Posts Tagged ‘Curl’

Monday Math 135

September 13, 2010

Suppose we have a three-dimensional scalar field a(r) and three-dimensional solenoidal (divergence-free) vector field V(r), both going to zero (sufficiently fast) as |r|→∞. Can we then construct a vector field F(r) such that \mathbf{\nabla}\cdot\mathbf{F}=a and \mathbf{\nabla}\times\mathbf{F}=\mathbf{V}; that is to say, can we construct a field with a specified divergence and curl?

As noted in final remarks here, the answer is yes, we can, and given our boundary conditions, it is unique. More specifically, Helmholtz’s theorem, also known as the “fundamental theorem of vector calculus,” detailed here, tells us our field F may be decomposed into irrotational (zero curl) and solenoidal (zero divergence) components. We further noted that the irrotational component may be expressed as the gradient of a scalar field, and the solenoidal term as the curl of a vector field; we thus expressed the decomposition as
\mathbf{F}=-\mathbf{\nabla}\phi+\mathbf{\nabla}\times\mathbf{A}
Since \mathbf{\nabla}\times\mathbf{A} is the solenoidal term, the divergence of F is just that of the irrotational term, and so our divergence condition is
\begin{array}{rcl}\mathbf{\nabla}\cdot\mathbf{F}&=&a\\\mathbf{\nabla}\cdot\left(-\mathbf{\nabla}\phi\right)&=&a\\\nabla^2\phi&=&-a\end{array},
Poisson’s equation.
Next, we noted here, that in three-dimensions, the Green’s function for the Laplacian with our boundary conditions is G(\mathbf{r},\mathbf{r}')=-\frac{1}{4\pi\left|\mathbf{r}-\mathbf{r}'\right|}. So, we see that
\phi(\mathbf{r})=\iiint{G(\mathbf{r},\mathbf{r}')a(\mathbf{r}')}\,d^3r'=\frac{1}{4\pi}\iiint\frac{a(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r',
since
\begin{array}{rcl}\nabla_r^2\iiint{G(\mathbf{r},\mathbf{r}')a(\mathbf{r}')}d^3r'&=&\iiint\nabla_r^2\left(G(\mathbf{r},\mathbf{r}')\right)a(\mathbf{r}')d^3r'\\&=&\iiint\delta(\mathbf{r}-\mathbf{r}')a(\mathbf{r}')d^3r'\\&=&a(\mathbf{r})\end{array}.

Now, let us consider the curl of our field. As the first term is irrotational, we then get curl equation
\begin{array}{rcl}\mathbf{\nabla}\times\mathbf{F}&=&\mathbf{V}\\\mathbf{\nabla}\times\left(\mathbf{\nabla}\times\mathbf{A}\right)&=&\mathbf{V}\\\mathbf{\nabla}\left(\mathbf{\nabla}\cdot\mathbf{A}\right)-\nabla^2\mathbf{A}&=&\mathbf{V}\end{array}.
Next, let us consider if our solution A is solenoidal; then the above would reduce to
\nabla^2\mathbf{A}=-\mathbf{V},
the vector Poisson’s equation, whose x, y, and z components are each (scalar) Poisson’s equations analogous to the one for φ. So, we then try solutions like in our previous part:
A_x(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_x(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'
A_y(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_y(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'
A_z(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_z(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r',
or combining into a vector equation,
\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'.
Now, we are just left to confirm that this is solenoidal.
First, we note that \mathbf{\nabla}\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)=-\mathbf{\nabla}'\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right), where \mathbf{\nabla}' is the del operator with respect to the primed coordinates.
Then, we see
\begin{array}{rcl}\mathbf{\nabla}\cdot\mathbf{A}&=&\mathbf{\nabla}\cdot\left(\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'\right)\\&=&\frac{1}{4\pi}\iiint\mathbf{V}(\mathbf{r}')\mathbf{\nabla}\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'\\&=&-\frac{1}{4\pi}\iiint\mathbf{V}(\mathbf{r}')\mathbf{\nabla}'\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'\end{array}.

Now, recall vector integration by parts; specifically, for scalar function φ(r) and vector function v(r), then one has on volume V with bounding surface S:
\iiint_{V}\mathbf{\nabla}\phi\cdot\mathbf{v}\,dV=\iint_{S}\phi\mathbf{v}\cdot\mathbf{\hat{n}}dS-\iiint_{V}\phi\mathbf{\nabla}\cdot\mathbf{v}\,dV.
Here, our vector field is V(r‘), and our scalar field is \frac1{\left|\mathbf{r}-\mathbf{r}'\right|}, also as a function of r‘. Then, we see that

\begin{array}{rcl}\iiint_{V'}\mathbf{V}(\mathbf{r}')\mathbf{\nabla}'\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'&=&\iint_{S'}\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\cdot\mathbf{\hat{n}}'dS'-\iiint_{V'}\frac{\mathbf{\nabla}'\cdot\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'\\&=&\iint_{S'}\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\cdot\mathbf{\hat{n}}'dS'\end{array},
as V is solenoidal, so the numerator in the integrand of the right-hand-side volume integral is zero. Now, if V(r) goes to zero faster than |r|-1 as |r|→∞, then the surface integral goes to zero as the volume integral is expanded to all space, and thus, our solution is solenoidal, as we assumed, and so satisfies \mathbf{\nabla}\times\left(\mathbf{\nabla}\times\mathbf{A}\right)=\mathbf{V}.
Thus, we have (unique) solution
\mathbf{F}=-\mathbf{\nabla}\phi+\mathbf{\nabla}\times\mathbf{A},
where
\phi(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{a(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r',
and
\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'.

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Physics Friday 123

June 11, 2010

Part 3: Vector Potential of a Solenoid

Consider the ideal, infinitely long solenoid of radius R, n turns per unit length, and carrying a current I. Let us use cylindrical coordinates (ρ, φ, z), with our z-axis on the axis of the solenoid, so that the field inside the solenoid is in the positive z direction (the current is in the positive φ direction). Then consideration of the symmetries and the right-hand rule indicates that the magnetic field is purely in the z direction. Consideration of Ampère’s Law in the integral form, , as described here over rectangular loops in a ρz plane indicates that the field is a uniform inside the solenoid, and vanishes outside the solenoid (see here). Now, what then is the magnetic vector potential (in the Coulomb gauge)?
(more…)

Physics Friday 121

May 28, 2010

Part 1: Basic Magnetostatics

The basic law of electrostatics is Coulomb’s Law. When working with charge densities rather than discrete points, the more useful form of this is Gauss’ Law (see here and here for previous uses). In integral form, Gauss’ Law reads:
.
Using the divergence theorem, this can be converted into the differential form
.
Comparably, the absense of magnetic monopoles gives the analogous equation
.

Extending beyond electrostatics into magnetostatics, we begin with the conservation of charge in relation to current. The continuity equation is

(as mentioned here). For a steady-state magnetic system, we need no change in the net charge density anywhere, and thus .

For a steady current in an infinitely thin wire, the (static) magnetic field is described by the Biot-Savart Law
,
where I is the current, dl is the differential element vector of the wire oriented in the direction of the current, is the displacement unit vector pointing from the wire element to the point where we are measuring the field, and r is the distance from the wire element to the point where the field is computed; the integral is computed over the wire, which is either a loop, or else extends to infinity (no endpoints). Using the knowledge that the displacement vector is , then we can rewrite it as
.

Let us compute the magnetic field from an infinitely long, straight wire with current I. We let our wire be the z axis, and let z‘ be the coordinate of our current element, and let us choose our z=0 plane to be the plane containing the point we want to measure our field, which due to symmetry should be independent of z, and dependent only on the distance from the wire. Then , , and so
,
which in cylindrical coordinates is . Thus, we have
.

Extending the Biot-Savart law from an infinitely thin wire to a non-zero thickness, we replace the current differential element with , where J is the current density, and the integral is over space; using r’ for the position vector of current elements and r as the position vector for the location at which we compute the field, we obtain

Viewing as a function of r (inverse-square vector field from the point r’), we can use that it is the negative gradient of the inverse distance scalar:
,
where the gradient involves r, not r’. Now, for any vector field a and scalar field ψ, it is an identity that
,
thus
,
and so
.
Now, since our curl is with respect to r, and the current density is a function of r’ only, , and so
, and since the integral is over r’, it will commute with the above curl over r, so we obtain
.
Note that since the divergence of a curl is always zero, this automatically gives us .

Next, let us take the curl of our magnetic field:
.
Here, we need to use the identity for curl of a curl: for any vector field a,
,
giving us
.
Now, the laplacian of the scalar field is equal to the divergence of it’s gradient, which is an inverse square vector field, and so is a three-dimensional Dirac delta:
, meaning that the rightmost integral term becomes
.
We can also use the fact that
,
where indicates the gradient as a function of r’; this gives us
;
this latter integral may be transformed via vector integration by parts; since
, and since our integral being over all space means the bounding surface term vanishes, we obtain
,
so
.

Now, recall that for a magnetic steady-state, we needed , and so the numerator of the integrand in the above is zero, giving us . This is the differential form of Ampère’s law. The corresponding integral form may be obtained from this (or vice-versa) via Stokes’ theorem, and says
,
or in words, the line integral of the magnetic field over a closed curve is equal to the total current I passing through the curve.

Physics Friday 96

November 6, 2009

Suppose, as in this post, we have a fluid of density ρ, with some arbitrary three-dimensional object immersed in the fluid, occupying a volume V. This object has surface ∂V. We saw that for any point on the surface, the pressure is , and the force on an area element is:
. Thus, the torque on that element is
,
where r is the coordinate vector to the element. Recalling that for vectors v and w and scalar a,
, so since P(x,y,z) is a scalar, we see

and so the total torque (about the origin) is
;
and one form of the divergence theorem tells us that for vector field A,
.
Thus

Now, the product rule for the curl of the product of a scalar field ψ and a vector field a is
.
Now, applying that to , and noting that (as it is a radial vector field), and , we see
.
Now, since is a constant vector, it can be “factored out” of the integral:
.
We recall that the total buoyant force on the object is ; plugging this into the above, we see
.

Now, recall that the center of mass of a region with density function ρ(x,y,z) is . If the density is a constant, it factors out of the integrals, and one gets . Thus, if we consider the volume as if it were filled with the fluid; that is to say, the volume of fluid displaced, its center of mass would be , and then we see that
,
which is equivalent to the torque if the buoyant force acted entirely on the point rb; this point is called the center of buoyancy. Just as the force of gravity on an extended object can be treated as if it acts entirely on the center of mass, the buoyant force can be treated as if it acts entirely on the center of buoyancy.