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



On 09/29/2010 07:06 PM, Moses Fayngold wrote:
The same definition P:=mU works perfectly for a photon as well, if you use m
as the notation for the rest (invariant) mass of an object (which you obviously
do). For a photon, |U| = 1 (just as for any massive particle!) and since its
invariant mass is zero, m = 0, the magnitude of its 4-momentum is always zero.

Let's explore that a little bit. In the spirit of proof
by contradiction we very hypothetically accept what was
said above, namely that the photon has 4-velocity U such
that its 4-momentum is P = m U. This hypothesis applies
to the following indented paragraphs.

Suppose that in addition to the photon we also have a
particle. Photons are not very interesting unless they
have a particle to interact with. 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.

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.

By repeatedly calculating P•v for many different particles,
we find that the photon energy is zero in every frame.

We have just "proved" that photons never have any energy,
subject to the hypothesis that the photon has 4-momentum
P = m U. We also used some basic laws of physics and
mathematics (such as the axiom that says zero times x is
zero for all x).

Therefore 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.