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In summary, the electric field depends on both vector and scalar potential. However, the contribution of the scalar potential can be comparatively ignored in radiation region.

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Observable said:

well, radiation is nothing more than oscillating electric and magnetic fields, so let's look at maxwell's equations in vacuo...

div

div

since the divergence of the electric field is zero (in vacuum), it is "solenoidal," that is, it's purely rotational, and in that situation, you can't use scalar potential--you have to use vector potential.

now it's in the same boat as the B-field! it's divergence is always zero (so long as we don't find any magnetic monopoles!), so it can't have an associated scalar potential.

...did that help?

now...i'm unsure of how the situation changes once charges are introduced. ...is that your question?

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Brad Barker said:since the divergence of the electric field is zero (in vacuum), it is "solenoidal," that is, it's purely rotational, and in that situation, you can't use scalar potential--you have to use vector potential.

Just to point what seemed to me as an incorrect reasoning:

Uniform electric field has zero divergence and is consequence of a well defined scalar potential V(X,Y,Z) = kX, for instance.

DaTario

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DaTario said:Just to point what seemed to me as an incorrect reasoning:

Uniform electric field has zero divergence and is consequence of a well defined scalar potential V(X,Y,Z) = kX, for instance.

DaTario

right! my mistake.

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from pg. 43 of arfken and weber...

"If we have the special case of the divergence of a vector vanishing, the vector...is said to be solenoidal... When a vector is solenoidal it may be written as the curl of another vector known as the vector potential."

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Observable said:

This is

The separation of E in dV/dr and dA/dt is unfortunately often not given.

(Jackson doesn't show it for instance in Chapter 14 on Radiation by moving charges)

You can find the formulas online here:

http://fermi.la.asu.edu/PHY531/larmor/

see formula 15 for dA/dt and formula 16 for dV/dx. The dotted beta (v/c) is

the accelaration (a/c) which gives the radiation terms. You can simplify the

formula by setting beta itself to zero (v<<c).

You also should be able to find them here:

www.pas.rochester.edu/~dmw/phy218/Lectures/Lect_67b.pdf

At least when rochester.edu is back up again (maintanance?) at page 28.

Regards, Hans

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I once used the radiation gauge and, as I understand it, it represents one possible way of expressing field situations. V and A concepts provide some amount of redundancy, and this gauge choice is a way of fixing this redundant system of language.

Best Regards,

DaTario

A scalar potential problem is a type of problem in physics where the goal is to find the solution to a scalar potential function, which is a mathematical representation of the potential energy of a system. This function is defined in terms of scalar variables, meaning it only has a magnitude and no direction.

A scalar potential problem is different from a vector potential problem in that the former only involves scalar variables while the latter involves both scalar and vector variables. In simpler terms, a scalar potential problem only has a single value for potential energy at any given point, while a vector potential problem has both magnitude and direction for potential energy.

Scalar potential problems have various applications in physics, such as in the study of electric and magnetic fields, fluid dynamics, and quantum mechanics. They are also used in engineering, specifically in the design and analysis of electric circuits and electronic devices.

Scalar potential problems are typically solved using mathematical techniques such as calculus and differential equations. The goal is to find the scalar potential function that satisfies the given conditions and constraints of the problem. This function can then be used to calculate the potential energy at any point in the system.

One of the main challenges in solving scalar potential problems is determining the appropriate boundary conditions and constraints. These conditions can significantly affect the solution and may require additional information or assumptions. Another challenge is that some problems may not have an analytical solution and may require numerical methods for approximation.

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