# A Electromagnetic strength tensor

1. Sep 14, 2016

### spaghetti3451

The antisymmetric 2-tensor $F_{ij}$ is given by $F_{ij}\equiv \partial_{i}A_{j}-\partial_{j}A_{i}$

so that $F_{ij}={\epsilon_{ij}}^{k}B_{k}$ and $B_{i}=\frac{1}{2}{\epsilon_{i}}^{jk}F_{jk}$.

I was wondering if the permutation tensor with indices upstairs is different from the permutation tensor with indices downstairs.

2. Sep 14, 2016

### Orodruin

Staff Emeritus
As long as you are doing rotations only (i.e., changing between different Cartesian coordinate systems and no boosts), it does not matter whether the indices are covariant or contravariant.

If you include general coordinate transformations, it does matter, but there is also an ambiguity in how you define the permutation symbol with indices in different places (it is not a tensor, it is a tensor density) and if you want B and F to transform as a pseudo-vector and tensor, respectively, you need to use a tensor instead of the permutation symbol.

3. Sep 14, 2016

### spaghetti3451

So, is the basic idea that it does not really matter if the indices on the permutation symbol are upstairs or downstairs?

4. Sep 14, 2016

### Orodruin

Staff Emeritus
If you restrict yourself to Cartesian coordinate systems and proper rotations, it never does.

5. Sep 14, 2016

### spaghetti3451

I see!

I guess this means that, for general coordinate transformations, it is not recommended to use the permutation symbol (but rather some tensor constructed out of the permutation symbol) to transform the magnetic field and field strength tensor in a tensor equation.

Is that so?

6. Sep 14, 2016

### Orodruin

Staff Emeritus
Right. This tensor would be the tensor $\eta_{ijk} = \sqrt{g} \epsilon_{ijk}$, where $g$ is the metric determinant (which happens to be a scalar density of the appropriate weight to make that thing a tensor). I have here defined the (covariant) permutation symbol $\epsilon_{ijk}$ to have components 1, -1, or 0 in any coordinate system, depending on the indices. Note that the corresponding definition using the contravariant permutation symbol (i.e., the permutation symbol with components 1, -1, or 0 and contravariant indices) would be $\eta^{ijk} = \sqrt{g}^{-1} \epsilon^{ijk}$.

7. Sep 14, 2016

### spaghetti3451

Does this mean that ${\eta_{ij}}^{k}=\sqrt{g}{\epsilon_{ij}}^{k}=-\eta_{ijk}=-\sqrt{g}\epsilon_{ijk}$ in Minkowski space (with mostly negative signature), which means that ${\epsilon_{ij}}^{k}=-\epsilon_{ijk}$ in Minkowski space (with mostly negative signature)?

I'm trying to understand if the raising and lowering of indices (using the MInkowski metric) on $\eta$ follow the usual rules.

8. Sep 14, 2016

### Orodruin

Staff Emeritus
In Minkowski space you are not working with a rank three anti-symmetric tensor and/or permutation symbol. You need to work with a rank 4 tensor instead. The corresponding tensor $\eta$ has three spatial indices for all of its non-zero components and it will therefore matter if you define the tensor using the covariant or contravariant permutation symbol. I strongly suggest not writing upper and lower indices on the same permutation symbol because of the ambiguity in whether you are referring to the covariant or contravariant permutation symbol. If you have a single text where it is well defined what is intended (for example, when it is first introduced you could say that it is the only one you are going to use and then use all of the raised and lowered indices with that definition as the basic object). However, this can cause problems when comparing texts that adopt different definitions and you will then be dealing with signs appearing and disappearing.

9. Sep 14, 2016

### spaghetti3451

I understand that covariant and contravariant permutation symbols are distinct objects in Minkowski space and also in general curvilinear coordinates in Euclidean space.

But aren't permutation symbols with mixed indices also well-defined objects, and so do we not have to know how each index on the permutation symbol (with mixed indices) transforms independently of the other indices?

10. Sep 14, 2016

### Orodruin

Staff Emeritus
Introducing permutation symbols with mixed indices as basic objects is not very well defined. The main reason for this is that the property you want from your permutation symbols is complete anti-symmetry. This becomes a problem if you want to mix covariant and contravariant indices (and are not restricting yourself to Cartesian coordinates). An anti-symmetry between two covariant or between to contravariant indices is well defined, an anti-symmetry between a covariant and a contravariant index is not. If you see a permutation symbol with mixed indices, it is most likely defined as either the contravariant or covariant permutation symbol with the lowered/raised indices contracted with the metric/inverse metric.

11. Sep 14, 2016

### spaghetti3451

Thanks for the explanation.

I was wondering if the electric field vector $\vec{E}$, when written in index notation becomes, $E_{i}$ or $E^{i}$.

Or does it not matter in either Cartesian coordinates or Minkowski space because the object is not a four-vector?

Does a similar conclusion apply for the magnetic field pseudo-vector as well?

12. Sep 14, 2016

### Orodruin

Staff Emeritus
In Cartesian coordinates it does not matter. In Minkowski space with general Lorentz transformations, the object mixes with the magnetic field vector.

The conclusion is similar for the magnetic field, yes. As long as you restrict yourself to rotations (no general coordinates in space or boosts), it does not matter.

13. Sep 14, 2016

### spaghetti3451

The index-free equation $\vec{E}=-\nabla\phi-\frac{\partial\vec{A}}{\partial t}$ becomes the index-full equation $E_{i}=-\partial_{i}\phi-\partial_{t}A_{i}$.
Similarly, the index-free equation $\vec{B}=\nabla\times\vec{A}$ becomes the index-full equation $B_{i}=\epsilon_{ijk}\partial_{j}A_{k}$.

Since this is in Cartesian coordinates, I presume that the position (upstairs of downstairs) of the indices does not matter, so that $E_{i}=-\partial_{i}\phi-\partial_{t}A_{i}$ is the same as $E^{i}=-\partial^{i}\phi-\partial_{t}A_{i}$, where the indices are raised and lowered without care.

Is this true?

Now, in Minkowski space, $F^{i0}=\partial^{i}A^{0}-\partial^{0}A^{i}=-\partial_{i}\phi-\partial_{t}A^{i}$. Now, I am not sure if this expression is equal to $E_{i}$, since in Minkowski space $\partial^{i}=-\partial_{i}$ and $A^{i}=-A_{i}$ but in Euclidean space in Cartesian coordinates $\partial^{i}=\partial_{i}$ and $A^{i}=A_{i}$.

Last edited: Sep 14, 2016
14. Sep 14, 2016

### Orodruin

Staff Emeritus
If you are using the Cartesian metric, yes. (I.e., if you are considering space and time separately.) Beware that if you use the Minkowski metric, raising the index changes the sign of the derivative. You just need to be careful with what you mean when writing down your equations.

Alternatively you can use a metric with -+++ signature and not have this problem for raising and lowering the spatial indices. Some minus signs will pop up in other places though.

15. Sep 14, 2016

### vanhees71

Another point of view is that neither $\vec{E}$ nor $\vec{B}$ alone have a well-determined behaviour under Lorentz transformations but only when brought together in terms of $F_{\mu \nu}$. Particularly neither $\vec{E}$ nor $\vec{B}$ are not spatial components of a Minkowski four-vector but components of the anti-symmetric Faraday tensor.

Of course, on the from the Faraday tensor $\vec{E}$ and $\vec{B}$ have a well-defined transformation behavior under Lorentz transformations.
$$F^{\prime \mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x) = {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(\Lambda^{-1}x). \qquad (*)$$
As turns out the only two scalars (under proper orthochronous Lorentz transformations) are $F_{\mu \nu} F^{\mu \nu}$ and $\epsilon_{\mu \nu \rho \sigma} F^{\mu \nu} F^{\rho \sigma}$ which are $\propto \vec{E}^2-\vec{B}^2$ and $\propto \vec{E} \cdot \vec{B}$. This suggests further that the complex vector (the Hilbert-Silberstein vector)
$$\vec{F}=\vec{E} + \mathrm{i} \vec{B}$$
are objects with a proper transformation behavior under proper orthochronous Lorentz transformations (LTs), because with the normal bilinear scalar product (not the sesquilinear product of a unitary space!) you get
$$\vec{F} \cdot \vec{F}=\vec{E}^2-\vec{B}^2 + 2 \mathrm{i} \vec{E} \cdot \vec{B},$$
which is invariant under LTs. Indeed the transformation properties from (*) lead to the transformation
$$\vec{F'}(x')=D(\Lambda) \vec{F}(x),$$
where $D(\Lambda) \in \mathrm{SO}(3,\mathbb{C})$ builds a proper representation of the LTs. The subgroup $\mathrm{SO}(3,\mathbb{R})$ corresponds to the rotations of course, because indeed $\vec{E}$ and $\vec{B}$ are transforming as vector fields under rotations. A rotation matrix with a purely imaginary rotation angle $\mathrm{i} \eta$ is also in $\mathrm{SO}(3,\mathbb{C})$, and these refer to the pure boosts in the specified direction. The parameter $\eta$ is the rapidity, related to the velocity of the boost by $\beta=v/c=\tanh \eta$.

16. Sep 14, 2016

### Orodruin

Staff Emeritus
While this is true, I would like to remark that the information contained in the E field of an observer is contained in a 4-vector orthogonal to its 4-velocity, namely $F_{\mu\nu} V^\nu$. Similarly $\tilde F_{\mu\nu} V^\nu$ contains exactly the same information as the B field. These vectors are clearly both space-like and have time-component zero and the E and B components, respectively, as spatial components.