# Mass of a photon

1. Apr 19, 2006

### row

i dont know if this is the right section for this but here goes my doubt,I read that the rest mass of a photon is 0,but how can a photon have a rest mass if speed of light is constant and about 3*10^8m/s ?i mean if light cannot be slowed how can we define mass of a photon at 0 speed?

2. Apr 19, 2006

### Garth

You can't, the concept of rest mass for a photon is meaningless, that is why it doesn't have it.

If you like, it is similar to the two different situations: having a bank account with nothing in it, or not having any bank account at all.

The question of a photon's rest mass is like not having any bank account at all.

In either case you may say: "I have no money in the bank", similarly it is often said: "The photon has no rest mass."

Garth

Last edited: Apr 19, 2006
3. Apr 19, 2006

### Staff: Mentor

This is one reason why many people prefer to say "invariant mass" rather than "rest mass".

4. Apr 19, 2006

### pmb_phy

The term "rest mass" is a poor term to use. The better term to use is "proper mass." The proper mass of a photon is often not measured in the same was as that of, say, an electron. The expression for proper mass m0 in terms of energy and momentum is usually given by

$$E^2 - (pc)^2 = m_0^2 c^4$$

For a photon E = pc so you can see how the proper mass, m0, is zero. Recall that the mass of anything is defined by the relation p = mv. If you substitute this into E = pc and arrange terms you get E = mc2 or solving for mass m = E/c2

Pete

5. Apr 19, 2006

### Giulio B.

If photon has an energy, (and it has...when we are exposed to sun we feel hot), it must necessarily have a rest mass...or proper mass anyway, otherwise E=mc² craps...i don't understand what you are sayng about the bank account...for me seems obviusly that it has a mass.

is possible that a photon goes to a bit less than speed of light to do that? (a matematical infinity bit less...)

6. Apr 19, 2006

### Hootenanny

Staff Emeritus
Just to paraphrase what you said. You are saying that light doesn't travel at the speed of light?

~H

7. Apr 19, 2006

### Hootenanny

Staff Emeritus
If a particle has energy, doesn't mean that it must have mass;

A general energy expression in terms of momentum is given by;

$$E = \sqrt{p^{2}c^{2} + (m_{0}c^{2})^{2}}$$

Now, if we assume that a photon has zero rest mass ($m_{0} = 0$), then the term $(m_{0}c^{2})^{2}$ drops out leaving us with;

$$E = \sqrt{p^{2}c^2{2}} = pc$$

Using a manipulation of the above equation

$$p = \frac{E}{c}$$

And Plank's relationship;

$$E = hf \Leftrightarrow E = \frac{hc}{\lambda}$$

The momentum of a photon is given by

$$p = \frac{h}{\lambda}$$

Thus, a particle with no mass can have energy.

~H

8. Apr 19, 2006

### Staff: Mentor

You have to be careful about how you apply $E=mc^2$, depending on whether you consider $m$ to be "invariant mass" or "relativistic mass."

If $m$ is "invariant mass", then $E$ is the "rest energy" of the particle, that is, the energy that the particle has when it's at rest. Photons don't have "rest energy", only kinetic energy, and their "invariant mass" is zero, so $E=mc^2$ reduces to 0 = 0.

If $m$ is the "relativistic mass", then $E$ is the total energy of the particle, that is, rest energy plus kinetic energy. In this case you can use $E=mc^2$ to calculate the "relativistic mass" of a photon, which is obviously not zero.

When most physicists talk about "mass", they usually mean "invariant mass." Most of these physicists prefer not to use the concept of "relativistic mass" at all. Some do use "relativistic mass", and you occasionally see arguments here about which kind of mass is "better."

The basic problem is that there is no single quantity in relativity which has all the properties of the classical "mass." People disagree on which of those properties are more important, and so they disagree on which kind of mass is more important or useful in relativity.

Last edited: Apr 19, 2006
9. Apr 19, 2006

### clj4

10. Apr 19, 2006

### pmb_phy

There is absolutely no reason to make such an assumption, expecially since its wrong.
E is inertial energy, not rest energy. You can have one without the other. Some people call E the total energy when it's really the energy of a free particle. If the particle has a non-zero potential V then the total energy W is W = E + V. I call E inertial energy so one doesn't fall into this trap. The term appears in the American Journal of Physics in the same year they nuked Japan.
Not if the proper mass of the photon is exactly zero.

Pete

Last edited: Apr 19, 2006
11. Apr 19, 2006

### pmb_phy

Take a look at MTW. At times they use "mass" to mean "relativistic mass" (references provided upon request).

One has to be careful in such arguements because taking one side might imply to others that the other side is wrong somehow. I've always argued that the correct response to the question "does mass depend on velocity" is to ask "by mass do you mean proper mass or inertial mass?" Then once that is cleared up there is no arguement. I do hold that relativistic mass has all the properties associated with the term "mass" as usually assumed and that proper mass is very important as well. If I had it my way then $\mu$ would stand for proper mass and m for inertial mass. If we use the greek symbol $\tau$ for proper time and "t" for "coordinate time" then we should use $\mu$ for proper mass and "m" for inertial mass (what one might call coordinate mass. - This is just my very humble opinion which I believe I can strongly support.
Sure they do - relativistic mass. Why do you disagree on this point?

Pete

12. Apr 19, 2006

### bernhard.rothenstein

13. Apr 19, 2006

### rbj

i disagree. i think that it's more like the saying that there is a bank account, but the balance is zero.

https://www.physicsforums.com/showpost.php?p=957440&postcount=24

and some guy, posted something about NIST specifying a non-zero upper bound for the rest mass of a photon:

https://www.physicsforums.com/showpost.php?p=957736&postcount=27

which i find very hard to swallow.

the reason why the photon (or any other "massless particle") has a concept of "rest mass", but such rest mass must be zero is that the speed of the particle is the speed of E&M propogation (light) and that the rest mass is:

$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}} = \frac{h \nu}{c^2} \sqrt{1 - \frac{v^2}{c^2}}$$

plug in $v = c$ and you get $m_0 = 0$.

14. Apr 19, 2006

### pmb_phy

A very odd way of looking at it. Something seems wrong but I can't quite put my finger on it.

Pete

Last edited: Apr 19, 2006
15. Apr 19, 2006

### Tom Mattson

Staff Emeritus
Cute. But the suggestion isn't necessarily as silly as it sounds. It depends on the sense in which the term "speed of light" is used.

The "speed of light" in SR need never have been identified with the "speed of light" from Maxwell's equations. That is just a leftover relic from the discovery of SR from electrodynamics. SR only requires an invariant speed, not some special one.

I think that when Giulio says, "is possible that a photon goes to a bit less than speed of light", he really means "is possible that a photon goes to a bit less than the invariant speed of SR".

16. Apr 19, 2006

### Hootenanny

Staff Emeritus
Ah, I feel rather stupid now(and a bit guilty) :shy: Apologies Giulio.

~H

17. Apr 19, 2006

### Staff: Mentor

Classical mass is an invariant property of a particle or object, leaving aside non-closed systems such as a rocket expelling fuel. Relativistic mass is not.

That upper bound is a statement about the precision of experimental data relating to the (invariant) photon mass. It means that experiments done so far cannot detect a invariant photon mass less than $6 \times 10^{-17}$ eV. Or, to put it another way, if that mass were greater than $6 \times 10^{-17}$ eV, some experiments would have seen its effects by now. It's like the upper end of an error bar on a data point on a graph.

Last edited: Apr 19, 2006
18. Apr 19, 2006

### pmb_phy

Something being coordinate dependant is not a problem in SR. In fact that's all we are able to measure. Therefore it is not a problem.
Yeah. Jackson covers all of that.

Thanks

Pete

19. Apr 19, 2006

### robphy

With this clarification, the correct statements are:

• "proper mass" or "rest mass" is an invariant property of a particle alone.
• "relativistic mass" (being an observer-dependent/coordinate-dependent quantity) is not an invariant property of the particle alone. Instead, it is a property of the "pair consisting of the particle and an observer". Both need to be specified. If you keep the qualification of "a particle and an observer", then the property is actually an invariant [since it's a scalar formed from two distinct tangent vectors (that of the particle and that of the observer) and the metric].

An analogous set of statements is:
• the "norm" or "magnitude" of a vector is an invariant property of a vector alone.
• the "projection of a vector onto an axis" (being an observer-dependent/coordinate-dependent quantity) is not an invariant property of the vector alone. Instead, it is a property of the "pair consisting of the vector and an axis". Both need to be specified. If you keep the qualification of "a vector and an axis", then the property is actually an invariant [since it's a scalar formed from two distinct vectors (that of the vector of interest and that of (say) a unit vector along an axis) and the metric].

20. Apr 19, 2006

### Staff: Mentor

I wasn't addressing the question of whether it is a problem or not. I was addressing the question of whether classical mass has a property that relativistic mass does not, which I thought was the point of the exchange at the bottom of posting #11, to which I was responding.