Archive for May, 2010

Monday Math 121

May 31, 2010

Prove that (with a>0).
Solution:

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.

Monday Math 120

May 24, 2010

Given a (non-degenerate) conic section, other than a parabola, draw two parallel lines, each of which intersects the conic at two points. Next, find the midpoints of these two parallel chords. Then the center of the conic section lies on the line connecting these two midpoints. This can be used to find the center (and thus the axis and focus/foci) of any conic with only straightedge and compass.

Here, we will give an analytic proof. Let our conic be written as . If p and q are both positive, we have an ellipse (or a circle, if p=q); if they differ in sign, we have a hyperbola. In all cases, the center is at the origin. Given the line , we substitute to find the intersections:

;
If the line has two intersections with the conic, then the discriminant of the above quadratic is positive. The x-coordinate of the midpoint is the average of the solutions to the above quadratic; considering the quadratic formula, this is
;
since the midpoint lies on the line, the y-coordinate is
.
The line connecting this to the origin has slope
.
Note that this is independent of the intercept b of the intersecting line. Thus, a parallel line will give a midpoint that gives the same slope, and thus lies on the same line through the center.

For the parabola, we note that it can be considered the limiting case of an ellipse, or hyperbola, as the eccentricity approaches 1 while the center goes to infinity; hence, a line through the midpoints of paralell chords, and thus through the center, will in the limiting case of the parabola will be parallel to the axis of the parabola (the path of the center as it moves toward infinity).

Physics Friday 120

May 21, 2010

Here is another classic problem: A ladder (of negligible thickness) has length l and mass m, distributed symmetrically enough so that the center of mass is at the geometric center. The ladder is set on a frictionless floor, and leaned against a frictionless wall. The base of the ladder is a small but negligible distance from the wall, so that the ladder, initially at rest, will have the base slide away from the wall, while the top end slides away from the wall. At what point does the ladder lose contact with the wall? What is the maximum force exerted on the ladder by the wall?
Solution:

Monday Math 119

May 17, 2010

Very few of the calculus textbooks I have used give a rigorous derivation of the derivatives of sine and cosine, instead using little more than the graphs of the functions as justification. Here, I will demonstrate a rigorous derivation, starting with a clear derivation of the limit .

[Click on figure for full size image]

Start with the unit circle diagram above, with small, positive θ.
Now, the area of the triangle ▵OAB is . Similarly, the area of the circular sector between OA and OB is . Lastly, the right triangle ▵OAD has area .
Comparing these areas, we have inequality , which means

Multiplying this by the positive quantity , we get
,
which, using , becomes
.
Therefore, the inverses obey the inequality
,
and since , by the squeeze theorem, we see
; and since is an even function, this must also be the left-hand limit, and so we have
.

Now, consider . We can find this limit using the above limit and a little trigonometry:
.

Using these two limits and the addition formulas for sine and cosine, we compute the derivatives from the definition .
First,
,
and second,
.

Happy 50th Birthday…

May 16, 2010

…to the laser.

Physics Friday 119

May 14, 2010

A skydiver of mass 125 kg (including gear), jumps from a plane, travelling with an air speed of 40 m/s at an altitude of 4000 meters. By the time they have descended to 3400 meters, his velocity is sufficiently close to the terminal velocity of 55 meters per second that the further aymptotic approach may be neglected, and velocity approximated by that 55 m/s. After another 45 seconds, he pulls the cord for the parachute. What is the total heat dissipated into the environment (and the skydiver) by the air resistance in the time period between leaving the plane and pulling the cord?
Solution:

Monday Math 118

May 10, 2010

The Laplace Transform
Part 19: Inverse Laplace Transform

Recall from here the Fourier transform
g(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-\imath\omega{t}}\,dt
and its inverse
f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\omega)e^{\imath\omega{t}}\,d\omega.
Combining these, we see
f(\tau)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}f(t)e^{-\imath\omega{t}}\,dt\right]e^{\imath\omega{\tau}}\,d\omega.
Now, letting f(t)=0 for t<0, the above becomes
f(\tau)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{0}^{\infty}f(t)e^{-\imath\omega{t}}\,dt\right]e^{\imath\omega\tau}\,d\omega.

Now, suppose this f(t) has a Laplace transform F(s) with region of convergence \Re(s)>\sigma. Thus, for a real number c>σ, the line \Re(s)=c lies entirely in the region of convergence. So let us make in the change of variables s=c+\imath\omega in the above integral; then ds=\imath\,d\omega, so d\omega=\frac{ds}{\imath}, and we get, with some rearranging, that
\begin{array}{rcl}f(\tau)&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{0}^{\infty}f(t)e^{-\imath\omega{t}}\,dt\right]e^{\imath\omega\tau}\,d\omega\\&=&\frac{1}{2\pi}\int_{c-\imath\infty}^{c+\imath\infty}\left[\int_{0}^{\infty}f(t)e^{(c-s)t}\,dt\right]e^{(s-c)\tau}\,\frac{ds}{\imath}\\f(\tau)&=&\frac{1}{2\pi\imath}\int_{c-\imath\infty}^{c+\imath\infty}\left[\int_{0}^{\infty}f(t)e^{-st}\,dt\right]e^{s\tau}\,ds\end{array},
and thus, the inverse of the Laplace transform is given by
f(t)=\mathcal{L}^{-1}[F(s)]{t}=\frac{1}{2\pi\imath}\int_{c-\imath\infty}^{c+\imath\infty}F(s)e^{st}\,ds,
where the contour of integration is over the line \Re(s)=c. This is known as the Bromwich integral. However, actually evaluating this integral is generally very difficult.

Physics Friday 118

May 7, 2010

Consider the infinite Atwood machine below, with massless, frictionless pulleys and and massless strings.

Letting positive acceleration be upward, what then is the acceleration a1 of the top mass when the masses are all equal, e.g.
m_1=m_2=m_3=\ldots?
What if, instead, successive masses are reduced by multiplication by a factor r, \frac14<{r}<1, e.g.
m_2=rm_1,\;m_3=rm_2,\ldots,\;m_{n+1}=rm_n,\ldots?
Solution:

Monday Math 117

May 3, 2010

The Laplace Transform
Part 18: The Bessel Functions

We can use the differential equation technique demonstrated in the previous post to find the Laplace transforms of the Bessel function of the first kind Jn(t) for nonnegative integer order n. These are defined as the solutions to the Bessel differential equation:

which are non-singular at the origin. Specifically, for integer n≥0, .
First, let’s find the transform of J0(t). This is a solution to with n=0, so

,
or dividing by t,
.
Taking the Laplace transform, and using we have
,
and we have a first order linear differential equation for the transform Y(s). In fact, this equation is separable:
,
where C is a constant, arising from our integration. To find its value, consider the initial value theorem:
.
We see
,
and so . Thus, we see that the transform of the Bessel function of the first kind of order zero is
.

Now, to find n=1, we use a derivative identity for the Bessel functions of the first kind (equation number 59 here):
.
Using the product rule on the left-hand side, we see
.
Taking the Laplace transform, and letting , we have
.
Letting n=1, we then get
.
Integrating,
.
To find the constant of integration, we consider the limit as s→∞; we saw earlier that , so to get , required for J1(t) to be nonsingular for t=0, we must have C=1, and so
.
In general, one can use the relation and proof by induction to show that the general formula is
,
for integer n≥0.
Using the scaling rule and some algebra, this generalizes to
.