# Homework Help: Potential Theory

1. Jan 6, 2005

### Raparicio

Hello,

I have a problem with an exercice of potential theory, and dont know how to continue.

The ecuations are this:

$$\Phi {\left ( \nabla \frac{{\partial }}{{\partial t}} + \nabla (c· \nabla) \right ) = {\left ( - \frac{{\partial (c· \nabla)}}{{\partial t}} - \frac{{\partial^2 \epsilon \nabla}}{{\partial^2 t}}\right ) \vec {A}$$

$$- \nabla \Phi \mu \sigma + \nabla \Phi \mu \epsilon \frac{{\partial }}{{\partial t}} = \nabla^2 \vec {A} - \left ( \nabla ( \nabla \vec {A}) - \left ( \mu \sigma \frac {\partial} {\partial t} \vec {A} + \mu \epsilon \frac{\partial^2 \vec {A}} {\partial^2 t} \right ) \right )$$

My question in: is the problem resolved, or I must continue? What must I do to end the problem?

2. Jan 6, 2005

### marlon

Well, what exactly is your question ? What are you trying to prove here ??? Are you trying to rewrite the Maxwell equations in terms of the vectorpotential A and the scalar potential phi ???

regards
marlon

3. Jan 6, 2005

### dextercioby

Marlon,u should have told him that the eq.was wrong wrt to mathematics involved.
In the LHS,u have a diff.operator.Because "c=constant",
$$\phi(\nabla\frac{\partial}{\partial t}+c\nabla^{2})$$ (1)
In the expression (1),the function on which the operator's acting is missing.
In the RHS:Again,"c" is constant,so th first terms should be
$$-\frac{\partial}{\partial t}\nabla\cdot \vec{A}-\frac{\partial^{2}\epsilon}{\partial t^{2}}\nabla\cdot \vec{A}$$ (2)
In the second eq.,in the LHS,again u're having the eqlaity between an operator and a fbunch of functions.
In the RHS,u can reduce the laplace-ian of 'A'.
In the first eq.,in the LHS,the missing function should be a vector,but u need to modity it,as the tensor ranks of the 2 diff.operators (nabla & nabla squared is not the same).The tensor rank of the two sides must be equal.
In the second eq.,in the LHS,the missing function is a scalar (again,the tensor rank of the 2 sides must be equal).

Did u make it up??To me it looks like that.

Daniel.

Last edited: Jan 6, 2005
4. Jan 6, 2005

### marlon

Not at all...

This is incorrect and incomplete. It is a classical mistake to assume that the term $$\nabla (c· \nabla)$$ will yield a laplacian. This is however certainly not the case because you need to know on which coordinates the operators will act and on which variable exactly. Besides this is NOT a dot-product, yet this term merely defines a new derivative-operator. This operator is constructed not via the inproduct but via the external vector product, yielding the socalled diade : $$A_{i}B_{j} = (AB)_{ij}$$

besides : i never said the equations were right or wrong...

marlon

Last edited: Jan 6, 2005
5. Jan 6, 2005

### Raparicio

Noooooo... I'm trying to learn Physics, and practising all disciplines. My target is to be more prepared.

If I could do what you say... don't you think my level of mathmatics and physics must be more elevated?

Only a student... with a very great ilusion to learn... and Dextercioby is a great master.

6. Jan 6, 2005

### marlon

I don't think you got my point. I was not trying to be insulting here. The vectorpotential and scalar potential are indeed used in order to rewrite the Maxwell equations into two coupled differential equations. Via the introduction of the gauge-conditions you can decouple these two equations which makes it a lot easier to solve them. You get two equations : one for V and one for A. There will be no V in the equation for A and the other way around. That is what i mean by decoupling...

The actual decoupling happens via the gauge-transformations that you can introduce because the equations for A and V are NOT unique. So you can add a gradient to A and still keep on doing the right calculations. What i mean is this $$B = \nabla \times A$$

But you may as well write (P is a chosen scalar field) $$A = A + \nabla P$$ Then you get $$B = \nabla \times A + \nabla \times (\nabla P)$$ then last term in the RHS is ZERO due to the mixed product...

:uhh: what ever...trust me, even he is not always right and somethimes a bit too willing to correct everybody...

regards
marlon

7. Jan 6, 2005

### dextercioby

Magister dixit: :tongue2:

Even if so,if it ain't a contracted tensor product (resulting in the scalar laplace-ian),the eq.would still not be balanced:
The Hessian has tensor rank 2 (it's a square matrix,whose trace is the laplace-ian),while the nabla (in the same side) has a tensor rank one.

Daniel.

PS.If i wasn't wrong from time to time,i'd be really boring,right...??