### Special Relativity

Einstein’s postulates are where it all began.

1. The laws of physics are the same in all inertial frames.
2. Light propagates in a vacuum rectilinearly with the same speed at all times, in all directions and in all inertial frames.

Time dilation: proper time (τ) is the time measured in the rest frame of an object. The time taken in any other frame is given by
t=γτ
where $\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$
as always.

Length contraction: the proper length (L) between two events (x1, t1) and (x2, t2) is the distance between them in the frame where t1=t2; i.e., the frame where the two events occur simultaneously. The length in any other frame is given by $l=\frac{l}{\gamma}$

Lorentz transformations for going from one frame (S) to another frame (S’) travelling at speed v relative to S.

$t'=\gamma(t-\frac{vx}{c^{2}})$

$x'=\gamma(x-vt)$

$y'=y$

$z'=z$

or, in matrix form

$\left[ \begin{array}{ c } ct' \\x' \\y' \\z' \end{array} \right]$ = $\left[ \begin{array}{ cccc} \gamma&-\beta\gamma&0&0 \\-\beta\gamma&\gamma&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array} \right]$$\left[ \begin{array}{ c } ct \\x \\y\\z \end{array} \right]$

where $\beta=\frac{v}{c}$

Lorentz transformations of energy and momentum

$E'=\gamma(E-pc\beta)$

$p_{x'}=\gamma(p_x-\frac{E\beta}{c})$

$p_{y'}=p_y$

$p_{z'}=p_z$

or, in matrix form

$\left[ \begin{array}{ c } E' \\p_{x'} \\p_{y'} \\p_{z'} \end{array} \right]$ = $\left[ \begin{array}{ cccc} \gamma&-\beta\gamma&0&0 \\-\beta\gamma&\gamma&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array} \right]$$\left[ \begin{array}{ c } E \\p_x \\p_y\\p_z \end{array} \right]$

Natural units operate on the principle that $c=\hbar=1$. The reason for doing this is basically because it involves less writing and makes equations look tidier- right up to the point where you need to get a meaningful number out. Then you need to look at the dimensions of the quantities you are working with and multiply by the appropriate powers of c and $\hbar$ to get an answer that makes physical sense.

Don’t worry if natural units sound pointless and difficult to use at first- the more you work with this field of physics, the more you get used to them.

4-vectors are just vectors with four components. Below are some common 4-vectors in natural units:

Position: $\underline{X} = (t, \underline{x})$

Momentum: $\underline{P} = (E, \underline{p})$

Wavevector: $\underline{K} = (\omega, \underline{k})$

Velocity: $\underline{V} = (1, \underline{v})$

Force: $\underline{F} = (\frac{\partial E}{\partial t}, \underline{f})$

Magnitude of 4-vectors: when taking the dot product of a 4-vector, the fourth component has the opposite sign to the other three, e.g. $|\underline{P}|^2 \equiv E^2-|\underline{p}|^2c^2$. We could represent this with a factor of i in the 4-vector, but that’s just a convention that is not universally used.

Compton Scattering with 4-vectors: scattering of high energy photons off electrons.

Photon: initial 4-momentum $\underline{P}=\frac{h\nu}{c}(1, \underline{\hat{n}})$ where $\underline{\hat{n}}$ is the unit vector in the direction of motion.

final 4-momentum $\underline{P}'=\frac{h\nu'}{c}(1, \underline{\hat{n}'})$

$\underline{P}^2=\underline{P}'^2=0$

Electron: initial 4-momentum $\underline{Q}=(m_ec,0)$ as it starts at rest.

final 4-momentum $\underline{Q}'$– we don’t know or care about this, so it will be eliminated later.

$\underline{Q}^2=\underline{Q}'^2$

Conservation of 4-momentum

$\underline{P}+\underline{Q}=\underline{P}'+\underline{Q}'$

$(\underline{P}+\underline{Q}-\underline{P}')^2=\underline{Q}'^2$

$\underline{P}^2+2. \underline{P}.\underline{Q}-2\underline{P}.\underline{P}'+\underline{Q}^2-\underline{Q}.\underline{P}'+\underline{P}'^2=\underline{Q}'^2$

$\underline{P}.\underline{Q}-\underline{P}.\underline{P}'-\underline{Q}.\underline{P}'=0$

Now let’s work out what these dot products actually are

$\underline{P}.\underline{P}'=\frac{h^2\nu\nu'}{c}(1-\cos\theta)$ where θ is the scattering angle.

$\underline{P}.\underline{Q}=hm_e\nu$

$\underline{P}'.\underline{Q}=hm_e\nu'$

Substitute back into conservation of momentum equation

$m_e\nu-\frac{h\nu\nu'}{c}(1-\cos\theta)-m_e\nu'=0$

$\frac{\nu-\nu'}{\nu\nu'}=\frac{h}{m_ec^2}(1-\cos\theta)$

$\frac{1}{\nu'}-\frac{1}{\nu}=\frac{h}{m_ec^2}(1-\cos\theta)$

But of course, $\nu\lambda=c$

$\lambda'-\lambda=\frac{h}{m_ec}2\sin^2\frac{\theta}{2}$

$\lambda'-\lambda=2\lambda_c\sin^2\frac{\theta}{2}$

Where $\lambda_c=\frac{h}{m_ec}$ is the Compton wavelength.

Invariance: the key to solving most problems in this subject is remembering that for a system of particles, $|P_{tot}|^2=E_{tot}^2-p_{tot}^2c^2$ is invariant.

Remember, for a single particle $E^2-p^2c^2=m^2c^4$, the rest mass (invariant).

For a system of particles, it just means that there is always some frame we can find where the sum of their momenta is zero (i.e. there is always a centre-of-mass frame).