## Archive for August, 2010

### Monday Math 133

August 30, 2010

Find the limit  as a function of positive, real x; where ζ(s) is the Riemann zeta function.
Solution:

### 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.

### Monday Math 132

August 23, 2010

Continuing from last week, let us have a central vector field  in n dimensions.

Using, as with this post the geometer’s definition of the n-sphere; let us then find the flux of our vector field over the n-sphere of radius R with center at the origin. Denoting this n-sphere as Σ, with outward normal  and differential element , the flux is then
.
But since Σ is an n-sphere centered on the origin, the outward normal  is the unit radial vector , and so . Thus, the flux is
. But r=R for all points on Σ, so the flux is then , where M(Σ) is the n-1 dimensional measure of Σ.
We recall from this post that the n-ball of radius R has n dimensional measure , where ; and that the n-sphere of radius R has n-1 dimensional measure . Thus, we see the flux is 

Next, let . Our vector field is then undefined at the origin if k is negative. Our flux through the n-sphere Σ is then . Now, note that if k=1-n, then this is constant with respect to R: the flux becomes simply .

For any point r>0, the divergence of our vector field is found, using our result from last week, to be
.
Note, however, if k=1-n, then the term differentiated is a constant, and the divergence is then zero for all r>0. Thus, in this case, for any closed n-1 dimensional “surface” not enclosing the origin, so that the divergence theorem is valid, the flux is thus zero. Similarly, we combine this with the above result about the n-sphere Σ to see that if a closed n-1 dimensional “surface” encloses the origin, the flux through it is nCn. Thus, if we want to extend the validity of the divergence theorem as applied to F for all regions Ω of n-dimensional space, we need
,
with  for x0; thus, we must have the divergence of F take the form of an n-dimensional Dirac delta distribution:
, and so
the function  obeys
.
Now, we recall that there is a scalar function  such that ; so long as gn‘(r)=f(r). Then, we see that
 becomes
.
Thanks to the translation symmetry of the Laplacian, we see then that
 satisfies

and so G(x,y)=Gn(xy) is a Green’s function for the n-dimensional Laplacian. This function Gn(x) is also sometimes called the Newtonian kernel. Let us now examine it for specific numbers of dimensions.
For n=1, nC_n=2, , so ; and since the distance from the origin in one-dimension is , we have .
For n=2, nC_n=2π, , so , and so .
For n≥3,  has n-1>1, so it easily integrates to find , and so . Specifically, for n=3, nC_n=4π, so
 (So  is a Green’s function for the three-dimensional Laplacian and the only one which goes to zero as , due to the uniqueness theorem).
For n=4, nC_n=2π2, so .

### 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.]

### Monday Math 131

August 16, 2010

Let us consider a (real) vector field F(x) in an n-dimensional Euclidean space, and let it be a central field. That is to say, that for any point in the space (save possibly the origin, where it may be undefined), the vector field points in the radial direction, directly away from or directly toward to the origin; and the magnitude of the vector depends only on the radial coordinate r, the distance between the point an the origin. Alternately, this may be stated as the field being symmetric under all rotations, proper and improper, about the origin (it has symmetry group O(n)).

Specifically, there is a scalar function f(r) such that
, where  is the radial coordinate, and  is the unit radial vector. The magnitude of F(x) is thus |f(r)|, with positive f(r) indicating an outward vector and negative f(r) an inward vector.

Now, let us consider a scalar central field G(x); this means there is function g(r) such that G(x)=g(r), where  is the radial coordinate. Due to the symmetry, the gradient of such a scalar field must be a central vector field. Denoting the cartesian coordinates by xi, i=1,2,…,n, and letting  denote the unit vector in the direction of xi, then
.
Now, , then
.
Thus,
;
so if g(r) is an antiderivative of f(r), so that g‘(r)=f(r), then
. Thus, as long as f(r) is well-behaved enough to be integrable, then the central vector field F(x) is the gradient of a scalar function.

Next, we note that due to the symmetry, the divergence of the vector field F(x) must be a scalar central field. Specifically, using our notation as before,

, where
Fi(x) is the ith component of F(x). Since , the ith component is thus
, and so
.
Now, noting that , we see then that
.

### Physics Friday 131

August 13, 2010

Part 11: Electromagnetic Energy and Poynting’s Theorem

We previously noted that the rate of work done per unit times by electric fields (magnetic fields do no work) in a volume V with current density J is
.
This represents the power being transfered out of the electromagnetic fields and into mechanical or thermal energy. Now, let us examine further the conservation of energy.

First, we can use Ampère’s law to eliminate J in the above, working only in terms of the fields instead. In SI units, we have
,
so
,
and so

Now, ,
so
,
and we recall from here that the term in the parentheses above is the energy density of the electric field UE,
so
.

Next, for the dot product with the curl, we use the product rule for divergence of a cross product (similar to our usage here, albeit with E instead of A):
.
Now, Faraday’s Law gives us the curl of the electric field:
; thus
,
and .
Plugging these in to our work equation,
;
and
,
where  is the energy density of the magnetic field, which we found here. Thus, we have
;
assuming that the total electromagnetic energy density U is the sum of the electric and magnetic energy densities, this is
;
defining , this is written
.
This has the form of a general continuity equation for electromagnetic energy:  represents the rate of removal of energy from the fields. The vector S, then, describes the flux of electromagnetic energy from one location to another. This equation of energy conservation is called Poynting’s theorem, and S is called the Poynting vector, and has units of intensity (or power per unit area). Since only the divergence of the Poynting vector appears in the theorem, one might think that it is arbitrary in that adding the curl of any vector field will not change that divergence; however, relativistic considerations confine the Poynting vector to the specific value .

### Monday Math 130

August 9, 2010

Suppose we have a twice-differentiable vector field F(x) in three dimensions, which as $r=|\mathbf{x}|\to\infty$ goes to zero faster than $\frac{1}{r}$. Then we can take the Fourier transform of F(x), its divergence, and its curl. With
$\mathbf{\tilde{F}}(\mathbf{k})=\mathcal{F}\left[\mathbf{F}(\mathbf{x})\right](\mathbf{k})$ the Fourier transform of F, then our results from last week tell us that
$\mathcal{F}\left[\mathbf{\nabla}\cdot\mathbf{F}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\cdot\mathbf{\tilde{F}}(\mathbf{k})$.
and
$\mathcal{F}\left[\mathbf{\nabla}\times\mathbf{F}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\times\mathbf{\tilde{F}}(\mathbf{k})$.

Now, the ability to decompose a vector into components parallel and perpendicular to a given vector means that an arbitrary vector field in k-space may be decomposed (uniquely) into components parallel and perpendicular to k:
$\mathbf{\tilde{F}}_{||}(\mathbf{k})=(\mathbf{\hat{k}}\cdot\mathbf{\tilde{F}})\mathbf{\hat{k}}=\frac{\mathbf{k}\cdot\mathbf{\tilde{F}}}{\mathbf{k}\cdot\mathbf{k}}\mathbf{k}$
is the component parallel to k. The perpendicular component is then $\mathbf{\tilde{F}}_{\perp}(\mathbf{k})=\mathbf{\tilde{F}}(\mathbf{k})-\mathbf{\tilde{F}}_{||}(\mathbf{k})$.
(See “vector projection” here.)
So we have
$\mathbf{\tilde{F}}(\mathbf{k})=\mathbf{\tilde{F}}_{||}(\mathbf{k})+\mathbf{\tilde{F}}_{\perp}(\mathbf{k})$ as our decomposition. Now, since $\mathbf{\tilde{F}}_{||}$ is parallel to k, their cross product is zero; and since $\mathbf{\tilde{F}}_{\perp}$ is perpendicular to k, their dot product is zero:
$\mathbf{k}\times\mathbf{\tilde{F}}_{||}(\mathbf{k})=0$
$\mathbf{k}\cdot\mathbf{\tilde{F}}_{\perp}(\mathbf{k})=0$.

Let $\mathbf{F}_{||}(\mathbf{x})$ and $\mathbf{F}_{\perp}(\mathbf{x})$ be the inverse Fourier transforms of $\mathbf{\tilde{F}}_{||}(\mathbf{k})$ and
$\mathbf{\tilde{F}}_{\perp}(\mathbf{k})$, respectively. Then, due to the linearity of the Fourier transform and its inverse,
$\mathbf{F}(\mathbf{x})=\mathbf{F}_{||}(\mathbf{x})+\mathbf{F}_{\perp}(\mathbf{x})$.
Then from our results from last week,
$\mathcal{F}\left[\mathbf{\nabla}\times\mathbf{F}_{||}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\times\mathbf{\tilde{F}}_{||}(\mathbf{k})$,
and
$\mathcal{F}\left[\mathbf{\nabla}\cdot\mathbf{F}_{\perp}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\cdot\mathbf{\tilde{F}}_{\perp}(\mathbf{k})$.
But $\mathbf{k}\times\mathbf{\tilde{F}}_{||}(\mathbf{k})=0$ and $\mathbf{k}\cdot\mathbf{\tilde{F}}_{\perp}(\mathbf{k})=0$, so these are both zero, and a field’s Fourier transform is zero if and only if it is zero;
thus
$\mathbf{\nabla}\times\mathbf{F}_{||}(\mathbf{x})=0$
and
$\mathbf{\nabla}\cdot\mathbf{F}_{\perp}(\mathbf{x})=0$.

This means that the vector field F(x) may be uniquely decomposed into the sum of two fields, $\mathbf{F}_{||}(\mathbf{x})$ and $\mathbf{F}_{\perp}(\mathbf{x})$, the first of which is irrotational (zero curl), and the other is solenoidal (zero divergence). This decomposition is known as the Helmholtz decomposition. The statement that any sufficiently smooth and decaying three-dimensional vector field may be decomposed in this way is Helmholtz’s theorem, also known as the “fundamental theorem of vector calculus.” In physics, the irrotational and solenoidal components are known as the longitudinal and transverse components, respectively.

Since $\mathbf{F}_{||}(\mathbf{x})$ is irrotational, it may be expressed as the gradient of a scalar field, and since $\mathbf{F}_{\perp}(\mathbf{x})$ is solenoidal, it may be expressed as the curl of a vector field. Thus, we can also express the Helmholtz decomposition as
$\mathbf{F}=-\mathbf{\nabla}\phi+\mathbf{\nabla}\times\mathbf{A}$
(where the minus sign arises from the usage in physics, particularly electrodynamics).

There is a related theorem which is also sometimes referred to as “Helmholtz’s theorem,” and which states that a vector field can be constructed with a specified divergence and a specified curl; and that if these divergence and curl are sufficiently smooth and vanish fast enough at infinity, the vector field is uniquely specified by them. This is crucial in electrodynamics, as Maxwell’s equations specify the divergence and curl of the electric and magnetic fields; this theorem tells us that (given physically appropriate boundary conditions at infinity) these are enough to determine the fields.

### Physics Friday 130

August 6, 2010

Part 10: Electrodynamic Potentials and Gauge Symmetry

As noted here, the introduction of Faraday’s Law means that in electrodynamics, the electric field is not irrotational, and we no longer have the scalar electric potential φ with . But we still have no magnetic monopoles, so we can still use a magnetic vector potential A, with .
;
moving the right-hand term over, we can put this as
.
Rewriting in terms of A,
.
The spatial and time derivatives commute, so we have

.
So, while the electric field is no longer irrotational, the quantity  is, and so can be written as the gradient of a scalar function:
,
or, solving for the electric field,
.
This and  thus satisfy the two of Maxwell’s equations that are homogeneous: Faraday’s Law and the absence of magnetic monopoles.
Note that when the magnetic field is static, , and the above reduces to our electrostatic potential.

Recall that when we first discussed the vector potential, it was noted that the vector potential is not unique, and that adding the gradient of any scalar field to A does not change the resulting field. We called such a transformation a gauge transformation, and noted that we could choose a potential such that , calling that choice the Coulomb gauge.
Now, for electrodynamics, we can again add the gradient of a scalar function Λ to A without changing the magnetic field. However, if we perform the transformation from A to , we see that
. With , we then see that to keep the electric field unchainged, we must simulateously change the scalar field from φ to  (so that we have cancelling  terms).
This is the gauge transformation for electrodynamics, and the invariance of the electric and magnetic fields under these is called gauge invariance.
As before, the Coulomb gauge (also known as the “transverse gauge”) is that where the potentials are chosen so that . In this case, we plug our expression for the electric field due to potentials into Gauss’ Law:

.
.
Since the time and space terms commute, and noting that the divergence of the gradient is the Laplacian,
.
In the Coulomb gauge, the second term on the left is zero, and so our scalar potential obeys the Poisson’s equation
,
the same equation the electric potential obeys in electrostatics; though, we can no longer compute the electric field from this potential alone, but must find the magnetic vector potential. As in electrostatics, the solution to this is given by
, the instantaneous Coulomb potential; hence the name.

Another convienient choice of gauge is done by expressing both Gauss’ Law and Ampère’s law in terms of the potentials. For Gauss’ Law, we have
,
as noted above. Rearranging Ampère’s law so that the field terms are on the same side, we have
.;
using the fact that ,
.
Plugging in  and ,
.
Now, the formula for curl of a curl tells us that
; and our time and space derivatives commute, so
, and so we have

.
.
Now, to remove φ from the second equation, we use the freedom of gauge transformations to choose potentials that obey
,
this is known as the Lorenz gauge condition, and the resulting gauge the Lorenz gauge. It is named for Ludvig Lorenz (not to be confused with Hendrik Lorentz).
If we make this choice, the second equation (Ampère’s Law) becomes
,
an inhomogeneous (vector) wave equation with wave velocity c. For the first equation (Gauss’ law) , the Lorenz condition means
; making this substitution,

,
and we have an inhomogeneous wave equation for the scalar potential; this choice has uncoupled the equations for φ and A.
Note that if we have φ and A obeying the Lorenz condition, and we make the gauge transform

,
with a scalar field Λ, then
,
so if Λ obeys the homogeneous wave equation , then the transformed fields φ‘ and A‘ also obey the Lorenz condition. Thus, the Lorenz gauge is not a single choice of potentials, as is the Coulomb gauge, but instead a restricted class of choices, with all potentials in it belonging to the Lorenz gauge.
The Lorenz gauge places φ and A on more equal and symmetric footing, and is also independent of the choice of coordinate system, making it the natural choice for dealing with electromagnetism in special relativity.

### Monday Math 129

August 2, 2010

Continuing from last week, we now consider the Fourier transform of the divergence of a vector field in three dimensions. First, vector integration by parts tells us that for vector field v and scalar field φ
$\iiint_{V}\phi\mathbf{\nabla}\cdot\mathbf{v}\,dV=\iint_{S}\phi\mathbf{v}\cdot{d}\mathbf{A}-\iiint_{V}\mathbf{\nabla}\phi\cdot\mathbf{v}\,dV$.
Letting $\phi(\mathbf{x})=e^{-\imath\mathbf{k}\cdot\mathbf{x}}$, this means
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{\nabla}\cdot\mathbf{v}\,dV=\iint_{S}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\cdot{d}\mathbf{A}-\iiint_{V}\left(\mathbf{\nabla}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\right)\cdot\mathbf{v}\,dV$,
and since $\mathbf{\nabla}e^{-\imath\mathbf{k}\cdot\mathbf{x}}=-\imath\mathbf{k}e^{-\imath\mathbf{k}\cdot\mathbf{x}}$, as before, we then have
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{\nabla}\cdot\mathbf{v}\,dV=\iint_{S}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\cdot{d}\mathbf{A}-\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}(-\imath\mathbf{k}\cdot\mathbf{v})\,dV$
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{\nabla}\cdot\mathbf{v}\,dV=\iint_{S}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\cdot{d}\mathbf{A}+\imath\mathbf{k}\cdot\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\,dV$.
Now, if v goes to zero sufficiently fast as r goes to infinity (|v| goes to zero faster than 1/r), then the surface integral will vanish as the volume is expanded to all space, and so
$\iiint{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{\nabla}\cdot\mathbf{v}\,dV=\imath\mathbf{k}\cdot\iiint{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\,dV$,
and so
$\mathcal{F}\left[\mathbf{\nabla}\cdot\mathbf{v}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\cdot\mathcal{F}\left[\mathbf{v}(\mathbf{x})\right](\mathbf{k})$.

Now, to find the Fourier transform of the curl of a vector field, we use a form of the divergence theorem which states that
$\iiint_{V}\mathbf{\nabla}\times\mathbf{P}\,dV=\iint_{S}\mathbf{\hat{n}}\times\mathbf{P}\,dA$,
where $\mathbf{\hat{n}}$ is the outward normal to the surface S bounding the volume V.
Now, the product rule for the curl of the product of a scalar field and a vector field is
$\mathbf{\nabla}\times(\psi\mathbf{v})=\mathbf{\nabla}\psi\times\mathbf{v}+\psi(\mathbf{\nabla}\times\mathbf{v})$,
so letting P=ψv, we see
$\iiint_{V}\mathbf{\nabla}\psi\times\mathbf{v}\,dV+\iiint_{V}\psi(\mathbf{\nabla}\times\mathbf{v})\,dV=\iint_{S}\mathbf{\hat{n}}\times(\psi\mathbf{v})\,dA$,
so
$\iiint_{V}\psi(\mathbf{\nabla}\times\mathbf{v})\,dV=\iint_{S}(\mathbf{\hat{n}}\times\mathbf{v})\psi\,dA-\iiint_{V}(\mathbf{\nabla}\psi)\times\mathbf{v}\,dV$.

Now, letting $\psi(\mathbf{x})=e^{-\imath\mathbf{k}\cdot\mathbf{x}}$, and using
$\mathbf{\nabla}\psi=\mathbf{\nabla}e^{-\imath\mathbf{k}\cdot\mathbf{x}}=-\imath\mathbf{k}e^{-\imath\mathbf{k}\cdot\mathbf{x}}$,
we get
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}(\mathbf{\nabla}\times\mathbf{v})\,dV=\iint_{S}(\mathbf{\hat{n}}\times\mathbf{v})e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,dA-\iiint_{V}(-\imath\mathbf{k}e^{-\imath\mathbf{k}\cdot\mathbf{x}})\times\mathbf{v}\,dV$,
and thus,
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}(\mathbf{\nabla}\times\mathbf{v})\,dV=\iint_{S}(\mathbf{\hat{n}}\times\mathbf{v})e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,dA+\imath\mathbf{k}\times\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\,dV$.
Again, we can expand the volume to all space, and if v goes to zero fast enough as r goes to infinity, then the surface integral vanishes, so
$\iiint{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}(\mathbf{\nabla}\times\mathbf{v})\,dV=\imath\mathbf{k}\times\iiint{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\,dV$,
and thus
$\mathcal{F}\left[\mathbf{\nabla}\times\mathbf{v}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\times\mathcal{F}\left[\mathbf{v}(\mathbf{x})\right](\mathbf{k})$.