# Why does Lorentz factor not hold for relativistic mass when we apply it to photons? [duplicate]

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: $$m=\frac{m_0}{\sqrt{1-v^2/c^2}}$$ We already do know some things: $m_0 = 0$, $v = c$. $$m=\frac{0}{\sqrt{1-c^2/c^2}}$$ $$m=\frac{0}{0}$$

Expression $\frac{0}{0}$ is not equal to 0. So what is this?
I'll remind you of those weird clock that are run by light - I failed to find any images, but the clock principle was similar to windmill. And it works.
Also there was an idea to make long-distance autonomic spaceships that are pushed by light rays from stars.

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## marked as duplicate by John Rennie, Waffle's Crazy Peanut, Qmechanic♦May 9 '13 at 18:50

It's not based on mass by which photon curves spacetime. It's based on its own energy (specifically, the stress energy tensor) which is blamed for the curvature. One more thing, the relativistic mass isn't applicable for photons (directly as you've done), instead the energy-momentum relation $E^2=p^2c^2+m^2c^4$... – Waffle's Crazy Peanut May 9 '13 at 16:07
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 $\nu$ of the light: $E = h\, \nu$, where $h$ is Planck's constant. The momentum is $p = h\, nu / c$, in whatever direction the photon is traveling.
The formula $m = m_0 / \sqrt{1-v^2/c^2}$ 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.