Maxwell's Equations and Potentials

[Hornbein]

I have an understanding of Maxwell's equations and a vague grasp on potentials. I'm trying to do something different with the potentials. I'm using the Feynman Lectures on physics, http://www.feynmanlectures.caltech.edu/II_21.html#mjx-eqn-EqII2113, using the equations an potentials in a box about a third of the way in.

I vaguely understand what they are doing with the potentials and think there might be an easier way. Let's say our scalar potential is defined on a 4-vector of t,x,y,z and our grad operator is defined on that as well. My naive question is, what more information is there? It would seem like this would be enough, since if we have a potential like that then everything the particle does in its past and future is included. That is, the problem is already solved, we are just massaging it into a more useful form.

Anyway, that's the best I can do. If there is something like this, surely it has already been done. So I'm hoping someone knows the name of such a thing, or something of that sort. Another way to look at it is it seems to me there should be someway to use one potential instead of two. Maybe.

 

[BvU]

A third of the way into the box ? Richard has done his best to number the equations properly ! Where do you think this smart guy does something that can be done easier ?

 

[jasonRF]

I'm not sure what you are asking. But it is possible to use one vector potential (with 3 components) instead of one scalar potential and one vector potential that combined have 4 components. The Hertz vector potentials are examples. For example, look in chapter 2, section 2.3.2, of
http://space.fmi.fi/~viljanea/eds2005/

These potentials are sometimes used by electrical engineers, which is where I learned them.

Is this helpful?
jason

 

[Hornbein]

A third of the way into the box ? Richard has done his best to number the equations properly ! Where do you think this smart guy does something that can be done easier ?

Thanks a lot. buddy. How might I ever properly express my gratitude for your gracious assistance in a time of need?

 

[Hornbein]

I'm not sure what you are asking. But it is possible to use one vector potential (with 3 components) instead of one scalar potential and one vector potential that combined have 4 components. The Hertz vector potentials are examples. For example, look in chapter 2, section 2.3.2, of
http://space.fmi.fi/~viljanea/eds2005/

These potentials are sometimes used by electrical engineers, which is where I learned them.

Is this helpful?
jason

Yes, it gives a lot more detail about what those potentials are. I think it just might do it.

 

[vanhees71]

In charge-current free regions you can do with two (3D-)scalar Debye potentials. All this freedom is due to the gauge invariance of electromagnetics, which is the most important concept to understand when learning electrodynamics. It's also much simplified by using the relativistic covariant formalism with four-potentials and the Faraday tensor than the old-fashioned 1+3 decomposition. The latter is of course more handy for solving concrete problems.

 

[marcusl]

I thought that Hertzian vectors were valid only in source-free regions.

 

[jasonRF]

I thought that Hertzian vectors were valid only in source-free regions.

Hertz vectors are definitely used in radiation problems with sources - both Sommerfeld and Stratton use them this way, as do other authors. Is that what you were referring to?.
For example, for a potential such that,
<br /> \mathbf{A} = mu_0 \epsilon_0 \frac{\partial }{\partial t} \mathbf{\Pi}<br />
and
<br /> \phi = - \nabla \cdot \mathbf{\Pi}<br />
the retarded potential solution is,
<br /> \mathbf{\Pi}(\mathbf{r},t) = \frac{1}{4 \pi \epsilon_0}\int_0^t dt^\prime \, \int \frac{d^3\mathbf{r}^\prime \, \, \mathbf{J}\left(\mathbf{r}^\prime, t^\prime-\left|\mathbf{r}-\mathbf{r}^\prime\right|/c\right)}{ \left|\mathbf{r}-\mathbf{r}^\prime\right|}<br />

jason

EDIT: would be more informative to note
<br /> \mathbf{E} = \nabla \nabla \cdot \mathbf{\Pi} - \mu_0 \epsilon_0\frac{\partial^2 }{\partial t^2} \mathbf{\Pi}<br />
<br /> \mathbf{B} = \epsilon_0 \nabla \times \frac{\partial }{\partial t} \mathbf{\Pi}<br />
and,
<br /> \nabla^2 \mathbf{\Pi} -\mu_0 \epsilon_0 \frac{\partial^2 }{\partial t^2} \mathbf{\Pi} = -\frac{\mathbf{p}}{\epsilon_0}<br />
where
<br /> \mathbf{J} = \frac{\partial}{\partial t}\mathbf{p}<br />

 

[marcusl]

Very good, thanks for clarifying this for me.