# Maxwell's Eqs. & Tensor Notation

• I
• Incand
In summary, the conversation discusses the use of Maxwell's equations to recreate classic equations, such as Gauss's law and the continuity equation. The discussion also touches on the use of the Minkowski metric and its implications for the sign of certain terms in equations. The conversation concludes with a clarification on how to handle the signs in the pseudo-scalar-product of Minkowski four-vectors.
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?

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.

Incand said:
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
Nugatory said:
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.

The Minkowski product is a bilinear form, and it is always good to distinguish vectors and their components although physicists 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!

## 1. What are Maxwell's equations?

Maxwell's equations are a set of four fundamental equations that describe the behavior of electric and magnetic fields in space. They are named after physicist James Clerk Maxwell and are considered to be one of the cornerstones of classical electromagnetism.

## 2. What are the four equations in Maxwell's equations?

The four equations in Maxwell's equations are Gauss's law, Gauss's law for magnetism, Faraday's law of induction, and Ampere's law. Together, these equations describe the relationships between electric and magnetic fields in space.

## 3. What is tensor notation?

Tensor notation is a mathematical notation used to represent tensors, which are mathematical objects that describe the relationships between vectors and scalars in a multi-dimensional space. It is commonly used in mathematical physics and engineering to simplify and generalize equations.

## 4. How is tensor notation used in Maxwell's equations?

In Maxwell's equations, tensor notation is used to represent the relationships between the electric and magnetic fields in three-dimensional space. For example, the electric and magnetic fields are represented as the components of a second-order tensor, known as the electromagnetic field tensor.

## 5. Why is tensor notation important in physics?

Tensor notation is important in physics because it allows for the concise representation of complex mathematical relationships between vectors and scalars in multi-dimensional space. It also allows for the generalization of equations, making them applicable in different coordinate systems and dimensions.

Replies
1
Views
543
Replies
10
Views
912
Replies
1
Views
1K
Replies
7
Views
527
Replies
2
Views
1K
Replies
6
Views
1K
Replies
4
Views
686
Replies
4
Views
3K
Replies
19
Views
3K
Replies
3
Views
2K