Discrepancies with Maxwell's Eqns - vector potentials

[VictorVictor5]

Greetings all,

Trying to resolve a discrepancy with vector and scalar potentials with Maxwell's Equations, specifically Ampere's law.

In my E&M textbook (Balanis, 1989, Eqn 6-17), Ampere's law with a magnetic vector potential and electric scalar potential can be expressed as

$$E= -\nabla\phi-j \omega A$$

where $$\phi$$ is the electric scalar potential, and A is the magnetic vector potential.

Now, in a paper I am referencing in my work, I see Ampere's expressed as the following:

$$E=-j \omega(A- \nabla \phi)$$

When you distribute this equation, you get the $$-j \omega A + j \omega \nabla \phi$$

where now the scalar potential is positive, and also has a $$j \omega$$ in front of it, where the first equation doesn't.

Is it because of the scalar potential being arbitrary since it's a function of position? Or is there something else?

I also checked Harrington, but no luck there either.

Thanks!
VV5

 

[DuckAmuck]

My guess is the notation in the paper, compared to the notation you're used to is related by $$\phi = -\frac{\phi}{jw}$$

So it's just a unit conversion basically. You see this a lot in EM. There's several different conventions.

 

[VictorVictor5]

DuckAmuck,

First, thanks for the reply.

Question for you. While the ratio you provided would work, and given that -j * -j = well, j^2, = -1 and the equation would work, but quick question. The ratio you provided - physically what would that mean?

Thanks again!
VV5

 

[DuckAmuck]

DuckAmuck,

First, thanks for the reply.

Question for you. While the ratio you provided would work, and given that -j * -j = well, j^2, = -1 and the equation would work, but quick question. The ratio you provided - physically what would that mean?

Thanks again!
VV5

It's just a unit convention. For example, in particle physics we like to set speed of light equal to 1, for simplicity, so we're not writing "c" over and over. So E=mc^2 becomes E=m. The consequence of this is that energy and mass are in the same units, which is okay, just something to keep in mind when doing problems.

It doesn't really mean much *physically*, it's just a writing convention. And actually, the convention I'm used to doesn't even use j, all components are real. :)

 

[VictorVictor5]

Great - thanks so much for your help!

VV5

 

[DuckAmuck]

Great - thanks so much for your help!

VV5

What I generally try to do is work through what the paper does in the convention I'm used to.