## Posts Tagged ‘Tensor’

### Physics Friday 133

August 27, 2010

Part 13: Electromagnetic FIelds and Angular Momentum

Last week, we considered momentum conservation in electromagnetism, finding that the momentum density is equal to , where  is the Poynting vector; and that the momentum flux density tensor is , where  is the Maxwell stress tensor, a symmetric rank-two tensor with components . Today, we consider angular momentum.
Recalling that for a particle of momentum p at displacement x from the origin, the angular momentum about the origin is
, and the force on a particle of charge q is , so the torque about the origin on the particle is
.
Converting to charge and current densities, if  is the total (mechanical) angular momentum of the charges in a volume V, then
.
However, we showed last week that
,
where  is the divergence of the Maxwell stress tensor , which is a vector with ith component .
Thus, we plug this in to get:
.
Note that since , ,
and the order can be interchanged between time derivatives and volume integrals, so the above becomes:
,
and so we see  is the angular momentum of the fields in the volume, and so  is the angular momentum density of the electromagnetic fields.
Now, let us consider the flux term. The ith component of the cross product of vectors a and b is , where  is the Levi-Civita symbol. Thus, the ith component of  is
.
Now, using the product rule, , and , so
,
and so
,
since  due to the symmetry of  and the antisymmetry of the Levi-Civita symbol (exchange the labels on the dummy indices j and k, then reverse orders on , with no sign change, and on , with sign change, to find that the sum is its own opposite, and therefore zero).
Note that the cross product of vectors a and b is the vector with ith component . Similarly, the cross product of the vector a with the tensor  is the tensor with components
.
In this vein, we see then that , and therefore, that
,
the ith component of the divergence of the tensor (or, more accurately, pseudotensor) . Thus, since the corresponding components are equal,
,
and so
,
giving us our integral continuity equation. Just as  is the linear momentum flux density tensor of electromagnetic fields, so we see that  is the angular momentum flux density tensor of electromagnetic fields. Thus, the differential form for our continuity equation is
.
And just as one may find the total force the fields exert on the charges in a volume by integrating the total force via , one can do the same with  to find the total torque.

### Physics Friday 132

August 20, 2010

Part 12: Electromagnetic Fields and Momentum

Last week, we considered energy conservation of electromagnetic fields, and found the continuity equation, which told us that the Poynting vector  gives the flow of energy in electromagnetic fields.
Now, let us consider linear momentum. Since electric and magnetic fields exert forces on charged particles, they change the momentum of these particles; therefore, momentum conservation means that the fields must themselves carry momentum. The total momentum in a volume is then the sum of the total mechanical momentum of the charged particles in that volume, , and the total momentum of the electromagnetic fields in the volume, . This sum, , must then obey a continuity equation; as momentum is conserved, the time change in this quantity is entirely due to the flux out of the volume.
The standard (differential form) continuity equation is
,
where v is the flux of the quantity φ. When φ is a scalar, v is a vector quantity, as the flow has both magnitude and direction. However, momentum is a vector quantity. So, letting Ptot be the momentum density, the flux  is not a vector, but a rank-two tensor. The component Gij of this tensor represents the flux of the ith component of electromagnetic momentum in the j direction. For reasons similar to those for the mechanical stress tensor, this tensor must be symmetric: Gij=Gji, so we have six independent components.

Thus, for a volume V bounded by surface S with outward normal , we have
,
where  is the vector whose ith component is  (the rank-two tensor acts as a function mapping vectors to vectors, here applied to ). Splitting the momentum into mechanical and electromagnetic components,
.
Now, let us consider the change in mechanical momentum. The force on a particle with charge q is . The total change in momentum for the particles in a volume is the sum of the forces on the particles; for discrete charges qi, this is
;
converting to a continuous charge distribution, we have , , and the sum becomes a volume integral:
.
Now, we can use Maxwell’s laws to eliminate the charge and current densities in favor of the fields. From Gauss’ Law, ; and from Ampère’s Law, .
Thus,
.
Now, the product rule for the partial derivative of the cross product tells us
,
so
,
and we have
.
Now, Faraday’s Law tells us , so we have
.
Or, using , then
,
so
.
Note the near symmetry between the electric term  and the magnetic term . Note, however, the term to make them symmetric would be , but since , this term is zero, and may be freely added:
so
. Thus, we have
.

Now, the product rule for the gradient of a dot product is
.
Letting a=b,
,
so
.
Thus,
.
Examining the components,

and
.

.
And since 
So, using the Kronecker delta , we have
.
This is the ith component of the divergence of the tensor with components . Thus, we define the tensor  with components
. This is called the Maxwell stress tensor. Then, we have
.
Note that  has units of momentum flow per area, or momentum/(area)(time), which is equivalent to force/area, the units of stress; hence the name Maxwell stress tensor.

So, we then identify the term  as the time derivative of the electromagnetic momentum in the volume. We then see the electromagnetic momentum density is ; the momentum density is parallel and proportional to the energy flux density with proportionality constant . We thus see that the momentum flux density tensor of electromagnetic fields is
. Thus, if  is the mechanical momentum density, then the differential continuity equation for electromagnetic momentum is
.

Or, using the integral form
,
and applying the divergence theorem,
.
Note that the flux per unit area of momentum across the surface S, is the force per unit area transmitted across the surface S and acting on the fields and particles within V, and is , which has ith component
. One can thus consider the force on a material object in electromgnetic fields by considering a boundary surface S enclosing the object, and integrating up the total force via .

[Within dielectric media, the issue of momentum of the electromagnetic fields is more complicated; see the Abraham–Minkowski controversy.]

### Physics Friday 62

March 6, 2009

[Part 1 of ?]

Let us consider a rotating oblate spheroid with equatorial radius a and polar semi-axis c. We choose an inertial frame (x,y,z) and body coordinate system (x’,y’,z’), both with origin at the ellipsoid’s center of mass. Suppose we have a rotation about a principal axis with angular momentum , where Iz’ is the moment of inertia along the z’ principal axis and  is the z’ unit vector).

Now, let us put a point mass M at a displacement d from the center of our ellipsoid. Then a mass element of our ellipsoid  at position r ( is the density at that point in the object, not necessarily uniform). experiences a gravitational force due to the mass M of

Integrating the torque element  due to this force over the volume V of the ellipsoid gives a total torque of
.

Now, suppose our point mass is far from our body, compared to its size, so that , where  and . Now using
, we see that approximating to first order in r/d,
, so we can in this situation approximate our torque by:
.

Now, to find the first integral on the right, we note that
.
For a an object in volumeV of total mass m and density , the center of mass has position (vector)

(see here); as our origin is the center of mass, we thus see that the integral , and the first integral in our approximation is the zero vector, and
.

Consider the moment of inertia tensior I of our object (see here):
, where E3 is the 3×3 identity matrix and  is the outer product of r with itself. Applying the tensor to d, we use  and  to get
.
Next, consider the cross product of d and this vector:
,
which is our remaining integral in the torque approximation:
.