Maxwell's equations and tensor notation

• I
In one of our lectures we wrote Maxwell's equations as (with ##c=1##)
##\partial_\mu F^{\mu \nu} = 4\pi J^\nu##
##\partial_\mu F_{\nu \rho} + \partial_\nu F_{\rho \mu} + \partial_\rho F_{\mu \nu} = 0##

where the E.M. tensor is
##
F^{\mu \nu} = \begin{pmatrix}
0 & -B_3 & B_2 & E_1\\
B_3 & 0 & -B_1 & E_2\\
-B_2 & B_1 & 0 & E_3\\
-E_1 & -E_2 & -E_3 & 0
\end{pmatrix}##

and ##J^\nu = (\mathbf J ,\rho)##,
##\partial_\mu = (\nabla, \frac{\partial}{\partial t} )##
(allowing the notation with the equal sign between the component and the 4-vector.)

Now from this I'm trying to recreate our classic Maxwell's equations.
If I set ##\nu = 4## in the first equation we get the L.H.S. as ##4\pi \rho##
The R.H.S. becomes
##\partial_\mu F^{\mu 4} = -\frac{\partial}{\partial x}E_1-\frac{\partial}{\partial y}E_2-\frac{\partial}{\partial z}E_3 = -\nabla \cdot E##.
Now this is a sign error since we should get ##\nabla \cdot E = 4\pi \rho##.

So I guess maybe I shouldn't put a minus sign before those but that goes against what I learned by using the Minkowski metric, ##g_{\mu \nu} = diag(-1,-1,-1,1)##. Similarly I get the same sign error in the continuity equation. Should I always get + signs?

vanhees71
Gold Member
You don't need the metric here, because the index of the four-gradient is already a lower index (covariant components of a vector), i.e.
$$\partial_{\mu}=\frac{\partial}{\partial x^{\mu}}.$$
So you get the correct Gauss law (in Gaussian units)
$$\vec{\nabla} \cdot \vec{E}=4\pi \rho.$$
The same is true for the continuity equation:
$$\partial_{\mu}j^{\mu}=\vec{\nabla} \cdot \vec{j}+\partial_t \rho=0.$$

Incand and Dale
Then I end up with another question. Consider 4-momentum vectors
##P^\mu = (P_x,E_1)##
##Q^\mu=(Q_x,E_2)##
Then ##(P^\mu+Q^\mu)^2= (E_1+E_2)^2-(Q_x+P_x)^2 = E_1^2+E_2^2-Q_x^2-P_x^2+2E_1E_2-2Q_xP_x##.
However we also have
##(P^\mu+Q^\mu)^2 = P^2+Q^2+2P_\mu Q^\mu = E_1^2-P_x^2+E_2^2-Q_x^2+2(E_1E_2+P_xQ_x)##

So In this case it's obviously a minus sign. Why?
So it should be something like this i guess
##(P+Q)^2 = P^2+Q^2+2g_{\mu \nu}P^\nu Q^\mu##?
But ##g_{\mu \nu}P^\nu = P_{\mu}## by lowering of index.

Nugatory
Mentor
So In this case it's obviously a minus sign. Why?
Set ##U=(P+Q)## and remember that ##(P+Q)^2=U^2=g_{\mu\nu}U^{\mu}U^{\nu}##

Write out the summation explicitly if that doesn't make it clear.

Incand
Set ##U=(P+Q)## and remember that ##(P+Q)^2=U^2=g_{\mu\nu}U^{\mu}U^{\nu}##
Yes that's what I did in the "first calculation".

However a lot of calculations rely on using the second approach looking at ##(P+Q)^2 = P^2+Q^2+\dots##. This approach is something that showed up a lot in our class for solving dynamics problem and when deriving things like Compton scattering or threshold energies.

Perhaps this can be thought of as it's always the normal scalar product but we have
##U^\mu = (\mathbf p , E)## and with lowered index
##U_\mu = g_{\mu \nu} U^\nu = (-\mathbf p,E)##.
In that case the normal scalar product of these two always gives the Lorentzian scalar product.

So my confusion is only that every other 4-vector I've seen I'm used to seeing ##U^\mu = (\mathbf p , E)## and ##U_\mu = (-\mathbf p,E)## while for the differentiation operator we defined it the 3-vector part positive for the lowered index.

I guess the problem I had is that in class we always spoke of
##U_\mu U^\mu## or ##A_\mu B^\mu## as the Lorentz scalar product.

vanhees71
Gold Member
The Minkowski product is a bilinear form, and it is always good to distinguish vectors and their components although physists always talk about ##V^{\mu}## is a vector, but really right is to say that these are tensor components with respect to a pseudo-orthogonal basis with four basis vectors ##\boldsymbol{e}_{\mu}##. The vector is
$$\boldsymbol{V}=V^{\mu} \boldsymbol{e}_{\mu}$$
with
$$\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}=\eta_{\mu \nu}.$$
You have the somewhat unusual convention to have the spatial components as the first three and the temporal component as the fourth component. So in your case, written in a matrix notation you have ##(\eta_{\mu \nu})=\mathrm{diag}(-1,-1,-1,1)## (west-coast convention).

A Lorentz transformation describes the change of one pseudo-orthogonal basis to another, i.e.,
$$\boldsymbol{e}_{\mu} = {\Lambda^{\nu}}_{\mu} \boldsymbol{e}_{\nu}',$$
which implies
$$\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\rho}=\eta_{\mu \rho} ={\Lambda^{\nu}}_{\mu} {\Lambda^{\sigma}}_{\rho} \boldsymbol{e}_{\nu}' \cdot \boldsymbol{e}_{\sigma}'={\Lambda^{\nu}}_{\mu} {\Lambda^{\sigma}}_{\rho} \eta_{\nu \sigma}.$$
Now it is conventient to introduce also the inverse metric components via
$$(\eta^{\mu \nu})=(\eta_{\mu \nu})^{-1}=\mathrm{diag}(-1,-1,-1,1)$$
and covariant vector components and contravariant basis vectors via
$$V_{\mu} = \eta_{\mu \rho} V^{\rho}, \quad \boldsymbol{e}^{\mu}=\eta^{\mu \rho} \boldsymbol{e}_{\rho},$$
and then you have, e.g.,
$$\boldsymbol{V}=\boldsymbol{e}^{\mu} V_{\mu}=\eta^{\mu \nu} \boldsymbol{e}_{\nu} V_{\mu}$$
and
$$\boldsymbol{V} \cdot \boldsymbol{W}=\eta_{\mu \nu} V^{\mu} W^{\nu}=V^{\mu} W_{\mu}=V_{\mu} W^{\mu},$$
etc.

For the Lorentz transformation properties of the components you now get in a straight forward way
$$\boldsymbol{V}=V^{\mu} \boldsymbol{e}_{\mu} = V^{\mu} {\Lambda^{\nu}}_{\mu} \boldsymbol{e}_{\nu}',$$
and thus since the decomposition of a vector in terms of a basis is unique (i.e., the basis vectors are a linearly independent set of vectors)
$$V^{\prime \nu}={\Lambda^{\nu}}_{\mu} V^{\mu}.$$
For the covariant components you find
$$V_{\rho} ' =\eta_{\rho \nu} V^{\prime \nu}=\eta_{\rho \nu} {\Lambda^{\nu}}_{\mu} V^{\mu} = \eta_{\rho \nu} {\Lambda^{\nu}}_{\mu} \eta^{\mu \sigma} V_{\sigma} = {\Lambda_{\rho}}^{\sigma} V_{\sigma}.$$
I hope this clarifies your confusion about how to handle all the signs from the pseudo-scalar-product of Minkowski four-vector space!

Incand
Cheers! That was very well explained and helped me a lot!