## Sunday, 27 April 2008

### Relativistic Quantum Mechanics: The Klein-Gordon Equation

The Schrödinger equation, used in non-relativistic quantum mechanics, is not Lorentz invariant. This can be seen upon inspection since it is first order in time derivatives, and second order in spatial derivatives. Special relativity must treat space and time on an equal footing, i.e. they must be of the same order. Here's the Schrödinger equation for reference:
i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V(\mathbf{r})\psi

In order to treat quantum mechanics in a Lorentz invariant framework, we must begin with the relativistic energy-momentum relationship:
E^2 = p^2c^2 + m_0^2c^4

Quantum mechanical operators for the energy and momentum can be defined as:
\begin{align*}\hat{E}=-i\hbar\frac{\partial}{\partial t} \\\hat{p}=-i\hbar\nabla\end{align*}

For a plane wave solution of the form
\psi(\mathbf{x},t)=A\exp(i\mathbf{k}\cdot\mathbf{x}-i\omega t)
, these operators give the expected relationships for energy and momentum as their eigenvalues:
\begin{align*}\hat{E}\psi = -i\hbar i\omega\psi = \hbar\omega\psi \\\hat{p}\psi = -i\hbar i\mathbf{k} = \hbar\mathbf{k}\end{align*}

These operators (squared) can be substituted directly into the relativistic energy-momentum relationship. From now on, I'm going to simplify matters by using natural units
(\hbar=c=1)
and writing the rest mass
m_0=m
.
\begin{align*}\hat{E}^2\psi = \hat{p}^2\psi+m^2\psi\\\Rightarrow -\frac{\partial^2}{\partial t^2}\psi = -\nabla^2\psi + m^2\psi \\\Rightarrow \left(-\frac{\partial^2}{\partial t^2} + \nabla^2 -m^2\right)\psi = 0 \\\Rightarrow \left(\partial_\mu\partial^\mu + m^2\right)\psi = 0 \\\Rightarrow \left( \Box +m^2\right)\psi(\mathbf{x},t) = 0\end{align*}

Where
\Box = \partial_\mu\partial^\mu = \frac{\partial^2}{\partial t^2}-\nabla^2
.
This is the Klein-Gordon Equation, and has plane-wave solutions of the form:
\begin{align*}\psi(\mathbf{x},t)=N\exp(-iEt+i\mathbf{p}\cdot\mathbf{x}) \\= N\exp(-ip\cdot x) \\\mathrm{where~}p\cdot x = p_\mu x^\mu = Et - \mathbf{p}\cdot\mathbf{x}\end{align*}

Since
E^2=p^2+m^2
, the energy solutions are:
E = \pm\sqrt{\mathbf{p}^2+m^2}

i.e. there are positive and negative energy states associated with the Klein-Gordon equation. Before discussing the meaning of these, I'd like to first obtain the probability current for the Klein-Gordon equation. For the Schrödinger equation, the probability density is the square of the wavefunction,
\rho = |\psi|^2 = \psi\psi^\dagger
. If we take the Klein-Gordon equation as:
\frac{\partial^2\psi}{\partial t^2}-\nabla^2\psi + m^2\psi = 0

We can multiply by
\psi^\star
and subtract
\psi
the complex conjugate of the Klein-Gordon equation. This results in:
\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{j}=0

Where:
\begin{align*}\rho=i\left[\psi^\star \frac{\partial\psi}{\partial t}-\left(\frac{\partial\psi^\star}{\partial t}\right)\psi\right] \\\mathbf{j} = \frac{1}{i}\left[\psi^\star\nabla\psi - (\nabla\psi^\star)\psi\right]\end{align*}

In four-vector notation, one can write the above as:
\partial_\mu j^\mu = 0

with
j^\mu = (\rho, \mathbf{j}) = i\left[\psi^\star\partial^\mu\psi - (\partial^\mu\psi^\star)\psi\right]

Since
\psi
is Lorentz invariant, and
\partial^\mu
is a contravariant four-vector,
j^\mu
is also contravariant.

The spatial current j is identical to the Schrödinger current, but for the Klein-Gordon equation, the probability density contains time derivatives since K.G. is second-order in time derivatives. The probability density
\rho
is not, therefore, constrained to be positive definite. For plane wave solutions:
\begin{align*}\psi=N\exp(-iEt-i\mathbf{p}\cdot\mathbf{x}) \\\rho = 2|N|^2E \\E = \pm\sqrt{\mathbf{p}^2+m^2} \\\Rightarrow \rho = \pm 2|N|^2\sqrt{\mathbf{p}^2+m^2}\end{align*}

The probability density is therefore positive for positive-energy states and negative for negative energy states. The Klein-Gordon equation was abandoned for some time as a result of this, but the negative energy states may be interpreted as representing particles moving backwards in time, corresponding to anti-particles moving forwards in time! (The Feynman interpretation). The Klein-Gordon equation describes the relativistic quantum mechanics of (massless) spin-zero particles. (And no such fundamental particles exist within the Standard Model, but the neutral pion, composed of
\frac{1}{\sqrt{2}}(u\bar{u}-d\bar{d})
can be considered at low energies. In order to deal with fermions (spin-1/2 particles) we need to look at the Dirac equation, which I might write about another time.

## Saturday, 26 April 2008

### Classical Electrodynamics

(Without proof) Maxwell's equations in Integral form are:
\oint \mathbf{E}\cdot d\mathbf{S} = \frac{\sum{Q}}{\epsilon_0} ~~~~\mathrm{(Gauss)}
\oint \mathbf{E}\cdot d\mathbf{L} = -\frac{\partial\Phi}{\partial t} = \int -\frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{S} ~~~~\mathrm{(Faraday)}
\oint\mathbf{B}\cdot d\mathbf{S} = 0 ~~~~ \mathrm{(No~magnetic~monopoles)}
\oint \mathbf{H}\cdot d\mathbf{L} = \mathbf{I}_c + \mathbf{I}_d ~~~~\mathrm{(Ampere)}

Here,
\mathbf{H}=\frac{\mathbf{B}}{\mu_0}-\mathbf{M}
and Ampere's law has been written with the Displacement Current.

In differential form (again without proof), these can be written:
\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0} ~~~~ \mathrm{(Gauss)}
\nabla\cdot\mathbf{B} = 0 ~~~~ \mathrm{(No~magnetic~monopoles)}
\nabla\times\mathbf{E} + \frac{\partial\mathbf{B}}{\partial t} = 0 ~~~~\mathrm{(Faraday)}
\nabla\times\mathbf{B} = \mu_0\mathbf{j} + \mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}

The continuity equation:
\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{j} = 0

follows from the Inhomogeneous Maxwell equations (
\nabla\cdot\mathbf{E}~,~\nabla\times\mathbf{B}
).

Electrostatics
In electrostatics, one can ignore the time-derivates, leaving:
\begin{align*}\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}\\\nabla\times\mathbf{E}=0\\\nabla\cdot\mathbf{B}=0\\\nabla\times\mathbf{B}=\mu_0\mathbf{j}\end{align*}

\nabla\times\mathbf{E} = 0 \Rightarrow \mathbf{E} = -\nabla V
and V satisfies
\nabla^2 V = -\frac{\rho}{\epsilon_0}~~~(\star)

\nabla\cdot\mathbf{B}=0\Rightarrow \mathbf{B}=\nabla\times\mathbf{A}
and A satisfies
\nabla^2\mathbf{A} = -\mu_0\mathbf{j}

(\star)
is the Poisson equation, and the steady field solutions for V and A are:
\begin{align*}V=\int\frac{\rho\,d\tau}{4\pi\epsilon_0 r}\\\mathbf{A}=\int\frac{\mu_0\mathbf{j}\,d\tau}{4\pi r}\end{align*}

Electromagnetic Waves
In vacuum, without sources, the Maxwell equations are:
\begin{array}{ll}\displaystyle \nabla\cdot\mathbf{E} = 0 & \nabla\cdot\mathbf{E} = 0 \nonumber\\\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t} & \nabla\times\mathbf{B}=\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t} \nonumber\end{array}

Now,
\begin{align*}\nabla\times\nabla\times\mathbf{E}=-\nabla\times\frac{\partial\mathbf{B}}{\partial t} \\= -\frac{\partial}{\partial t}(\nabla\times\mathbf{B}) \\= -\epsilon_0\mu_0\frac{\partial^2\mathbf{E}}{\partial t^2} \\\Rightarrow \nabla^2\mathbf{E} - \mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2} = 0\\\nabla^2\mathbf{B} - \mu_0\epsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2} = 0\end{align*}

The last two lines are three-dimensional wave equations with velocity
c=\frac{1}{\sqrt{\epsilon_0\mu_0}}
.

With Sources: Retarded Potentials
Including source terms, the Maxwell equations are:
\begin{array}{ll}\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0} & \nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t} \nonumber \\\nabla\cdot\mathbf{B} = 0 & \nabla\times\mathbf{B} = \frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t}+\mu_0\mathbf{j} \nonumber\end{array}

One can still define
\mathbf{B}=\nabla\times\mathbf{A}
since
\nabla\cdot\mathbf{B}=0
, but now
\mathbf{E}=-\frac{\partial\mathbf{A}}{\partial t} - \nabla V
.
\begin{align*}\frac{\partial}{\partial t}\nabla\cdot\mathbf{A} + \nabla^2V = \frac{\rho}{\epsilon_0} \\\Rightarrow \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} \\= -\frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2}-\frac{1}{c^2}\nabla\left(\frac{\partial V}{\partial t}\right) + \mu_0\mathbf{j}\end{align*}

If we choose
\nabla\cdot\mathbf{A} = -\frac{1}{c^2}\frac{\partial V}{\partial t}
(i.e. the Lorenz Gauge), then the equations uncouple to:
\begin{align*}\nabla^2 V = -\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\epsilon_0}\\\nabla^2 \mathbf{A} = -\frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0\mathbf{j}\end{align*}

These can be thought of as either Poisson equations with the addition of
\frac{\partial^2}{\partial t^2}
, or as wave equations with sources. The general solution of the Poisson equation for steady
\rho
(as has already been given) is:
V = \int\frac{\rho\,d\tau}{4\pi\epsilon_0 r}

Replace
\rho
by
\rho\left(t-\frac{r}{c}\right)
, rather than by
\rho(t)~\Rightarrow
Retarded Potentials (i.e. the potential lags the source by r/c).
\begin{align*}V=\int\frac{\rho\left(t-\frac{r}{c}\right)\,d\tau}{4\pi\epsilon_0 r} \\\mathbf{A}=\int\frac{\mu_0\mathbf{j}\left(t-\frac{r}{c}\right)\,d\tau}{4\pi\epsilon_0 r}\end{align*}

Electromagnetic Energy: The Poynting Vector
For an unbounded plane wave,
\begin{align*}\frac{\partial E_y}{\partial x}=-\frac{\partial B_z}{\partial t} \\\frac{\partial B_z}{\partial x}=-\frac{1}{c^2}\frac{\partial E_y}{\partial t}\end{align*}

The energy density transported by an electromagnetic wave is:
u_{EM}=\frac{1}{2}\epsilon_0E^2 + \frac{B^2}{2\mu_0}

The transported energy is thus
U_{EM} = \int u_{EM} \,dx\,dy\,dz

The rate of energy loss (e.g. from an aerial) is
-\frac{\partial U_{EM}}{\partial t}
.
\begin{align*}-\frac{\partial U_{EM}}{\partial t} = \int -\frac{\partial}{\partial t}\left( \frac{B_z^2}{2\mu_0}+\frac{\epsilon_0 E_y^2}{2} \right)\,dx\,dy\,dz \\= \int\left[ -\left(\frac{B_z}{\mu_0}\right)\frac{\partial B_z}{\partial t} - (\epsilon_0 E_y)\frac{\partial E_y}{\partial t}\right] \,dx\,dy\,dz \\= \int\left[ \frac{B_z}{\mu_0}\frac{\partial E_y}{\partial x}+\frac{E_y}{\mu_0}\frac{\partial B_z}{\partial x}\right]\,dx\,dy\,dz\end{align*}

Since
H_z=\frac{B_z}{\mu_0}
, then:
\begin{align*}\int_{\mathrm{volume}}\frac{\partial}{\partial x}(E_yH_z)\,dx\,dy\,dz \\= \int_{\mathrm{surface}}\left[(E_yH_z)_{x_1}-(E_yH_z)_{x_2}\right]\,dy\,dz\end{align*}

This suggests that a vector N of the form below correctly describes both the magnitude and direction of energy flow. This is the Poynting Vector:
\mathbf{N}=\mathbf{E}\times\mathbf{H}

## Tuesday, 15 April 2008

### Four-Vector Relativistic Velocity Transformations

Following on from my earlier post, I decided to write about the relativistic velocity transformations using a four-vector approach. We can define a space-time four-vector (or position-time) as follows:
x^\mu = \left(\begin{array}{c} ct \\ x \\ y \\ z \end{array}\right)

The Lorentz transform can be written in matrix form, using
\beta=\frac{u}{c}
and
\gamma=\frac{1}{\sqrt{1-\beta^2}}
:
\mathbf{L}=\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)

Now, one can write a Lorentz transform in the x-direction from a frame S to a frame S' moving with velocity u relative to S as:
x^{\prime\nu} = \mathbf{L}^\nu_\mu x^\mu

Where I've used the Einstein summation convention, in which one sums over repeated indices... in this case from 0 to 3:
\begin{align*}x^{\prime 0} = \mathbf{L}^0_0 x^0 + \mathbf{L}^0_1 x^1 + \mathbf{L}^0_2 x^2 + \mathbf{L}^0_3 x^3 \\x^{\prime 1} = \mathbf{L}^1_0 x^0 + \mathbf{L}^1_1 x^1 + \mathbf{L}^1_2 x^2 + \mathbf{L}^1_3 x^3 \\x^{\prime 2} = \mathbf{L}^2_0 x^0 + \mathbf{L}^2_1 x^1 + \mathbf{L}^2_2 x^2 + \mathbf{L}^2_3 x^3 \\x^{\prime 3} = \mathbf{L}^3_0 x^0 + \mathbf{L}^3_1 x^1 + \mathbf{L}^3_2 x^2 + \mathbf{L}^3_3 x^3 \end{align*}

Since we're dealing with a Lorentz transform in the x direction, we'll ignore the 2 and 3 indices; they correspond to the y and z directions. This leaves the 0 (time) index, and the 1 (x-direction) index to deal with:
\begin{align*}x^{\prime 0} = \gamma ct -\beta\gamma x\\x^{\prime 1} = -\beta\gamma ct + \gamma x\end{align*}

Expanding
\beta
gives:
\begin{align*}x^{\prime 0} = \gamma\left(ct - \frac{ux}{c}\right) = c\gamma\left(t - \frac{ux}{c^2} \right)\\x^{\prime 1} = \gamma(x - ut)\end{align*}

From the Lorentz transforms I discussed in another earlier post:
\begin{align*}t^\prime = \gamma\left(t - \frac{ux}{c^2}\right) \\x^\prime = \gamma(x - ut)\end{align*}

It should therefore be clear to see that our transformed four-vector,
x^{\prime \nu}
has the following form:
x^{\prime \nu} = \left(\begin{array}{c} ct^\prime \\ x^\prime \\ y \\ z \end{array}\right)

We have, therefore, managed to transform a four-vector and equate its components to the expected forms for the Lorentz transformation of x and t coordinates, bearing in mind that the zeroth component of a four-vector is not just time t, but ct. It is then a simple case of dividing the 1st component by the 0th, having removed the c from component 0:
\begin{align*}v^\prime = \frac{x^\prime}{t^\prime} = \frac{vt - ut}{t-\frac{uvt}{c^2}} \\= \frac{t(v-u)}{t(1-\frac{uv}{c^2})} \\=\frac{v-u}{1-\frac{uv}{c^2}}\end{align*}

But there's also another way to approach this problem. We can define a four-velocity
u^\mu = \gamma \left(\begin{array}{c}c\\v\end{array}\right)
. This can be transformed in exactly the same way, using the L matrix:
\begin{align*}u^{\prime \nu} = \mathbf{L}^\nu_\mu u^\mu \\= \gamma\left(\begin{array}{c}\gamma c - \beta\gamma v \\-\beta\gamma c + \gamma v\end{array}\right) \\= \gamma\left(\begin{array}{c}\gamma c\left(1-\frac{uv}{c^2}\right) \\\gamma(v-u)\end{array}\right)\end{align*}

Now, dividing the "space-like" component by the "time-like" one and multiplying by c, i.e.
c\frac{u^{\prime 1}}{u^{\prime 0}}
gives:
\begin{align*}v^\prime = \frac{\gamma(v-u)}{\gamma\left(1-\frac{uv}{c^2}\right)} \\v^\prime = \frac{v-u}{1-\frac{uv}{c^2}}\end{align*}

We have recovered the expression for relativistic velocity transformations!

### Relativistic Velocity Transformations

If a body is moving with speed v in an inertial frame S, what is the speed v' measured in an inertial frame S' moving at speed u relative to S?

In frame S:
v = \frac{x}{t}

In S':
v^\prime = \frac{x^\prime}{t^\prime}

= \frac{\gamma(x-ut)}{\gamma(t-\frac{ux}{c^2})}

= \frac{vt-ut}{t-\frac{uvt}{c^2}}

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

It is possible, from here, to prove that the speed of light is constant under velocity transformations (and thus that c is a limiting speed). Setting v = c:
v^\prime = \frac{c - u}{1-\frac{u}{c}} = c\frac{c-u}{c-u} = c

Cool, huh?

### Relativity Confusion

I've just been working through some relativity questions, and most of the stuff I can just about manage, when prompted by the solutions. Indeed, figuring out which Lorentz transformation to use, and the rough procedure to solve a problem, aren't actually too bad, but today I'm having problems with basic mathematics.

The solutions given are in quite a brief format, where huge swathes of steps have been missed out. Often, it's just a few lines of algebra which is fairly straightforward, but today I don't seem to be able to "see" these, and have to work through them, blind.

One particular bit, which I've tried for the best part of 15 minutes, and have now given up on temporarily, is convincing myself that
\frac{1-v/c}{\sqrt{1-v^2/c^2}} = \sqrt{\frac{1-v/c}{1+v/c}}

This is a fairly fundamental step in calculating Relativistic Doppler shifts, since:
\nu^\prime = \nu\sqrt{\frac{1-v/c}{1+v/c}}

If anyone can tell me how those two things are equal, I'd love to know! Of course, maybe tomorrow it'll be clear. Today is not a good day to revise, but I have to do it anyway.

## Monday, 14 April 2008

### Squirrel at Castleton

[ Click image for larger version ]

This was taken using my 70-300 zoom lens, from a distance of a couple of metres. I was sat in a caravan, watching the rain fall, and this squirrel persistently climbed the bird feeder and ate the nuts contained within!

I like the way (s)he's looking right at me in this photo!

### An Introduction to Relativity

Since I'm supposed to be revising for an exam in Relativity and Electrodynamics, I thought I'd describe a bit of relativity here. I won't go into much of the background, for instance the reasons that Einstein's theory of relativity was required, or other such complications. Wikipedia and other similar resources have more than enough on the history of Einsteinian Relativity. I thought, instead, I'd describe the Lorentz transformation.

Einstein's Principles of Relativity

1. All inertial frames are equivalent with respect to the laws of physics (Principle of Relativity)

2. The speed of light in empty space is independent of the state of motion of its source (Constancy of the speed of light)

Assuming that there is no change of direction of motion, and we have a linear transformation, i.e.
(\Delta x)^\prime = (x_2 - x_1)^\prime = (x_2^\prime - x_1^\prime)

we can choose two frames of reference which are synchronised at the start of a light signal (i.e.
x=x^\prime=0~,~t=t^\prime=0
) and describe a light signal in these two frames, travelling in 1-dimension.

Direction is preserved, hence:
\frac{\Delta x}{\Delta t} = \pm c \rightarrow \frac{\Delta x^\prime}{\Delta t^\prime} = \pm c

Combining these gives:
(\Delta x)^2 - c^2(\Delta t)^2 = 0

and, with the synchronisation of the two reference frames,
x^2 - c^2t^2 = x^{\prime 2} - c^2t^{\prime 2} = 0

for any interval,
(\Delta x)^2 - c^2(\Delta t)^2 = (\Delta x^\prime)^2 - c^2(\Delta t^\prime)^2

We have a linear transformation: If a body K in reference frame S moves with constant velocity u, then K in frame S' moves with constant velocity u'.

x^\prime = a_{11}x + a_{12}t + \mathrm{const}_1 ~~~ (A)

t^\prime = a_{21}x + a_{22}t + \mathrm{const}_2 ~~~ (B)

If S and S' are synchronised at some point
t_0^\prime = t_0 = 0
;
x_0^\prime = x_0 = 0
then
\mathrm{const}_1 = \mathrm{const}_2 = 0
.

For 1D motion, the velocity
\mathbf{u}=u\cdot \mathbf{e}_x
. At the origin of S' (x' = 0):
a_{11}x+a_{12}t = 0 \Rightarrow -\frac{a_{12}}{a_{11}} = \frac{x}{t} = u

Substituting back into (A):
x^\prime = a_{11}(x-ut) = a_{11}(u)(x-ut)~~~(A1)

i.e.
a_{11}
can be a function of u, since u is constant for any given reference frame.

From the principle of Relativity, S and S' are equally valid descriptions of the object K, and S moves relative to S' with velocity -u.
\Rightarrow x = a_{11}(-u)(x^\prime - (-u)t^\prime)~~~(A2)

Invariance under time reflection gives:
t \rightarrow -t \Rightarrow u \rightarrow -u

with x unchanged:
\begin{array}{cc}x^\prime = a_{11}(u)x+a_{12}(u)t & x^{\prime\prime} = a_{11}(-u)x-a_{12}(-u)t \\t^\prime = a_{21}(u)x+a_{22}(u)t & t^{\prime\prime} = a_{21}(-u)x-a_{22}(-u)t\end{array}

\begin{array}{cc}a_{11}(-u) = a_{11}(u) & a_{12}(-u)=-a_{12}(u) \\a_{21}(-u)=-a_{21}(u) & a_{22}(-u)=a_{22}(u)\end{array}

This implies that the coefficients of
a_{11}~\mathrm{and}~a_{22}
should be functions of
u^2
.

For a light signal,
x=ct~;~x^\prime=ct^\prime
from Einstein's 2nd principle. Substituting these into (A1) and (A2) gives:
x^\prime = a_{11}(u^2)x\left(1-\frac{u}{c}\right)

x = a_{11}(u^2)x^\prime\left(1+\frac{u}{c}\right)

Transforming from S to S' and back again should give the identity transformation, i.e.
1 = a_{11}(u^2)a_{11}(u^2)\left(1-\frac{u}{c}\right)\left(1+\frac{u}{c}\right)

\Rightarrow a_{11}^2(u^2) = \frac{1}{1-\frac{u^2}{c^2}}

\Rightarrow a_{11} = \gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} = \frac{1}{\sqrt{1-\beta^2}}

With
\beta =\frac{u}{c}
. These are the Lorentz
\beta, \gamma
factors.

From
-\frac{a_{12}}{a_{11}} = \frac{x}{t}=u
, we get
a_{12} = \frac{-u}{\sqrt{1-\beta^2}}

Hence
x^\prime = \gamma(x-ut)
.

For
a_{21}, a_{22}
, take (A2) and solve for t':
x = a_{11}(x^\prime+ut\prime) \Rightarrow t^\prime = \frac{1}{u}\left(\frac{x}{a_{11}}-x^\prime\right)

t^\prime = \frac{1}{u}\left[\frac{x}{\gamma}-\gamma(x-ut)\right] = \frac{1}{u}\left[\left(\frac{1}{\gamma}-\gamma\right)x+\gamma ut\right]

\sqrt{1-\beta^2}-\frac{1}{\sqrt{1-\beta^2}} = -\gamma\beta^2

Hence:
t^\prime = \frac{1}{u}(\gamma u t - \gamma\beta^2x) = \gamma(t-\frac{ux}{c^2})

a_{22} = \gamma ~;~ a_{21}=-\frac{u\gamma}{c^2}

Thus, the 1-dimensional Lorentz transformations between two inertial frames moving with a relative velocity u are:
x^\prime = \frac{x-ut}{\sqrt{1-\frac{u^2}{c^2}}}

t^\prime = \frac{t-\frac{ux}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}

The inverse transformations are:
x = \frac{x^\prime+ut^\prime}{\sqrt{1-\frac{u^2}{c^2}}}

t = \frac{t^\prime+\frac{ux^\prime}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}

## Thursday, 10 April 2008

### Mobius Battle

The xkcd strip might have been a while ago, but today I was sufficiently bored to actually make one...

## Wednesday, 2 April 2008

### Lake Windermere

I went to the English Lakes for a long weekend, just over a week ago. The photo below is looking along Lake Windermere, from Bowness towards Ambleside, with the Fairfield Horseshoe (central) dominating the scene.