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

Derek McKenzie wrote on Mon, September 27, 2010 7:30:36 PM:

... a massive particle (i.e. one such that m > 0), has a well-defined
(time-like) world line in absolute space-time (let's restrict ourselves to
Minkowski >Space here), and at each event of its world-line we can define
(intrinsically - using differential geometry) a unit tangent vector U, from
which we can >construct the vector P := mU and call it the (absolute)
four-momentum of the particle.


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.

But then we come to massless particles - the photon being our prototype, of
course...
...the frame-dependent four momentum of such a particle can be defined as
(|p|,p), where p is the three-momentum of the particle.

This equivalent definition is 100% consistent with my first statement that for
a photon |P| =0.

But I now want to express this vector as an abstract vector in Minkowski Space,
just like I did with mU.

So do exactly the same with the photon! This is one of the beautiful features of
Relativity that in this respect it does not discriminate between m > 0 and m = 0
(and even between real m and imaginary m for hypothetical superluminal
particles).

Since none of the standard relativity tomes on my shelf (Rindler, Misner et al,
Wald, etc.) seem to describe the idea of an absolute four-momentum >vector for
massless particles, even though they do discuss photons, I'm wondering if this
issue points to something deep.

This idea has been addressed in my book "Special Relativity and How It Works"
(Sec. 5.1, pp. 125 - 130)

Moses Fayngold,
NJIT

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