Do Photons Have Mass? Confused Student Seeks Answers

[Rasalhague]

The phrase "rest mass is conserved" is somewhat ambiguous, which is why there has been some confusion in this thread.

1. The invariant mass of the whole system, $\sqrt{[(\Sigma E)^2 - (\Sigma p)^2 ]}$ (in units where c=1), is conserved.
2. The sum of the individual particles' rest masses is not conserved.

"Conservation of rest mass" is often taken to relate to the second statement, and therefore isn't true.

In fact some authors avoid this confusion by using "rest mass" to refer only to individual particles. For a whole system they use the phrase "invariant mass" or "system mass".

Thanks, DrGreg, that clears it up!

And apologies to Ansgar for my misunderstanding in #20. I wasn't aware of this usage. As is probably apparent by now, I've been using "rest mass" to mean what these authors refer to as "invariant mass". It's good there are different names to save us confusion, but a shame some of these names aren't very transparent, in particular that some call it "invariant" (suggestive of frame-invariance) rather than "conserved" or "system" mass.

Since "rest mass" alone is ambiguous in the context of such a thread as this, and since "rest" is also used to distinguish between "rest" and "relativistic" mass, we'd better spell out at the beginning of any discussion what names we're giving to

(1) (a) Sum of rest masses of particles; (b) sum of relativistic masses (energies in units where c=1) of particles = 2.b.

(2) (a) System rest mass; (b) system relativistic mass (energy in units where c=1) = 1.b.

Obviously, because of this variety of terminology, questions from someone we don't know along the lines of "is mass conserved / not conserved?" or "does mass depend on speed?" or "do photons have mass?" can't be answered with a simple "yes/no" (or a barrage of simple yeses and nos from different people).

 

[Rasalhague]

"Invariant mass" after annihilation is (Eph1+Eph2)/c^2 in mass centered frame of reference where P1+P2=0. We easily find that the mass centered frame of reference and "invariant mass" is conserved before and after annihilation process.

Sum of mass before annihilation is 2m, that of after is 0. Sum of mass over particles are not conserved quantity of the system.

This agrees with what I've read. Taylor and Wheeler call it the "center of momentum frame".

 

[Rasalhague]

rest mass is frame dependent.

I don't think either of the meanings of rest mass that have been made explicit so far in this thread are frame dependent, are they? That's meanings 1.a and 2.a in #36. Are you using it in a third sense, or did you just mean to say 1.a is not conserved? Or are you talking about general relativity or quantum mechanics, in which case I don't know enough to comment. If we are talking about special relativity, I take rest mass to mean the energy of something with respect to an inertial reference frame in which it's at rest (in contrast to relativistic mass = the energy of something wrt an arbitrary inertial reference frame, in units where c=1), except in the case of a massless particle, for which there is no such reference frame, and whose rest mass is defined as zero. I take a frame dependent value to mean a value not necessarily unchanged by a Lorentz transformation.

 

[Dead Boss]

I thought that the photon could not have mass mathematically because since it is traveling at the cosmological constant it would have infinite mass.

You mean Einstein's lambda? What does it have to do with the speed of light? I am really curious here, because I've never heard of it before.

 

[George Jones]

You mean Einstein's lambda? What does it have to do with the speed of light? I am really curious here, because I've never heard of it before.

I think that filegraphy meant the constant speed of light.

 

[Phrak]

Phrak, quantum mechanics is outside the scope of special and general relativity and is dealt with in quantum field theory (QFT) instead. The discussion of photons in this thread refers to "classical photons" that are assumed to have a definite energy and momentum. As I've never studied QFT, I've no idea how the concept of mass is treated in QFT, but it's probably best discussed in the the Quantum Physics forum of this site rather than here.

Dear DrGreg. That's fair enough. In general relativity what we would really like to see are continuous fields. I'm not prepared at this time to support, what I think is obvious, with solid mathematics which should apparently deal with momentum density.

 

[sweet springs]

Hi, just to become accustomed to kinds of mass, let me say on a few cases.
#1
There are two apparently same balls. A is made of some material. B itself is weightless, hollow, mirrored inside and photons are packed. They have same weight, inertia and source of gravity, say M.
How is Mass? Ans. A=M B=0
How is Energy? Ans. A=B=Mc^2
How is Invariant mass? Ans. A=B=M
#2
Does an electron have mass? Yes, it does. /No, it is massless. Electron is in Zitterbewegung motion at velocity of light. Energy/c^2 measured in the frame of reference where p=0 gives invariant mass. In daily physics this invariant mass is regarded as mass of electron.
#3
Do quarks and leptons have mass? Yes, they do. /No, they are massless and usually moving at velocity of light. Only in case they are in Higgs field when the universe is cooled down, we call the indicator of interaction energy "mass". So mass is not more than a kind of energy, like kinetic energy, potential energy, thermal energy and elastic energy are. Total energy conserves, each kind of energy including mass does not.
Regards.

 

[starthaus]

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.

Easy (to put DrGreg's post in a formal way):

$$\Sigma m_i$$ is not conserved
$$E=c^2\Sigma \gamma(v_i)m_i$$ is conserved
$$\vec{P}=\Sigma \gamma(v_i})m_i \vec{v_i}$$ is conserved
Therefore:
$$E^2-(c\vec{P})^2$$ is conserved as well

 

[Angaddegratis]

Photons do have mass.Depends on how you want to define mass. A new identity needs to be created to understand this.

 

[Count Iblis]

Photons do have mass.Depends on how you want to define mass. A new identity needs to be created to understand this.

http://en.wikipedia.org/wiki/0_(number)"

 

[Phrak]

Invariant mass is not a scalar quantity in Minkowski spacetime. Requiring conservation of energy and momenum, it doesn't add as a scalar.

If (m₁ =0 & m₂=0), in general m₁+m₂ =/= 0.

What is it?

 

[DrGreg]

Invariant mass is not a scalar quantity in Minkowski spacetime. Requiring conservation of energy and momenum, it doesn't add as a scalar.

If (m₁ =0 & m₂=0), in general m₁+m₂ =/= 0.

What is it?

I'm not sure what your concept of "scalar" is here. Usually "scalar" is taken to mean "invariant", not "conserved". So (invariant) mass is a scalar.

See the thread "Is time scalar?"; later in that thread, several interpretations of "scalar" are discussed.

 

[Ich]

Invariant mass is not a scalar quantity in Minkowski spacetime.

Of course it is. The invariant mass of an object is the length of the object's four momentum.

it doesn't add as a scalar.

How do scalars add?
m(Object1) + m(Object2) = m(Object1) + m(Object2) (that's addition), but
m(Object1) + m(Object2) != m(Object1+Object2) (tht's nonlinearity of the mass function).

 

[Phrak]

OK, you want the mass of a single object to be a rank 0 tensor, right? But also the sums don't add as scalars. Do sums of your rank 1 object for momentum transform as a tensor under a coordinate transformation?

 

[Rasalhague]

It's obviously a scalar in the mathematical sense of an element of the underlying set of the field over which spacetime vectors are defined. The other sense of scalar that I've met is, in a physics context, as a shorthand way of saying scalar field: a function that associates a unique real number with each point in spacetime, S:M-->R, or each point in an open subset of spacetime, S:U-->R (with frame invariance implied by the lack of any mention of coordinates in this definition). Is it possible to define a mass field like this; is that what the discussion is about?

I wasn't aware of any addition rule as part of the definition of a scalar field. Thinking of the archetypal examples of scalar fields in introductory texts, suppose we have a Newtonian gravitational potential field in Euclidean space, or a temperature field whose values are the limit of temperature in regions centred on each point as the volume goes to zero. What is the physical meaning of adding together the temperature at two points? The fact that the temperature of a system consisting of two objects isn't generally the sum of the temperature of each object doesn't seem to disqualify this from being a scalar field.

 

[Phrak]

Re: scalars in classical physics. What we would like is a quantity that is independent of coordinates, such as a rank zero tensor. In this case, in relativity physics, a scalar is an element of a vector space over the field of reals, and so should obey the rule of vector addition as well as the other nine(?) axioms of a vector field. With this definition of a scalar, some things can be scalars in Minkowski space, though fail to be in general relativity. Preferably it should also be a scalar in general relativity. Under a coordinate transformation it should remain constant, so that scalar densities are not true scalars.

If we don't like the name scalars, call them csalars, but these things have very nice symmetrics. So I am curious as to what quantity has these invariances that are better to talk about and utilize than intrinsic mass.

 

[DrGreg]

In this case, in relativity physics, a scalar is an element of a vector space over the field of reals, and so should obey the rule of vector addition as well as the other nine(?) axioms of a vector field.

No, a scalar (in this sense) isn't a member of a vector space; a member of a vector space is called a "vector". Over the field of reals, a scalar is simply any real number. The additivity is simply the fact that the sum of two reals is a real. If we are ignoring the issue of invariance, a scalar is simply a real number. Mass is a real number and therefore a scalar.

 

[filegraphy]

The mass of a photon is 0. The mass of a system of multiple photons may be non-zero.

How could this be? It is like multiplying zero by a certain number. So the mass of multiple photons would be zero because the mass of each photon is zero. Am I correct or mistaken?

 

[ansgar]

How could this be? It is like multiplying zero by a certain number. So the mass of multiple photons would be zero because the mass of each photon is zero. Am I correct or mistaken?

just do the adding of their four momentum and compute the invariant mass

 

[starthaus]

How could this be? It is like multiplying zero by a certain number. So the mass of multiple photons would be zero because the mass of each photon is zero. Am I correct or mistaken?

Two photons of energies $$E_1$$ and {tex]E_2[/tex] (they have different frequencies.
The energy of the system of two photons is $$E=E_1+E_2$$

Case A: the photons move in the same direction, the momentum of the system is:
$$\vec{P}=(E_1+E_2)/c$$

The invariant mass of the system made up by the two photons is:

$$\sqrt{E^2-(\vec{P}c)^2}=0$$

Case B: the photons move in opposite directions, the system momentum is:

$$\vec{P}=(E_1-E_2)/c$$

The invariant mass of the system made up by the two photons is:

$$\sqrt{E^2-(\vec{P}c)^2}=\sqrt{(E_1+E_2)^2-(E_1-E_2)^2}$$

 

[Dale]

How could this be? It is like multiplying zero by a certain number. So the mass of multiple photons would be zero because the mass of each photon is zero. Am I correct or mistaken?

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
$$m_{s}=|p_{A}+p_{B}|=|(1,1,0,0)+(1,-1,0,0)|=|(2,0,0,0)|=2$$

 

[D H]

How could this be? It is like multiplying zero by a certain number. So the mass of multiple photons would be zero because the mass of each photon is zero. Am I correct or mistaken?

You are mistaken. You are implicitly assuming here that the invariant mass of a collection of particles is the simply the sum of the invariant masses of the individual particles that comprise the system. The invariant mass of a collection of particles is not the sum of the invariant masses of the individual particles that comprise the system.

By way of analogy, suppose the velocity of object B with respect to object A is v_A:B and the velocity of object C with respect to object B is v_B:C. You certainly can add those velocities, but that sum doesn't have any physical meaning. The velocity of object C with respect to object A is not the vector sum of v_A:B and v_B:C.

 

[filegraphy]

I am not getting this. A photon does not have mass. Three photons have mass? What difference does it make just by increasing the quantity of photons if their mass is zero. So for three photons, the mass would be: 0+0+0? You guys are out of my league. I cannot grasp this concept. Can I get some help and please do not include all of this mathematics. Sorry for the disruption. I appreciate all help. Thanks.

 

[starthaus]

I am not getting this. A photon does not have mass. Three photons have mass? What difference does it make just by increasing the quantity of photons if their mass is zero. So for three photons, the mass would be: 0+0+0? You guys are out of my league. I cannot grasp this concept. Can I get some help and please do not include all of this mathematics. Sorry for the disruption. I appreciate all help. Thanks.

The proper mass of a system of particles is NOT equal to the sum of the proper masses of the composing particles.
You need to read post 55 or 56. Both posts explain your confusion.

 

[closet mathemetician]

At the scale of a photon, we are using quantum mechanics. So there, the momentum of a photon is given by the product of its frequency and (reduced)Planck's constant. Therefore, even though it has no mass, it has momentum. p=h-bar*k.

 

[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.)