# Proving Green's Reciprocity theorem

This is problem number 50 from the third chapter of potentials from Griffiths:

Two charge distributions, $$\rho_1(\textbf{r})$$ produces a potential $$V_1(\mathbf{r})$$ and some other charge distribution $$\rho_2(\mathbf{r})$$ produces a potential $$V_2(\mathbf{r})$$. We are asked to prove $$\int_\limits{\mathit{all \,space}} \rho_1 V_1\, d\tau = \int_\limits{{all \, space}} \rho_2 V_1 \, d\tau$$ Which is known as "Green's reciprocity theorem"

Here's what I did and the author suggested

We can write$$\int \vec{E_1}\cdot\vec{E_2} \,d\tau$$ in two different ways by writing $$\vec{E} = -\nabla{V}$$

So using the product rule and integrating by parts I arrive at $$\ -\oint_\mathcal{S} V_1 \vec{E_2} \cdot d \vec{a} \, + \int_\mathcal{V} V_1 \frac{\rho_2}{\epsilon_0} d\tau=\ -\oint_\mathcal{S} V_2 \vec{E_1} \cdot d \vec{a} \, + \int_\mathcal{V} V_2 \frac{\rho_1}{\epsilon_0} d\tau$$

Now I make this argument: "Integrating over all space will make the surface integrals vanish as both $$V_1, V_2 \to 0$$ for an arbitrarily large volume." I'm unsure about this as the charge distribution may themselves go on to infinity(?) In that case the integrals won't vanish, I think. Is this argument correct?

Also I would like to ask if Green's reciprocity theorem is simply a mathematical coincidence (which seems unlikely to me) or does it also have any physical significance as well.

Also I'm a high school senior graduating in a few months aspiring to be a physicist. I'm reading EM by Griffiths and was wondering if there are any other good reads that I can check out. I'm willing to work hard to understand the subject. I have read vector calculus only up to Griffiths level but I will brush that up if needed.

Since $$\rho(r)$$ will always be a "well behaved" function, your argument for the surface integral is good.

Moreover we can see the physical meaning of Green's reciprocity theorem looking at the following situation: suppose that we have only one charge in a region $$a$$ such that

$$Q_a = \int_a\rho_1d\tau = Q\qquad Q_b = \int_b\rho_2 d\tau=0$$

now the charge $$Q_a=Q$$ produces a potential where the charge $$Q_b$$ would be placed $$V_{1b}\equiv V_{ab}$$. Now we do the converse, turn off charge $$Q_a$$ and turn on charge $$Q_b$$ such that they produce the same charge $$Q$$

$$Q_a = \int_a\rho_1d\tau = 0 \qquad Q_b = \int_b\rho_2d\tau = Q$$

This charge $$Q_b=Q$$ produces a potential where the charge $$Q_a$$ would be placed $$V_{2a}\equiv V_{ba}$$ since they are generated by the same amount of charge $$Q$$. Using Green's reciprocity theorem we conclude that

$$QV_{ab} = QV_{ba}\implies V_{ab}=V_{ba}$$

which is exactly what we would physically expect.

For the second question, although usually you shouldn't ask more than one question, i would recommend Jackson's Classical Electrodynamics. It's the bible of the classical theory of EM. As well as Landau's books Classical theory of fields and Electrodynamics of Continuous Media. All of them, especially Jackson's book, are quite heavy with the mathematics, they go far beyond vector calculus. But give them a look and judge by yourself.

• thank you for the help! Mar 12, 2020 at 15:55
• @U.Basumatary You're welcome! If you feel like this answered your question you should accept it with the check-mark! Mar 12, 2020 at 15:57