Archive for June, 2010

Monday Math 125

June 28, 2010

Consider two bins, each with N items, with N a large number. Let us randomly select one of the two bins, with equal probability (such as by flipping a fair coin), and remove an item from the selected bin. We repeat this procedure of removing items from the bins by random selection until one bin is empty. What, then, is the expected value n of the number of items remaining in the non-empty bin?
Solution:

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Physics Friday 125

June 25, 2010

Part 5: Work Done on Currents

Consider a point charge of charge q moving from position r0 at time t0 to position r1 at later time t1, and let the curve C be the path it takes between these points. Further, let their be an external electric field E(r). What, then, is the total work done on the charge by this field in the time interval from t0 to t1?

The work done by a force field F on a particle traveling on a curve C is the line integral of F on C:
.
For an electric force, we have F=qE; and with position as a function of time r(t), then the line element is , where v(t) is the velocity. Thus, our work becomes:

Thus, the work done per unit of time, the power transferred to the particle by the field, for a given moment of time is thus

(by the fundamental theorem of calculus).

Now, let us have a current I flowing along the curve C, with external field E. Consider a line element ds of the current. Let the time for a bit of charge in the current to traverse this length be dt. Since the current is the amount of charge flowing per unit time, in this (infinitesimal) time interval, we have a charge of passing through. And since the velocity of this charge is , we have , and so the work done per unit time on this piece of the current is
, and so the total power delivered to the current by the field is
.
Now, if the electric field corresponds to an electric potential φ, then , and so we see
,
and so by the gradient theorem (fundamental theorem of calculus for line integrals):
.
Thus, if we have a drop in potential, a voltage drop, of , then the power delivered to the current by the source of this voltage difference (and the corresponding electric field) is given by the above ; power is current times voltage. This law is a key tool in the analysis of circuits.

Lastly, let us replace our current in a loop with a general current density distribution . Consider within this distribution a small segment ds parallel in direction to the local current density vector J. So further, let us consider a small path C, with a small cross section Δσ perpendicular to this curve. Then the current in this elemental “tube” is , with , and so the power imparted to this element by the field is

But since ds is in the direction of J, then , and so we have
.
Now, considering the entire current density distribution as broken up into such elements, we see , and so our total power is given by the volume integral
.

For magnetic fields, the force exerted on a point charge q with velocity v is the Lorentz force
.
Thus, the magnetic force is always perpendicular to the path of the particle, , and so the magnetic field does no work.

Monday Math 124

June 21, 2010

The central limit theorem tells us that for sufficiently large n, the mean of n samples of any random variable for which the first and second moments (and thus mean and variance) exist has a distribution that approaches, and can thus be approximated by, a normal distribution (Gaussian distribution).

Now, consider the case of the probability of getting exactly k heads in n (fair) coin flips. This is described by the binomial distribution with p=q=1/2
,
where
.
The mean is .

Now, suppose we have 2n flips (so that the mean is n), and we consider the probability of getting heads (so that x is the difference from the mean). What is the Gaussian approximation for the probability distribution when n is large, and what are the limits on x for which the approximation is valid?
Solution:

Physics Friday 124

June 18, 2010

Part 4: Distant Fields

Let us consider the first-order magnetic potential and field far from a localized current density . The first step is to consider the denominator of the integral .
In particular, choosing an origin within the localized distribution, so that , we expand the fraction to first order in r’, obtaining
,
giving us the expansion
.

Now, we need to use a bit of vector calculus. For any well-behaved scalar field u(r) and vector field v(r), then vector integration by parts over a volume V bounded by simple surface S
.
Now, letting our vector field be the current density J, if our volume V completely encloses (and extends beyond) the region to which the current is localized, then J, and thus uJ, is thus zero on the surface S, and so that integral remains zero as the volume is expanded to all space, leaving us
,
or combining,
.
Now, if we let , and use the magnetostatic requirement that , then we see that
,
,
where Jx is the x component of the current density vector field. Note that this integral is the x component of the integral first integral in our expansion, . Similarly, using and each give us

and
,
respectively. Thus, the initial term in our expansion, the monopole term, is zero. Now, indexing the cartesian components of our vectors with i=1,2,3, we now have
.
Now, we return to our integration by parts formula. We use , and
(where is the unit vector for the ith coordinate); together with our requirement that the current be divergence-free, we obtain

Now, taking our integral term , and expanding the dot product into the sum of coordinate products via index j, we have
.
Now, since , we can subtract half of this integral from each integral in the above sum, and
,
giving us


Now, for i=j, the above integral is zero. Similarly, we note that ; and similar expressions, which allow us to write
,
where is the Levi-Civita symbol. This means
;
but this is the ith component of a cross product (see here), so
we have
.
Now, we define the magnetic moment density by , and it’s integral . Then, in terms of this, we see
.
Taking the curl of this, and using the product rule for curl,
, along with the fact that μ is a constant vector, we see
,
where r=|r| is the distance from the origin, and is the unit vector in the direction of r. Note that the field of an electric dipole of dipole moment p is (see here). Thus, our field is that of a magnetic dipole, and μ is our magnetic moment (thus justifying our terming as magnetic moment density).
Note that if our current is confined to a current I in a closed plane curve C with line element , then our integral for magnetic moment becomes
,
and via Green’s theorem (reversing our use of it here), we see
, where n is the normal to the plane of the loop (with direction via the right-hand rule), and the A over which we integrate is the region enclosed by the loop. Thus, we obtain , where is the area vector. This is the same result that we found here by considering the torque a constant magnetic field exerts on a current loop.

Similarly, if we have point charges with masses mi and charges qi, in motion with positions ri and velocities vi, then (approximating it as a static current distribution), we have
, and so the magnetic moment integral becomes a sum via the delta functions:
. However, the angular momentum of the ith particle is
,
so we obtain
.
If the particles all have the same charge-to-mass ratio , then it factors out of the sum, and
,
where L is the total angular momentum. Note that for electrons, this then gives , the same result we obtained here for a single electron in a circular orbit.

Monday Math 123

June 14, 2010

What is the smallest number of integers needed, such that one is guaranteed to have a subset of three integers whose sum is divisible by three?
Solution:

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…)

Monday Math 122

June 7, 2010

Find .
Solution:

Physics Friday 122

June 4, 2010

Part 2: The Vector Potential and Gauge Transformation.

In electrostatics, one often works not with the electric field, but the electric potential. Since the electric field is irrotational in electrostatics ( for all space), it must be the gradient of some scalar function. We choose the negative of this function as the potential, so that ; note that the potential is unique only up to a constant.

For magnetostatics, we have


,

Here, we use the first equation; since the magnetic field is divergence-free, it must be the curl of some vector field, which we call the vector potential:
. In fact, last week, we saw that , already showing that the magnetic field is the curl of a vector field, specifically .
However, just as the scalar electric potential may be transformed by addition of an arbitrary constant without changing the physics, the vector potential is not unique. Noting that the curl of the gradient of any scalar function is identically zero, adding the gradient of an arbitrary scalar field to the vector potential produces the same physics; this transformation is called a gauge transformation. Thus, our general form for the vector potential is
.

Substituting the definition of A into the equation for the curl of the field, we get
,
and using the identity for curl of a curl, we see
.
Thanks to the freedom of gauge transformation, we can choose a scalar φ so that . This is known as the Coulomb gauge. With this choice, we see the vector potential satisfies the (vector) Poisson equation .
Taking the divergence of
,
we see that the Coulomb gauge reduces to , since the integral term has zero divergence. For an unbounded space and the absence of sources at infinity, this reduces to a constant φ, which in turn gives us
.

Also of note, consider an orientable surface S bounded by a simple closed curve C, oriented via the right-hand rule. Then the magnetic flux through S is ; in terms of the vector potential, this is
;
and by Stokes’ theorem, the latter integral can be turned into an integral over the bounding curve C:
;
and the magnetic flux through a simple loop is equal to the line integral of the vector potential over the loop.