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

The limit of 5x * 1/x is 5. The point is that the limit of 0 * infinity is indeterminant.

Bob at PC

From: [] On Behalf Of Moses Fayngold []
Sent: Friday, October 01, 2010 9:28 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Absolute four-momentum of massless particles

On Fri, October 1, 2010 1:17:27 PM, John Denker wrote:

I would like to emphasize that understanding arithmetic is a
prerequisite for understanding relativity.

How many people on this list -- or their students -- are
aware of any axioms of arithmetic that allow for multiplying
infinity by zero? Not many, I would wager.

Just now I tried googling for such an axiom, and found nothing
... for good reason.

As far as I remember from my student days, it is a common knowledge
that multiplying infinity by zero is defined as a limiting procedure, e.g.,
"0 * infinity" is a limit at |x| --> 0 of |x| * (1/|x|) .
In this case, the product is 1 even though one factor is zero.

Let's do another little calculation. Again we hypothesize that
P = m U
m = 0

We multiply both sides by 2.
2 P = 2 m U

Since another of the axioms says that multiplication is associative,
we can write this as
2 P = (2 m) U

Now we also have
2 m = m + m
= m + 0
= m
where the last step follows from the axiom that says 0 is the additive

Collecting results we have
2 P = (2 m) U
= m U
= P

This in turn guarantees that P = 0.

So once again we have "proved" that every photon has zero energy and
zero momentum in every frame (subject to the hypothesis that P = m U
and m = 0).

Again we must reject the hypothesis.

Once we started talking about axioms of arithmetic, we should recall
even a more fundamental axiom: "Do not substitute terms during discussion."
There is an obvious substitution of terms in your argument: one and the same
symbol stands for two totally different characteristics.
1) The symbol U is used as the norm |U| of 4-velocity (and the same for P).
In this case |P| = m |U|, and since |U| = 1 and m = 0, we have |P| = 0 (Eq. 1)
This means that temporal and spatial parts of photon's 4-momentum are
equal to one another. The result obtains immediately, in one simple step.
If your U means 1), you get the same result, but in a needlessly convoluted
way going from Brooklyn, NY to Manhattan, NY via San Francisco.
Needless to say that in this case the conclusion that "every photon has
zero energy and zero momentum" DOES NOT follow from P = 0 (that is,
from |P| = 0!). Energy and momentum are, respectively, temporal and spatial
COMPONENTS of 4-momentum. In Lorentzian geometry which you,
doubtlessly, know, a 4-vector A may have the zero norm |A| = 0, and non-zero
components, e.g., A0 = A1, both non-zero, so that |A|^2 = A0^2 -- A1^2 = 0.
Photon's 4-momentum, or a 4-displacement along photon's world line
are well-known physical examples.
2) The symbol U stands for the whole 4-vector, e.g., P = (P0, P1) and
accordingly U = (U0, U1) (using, for simplicity, only one spatial dimension).
In this case your equations, first, are vector equations; and secondly,
IN THIS CASE, with all sophistication of your treatment, your result is wrong.
The fact that P0 = m U0 or P1 = m U1 with m = 0 does NOT make either of
the P-components equal to zero, since both U-components are infinite for
a photon.
The conclusion that definition P = mU (vector equation!), when applied
to a photon, leads to |P| = 0 (scalar equation!) is correct.
The conclusion that this definition leads to P = 0 (vector equation!)
is downright wrong.
Both conclusions together in one argument "obtain" only if the same symbol
stands for the norm of a 4-vector in one part of the argument, and for its
components in the other part. I think, this is one of the most elusive
and thereby most dangerous errors, especially when taught to the students.

Moses Fayngold,

Forum for Physics Educators