# I Maxwell's equations and tensor notation

1. Dec 7, 2016

### Incand

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?

2. Dec 7, 2016

### vanhees71

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:
$$\partial_{\mu}j^{\mu}=\vec{\nabla} \cdot \vec{j}+\partial_t \rho=0.$$

3. Dec 7, 2016

### Incand

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.

4. Dec 7, 2016

### Staff: Mentor

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.

5. Dec 7, 2016

### Incand

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.

6. Dec 8, 2016

### Incand

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.

7. Dec 8, 2016

### vanhees71

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!

8. Dec 8, 2016

### Incand

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