Can the Dot Product of Electric and Magnetic Fields be Proven as Invariant?

[Ed Quanta]

If I were to attempt to prove that the dot product of an electric and magnetic field is invariant under the conditions of Einstein's Special Theory of Relativity, how would I do this? Would the proof be very involved and complicated? Or should I just use hypothetical magnetic and electric fields and demonstrate how the dot product is unchanged when dealing with relativistic frames of reference?

 

[turin]

Originally posted by Ed Quanta
If I were to attempt to prove that the dot product of an electric and magnetic field is invariant under the conditions of Einstein's Special Theory of Relativity, how would I do this?

First of all, what makes you think that the dot product is invariant? I don't know for sure that it is not, but I would imagine that it is not. Did you hear or read this somewhere, or is this your own idea?

Originally posted by Ed Quanta
Would the proof be very involved and complicated?

I would personally use the Faraday tensor (in fact, after I'm done typing, I will do this and see for myself if the dot product is invariant). It shouldn't be too complicated. I will just consider some arbitrary Faraday tensor, Lorentz transform it in an arbitrary direction at an arbitrary speed, extract the components for the E & H fields to calculate the dot products, and then compare the results. I have a feeling the dot product will be inconstent from one frame to the next.

Originally posted by Ed Quanta
Or should I just use hypothetical magnetic and electric fields and demonstrate how the dot product is unchanged when dealing with relativistic frames of reference?

Do you have formulas for how the field vectors change from one frame to the other?

 

[Ambitwistor]

Originally posted by turin
First of all, what makes you think that the dot product is invariant?

$\vec{E}\cdot\vec{B}$ and $E^2-B^2$ are the only quadratic invariants you can build out of the electric and magnetic fields.

 

[lethe]

Originally posted by Ambitwistor
$\vec{E}\cdot\vec{B}$ and $E^2-B^2$ are the only quadratic invariants you can build out of the electric and magnetic fields.

i didn t know this but it seems believable... basically, to get a lorentz scalar, we would have to contract $F^{\mu\nu}F^{\rho\sigma}$ with something. i think there are only two choices available $\epsilon_{\mu\nu\rho\sigma}$ and $\eta_{\mu\rho}\eta_{\nu\sigma}$

the first choice will give you $E^2-B^2$ and the second $E\cdot B$ (i would imagine. i haven t checked it)

i had a question come up the other day... i was trying to remember the formula for the energy density in terms of the electromagnetic 2 form. i know the lagrangian is $F\wedge*F=(E^2-B^2)\text{vol}$, and since energy density is nearly the same thing, with a plus instead of a minus, i thought i could make a similar formula.

of course i was mistaken, energy density is not a lorentz covariant object, so there will be no way to write down the energy density from the field strength in a covariant way.

but can you write the stress energy tensor? i have seen the Noether formula for the stress energy tensor, but i was hoping for a concise formula from the field strength 2 form. i am suspecting that that doesn t exist either, since the stress tensor is symmetric...

 

[Ambitwistor]

As you say, you can write the electromagnetic stress-energy tensor in terms of the field strength tensor, but not solely by means of operations on differential forms, because the stress-energy tensor isn't a differential form.

 

[turin]

Originally posted by lethe
i think there are only two choices available $\epsilon_{\mu\nu\rho\sigma}$ and $\eta_{\mu\rho}\eta_{\nu\sigma}$

the first choice will give you $E^2-B^2$ and the second $E\cdot B$ (i would imagine. i haven t checked it)

As I started writing to work things out, things started to look like the dot product may be invariant. That does surprise me, given the way the magnetic field arises from the Faraday tensor. From my cursory inspection, though, I would rather have expected contraction with $\epsilon_{\mu\nu\rho\sigma}$ to give the dot product. I have:

$E\cdot B$ = (1/c)(F⁰¹F³² + F⁰²F¹³ + F⁰³F²¹).

(Sorry, I don't know how to make that math looking text.)

Anyway, there are clearly permutations of the indices that suggest nontrivial elements would arise only in the case of a permutation. Are you sure you have the appropriate contractions identified with the corresponding invariants?

Originally posted by lethe
i know the lagrangian is $F\wedge*F=(E^2-B^2)\text{vol}$, and since energy density is nearly the same thing, with a plus instead of a minus, i thought i could make a similar formula.

That is certainly above my level. I still don't have a satisfactory understanding of what a Lagrangian is.

 

[lethe]

Originally posted by turin
Are you sure you have the appropriate contractions identified with the corresponding invariants?

no, i had them backwards. you are correct... the Levi-Civita contraction gives the dot product, and the metric contraction gives the Lagrangian.

 

[lethe]

Originally posted by lethe
no, i had them backwards. you are correct... the Levi-Civita contraction gives the dot product, and the metric contraction gives the Lagrangian.

actually, i should have guess more cleverly. the lagrangian should be invariant under the full Lorentz group (i.e. parity transformations, as well as Lorentz rotations), whereas the dot product, since if involves a pseudovector, should transform nontrivially under parity transformations.

the Levi-Civita tensor reverses sign under parity transformations, so i should have been able to guess which is which...

silly me.

 

[Ed Quanta]

I was told to prove the following in my electromagnetic theory class:
1) that the dot product of E and B would be relatvistically invariant
2) that the quantity E^2 - c^2B^2 is relativistically invariant.


We will assume the two relativistic frames S and S' where x'1=x1, x'2=x2,x'3= j(x3-vt) where j = 1/(square root of 1 - B^2) where B= v/c, and t'=j(t-B/cx3).

When defining the electric and magnetic fields in relativistic frames, current density J and charge density p are not separable nor distinct since a static charge distribution in one frame of reference will be a current density in a moving frame.

In the static frame of reference S we will assume a charge density po in a certain volume, were the change in charge dq=po(dV). Thus in the moving frame of reference S', p=poj. This lorentz contraction and thus change of charge density only occurs in the x3 direction since this is the only dimension which is moving in relativistic frame S'. This is all the information we need to calculate E in the frame of reference S.

Since reference S is static, B=0 because there is no current density. So now we can define the current density J (in frame S' only) as (pu,icp) pu corresponds to the three spatial dimensions, while icp corresponds to the dimension of time. My question is this. How do we calculate E and B in the moving frame of reference S' so I can begin to take the dot products of E and B in both references and see if it is indeed invariant. Sorry if I was confusing or unclear.

 

[turin]

Originally posted by Ed Quanta
When defining the electric and magnetic fields in relativistic frames, current density J and charge density p are not separable nor distinct since a static charge distribution in one frame of reference will be a current density in a moving frame.

In the static frame of reference S we will assume a charge density po in a certain volume, were the change in charge dq=po(dV). Thus in the moving frame of reference S', p=poj.

...

This is all the information we need to calculate E in the frame of reference S.

Since reference S is static, B=0 because there is no current density. So now we can define the current density J (in frame S' only) as (pu,icp) pu corresponds to the three spatial dimensions, while icp corresponds to the dimension of time.

I wouldn't do it this way.

Originally posted by Ed Quanta
This lorentz contraction and thus change of charge density only occurs in the x3 direction since this is the only dimension which is moving in relativistic frame S'.

It doesn't matter what direction; density is density.

Originally posted by Ed Quanta
My question is this. How do we calculate E and B in the moving frame of reference S' so I can begin to take the dot products of E and B in both references and see if it is indeed invariant.

Are you allowed to use the Faraday tensor? Just transform that, take the appropriate components, and compare dot products.

E_i = Fⁱ⁰, cB₁ = F³², cB₂ = F¹³, cB₃ = F²¹.

E'_i = F'ⁱ⁰, cB'₁ = F'³², cB'₂ = F'¹³, cB'₃ = F'²¹.

F'^μν = a^μ_αa^ν_βF^{αβ}.

where a^μ_α are the components of the lorentz transformation matrix.

 

[lethe]

Originally posted by turin

Are you allowed to use the Faraday tensor? Just transform that, take the appropriate components, and compare dot products.

if you are allowed to use relativistic tensors, then it can be even easier.

it is a fact that any tensor expression with all indices contracted must be a (restricted) lorentz scalar. thus, just take the tensor expression above, check that it is equal to E*B, and you re done. you don t even have to check how it transforms, since it is guaranteed to be a scalar.

 

[chroot]

Originally posted by lethe
if you are allowed to use relativistic tensors, then it can be even easier.

it is a fact that any tensor expression with all indices contracted must be a (restricted) lorentz scalar.

This makes sense to me -- if you contract with the metric, you get the Langrangian; if you contract with the Levi-Civita tensor, you get the dot product.

I do have to ask, though -- what exactly is a "Lorentz scalar?" What properties does it have beyond those of the usual scalar?

- Warren

 

[lethe]

Originally posted by chroot

I do have to ask, though -- what exactly is a "Lorentz scalar?" What properties does it have beyond those of the usual scalar?

the (restricted) lorentz scalar is invariant under (restricted) lorentz transformations. and, i could compare this with "the usual scalar", as soon as you tell me which usual scalar you mean.

 

[chroot]

Originally posted by lethe
the (restricted) lorentz scalar is invariant under (restricted) lorentz transformations. and, i could compare this with "the usual scalar", as soon as you tell me which usual scalar you mean.

Hmmm... I guess I always thought that ANY scalar was unchanged by Lorentz transforms.

And I suppose I mean scalar, as per MathWorld: "A one-component quantity which is invariant under rotations of the coordinate system."

- Warren

 

[lethe]

Originally posted by chroot

And I suppose I mean scalar, as per MathWorld: "A one-component quantity which is invariant under rotations of the coordinate system."

yeah, that s about right. so Lorentz scalars are scalars that are invariant under rotations in Minkowski space (the set of all such rotations is called the Lorentz group. hence the name)

 

[lethe]

Originally posted by chroot

And I suppose I mean scalar, as per MathWorld: "A one-component quantity which is invariant under rotations of the coordinate system."

- Warren

note that according to this definition, you might think that energy was a scalar.

and it is, under euclidean rotations. but it is not a Lorentz scalar.

once you choose some coordinate system, i can talk about scalars that are invariant under rotations of that coordinate system.

 

[chroot]

Not necessary, I understand the distinction now. Energy is not a scalar when the group of coordinate transformations in question is the Lorentz group. The best you can say about it that it is the zeroth component of the 4-momentum. Not all "one-component quantities" are scalars -- only those that are invariant under the relevant coordinate transformations.

Thanks as always,

- Warren

 

[Ed Quanta]

I guess my main problem is calculating the 16 components of the electromagnetic field tensor. Once I calculate this, I can just multiply it by the matrix |1 0 0 0 | and |1 0 0 0 |.
|0 1 0 0 | |0 1 0 0 |
|0 0 j iB| |0 0 j-iB|
|0 0-iB j| |0 0 iB j|

This is in accordance to the equation F'= lambda(F)lambda'. Once again I apologize for lack of notation. So in other words, once I can calculate electromagnetic field tensor F, I can calculate the relativistically transformed field tensor F'. And then I should be able to take the dot product of E and B in both relativistic frames,correct?

 

[lethe]

Originally posted by chroot
Not necessary, I understand the distinction now. Energy is not a scalar when the group of coordinate transformations in question is the Lorentz group. The best you can say about it that it is the zeroth component of the 4-momentum. Not all "one-component quantities" are scalars -- only those that are invariant under the relevant coordinate transformations.

sometimes a photon with a time-like polarization is called a scalar photon, not because its a Lorentz scalar, but because its a euclidean scalar.

or, when you do Kaluza-Klein reductions, things that are tensors under SO(4,1) reduce to a tensor, a scalar and a vector under SO(3,1). so one mans scalar is another mans zeroth component. it just depends on which coordinates you feel like working with.

 

[chroot]

Originally posted by lethe
or, when you do Kaluza-Klein reductions, things that are tensors under SO(4,1) reduce to a tensor, a scalar and a vector under SO(3,1). so one mans scalar is another mans zeroth component. it just depends on which coordinates you feel like working with.

Yup.

- Warren

 

[turin]

Originally posted by lethe
thus, just take the tensor expression above, check that it is equal to E*B, and you re done.

Wouldn't this be just as complicated?

Another thing: How do you know that this is going to give you the dot product? I mean, I can see that going throught the calculations would show this, but, how do you know to start with the Levi-Civita contraction in the first place? It seems kind of out-of-the-blue to me. Is there some theorem or something?

 

[turin]

Originally posted by Ed Quanta
... once I can calculate electromagnetic field tensor F, ...

If you can use the Faraday tensor, then you should be able to just say that the corresponding components are the components of the electric and magnetic field, without worrying about the sources. There is no need to calculate these components, you just let them be arbitrary.

Originally posted by Ed Quanta
... I can calculate the relativistically transformed field tensor F'. And then I should be able to take the dot product of E and B in both relativistic frames,correct?

This is basically what I had in mind, but I think lethe has a better idea. I don't quite understand how you can justify starting with the contraction with the Levi-Civita tensor, but apparently it does give the desired result.

 

[lethe]

Originally posted by turin
Wouldn't this be just as complicated?

you be the judge

Another thing: How do you know that this is going to give you the dot product? I mean, I can see that going throught the calculations would show this, but, how do you know to start with the Levi-Civita contraction in the first place? It seems kind of out-of-the-blue to me. Is there some theorem or something?

let s have a look:

the Levi-Civita tensor will select all the terms with even permutations of 1234 with a plus sign, and all the terms with odd permutations with a minus sign


$$\begin{multline*} F^{\mu\nu}F^{\rho\sigma}\epsilon_{\mu\nu\rho\sigma}\\ =4(F^{01}F^{23}+F^{02}F^{13}+F^{03}F^{12})=4\mathbf{E}\cdot\mathbf{B} \end{multline*}$$

and that s all i need to do, it is guaranteed to be lorentz invariant.

 

[Ed Quanta]

Thank you Turin and everybody. I think I have got it now.

 

[turin]

Originally posted by lethe
the Levi-Civita tensor will select all the terms with even permutations of 1234 with a plus sign, and all the terms with odd permutations with a minus sign

Yeah, I'm with you on this one. It just wouldn't have occurred to me to use the Levi-Civita until after I had written it out once. I was just wondering if there was some other reason behind it besides seeing that it does in fact work. Like, "given a ..., contraction with the Levi-Civita tensor ..." I don't really know exactly what I want to hear about this. I just don't like proofs to come out of the azz.

 

[lethe]

Originally posted by turin
Yeah, I'm with you on this one. It just wouldn't have occurred to me to use the Levi-Civita until after I had written it out once. I was just wondering if there was some other reason behind it besides seeing that it does in fact work. Like, "given a ..., contraction with the Levi-Civita tensor ..." I don't really know exactly what I want to hear about this. I just don't like proofs to come out of the azz.

well, this is pretty much how proofs with tensors go. you start with some tensor expression of some rank and you want some other tensor expression of some other rank. there are only a few valid manipulations you can do to these tensors to accomplish this. contracting indices with another tensor is always something you can do. if the space has a metric, then the metric tensor is available for this purpose. if the space has an orientation, then the Levi-Civita tensor is also available. if the space has a connection, then you can also take the covariant derivative.

there are some other stuffs too, but the problem didn t call for any more involved manipulations. maybe you are very familiar with raising and lowering indices with the metric tensor, and not so familiar with raising and lowering indices with the Levi-Civita. most people are. well, let this thread be a lesson to you.

it is worth noting that when i was considering what possible quadratic scalars i can make out of $F^{\mu\nu}$, i also had to consider $F^{\mu\nu}F^{\rho\sigma}\eta_{\mu\nu}\eta_{\rho\sigma}$. this is another possible scalar that i could make out of the field strength, but i didn t mention it above because it vanishes (the field tensor is antisymmetric and hence traceless)

 

[lethe]

Originally posted by turin
but, how do you know to start with the Levi-Civita contraction in the first place? It seems kind of out-of-the-blue to me. Is there some theorem or something?

and, as you saw, i didn t know to start with the Levi-Civita in the first place. in fact, i made a mistake and thought the answer would be to start with the contraction with the metric tensor instead.

it wasn t until you pointed out that i was wrong that i realized this.

even so, i had from the beginning recognized the contraction with the Levi-Civita as one possible way of obtaining a scalar.

and in fact, as i argue above, there is evidence why i should have known before hand to start with Levi-Civita (Levi-Civita changes sign when you reflect it in the mirror, and so does E*B)

 

[lethe]

and i ll just say one more reason why i like my proof a lot:

i have never been able to once and for all put to memory exactly what the elements of the Lorentz transformation look like. if i don t know the elements of the Lorentz transformation, i can t figure out how the electric field and magnetic field transform.

so i can t compare the dot product in two frames, without either going to my book to look up some formulas, or else going to first principles and deriving the form of a Lorentz tranformation ($\Lambda^Tg\Lambda=g$ is how it goes, i think?)

this is a very nice feature of the tensor index notation. all you have to do is figure out how to get rid of all the indices, and you are guaranteed to have a Lorentz scalar.

 

[selfAdjoint]

Lethe, this thread is a lesson to us all.

This

actually, i should have guess more cleverly. the lagrangian should be invariant under the full Lorentz group (i.e. parity transformations, as well as Lorentz rotations), whereas the dot product, since if involves a pseudovector, should transform nontrivially under parity transformations.

is so clear, and so handy, and I never would have thought of it by myself. I have now stored it in my brainbox, and I hope I can find it again when I need it.

Thanks.

 

[turin]

Originally posted by selfAdjoint
Lethe, this thread is a lesson to us all.

This ... is so clear, and so handy, ...

Did you ever expect any less from the man himself?

 

[turin]

Originally posted by lethe
there are only a few valid manipulations you can do to these tensors to accomplish this.

That is the first I have heard of this. What about:

$\eta_{\alpha\sigma}\eta_{\beta\tau}...\eta_{\kappa\epsilon}\eta_{\mu\eta}F^{\alpha\beta}F^{\gamma\delta}...F^{\kappa\lambda}F^{\mu\nu}$

?

Is this not considered distinct from other contractions involving the metric, or is it considered just another version of the same type of contraction. I don't understand how there can only be a few (or even a finite number of) valid manipulations.

(For some reason, It won't let me edit that itex stuff. There are a few typos in it.)

Originally posted by lethe
if the space has an orientation, then the Levi-Civita tensor is also available. if the space has a connection, then you can also take the covariant derivative.

What does it mean for a space to have an orientation or connection?

Originally posted by lethe
maybe you are very familiar with raising and lowering indices with the metric tensor, and not so familiar with raising and lowering indices with the Levi-Civita.

I am not really "familiar" with anything that has to do with tensors. My major prof is so hung up on the facility of the notation that it has been slow going trying to understand the meaning behind any of this stuff. For instance, I thought that raising and lowering (what he calls "index gymnastics") was defined by contraction with the covariant and contravriant components of the metric tensor. This is the first I've heard to the contrary, and now I am utterly confused. I thought that I understood raising and lowering of the indices of a component to be changing that component between contravariant and covariant. I don't understand how the Levi-Civita symbol/tensor can do this, since it doesn't contain any dot products (or does it?).

Originally posted by lethe
well, let this thread be a lesson to you.

As always.

 

[turin]

Originally posted by lethe
(Levi-Civita changes sign when you reflect it in the mirror, and so does E*B)

I got a hold of an undergrad math meth book by Arfken. In the chapter on tensors, it is said that the Levi-Civita symbol isn't really a tensor, but a pseudo-tensor. Any comments on this, and how it relates to the discussion? Does this mean that the dot product in question isn't really a scalar, but a pseudo-scalar? Is this only the case for the Levi-Civita symbol in 3-D, and in 4-D it is a true tensor?

 

[turin]

Originally posted by lethe
i have never been able to once and for all put to memory exactly what the elements of the Lorentz transformation look like. if i don t know the elements of the Lorentz transformation, i can t figure out how the electric field and magnetic field transform.

I do recognize that your preference is your own and that I shouldn't really concern myself, but I just can't resist my curiosity.

Are you saying that you are willing to commit the components of the Faraday tensor to memory before you are willing to commit the elements of the Lorentz transformation? Am I missing something here? It seems like, either way you do this, if you don't have the Faraday tensor memorized, then you're going to have to look it up. And, I don't understand how it would be so much more effort to look up the Lorentz transformation while you're at it.

But, like I said, your preference isn't my business. I'm just curious. Can you fill me in if I'm missing something here?

 

[lethe]

Originally posted by turin
I do recognize that your preference is your own and that I shouldn't really concern myself, but I just can't resist my curiosity.

Are you saying that you are willing to commit the components of the Faraday tensor to memory before you are willing to commit the elements of the Lorentz transformation?

yeah, you bet! but the Faraday tensor is really easy to remember. here it is: any part of the Faraday field tensor with a 0 in it is electric field, and the other nonzero components are magnetic field. that s all i have to know, and its real easy to remember. i ll never have to look that up in a book!

i think there are some plus or minus signs thrown in there for good measure, but who cares about them?

 

[turin]

Originally posted by lethe
any part of the Faraday field tensor with a 0 in it is electric field, and the other nonzero components are magnetic field. that s all i have to know, and its real easy to remember.

Every time I use it, I have to look up which magnetic field component is which Faraday component. What's your trick for remembering this?