Field Strenght Tensor and its Dual (in SR)

[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

 

[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...?

 

[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).

 

[Zag]

Thank you for your replies, everyone.

"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...?

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

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).

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

 

[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.

 

[Bill_K]

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).

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

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"?

 

[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...

 

[samalkhaiat]

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...

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 }).

 

[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.

 

[TrickyDicky]

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.

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

 

[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.

 

[robphy]

Possibly helpful...

By googling your quoted sentence without the quotes, the second link leads to
http://arxiv.org/pdf/1404.0409 (see page 2)

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

 

[Bill_K]

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.

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.

 

[samalkhaiat]

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.

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 }.

 

[Matterwave]

Possibly helpful...

By googling your quoted sentence without the quotes, the second link leads to
http://arxiv.org/pdf/1404.0409 (see page 2)

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

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

 

[Zag]

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.

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.

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