Do Photons Have Mass? Confused Student Seeks Answers

[Rasalhague]

Say in some units where c=1 that an electron and a positron have four-momenta of
p_{e^-}=p_{e^+}=(1,0,0,0)

The mass of each particle is:
m_{e^-}=m_{e^+}=|p_{e^-}|=|p_{e^+}|=|(1,0,0,0)|=1
And the mass of the system is
m_{s}=|p_{e^-}+p_{e^+}|=|(1,0,0,0)+(1,0,0,0)|=|(2,0,0,0)|=2

The electron and positron anhilate and produce two photons of four-momenta:
p_{A}=(1,1,0,0)
p_{B}=(1,-1,0,0)

The mass of each particle is:
m_{A}=m_{B}=|p_{A}|=|p_{B}|=|(1,1,0,0)|=|(1,-1,0,0)|=0
And the mass of the system is
m_{s}=|p_{A}+p_{B}|=|(1,1,0,0)+(1,-1,0,0)|=|(2,0,0,0)|=2

Would one or both of the electron and positron need to have some 3-momentum to begin with if they're to collide and annihilate, or does the scenario begin, in some sense, after they've collided and stopped moving towards each other but before they annihilate?

 

[Rasalhague]

filegraphy, does this Minkowski diagram help? It shows the 4-momenta of a pair of photons of equal energy traveling in opposite directions in a frame of reference in which the vector sum of their 3-momenta is zero. The timelike axis is vertical, and the spacelike axis horizontal.

The mass, m₁, of Photon 1 is

m_1=\sqrt{(E_1)^2-(p_1)^2} = 0

where E₁ is its energy and p₁ is the Euclidean norm (magnitude) of its 3-momentum, and E₁ = p₁. Similarly for Photon 2. In this case, E₁=E₂, and p₁ = p₂. The only difference between the photons is that their 3-momenta are in opposite directions.

The mass of the system of two photons is

m_s=\sqrt{(E_1+E_2)^2-\left \| \vec{\textbf{p}}_1+\vec{\textbf{p}}_2 \right \|^2}

=\sqrt{(E_1+E_2)^2-(\vec{\textbf{p}}_1+\vec{\textbf{p}}_2) \cdot (\vec{\textbf{p}}_1+\vec{\textbf{p}}_2) }

The reason it's not zero is that 3-momentum is a 3-vector, and non-zero vectors can add to zero. Here, the 3-momenta of the two photons are equal in magnitude but opposite in direction, so they sum to 0. On the other hand, energy doesn't care about direction in space; here it's just a sum of two positive numbers. There's no 3-momentum to cancel it out, so the mass of the system is sum of the energies, and thus not zero.

 

[Rasalhague]

Another diagram. You can visualise addition of 4-momenta like this on a Minkowski diagram. Put the head of one vector to the tail of the other and draw the vector representing their sum from the tail of the first to the head of the second. See how the 4-momentum of the system, unlike those of its individual constituents, has no spacelike component (3-momentum) to cancel out the timelike component (energy) in the equation for mass (which is the magnitude of 4-momentum). The energy is the system in this particular reference frame, its center-of-momentum frame (that is, one where it's 3-momentum is zero), is equal to its rest mass.

 

[Phrak]

Dr Greg and Rasalhague,

OK, I'm properly chastised. Not all scalars are rank zero tensors. This is a side issue in my view.

Velocity doesn't combine through addition. Neither does intrinsic mass combine through addition. Perhaps I'm alone in this, but shouldn't we be interested in better objects than these. As we should prefer, on a four dimensional manifold to define velocity as 3 components of a four velocity, intrinsic mass should have no special place, but a derived quantity from four dimensional objects.

 

[DrGreg]

As we should prefer, on a four dimensional manifold to define velocity as 3 components of a four velocity, intrinsic mass should have no special place, but a derived quantity from four dimensional objects.

That is a good way of looking at it. Each particle has a 4-momentum, which is a tensor and therefore a coordinate-independent concept, and (in special relativity, in a "closed" system) 4-momentum is additive.

(For continuous distributions you need to consider the stress-energy-momentum tensor field instead of particles' 4-vectors, and, in "non-static" GR, conservation has to be expressed as a (local) differential equation as you can't "add" over a non-local region of spacetime.)

 

[Phrak]

That is a good way of looking at it. Each particle has a 4-momentum, which is a tensor and therefore a coordinate-independent concept, and (in special relativity, in a "closed" system) 4-momentum is additive.

I realize that the 4-momentum as the fundamental object has been the theme in the last few pages of this thread. But is the 4-momentum in units of MD/T a tensor under a general coordinate transformation or only under the Poisson group? That is: Is it a tensor or a tensor density? Back to the books for me--when I can manage. In the mean time do you happen to know how this is resolved?

(For continuous distributions you need to consider the stress-energy-momentum tensor field instead of particles' 4-vectors, and, in "non-static" GR, conservation has to be expressed as a (local) differential equation as you can't "add" over a non-local region of spacetime.)

Many, many interesting things come out of examining this simple mass-energy-momentum equation. One is as you seem to mention now, the idea that a photon is not really a point object, or a world line in spacetime, but classically an extended object, so we have to ask if assigning m²=E²-p² properly applies over the whole class of extended fields for light upon Minkowski space, let alone GR. More, is it inapplicable only to confined electromagnetic radiation such as a cavity resonator? Next, we might want to push a little beyond this comfortable equation of 4-momentum, just slightly, and look at 4-angular momentum. This one doesn't have the visceral attraction of m²=E²-p² but deserves better attention in my mind. Another is the interesting Lorentz group duality that we can see implied by Rasalhague's graph of energy and momentum. We can equally presume that the vector space of the graph is space-time, or momentum-energy.

However, do you happen to know how to resolve whether p^mu, in units of momentum, is a tensor or a tensor density?

 

[DrGreg]

I have to confess I'd never encountered the concept of "tensor density" before, so I had to look it up. Given my lack of experience, all I can say is, as far as I know, 4-momentum is a tensor, and it applies to discrete point-particles, and, via summation, to collections of discrete particles. As I understand it, you can't use a rank-1 tensor (i.e. vector) when you move on to continuous distributions, you need a rank-2 tensor.

By the way, in special relativity the angular momentum of a point-particle (without spin) is described, not by a rank-1 4-vector, but by an antisymmetric rank-2 tensor

M^{ab}=x^a p^b - x^b p^a​

Roger Penrose calls it "6-angular momentum" as it has 6 independent components, 3 of which correspond to 3-angular momentum, the other 3 forming the conserved 3-vector tp - Ex.

 

[Phrak]

Thanks, Dr Greg. Your reply tells me that my concern over densites is less a matter than I had thought, though I'll continue to learn about them.

And I see the angular momentum, being antisymmetric, may have an a particularly simple form in lower indeces with a form unchanged over displacements on a curved manifold as:

M_{ab}=x_a p_b - x_b p_a = x_a \wedge p_b​

I say 'may have', because raising and lowering indeces involves the metric, that itself, is a tensor density.

 

[Phrak]

I say 'may have', because raising and lowering indeces involves the metric, that itself, is a tensor density.

I don't like leaving this error hanging at the end of a thread. The determinant of the metric is a tensor density. The metric itself is a well behaved tensor (with a density weight of zero).

 

[DrGreg]

I don't like leaving this error hanging at the end of a thread. The determinant of the metric is a tensor density. The metric itself is a well behaved tensor (with a density weight of zero).

That makes more sense!

I was careful to define

M^{ab}=x^a p^b - x^b p^a​

in special relativity only, because on a curved manifold p^a is a vector (in the tangent space) but x^a is not, so the expression doesn't make sense as a tensor. I've never really looked into the question of angular momentum on curved manifolds. (In fact I think I read somewhere it's only conserved when the metric itself has circular symmetry, but I could be wrong.)

 

[Phrak]

That makes more sense!

I was careful to define

M^{ab}=x^a p^b - x^b p^a​

in special relativity only, because on a curved manifold p^a is a vector (in the tangent space) but x^a is not, so the expression doesn't make sense as a tensor. I've never really looked into the question of angular momentum on curved manifolds. (In fact I think I read somewhere it's only conserved when the metric itself has circular symmetry, but I could be wrong.)

Sorry to have mislead you.

It took me a little while to properly notice this:

"Roger Penrose calls it "6-angular momentum" as it has 6 independent components, 3 of which correspond to 3-angular momentum, the other 3 forming the conserved 3-vector tp - Ex."

I think there is an angular momentum continuity equation lurking in this. Very interesting. Taking partial derivatives with respect to coordinates, and grouping all the derivatives of XP terms on one side will leave the enigmatice TP and XE terms on the other side. Physically, I have no idea what TP and XE should mean! Eventially we should be able to say that the change in angular momentum at a point p, is equal to the inflow of something like TP+XE, more or less. This will take some time to sort out.

By the way, are you using lower case Latin to indicate non-coordinate bases?

This is a lot of stuff all at once. I'm not prepared to comprehend the most of it. As my references I have Sean Carroll, Wald and MWT. Where did you find mention of angular momentum?

 

[DrGreg]

I got this definition from Penrose's The Road to Reality, p.437 (curiously not listed under the index entry for angular momentum!) He comments that the conservation of tp − Ex, when summed over multiple particles, amounts to saying that the centre of (relativistic) mass \Sigma(E\textbf{x})/\Sigma E moves with constant velocity.

(Don't overlook the fact that the tensor has another 3 components that form the 3-angular momentum \textbf{x} \wedge \textbf{p}.)

To be honest I've never studied this in any depth, I just remember reading about it.

There's no significance in my use of a,b as indexes; I'm just copying Penrose. They range over all 4 dimensions.

 

[Rasalhague]

There's no significance in my use of a,b as indexes; I'm just copying Penrose. They range over all 4 dimensions.

Besides the convention Phrak mentioned, I've seen some authors use Latin indices to mean only spatial components: 1,2,3. I've also come across a convention, which Benjamin Crowell attributes to Roger Penrose ( http://www.lightandmatter.com/html_books/genrel/ch03/ch03.html ), whereby "Latin superscripts and subscripts indicate that an equation is of general validity, without regard to any choice of coordinate system, while Greek ones are used for coordinate-dependent equations." (I wonder what people who follow this convention use as a default if they don't know or don't want to specify whether the equation is coordinate dependent; would they use Latin indices in that case too?) I don't think Penrose follows this convention in The Road to Reality though; he doesn't mention it in the Notation guide at the beginning, and from what I've read, Latin indices seem to be used throughout.

 

[Phrak]

The mass of a photon is 0.

I've been meaning to ask you this question. On what do you base this conclusion?

 

[jndbrn]

I’m a late entry to this discussion. Hope the following hasn’t been said in some form before and I’ve missed it.

Cox and Forshaw explain in ‘Why does E = mc2’ that massless particles must travel at the cosmic speed limit c, and that if photons are massless this is what they are forced to do. Relative to the observer that is. According to themselves (if they were able to make observations) time stands still.

Cox and Forshaw state that, if ever it is found that photons do have a tiny mass (so small that it is impossible with currently available tools to measure it), then nothing will change to c, but photons will travel at speeds less than c. The difference will have to be extremely small or it would have been measured by now.

I find their explanation very clear. It means that so far we know, a photon has zero mass, but there is no guarantee that this will not be overruled by future experiments. All we can say is that its mass must be either zero or extremely small. We can probably figure out an upper limit for it from experimental results.

It’s interesting that space shrinks in the direction in which something moves (not for the observer but for the thing that moves), and that for a photon traveling at speed c space shrinks all the way to zero. This means that the concepts of time and space do not mean anything for a photon. This is strange because photons do seem to have something to do with the fact that we experience this thing called spacetime in the first place.

Physics is fascinating. You can’t really understand it but you can gradually begin to accept it, and admire it for its consistency.

 

[Dale]

I've been meaning to ask you this question. On what do you base this conclusion?

Because experimental evidence places a very small upper limit on the mass of a photon of about 10^(-16) eV. See the FAQ
http://www.phys.ncku.edu.tw/mirrors/physicsfaq/ParticleAndNuclear/photon_mass.html

Perhaps I will be proven wrong in the future as experimental techniques improve, but with today's state of the art I won't get the wrong result if I assume 0. I am OK with that.

 

[Phrak]

Because experimental evidence places a very small upper limit on the mass of a photon of about 10^(-16) eV. See the FAQ
http://www.phys.ncku.edu.tw/mirrors/physicsfaq/ParticleAndNuclear/photon_mass.html

Perhaps I will be proven wrong in the future as experimental techniques improve, but with today's state of the art I won't get the wrong result if I assume 0. I am OK with that.

I'm also assuming exactly zero in this sense. So I don't think we are really disagreeing. The one experiment I looked into was to measure Coulomb force. In the particle theories that address both the range of massive and non-massive force-carrying bosons, the Coulomb force is mediated by offshell photons. These are photons where |p|=E doesn't hold. They are off the null surface in momentum space.

So the theoretical framework required to indirectly establish the mass of the photon requires that not all photons have null mass, but does require the shell to be degenerate. The hyperbolic surface is a conical surface.

Here is a thought experiment.

Take a laser capable of generating pulses measured in picoseconds—a few millimeters in length. The output is attenuated so that photons will pass, nominally one at a time, through the attenuating filter. This is followed by a beam splitter. Through a system of mirrors, the two beam paths are redirected head-on. In some region, the split wave packets will intersect with opposite momentum.

How do P_a=(E_a, p_a) and P_b=(E_b, p_b) superimpose? Adding these vectors together I get non-zero mass.

Instead of looking only at the energy momentum equation of a particle in this experiment we can look at the stress energy tensor. In the region where the wave packets recombine does the stress energy tensor have the form given for matter or light?


As I see it, not all photons, at all times are non-massive, though we limit our observations to local, nominally flat regions of spacetime.

 

[Dale]

So I don't think we are really disagreeing.

I didn't realize we had even a semblance of a disagreement.

As I see it, not all photons, at all times are non-massive

That doesn't fit with the standard model or really any particle model of which I am aware. A given type of fundamental particle only has one mass. Two different particles with different masses would be given different names. For example, the electron, muon, and tau, all have charge -1 and spin 1/2 and they differ in their mass. I could conceive of a model where the bosons also had multiple generations just like the fermions, but if we found both massive and massless bosons that mediated the electromagnetic force we would certainly call the massive boson something other than a "photon".

 

[Phrak]

OK. What do you obtain for the stress energy tensor?

 

[Dale]

Huh? I don't know what you are referring to. I don't think I have made any comments about the stress energy tensor in this thread.

 

[Phrak]

Huh? I don't know what you are referring to. I don't think I have made any comments about the stress energy tensor in this thread.

No, I did in post 77.

 

[Dale]

If you were to measure the energy and momentum of the photon in the region where its wavefunction crosses itself like you describe then you would always find that the energy is proportional to the momentum (0 mass) with a 50% chance of the momentum being in either direction. As far as the stress-energy tensor goes, I don't know, I think that would require a quantum theory of gravity.

 

[jndbrn]

If a photon does have a mass, however small, should we then not according to theory observe photons traveling at much lower speeds than c? In fact, should not the whole range from zero to (almost) c be available?

Is it correct to say that, since we don't see that, and if we assume the theory (special and general relativity) to be correct, it follows that photons have no mass?

Then it follows, I think, that if a tiny mass is ever found for a photon, the theory will have to be revised.

 

[Count Iblis]

If a photon does have a mass, however small, should we then not according to theory observe photons traveling at much lower speeds than c? In fact, should not the whole range from zero to (almost) c be available?

Is it correct to say that, since we don't see that, and if we assume the theory (special and general relativity) to be correct, it follows that photons have no mass?

Then it follows, I think, that if a tiny mass is ever found for a photon, the theory will have to be revised.

This reasoning does not hold because of screening effects. If the photon had a tiny mass, then photons that move significantly slower than c would have a a very small energy, hence a very long wavelength. Now, in a perfect vacuum we could then see the difference between such massive photons and massless photons (e.g. we could could generate extremely long wavelength radio waves and measure the propagation speed).

But in reality we don't have access to a perfect vacuum, even in interstellar space there is a diffuse plasma. When electromagnetic waves propagate through a plasma they behave in a similar way as a massive electromagnetic field would propagate in a perfect vacuum. There is a so-called Debye screening length which translates to an effective photon mass. Below this wavelength, electromagnetic waves cannot propagate. For the interstellar medium, the minimum frequency is a few hundred KHz, I think.

So, since we don't see the extremely low energy photons propagate freely, we cannot draw conclusions about the photon mass in this way.

 

[Phrak]

If you were to measure the energy and momentum of the photon in the region where its wavefunction crosses itself like you describe then you would always find that the energy is proportional to the momentum (0 mass) with a 50% chance of the momentum being in either direction. As far as the stress-energy tensor goes, I don't know, I think that would require a quantum theory of gravity.

This could lead into a fruitless quantum mechanical quagmire over what the wave equation 'n stuff describes. That's something I'd rather avoid, and I'm pretty sure you do to. So I'm assuming without justification that a photon is something with quantized energy, and not localized, but a continuous field.

Assuming this, I obtained the following. A single picosecond pulse taveling in the x direction, where we aren't too concerned about the edges, looks like a planar wave. The normalized stress energy tensor is

T(x,t) = |sin(kx-\omega t)| \left[ \begin{array}{cccc} <br /> 1&amp;1&amp;0&amp;0 \\ 1&amp;-1&amp;0&amp;0 \\ 0&amp;0&amp;-1&amp;0 \\ 0&amp;0&amp;0&amp;-1<br /> \end{array}\right]

A perfect fluid at rest with density rho and pressure, p.

T = \left[ \begin{array}{cccc} <br /> \rho&amp;0&amp;0&amp;0 \\ 0&amp;p&amp;0&amp;0 \\ 0&amp;0&amp;p&amp;0 \\ 0&amp;0&amp;0&amp;p<br /> \end{array}\right]

The superposition of two equal halves of a pulse traveling in the x and -x directions.

T(t) = |sin(\omega t)| \left[ \begin{array}{cccc} <br /> 1&amp;0&amp;0&amp;0 \\ 0&amp;1&amp;0&amp;0 \\ 0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;1<br /> \end{array}\right]

The momentum density has disappeared to look like normal matter at rest, but I'm suprised to see the tension has become pressure. In fact, the magnitude of the pressure looks suspect. I don't know how to interpret it.

 

[Dale]

I would interpret it as a probability or an expectation value. Basically, QM uses Maxwell's equations and just interprets them as probabilities. On average you will indeed measure a pressure since half of the photons will be going either direction. Each individual photon that you measure will have a net momentum which is proportional to the energy.

 

[Phrak]

I would interpret it as a probability or an expectation value. Basically, QM uses Maxwell's equations and just interprets them as probabilities. On average you will indeed measure a pressure since half of the photons will be going either direction. Each individual photon that you measure will have a net momentum which is proportional to the energy.

I think we've tread this ground before. The keyword I've been looking for is Komar mass. It is, apparently, what you have, before a photon is measured. But I can't interpret the relative size of rho and p; rho=p. It seems larger than is physically possible with normal matter.