How can Green's Theorem be used to derive Maxwell's equations?

Question

I've learned how to prove Green's Theorem and I read that it contributed to deriving Maxwell's equations. How can Green's Theorem be used to derive any of four Maxwell's equations? What else do I have to learn in order to derive Maxwell's equations?

Answer 1

This is likely not the answer you're looking for, but you can't really derive physics equations. At best, math helps you reformulate physical principles and derive consequences of them. In the case of Maxwell's equations, Green's Theorem helps you pass between the integral and differential forms of Maxwell's equations.

For example, one often sees the integral equation $$ \int_{\partial \Omega} \mathbf{E} \cdot d\mathbf{l} = -\frac{d}{dt} \int_{\Omega} \mathbf{B} \cdot d \mathbf{S} $$ In words, this says that the line-integral of the electric field around the boundary of a region $\Omega$, denoted $\partial \Omega$, is equal to the opposite of the rate of change of magnetic flux through $\Omega$. This is Faraday's law of induction, one of the four equations referred to collectively as Maxwell's equations. If $\Omega$ is a static region in the $xy$ plane, then Green's theorem can help us pass to the differential form of Faraday's law. Green's theorem tells us that $$ \int_{\partial \Omega} \mathbf{E} \cdot d\mathbf{l} = \int_{\partial \Omega} E_x dx + E_y dy = \int_{\Omega} \left(\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \right) dxdy $$ Thus $$ \int_{\Omega} \left(\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \right) dxdy = -\frac{d}{dt} \int_{\Omega} \mathbf{B} \cdot d \mathbf{S} = -\frac{d}{dt} \int_{\Omega} B_z dxdy = \int_{\Omega} -\frac{\partial B_z}{\partial t} dx dy $$ This suggests, though does not prove (note the assumptions we were forced to make on $\Omega$) the differential form of Faraday's law: $$ \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} = - \frac{\partial B_z}{\partial t} $$

Using Green's theorem, we passed from the integral form of Faraday's law, which is stated, albeit less technically, in most high school physics courses, to the differential form of Faraday's law, which is often easier to work with.

There is a stronger result, known as Stokes's theorem, which actually does suffice for the proof outlined above by allowing us to consider general regions $\Omega$.

If you are interested in multivariable calculus and its connections to physics, I can recommend Schey's Div, Grad, Curl, and All That.

Answer 2

Stoke's theorem which states that $$\iint_S \vec{F}\cdot \hat{n}\,dA=\int_{\partial S}\vec{F}\,d\vec{l}$$ and Gauss's theorem $$\iiint_V \text{div}{F} \,dA=\iint_{\partial V} \vec{F}\hat{n}dS$$ can be used to go from differential form to integral form of Maxwell's equations and Gauss's law can be used to derive the first and second Maxwell equation. Any introductory book on electrodynamics shows this. See for example "Introduction to electrodynamics" by Griffiths.

Answer 3

Consider a time dependent contour $\partial \Sigma$ in the $xy$-plane. Faraday's law for this contour is $$\oint_{\partial \Sigma}(E_x dx + E_y dy) = \iint_\Sigma \frac{\partial B_z}{\partial t}d \Sigma$$

Using Green's theorem on can re-write the LHS as $$\oint_{\partial \Sigma}(E_x dx + E_y dy) = \iint_\Sigma \left(\frac{\partial E_y}{\partial x}- \frac{\partial E_x}{\partial y} \right)d\Sigma$$

Which clearly implies for any surface in the plane $$\iint_\Sigma \left(-\frac{\partial E_y}{\partial x}+\frac{\partial E_x}{\partial y} +\frac{\partial B_z}{\partial t}\right)d \Sigma=0$$

This integral vanishes iff $$-\frac{\partial E_y}{\partial x}+\frac{\partial E_x}{\partial y} +\frac{\partial B_z}{\partial t}=0$$

Whence $$\nabla \times \overrightarrow{E} = -\frac{\partial \overrightarrow{B}}{\partial t}$$

Giving the differential form of Faraday's law.

Answer 4

If you consider Stokes' theorem and Gauss's theorem to be versions of Green's theorem, they allow to derive Maxwell equations from their integral form to the local form.

For example, it was known (Gauss's Law) that the flux of the electric field through a closed surface is proportional to the total electric charge enclosed in that surface, specifically $$ \int_{\partial V} \vec E\cdot d\vec S = \frac{1}{\epsilon_0}\int_V \rho \,dV$$ but from the Gauss's theorem $$ \int_{\partial V} \vec E\cdot d\vec S = \int_V \vec\nabla\cdot\vec E\,dV$$ Since it is true for any volume $V$ it means that $$ \vec\nabla\cdot\vec E = \frac{1}{\epsilon_0} \rho$$ which is one of Maxwell's equations.

Similiarily, it was discovered that the magnetic field changing in time creates circulating electric fieds (Faraday's Law). Specifically, the circulation of the electric field along a given closed curve is equal to the minus the rate of change of the flux of magnetic field through the surface bounded by that curve, that is $$ \oint_{\partial\Sigma} \vec E\cdot d\vec r = -\frac{d}{dt}\int_\Sigma \vec B\cdot d\vec S$$ But, from Stokes's Theorem $$ \oint_{\partial\Sigma} \vec E\cdot d\vec r = \int_\Sigma (\vec\nabla \times \vec E)\cdot d\vec S$$ Since these equations are true for any surface $\Sigma$, it means that $$ \vec\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$$

The remaining Maxwell's equations are derived in a similar way.