Another brick in the wall: There's no relativistic mass

[John C. Polasek]

This business of shrugging off of relativistic mass is an embarassment
for relativity because it raises an obvious question that I have never
seen answered.
E^2 = m^2v^2c^2 + m^2c^4

No it's not, since p is not equal to mv.
- Randy

Then define it. Don't let it hang, like "no relativistic mass".

How many ways can you be stupid? By definition, the momentum is
what generates a translation along a spatial direction. What makes
you think that is equal to mv, apart from the fact that you never
realized the definition depends upon the group of transformations
that define spacetime? Armed with that new insight, you ought to
be able to figure out that p = mv comes from assuming invariance
under a galilean boost. Why would you expect p = mv if you don't
assume invariance under galilean boosts?

And, as I already pointed out, I would be willing to incorporate that
improvement except that it makes things so much worse and further
amplifies my assertion you already have one too many gammas. If you
think that one more gamma will clarify anything, run off a couple of
lines of algebra to prove it.

Huh? This has nothing to do with ``gammas.'' It's a simple fact
that different groups have different generators -- otherwise they
wouldn't be different groups. Here we have the G-A-L-I-L-E-A-N
and P-O-I-N-C-A-R-E G-R-O-U-P-S. I'm willing to bet you can't
find the invariants of the G-A-L-I-L-E-A-N G-R-O-U-P, so I'm not
surprised you don't understand where any of this stuff comes from.

Sarcasm and casuistry won't get us anywhere. Why not let's stay on
topic and factor the total energy equation, even including your
much-touted p = Gmv with G = gamma
E^2 = m^2c^2(G^2c^2 + v^2)
E = mc^2*sqrt(G^2 + beta^2)
Either m or c^2 are all pumped up with the bracket quantity. It's
agreed tacitly that m = m0. So it must be c^2 that's larger than
life. It's plain algebra.
John Polasek

[John C. Polasek]

This business of shrugging off of relativistic mass is an embarassment
for relativity because it raises an obvious question that I have never
seen answered.
E^2 = m^2v^2c^2 + m^2c^4

No it's not, since p is not equal to mv.
- Randy

Then define it. Don't let it hang, like "no relativistic mass".

How many ways can you be stupid? By definition, the momentum is
what generates a translation along a spatial direction. What makes
you think that is equal to mv, apart from the fact that you never
realized the definition depends upon the group of transformations
that define spacetime? Armed with that new insight, you ought to
be able to figure out that p = mv comes from assuming invariance
under a galilean boost. Why would you expect p = mv if you don't
assume invariance under galilean boosts?

And, as I already pointed out, I would be willing to incorporate that
improvement except that it makes things so much worse and further
amplifies my assertion you already have one too many gammas. If you
think that one more gamma will clarify anything, run off a couple of
lines of algebra to prove it.

Huh? This has nothing to do with ``gammas.'' It's a simple fact
that different groups have different generators -- otherwise they
wouldn't be different groups. Here we have the G-A-L-I-L-E-A-N
and P-O-I-N-C-A-R-E G-R-O-U-P-S. I'm willing to bet you can't
find the invariants of the G-A-L-I-L-E-A-N G-R-O-U-P, so I'm not
surprised you don't understand where any of this stuff comes from.

Sarcasm and casuistry won't get us anywhere. Why not let's stay on
topic and factor the total energy equation, even including your
much-touted p = Gmv with G = gamma
E^2 = m^2c^2(G^2c^2 + v^2)
E = mc^2*sqrt(G^2 + beta^2)
Either m or c^2 are all pumped up with the bracket quantity. It's
agreed tacitly that m = m0. So it must be c^2 that's larger than
life. It's plain algebra.
John Polasek