# Field Strenght Tensor and its Dual (in SR)

1. Apr 10, 2014

### Zag

Hello everyone,

I have recently read a puzzling statement on my Electromagnetism (Chapter on Special Relativity) material regarding the Field Strength Tensor, $F^{\mu\nu}$, and its dual, $\tilde{F}^{\mu\nu}$. Since I've been thinking about this for a while now, and still can't understand it, I was hoping to hear your thoughts about it.

I believe we all agree that a possible definition for the Dual Tensor in terms of the original Field Strength Tensor is the following:

$\tilde{F}^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\sigma\rho}F_{\sigma\rho}$

Having that in mind, the text states the following:

"It can be shown that this is the only way in which we can construct a Lorentz-invariant four-tensor involving the fields that is independent of the original field strength tensor."

I can't understand why should this be the case, nor can I come up with a proof for this statement. Any thoughts on this matter would be greatly appreciated!

Thank you very much,
Zag

2. Apr 10, 2014

### Matterwave

"Independent of the original field strength tensor" doesn't sound right to me...obviously the field strength and its dual hold the exact same information, and there is a 1-1 correspondence between the two with a simple way to go from one to the other. Perhaps the book is using the word "independent" in a way that I'm missing right now...?

3. Apr 10, 2014

### dextercioby

I'm sure you can prove it: $\tilde{F}^{\mu\nu} = A^{\mu\nu\rho\sigma}F_{\rho\sigma}$. $A=A\left(\delta,\epsilon,g\right)$.

4. Apr 10, 2014

### Zag

Thank you for your replies, everyone.

Maybe the author meant "linearly independent"? I'm not sure.

Could you elaborate more on this dextercioby? I am not sure if I am able to follow your notation here.

5. Apr 10, 2014

### Ben Niehoff

I've always thought this style of reasoning somewhat bizarre. "What tensors can you make out of X, Y, and Z?". I think it misses the point. We're not interested in tensors because they have such-and-such transformation properties; we're interested in tensors because they represent real, measurable objects and not coordinate artifacts.

The physical significance of the dual field tensor is quite clear: it exchanges the roles of $\vec E$ and $\vec B$. So if $F$ describes a bunch of electric charges, then $\tilde F \equiv \star F$ describes a bunch of magnetic monopoles.

6. Apr 10, 2014

### Bill_K

I believe he's pointing out that this is not so much a property of the electromagnetic field as it is a property of four dimensions. He's saying, "*F must contain the same components as F, rearranged in some linear fashion, hence it can be obtained from F by multiplication with some constant tensor A," and asking, "What are the only rank four tensors A which have the desired symmetry properties [antisymmetric on the first pair of indices and the last pair] and can be built using only the Kronecker delta δ, the epsilon tensor ε and the metric g"?

7. Apr 10, 2014

### Matterwave

Anyone know what the author meant by "independent"? It's irking me that I can't figure out that statement...it can't be linear independence right since any symmetric tensor is linearly independent from an anti-symmetric tensor...

8. Apr 10, 2014

### samalkhaiat

They have different transformation properties under the Lorentz group $SL(2, \mathbb{C})$. In terms of the (self-dual) $(1,0)$ and the (antiself-dual) $(0,1)$ representations, they can be expressed as
$$F_{ \mu \nu } = ( F^{ ( 1 , 0 ) } + F^{ ( 0 , 1 ) } )_{ \mu \nu } ,$$
$$^{*}F_{ \mu \nu } = ( F^{ ( 1 , 0 ) } - F^{ ( 0 , 1 ) } )_{ \mu \nu } .$$
They also are linearly independent: $a_{ 1 } F_{ \mu \nu } + a_{ 2 } \epsilon_{ \mu \nu \rho \sigma } F^{ \rho \sigma } \neq 0$ for non-zero $(a_{ 1 }, a_{ 2 })$.

Last edited: Apr 10, 2014
9. Apr 10, 2014

### Matterwave

I'm not saying that they aren't linearly independent, I'm just saying that the dual tensor can't possibly be the only rank 2 tensor that's linearly independent... there should be 14 other ones from a simple counting of components.

10. Apr 11, 2014

### TrickyDicky

The quote alluded to the independence of the rank 4 tensor.

11. Apr 11, 2014

### Matterwave

The totally anti-symmetric tensor is the only totally antisymmetric rank 4 tensor in 4 dimensions, but there are still 253 linearly independent other rank 4 tensors in 4 dimensions...? I'm still so confused by that statement lol.

12. Apr 11, 2014

### robphy

http://arxiv.org/pdf/1404.0409 (see page 2)

(What is the original source of your quote? What is the context?)

13. Apr 11, 2014

### Bill_K

The point is that the rank-4 tensor must be built from quantities intrinsic to the 4-geometry. As stated above, these are only the Kronecker delta δ, the metric g, and the epsilon tensor ε. Yes there are many other possible rank-4 tensors, but the additional structure necessary to build them would represent some physical quantity occupying the spacetime and breaking Lorentz invariance.

14. Apr 11, 2014

### samalkhaiat

Can you state a QUESTION instead of throwing meaningless numbers at me? In 4-dimensional Minkowski space-time, there are two Lorentz-invariant tensors, the metric $\eta_{ \mu \nu }$ and the totally antisymmetric tensor $\epsilon_{ \mu \nu \rho \sigma }$. From these you can show that the only rank-2, antisymmetric tensor that can be constructed from $F_{ \mu \nu }$ and linearly independent of it, is the dual tensor $\epsilon_{ \mu \nu \rho \sigma } F^{ \rho \sigma }$.

15. Apr 12, 2014

### Matterwave

Thanks. The quote in that paper makes sense to me.

16. Apr 12, 2014

### Zag

Thank you for your reply, guys. It makes more sense now thinking in these terms. I really appreciate your help. :)