A mathematical doubt in the expression for the energy of a continuous Charge Distribution.

Sep 2019
16
2
I'm studying Introduction to Electrodynamics by David J.Griffiths and here I'm gonna present his method of calculating the the energy stored (or work done in assembling) of continuous charge distribution.
\(\displaystyle W = \frac{1}{2} \int \rho V~d\tau\)
\(\displaystyle d\tau\) stands for volume element.
By Gauss' Law \(\displaystyle \frac{\rho}{\epsilon_0} = \mathbf{\nabla} \cdot \mathbf{E} \) so, \(\displaystyle W= \frac{\epsilon_0}{2} \int (\mathbf{\nabla}\cdot\mathbf{E})V~~d\tau\)
Now, using integration by parts and considering first function \(\displaystyle (\mathbf{\nabla} \cdot \mathbf{E})\) and second function \(\displaystyle V\) we can write things like this
\(\displaystyle W = \frac{\epsilon_0}{2} \int (\mathbf{\nabla}\cdot\mathbf{E}~~d\tau - \int \mathbf{\nabla}V ~~\int(\mathbf{\nabla}\cdot\mathbf{E})~d\tau ~~~~~~~~d\tau \)
Now, by The Fundamental Theorem For Divergences we have \(\displaystyle \int (\mathbf{\nabla}\cdot\mathbf{E})~~d\tau ~~~~~~= ~~~~~~~\oint_S \mathbf{E} \cdot d\mathbf{a} \) and we can write the above equation as
\(\displaystyle W = \frac{\epsilon_0}{2} V \oint_S \mathbf{E}\cdot d\mathbf{a} - \int \mathbf{\nabla}V \oint_S \mathbf{E} \cdot d\mathbf{a}~~~~~~~d\tau ~~~~~~~~~~~~~~~~~ (1) \)
But in the book the équation (1) is written as \(\displaystyle W = \frac{\epsilon_0}{2} \left [ - \int \mathbf{E}\cdot(\nabla V) ~~d\tau + \oint_S V\mathbf{E}\cdot d \mathbf{a} \right ]\)
The problem is how he has converted \(\displaystyle \int \nabla V \oint_S \mathbf{E} \cdot d\mathbf{a}~~~~~d\tau\) into \(\displaystyle \int \mathbf{E} \cdot (\nabla V) ~~~~~~d\tau \).
I mean how converted that surface integral to get \(\displaystyle \mathbf{E}\). I'm attaching an image of the book so that you can his arguments.

Thank you. Any help will be much appreciated.
 

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topsquark

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All right. You are having some problems with your integration by parts. I often do too, so let's do this a slightly different way.

Let's look at
\(\displaystyle W = \dfrac{\epsilon _0}{2} \int V \left ( \vec{ \nabla } \cdot \vec{E} \right ) d \tau\)

Note that \(\displaystyle \vec{ \nabla } \cdot \left ( V \vec{E} \right ) = V \left ( \vec{ \nabla } \cdot \vec{E} \right ) + \vec{E} \cdot \left ( \vec{ \nabla } V \right )\)

So the integral becomes
\(\displaystyle W = \dfrac{ \epsilon _0}{2} \int \left ( \vec{ \nabla } \cdot \left ( V \vec{E} \right ) - \vec{E} \cdot \left ( \vec{ \nabla } V \right ) \right ) ~ d \tau\)

By Gauss' Law \(\displaystyle \int _{vol} \vec{ \nabla } \cdot \left (V \vec{E} \right ) ~d \tau = \oint _{ \sigma } V \vec{E} \cdot d \vec{a}\)

So we finally get
\(\displaystyle W = \dfrac{\epsilon _0}{2} \int V \left ( \vec{ \nabla } \cdot \vec{E} \right ) d \tau = \dfrac{ \epsilon _0}{2} \oint _{ \sigma } V \vec{E} \cdot d \vec{a} - \dfrac{ \epsilon _0}{2} \int \vec{E} \cdot \left ( \vec{ \nabla } V \right ) d \tau\)

-Dan
 
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Sep 2019
16
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All right. You are having some problems with your integration by parts. I often do too, so let's do this a slightly different way.

Let's look at
\(\displaystyle W = \dfrac{\epsilon _0}{2} \int V \left ( \vec{ \nabla } \cdot \vec{E} \right ) d \tau\)

Note that \(\displaystyle \vec{ \nabla } \cdot \left ( V \vec{E} \right ) = V \left ( \vec{ \nabla } \cdot \vec{E} \right ) + \vec{E} \cdot \left ( \vec{ \nabla } V \right )\)

So the integral becomes
\(\displaystyle W = \dfrac{ \epsilon _0}{2} \int \left ( \vec{ \nabla } \cdot \left ( V \vec{E} \right ) - \vec{E} \cdot \left ( \vec{ \nabla } V \right ) \right ) ~ d \tau\)

By Gauss' Law \(\displaystyle \int _{vol} \vec{ \nabla } \cdot \left (V \vec{E} \right ) ~d \tau = \oint _{ \sigma } V \vec{E} \cdot d \vec{a}\)

So we finally get
\(\displaystyle W = \dfrac{\epsilon _0}{2} \int V \left ( \vec{ \nabla } \cdot \vec{E} \right ) d \tau = \dfrac{ \epsilon _0}{2} \oint _{ \sigma } V \vec{E} \cdot d \vec{a} - \dfrac{ \epsilon _0}{2} \int \vec{E} \cdot \left ( \vec{ \nabla } V \right ) d \tau\)

-Dan
Thank you sir for replying. I don't know about this identity
\(\displaystyle \vec{ \nabla } \cdot \left ( V \vec{E} \right ) = V \left ( \vec{ \nabla } \cdot \vec{E} \right ) + \vec{E} \cdot \left ( \vec{ \nabla } V \right )\)
Can you please give me some links which explains this?
 

topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
It's simply the product rule as used by the divergence. Consider the 1-D case:

\(\displaystyle \dfrac{d}{dx} ( f(x) g(x) ) = g(x) \dfrac{df}{dx} + f(x) \dfrac{dg}{dx}\)

We can look at this in terms of components. Let \(\displaystyle \partial _i\) be the derivative in the ith direction (just like x, y, z except I'm using an i instead.)

This all looks a lot worse than it is. Don't try to inhale the formulas all at once...take it one step at a time.

The gradiant operator is \(\displaystyle \vec{ \nabla } = \hat{x} \partial _x + \hat{y} \partial _y + \hat{z} \partial _z = \sum_i \hat{i} \partial _i\). (Where \(\displaystyle \hat{i}\) is the unit vector in the "ith" direction.)
So the gradiant of a scalar B is
\(\displaystyle \vec{ \nabla } B = \sum_i \hat{i} \partial _i B\).

The divergence of \(\displaystyle \vec{A}\) would be writen as
\(\displaystyle \vec{ \nabla } \cdot \vec{A} = \left ( \sum _i \hat{i} \partial _i \right ) \cdot \left ( \sum_i \hat{i} A_i \right ) = \sum_i \partial _i A_i\)

So putting it together:
\(\displaystyle \vec { \nabla } \cdot \left ( V \vec{E} \right ) = \left ( \sum_i \hat{i} \partial _i \right ) \cdot \left ( \sum_i \hat{i} V E_i \right ) = \left ( \sum_i \hat{i} \partial _i V \right ) \cdot \left ( \sum_i \hat{i} E_i \right ) + V \left ( \sum_i \partial _i E_i \right )\)

\(\displaystyle = \left ( \vec{ \nabla } V \right ) \cdot \vec{E} + V \left ( \vec{ \nabla } \cdot \vec{E} \right )\).

Here's what I could find on the internet. Focus on the sub-section "Cartesian Coordinates" (in the "Definition in Coordinates" section) and the "Properties" section. However I'd advise looking in a Calculus III textbook.

-Dan
 
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Sep 2019
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Sir your little note that it’s just a product rule has solved all my problem at once. I didn’t even look at the link.
 
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topsquark

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Apr 2008
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On the dance floor, baby!
Sir your little note that it’s just a product rule has solved all my problem at once. I didn’t even look at the link.
You mean I had to type all of that and for nothing?! You ungrateful little..

Just kidding. I'm glad it helped. :)

-Dan
 
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