Derivative of Electromagnetic Field tensor

In summary, the user is asking for the algebraic form of the derivative of the electromagnetic field tensor, specifically for the electric and magnetic fields, in relation to the spin precession and Thomas precession. They have provided equations for the expansion of SxB(z) and PxE(z) and are asking for clarification on how this translates to the electromagnetic field tensor. They also request for the expression to be written in matrix form. The conversation also touches on the covariant form of the electric and magnetic fields, and the user expresses difficulty in explaining their question.
  • #1
Theraven1982
25
0
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 (~) into 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 ;).
Thanks in advance,
 
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  • #2
Are u looking for this?

[tex] \partial^{\mu}F_{\mu\nu}=j_{\nu} [/tex]

in matrix form...?

[tex] \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) [/tex]

Daniel.
 
  • #3
Of course

[tex] \partial^{i}=-\partial_{i} [/tex] (1)

[tex] \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) [/tex] (2)

[tex] \partial^{0}=\partial_{0}=:\frac{\partial}{\partial t} [/tex] (3)

[tex] j_{i}=-j^{i} [/tex] (4)


Daniel.

NOTE:In both posts i used Heaviside-Lorentz units.
 
Last edited:
  • #4
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
Theraven1982 said:
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,

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

Theraven1982 said:
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.

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

Theraven1982 said:
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).

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

Theraven1982 said:
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)).

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

Theraven1982 said:
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?

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

Theraven1982 said:
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,

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

Daniel.
 
  • #6
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
Thanks again. I'll try to find the book and read it...
Thanks for your time.
 

Related to Derivative of Electromagnetic Field tensor

1. What is the Derivative of Electromagnetic Field Tensor?

The derivative of electromagnetic field tensor refers to the rate of change of the electromagnetic field tensor with respect to a given variable or set of variables. It is a mathematical concept used in physics to describe how the electromagnetic field changes over time or in different conditions.

2. How is the Derivative of Electromagnetic Field Tensor Calculated?

The derivative of electromagnetic field tensor is calculated using calculus techniques such as partial differentiation and vector calculus. It involves taking the partial derivatives of the four components of the electromagnetic field tensor with respect to the four coordinates in spacetime.

3. What Does the Derivative of Electromagnetic Field Tensor Represent?

The derivative of electromagnetic field tensor represents the change in the electromagnetic field over time or in different conditions. It allows us to analyze how the electromagnetic field behaves and evolves in different situations, such as in the presence of electric charges or moving objects.

4. What Are Some Applications of the Derivative of Electromagnetic Field Tensor?

The derivative of electromagnetic field tensor has many applications in physics, particularly in the fields of electromagnetism and relativity. It is used to study the behavior of electromagnetic waves, the interaction between electric and magnetic fields, and the effects of gravity on the electromagnetic field.

5. How Does the Derivative of Electromagnetic Field Tensor Relate to Maxwell's Equations?

The derivative of electromagnetic field tensor is closely related to Maxwell's equations, which are a set of fundamental equations that describe the behavior of electric and magnetic fields. In fact, Maxwell's equations can be written in terms of the electromagnetic field tensor, making it a powerful tool for understanding and solving problems in electromagnetism.

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