Comments about the FAQ: What is the mass of a photon?

Following is a discussion about this Faq
In the last paragraph I explain my own opinion.


[Physics FAQ] [Copyright]

Updated 2008 by Don Koks.
Updated 1998 by Phil Gibbs.
Updated 1992 by Scott Chase.
Original by Matt Austern.

What is mass of a photon?

This question falls into two parts:
In order to answer this question you have to perform an experiment. In fact the second part should be raised before the first part.

Does the photon have mass?  After all, it has energy and energy is equivalent to mass.

How do you know that a photon has energy ? In fact the whole discussion starts by performing an experiment.

Photons are traditionally said to be massless.  This is a figure of speech that physicists use to describe something about how a photon's particle-like properties are described by the language of special relativity.

The logic can be constructed in many ways, and the following is one such.  Take an isolated system (called a "particle") and accelerate it to some velocity v (a vector).  Newton defined the "momentum" p of this particle (also a vector), such that p behaves in a simple way when the particle is accelerated, or when it's involved in a collision.  For this simple behaviour to hold, it turns out that p must be proportional to v.  The proportionality constant is called the particle's "mass" m, so that p = mv.

Considering Newton mechanics, two laws should be considered: "The Law of conservation of momentum" and "The law of conservation of energy"
The first law states that: ∑i mivi=∑j mjvj
The second law states that: ∑i miv2i=∑j mjv2j
In both cases with mi and vi the masses and velocities of the particles involved before the collision and mj and vj the masses and velocities of the particles involved after the collision.

In special relativity, it turns out that we are still able to define a particle's momentum p such that it behaves in well-defined ways that are an extension of the newtonian case.  Although p and v still point in the same direction, it turns out that they are no longer proportional; the best we can do is relate them via the particle's "relativistic mass" mrel.  Thus

           p = mrelv .
The next question to answer is: what is this relativistic mass mrel
IMO
mrel = mrest/sqrt(1-v2/c2)
using: E = mrelc2
We get: E2 = m2rel c4 = m2restc4/(1-v2/c2)
Or: E2*(1-v2/c2) = m2restc4
Or: E2 - m2relv2c2 = m2restc4           (2)
Or: E2 = p2c2 + m2restc4           (1)

When the particle is at rest, its relativistic mass has a minimum value called the "rest mass" mrest.  The rest mass is always the same for the same type of particle.  For example, all protons, electrons, and neutrons have the same rest mass; it's something that can be looked up in a table.  As the particle is accelerated to ever higher speeds, its relativistic mass increases without limit.

The remark that the mass can be looked up in a table is not very scientific. What should be explained how the mass of a proton, neutron and elektron (with v=0) is calculated by means of an experiment.

It also turns out that in special relativity, we are able to define the concept of "energy" E, such that E has simple and well-defined properties just like those it has in newtonian mechanics.  When a particle has been accelerated so that it has some momentum p (the length of the vector p) and relativistic mass mrel, then its energy E turns out to be given by

           E = mrelc2 ,   and also    E2 = p2c2 + m2restc4 .           (1)

There are two interesting cases of this last equation:

  1. If the particle is at rest, then p = 0, and E = mrestc2.
  2. If we set the rest mass equal to zero (regardless of whether or not that's a reasonable thing to do), then E = pc.

In fact the above equation is not interesting at all. The reason is much better explained by equation 2. (which is the same as equation 1) Equation 2 contains two unknowns mrel and mrest which both have to be calculated first in order to calculate E.

Using Equation 2 you can also see what happens when you set mrest equal to zero. In that case E = mrelvc. This may be mathematical correct, but physical it is not because mrel and mrest are physical parameters of the same particle under different conditions.

In Special relativity the only two equations that are interesting are "The Law of conservation of momentum" and "The law of conservation of energy". THey are:

The first law states that: ∑i miviγi=∑j mjvjγj
The second law states that: ∑i miv2iγi=∑j mjv2jγj
The question to answer is (decided by experiment) are those equations with γ more accurate than the same equations without γ

In classical electromagnetic theory, light turns out to have energy E and momentum p, and these happen to be related by E = pc.  Quantum mechanics introduces the idea that light can be viewed as a collection of "particles": photons.  Even though these photons cannot be brought to rest, and so the idea of rest mass doesn't really apply to them, we can certainly bring these "particles" of light into the fold of equation (1) by just considering them to have no rest mass.  That way, equation (1) gives the correct expression for light, E = pc, and no harm has been done.  Equation (1) is now able to be applied to particles of matter and "particles" of light.  It can now be used as a fully general equation, and that makes it very useful.

Is there any experimental evidence that the photon has zero rest mass?

Alternative theories of the photon include a term that behaves like a mass, and this gives rise to the very advanced idea of a "massive photon".  If the rest mass of the photon were non-zero, the theory of quantum electrodynamics would be "in trouble" primarily through loss of gauge invariance, which would make it non-renormalisable; also, charge conservation would no longer be absolutely guaranteed, as it is if photons have zero rest mass.  But regardless of what any theory might predict, it is still necessary to check this prediction by doing an experiment.

Only the last part makes any sense and requires a more detailed explanation. How do you calculate by means of an experiment the Energy of a photon. Also a second question has to be answered: Do all photons have the same energy.

It is almost certainly impossible to do any experiment that would establish the photon rest mass to be exactly zero. The best we can hope to do is place limits on it. A non-zero rest mass would introduce a small damping factor in the inverse square Coulomb law of electrostatic forces. That means the electrostatic force would be weaker over very large distances.

Likewise, the behavior of static magnetic fields would be modified. An upper limit to the photon mass can be inferred through satellite measurements of planetary magnetic fields. The Charge Composition Explorer spacecraft was used to derive an upper limit of 6 10-16 eV with high certainty. This was slightly improved in 1998 by Roderic Lakes in a laboratory experiment that looked for anomalous forces on a Cavendish balance. The new limit is 7 10-17 eV. Studies of galactic magnetic fields suggest a much better limit of less than 3 10-27 eV, but there is some doubt about the validity of this method.


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Created: 21 september 2008

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