# Rest Mass of Photon.

1. Jun 5, 2012

### aleemudasir

How did we come to the conclusion photon has no rest mass? Does photon have any mass(not talking about rest mass) while moving at usual c?

2. Jun 5, 2012

### phinds

photon has no rest mass because it cannot be at rest so the question is meaningless.

At c, photon has energy equivalent of mass

3. Jun 5, 2012

### Staff: Mentor

The "rest" mass, energy and momentum of an object are related by

$$E^2 = (pc)^2 + (mc^2)^2$$

We know to some degree of precision that E = pc for photons. For example, we derive the equations for Compton scattering by assuming that E = pc for both the incoming and outgoing photon, and measurements of Compton scattering agree with those equations.

4. Jun 5, 2012

5. Jun 5, 2012

### jnorman

is the photon not at rest when it is absorbed by an atom? the mass increase of the atom should represent the rest mass of the photon... (i know this is wrong, but why?)

6. Jun 5, 2012

### Matterwave

The mass increase of the atom corresponds to the energy of the photon (divided by c^2). The photon does not "come to rest" inside the atom, it simply no longer exists.

7. Jun 5, 2012

### M Quack

When the photon is absorbed, it ceases to exist.

Similarly, when an atom drop from an excited state to a lower one and emits a photon, that photon is created. It just appears, observing a number of conservation laws (energy, momentum, etc) for the entire (multi)atom-photon system

There is no conservation law for the number of photons. You can just pull them out of your hat, provided that it is not completely black.

8. Jun 5, 2012

### danmay

It may help to think of a photon is a packet of kinetic energy that's always at c. That's all. When it's absorbed, it's being converted into other forms of energy.

9. Jun 5, 2012

### Darwin123

"Does photon have any mass(not talking about rest mass) while moving at usual c?"
Yes. A photon moving at the usual speed of c has a relativistic mass given by:
m=E/C^2
where "E" is the energy of the photon and "m" is the relativistic mass of the photon.
The relativistic mass, "m", is not zero. So a photon does have a relativistic mass.
Photons are considered "massless" because their rest mass is zero. However,
a photon can't stand still. A photon moves at a speed "c" in all inertial frames. So
the "rest mass" of a photon is not measurable, anyway. The relativistic mass of a
photon is measurable. However, it will vary with inertial frame.

10. Jun 5, 2012

### aleemudasir

If the rest mass of photon is zero then what will happen when we will use the equation
m= m(0)/ sqrt(1-v^2/c^2), which comes upto 0/0. Please explain this situation. And what will happen in this situation when we will go further to use m=E/c^2.
Thanks!

11. Jun 6, 2012

### Drakkith

Staff Emeritus
You can't use 0 as the mass in those equations, as it comes out to nonsense. (Dividing zero by zero for instance) You must use the equations that apply to a photon, such as e=pc.

12. Jun 6, 2012

### Nabeshin

You cannot go around willy-nilly applying equations. These equations were derived under certain assumptions, among them being that the object is traveling below c (or equivalently, has mass).

13. Jun 6, 2012

### aleemudasir

Even after using E=pc, the momentum p is defined as, p=mv, so, what now? There still is m!
Correct me if I am wrong.
Thanks.

14. Jun 6, 2012

### aleemudasir

Which equation do we use to infer that speed of light is the limit?

15. Jun 6, 2012

### Drakkith

Staff Emeritus
I believe it is p=h/wavelength, where h is Plancks constant.

16. Jun 6, 2012

### pervect

Staff Emeritus
People have been trying to correct you all along - you don't give the apperance of actually listening to the corrections, however...

17. Jun 6, 2012

### aleemudasir

I listen, I just want to clear my doubts. I hope that is the purpose of this forum!

18. Jun 6, 2012

### aleemudasir

OK, but it is just the rearrangement of equation λ=h/p, where p=mv.

19. Jun 6, 2012

### M Quack

No, you are wrongly simplifying a general equation to a very special case.

$p = \hbar k$ where k is the wave vector ($\frac{2\pi}/\lambda$) is used pretty much everywhere in QM, including massive particles such as electrons. Note that momentum is conserved, but not velocity, so momentum is a much more fundamental property than velocity.

In the non-relativistic limit for massive particles you get $E = \frac{1}{2m}p^2$.

For a photon, the non-relativisitc limit obviously does not make any sense. The most reasonalbe thing you can do is (as jtbell pointed out) use $E^2 = (m_0 c^2)^2 + (pc)^2$.

Now all experimental observations over a huge huge range of wavelengths/energies/wave numbers match $E=pc = \hbar \omega = \frac{h}{\lambda}$, i.e. they match the above equation with $m_0 = 0$. In other words, all available experimental data supports the hypothesis that the photon is massless.

But then again that is just a theory, just as evolution is just a theory.

20. Jun 6, 2012

### vin300

The magnitude of photon four momentum is zero, and the magnitude equals -m2c2, that implies the photon rest mass, m, is zero.