## Archive for April, 2010

### Physics Friday 117

April 30, 2010

What is the net force exerted by a uniform pressure P on the interior of a hemispherical surface of radius R?

One could find the force via the surface integral
$\mathbf{F}=\iint_{\Sigma}P\mathbf{\hat{n}}\,dA$; more specifically, one could use the rotational symmetry, which requires that the net force must be in the direction of the axis, to show one only needs the component along the axis:
$\mathbf{F}=P\mathbf{\hat{z}}\iint_{\Sigma}\cos\theta\,dA$ (where θ is the angle between the surface normal and the axial direction, so $\cos\theta=\mathbf{\hat{n}}\cdot\mathbf{\hat{z}}$).

However, a simpler method is simply to add the flat disk with the same edge as the hemisphere, to create a closed, hemispherical cavity. Now (since we are neglecting gravity), a gas contained within the cavity should exert no net force on its container. Thus, the force on the hemispherical surface must be equal in magnitude and opposite in direction to that on the flat surface. For the flat surface though, the force is simply the pressure times the area. As the surface is a disk of radius R, it has area $A=\pi{R^2}$, so the force on the flat face is $F=\pi{R^2}P$, and thus the force on the hemispherical surface is also of magnitude $F=\pi{R^2}P$ in the direction of the symmetry axis.

Note: we can confirm this by performing the integral outlined above:
$\begin{array}{rcl}\mathbf{F}&=&P\mathbf{\hat{z}}\iint_{\Sigma}\cos\theta\,dA\\&=&P\mathbf{\hat{z}}\int_{0}^{\pi/2}\int_0^{2\pi}\cos\theta\cdot{R^2}\sin\theta\,d\phi\,d\theta\\&=&R^2P\mathbf{\hat{z}}\int_{0}^{\pi/2}\int_0^{2\pi}\cos\theta\sin\theta\,d\phi\,d\theta\\&=&R^2P\mathbf{\hat{z}}\int_{0}^{\pi/2}2\pi\cos\theta\sin\theta\,d\theta\\&=&\pi{R^2}P\mathbf{\hat{z}}\int_{0}^{\pi/2}\sin(2\theta)\,d\theta\\&=&\pi{R^2}P\mathbf{\hat{z}}\left[-\frac{\cos(2\theta)}2\right]_{0}^{\pi/2}\\\mathbf{F}&=&\pi{R^2}P\mathbf{\hat{z}}\end{array}$.

### Monday Math 116

April 26, 2010

The Laplace Transform
Part 17: Differential Equations with Non-constant Coefficients

The Laplace transform can also be used to solve some linear differential equations with non-constant coefficients. For example, let us consider the problem of solving, for t≥0, the differential equation  with initial conditions .

Taking the Laplace transform of both sides, again with ,
.

Now, recall our frequency differentiation formula:

Thus, we see
,
and
.
Thus, our transformed equation becomes:
,
and we have turned our second-order differential equation into a first order one. Further, it is a separable equation:
,
where C is a constant of integration. To find the value of this constant, we turn to the initial value theorem:
.
Here, we see
,
so we find that , and we have . Recalling that , we see that our solution is simply , t≥0.

### Physics Friday 116

April 23, 2010

Consider an inward-sloping surface of revolution with a vertical axis (a “bowl” or “funnel”). Let an object slide frictionlessly on this surface. Let it be at a height where the surface has radius r, and inclination θ from the horizontal; what must the (horizontal, tangential) velocity v be so that the object remains at that height? What is the period T of the object’s “orbit”? What function r(z) gives a surface for which the square of the period is directly proportional to the radius cubed at all heights (so that it mimics Kepler’s third law)?

The object experiences two forces; gravity downward, and a normal force at an angle θ inward from the vertical. For a horizontal orbit, we need the net force to be purely horizontal. Letting N be the magnitude of the normal force, and m be the mass of the object, the balance of vertical force components means , and so . The net force is then the horizontal component of the normal force:

This must be the centripetal force for the uniform circular motion:
.

The length of the object’s path is 2πr, so the period is

Thus, the period squared is
,
so for this to be proportional to r3 (with proportionality constant c), we have
,
where . Now, if we have function r(z), then the derivative r‘(z) gives the outward slope relative to the vertical, and so equals the tangent of the angle of the surface from the vertical, which is ; thus
, giving us the differential equation
;
this is a separable differential equation with solution
, (, ); we have z<b. We can alternately describe this by the inverse function ; below is a graph of the resulting funnel shape.

[Note: this funnel shape gives a height, and thus potential energy, inversely proportional to the radius of the orbit; compare to the inverse proportionality to radius of gravitational potential energy in actual Keplerian orbits. Thus, this last step could also have been determined by considering the energy of the orbiting object.]

### Monday Math 115

April 19, 2010

The Laplace Transform
Part 16: Nonhomogeneous Differential Equations

Another advantage of using Laplace transforms on differential equations is when the equation is nonhomogeneous. For example, consider the problem
$\frac{d^2y}{dt^2}+4\frac{dy}{dt}+9y(t)=H(t)f(t)$, $y(0)=y'(0)=0$,
where H(t) is the Heaviside step function, and f(t) is the square wave $f(t)=\mathrm{sgn}\left(\sin\left(\frac{t}{3}\right)\right)=\left\{\begin{array}{rcl}1&\,\text{if}&\;\sin\left(\frac{t}{3}\right)>{0}\\-1&\,\text{if}&\;\sin\left(\frac{t}{3}\right)<{0}\end{array}\right.$. Note that due to our initial conditions, and the presence of the Heaviside step function on the right-hand side, y(t)=0 for t<0.

Taking the Laplace transform of our differential equation, letting $Y(s)\equiv\mathcal{L}\left[y(t)\right](s)$,
$\begin{array}{rcl}\mathcal{L}\left[\frac{d^2y}{dt^2}+4\frac{dy}{dt}+9y\right](s)&=&\mathcal{L}\left[H(t)f(t)\right](s)\\\mathcal{L}\left[y''\right](s)+4\mathcal{L}\left[y'\right](s)+9\mathcal{L}\left[y\right](s)&=&\mathcal{L}\left[f(t)\right](s)\\\left(s^2Y-sy(0)-y'(0)\right)+4\left(sY-y(0)\right)+9Y&=&\mathcal{L}\left[f(t)\right](s)\\s^2Y+4sY+9Y&=&\mathcal{L}\left[f(t)\right](s)\\\left(s^2+4s+9\right)Y(s)&=&\mathcal{L}\left[f(t)\right](s)\end{array}$

Recall from here that we found, using the periodic function rule, that the square wave $\mathrm{sgn}(\sin{t})=\left\{\begin{array}{rcl}1&\,\text{if}&\sin{t}>{0}\\-1&\,\text{if}&\;\sin{t}<{0}\end{array}\right.$ has Laplace transform $\frac1s\tanh\left(\frac{\pi}2s\right)$, so using the scaling rule, we see that the Laplace transform of our function f(t) is $F(s)=\frac{1}{1/3}\left(\frac{1}{s}\tanh\left(\frac{\pi}{2}\frac{s}{1/3}\right)\right)=\frac{3}{s}\tanh\left(\frac{3\pi}2s\right)$.
So, to continue with our solution,
$\begin{array}{rcl}\left(s^2+4s+9\right)Y(s)&=&\mathcal{L}\left[f(t)\right](s)\\\left(s^2+4s+9\right)Y(s)&=&\frac3s\tanh\left(\frac{3\pi}2s\right)\\Y(s)&=&\frac{1}{s^2+4s+9}\frac3s\tanh\left(\frac{3\pi}2s\right)\\&=&\frac{1}{(s+2)^2+5}\frac3s\tanh\left(\frac{3\pi}2s\right)\\&=&\frac{1}{\sqrt5}\frac{\sqrt5}{(s+2)^2+(\sqrt5)^2}\frac3s\tanh\left(\frac{3\pi}2s\right)\\&=&\mathcal{L}\left[e^{-2t}\sin\left(\sqrt5t\right)\right](s)\mathcal{L}\left[H(t)f(t)\right](s)\end{array}$,
where we have used the frequency shift formula and the transform for the sine. Now, recalling the convolution rule, we see then that
$\begin{array}{rcl}y(t)&=&\left(e^{-2t}\sin\left(\sqrt5t\right)\right)*f(t)\\&=&\int_0^tf(\tau)e^{-2(t-\tau)}\sin\left(\sqrt5(t-\tau)\right)\,d\tau\\&=&\int_0^t\mathrm{sgn}\left(\sin\left(\frac{\tau}{3}\right)\right)e^{-2(t-\tau)}\sin\left(\sqrt5(t-\tau)\right)\,d\tau\end{array}$.

For those who wish to know what this convolution looks like, we continue by breaking up the region of integration piecewise via the square wave:
$\begin{array}{rcl}y(t)&=&\int_0^tf(\tau)e^{-2(t-\tau)}\sin\left(\sqrt5(t-\tau)\right)\,d\tau\\&=&\int_0^{3\pi}e^{-2(t-\tau)}\sin\left(\sqrt5(t-\tau)\right)\,d\tau\\&&\;-\int_{3\pi}^{6\pi}e^{-2(t-\tau)}\sin\left(\sqrt5(t-\tau)\right)\,d\tau\\&&\;+\cdots\\&&\;+(-1)^{\left\lfloor\frac{t}{3\pi}\right\rfloor}\int_{3\pi\left\lfloor\frac{t}{3\pi}\right\rfloor}^{t}e^{-2(t-\tau)}\sin\left(\sqrt5(t-\tau)\right)\,d\tau\\ &=&\left(\sum_{n=0}^{\left\lfloor\frac{t}{3\pi}\right\rfloor-1}(-1)^n\int_{3\pi{n}}^{3\pi(n+1)}e^{-2(t-\tau)}\sin\left(\sqrt5(t-\tau)\right)\,d\tau\right)\\&&\;+(-1)^{\left\lfloor\frac{t}{3\pi}\right\rfloor}\int_{3\pi\left\lfloor\frac{t}{3\pi}\right\rfloor}^{t}e^{-2(t-\tau)}\sin\left(\sqrt5(t-\tau)\right)\,d\tau\end{array}$
$\begin{array}{rcl}y(t)&=&\Bigg[\sum_{n=0}^{\left\lfloor\frac{t}{3\pi}\right\rfloor-1}\frac{(-1)^ne^{6n\pi-2t}}{9}\bigg(e^{6\pi}\Big(2\sin[\sqrt5(t-3(n+1)\pi)]\\&&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\sqrt{5}\cos[\sqrt5(t-3(n+1)\pi)]\Big)\\&&\;\;\;\;-2\sin(\sqrt5(t-3n\pi))+\sqrt{5}\cos(\sqrt5(t-3n\pi))\bigg)\Bigg]\\&&\;+(-1)^{\left\lfloor\frac{t}{3\pi}\right\rfloor}\left[\frac{\sqrt5-e^{6\pi\left\lfloor\frac{t}{3\pi}\right\rfloor-2t}\left(2\sin\left(\sqrt5\left(t-3\pi\left\lfloor\frac{t}{3\pi}\right\rfloor\right)\right)+\sqrt{5}\cos\left(\sqrt5\left(t-3\pi\left\lfloor\frac{t}{3\pi}\right\rfloor\right)\right)\right)}{9}\right]\end{array}$
Below is a graph of this function for over two periods of f(t),

And here is a close-up for small t, showing more clearly that y(0)=y‘(0)=0.

### Physics Friday 115

April 16, 2010

Consider a set of n point masses, with the ith mass having position vector ri and mass mi. Then the total scalar moment of inertia of this collection about our origin is
,
now, using the fact that , we see that, assuming the masses are constant, the time derivative
,
where pi=mivi is the momentum of the ith mass. Let us now define a quantity equal to half of the above:

The quantity G is known as the virial.
Let us now take the time derivative of the virial:
,
where T is the total kinetic energy of the masses, and Fi is the force on the ith mass.

Now, let us denote by Fij the force exerted on mass i by mass j. If we have no external forces, then . Note also that  for all i and j. (For ij, this is Newton’s third law of motion. For i=j, this says Fii=0; no object exerts a force on itself.)
Then we see that


Now, suppose that the system is a stable, bound system. Then, we expect the virial to be bounded to some finite range; thus, when one takes the time average of the derivative, , one should expect it to go to zero when the averaging is over a sufficiently long time scale; thus, since , for such a system, we get
. This is known as the virial theorem.

Now, suppose that the force between the masses is a central force; the force between two masses depends only upon the distance between them, and is directed along the line between them. Then we have a potential V(r), a function of the distance only, of which the force is the negative gradient .
Then, the force on mass i due to mass j is given by
, where  is the distance between mass i and mass j.
Thus,
,
and so
,
and the virial theorem becomes
.

Now, if the potential is given by a power law, , then , and . Therefore, in the power law case, , and so the virial theorem becomes
.
For systems bound by gravity (or for electrostatic systems), we have an inverse square force, so k=-1, and we obtain .

It was in applying the virial theorem to the Coma galaxy cluster that astronomer Fritz Zwicky inferred the existence of what we now call dark matter.

### Monday Math 114

April 12, 2010

The Laplace Transform
Part 15: Applying the Laplace Transform – Differential Equations

As was briefly noted in this part, the primary use of Laplace transforms is in solving linear, ordinary differential equations, as the transform turns the derivatives into algebraic operatons. For a more in-depth example, let us consider the initial value problem
, .
As this has constant coefficients, we could find the indicial equation, solve for its roots, and obtain the general solution. However, to find the specific solution with the above initial conditions would then require differentiating the general solution twice, and inserting values to obtain three linear equations for the three constants. With the Laplace transform, we will directly insert the values for y(0), y‘(0), and y”(0); and so, obtain the particular solution we want with less steps.

Taking the Laplace transform of our differential equation, and letting  to simplify our notation,

,
and so our solution is the inverse Laplace transform of the above.
Since , we can use partial fractions on the above to find that
,
and so we see that
,
and using from here that the Laplace transform of an exponential is
, and from here that the frequency-shifted sine and cosine formulas are

and
,
we see that we have
,
and we have our solution.

### Physics Friday 114

April 9, 2010

Consider a semicircle formed from n distinct, identical, equally spaced masses, with total mass M. Now, let us send in a mass m with initial speed v toward one end of the semicircle, so that it undergoes successive perfectly elastic collisions with each and every one of the n masses. What restrictions are there on m that this is possible, and what is the total momentum of the n masses after the collisions in the limiting case on m? In particular, what are both of these in the limit of large n?

Note that after all n collisions, the mass m will have been deflected by a total angle π, by symmetry (or via geometry), we see then that each collision must deflect this mass by an angle of . Note that each of the masses of our semicircle, the deflecting masses, must have mass . We found here the maximum angle by which a larger mass can be deflected by a smaller one in a perfectly elastic collision, giving us , and thus, we see
,
and thus
,
with equality being our limiting case. When n is large, θ is small, so , and with this approximation,
.

Now, we found here that in the maximum angle deflection case, the velocity post-collision has magnitude ,
and recall that in the mass limit, we have , so after our first collision, mass m has speed

each collision is similar, so each reduces the velocity by the factor , and so the final speed of the mass m after all the collisions is
,
in a direction opposite that of the original velocity; thus, the momentum has undergone a change of , which is equal to the total combined momentum imparted to the n masses.
For small θ, we see that to first order,
, giving us a large n approximation of
,
and recalling that , we see that in the large n limit,
,
and so the net momentum transfer to the masses is
.