What are the names of these formulas in EM

[sokratesla]

# Hi
# My instructor of EM course wrote two formulas on the blackboard and asked whether they are correct and what their names are. (He does not know the answers too)
# These formulas give V and \vec{A}, given \vec{E} and \vec{B}

V(\vec{r},t) = - \vec{r} \cdot \int_{0}^{1} d\lambda \vec{E}(\lambda\vec{r},t)

\vec{A}(\vec{r},t) = - \vec{r} \times \int_{0}^{1} \lambda d\lambda \vec{B}(\lambda\vec{r},t)

# If we'll learn their names we can continue to investigate the subject further.
# Thanks.

 

[Gokul43201]

The things on the left are called (in order) the scalar and vector potentials of the E- and B-fields.

 

[sokratesla]

The things on the left are called (in order) the scalar and vector potentials of the E- and B-fields.

# Thanks. But I know the names of V and A. :-) We need the names of these formulas which involve the variable \lambda.

 

[Gokul43201]

I didn't read it carefully. I thought \lambda was just a scalar multiplier of the unit vector along the position direction that produces the position vector: \vec{r} = |\vec{r}| \hat{r} \equiv \lambda \hat{r}. That is not the case.

 

[blechman]

V(\vec{r}) = - \int_{\vec{r}_0}^{\vec{r}} d\vec{r}' \cdot \vec{E}(\vec{r}')

\vec{A}(\vec{r}) = - \int_{\vec{r}_0}^{\vec{r}} d\vec{r}' \times \vec{B}(\vec{r}')


These formulas are not what you gave, but they are close. They don't have names. They are simply the definition of the time-independent scalar and vector potential (in integral form rather than the usual differential form) in a particular gauge. It's easy enough to check if they're correct: take the gradient of the first, curl of the second and see if they repoduce the correct answer. Your formulas are the same as mine with the additional constraint that \vec{r}'=\lambda\vec{r} with \vec{r}_0=\vec{0}. This is fine, since changing the path in the contour integral is the same as a gauge transformation.

The time-dependence is a little trickier. I'm not sure if what you wrote down was correct since there's a \frac{d\vec{A}}{dt} term you're not taking into account. Although I admit it's been a while...

But to answer your question: No, they don't have any special names. They're just the definition of potential written in integral form rather than differential form.

 

[Gokul43201]

Your formulas are the same as mine with the additional constraint that \vec{r}'=\lambda\vec{r} with \vec{r}_0=\vec{0}.

This is what I first thought, but it appears the second integral has an extra \lambda in it, no?

Also, the path chosen appears to be a straight line.

 

[blechman]

This is what I first thought, but it appears the second integral has an extra \lambda in it, no?

oh, you're right. That extra \lambda shouldn't be there!

Also, the path chosen appears to be a straight line.

that's okay - it's just a gauge choice.

 

[pmb_phy]

oh, you're right. That extra \lambda shouldn't be there!



that's okay - it's just a gauge choice.

That is not what a gauge choice is. The choice of gauge changes the potential. What curve you evaluate it over is something altogether different. If the curve is closed then the integral for V vanishes regardless of the gauge you choose.

What exactly do you think the term gauge means anyway?

Pete

 

[blechman]

That is not what a gauge choice is. The choice of gauge changes the potential. What curve you evaluate it over is something altogether different. If the curve is closed then the integral for V vanishes regardless of the gauge you choose.

What exactly do you think the term gauge means anyway?

Pete

Go back and review your vector calculus. If you perform the integration along different paths (for example, paths that start at different \vec{r}_0; or paths that wrap around line charges), you will get different potentials. However, you can show that the difference between the integrals along path one and path two are in fact nothing more than a (possibly but not necessarely vanishing) total gradient. This is what I, and everyone else, means when we say gauge.

Perhaps a better way to say it is: if you change your contour of integration, you can at most only change your gauge choice. Is that better?