About light, photon mass and E=mc^2

1. Mar 12, 2015

akashpandey

Light is form of energy.
Accordingto einstine theory if we concentrate the energy of light we have to get mass of light.

But their is no mass of light.

Acc.to E⇒mc2.

2. Mar 12, 2015

Staff: Mentor

That's not what E=MC^2 says.

It says mass is a form of energy, like chemical energy is a form of energy, or gravitational potential energy is a form of energy. It does not say energy is mass.

Thanks
Bill

3. Mar 12, 2015

TangledMind

Which mass? Rest mass? Show me a photon at rest :D

The concept of mass is a little bit more complicated than "m". And no, I do not fully understand it. Not even nearly so.

You may also be interested in: Invariant mass, mass in special relativity, gravitational lens, mass in general relativity, ...

Last edited: Mar 12, 2015
4. Mar 12, 2015

DennisN

See the last equation on this page: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4
Photons have no mass (rest mass), so plug in m0 = 0 into that equation and see what the energy expression becomes...
E = ? (insert equation here )

5. Mar 14, 2015

vanhees71

In modern physics (i.e., since 1908, when Minkowski discovered the mathematical structure behind the special theory of relativity) we are used to call mass the quantitity that's more accurately called "invariant mass". This is a quantity that is independent of the object's velocity and thus the same in any inertial frame of reference.

For a classical point particle, this invariant mass (which is the only sensible definition of mass one can think of) is defined as
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
where $x^{\mu}$ is the particle's position in four-dimensional spacetime, i.e., $x^0=c t$ ($c$: speed of light, $t$ coordinate time wrt. to the frame of reference we work in, and $\tau$ the proper time of the particle). Since by definition of proper time
$$\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x_{\mu}}{\mathrm{d} \tau}=c^2$$
we have
$$p_{\mu} p^{\mu}=m^2 c^2,$$
which shows that $m^2$ is a scalar quantity. Since $p^0=E/c$ this expression, split in temporal and spatial components, reads
$$\frac{E^2}{c^2}-\vec{p}^2=m^2 c^2.$$

A photon is somewhat tricky. You cannot fully understand it without quantum field theory. It is described by a massless quantum field with spin 1, the electromagnetic field. An energy-momentum eigenmode of the field is characterized by the three-momentum eigenvalues $\vec{p}$, and the energy is given by the relation
$$\frac{E^2}{c^2}-\vec{p}^2=0.$$
This explains, why the field is called "massless", because that's a similar relation as for classical particles with an invariant mass of 0.

Of course, one shouldn't call the "invariant mass" "rest mass", because a massless quantum can not be at rest. It always moves with the speed of light wrt. any inertial observer.

6. Mar 14, 2015

DennisN

Good point, and akashpandey, I hesitated at first, but wrote (rest mass) above since the article I linked to used the word "rest mass".