# Photons have no mass?

1. May 31, 2010

### Tach

Hello, I'm quite confused about something. Maybe you all can help :)

I'm not sure I understand why the science community seems to assume photons have no mass. I can understand how it wouldn't make sense to say a photon has mass by the mathematics that is used to describe the energy of a particle, because it has no rest mass, but if a positron can interact with an electron and annihilate both particles, creating photons, then if photons have no mass, we have just destroyed mass. And if photons result from 'destroying mass' then shouldn't we also be able to 'create mass' from photons?

And in relation to the above paragraph and following from the same train of thought, then since photons travel at a constant speed and can create mass, then wouldn't it be wise to assume a constant mass for photons? Since relativistic mass wouldn't apply because of non-changing speeds.

I'm quite confused. Or perhaps the whole concept of mass is flawed to begin with? Or am I not understanding the basics right?

2. May 31, 2010

### nicksauce

Sure, you can create and destroy mass. The "law of conservation of mass" is a classical approximation that is not valid in quantum theory.

3. May 31, 2010

### ansgar

according to relativity, mass is just a type of energy.

this has nothing to do with quantum

4. May 31, 2010

### Char. Limit

It actually has to do with mass-energy equivalence, I believe. In the creation of a photon, mass is converted to energy (and multiplied by c^2 in the process), creating the familiar packet of energy we call a photon, and getting rid of the massive particle in the process.

5. May 31, 2010

### Tach

Thanks for your responses. I guess I'm having a hard time accepting this since I can't think of all the energy of the world around me in terms of a 'unit of space' or 'unit of some physical size'. I don't know how to imagine photons if they don't occupy a space and thus have some kind of mass; although I associate mass with particles occupying a space.

So it's as if photons are just a 'magical particle' because they don't actually exist in space? Such an idea really bugs me and seems extremely counter-intuitive and that there is some kind explanation of how photons do 'occupy a space', but that's assuming a lot, I suppose. I guess it's possible that some things can exist without making any rational sense.

6. May 31, 2010

### pallidin

That an electron, for example, can exist and yet have no definitive spacial representation is a problem. Welcome to the quantum world.

Last edited: May 31, 2010
7. May 31, 2010

### Passionflower

Two photons going in opposite directions certainly have mass.

8. May 31, 2010

### Naty1

"So it's as if photons are just a 'magical particle' because they don't actually exist in space?"

Perhaps you could say that, but nobody understands exactly what ANY fundamental "particle" is.....photons are not necessarily more 'magical' than, say, an electron, that "doesn't actually exist in space" either but is rather an "electron probability cloud". For "heaven's sake" why can't we precisely locate a simple electron?? Oh yes, Heisenberg uncertainty.....

We have lots of mathematic behavior descriptions, but alas, no really good ones of fundamental origins of particles.

".. Such an idea really bugs me and seems extremely counter-intuitive and that there is some kind explanation of how photons do 'occupy a space', but that's assuming a lot, I suppose. I guess it's possible that some things can exist without making any rational sense..."

Richard Feynman said "Nobody understands quantum mechanics"...the science of the very small....and that's still true today. But then general relativity is hardly "intuitive" either.

Another way to think about all the above is that everything is a vibrating string...a resonance of energy and what you THINK you see as a 'solid' mass is 99.999999999% empty space....gravitons are also massless and move at lightspeed like photons...so don't pick on the poor photon....and maybe you won't like black holes either as their singularity is also a point of no volume...

We apparently have a LOT more to learn about the relationship between space, mass, energy and time.

9. May 31, 2010

### Phrak

This would make the PF physics FAQ fairly meaningless without a proper make-over. Where two photons can 'go in opposite directions' so can a single photon, so that anything other than an ideal planar wave endows regions of spacetime with nonzero intrinsic mass.

10. May 31, 2010

### Rasalhague

That's how I understand it. Mass (i.e. rest mass), energy (i.e. total energy) and momentum are each conserved in special relativity. In this case, the mass of the system of particles that annihilated is the same as the mass of the system of photons they produce, isn't it? I have trouble understanding the concepts of mass-energy equivalence and mass turning into energy or vice-versa, but, following previous discussions here, I think these viewpoints depend on implicitly shifting the reference of the word. For example, Taylor/Wheeler say in Spacetime Physics

"Does the invariance of rest mass mean that rest mass cannot change as a consequence of a collision? No. Rest mass often changes in an inelastic encounter. Example 1: Collision between two balls of putty--hotter and therefore very slightly more massive after the collision than before."

But if I've got this right, then the "change" only comes from the fact that they start out talking about the sum of the rest masses of the balls (not the rest mass of the system), then switch to talk about the rest mass of the combined ball which latter is equal to the mass of the system throughout.

But I suppose it just depends how the system is defined, for conservation of mass, as for conservation of energy and conservation of momentum.

Last edited: May 31, 2010
11. May 31, 2010

### Rasalhague

But if they're going in opposite directions, the vector sum of their 3-momenta will be zero, so the magnitude of the total 2-momentum is zero. This being the case, the rest mass of the system of two photons is just the sum of their energies.

12. May 31, 2010

### D H

Staff Emeritus
Rest mass is not a conserved quantity. Think of an annihilation event.

13. May 31, 2010

### Rasalhague

Taylor/Wheeler, excercise 97, deals with a positron and electron annihilating to produce radiation: "By considering the center-of-momentum frame (the frame of reference in which the total momentum of the initial particles is equal to zero), show that it is necessary for at least two gamma rays (rather than one) to result from the annihilation." The argument they give is that 3-momentum is conserved, therefore if 3-momentum is initially zero, it must remain zero. The 3-momentum of one gamma ray is equal to its energy and not zero, therefore there must be two (in opposite directions, right?).

But if 3-momentum is conserved and energy is conserved, rest mass must also be conserved, since it's a function only of momentum and energy.

Presumably there's a different way to describe this event, defining--or redefining--the system in such a way that it doesn't conserve energy or rest mass. Could you explain it from this viewpoint; and would you see 3-momentum as being conserved? How would you answer the question about the number of rays emitted from this viewpoint?

Last edited: May 31, 2010
14. May 31, 2010

### D H

Staff Emeritus
You cannot treat an ensemble as one. Just because the energy and momentum of one photon are directly related to one another does not mean there is any direct relationship between the total energy and total momentum of a set of more than one photon.

15. May 31, 2010

### Rasalhague

Could you elaborate? If by "no direct relationship" you mean that energy of an ensemble of massless particles doesn't necessarily equal their total momentum, that agrees with what Passionflower wrote, as I understand it.

The message I get from Spacetime Physics is that the equation

$$m^2 = E^2 - p^2$$

applies to any system (c=1, inertial frame), whether composed of one or more particles. Is that right?

16. May 31, 2010

### Tach

Ah, okay. I think I understand now. So I shouldn't think of mass as 'occupying space', but as a construct to describe a mathematically observed difference between differing objects due to gravitational forces.

So when I think of photons not occupying space, that is incorrect to say, in this sense. And in that regard I'm going to imagine photons as occupying space with all other matter then. That seems to be the only way to make sense of this for me.

17. May 31, 2010

### Staff: Mentor

Passionflower is correct. The invariant mass of a system is preserved even across an annihilation event. The invariant mass of a system of particles is the norm of the sum of the individual particle's four-momenta. This can be different from the sum of the norms of the individual particle's four-momenta, which is not conserved. When people speak about mass not being conserved they are almost always erroneously identifying the latter quantity with the system's mass.

Last edited: May 31, 2010
18. May 31, 2010

### Count Iblis

Yes, or they implicitely redefine the system. So, when an atom emits a photon, you can focus on the atom, instead of taking your system to be the atom-photon system.

19. May 31, 2010

### ansgar

No they are not, only the invariant mass, momentum and total energy is conserved. Besides this, energy and momentum are frame dependent.

20. May 31, 2010

### Rasalhague

That's what I said; rest mass = (frame)-invariant mass. Did you mean "yes they are...are conserved" or "no they are not...are not conserved"?

Yes.

21. May 31, 2010

### Rasalhague

This must be what's going on in the Taylor/Wheeler inelastic-collision example I quoted in #12. That implicit redefinition threw me at first because it seemed inconsistent in the context of a chapter that discusses the conservation of energy and 3-momentum. It seemed to be saying that rest mass, unlike these quantities, is not conserved. But if the system is redefined so that mass is not conserved, then presumably energy and 3-momentum can't both be conserved either.

22. May 31, 2010

### starthaus

True. Interestingly, due to the fact that total energy is conserved:

$$\Sigma(\gamma(v_i)*m_i*c^2)=constant$$

it follows that relativistic mass is conseved since the above can be written as:

$$c^2\Sigma(\gamma(v_i)*m_i)=constant$$ !!!

Both energy and momentum are conserved:

$$\frac{d}{dt}\Sigma(\gamma(v_i)*m_i*\vec{v_i}=\Sigma \vec{F_i}$$

If the sum of the forces acting on the system is zero, then:

$$\Sigma(\gamma(v_i)*m_i*\vec{v_i}=constant$$

Last edited: May 31, 2010
23. May 31, 2010

### Phrak

This is why I said the physics forums FAQ, found https://www.physicsforums.com/showthread.php?t=104715", needs a slight make-over. There are few scenarios to examine.

One is the electron-positron annihilation into two gamma particles. The wave functions are even upon interchange of the two particles. There's no way to distinguish one from the other. Each photon is equally likely to have momentum to the left as to the right until measured.

The second scenario is a confined photon such as found within a laser cavity, as we've talk about before, with an uncertainty in position in the order of the cavity length. Some might argue that the photon cannot be treated in isolation without considering the cavity walls, so...

number three is the case of self interference. If you like, the two slits of an interference experiment. Withing some given region of space the momentum contributions from each slit add vectorially. So that the energy-momentum equation is satisfied, the photon must have intrinsic mass. Only under the ideal condition of a planar propagating wave is the momentum constant and the photon massless.

The best I can discern, is that in general, most photons have mass, where to say the photon has zero intrinsic mass refers to an idealized state of the particle.

Last edited by a moderator: Apr 25, 2017
24. Jun 1, 2010

### Rasalhague

Are you agreeing that "rest mass, unlike these quantities, is not conserved" (and redefining the system, as Count Iblis puts it), or just agreeing that the quote seems to be saying this? If the former, how does this tally with

How can rest mass--which is defined sqrt(E2-p2)--not be conserved if E and p are? The square root of the square of a constant minus the square of another constant is also a constant.

25. Jun 1, 2010

### Char. Limit

I thought that for light, E=pc, and thus when c=1, then E=p and m^2=E^2-p^2 cancels to zero... am I wrong?