## Monday, 10 November 2008

### Basic Relativity

I realise I haven't made any posts here recently; I've been far too busy with my Ph.D. and various other commitments. In lieu of the post I wanted to make on special relativity, defining everything carefully and introducing the tensor formulation, I'm going to present some basic concepts of Special Relativity, almost exactly as I taught them to a first year undergraduate examples class. These people had lectures on relativity already, so the stuff below is really a recap and a worked example.

Lorentz Transforms
Consider a coordinate frame S' moving with velocity v in the positive x-direction of a frame S. The Lorentz transforms between the two frames (x, y, z, t in S and x', y', z', t' in S') are given by:
\begin{align*}x^\prime &= \gamma(x-vt) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x - vt)\\y^\prime &= y\\z^\prime &= z\\t^\prime &= \gamma\left(t - \frac{vx}{c^2}\right) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left(t - \frac{vx}{c^2}\right)\end{align*}

The reverse transforms, from S' back to S, are given by:
\begin{align*}x &= \gamma(x^\prime+vt^\prime) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x^\prime + vt^\prime)\\y &= y^\prime\\z &= z^\prime\\t &= \gamma\left(t^\prime + \frac{vx^\prime}{c^2}\right) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left(t^\prime + \frac{vx^\prime}{c^2}\right)\end{align*}

Velocity Transformation
If an object is moving with a velocity u' in frame S' (as defined above), we need to determine a transform for the velocity in frame S. Start out with the fact that the velocity in S is
u = \frac{\Delta x}{\Delta t}
and substitute in the transforms above:
u = \frac{\Delta x}{\Delta t} &= \frac{\gamma(\Delta x^\prime + v\Delta t^\prime)}{\gamma\left(\Delta t^\prime + \frac{v\Delta x^\prime}{c^2}\right)}

Divide through by
\Delta t^\prime
:
u= \frac{\frac{\Delta x^\prime}{\Delta t^\prime} + v}{1+\frac{v\Delta x^\prime}{\Delta t^\prime c^2}}

Since
\frac{\Delta x^\prime}{\Delta t^\prime}=u^\prime
,

u = \frac{u^\prime + v}{1+\frac{u^\prime v}{c^2}}

which is the velocity transform between u and u'!

Time Dilation
Consider a particle with a lifetime
\tau
in its rest frame (i.e. in an inertial frame in which the particle has zero momentum). The lifetime t, in a frame moving with velocity v is given by:
t = \gamma\tau

Since
\gamma
is always greater than or equal to 1, t is always as large as, or larger than,
\tau
. This is the phenomenon of time dilation and the relationship can be used to determine the time dilation in any inertial frame.

Length Contraction
If the length measured in the lab frame is given by
L_0
, the length seen by a particle moving at speed v in the lab frame is given by:
L = \frac{L_0}{\gamma}

The length is therefore always smaller than (or exactly the same as) the length in the lab frame, and this is the phenomenon of length contraction.

Be careful never to use both time dilation and length contraction in the same (part of a) physics problem! If you're working in the lab frame, use time dilation to alter the particle lifetime, and leave the length as measured in the lab frame. If you're in the particle rest frame, use the particle lifetime as given (since they are defined as lifetime in a particle rest frame) and use length contraction to determine the length seen by the particle from the length as measured in the lab.

The classic atmospheric muon problem, below, illustrates how to use these concepts.

The Atmospheric Muon Problem
Muons are produced in cosmic-ray interactions high in the Earth's atmosphere (at, say 8000m). The muon lifetime is
2.2\times 10^{-6}~\mathrm{s}
.

Muons produced in this way are detected at the Earth's surface. For a muon travelling at 0.998 times the speed of light,
(a) Calculate how far the muon would travel using only classical (non-relativistic) physics.

Here, we use the classical
s = ut
relationship. The maximum distance the muon can travel is given by:
s = 0.998c \tau = 0.998\times 3\times 10^{8} \times 2.2\times 10^{-6} = 658.68~\mathrm{m}

In this situation, the muon cannot reach the Earth's surface.

(b) In the rest frame of the muon, use relativistic physics to show that the muon can indeed reach the surface.

In the muon rest frame, the lifetime is still
2.2\times 10^{-6}~\mathrm{s}
, but the length it has to travel is shortened by length contraction:
L_\mu = \frac{1}{\gamma}L_E

The Lorentz factor is given by:
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = 15.8

giving the length seen by the muon as:
L_\mu = \frac{8000}{15.8} = 506.32~\mathrm{m}

We already worked out that a muon travelling at 0.998c can travel over 600m before it decays using its lifetime as stated, in part a, so now the muons will make it to the surface of the Earth!

(c) In the Earth frame, use relativistic physics to show that the muon can make it to the surface.

Here, the length is as measured, 8000m, but the lifetime of the muon must be dilated:
t = \gamma \tau

We worked out the Lorentz factor above as 15.8, so now we have:
t = 15.8 \times 2.2\times 10^{-6} = 3.48 \times 10^{-5}

The maximum distance such a muon can travel is then:
s = u t = 0.998\times 3\times 10^8 \times 3.48\times 10^{-5} = 10419~\mathrm{m}

The muon can now travel over 10 km in the Earth frame, well over the 8 km it needs to hit the surface, so again we can detect such muons on the Earth.

This example demonstrates several interesting aspects of relativistic physics. Firstly, that it works! We can detect muons produced in this way, which we wouldn't be able to detect if those relativistic effects didn't occur! Secondly, you can analyse a problem using either length contraction or time dilation, but you need to choose your frame carefully. Don't contract or dilate quantities that were measured in the frame you're using; only alter those measured in other frames. Note that lengths are often (but not always) measured in a lab frame, and that particle lifetimes are always given in the rest frame of the particle concerned!