Conservation of Energy & Momentum | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) PDF Download

Coulomb Gauge and the Potential formulation of Maxwell’s equations

The Maxwell’s equations in their final form are

We had, in the last lecture, made a reformulation of these equations in terms of scalar and vector potentials. This gave us two “coupled” equations for four quantities, i.e. 3 components of the vector potential and one component of scalar potential. We had seen that these equations get decoupled in Lorentz gauge.

We had a lot of discussion on Coulomb gauge in which the divergence of the vector potential is zero. Is this gauge any good to be used now?

Recall the pair of equations,

Note that first equation gets decoupled in Coulomb gauge and becomes a Poisson’s equation for the scalar potential with the formal solution,

The second equation is not straightforward and requires a lot of mathematical mannipulation before the decoupling can be seen.

To avoid repeating the same expression unnecessarily, we will concentrate on the right hand side of eqn. (2). In Coulomb gauge the term with the divergence drops out and the right hand side of (2) becomes,

We insert the formal solution of the scalar potential into this equation,

Now the term   is a current and we replace it by   using the equation of continuity and the term becomes

We will do some further simplification to (I). But first let us recall that we have learnt that a vector is completely determined when its divergence and curl are specified. This allows us to write the current density vector inside the integral as a sum of one part   whose divergence is zero and another part   whose curl is zero,

Using the identity,

we can write,

where we have used   in arriving at both these relations. We will shortly return to using these relations.

Let us get back to the relation (I). Note that the gradient outside is taken with respect to the point of observation. We can take it inside the integration and it will act only on   However, since this depends only on the difference of   and   we can replace  by 'by incorporating a minus sign. With this (I) becomes

We will simplify this further by using chain rule differentiation,

Substituting this into the preceding term, we have two integrals, one of which is a volume integral of a gradient. This term can be converted to a surface integral like the way we do for the divergence theorem and take to surface to infinity to make this term zero. This implies the remaining integral is over all space and we get

in obtaining this step, we have used   and used the fact that   can be written as   because the transverse part has zero divergence. Further, we can use the fact that   is irrotational to write,

With this (I) takes the form

At this stage, we will use the Green’s identity for fields T and U according to which

As our fields vanish at infinity, we have,

Using this, we get,

Thus the original equation for the vector potential has now been completely decoupled from the scalar part and we have , instead, an inhomogeneous wave equation,

Electromagnetic Momentum

We have seen that the electromagnetic field carries energy. A natural question arises as to whether it carries momentum as well. The answer is affirmative and we will illustrate this by a simple procedure. A more rigorous derivation requires use of the theory of relativity. Suppose, we have two charged particles, q₁ and q₂, the former moving along the x axis while the latter moves along the y axis.

The force on q₁ just as it passes by the origin is purely electrical as it lies along the direction of motion of q₂, and is given by 

where d is the distance between the charges at that instant. This is also the magnitude of the electric force on q₂ due to q₁. However, if we look at the force on q₂ due to q₁, there is also a magnetic force. This is because the moving charge q₁ creates a magnetic field in the z direction which exerts a force on q₂. This is certainly anomalous and is in apparent violation of the third law. We say it is an apparent violation because the third law is essentially a statement of conservation of momentum and it is the total momentum of the system that needs to be conserved. In this case, in addition to the two charges, there exists the electromagnetic field and if the field itself carries momentum, there is no violation. This is intuitive but the fact that electromagnetic field carries momentum is a fact.

Let us look at the force exerted on a system of sources (charges and currents) and electromagnetic field. Let   represent the moment associated with the sources. The force exerted on the system of sources is then given by Lorentz force equation

Once again, for simplicity, we will consider linear electric and linear magnetic material. We will use Maxwell’s equations to cast these equations in in terms of field variables. We replace ρ sing Gauss’s law and the current density from Ampere’s law,

and

Substituting these in the force equation, we get,

Let us simplify some of the terms,

In the last expression we have used Faraday’s law. Thus we have,

In the second term we have changed the order of cross product and hence a minus sign.

The last term on the right can be seen to be   where   is the Poynting vector. Since the expression has the dimension of force,   has the dimension of momentum density and we will identify this term as the rate of change of momentum associated with the electromagnetic field (radiation field) and take it to the left hand side to be added to the rate of change of momentum of the sources P_rad.

The expression on the right looks asymmetric in the electric and magnetic quantities which can be rectified easily. Let us look at the electric field terms represented by the first and the third terms on the right.

The integrand is   other than for a factor of ∈₀ In order to simplify this to desired form, we will calculate the Cartesian components of this. Let us find what its x component is and then we will add up the three components.

From symmetry, one can write the y and z components

Let us add the three components. The second term of each of the expressions add to give us

The remaining terms look messy and we will return to them shortly.

Let us consider the magnetic field terms. In order to make it symmetrical with the magnetic field, we need to add a term   to the triple vector product term. This is simply adding zero because   However, this will make the electric and the magnetic fields at par and we would get a contribution of   similar to the case of the electric field. This would give us a term

where u is the energy density of electromagnetic field. Thus this term (i.e. the gradient of the energy density) represents the momentum density term.

We will return to a discussion of the remaining terms in the next lecture.

In the last lecture, we had written down the expression for the rate of change of momentum of the sources and fields as given by,

We could express this equation in a symmetric fashion by adding a term prooportional to the divergence of the magnetic field (which is identically zero) and rewrite the above as

We considered the pairs of term in parenthesis on the right and simplified their components as

Adding the second term of each of the components, we wrote it as   and a similar contribution from the magnetic field gives us   Inserting the associated constants ∈₀ and   we get this contribution as the gradient of energy density   which is the momentum density term.

We will now try to write the remaining terms in a compact fashion. We will simplify only the electric field terms and write down the corresponding magnetic field term by analogy. The remaining term are

Since the left hand side is a force, we would have liked to express it as a gradient of a scalar which would act like a potential. However, looking at the terms does not seem to suggest such a possibility because the components seem to have been mixed.

We , therefore, bring back the terms which we have been able to express as a gradient of the energy density and see whether there can be alternative ways of expressing them.

We note here that in expressing it as a gradient of a scalar, we started with a single function and obtained three quantities, the components of a vector. we are aware that if we start with a vector and take a divergence, we would get a scalar. Thus in some sense, the gradient increases the space but a divergence decreases the same. We generalize the definition of a vector to a “Tensor” of arbitrary rank n as a collection of 3ⁿ quantities. Thus a scalar is a tensor of rank zero, a vector a tensor of rank 1. In this case, we take a tensor of rank 2, denoted by nine components   A divergence of a tensor of rank 2 will reduce its rank by 1, giving us a vector which has 3 components. 

We will first state the result and then show that this can be actually done. We claim that the right hand side of our force equation can be expressed as a divergence of a tensor of rank 2 denoted by   whose components are given by

To fix our ideas, let us once again, take B=0. We can express  as a 3 x 3 matrix as follows:

Since the divergence of the tensor is a vector, it has three components and we define the components as follows

Substituting the components given in the matrix above, we find,

which is precisely the x component of the electric field term that we had found. Similar statement can be proved for the y and z components and also for the magnetic field terms. Thus we have,

where, we have written da for the area element instead of dA so as not to confuse with the Poynting vector which we have denoted by   The last relation is obtained by a procedure similar to the divergence theorem.

 is the momentum flux nornmal to the bounding surface of the volume. Since the relation is true for arbitrary volume, the relation can be stated in a differential form in terms of the momentum densities, indicated by   and   as

This is a statement of the conservation of linear momentum for the electromagnetic field which sates that the rate of change of momentum of a closed system containing sources and fields can occur only through a transport of momentum through the bounding surface.

Examples 

Force on one of the plates of a parallel plate capacitor

Consider a parallel plate capacitor with charge densities   Let the plates be in the yz plane so that the electric field is directed along the x-direction. we have,   Thus the Stress tensor in this case is diaginal and can be written as 

The force acting on the negative plate is (assuming it is to the right of the positive plate) is along the negative x direction as  is directed along – x direction. We have,

Force on the Northern Hemisphere of a spinning charged sphere by the Souther Hemisphere 

Though solving this problem can be done by more straightforward method, we are doing it by the method of Stress Tensor as an illustration.

First, let us consider the spinning shell. As the angular velocity is constant, the spinning shell is equivalent to a magnetic moment directed along the direction of the angular velocity vector. Let us take the angular velocity vector along the z direction.

The moving charges on the surface are equivalent to circulating current in the   direction. The surface current density is given by, in spherical polar,

where r=R  on the surface of the sphere. Consider a thin circular strip lying between angles θ and  θ + d θ.The radius of the circle is R sin θ.The width of the strip (which is crossed by the linear current density) is Rd θ so that the current flowing in the strip is   The circulating loop is equivalent to a magnetic moment (πR² sin²θ)dl in the z- direction. Since all such loops are parallel, the net magnetic moment of the spinning shell is

The magnetic field due to the shell is given by

Since the fields are constant, so is the Poynting vector. Thus the force is given by

By symmetry, the force is in z direction. Thus we will need to calculate only the z component of the normal component of the stress tensor. In this case, we consider only the magnetic forces (the electric force is also there as the shell has charges, but we restrict ourselves to magnetic case only. Since we are interested only in the z components, the only components of the stress tensor that we requires are

In terms of these, we can write down the force as

As we are interested in calculating force on the northern hemisphere, there are two surfaces to take into account. The equatorial plane inside the sphere for which the normal direction is along the negative z direction. Since we need to consider only the inside field, the field to be considered here is constant.

The field is given by   Thus only term in stress tensor is   and as the field is constant, the force is

That leaves us with the outside hemisphere. The normal to the surface is along the radial direction. We need to compute the integral   Let us compute each term of the integrand using the expression for the field outside the shell, which is given by

Since   we have,

Using these,

the limits of integration are from 0 to π/2 as it is a hemisphere.

The second integral is   and is given by, using  

Adding the two contributions, the force on the outer surface is

Substituting the value of the magnetic moment,  

If we add both the inside and the outside contributions, the magnetic force on the northern hemisphere is

the negative sign shows that the force is directed towards the southern hemisphere and is attractive in nature.