Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

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



Bob LaMontagne wrote on Sat, October 2, 2010 5:05:37 PM

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


I would put it this way. The value P = 0*Inf is indeterminable as long as we
have an explicit Inf in its expression. But we cannot say that P as the LIMIT of
0*Inf is always indeterminable if we have 0 = limit of Q and Inf = limit of R,
and can determine the way each factor approaches its respective limit. There are
an infinite number of ways leading to Inf and an infinite number of ways
approaching 0. Specifying the way to each limit can determine P.
E.g. (below we always assume x --> 0):
P = 5x*(1/x) = 5;
P = ax*(1/x) = a, and gets completely determinate when we specify
the value of a;
P = sqrt{x}*(a/x) = Inf (at any finite a);
P = x*(a/sqrt{x}) = 0 (at any finite a);
P = sqrt{x}*(a/xSin(1/x)) = What???????
In the last example P is intrinsically indeterminable. There are infinitely many
such cases, but also, as we can see, infinitely many cases of exactly
determinable P. And in the overwhelming amount of such cases P comes up
non-zero, and sometimes even infinite, despite one zero factor.
BTW, in all these cases, as Chuck Britton noted quite correctly, we can always
represent 0*Inf as the limit 0/0 and apply L'Hospital rule.
As a physical example, we can use the general definition of 4-momentum P = mU
in terms of m and 4-velocity U and apply it to a photon, which has definite m =
0 and definite U with at least 2 components infinite (I call these components
definite in the sense that we know either of them is not 0, not 1.7, ...not any
finite number). In this case we get P0* = mc*gamma(c) = 0*Inf. The only way to
find P0 is to get back to general expression P0 = mU0 = mc*gamma(v) and see the
behavior of its factors at m --> 0, and v --> c. Recalling that at ANY m and v
we have m*gamma(v) = M(v) = E/c^2, where M is relativistic mass and E is the
corresponding energy, we see that P0 for a photon is finite and exactly defined.
This also gives another example of the relevance and usefulness of the concept
of relativistic mass. It works in calculations in which straightforward use of m
does not.

Moses Fayngold,
NJIT