Stupid question about invariance of maxwell equations

[kron]

Hi,

as you all know one can write the Maxwell-equations in covariant form, namely:

\partial_a F^{ab} = \frac{4\pi }{c} j^{b}

and

\partial_a G^{ab}=0

where \textbf{G} is the dual Tensor to \textbf{F}.

Now the two simple questions.
I can see that they are invariant, because I have a 4-Vector on both sides, and so the rhs and lhs
will transform in the same way, right ?
So the equation will have in another frame exactly the same form.

But on the other hand this equations would be invariant under all such transformations, not only
Lorentztransformations ?

I don't get it..

Thanks

 

[Fredrik]

All "such" transformations? Do you mean all linear transformations?

\partial'_a F'^{ab}=\Lambda_a{}^c\Lambda^a{}_d\Lambda^b{}_e \partial_c F^{de}=\delta^c_d\Lambda^b{}_e \partial_c F^{de}=\Lambda^b{}_e \partial_c F^{ce}=\Lambda^b{}_e j^e=j'^b

The second equality only holds for Lorentz transformations.

 

[malawi_glenn]

What do you mean by "all such" transformations?

What is meant by "such"? What do you refer to? example?

Lorentz transformations is boosts in space-time, rotations, parity and time-reversal.

Secondly, the quantities you wrote are not invariant under LT, they transforms as Lorentz 4vectors. The invariants under Lorentz transformation are Lorentz scalars, such as:
F^{\mu\nu}F_{\mu \nu} and
x^{\mu}x_{\mu} .

Those are invariants under Lorentz transformations. A 4vector such as the 4current
j^\alpha transforms:
j'^\alpha = \Lambda ^\alpha_\beta j^\beta \neq j^\alpha
So the 4-current is not a Lorentz invariant.

But you are correct that the "equations are invariant".

 

[robphy]

But on the other hand this equations would be invariant under all such transformations, not only
Lorentztransformations ?

I don't get it..

Thanks

I've been trying to sort this issue out as well.

Some facts that play a role in this puzzle.
- The above equations (or something close to it) are associated with terms like "Generally Covariant Form" or "pre-metric".
- The metric plays a role in defining the "duality" relationship between F and G, which is associated with the constitutive relations. (Do you use E,B or E,B,D,H?)

 

[kron]

All "such" transformations? Do you mean all linear transformations?

\partial'_a F'^{ab}=\Lambda_a{}^c\Lambda^a{}_d\Lambda^b{}_e \partial_c F^{de}=\delta^c_d\Lambda^b{}_e \partial_c F^{de}=\Lambda^b{}_e \partial_c F^{ce}=\Lambda^b{}_e j^e=j'^b

The second equality only holds for Lorentz transformations.

Yes I mean linear transformations, sorry.
My problem is, what can I say about the \Lambda^b{}_e-matrices ?
I can put anything I want in the lambda's, because the equations (not the 4-vectors) will
remain invariant.

I'm not sure if the second equality just holds for LT. Isn't this just a general property
of tensor-transformation ?

What do you mean by "all such" transformations?

What is meant by "such"? What do you refer to? example?

Lorentz transformations is boosts in space-time, rotations, parity and time-reversal.

Secondly, the quantities you wrote are not invariant under LT, they transforms as Lorentz 4vectors. The invariants under Lorentz transformation are Lorentz scalars, such as:
F^{\mu\nu}F_{\mu \nu} and
x^{\mu}x_{\mu} .

Those are invariants under Lorentz transformations. A 4vector such as the 4current
j^\alpha transforms:
j'^\alpha = \Lambda ^\alpha_\beta j^\beta \neq j^\alpha
So the 4-current is not a Lorentz invariant.

But you are correct that the "equations are invariant".

Sorry for my sloppiness, yes I mean linear transformations and of course the invariance of
the equations and not 4-vectors or the field-strength tensor or something like that.

I've been trying to sort this issue out as well.

Some facts that play a role in this puzzle.
- The above equations (or something close to it) are associated with terms like "Generally Covariant Form" or "pre-metric".
- The metric plays a role in defining the "duality" relationship between F and G, which is associated with the constitutive relations. (Do you use E,B or E,B,D,H?)

I use E and B like in Jackson. It's just all a little bit too short in Jackson (for my purpose).
So would it be right to say the above two equations are invariant under all linear transformations which
include the Lorentz transformations ?

Thanks for all your help !

 

[Fredrik]

I'm not sure if the second equality just holds for LT. Isn't this just a general property
of tensor-transformation ?

No, what I used there is the definition of a Lorentz transformation:

\Lambda_a{}^c\Lambda^a{}_d=\delta^c_d

This might look more familiar to you if you multiply by \eta_{bc}

\Lambda_a{}^c\eta_{bc}\Lambda^a{}_d=\delta^c_d\eta_{bc}=\eta_{bd}

and then rewrite the left-hand side as

\Lambda_a{}^c\eta_{bc}\Lambda^a{}_d=\Lambda_a{}_b\Lambda^a{}_d=\Lambda^c{}_b\eta_{ca}\Lambda^a{}_d

So now we have

\Lambda^c{}_b\eta_{ca}\Lambda^a{}_d=\eta_{bd}

which clearly is just the bd component of the matrix equation

\Lambda^T\eta\Lambda=\eta

 

[kron]

Ok, thanks Fredrik.

I still could say one sums over the index 'a' so I don't have to transform that as a
Tensor with three indices but as a "normal" 4-vector.

\partial'_a F'^{ab}=j'^b

But if I would take an arbitray linear transformation for that I wouldn't get the desired result,
because the primed quantities on both sides wouldn't have terms that cancel (if I would expand them into
their relations with the unprimed ones), so that one would
get the unprimed quantites on both sides as expected.
That was pretty much where I got stuck. And I think that this is
what is really meant by an equation is lorentz invariant.

But how is it possible to see, that this transformation will turn out, that the
extra terms on both sides cancel ? I mean on the right I have a 4-vector, on the left I have the
divergence of a second rank tensor, which is also (a different) 4-vector.

Edit:

I mean I can not see that the divergence of the field strength tensor on the left
and the 4-vector on the right have a priori this transformation relation.

Thanks