"Next year is the 100th anniversary of spacetime. All the
students have heard of spacetime; it's well established as
part of pop culture. Isn't it about time we taught them
what spacetime really is, and how to use it?"
Before answering John's other comments, let me derive E = mc^2
from the concept of spacetime, using the Feynman approach.
My notations will by necessity be different from those
of John, - my "m" stands for relativistic mass, and his "m"
stands for the invariant mass. I will denote the invariant mass
as m0.
Define 4-momentum as p^j = m0u^j, where index j = 0,1,2,3,
m0 is the rest (invariant) mass and u^j is 4-velocity. Since
the norm of 4-velocity is c, the norm of 4-momentum is m0c.
According to definition, the temporal component of the
4-momentum is given by
p0 = m0u0 = m0c gamma (v) = mc (1)
where
m = m0 gamma (v) (2)
(The label "0" in P stands for the "ct"-component of momentum.
Do not confuse it with the similar label in m0 to indicate the
rest mass).
Before doing any math, we can come up with a tentative answer
about the meaning of (1) by analogy. We have two counterparts of
reality - space and time - combined into spacetime. Out of 4
components of a space-time interval, the component x0 = ct
represents time. Similarly, we can expect that in relativistic
mechanics the two conserving characteristics of motion - momentum
and energy - must combine into 4-momentum in the 4-dimensional
momentum space. If this is true, then the component P0 must
represent the energy of the system.
We can confirm this expectation by the following argument.
Denote (v/c) = b and expand the mass (2) into the Taylor series:
m = m0 + (1/2)m0 b^2 + ... (3)
Then multiply by c^2:
mc^2 = moc^2 + (1/2)m0 v^2 + ... (4)
The second term on the right is the familiar (non-relativistic!)
expression for kinetic energy. Therefore the following terms must
be the relativistic corrections, and the whole sum of all the terms
depending on speed must represent relativistric kinetic energy K
of a body moving with a speed v. Thus, we can write
mc^2 = m0c^2 + K (5)
Since energy can be added only to energy, the first term on the right
must also represent energy, and the same therefore can be said about
the term on the left. What kind of energy are these? Now we have
only one possible interpretation: since m0 is the rest mass of the
system, the product m0c^2 must be its rest energy; similarly, the
product on the left must be the total relativistic energy of the
moving system. Naturally, the total turns out to be the sum of the
rest energy and the kinetic energy. Denoting this total as E, we come
to the mass-energy relationship:
E = mc^2 (6)
Note that this is much more general than just
E0 = m0c^2 (7)
for the rest energy only.
"I don't know what Sciama said about this, but my usage agrees
with Misner/Thorne/Wheeler, with Weinberger, ..... and with
Einstein"
John, if you are with Einstein, you must know that Einstein's
celebrated equation is (6), rather than its truncated version (7).
And if you are with Misner/Thorne/Wheeler, then you should not be
against the concept of relativistic mass (2), since the same
Wheeler has admitted that this concept can be used consistently
(See, e.g., Taylor, Wheeler, "Exploring Black Holes").
And if I am with Feynman (and many others, too many to mention here),
then I am in a good company, too.
As to Oas (and his referred paper rejected by the AJP), - we all
know that good papers are rather frequently rejected and bad papers
frequently accepted. Judging Oas' paper on its merit, I would say,
that it is not about Physics, but about statistics - about how
the fashionable trend of using ONLY the invariant mass m0 kreeps in
into Physics. As far as I remember the numbers in that paper, the
majority is still very comfortable with the relativistic mass (2)
as well as with the invariant m0. And even if I am wrong with those
numbers, the truth is not always resting with a dominant fashion.
"In particular, if we define "mass" the way thoughtful experts
do (and have done for many decades), then photons are massless
and it is trivial to come up with examples where the center of
mass hops around, even in the absence of externally-applied
forces. Just convert massive electron-positron pairs to
massless photons"
I am afraid the experts your refer to were not very thouhtful.
If we define the center of mass consistently, it is as useful
a concept in relativity (especially in collision problems) as it
is in Newtonian physics; it approaches its usual non-relativistic
expression in the limit of slow motions; and it does NOT hop around
even under e - p conversion into massless (having zero REST mass)
photons. Ironically, John came up here with a very compelling
argument in favor of relativistic mass, because only using the
relativistic mass (2) (as the really thoughtful experts have done)
gives the consistent definition of the center of mass.
"This (my mentioning of a photon in a superposition of different
momentum eigenstates - MF) is not even remotely relevant to the
article in question ... so why bring it up?"
Only to show that the "thoughtful experts" insisting that ONLY
rest mass is entitled to be called mass, were not thoughtful enough
when saying that a single free photon is always massless. It is not
generally true, even within the framework of the rest mass only.
Moses Fayngold,
MJIT
======================
In any case,
A) There are lots easier ways of getting to E=mc^2, and
B) That's not where you want to get to, anyway.
The recommended modern (i.e. post-1908) approach is to write
m^2 c^4 = E^2 - ps^2 c^2 [1]
or simply
m^2 = E^2 - ps^2
where m is "the" mass (which is an invariant Lorentz scalar)
and where ps is the 3-momentum i.e. the spatial part of
the 4-momentum.
We see that -m^2 is just p·p i.e. the dot product of the
4-momentum with itself. The 4-momentum is also known as
the [energy,momentum] 4-vector.
Equation [1] has the advantage that it includes E=mc^2 as a
special case, and makes it clear that mc^2 is the /rest energy/
(not the total energy).
Another advantage is that it builds upon and reinforces what
the students know about vectors.