Tensor gradient and maxwell's equation

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The discussion focuses on the challenges of deriving Maxwell's equations using the gradient of a tensor representing field strength. The tensor provided is analyzed, and the concept of the covariant derivative operator is introduced, which transforms a tensor of rank p into a tensor of rank p+1. The importance of orthonormal coordinate systems is emphasized, where the covariant derivative simplifies to the ordinary derivative. The conversation also touches on the relationship between covariant and non-covariant forms of the equations, specifically how to express Maxwell's equations using the tensor notation. Overall, the participants seek clarity on applying tensor gradients to derive fundamental electromagnetic equations.
steveurkell
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Hi everyone,
I now have difficulties in using formula for gradient of tensor. The following is
tensor for field strength
Fsubscripts_alpha_beta =
[ 0 -Ex -Ey -Ez
Ex 0 Bz -By
Ey -Bz 0 Bx
Ez By -Bx 0 ]
My question is, how do we derive the Maxwell equations div(B) = 0 and curl(E)+dB/dt using gradient of tensor of the above field tensor?
I myself not quite sure how to use formula for tensor gradient, especially due to the additional term apart from the vectors and 1-forms that the tensor has
e.g.
del_zeta (S) = (dS/dx_superscript_delta) zeta_superscript_delta
What does zeta term here do?
I will be indebted if someone here could give a real example of how to find the gradient of a tensor with the tensor given in the example.
Thank you for any help
regards,
 
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This borad has "tex" notation, you can click on any of the equations below to see the "tex" source, which makes it a lot easier to communicate.

I think you are asking about the covariant derivative operator

\nabla_a

What this does is it takes any tensor of rank p, and produces a tensor of rank p+1. A scalar counts as a tensor of rank 0 for this purpose.

If you think about it this should make sense - the partial derivative of a scalar is a vector. The 'a' represents the extra index in the new tensor, you start with a rank p tensor, you end up with a rank p+1 tensor, this means you add one index to the tensor, this added index is represented by the symbol 'a'.

If you have a nice orthonormal coordinate system, the covariant derivative operator reduces to the ordinary derivative.

Thus

\nabla_a f = (\frac{\partial f}{\partial t}, \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})

where I have replaced a_0, a_1, a_2, a_3 with t,x,y,z to make the point clearer.

Note that we started out with a scalar, and wound up with a one-form. The index a represents the components of that one form (the scalar didn't have any indices at all).

The covariant derivative is only equivalent to the partial derivative in orthonormal coordinate systems where the Christoffel symbols are all zero though.

I hope I've understood your question properly, I'm not positive I have. I realize I haven't answered your question about Maxwell's equations yet, but I want to see if the notation I'm using is the notation you are using.

Sometimes people use other notations and notions (like the exterior derivative) - but that would be symbolized by

dF
 
IIUC,you're trying to get from the covariant formalism to the noncovariant one.The equations:

\nabla\cdot\vec{B}=0(1)

\nabla\times\vec{E}+\frac{\partial\vec{B}}{\partial t}=\vec{0} (2-4)...

can be found by simply giving all 4 values to the subscripts in the field equations

\partial_{[\mu}F_{\nu\rho]}=0

(Pervect mentioned covariant derivative,but in this "fortunate" case

\partial_{[\mu}F_{\nu\rho]}\equiv \nabla_{[\mu}F_{\nu\rho]}=0 )

So do it...

Daniel.

P.S.Report any problems.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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