# Do photons possess energy?

1. Aug 6, 2007

### Seele

This question has been plaguing me for a few days now, I am fully aware that light quanta are massless, though through that, I have been scratching my head to figure out if photons have energy. My reference is Einstein's E=mc2, where it states that mass and energy are interchangeable, and thus linked to one another. So since light is without mass, wouldn't it also be without energy? Or is the frequency of a photon its definition of energy?

2. Aug 6, 2007

### chroot

Staff Emeritus
When you stand in the sunlight, you feel warmth. This obviously implies that photons do indeed carry energy. The equation $E = mc^2$ only indicates that matter and energy are "interchangeable," not that the presence of one implies the presence of the other.

- Warren

3. Aug 6, 2007

The photon is the particle that is moved by the E.S. waves and also is a wave, so it would seem that it has to have energy.

Did I get that right?

4. Aug 6, 2007

### chroot

Staff Emeritus
I have no idea what an "E.S. wave is," nor what it means to be "moved" by one. So no, I don't think you've got it right.

- Warren

5. Aug 6, 2007

Sorry, I'm lazy. Electromagnetic Spectrum. But it is a wave in any regard.
And moved probably wasn't the right word. It is a wave, in essence.

Correct?

6. Aug 6, 2007

### ZapperZ

Staff Emeritus
This is rather puzzling especially if you have read a bit about the history of physics. A "photon" or a "corpuscular of light" (as it was originally known) was defined as a quanta of energy. This is how Einstein was treating it when he tackled the theory of Photoelectric Effect. So what you are asking here is puzzling because that was HOW a photon was defined in the first place, as a quanta of energy. So of course it has energy since that is how its existence was defined.

You may want to read the FAQ in the General Physics section with regards to how you're using the E=mc^2 equation wrongly in this case.

Zz.

7. Aug 6, 2007

### ZapperZ

Staff Emeritus
That is even MORE wrong than before. You are saying that a photon is " moved by the E.S. waves", which means that the EM wave is external to the photon. This is highly incorrect. There aren't two separate entities here.

Zz.

8. Aug 6, 2007

No, of course not! That wasn't what I was saying. The photon is a particle OF the wave, not in it.

Sorry for not being clear.

9. Aug 6, 2007

### ZapperZ

Staff Emeritus
Does that actually make any sense to you? ".. a particle OF the wave..."?

What exactly does that mean? Are you imagining that this is like the water molecules being the "particle" of the "water wave"? If you are, this isn't correct either.

Zz.

10. Aug 6, 2007

### Gokul43201

Staff Emeritus
This only gives you the rest energy of a photon. The total energy is given by $E=\gamma mc^2$, but $\gamma$ for a photon is infinitely large, leaving you with an indefinite form: $0 \cdot \infty$. So really, you have to find some other way to compute the energy of the photon, and that is exactly what is achieved through $E=h \nu$.

11. Aug 6, 2007

### nrqed

E=mc^2 is a special case of a more general equation!!

The real equation is $E = \sqrt{ m^2 c^4 + p^2 c^2}$ where p is the magnitude of the momentum vector. Only for massive particles can one take p=0 and then recover E = mc^2. For light, m =0 but p cannot be set to zero (light is always moving) so that for light E = cp.
Notice that people sometimes then argue that it does not make sense fo rlight to have momentum since (they say) p = mv and m is zero. but again, p=mv is a special case that cannot be used for light.

12. Aug 11, 2007

### premagg

see,photons have rest mass of zero but as soon as they travel with the speed of light their mass is given by m=m0/(a-v^2/c^2)0.5,this is 0/0 from hence can have a finite value.As the mass is now not zero,the energy is given by E=mc^2

13. Aug 11, 2007

### ZapperZ

Staff Emeritus
Er.. I think you need to read the FAQ as well.

Note that photons ALWAYS travel at c. So it is meaningless to say "... but as soon as they travel with the speed of light..." because they never do anything else.

Zz.

14. Aug 13, 2007

### Gza

seriously?

15. Aug 13, 2007

### Nesk

The question is pretty well answered here:

"[url [Broken]"]http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html [Broken][/URL]

Last edited by a moderator: May 3, 2017
16. Aug 14, 2007

### Dahaka

first, let's set it straight that a light is neither a particle nor a wave, it is BOTH, whenever experiments are carried out, it is OBSERVED as either or, but it is still a wave-particle under the de broglie duality, the heisenberg uncertainty principle prevents us from observing it as both, as the more definite its position gets, it becomes a particle and not a wave (since a wave cannot occupy a single point) and vice versa, we simply use wave packets, determine both conjugate variables to lesser certainties and predict their values through percentages, light does not "possess" energy, it "is" energy, since it has no mass it is composed entirely of kinetic energy, but the gauge field equations show how photons do not have mass, i have questioned the why of the "masslessness" too, but i have not found a good answer to it other than those of special relativity

17. Aug 14, 2007

### orange juice

1) light does have energy;
2) you can get the E by both E=m*(sqr c) and E=h(bar)*w, BUT you should notice that m=m0/{1-[(v*v)/(c*c)]}, where m0 is the static mass and for photon, m0=0 (it's what you meant by "massless"); but for photon, v=c, so you get m=0/0, and you can't get E by the first way.
3) so for the photon, you could only use E=h(bar)*w.(but that doesn't mean E=m*(sqr c) is wrong, it's just that you can't get E via it in this case).