# Derivative of Electromagnetic Field tensor

1. May 17, 2005

### Theraven1982

Hello,

i had a question (as many do on these forums, it appears ;).
I know E_{k}=F_{0k}
I also know B_{k}=(1/2)*(\epsilon_{klm}*F_{lm})

EM field tensor F^{uv} defined as:
(I put tildes (~) in to make it more like a matrix form)

0~~~~E_x~~~E_y~~~E_z
-E_x~~~0~~~~-B_z~~B_y
-E_y~~B_z~~~~0~~~-B_x
-E_z~~-B_y~~~B_x~~~0

However, i want to know how i can write the derivative of this tensor:

0~~~~~~~~Dx(E_x)~~~~Dy(E_y)~~~~~Dz(E_z)
-Dx(E_x)~~~~~0~~~~~~-Dz(B_z)~~~~~Dy(B_y)
-Dy(E_y)~~~Dz(B_z)~~~~~~0 ~~~~~~-Dx(B_x)
-Dz(E_z)~~~-Dy(B_y)~~~~Dx(B_x)~~~~~~0

Where Dx means the derivative wrt x.

So what's the derivative of this tensor in algebraic form?
(like d_{p}F^{uv}*epsilon_{puv} )

I hope someone understands what i mean. And i even more hope someone has a solution: eternal gratitude to this person ;).

2. May 17, 2005

### dextercioby

Are u looking for this?

$$\partial^{\mu}F_{\mu\nu}=j_{\nu}$$

in matrix form...?

$$\left( \begin{array}{cccc}\partial^{0} & \partial^{1} & \partial^{2} & \partial^{3} \end{array}\right) \left (\begin{array}{cccc} 0 & F_{01} & F_{02} & F_{03} \\F_{10} & 0 & F_{12} & F_{13} \\F_{20} & F_{21} & 0 & F_{23} \\F_{30} & F_{31} & F_{32} & 0 \end{array} \right) =\left( \begin{array}{cccc} j_{0} & j_{1} & j_{2} & j_{3} \end{array}\right)$$

Daniel.

3. May 17, 2005

### dextercioby

Of course

$$\partial^{i}=-\partial_{i}$$ (1)

$$\partial_{i}=:\frac{\partial}{\partial x^{i}}=\left(\begin{array}{ccc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \end{array} \right)$$ (2)

$$\partial^{0}=\partial_{0}=:\frac{\partial}{\partial t}$$ (3)

$$j_{i}=-j^{i}$$ (4)

Daniel.

NOTE:In both posts i used Heaviside-Lorentz units.

Last edited: May 17, 2005
4. May 17, 2005

### Theraven1982

Hi,

Thanks for the reply, but this is not what i mean.

I'll try to be more clear en complete:

When you have a constant magnetic and electromagnetic field, the equation of motion (in rest frame) is

dS/dt=(ge)/(2mc)SxB. (S=spin)

However, this is only true when the magnetic field is constant.
If we want to add the influences of an electric dipole, we get a similar term:

ds/dt=(ge)/(2mc)SxB + (g'e)/(2mc)PxE

(P=electric dipole moment).
As said, this is only true when the magnetic and electric field are uniform/constant. What if we want to add gradients to these fields?
I did a Taylor / Mac Laurin's expansion of

SxB(z) (Cross product of S and B in point z)
PxE(z)

You get:

SxB(z)=SxB(0)+Sx(z.(divB))
PxE(z)=PxE(0)+Px(z.(divE)).

How does this translate to the Electromagnetic Field Tensor? How is this expressable with the EM tensor? If we want to do the same with the Electromagnetic Field Tensor F, what do we get?

In your first reply you take derivatives to all coordinates:time derivative of F_{01} and the space derivatives of F_{01}.
I only want for

Electric Field:
#F_{0k} --> Dk(F_{0k} ) (for example, derivative to x for E_x; derivative to y for E_y, etc. So that you get the dot-product of the E-field.)

Magnetic Field:
#F_{kj} --> Dm(F_{kj} ) (for example, derivative to x (coordinate 1) for F_{23}=B_x; derivative to y for F_{13}=B_y, etc.)

(write it in a matrix and you will see).
THIS i want to be expressed in an algebraic form as function of the tensor, and derivative.

Man, it's hard to explain. Don't think i will ever be a teacher ;).

Or am I asking something that doesn't make any sense? (or is impossible?)

Thanks again,

5. May 17, 2005

### dextercioby

That doesn't make any sense...The magnetic field can be constant,the electromagnetic field,not.

So you're interested in the spin precession,a.k.a.Thomas precession.Should have said it from the very beginning.

Never seen this equation.Could u supply a reference for it ?

Could u rewrite those? (it would be okay,if u used LaTex) I don't think they look okay...

Is that a covariant formulation of the Thomas precession,that u're interested in?I'll search for it.

This part doesn't make sense.The covariant form of $\vec{E}$ and $\vec{B}$ is $F_{\mu\nu}$.The only derivatives are the the ones in Minkowski space.

Daniel.

6. May 17, 2005

### dextercioby

Section 11.11,pages 561 pp. 565 from the 3-rd edition (1999,Wiley) of J.D.Jackson's "Classical Electrodynamics" is all you need.

Daniel.

7. May 17, 2005

### Theraven1982

Thanks again. I'll try to find the book and read it...