Why does Lorentz factor not hold for relativistic mass when we apply it to photons? We know that the photon itself is massless particle m_0=0. But we also know, that the mass of the objects does increase with their energy. And we know that under certain circumstances (gravity, collision with objects) the light does behave like a beam of particles that do have a mass.

Kamila Frye

Kamila Frye

Answered question

2022-10-29

Why does Lorentz factor not hold for relativistic mass when we apply it to photons?
We know that the photon itself is massless particle m 0 = 0. But we also know, that the mass of the objects does increase with their energy. And we know that under certain circumstances (gravity, collision with objects) the light does behave like a beam of particles that do have a mass.
Now, this is the equation to get the (relativistic) mass of the object with certain speed:
m = m 0 1 v 2 / c 2
We already do know some things: m 0 = 0, v = c.
m = 0 1 c 2 / c 2
m = 0 0
Expression 0 0 is not equal to 0. So what is this?

Answer & Explanation

Rene Jordan

Rene Jordan

Beginner2022-10-30Added 10 answers

Photons don't have mass, but they do have energy and momentum. And since they can be absorbed or reflected, they can transfer their momentum to whatever it is that reflects or absorbs. The amount of energy is proportional to the frequency v of the light: E = h ν, where h is Planck's constant. The momentum is p = h n u / c, in whatever direction the photon is traveling.
The formula m = m 0 / 1 v 2 / c 2 just doesn't work for particles traveling at the speed of light. That equation comes from the more general expression E 2 = p 2 c 2 + m 0 2 c 4 , where you make a substitution for p that only works with massive particles.
As for gravity, our best theory of gravity at the moment is general relativity. And the source of gravity is called the stress-energy-momentum tensor. Basically, since photons do have energy and momentum, they act as a source for gravity. (Though usually such a tiny source that you can disregard them.)

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