What's the way to convert n(r) to \rho(r) in case of a spherical cloud.
n(r) is the column density, \rho(r) is the density.
I tried (but didn't manage) to get it from their relation to the total mass:
M = 4 \pi \int r^2 \rho(r)dr = 2 \pi \int r n(r)dr (is it correct anyway?)
Thank you,
Thanks ChrisVer,
A^\mu has nothing to do with B_{\mu \nu}
I meant the number dof of the theory.
H_{\nu ρσ} is antisymmetric, so it has only \binom{4}{3}=4 dof, doesn't it? thus in total it's 3X4=12 dof, isn't it?
And more important for me is to know the action symmetries, both the global and...
S=\int d^4x\frac{m}{12}A_μ ε^{μ \nu ρσ} H_{\nu ρσ} + \frac{1}{8} m^2A^μA_μ
Where
H_{\nu ρσ} = \partial_\nu B_{ρσ} + \partial_ρ B_{σ\nu} + \partial_σ B_{\nu ρ}
And B^{μ \nu} is an antisymmetric tensor.
What are the global symmetries and what are the local symmetries?
p.s how...
I mean for example that I saw that for the scalar field it is :
j^{μσμ} = x^ρT^{\mu \sigma} - x^σT^{\mu ρ}
I don't know how to get there, so I'd like to see the full derivation.
E = - grad*phi - 1/c (dA/dt).
phi is the scalar potential, and is given. How do I calculate the vector potential = A ?
Is it A = (v/c) * phi ? If it is, then where is this equation coming from?
Thank you.