's either the analytic continuation of the in-medium Matsubara propagator to real time or equivalently the corresponding matrix element of the Schwinger-Keldysh contour propagator). If this spectral function is sharply peaked enough, you can use a quasi-particle description and define an invariant mass of the photon through the corresponding dispersion relation, otherwise this doesn't make sense und you have to use the broad spectral distribution to describe the corresponding plasmon excitations.

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