## Sunday, 29 April 2018

### Exercise 1.06 Differentiation

#### Question

In Euclidean three-space, let p be the point with coordinates (x,y,z )= (1,0,-1). Consider the following curves that pass through p:

(a) Calculate the components of the tangent vectors to these curves at p in the coordinate basis {∂x, y, ∂z}.

(b) Let  f = x2 + y2 - yz. Calculate df ⁄ dλ , df ⁄ dμ , df ⁄ dσ.

If like me you are a bit rusty (40 years of rust in my case), this was a very good exercise and the Differentiation rules on Wikipedia are a handy reference.

## Saturday, 28 April 2018

### Exercise 1.05 Particle accelerator

#### The question was

Particle physicists are so used to setting c=1 that they measure mass in units of energy. In particular, they tend to use electron volts (1 eV=1.6 x 10-12 erg=1.8 x 10-33 g), or, more commonly, keV, MeV, and GeV. The muon has been measured to have a mass of 0.106 GeV and a rest frame lifetime of 2.19 x 10-6 seconds. Imagine that such a muon is moving in the circular storage ring of a particle accelerator, 1 kilometer in diameter, such that the muon’s total energy is 1000 GeV. How long would it appear to live from the experimenter’s point of view? How many radians would it travel around the ring?

First all I worked this out for  a (rather long) linear accelerator and converted to a circle. The muon will  whizz around it about 1,973 times, subtending an angle of 12,400 radians with a velocity of 0.999 999 99438c.

I was then less feeble and did it in polar coordinates using angular momentum.

On this occasion I completely agreed with Petra Axolotl, who did it in four lines.

## Monday, 23 April 2018

### Exercise 1.11 Maxwell in Relativity

#### The question was

Verify that (1.98) is indeed equivalent to (1.97), and that they are both equivalent to the last two equations in (1.93).

The last two of these are Maxwell's electromagnetic field equations written as tensors.

We also had the electromagnetic field strength tensor, which was claimed to be

The first part of the question is easy: Expand ∂Fνλ] according to the recipe in (1.80).

The second part is more intimidating because, equation (1.98) looks like a monster. It is really 64 equations for all possible values of μ,ν,λ. However, when you work them out, due to the much redundancy and the antisymmetry of Fμν we are left with only four!

## Saturday, 14 April 2018

### Exercise 1.07 Tensors and Vectors

#### The Question was

To see my answer Click Here. The formulae are too complicated to put into this simple editor.

One thing that Sean Carroll had omitted to tell us was that an index can be raised or lowered by applying the metric tensor. So for example,
Xμν = ηνσ Xσμ
This is pretty vital for this exercise! The Wikipedia article is here.

X[μν] = X[μν]
I don't know if that is significant.

## Wednesday, 11 April 2018

### Exercise 1.04 Super luminal effects

#### The question was

Projection effects can trick you into thinking that an astrophysical object is moving “superluminally”. Consider a quasar that ejects gas with speed v at an angle θ with respect to the line-of-sight of the observer. Projected onto the sky, the gas appears to travel perpendicular to the line of sight with angular speed vapp/D, where D is the distance to the quasar and vapp is the apparent speed. Derive an expression for vapp in terms of v and θ. Show that, for appropriate values of v and θ, vapp can be greater than 1.

PSR J0108-1431, the closest known pulsar to the Earth, is 280 light years distant . If the ejected gas position was measured at times a year apart, the angular difference in position α would be 1/128 = 0.0078125. We remember that sin's and tan's of very small angles are very close to the angle.
I don't know why we have to bring relativity into this, so I haven't. Consider the diagram below, α is unrealistically big for visibility.

Our famous observer is at O and the pulsar is at P. The ejected gas arrives at G after, say a year, and it's velocity v is represented by PG. The apparent velocity vapp isis represented by the line PGapp. We added the line PQ at right angles to OGapp and passing thru P. By simple geometry, we get angles in this order:
Starting at O: PGO = π - α - θ then QGP = α + θ
Starting at Gapp: PGappO = π/2 - α then QPGapp = α
calculating  PQ in two different ways we get
PQ = vapp cos α   = v sin (α + θ)
Therefore
vapp   = v sin (α + θ) / cos α
This is not quite what was wanted, but for maximum leverage we obviously want v = π/2 - α so v lies along PQ. We then get
vapp   = v / cos α

Let's assume the gas comes from PSR J0108-1431 and that it's velocity is 0.99997, vapp is 1.000000517. If v is 0.99996, vapp is 0.999990517. It's a close run thing.

#### Embarassing

Sadly my answer was completely wrong as I discovered when I looked at the UCSB answer, which was very clear. They have a slightly different illustration (from Hartle). O is the Observer again and N is the pulsar. They also do not use relativity but they do take account of the fact that if the cloud is moving rapidly towards the observer, the time taken for the light to arrive will be shorter, making the apparent velocity larger. That was at least one of my mistakes. From Hartle via UCSB
The formula they arrive at is
vapp   = v sin θ / (1 - v cos θ)
I have omitted a c because, for us, it is always 1.

This gives much more exciting results. For example at v=0.99, the apparent velocity can reach nearly 7 times the speed of light! Here is a graph of vapp vs θ for a more modest v = 0.95. vapp vs θ for v = 0.95