SELF

Addition to the formula (4)

S.B. Karavashkin and O.N. Karavashkina

Newsgroup: sci.physics, sci.physics.electromag, alt.sci.physics.new-theories, fj.sci.matter, sci.physics.relativity
Date: 2004-02-09 01:48:31 PST

[snip]

Looking at your

http://selftrans.narod.ru/v4_1/grad/grad02/grad02.html

we immediately see that your equation (4) is wrong since

phi depends on theta in your equation (3).

In your case where alpha is constant and zero, you should

write:

grad(phi) = @phi/@r e_r + 1/r @phi/@theta e_theta

Your equation (5) for the curl is okay.

So your equation (6) is wrong.

Compare with the correct expressions for grad in eq (3)

and curl in eq (72) in spherical coordinates:

http://164.8.13.169/Enciklopedija/math/math/s/s571.htm

Note that:

your phi is their F

your alpha is their theta

your theta is their phi

You made a very elementary mistake.

Dirk Vdm

My response to this post by Dirk

Thank you, Dirk. At last I see that someone analyses our work, not trying to thoughtlessly squeeze it into the procrustean bed of dogmas. Though this inaccuracy which you have found does not effect on the conclusion that curl of gradient does not vanish, none the less, I'm very pleased. I fully agree with you, gradient of scalar potential has to contain not only radial but also tangential component. Our analysis that you can find some further in this paper, in the problem of field of oscillating potential source - formula (14) in the page 7 - corroborates this.

To show that the inaccuracy you found will not turn to zero the curl of gradient, we will proceed from the expression (3) of our discussed paper [1] for scalar potential, which you think correct:

(1)

and standard expression for gradient of scalar function in spherical coordinates [2, p. 183] which you seemingly do not question:

(2)

Taking into account (1), in the right-hand part of (2) only first and third summands will remain, as (1) does not depend on . This is because the problem of dipole radiation is axially symmetric.

Substituting (1) into (2), we yield

(3)

In (3) we can see such feature that the tangential component of gradient of scalar potential does not depend on the angle , whilst the radial component does depend on the cosine of this angle. This difference in regularities just causes that curl of gradient of potential does not turn into zero. To show it, determine the curl of vector (3). To make so, it will be sufficient to determine only the projection which does not vanish automatically:

(4)

Thus, Dirk, if you admit true the expression for scalar potential (1) and standard expressions for gradient of potential and curl of vector, then the curl of scalar potential does not turn into zero in dynamic fields, even if we take into account the tangential component of gradient. None the less, I'm very grateful that you found it. This shows again, how hard is it - to get over the inertia of habit and to understand, scalar potential has quite complicated regularity in dynamic fields. Due to this, it requires from us to be attentive in calculations of dynamic fields, and not always the gradient is in direction with the radius-vector from the source of field. But we so much used to... :-)

Kind regards,

Sergey

References:

1. S. B. Karavashkin and O.N. Karavashkina On gradient of potential function of dynamic field. SELF Transactions, 4 (2004), 1- 9,   http://selftrans.narod.ru/v4_1/grad/grad01/grad01.html

2. Korn, G. and Korn, T. Mathematical handbook for scientists and engineers. Mgraw Hill , New York- Toronto- London, 1961