9.12 Maxwell’s Equations Differential Form
Let’s recall Maxwell equations. In integral form, we have seen that the Maxwell equations were such that the first one was Gauss’s law for electric field and that is electric field dotted with incremental area vector dA integrated over a closed surface S is equal to net charge enclosed in the volume, surrounded by this closed surface S, divided by permittivity of free space ε0. Therefore this was Gauss’s law for E field.
The second Maxwell equation was the Gauss’s law for the magnetic field and that was B dot dA integrated over, again, a closed surface equal to 0. The reason for that was there cannot be any magnetic monopoles, therefore we cannot have, as a source of magnetic field, a single north pole, isolated north pole, magnetic north pole, or a magnetic south pole. This was the case of Gauss’s law for B field.
The third Maxwell equation was Faraday’s law of induction and it was such that integral of E dot dl over a closed contour C, closed loop, is equal to negative of the change in magnetic flux, dΦB over dt. And this was Faraday’s law of induction.
The last Maxwell equation was Ampere-Maxwell’s law and it was in the form of B dot dl integrated over a closed contour or closed loop, which was equal to μ0 time the net current passing through the area surrounded by this closed loop, plus μ0 ε0, change in electric field, with respect to time. Where ε0 times this change in electric flux was displacement current and it simply indicated that the change in electric field generates a magnetic field. In the case of Faraday’s law, change in magnetic field generates an electric field. Therefore this one was the Ampere-Maxwell’s law.
These four equations represented the integral form of the Maxwell’s equations. As you recall from calculus, taking a derivative is always much easier than taking an integral. So we are going to try to express these four equations in the form of differential equations. In order to do that, we’re going to first recall an operator that we have introduced earlier while we were studying the electric field and electric potential. We denoted this operator with delta and it is in calculus known as “del operator” (∇).
In rectangular coordinate system, or in Cartesian coordinate system, the explicit form of this operator was ∂ over ∂x times î unit plus ∂ over ∂y times ĵ unit plus ∂ over ∂z times k̂ unit, where î, ĵ and k̂ were the unit vectors in a rectangular coordinate system along x, y, and z directions.
As we can see from the definition of this operator, it simply takes the partial derivative of a function, whenever it acts on it, along these specific directions. As you recall, we have seen that as physical application of this operator as when we were calculating the electric field from the potential and it was equal to –∇ times the potential function. We called this process as “gradient”. In other words, whenever the ∇ operator was acting on a scalar function, like in the case of potential, it was called as the gradient of potential. So the negative gradient of potential gave us the electric field.
An interesting point over here from the mathematical angle is such that the operator acts on a scalar function here, ∇ operator, and converts it to a vector. As you recall, when we take the partial derivative of the potential function along specific directions, we obtain the electric field components along those directions.
If we look at this operator and we take its dot product with itself, then it will be equal to ∂ over ∂x î plus ∂ over ∂y ĵ plus ∂ over ∂z k̂ dotted with the same vector, which is ∂x î plus ∂ over ∂y ĵ and plus ∂ over ∂z k̂. Again as you recall, partial derivative is that if the function we are taking the derivative is a function of different coordinates, then when we take the partial derivative with respect to a specific coordinate, we keep the other ones as constant. In that sense, the process of taking the derivative is not any different than taking the total derivative of that function.
If we look over here and if we write this dot product in explicit form, we will have ∂ over ∂x times ∂ over ∂x will give us the second partial derivative with respect to x. That is ∂2 over ∂x2 and we will have î dot î for the vector part and we’re going to have ∂2 over ∂x ∂y, î dot ĵ for the second term, plus ∂2 over ∂x ∂z, î dot k̂. Then moving on, for the product of the second term, plus ∂2 over ∂y ∂x, ĵ dot î, plus ∂2 over ∂y2, ĵ dot ĵ, plus ∂2 over ∂y ∂z, ĵ dot k̂. And multiplying, or taking the dot product of all these three terms over here in the second bracket by ∂ over ∂z, will give us ∂2 over ∂z ∂x, k̂ dot î, plus ∂2 over ∂z ∂y, k̂ dot ĵ, and the last term will be ∂2 over ∂z2, k̂ dot k̂.
Here, if we look at î dot î, ĵ dot ĵ, and k̂ dot k̂, these vectors have the magnitude of 1 and the angle between them, since they are along the same direction, will be 0. Therefore, 1 times 1, the magnitude of unit vector î times cosine of 0, which is 1, this is going to give us just 1. And then similarly, ĵ dot ĵ is going to give us 1 and k̂ dot k will give us just 1.
When we look at these cross products, î dot ĵ, î dot k̂, ĵ dot î, ĵ dot k̂, there the magnitude of the first vector is 1, times the magnitude of the second vector, which is 1, and times the cosine of the angle between these two vectors and that is cosine 90 because î is perpendicular to ĵ, which is also perpendicular to k̂, and similarly, ĵ is perpendicular to k̂. So all these cross terms, since cosine 90 is 0, will give us just 0. Therefore, the only surviving terms are going to be ∂2 over ∂x2, ∂2 over ∂y2, and ∂2 over ∂z2.
∇ dot ∇ is represented as ∇2, therefore this operator is going to be equal to ∂2 over ∂x2 plus ∂2 over ∂y2 plus ∂2 over ∂x2. It’s another important operator in vector algebra. It is known as “Laplacian operator”. You are going to study these operators in detail later on when you take vector algebra course. So the first one is ∇ operator, the second one is ∇2 operator.
At this point, we are going to introduce again two important theorems in vector calculus that you are going to study later on in vector algebra course in detail. The first one is known as Stokes’ theorem. If we say let β be any vector, then Stokes’ theorem states that the closed loop integral of β dot dl, so integral of this displacement vector dl, integrated over a closed loop, is equal to ∇ cross β dot dA integrated over a surface S, and that is the surface enclosed by this closed loop C. So Stokes’ theorem is simply stating that if we take the closed loop integral of a vector, β dot dl, that will be equal to ∇ cross β dot dA.
The second important theorem is known as divergence theorem. Again, for a vectoral quantity β, this theorem can be expressed as integral of β dot dA over a closed surface S is equal to ∇ dot β times dV, integrated over a volume, which is surrounded by closed surface S. So in Stokes’ theorem, a loop integral is being converted into a surface integral or vice versa and in the case of divergence theorem, a closed surface integral is being converted into a volume integral.
Here, let’s introduce a couple of definitions. As I mentioned earlier, when ∇ operator acts on a scalar function, this process is called gradient of U and when ∇ operator is acting on a vectoral quantity, through dot product, it is called divergence of β of this vector. If it is acting on a vector through cross product, it is called curl of β. If we look at what these processes are associated with, the curl of a vector represents a sense of rotation and the divergence of a vector. If it is positive, it indicates a vector created at that point, and if it is negative, then it represents a vector destroyed at that point. These are interesting terminologies associated with vector calculus.
All right. now, let’s consider each one of the Maxwell’s equations, one by one, starting with the first one. That was Gauss’s law for the electric field, which was E dot dA integrated over a closed surface is equal to net charge enclosed in the volume surrounded by the surface divided by ε0 permittivity of free space. The net charge enclosed inside of the volume surrounded by this surface can also be expressed as, in terms of the charge density, the volume charge density ρ times dV over ε0 integrated over the volume surrounded by this surface S.
All right. So we have integral of E dot dA over a closed surface is equal to ρ times dV over ε0 integrated over the volume, which is surrounded by this closed surface S. Well, if we go back to divergence theorem over here, it says β dot dA integrated over a closed surface is equal to divergence of β integrated over the volume surrounded by the surface. If we apply this theorem for our case, for our vector we have electric field, instead of β we have E so E dot dA over a closed surface S is going to be equal to then integral over the volume of divergence of E times dV. That is, again, the volume surrounded by S.
So this is directly from the divergence theorem. But the right hand side is also equal to ρ times dV over ε0 integrated over the same volume, volume surrounded by S. Both of these integrals are taken over the same volume. We can get rid of the integrals. Then we end up with divergence of E is equal to ρ over ε0. That is the differential form of Gauss’s law for E field.
When we look at the second equation which was the Gauss’s law for magnetic field, B dot dA over a closed surface S was equal to 0, so applying the divergence theorem and following the similar type of procedure, we end up with here, divergence of B is equal to 0 as the second Maxwell’s equation in differential form, which is Gauss’s law for B field.
Now let’s consider the third Maxwell’s equation, which is Faraday’s law of induction, and that was given as E dot dl over a closed loop is equal to –dΦB over dt. Change in flux with respect to time. Here, the flux that we are talking about is the net flux flowing through the surface surrounded by this closed loop. So the right-hand side in explicit form can be expressed as minus d over dt of, and the flux then, in its explicit form, will be B dot dA integrated over the surface S and that is the surface surrounded by loop C. Since the integral is taken over space variables, then we can easily place the derivative operator inside of the integral. So if we do that, we end up with minus integral of d over dt of B dot dA integrated over this closed surface S, which is surrounded by loop C.
On the other hand, the left-hand side of the Faraday’s law can be expressed, again, in the form of a surface integral by applying Stokes’ theorem, which simply states that any vector dotted with a displacement vector integrated over a closed loop is equal to curl of that vector dotted with dA incremental surface area vector integrated over the surface enclosed by that loop C. Here, instead of β, in our case we have the electric field vector. Therefore this integral can be expressed as integral of ∇ cross E, or curl of E, dotted with incremental area vector dA integrated over the surface S, which is the surface surrounded by closed loop C. These two surfaces are the same surfaces so we can rid of the surface integrals.
Then that leaves us ∇ cross E or curl of E is equal to minus dB over dt. Now here we can express this in most general form as ∂B over ∂t because the magnetic field can be a function, not only of time, but can be a function of the space coordinates. Therefore negative partial derivative of the magnetic field becomes equal to ∇ cross E. From this expression, which is the differential form of Faraday’s law, we can easily see that the change in magnetic field is creating or generating an electric field. That’s the other way of actually recalling or expressing Faraday’s law of induction. That is also known as changing B field generates E field.
All right. Now if we consider the last Maxwell’s equation, which is Ampere-Maxwell’s law, and it is given as B dot dl integrated over a closed loop is equal to μ0 times i-enclosed plus μ0 times the displacement current, which is ε0, change in electric field flux with respect to time. Again, B dot dl integrated over this closed loop can be expressed as μ0 times, for the i-enclosed we can express this in terms of the current density, and that will be J dot dA integrated over the surface S, which is surrounded by this closed loop C, plus, for the second term, we have μ0 ε0, again, express the flux in its explicit form. Here we are talking about the electric field flux through the area surrounded by this loop C, so d over dt of E dot dA integrated over the surface S, again, this is the surface surrounded by loop C.
Here μ0 is constant, we can place it inside of this integral, as well as μ0 ε0 and time derivative can be placed inside of this integral because the integral is taken over the space variables. We can express B dot dl by applying Stokes’ theorem as we did in the previous case in the form of ∇ cross B dot dA, and this is from Stokes’ theorem, integrated over a surface S surrounded by this loop C will be equal to integral of μ0 J dot dA integrated over the same surface S plus integral of μ0 ε0 d over dt of E dot dA, again, integrated over the same surface S.
So all these integrals are taken over the same surface S, which is surrounded by closed loop C and we can then get rid of the integrals, leaving us a differential equation in the form of curl of B magnetic field is equal to μ0 J plus μ0 ε0. Again, using the most general form by considering that the electric field can be a function of both position as well as time, therefore using partial derivative here, ∂E over ∂t is going to give us the differential form of Ampere Maxwell’s law. Again, one can easily see that this expression can be interpreted as changing electric field, in this case, generating magnetic field. So here we can say that this is the Ampere-Maxwell’s law and it is also known as changing E field generates B field.
So once we obtain these differential forms of the Maxwell’s equations, we can express them over here right across from the integral forms of the equations. For the Gauss’s law, we have divergence of E is equal to ρ over ε0. For the Gauss’s law for magnetic field, we have divergence of B is equal to 0. For the Faraday’s law of induction, we have ∇ cross E is equal to -∂B over ∂t, changing magnetic field is generating electric field.
Finally, for the Ampere-Maxwell’s law, we have ∇ cross B is equal to plus μ0 J plus μ0 ε0 ∂E over ∂t, changing electric field is generating magnetic field. So these are the differential forms of the Maxwell’s equations. These four equations are the fundamental equations for the electromagnetic theory and one can analyze and explain every electromagnetic phenomenon through these four equations. Let’s say this is the differential form and these are the integral forms.