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, q_{1} and q_{2}, the former moving along the x axis while the latter moves along the y axis.
The force on q_{1} just as it passes by the origin is purely electrical as it lies along the direction of motion of q_{2}, 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 q_{2} due to q_{1}. However, if we look at the force on q_{2} due to q_{1}, there is also a magnetic force. This is because the moving charge q_{1} creates a magnetic field in the z direction which exerts a force on q_{2}. 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 3^{n} 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 xdirection. 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 (πR^{2 }sin^{2}θ)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.
11 videos45 docs73 tests

1. What is the law of conservation of energy? 
2. What is momentum and how is it conserved? 
3. How does the conservation of energy relate to the conservation of momentum? 
4. Can energy be converted into momentum or vice versa? 
5. What are some reallife examples of conservation of energy and momentum? 
11 videos45 docs73 tests


Explore Courses for Electrical Engineering (EE) exam
