In reply to
JdeBP
@JdeBP__dup_33984@mastodonapp.uk
This is a general, but non-politics, account covering everything from computer programming to supermarket meal deals; any politics will be taken up by @ JdeBP . For # senryu and # SlowLife tasks, see @ JdeBP . For the command-line and system tools (including # nosh and # djbwares ), specifically, see @ JdeBP .
mastodonapp.uk
JdeBP
@JdeBP__dup_33984@mastodonapp.uk
This is a general, but non-politics, account covering everything from computer programming to supermarket meal deals; any politics will be taken up by @ JdeBP . For # senryu and # SlowLife tasks, see @ JdeBP . For the command-line and system tools (including # nosh and # djbwares ), specifically, see @ JdeBP .
mastodonapp.uk
@JdeBP__dup_33984@mastodonapp.uk
·
Feb 28, 2026
@jmax@mastodon.social
You are likely going to regret that. (-:
When discussing the energy–momentum relation, mass *is* by convention rest mass, as the usual formulation E^2=m^2×c^4+(p×c)^2 is in terms of rest mass m.
The relation says that energy does not imply mass when m=0. Energy implies the momentum portion of the sum, which photons have, defined as p=h/λ.
With m=0 the full form reduces to E^2=(p×c)^2 which after substitution for photon momentum becomes E=c×h/λ=h×f .
But this does not become a statement about mass. It's fallacious to then substitute E=m×c^2 and solve for m to get m=h×f/c^2 .
E=m×c^2 is a different reduced case for massive stationary objects (m>0, p=0), neither of which is the case for photons. Furthermore, the maths yields divergent γ=∞ Taylor series sums when u=c so thinking of K.E. terms for photons is aphysical.
Energy-mass equivalence is a special form for the case of m>0, γ≠∞. Energy does not imply mass in the general case.
@cstross@wandering.shop
#physics #relativity
View full thread on mastodonapp.uk
0
0
0
Loading comments...