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Re: [Phys-l] Absolute four-momentum of massless particles

John Denker wrote in part on Thu, September 30, 2010 1:33:59 AM

Suppose that in addition to the photon we also have a
Suppose the particle is moving with instantaneous 4-velocity v.

Let's calculate the invariant quantity P•v. This is
easy, because P•v is just m times U•v and therefore
P•v is zero.

This is easy indeed, but wrong. You can write P*v = mU*v, but you cannot
conclude from this that P*v = 0 even though m = 0. Such conclusion is
based on unspoken assumption that U*v IS FINITE. But precisely this
assumption is wrong here. Even though the norm |U| = 1, its COMPONENTS
are infinite for the photon (they are all proportional to the Lorentz factor -
find its value
for the speed of light!). Therefore the scalar product U*v is INFINITE.
It becomes immediately evident in the line of your own argument below:

Now it turns out that the P•v is equal to the energy
>of the photon as measured by an observer comoving with
>the particle.

So let us calculate the invariant U*v in a frame comoving with the particle.
In this frame we have v = 0, so all spatial components of v are zeros,
and the temporal component is 1. The dot product thus reduces to
Inf *1 - Inf * 0 = Inf. Accordingly, all we can say is that
P*v = m U*v = 0 * Inf. Generally, it may be 0, finite, or infinite, depending
on the nature of both factors (the way they approach their respective limits.)
You cannot say that it is definitely zero when the other factor in the product
is entered as explicit infinity. Actually, you could calculate it using, e.g.,
limiting procedure and find the result finite (not zero!) and equal, as it
to the photon energy in the chosen frame. Thus, your "proof by contradiction"
does not prove anything, and definitely does not disprove my initial statement
P = m U for a photon. This definition remains useful to show immediately
that the norm of the photon's 4-momentum is zero, since |U| = 0.

We also used some basic laws of physics and
mathematics (such as the axiom that says zero times x is
zero for all x).

You forgot a "minor" detail in this axiom - that it holds for all FINITE x.
For x infinite as is our case, the axiom does not work.

So we must reject the hypothesis. Unless you believe
that photons never have any energy, and/or you want to start
repealing the axioms of arithmetic, then you'd better not
accept the idea that P = m U for photons.

Weigh both arguments and judge for yourself.

Moses Fayngold,