# Mass-energy and massless particles

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1. Jun 12, 2015

### Swetha.M.L

may i ask you something? if there is any wrong excuse me.
according to mass-energy eqation mass &energy are not different but two forms of the same.
photon ,graviton... are the mass less particles but photon is a form of energy. can you explain why photon is massless?

2. Jun 12, 2015

### Staff: Mentor

Mass has energy associated with it via E=mc2, but not all energy has mass associated with it via m = E/c2.

3. Jun 12, 2015

### Simon Bridge

Not everything has all forms of energy at the same time. Photons have kinetic energy but not mass-energy.

4. Jun 12, 2015

### wabbit

You're probably thinking of the relation $E=mc^2$ - but this applies only in a frame where the particle is at rest, and photons are never at rest since their velocity is $c$ in any frame.

The general equation is $E^2=m^2c^4+p^2c^2$ where $p$ is the momentum of the particle, and for a massless particle this reduces to $E=pc$. In other words photons do not have mass but they have momentum, hence (kinetic) energy, given by that formula.

5. Jun 12, 2015

### Staff: Mentor

It depends on what you would consider an "explanation". Experimentally, photons are known to be massless to a very high accuracy. Theoretically, photons are modeled as massless because that's the model that best matches experiments.

Last edited: Jun 12, 2015
6. Jun 12, 2015

### DaveC426913

Are you asking how a photon - which is energy - can be on one side of the mass-energy formula, yet not have any mass?

Isn't it kind of like asking how ice and steam are two forms of the same thing, yet only one of them will break your knuckles if you take a swing at it?

i.e. the very nature of the "two" in "two forms of one thing" is that they have different properties.

7. Jun 12, 2015

### Topolfractal

Well there are several reasons why a photon is considered is considered massless .
1. Yes through experimentation.
2. Special Relativity implies it as shown below.
- E = m(c^2)/ ( sqrt ( 1-(v^2/c^2))
Multiplying both sides by sqrt(1-(v^2/c^2))
E(1-(v^2/c^2))= m( c^2)
Okay when v= c the E side equals (0)
So therefore the speed of light not being zero implies the mass is zero.
Also this implies it does have no rest energy and therefore it's energy is entirely kinetic in nature.

8. Jun 12, 2015

### Staff: Mentor

This is not a valid argument, because your formula for $E$ is only valid for an object with $v < c$.

Also, it's obvious that there's a problem when your formula says $E = 0$ for $v = c$, since photons do not have zero energy. Saying that $m = 0$ makes the equation work out is irrelevant, because the equation still says $E = 0$. But photons have $E > 0$ with $m = 0$; your argument does not explain how that can be.

9. Jun 12, 2015

### Topolfractal

I agree and see my error