Thank you for the summary. I had one comment (really just one, this time!)
Horace Heffner wrote:
Mass in the
conventional spacetime metric is considered invariant.
There's a semantic problem here. "An invariant" is a well-defined
mathematical concept. However, it's just that -- a mathematical
concept. Saying "this is an _invariant_" doesn't mean it's some simple
physical property which always has the same value.
In this case, the square of the rest mass is the (negative) squared
magnitude of the 4-momentum, and so it is "an invariant"; no matter what
frame of reference is used, the squared magnitude of the 4-momentum is
still the square of the rest mass. (Math given below...) But can you
weigh it on a scale? No, not necessarily; that's not what it means.
Here's another "invariant": The quantity m0*gamma for a particular
object, as measured by a particular observer. It's the inner product
of the observer's 4-velocity with the object's 4-momentum. The inner
product of two 4-vectors is an invariant, and so this value -- commonly
called the "relativisitic mass" seen by a particular observer -- is also
an invariant.
It's currently not in style to use the term "relativistic mass" and in
fact you can sometimes start an argument going in a group of
relativityists (and get called a lot of names) just by pointing out that
relativistic mass as viewed by a particular observer is an invariant.
People don't seem to like that -- perhaps it makes it too obvious that
the property of invariance is a mathematical, not physical, property.
People like to believe that they know what's real and what's just an
illusion or artifact, and pointing out that the lines between the two
are fuzzy tends to upset them.
It's easy to forget that relativity theory says _nothing_ about what is
"real" and what is not. ALL it can do is predict what will be observed
in various situations. Anything more than that requires some additional
theory (or philosophy).
* * *
Justifications for what I just said:
4-momentum of an object:
m * gamma * (1, vx, vy, vz)
Squared magnitude of 4-momentum:
m^2 * gamma^2 * (v^2 - 1)
= -m^2 * (1 - v^2) / (1 - v^2)
= -m^2
So, the rest mass is "an invariant".
Now, let's say an object of rest mass M is viewed by an observer moving
at velocity V relative to the object. Plugging in values for an
arbitrary frame leads to a mess, but we can evaluate it trivially in the
observer's rest frame. In that case the object's 3-velocity is -V, and
for the inner product of the object's 4-momentum with the observer's
4-velocity, we have:
<M * gamma * (1, -V) , (1, 0, 0, 0,)>
= M * gamma * (-1 - V*0)
= -M * gamma
If you transform both 4-vectors to any other frame of reference, their
inner product doesn't change. So, relativistic mass as seen by a
particular observer is "an invariant", too.
With all that said, when someone refers to the "invariant mass" they
mean the rest mass.
