263 Math session #3 (12 April 2012)

Math  Stuff

BTM 4.3.4

Keeping the c's in place just because why not, the relativistic expressions for energy and momentum in terms of velocity are:

γ = 1/Sqrt[1-(v/c)^2]
e = m γ c^2
p = m γ v

It is claimed that (when c=1) e^2 = p^2 + m^2.  Ask Mathematica to do the algebra, to verify this.  That is,
Your job: construct the appropriate assertion (using == and putting in c's) and Simplify[] to get True.

BTM 4.3.4 asks you to verify this relationship between the series.
Your job: to warm up, ask for the series for the energy in powers of v (around v=0 of course) to as many terms as you can stomach.  Call the results, say seriesE.

seriesE =

Challenge: can you find the pattern and write the coefficients in terms of n! and things?

Click at right for the answer

a[n_] = 2^(-n)(2n-1)!!/n!

Table[a[n],{n,0,10}]

BTW what is the meaning of the double factorial symbol?

Your job:

What is the radius of convergence of this series?

OK then ask for the series for p to some high order and call it seriesP

seriesP =

Now compute

(seriesE)^2 - (seriesP c)^2

(Your job): do this by hand to order v^4 for the homework.

Complex numbers

A small part of the math homework asks you to start to become familiar with complex numbers.
Of course Mathematica knows all about these.
Your job: evaluate each of the following cells, first guessing what the output will be:

Sqrt[-1]

I^2

I*(2+3 I)

(1+I)*(1+I)

(1 + 2 I)*(3 + 4 I)

Conjugate[1+2I]

1/(1+2I)

Lastly, check problem 5.2.3:

Solve[(2+3 I)/(6+7 I)+ 2/z == 2+9 I,z]

The gravitational redshift

As I will have explained on the blackboard, the equivalence principle says that the experience of acceleration and the experience of gravity are the same.  Here will will tip-toe in to general relativity as we derive the surprising fact that a clock at the top of a building runs faster than a clock in the basement.

Let person A be at the back of a spaceship of proper length L, and person B be at the front.
We let the spaceship accelerate at g for a short time, short enough that v/c is always small.
At t=0 A fires photon 1.  At (small) time T, A fires photon 2.

Your job: write expressions for the positions zA, zB, z1 and z2, e.g.

zA[t_] = (1/2) g t^2

As a check, make a plot, choosing these parameters:

params = {c->4,g->1,T->1,L->1}

Plot[({zA[t],zB[t],z1[t],z2[t]}/.params),{t,0,1.5},PlotRange->{0,2}]

Next, find the time t1 at which photon 1 reaches B (i.e. z1[t]==zB[t]).
You will have to choose the appropriate root of course.

t1 = (fill this in!)

Likewise for photon 2.

t2 = (?!)

In the absence of gravity the time difference t2-t1 (the difference in reception times) should just be T (the difference in emission times).  Is that true? Ask for

Limit[t2-t1,g->0]

Let's take the ratio of times and make it a function:

f[T_] = (t2-t1)/T

Then the point is that even when T->0 this function is not just 1

Limit[f[T],T->0]

Your job: ask for the Series in powers of g or L, keeping the first two terms
(What is the dimensionless parameter we are really expanding in?)

This is the effect we are looking for: B receives the photons less often than A sends them,
or ... B says A's clock runs slow.

Suppose A starts out with synchronized clocks and accelerates for a short time T to
a speed v=gT.  In terms of v, by how much are the clocks out of sync, and which one
is ahead?  Is this in agreement with the mantra you have learned about moving trains?
i.e. that if a moving observer claims his clocks are in synch, then a ground based
observer says no, the (front/back--which?!) clock is ahead by (?! an amount I forget).

The moving square

Choose a speed:

v=.;γ=.;

v= .9; γ = 1/Sqrt[1-v^2];

Suppose an object is to be found at x= x0+vt,y=y0. If the closest corner of our square is at (1,1) at t=0,
here is where it *is*:

plot1=Show[ParametricPlot[{1,1+y},{y,0,1},PlotRange->{{-1,2},{0,2.5}},AspectRatio->2.5/3,PlotStyle->Dashed],ParametricPlot[{1+x,1},{x,0,1/γ},PlotStyle->Dashed],ParametricPlot[{1+1/γ,1+y},{y,0,1},PlotStyle->Dashed],ParametricPlot[{1+x,1+1},{x,0,1/γ},PlotStyle->Dashed]]

But the light arriving *now* (t = 0) at the origin was emitted when :

solnt = Solve[(x0 + v t)^2 + y0^2 == t^2,t]

We want the earlier one of course, at which time the particle is located at:

P[x0_,y0_] = {x0 + v t ,y0}/. solnt[[1]]

Here is a drawing of where the various photons were emitted, i.e. what we literally "see" at time t=0:

plot2=Show[ParametricPlot[P[1,1+y],{y,0,1},PlotRange->{{-1,2},{0,2.5}},AspectRatio->2.5/3],ParametricPlot[P[1+x,1],{x,0,1/γ}],ParametricPlot[P[1+1/γ,1+y],{y,0,1}],ParametricPlot[P[1+x,1+1],{x,0,1/γ}]]

Altogether now:

Show[plot1,plot2]

And now make it move:

Animate[Show[ParametricPlot[P[v t,1+y],{y,0,1},PlotRange->{{-1,2},{0,2.5}},AspectRatio->2.5/3],ParametricPlot[P[v t+x,1],{x,0,1/γ}],ParametricPlot[P[v t+1/γ,1+y],{y,0,1}],ParametricPlot[P[v t+x,1+1],{x,0,1/γ}],
ParametricPlot[{v t,1+y},{y,0,1},PlotRange->{{-1,2},{0,2.5}},AspectRatio->2.5/3,PlotStyle->Dashed],ParametricPlot[{v t+x,1},{x,0,1/γ},PlotStyle->Dashed],ParametricPlot[{v t+1/γ,1+y},{y,0,1},PlotStyle->Dashed],ParametricPlot[{v t+x,1+1},{x,0,1/γ},PlotStyle->Dashed]],{t,-1,2}]

Your job: watch the movie clips posted on the website.