# What are the differences between the differential and integral forms of (e.g. Maxwell's) equations?

I would like to understand what has to be differential and integral form of the same function, for example the famous equations of James Clerk Maxwell: How to know where to apply each way? Excuse the ignorance, but always confused me the head.

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Remembering that, I know that the concept of integration and derivation, is something like: • May 19, 2016 at 5:30
• Hi Marcelo, Proof of equality of the integral and differential form of Maxwell's equation seems to be asking the same thing as this question. Can you edit your question to explain how it differs? Otherwise I think this question should be closed as a duplicate. May 19, 2016 at 5:37
• Well, the question is different, he asked out of curiosity how to solve the integration to reach the derivative, happened to the answer to the question, he had posted a link that answered my doubts in parts rs May 19, 2016 at 5:53
• My doubts would be more related to when to use, but following the link and reading that passage had my doubts May 19, 2016 at 5:55

The equations are entirely equivalent, as can be proven using Gauss' and Stokes' theorems.

The integral forms are most useful when dealing with macroscopic problems with high degrees of symmetry (e.g. spherical or axial symmetry; or, following on from comments below, a line/surface integrals where the field is either parallel or perpendicular to the line/surface element).

The differential forms are strictly local - they deal with charge and current densities and fields at a point in space and time. The differential forms are far easier to manipulate when dealing with electromagnetic waves; they make it far easier to show that Maxwell's equations can be written in a covariant form, compatible with special relativity; and far easier to put into a computer to do numerical electromagnetism calculations.

I would think that these three points generalise to any system of differential vs integral forms in physics.

• I would say that the integral form is useful iff you have an obvious integration contour. May 19, 2016 at 10:40
• @DarioP An obvious curve is not enough. You also need a high degree of symmetry. But I guess it depends on what you mean by "an obvious integration contour". It's not obvious to me what that means. May 19, 2016 at 15:02
• @garyp I guess that Rob understood :) Basically what I meant is that since integration is so hard, I either see an outstanding contour that makes it (almost) trivial, or I leave my hopes. Symmetry helps. May 19, 2016 at 15:34

Your confusion lies in failing to recognize that they are exactly the same equations. Take for example Gauss's law:

$$\vec \nabla \cdot \vec E = \dfrac{\rho}{\epsilon_0}$$ You can see that there $$\rho$$ is the charge distribution, and in general, can be a function of the position.

Now consider a volume $$V$$, you can just integrate the density to obtain the total charge in the volume, but you can also integrate the electric field gradient to get a measure of the electric field $$\vec \nabla \cdot \vec E = \dfrac{\rho}{\epsilon_0} \hskip 30pt / \int_V dV$$

$$\int_V \vec \nabla \cdot \vec E dV = \int_V \dfrac{\rho}{\epsilon_0} dV$$

Now you can use the Divergence's Theorem on the left side. $$\vec \nabla \cdot \vec E = \iiint_V \vec \nabla \cdot \vec E = \iint_S E\cdot d\vec S$$

Here $$S$$ is the surface of the volume and $$d\vec S$$ is the area element of the surface pointing perpendicularly to the surface out from the volume so the equation final form is

$$\iint_S E\cdot d\vec S = \iiint_V \dfrac{\rho}{\epsilon_0} dV$$

Which convention's aside is the same equation as shown in Wikipedia. You can do exactly the same to each of the Maxwell's equations using the Divergence's Theorem and Stokes' Theorem

Now, which one is which. If you look at the equations you will see that every equation in the differential form has a $$\overrightarrow{\nabla}$$ operator (which is a differential operator), while the integral form does not have any spatial differential operator, but it's integrating the terms of the equations.

Finally, as to which one to use, it doesn't matter, because they are the exact same equations and to solve many problems you end up integrating the equations anyway. That being said I like to start from the differential form and if needed integrate the equations.

• This is a proof that the differential forms of the equations imply the integral forms of the equations. If we were being ultra-pedantic, we would also want to prove that the integral forms imply the differential forms. This can be done, but the argument is a bit more subtle; the key is to assume that all functions are continuous and that the integral equations hold for all volumes. May 19, 2016 at 13:45

All this tells you is that the fields satisfies both the inategral and the differential equations. The two are related by the mathematical identities called the divergence theorem and Stokes' theorem.

So which do you apply? Well, which ever one you want! If you run into an integral, you use the integral form, and if you're ever asked for the divergence or curl of the electromagnetic fields, you know exactly what it is, without even trying!

I would like to point out also, the differences between the Maxwell-Faraday equation, and Faraday's law itself ( emf = -d/dt(magnetic flux)). Faraday's law, combined with the Maxwell-Faraday equation, can show that the total emf due to ANY change in flux is equal to Emf =int (e+Vxb).dl which corresponds to transformer emf, and also motional emf (due to magnetic force)

The Maxwell-Faraday equation is only due to an induced electric, but Faraday's law is due to transformer + motional emf.

Sorry if this isn't relevant, but I thought it was worth mentioning since this also includes the Lorentz force which is just as important as Maxwell's equations. Actual derivation here: