Monday Math 107

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(τ-t) is clearly defined only over the range of the integral, 0≤τ≤t. 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≤τ≤t. 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

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!

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

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“).

Monday Math 106

The Laplace Transform
Part 7: Periodic functions

Let us consider a general periodic function f(x), with period T, and its Laplace transform. From the definition of the Laplace transform,
.
Now, dividing up the region of integration into (infinitely many) intervals of length T, we get
.
Next, on each integral term, we perform the substitution u=t-nT, which gives us
.

Physics Friday 106

Consider two parallel rings of radius R, separated by a perpendicular distance 2d, with a soap bubble (in equilibrium) stretching between them. What is the shape of the soap bubble?

Monday Math 105

The Laplace Transform
Part 6: Frequency shifting, time shifting, and scaling properties

Expanding upon last time’s work on the Laplace transform of the exponential, we can prove quickly the “frequency shifting” property of the Laplace transform. Let real function f(t), t≥0, have Laplace transform . Then, for real number a, we see that
,
and thus multiplication by an exponential in the time domain equals a translation in the s domain.
This can be combined with our results for sine and cosine to get

.
And our results for powers of t to get
.
Note that for all of these, the shift in the “frequency” domain entails an equal shift in the region of convergence; if for , then for .

Next, let’s consider what happens when we scale our time variable. Again, we have f(t), t≥0, with Laplace transform . Now, let us find the Laplace transform for f(at), with a>0.
.
Making u-substitution u=at, then
.
As with our other result, the region of convergence will be transformed along with the scaling.

We found what happens in the time domain when the transform is shifted in the “frequency” domain, but what happens when we shift in the time domain? Recall that our function f(t) is defined for t≥0; when we shift our function forward, we must set those portions that are at t<0 before our shift to zero, so that our translated function is f(t-τ)H(t-τ), where H(t) is the Heaviside step function, and τ>0 is our time delay.
Integrating as we did for the transform of the Heaviside step function,
.
Now, making u-substitution u=t-τ, we find
.
Thus, a time shift corresponds to multiplication by an exponential in the “frequency” domain.

Physics Friday 105

Suppose you have 30 identical 1 Ω resistors connected to form the edges of a regular icosahedron? What, then, is the effective resistance between adjacent vertices? What about the effective resistance between adjacent vertices of a regular dodecahedron formed of 30 identical 1 Ω resistors?

Monday Math 104

The Laplace Transform
Part 5: Exponential and trigonometric functions

Next ia a very simple Laplace transform: that of exponential functions. If , we use the definition of the Laplace transform to find that
,
with for convergence.
From this, and the linearity of the Laplace transform, we see
,
and
,
both with .

Now, in the integral we used for the exponential, we see that the integration still holds, for , if we replace a with ıω; we get
.
Thus, we use and to get
,
and
,
both with .

Phyiscs Friday 104

Consider a massive rope, with length L and mass per unit length of λ. Let this rope be held so that it hangs vertically, with a thin sheet of paper just below the bottom end of the rope. This sheet of paper can support without breaking a maximum weight equal to twice the weight of the rope. If one releases the top of the rope, will the paper stay unbroken until the rope is entirely atop it? If not, what will the height of the top of the rope (relative to the paper) be when the paper breaks?