## Archive for February, 2010

### Physics Friday 110

February 26, 2010

Today’s post is based on an important detail that came up in a physics problem I was discussing with a friend of mind, and which we both initially missed. Here, I will modify the situation from an earlier post to display this.

Given two masses, m1 and m2, connected to each other by a spring of spring constant k2, and with each mass connected to a fixed wall by a spring with spring constant k1, so that at (static) equilibrium position, the springs are all at their resting lengths.

So far, this differs from the problem here only by the fact that the masses now are not the same. With the horizontal displacements of the masses being, as before, x1 and x2, we have differential equations:

,
obtained via F=ma and Hooke’s Law.
Now, suppose that one of the masses, say m2, is decaying or evaporating, so that it obeys , so that it has mass m0 at t=0, and decays toward mf, with τ the time constant for this decay. So, to solve this situation, we just plug in this  into the second differential equation above, to get a linear ordinary differential equation with variable (instead of constant) coefficients, right?

Wrong.

This is where the key point comes in. Our differential equations were made using , but this assumes that the mass is constant. We might be tempted to use the  form, apply the product rule to take the derivative of the momentum  to get , but this would also be incorrect. As Plastino and Muzzio note, this equation violates Galilean invariance.
Specifically, we have to consider the momentum of the mass leaving the system. If the center of mass of the body has velocity v, the escaping mass has velocity u relative to the center of mass, and F is the net external force, then we have
.
Note, that if the mass is lost isotropically (relative to the evaporating body), then the
total effect of the  term(s) will be zero, and we do get
, and our original direct substitution would work. However, if the mass loss is not isotropic, then we will have another term in our differential equation.

### Monday Math 109

February 22, 2010

The Laplace Transform
Part 10: Initial and Final Value Theorems

Last time, we found the formula for the Laplace transform of the derivative of a function; if we have f(t), t>0, with Laplace transform F(s), then
.

Now, let us consider what happens to Laplace transforms as s→∞. By the definition of the Laplace transform, . Noting that as  for all t>0, we see that in the limit as s→∞, the integrand goes to zero throughout the entire region of integration, except the point t=0; so long as f(t) does not have a delta function spike at t=0, the integral has to go to zero in the limit. Thus,
.
So, let us take the limit as s→∞ of our derivative formula. So long as f(t) has no step discontinuities at t=0, f’(t) will not have a delta function spike at t=0, and the above applies to its transform, so we see that
.
This is called the initial value theorem (since it involves the initial value of f).
To find the version for when there is a discontinuity at t=0, we have to remove the point at zero from the integral (replace lower limit 0 with 0+, and so we get
.

Next, let us take the limit as s→0 of the Laplace transform of f’(t). When s=0, , and so
,
and so, if we take the limit as s→0 of our derivative formula,
,
this last being the final value theorem. This tells us that the behavior of f(t) as t→∞ is reflected in the location of the poles of F(s); specifically, we see four cases:
I. If all poles are on the left side () of the s complex plane, then F(0) exists and is finite, and so ; our function f(t) decays to zero exponentially (or faster) at large t.
II. If there are poles are on the right side () of the s complex plane, such as for the Laplace transforms of the hyperbolic sine and cosine, then f(t) contains terms growing exponentially, and f(∞) does not exist.
III. If there are complex conjugate poles on the imaginary axis () of the s complex plane, such as for the Laplace transforms of the sine and cosine, then f(t) contains sinusoidal terms, and f(∞) is undefined.
IV. If there is a simple pole at the origin of the s complex plane (and none in the right side), such as for the Laplace transform of the constant function, then  exists and is nonzero.

### Perspective

February 21, 2010

You wouldn’t expect there to be much in common between them, but both this recent Mark Steyn ariticle on NRO, and this comment at Pharyngula highlight the degree to which modern American culture’s sense of what is or isn’t a danger or a threat has become utterly messed up.

### Physics Friday 109

February 19, 2010

Continuing from last week, we consider the speed of the mass M after it undergoes a deflection by the maximum angle  in a collision with an initially stationary mass m<M.

Recalling the circle formed by the possible velocities, with the maximum angle representing the tangent vector, we can use the right triangle to find the velocity.

Specifically, the Pythagorean theorem tells us that
,
so that we see that the smaller the stationary mass m is in proportion to the mass M, the closer the maximum-angle post-collision velocity is to the pre-collision velocity in both direction and magnitude; and as the masses aproach equality, the maximum-angle post-collision speed goes to zero.
Lastly, we can find the speed as a function of the maximum angle via :
.

### Monday Math 108

February 15, 2010

The Laplace Transform
Part 9: Derivatives and Integrals

Now, let’s examine the Laplace transform’s action on derivatives and integrals. Let us have function f(t) with Laplace transform:
.
What, then, is the Laplace transform of f’(t)?

Using the definition of the Laplace transform,
.
Now, we perform integration by parts, with , , giving us , , and so we have
,
but for our region of convergence, the limit  must be zero, and we have that . Plugging into the above, we see
,

(Note, when there is a discontinuity in f(t) at t=0, then we replace f(0) in the above formula with ; contrast to here.)

Consider then the Laplace transform of the second derivative f”(t). We can apply the above rule twice; since ,
.
Extending on, we see the Laplace transform of the nth derivative of f(t) is
.

This gets us the prime use of the Laplace transform: it turns differential equations, particularly initial value problems, into algebraic equations. For example, the inhomogeneous differential equation , , where H(t) is the Heaviside step function (this corresponds to the charging of a previously uncharged RC circuit via a battery connected at t=0).
Naming the Laplace transform of y(t) as Y(s), we take the Laplace transform of both sides of our differential equation:
,
and our solution is simply a matter of finding the inverse transform of that last term. Using partial fractions,
,
and from here, we see , and so
.

Next, we consider integrals. Let us have, as before, f(t), t>0, with Laplace transform F(s). Now, let us define . Then g(0)=0 and g’(t)=f(t). Letting us write the Laplace transform of g(t) as G(s), we apply our derivative rule to the transform of g’(t) to get:
,
but g’(t)=f(t), so
. Solving for G(s), we then see
,
and we have the formula for integrals.

Note, we could also have found this by noting that , where the * denotes convolution, and by our result here regarding the Laplace transform of convolutions,
,
as above.

### Physics Friday 108

February 12, 2010

Last week, I illustrated an example of momentum and energy conservation under special relativity, with the example of Compton scattering. Now, let’s go back to classical mechanics, and examine perfectly elastic collisions (mechanical energy conserved) between a mass M, moving with velocity V in our laboratory frame, and a mass m stationary in our laboratory frame.

First, consider when the collision is direct enough so that the mass m has post-collision velocity in the direction of V, so that the problem is one-dimensional. Thus, we choose our direction so that V is positive. Then if the masses M and m have post-collison velocities Vf and vf, respectively, then momentum conservation tells us that
,
and mechanical energy conservation tells us that
.
Solving the first equation for the post-collision momentum of m and the second for the post-collision kinetic energy of m, we get
,
.
Squaring the first and multiplying the second by 2m, we get

and
.
Equating these,
.
The above is true (both sides equal zero) if V=Vf, but this is where there is no collision. Thus, for a collision, VVf, and we can cancel a factor of VVf from both sides to get
.

Note that this is negative when M<m; a lighter mass will bounce back when colliding elasticly with a larger one. Note also when the masses are equal, then Vf=0 (and vf=V), so that the motion transfers entirely from one mass to the other; consider Newton’s cradle.
What is most significant, though, is when the moving mass is the heavier one, M>m; then Vf is positive, and both masses will be moving forward post collision.

Now, consider the two-dimensional problem, where the collision can deflect the mass of at an angle. While for M<m, the mass can be deflected through any angle (except exactly zero), our above result indicates that when M>m, then the mass cannot be reflected directly back, and in fact there must be some angle θ<180° which is the maximum angle by which the collision can deflect M.

To determine this angle, let us first consider the center of mass. For two masses m1 and m2 at locations x1 and x2, respectively, then the center of mass is located at . If they are moving, then the velocity of the center of mass is
,
so for our colliding masses, the center of mass will have velocity  in the direction of our initial velocity V, here chosen to be the positive x direction. Note that this will be the velocity of the center of mass both before and after the collision.
So, now let us consider coordinates moving relative to our laboratory frame with velocity equal to vcm. This is our center of mass frame, in which the center of mass is stationary. We will denote quatities in this frame with primes. Since this is a classical problem, we will be using the Galilean transformation between frames.

Now, in the center of mass frame, our mass M has initial velocity in the x‘ direction (same as the x direction) of

and an initial momentum of ,
while the mass M has initial velocity

and an initial momentum of ,
so that the total momentum is zero in this frame, as expected for a center of mass frame.
If the mass M is deflected by an angle θ‘ in this frame, then we see that the mass m must also be deflected by this angle, as the momenta must be equal in magnitude and opposite in direction after the collision. Combining this with mechanical energy conservation, we see that in the center of mass frame, the post-collision velocities must have the same magnitudes after the collision as before; only the directions change.

Thus, in the velocity space, the tip of the  vector traces a circle of radius  about the origin. The velocities in the lab frame are , and so trace an off-center circle in the lab frame velocity space. If M>m, then  is greater than , and this circle does not contain the origin within its interior:

The largest possible deflection, which we see is not only less than 180°, but is less than 90°, is obtained when our lab velocity vector Vf is tangent to the circle.

Simple trigonometry then tells us that
.
So, for example, if M is twice as massive than m, then , and so θ=sin-1(1/2)=30°

Lastly, one should note that when M=m, then the circle of possible velocities in the lab frame passes through the origin. Secondly, since the momenta in the center of mass frame are opposite vectors, and the masses are the same, the center of mass frame velocity vectors  and  are equal in magnitude and opposite in direction; we see that they form a diameter of the circle. Thus, by the inscribed angle theorem (or more specifically, the special subcase known as Thales’ Theorem), we see that the lab frame post collision velocities  and  must form a right angle:

This result is well known in billiards.

### Monday Math 107

February 8, 2010

The Laplace Transform
Part 8: Convolution

Let’s consider two functions, f(t) and g(t), defined for t≥0. Next, we denote their Laplace transforms by F(s) and G(s), respectively. What, then, might we say about the product of those transforms, F(s)G(s)?

From the definition of the Laplace transform, we see

and
.
So then the product is
.
Renaming the variables of integration from t to u and v, we can combine the two into one double integral:
.
Now, we perform a change of variable substitution on the inner integral, defining new variable w=v+u, dw=dv; note that the lower limit v=0 becomes w=u, while the infinite limit remains infinite. We then get
.
Now, reversing the order of integration over the infinite triangular region, we get:
.
Now, to make this more clear, let us rename u with τ, and w with t:


Now, note that if we denote the term of the inner integral by defining the function , we see the above is
,
the Laplace transform of h(t). But what is this function? Note that we said that f(t) and g(t) were defined for t≥0, which means that f(τ)g(τt) is clearly defined only over the range of the integral, 0≤τt. However, if we extend the functions into negative t by zero; that is, define f(t)=g(t)=0 for t≤0, then the product is defined for all τ, but is zero for τ outside the range 0≤τt. With this extension in mind, we see that , and we can see that
,
where f*g denotes the convolution of f and g; thus the Laplace transform of the convolution of two functions is the product of the Laplace transforms of the functions, with the caveat that the functions be zero for negative time. Compare this to the very similar convolution theorem for the Fourier transform.

To demonstrate an example, let us try to find the function f(t) with Laplace transform . Using
, we see
.
Now, we recall from here and here that

and
, so we see that
,
and so we see that f(t) is  times the convolution of  and , where H(t) is the Heaviside step function, needed to make the functions zero for negative t:
,
which can be expressed in terms of Fresnel integrals.

### Physics Friday 107

February 5, 2010

The usage of four-vectors (vectors in Minkowski space) can simplify a number of problems where special relativity must be considered. For an example, we will consider the phenomenon of Compton scattering. Note, for the rest of this I will use the (+,-,-,-) signature; namely that the scalar product of four vectors is .

The energy-momentum 4-vector, also known as the 4-momentum, or sometimes by the (IMO atrocious) portmanteau “momenergy”), is the four-vector extension of momentum, the time component corresponding to energy. Specifically, I will use the convention
 (as seen here). In this form, we see
;
thus the (invariant) length of the four-vector is simply the rest energy of the object.

Now, in Compton scattering, a photon of wavelenth λ collides with an electron at rest in our laboratory frame, and is scattered by an angle θ from its original direction, with new wavelength λ’. We desire to derive a formula for λ’ in terms of λ and θ.
The key is to apply the conservation of energy and momentum, which the use of 4-vectors makes simple: the combined energy-momentum 4-vector must be the same both before and after the collision. We choose our spatial coordinates so that the photon is initially moving in the x direction, and scattering is in the xy plane, with θ in the usual direction in that plane.
As discussed here, the energy and momentum of a photon are given by , and , giving us an energy-momentum 4-vector for our photon pre-collision of .
And letting m be the rest mass of an electron, then our electron has pre-collision energy-momentum 4-vector .
After collision, we have new energy-momentum 4-vectors  and . From our energy and momentum relations for photons, along with the scattering angle θ, we have post-collision that .

Now, our conservation of energy and momentum is
.
Solving for the post-collision 4-momentum of the electron, we have
.
Taking the norm square of both sides, we see
.
But since the norm squared of a four-vector is its scalar product with itself, and the scalar product is a bilinear form, the right hand side of the above is
, and so
. But we recall that the length of an energy-momentum 4-vector is the rest energy; thus , as photons have no rest mass, and
. Thus we need only find the scalar products , , and . As for the first two, since all the spatial components of  are zero,  and . For the last scalar product,
.
Plugging in these, we see
.
Multiplying both sides by λλ’, we get
,
and then dividing both sides by 2hmc3 gives
,
where the quantity  is the Compton wavelength of the electron.

Note that our use of the 4-momentum meant not only could we combine energy and momentum conservation into a single equation, by taking the norm square as we did, we eliminated the need to consider the individual components of the post-collision 4-momentum of the electron.

### Smackdown!

February 3, 2010

Razib Khan at Gene Expression has an enjoyable, and substantial, response to fellw Scibling Martin Rundkvist’s recent post arguing that “for a person to produce more than two children is unethical” because of the environmental footprint, entitled “How to be more ethical than the Swedes.”

### TV Tropes Will Ruin Your Life

February 3, 2010

I rather enjoy the massive time-sink that is TV Tropes. It often has a number of pithy terms and interesting analyses. It has also helped me further recognise the reasons why I don’t like much of popular fiction (specifically, a number of tropes that irk me, such as just about everything listed on the “Double Standard” page, and the overused formulae that make so much of American television too predictable for me). In fact, though, I even find cases outside of fiction that will bring something from TV Tropes to mind.

Specifically, this morning, I read an article about the Tim Tebow/Focus on the Family Super Bowl ad controversy by Sally Jenkins: “Tebow’s Super Bowl ad isn’t intolerant; its critics are“. Though she herself is pro-choice, Jenkins finds fault with those who would silence the opposition:

As statements at Super Bowls go, I prefer the idea of Tebow’s pro-life ad to, say, Jim McMahon dropping his pants, as the former Chicago Bears quarterback once did in response to a question. We’re always harping on athletes to be more responsible and engaged in the issues of their day, and less concerned with just cashing checks. It therefore seems more than a little hypocritical to insist on it only if it means criticizing sneaker companies, and to stifle them when they take a stance that might make us uncomfortable.
I’m pro-choice, and Tebow clearly is not. But based on what I’ve heard in the past week, I’ll take his side against the group-think, elitism and condescension of the “National Organization of Fewer and Fewer Women All The Time.” For one thing, Tebow seems smarter than they do.

Where TV Tropes comes in, however, is with this portion later:

You know what we really need more of? Famous guys who aren’t embarrassed to practice sexual restraint, and to say it out loud. If we had more of those, women might have fewer abortions. See, the best way to deal with unwanted pregnancy is to not get the sperm in the egg and the egg implanted to begin with, and that is an issue for men, too — and they should step up to that.
“Are you saving yourself for marriage?” Tebow was asked last summer during an SEC media day.
“Yes, I am,” he replied.
The room fell into a hush, followed by tittering: The best college football player in the country had just announced he was a virgin. As Tebow gauged the reaction from the reporters in the room, he burst out laughing. They were a lot more embarrassed than he was.
“I think y’all are stunned right now!” he said. “You can’t even ask a question!”
That’s how far we’ve come from any kind of sane viewpoint about star athletes and sex. Promiscuity is so the norm that if a stud isn’t shagging everything in sight, we feel faintly ashamed for him.

My immediate thought went to “A Man is Not a Virgin” which, as an asexual, I’ve always found particularly bothersome (along with the related “All Men are Perverts,” and “I’m a Man, I Can’t Help It“).