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Archive for January, 2011

PHY450H1S. Relativistic Electrodynamics Tutorial 2 (TA: Simon Freedman). Two worked problems.

Posted by peeterjoot on January 30, 2011

What we will discuss.

\begin{itemize}
\item 4-vectors: position, velocity, acceleration
\item non-inertial observers
\end{itemize}

Problem 1.

Statement

A particle moves on the x-axis along a world line described by

\begin{aligned}ct(\sigma) &= \frac{1}{{a}} \sinh(\sigma) \\ x(\sigma) &= \frac{1}{{a}} \cosh(\sigma)\end{aligned} \hspace{\stretch{1}}(2.1)

where the dimension of the constant $[a] = \frac{1}{{L}}$, is inverse length, and our parameter takes any values $-\infty < \sigma < \infty$.

Find $x^i(\tau)$, $u^i(\tau)$, $a^i(\tau)$.

Solution

Parametrize by time.

First note that we can re-parametrize $x = x^1$ in terms of $t$. That is

\begin{aligned}\cosh(\sigma)&= \sqrt{1 + \sinh^2(\sigma)} \\ &= \sqrt{ 1 + (act)^2 } \\ &= a \sqrt{ a^{-2} + (ct)^2 }\end{aligned}

So

\begin{aligned}x(t) = \sqrt{ a^{-2} + (ct)^2 }\end{aligned} \hspace{\stretch{1}}(2.3)

Asymptotes.

Squaring and rearranging, shows that our particle is moving through half of a hyperbolic arc in spacetime (two such paths are possible, one for strictly positive $x$ and one for strictly negative).

\begin{aligned}x^2 - (ct)^2 = a^{-2}\end{aligned} \hspace{\stretch{1}}(2.4)

Observe that the asymptotes of this curve are the lightcone boundaries. Taking derivatives we have

\begin{aligned}2 x \frac{dx}{d(ct)} -2 (ct) = 0,\end{aligned} \hspace{\stretch{1}}(2.5)

and rearranging we have

\begin{aligned}\frac{dx}{d(ct)}&= \frac{c t}{x} \\ &= \frac{ct}{\sqrt{(ct)^2 + a^{-2}}} \\ &\rightarrow \pm 1\end{aligned}

Is this timelike?

Let’s compute the interval between two worldpoints. That is

\begin{aligned}s_{12}^2&= (ct(\sigma_1) - ct(\sigma_2))^2 - (x(\sigma_1) - x(\sigma_2))^2 \\ &= a^{-2} (\sinh \sigma_1 - \sinh \sigma_2)^2 - a^{-2} (\cosh\sigma_1 - \cosh\sigma_2)^2 \\ &= 2 a^{-2} \left( -1 - \sinh\sigma_1 \sinh \sigma_2 + \cosh\sigma_1 \cosh\sigma_2 \right) \\ &= 2 a^{-2} \left( \cosh( \sigma_2 - \sigma_1) -1 \right) \ge 0\end{aligned}

Yes, this is timelike. That’s what we want for a particle that is realistic moving along a worldline in spacetime. If the spacetime interval between any two points were to be negative, we would be talking about something of tachyon like hypothetical nature.

Reparametrize by proper time.

Our first task is to compute $x^i(\tau)$. We have $x^i(\sigma)$ so we need the relation between our proper length $\tau$ and the worldline parameter $\sigma$. Such a relation is implicitly provided by the differential spacetime interval

\begin{aligned}\left(\frac{d\tau}{d\sigma}\right)^2&= \frac{1}{{c^2}} \left(\frac{ds}{d\sigma}\right)^2 \\ &= \frac{1}{{c^2}} \left(\left( \frac{d(x^0)}{d\sigma}\right)^2-\left( \frac{d(x^1)}{d\sigma}\right)^2\right) \\ &= \frac{1}{{c^2}} \left( a^{-2} \cosh^2 \sigma - a^{-2} \sinh^2 \sigma \right) \\ &= \frac{1}{{a^2 c^2}}.\end{aligned}

Taking roots we have

\begin{aligned}\frac{d\tau}{d\sigma} = \pm \frac{1}{{a c}},\end{aligned} \hspace{\stretch{1}}(2.6)

We take the positive root, so that the worldline is traversed in a strictly increasing fashion, and then integrate once

\begin{aligned}\tau = \frac{1}{{ac}} \sigma + \tau_s.\end{aligned} \hspace{\stretch{1}}(2.7)

We are free to let $\tau_s = 0$, effectively starting our proper time at $t=0$.

\begin{aligned}x^i(\tau) = ( a^{-1} \sinh( a c \tau), a^{-1} \cosh( a c \tau ), 0, 0 )\end{aligned} \hspace{\stretch{1}}(2.8)

As noted already this is a hyperbola (or degenerate hyperboloid) in spacetime, with asymptote 1 (ie: approaching the speed of light).

Proper velocity.

The next computational task is now simple.

\begin{aligned}u^i= \frac{dx^i}{d\tau} = c ( \cosh( a c \tau ), \sinh( a c \tau ), 0, 0) \\ \end{aligned} \hspace{\stretch{1}}(2.9)

Is this light like or time like? We can answer this by considering the four vector square

\begin{aligned}u \cdot u \end{aligned} \hspace{\stretch{1}}(2.10)

Time like vectors.

What is a light like or a time like vector?

Recall that we have defined lightlike, spacelike, and timelike intervals. A lightlike interval between two world points had $(ct - c\tilde{t})^2 - (\mathbf{x} -\tilde{\mathbf{x}})^2 = 0$, whereas a timelike interval had $(ct - c\tilde{t})^2 - (\mathbf{x} -\tilde{\mathbf{x}})^2 > 0$. Taking the vector $(c \tilde{t}, \tilde{\mathbf{x}})$ as the origin, the distance to any single four vector from the origin is then just that vector’s square, so it logically makes sense to call a vector light like if it has a zero square, and time like if it has a positive square.

Consider the very simplest example of a time like trajectory, that of a particle at rest at a fixed position $\mathbf{x}_0$. Such a particle’s worldline is

\begin{aligned}X = ( c t, \mathbf{x}_0 )\end{aligned} \hspace{\stretch{1}}(2.11)

While we interpret $t$ here as time, it functions as a parametrization of the curve, just as $\sigma$ does in this question. If we want to compute the proper time interval between two points on this worldline we have

\begin{aligned}\tau_b - \tau_0 &=\frac{1}{{c}} \int_{\lambda = t_a}^{t_b} \sqrt{ \frac{dX(\lambda)}{d\lambda} \cdot \frac{dX(\lambda)}{d\lambda} } d\lambda \\ &=\frac{1}{{c}} \int_{\lambda = t_a}^{t_b} \sqrt{ (c, 0)^2 } d\lambda \\ &=\frac{1}{{c}} \int_{\lambda = t_a}^{t_b} c d\lambda \\ &= t_b - t_a\end{aligned}

The conclusion (arrived at the hard way, but methodologically) is that proper time on this worldline is just the parameter $t$ itself.

Now examine the proper velocity for this trajectory. This is

\begin{aligned}u = \frac{dX(\tau)}{d\tau} = (c, 0, 0, 0)\end{aligned} \hspace{\stretch{1}}(2.12)

We can compute the dot product $u \cdot u = c^2 > 0$ easily enough, and in this case for the particle at rest (but moving in time) we see that this four-vector velocity does have a time like separation from the origin, and it therefore makes sense to label the four-velocity vector itself as time like.

Now, let’s return to our non-inertial system. Is our four velocity vector time like? Let’s compute it’s square to check

\begin{aligned}u \cdot u = c^2 ( \cosh^2 - \sinh^2 ) = c^2 > 0\end{aligned} \hspace{\stretch{1}}(2.13)

Yes, it is timelike.

Spatial velocity.

Now, let’s calculate our spatial velocity

\begin{aligned}v^\alpha= \frac{dx^\alpha}{dt}=\frac{dx^\alpha}{d\tau} \frac{d\tau}{dt}\end{aligned} \hspace{\stretch{1}}(2.14)

Since $ct = \sinh( a c \tau )/a$ we have

\begin{aligned}c = \frac{1}{{a}} a c \cosh( a c \tau ) \frac{d\tau}{dt},\end{aligned} \hspace{\stretch{1}}(2.15)

or

\begin{aligned}\frac{d\tau}{dt} = \frac{1}{{\cosh( a c \tau) }}\end{aligned} \hspace{\stretch{1}}(2.16)

Similarly from 2.8, we have

\begin{aligned}\frac{dx^1}{d\tau} = c \sinh( a c \tau )\end{aligned} \hspace{\stretch{1}}(2.17)

So our spatial velocity is $\sinh/\cosh = \tanh$, and we have

\begin{aligned}v^\alpha = (c \tanh( a c \tau), 0, 0)\end{aligned} \hspace{\stretch{1}}(2.18)

Note how tricky this index notation is. Four our four vector velocity we use $u^i = dx^i/d\tau$, whereas our spatial velocity is distinguished by a change of letter as well as the indexes, so when we write $v^\alpha$ we are taking our derivatives with respect to time and not proper time (i.e. $v^\alpha = dx^\alpha/dt$).

Four-acceleration

From 2.9, we have

\begin{aligned}w^i (\tau) = \frac{ du^i }{d\tau} = a c^2 x^i(\tau)\end{aligned}

Observe that our four-velocity square is

\begin{aligned}w \cdot w = a^2 c^2 a^{-1} (-1)\end{aligned} \hspace{\stretch{1}}(2.19)

What does this really signify? Think on this. A check to verify that things are okay is to see if this four-acceleration is orthogonal to our four-velocity as expected

\begin{aligned}w \cdot u &= a c^2 ( a^{-1} \sinh( a c \tau), a^{-1} \cosh( a c \tau ), 0, 0 ) \cdot c ( \cosh( a c \tau ), \sinh( a c \tau ), 0, 0) \\ &=c^3 ( \sinh(a c \tau)\cosh(a c \tau) - \cosh(a c \tau) \sinh(a c \tau) ) \\ &=0\end{aligned}

Spatial acceleration

A last beastie that we can compute is the spatial acceleration.

\begin{aligned}a^\alpha &= \frac{du^\alpha}{dt} \\ &= \frac{dv^\alpha}{d\tau} \frac{d \tau}{dt} \\ &= \frac{a c^2}{\cosh^2(a c \tau)} \frac{1}{{\cosh(a c \tau) }} \\ &= \frac{a c^2}{\cosh^3(a c \tau)} \\ \end{aligned}

Summary

Collecting all results we have

\begin{aligned}x^i(\tau) &= \left( a^{-1} \sinh( a c \tau), a^{-1} \cosh( a c \tau ), 0, 0 \right) \\ u^i(\tau) &= c \left( \cosh( a c \tau ), \sinh( a c \tau ), 0, 0\right) \\ v^\alpha(\tau) &= \left( c \tanh(a c \tau), 0, 0 \right) \\ w^i(\tau) &= a c^2 x^i(\tau) \\ a^\alpha(\tau) &= \left( \frac{a c^2}{\cosh^3 (a c \tau)}, 0, 0 \right).\end{aligned} \hspace{\stretch{1}}(2.20)

Problem 2. Local observers.

Basis construction.

Observations are made of either the three-vector, or the time like components of four-vectors, since these are the quantities that we can measure from our local observer frame. This is something that can be viewed in an approximate sense as being inertial, provided that we ignore the earth’s rotation, the rotation around the solar system, the rotation of the solar system in the galaxy, the rotation of the galaxy in the local cluster, and so forth. Provided none of these are changing too fast relative to our measurements, we can make the inertial approximation.

Example. If we want to measure energy, it is the timelike component of the momentum.

\begin{aligned}E = c p^0\end{aligned} \hspace{\stretch{1}}(3.25)

PICTURE: Let’s imagine a moving worldline in three dimensions. We can setup a frame and associated basis along the worldline of the particle, as well as a frame and basis for the stationary observer.

In class Simon used notation like $\{ e_{\hat{o}}^i \}$, and $\{ e_{\hat{a}}^i \}$, but also used $e_{\hat{0}}^i$, $e_{\hat{1}}^i$, $e_{\hat{2}}^i$, $e_{\hat{3}}^i$. It was fairly clear by the context what was meant, but lets avoid any more than one index at a time, and write $\{ f^i \}$ for the frame moving along the worldline, and $\{ e^i \}$ for the stationary frame.

Constructing a basis along the worldline.

For any timelike four-vector worldline we have a four-vector velocity of magnitude $c$, so we are free to define a timelike basis vector for our moving frame as

\begin{aligned}f^0 = u / c\end{aligned} \hspace{\stretch{1}}(3.26)

going back to the first problem for $u^i$ we have

\begin{aligned}f^0 = ( \cosh( a c t ), \sinh( a c t), 0, 0 ) \end{aligned} \hspace{\stretch{1}}(3.27)

We are free to pick spatial unit vectors perpendicular to this, so for the $y$ and $z$ components it is natural to use

\begin{aligned}f^2 &= ( 0, 0, 1, 0 ) \\ f^3 &= ( 0, 0, 0, 1 )\end{aligned} \hspace{\stretch{1}}(3.28)

We need one more, that’s perpendicular to each of the above. By inspection one can pick

\begin{aligned}f^1 = ( \sinh( a c t ), \cosh( a c t), 0, 0) \end{aligned} \hspace{\stretch{1}}(3.30)

Did Simon use any other principle to define this last beastie? I missed it if he did. I see that this happens to be the unit vector proportional to $x^i$.

Consider the stationary observer.

For a stationary observer, our worldline and four velocity respectively, for some constant $\mathbf{x}_0$ is

\begin{aligned}X &= ( ct, \mathbf{x}_0 ) \\ \frac{dX}{d\tau} &= c ( 1, \mathbf{0}) \end{aligned} \hspace{\stretch{1}}(3.31)

Our time like unit vector is very simple

\begin{aligned}e^0 = \frac{1}{{c}} \frac{dX}{d\tau} = ( 1, \mathbf{0} ) \end{aligned} \hspace{\stretch{1}}(3.33)

For the spatial unit vectors we have many choices. One would be aligned from the origin to the position vector

\begin{aligned}e^1 = \left( 0, \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \right),\end{aligned} \hspace{\stretch{1}}(3.34)

with $e^2$ and $e^3$ oriented in any pair of mutually perpendicular spatial directions. Another option would be simply pick a $e^\alpha$ for each of the normal Euclidean basis directions

\begin{aligned}e^1 &= ( 0, 1, 0, 0 ) \\ e^2 &= ( 0, 0, 1, 0 ) \\ e^3 &= ( 0, 0, 0, 1 )\end{aligned} \hspace{\stretch{1}}(3.35)

Observe, that we have (no sum) $e^\alpha \cdot e^\alpha = -1$ (and $e^0 \cdot e^0 = 1$).

Consider an inertial observer.

Now consider a slightly more complex case, where an observer is moving with some constant velocity $\mathbf{V} = c \boldsymbol{\beta}$. Our worldline is

\begin{aligned}X = ( ct, \mathbf{x}_0 + \boldsymbol{\beta} c t) .\end{aligned} \hspace{\stretch{1}}(3.38)

Let’s calculate the four velocity. We have

\begin{aligned}\frac{dX}{dt} = c ( 1, \boldsymbol{\beta} ).\end{aligned} \hspace{\stretch{1}}(3.39)

From this our proper time is

\begin{aligned}\tau = \frac{1}{{c}} \int_0^t c \sqrt{ (1, \boldsymbol{\beta})^2 } dt = \sqrt{1 - \boldsymbol{\beta}^2} t.\end{aligned} \hspace{\stretch{1}}(3.40)

Our worldline and four-velocity, parametrized in terms of proper time, with $\gamma = (1 - \boldsymbol{\beta}^2)^{-1/2}$, are then

\begin{aligned}X &= ( \gamma c\tau, \mathbf{x}_0 + \gamma \boldsymbol{\beta} c \tau) \\ u &= \gamma c ( 1, \boldsymbol{\beta} )\end{aligned} \hspace{\stretch{1}}(3.41)

For this system, let’s label the basis $\{h^k\}$. From above our time like unit vector is

\begin{aligned}h^0 = \gamma ( 1, \boldsymbol{\beta} )\end{aligned} \hspace{\stretch{1}}(3.43)

We observe that this has the desired time like property, $(h^0)^2 = 1 > 0$.

Now, let’s try Gram-Schmidt, subtracting the projection of $h^0$ on $e^1$ from $e^1$ and see what we get. Our projection is

\begin{aligned}\text{Proj}_{h^0}(e^1) &= \frac{e^1 \cdot h^0}{h^0 \cdot h^0} h^0 \\ &= (0, 1, 0, 0) \cdot \gamma (1, \boldsymbol{\beta}) \gamma (1, \boldsymbol{\beta}) \\ &= -\gamma^2 \beta_x (1, \boldsymbol{\beta}).\end{aligned}

We should have a space like vector normal to $h^0$ once we take the Gram-Schmidt difference

\begin{aligned}e^1 - \frac{e^1 \cdot h^0}{h^0 \cdot h^0} h^0 =(0, 1, 0, 0) + \gamma^2 \beta_x (1, \boldsymbol{\beta})\end{aligned} \hspace{\stretch{1}}(3.44)

Let’s compute the norm of this vector and verify that it is space like. We should also verify that it is normal to $h^0$ as expected. For the norm we have

\begin{aligned}-1 + \beta_x^2 + 2 \beta_x \gamma^2 (0, 1, 0, 0) \cdot (1, \boldsymbol{\beta})&=-1 + \beta_x^2 + 2 \beta_x \gamma^2 (-\beta_x) \\ &=\beta_x^2 ( 1 - 2 \gamma^2 ) -1 \\ &=\beta_x^2 \frac{1 - \boldsymbol{\beta}^2 - 2 }{1 - \boldsymbol{\beta}^2} -1 \\ &=-\beta_x^2 \frac{1 + \boldsymbol{\beta}^2}{1 - \boldsymbol{\beta}^2} -1 \\ \end{aligned}

This is less than zero as we expect for a spacelike vector. Good. Our second spacelike unit vector is thus

\begin{aligned}h^1 = \left( \beta_x^2 \frac{1 + \boldsymbol{\beta}^2}{1 - \boldsymbol{\beta}^2} +1 \right)^{-1/2} \left( (0, 1, 0, 0) + \gamma^2 \beta_x (1, \boldsymbol{\beta})\right)\end{aligned} \hspace{\stretch{1}}(3.45)

Let’s verify that these two computed spacetime basis vectors are normal. Their dot product is proportional to

\begin{aligned}((0, 1, 0, 0) + \gamma^2 \beta_x (1, \boldsymbol{\beta})) \cdot (1, \boldsymbol{\beta}) &= -\beta_x + \gamma^2 \beta_x ( 1 - \boldsymbol{\beta}^2 ) \\ &= -\beta_x + \beta_x \\ &= 0 \qquad \square\end{aligned}

We could continue this, continuing the Gram-Schmidt iteration using $e^2$ and $e^3$ for the remainder of the initial spanning set.

Doing so, we’d have

\begin{aligned}h^2 \propto e^2 - \frac{e^2 \cdot h^1}{h^1 \cdot h^1} h^1 - \frac{e^2 \cdot h^0}{h^0 \cdot h^0} h^0.\end{aligned} \hspace{\stretch{1}}(3.46)

After scaling so that $h^2 \cdot h^2 = -1$, we’d then have

\begin{aligned}h^3 \propto e^3 - \frac{e^3 \cdot h^2}{h^2 \cdot h^2} h^2 - \frac{e^3 \cdot h^1}{h^1 \cdot h^1} h^1 - \frac{e^3 \cdot h^0}{h^0 \cdot h^0} h^0.\end{aligned} \hspace{\stretch{1}}(3.47)

Projections and the reciprocal basis.

Recall that for Euclidean space, when we had orthonormal vectors, we could simplify the Gram-Schmidt procedure from

\begin{aligned}e^{k+1} \propto f^{k+1} - \sum_{i=0}^k \frac{f^{k+1} \cdot e^i}{e^i \cdot e^i} e^i,\end{aligned} \hspace{\stretch{1}}(3.48)

to

\begin{aligned}e^{k+1} \propto f^{k+1} - \sum_{i=0}^k \left( f^{k+1} \cdot e^i \right) e^i.\end{aligned} \hspace{\stretch{1}}(3.49)

However, for our non-Euclidean space, we cannot do this. This suggests a nice intuitive motivation for the reciprocal basis. We can define, for any normalized basis $\{f^i\}$ in our Minkowski space (no sum)

\begin{aligned}e_i = \frac{e^i}{e^i \cdot e^i}\end{aligned} \hspace{\stretch{1}}(3.50)

Now our Gram-Schmidt iteration becomes

\begin{aligned}e^{k+1} \propto f^{k+1} - \sum_{i=0}^k \left( f^{k+1} \cdot e^i \right) e_i,\end{aligned} \hspace{\stretch{1}}(3.51)

and we identify, for a four vector $b$, the projection onto the chosen basis vector, as (no sum)

\begin{aligned}\text{Proj}_{e^i}(b) = (b \cdot e^i) e_i.\end{aligned} \hspace{\stretch{1}}(3.52)

In particular, we have for the resolution of identity (now with summation implied again)

\begin{aligned}b = (b \cdot e^i) e_i.\end{aligned} \hspace{\stretch{1}}(3.53)

This is nice and it allows us to work with four vectors in their entirety, instead of in coordinates. We have

\begin{aligned}x = x^i e_i = x_i e^i,\end{aligned} \hspace{\stretch{1}}(3.54)

where

\begin{aligned}x^i &= x \cdot e^i \\ x_i &= x \cdot e_i\end{aligned} \hspace{\stretch{1}}(3.55)

Also note that $e^\alpha = -e_\alpha$ and $e^0 = e_0$, just as the coordinates themselves vary sign with index raising and lowering dependent on whether they are time like or space like.

We’ve seen that the representation of the basis can be chosen to depend on the observer, and for the stationary observer, we had simply

\begin{aligned}e^0 &= (1, 0, 0, 0) \\ e^1 &= (0, 1, 0, 0) \\ e^2 &= (0, 0, 1, 0) \\ e^3 &= (0, 0, 0, 1),\end{aligned} \hspace{\stretch{1}}(3.57)

with a reciprocal basis $e^i \cdot e_j = {\delta^i}_j$

\begin{aligned}e_0 &= (1, 0, 0, 0) \\ e_1 &= -(0, 1, 0, 0) \\ e_2 &= -(0, 0, 1, 0) \\ e_3 &= -(0, 0, 0, 1).\end{aligned} \hspace{\stretch{1}}(3.61)

An alternate basis for the inertial frame.

Given the same $h^0$ as defined above for the inertial frame, let’s define an alternate $h^1$, subtracting the timelike component from the worldline of the particle itself. Let

\begin{aligned}X &= ( \gamma c \tau, \mathbf{x}_0 + \gamma \boldsymbol{\beta} c \tau ) \\ h^0 &= \gamma ( 1, \boldsymbol{\beta} ) \\ Y & = X - (X \cdot h^0) h^0 \end{aligned}

The dot product above is

\begin{aligned}X \cdot h^0&=( \gamma c \tau, \mathbf{x}_0 + \gamma \boldsymbol{\beta} c \tau ) \cdot \gamma (1, \boldsymbol{\beta}) \\ &=\gamma^2 c \tau - \gamma (\boldsymbol{\beta} \cdot \mathbf{x}_0) - \gamma^2 \boldsymbol{\beta}^2 c \tau \\ &=\gamma^2 c \tau ( 1 - \boldsymbol{\beta}^2) - \gamma (\boldsymbol{\beta} \cdot \mathbf{x}_0) \\ &=c \tau - \gamma (\boldsymbol{\beta} \cdot \mathbf{x}_0)\end{aligned}

Our rejection of $h^0$ from $X$ is then

\begin{aligned}Y &= ( \gamma c \tau, \mathbf{x}_0 + \gamma \boldsymbol{\beta} c \tau ) - (c \tau - \gamma (\boldsymbol{\beta} \cdot \mathbf{x}_0)) \gamma( 1, \boldsymbol{\beta}) \\ &= ( \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0), \mathbf{x}_0 + \gamma \boldsymbol{\beta} c \tau - c \tau \gamma \boldsymbol{\beta} + \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0) \boldsymbol{\beta} ) \\ &= ( \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0), \mathbf{x}_0 + \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0) \boldsymbol{\beta} ) \\ &= \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0)(1, \boldsymbol{\beta}) + (0, \mathbf{x}_0)\end{aligned}

We can verify that this is spacelike by computing the square

\begin{aligned}Y^2 &= \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0)^2 - \mathbf{x}_0^2 + 2 \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0) (1, \boldsymbol{\beta}) \cdot (0, \mathbf{x}_0) \\ &= \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0)^2 - \mathbf{x}_0^2 - 2 \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0)^2 \\ &= -\gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0)^2 - \mathbf{x}_0^2 \\ &< 0.\end{aligned}

A final normalization of this yields

\begin{aligned}h^1 = (\gamma^1 (\boldsymbol{\beta} \cdot \mathbf{x}_0)^2 + \mathbf{x}_0^2)^{-1/2} \left( \gamma^2 (\boldsymbol{\beta} \cdot \mathbf{x}_0)(1, \boldsymbol{\beta}) + (0, \mathbf{x}_0) \right)\end{aligned} \hspace{\stretch{1}}(3.65)

It’s easy enough to verify that we have $h^1 \cdot h^0$ as desired.

A followup note on the worldline basis.

Note that we can construct the spatial vector $f^1$ in 3.30 systematically without use of any sort of intuition. We get this by Gram-Schmidt directly

\begin{aligned}f^1 &\propto e^1 - (e^1 \cdot e^0) e_0 - (\underbrace{e^1 \cdot e^2}_{=0}) e_2 - (\underbrace{e^1 \cdot e^3}_{=0}) e_3 \\ &= (0, 1, 0, 0) - (0, 1, 0, 0) \cdot ( \cosh(ac\tau), \sinh(ac\tau), 0, 0) e_0 \\ &= (0, 1, 0, 0) + \sinh(ac\tau) ( \cosh(ac\tau), \sinh(ac\tau), 0, 0) \\ &= (\sinh(ac\tau) \cosh(ac\tau), 1 + \sinh^2(ac\tau), 0, 0 ) \\ &= (\sinh(ac\tau) \cosh(ac\tau), \cosh^2(ac\tau), 0, 0 ) \\ &\propto (\sinh(ac\tau), \cosh(ac\tau), 0, 0 ) \qquad \square\end{aligned}

It’s also noteworthy to observe that we have $f^i \cdot f^j = 0, i \ne j$, and $f^0 \cdot f^0 = 1$ and $f^\alpha \cdot f^\alpha = -1$, as desired.

Relating the Lorentz transformation and coordinate transformations.

We are familiar now with the tensor form of the Lorentz transformation. This takes coordinates to coordinates

\begin{aligned}{x'}^i = {L_j}^i x^j\end{aligned} \hspace{\stretch{1}}(3.66)

Specifying just the coordinates and not the basis associated with the coordinates leaves out some valuable seeming information. For instance, is the basis associated with the pre and post transformed coordinates the same?

For example, suppose that our basis for the primed coordinates is $\{f_i\}$, construction of the four vector (in its entirety) out of its coordinates and this basis requires the sum

\begin{aligned}X &= {x'}^i f_i \\ &= ({L_j}^i f_i) x^j \end{aligned}

This interior sum ${L_j}^i f_i$ is a linear combination of the primed basis vectors, but we see that these are in fact a set of vectors, and can be considered the basis for the unprimed coordinates. We could for example write

\begin{aligned}e_i = {L_j}^i f_i.\end{aligned} \hspace{\stretch{1}}(3.67)

With such a description, our Lorentz transformation becomes just a mechanism to map vectors in one basis into another. To make this clear, let’s work in the opposite order, and suppose that we have a pair of bases $\{e_i\}$ and $\{f_i\}$. For any vector $X$ we can calculate the coordinates utilizing the reciprocal frame.

\begin{aligned}X = (X \cdot e^i) e_i = (X \cdot f^j) f_j.\end{aligned} \hspace{\stretch{1}}(3.68)

Writing

\begin{aligned}x^i &= X \cdot e^i \\ {x'}^i &= X \cdot f^i.\end{aligned} \hspace{\stretch{1}}(3.69)

This is

\begin{aligned}{x'}^k f_k = x^j e_j.\end{aligned} \hspace{\stretch{1}}(3.71)

Dotting with $f^i$ we have

\begin{aligned}{x'}^i = x^j (e_j \cdot f^i).\end{aligned} \hspace{\stretch{1}}(3.72)

In this form we see explicitly that the Lorentz transformation is in fact the “direction cosines” associated with a change of basis. Specifically, we can write

\begin{aligned}{L_j}^i = e_j \cdot f^i\end{aligned} \hspace{\stretch{1}}(3.73)

I like this as a way to view the Lorentz transformation, since the explicit inclusion of the basis sets involved makes the geometry clear.

A coordinate calculation example.

I have gone to the effort of calculating some basis representations in a lot more detail than we covered in the tutorial, and explore some of the ideas further. This seemed important to get a feel for what we were discussing, and to see how the pieces fit together.

Let’s do one more simple example, where we look at the coordinates of a four vector in the coordinate system where the time like direction is the proper velocity, and also eliminate the the $y$ and $z$ coordinates from the mix to simplify it further. For such a system we have only two choices for our spatial basis vector (we can alter the sign).

For our spacetime point, consider the worldline for a particle moving at a constant velocity. That is

\begin{aligned}X = (ct, p_0 + \beta c t).\end{aligned} \hspace{\stretch{1}}(3.74)

As before our proper time is

\begin{aligned}\tau = \sqrt{1 - \beta^2} t,\end{aligned} \hspace{\stretch{1}}(3.75)

allowing us to re-parametrize the worldline, and have a proper time parametrized velocity

\begin{aligned}X &= (\gamma c \tau, p_0 + \beta \gamma c \tau) \\ u &= \gamma (1, \beta)\end{aligned} \hspace{\stretch{1}}(3.76)

Let’s utilize the standard basis for the stationary frame, and denote this $\{e^i\}$

\begin{aligned}e^0 &= (1, 0) \\ e^1 &= (0, 1)\end{aligned} \hspace{\stretch{1}}(3.78)

and calculate a basis $\{f^i\}$ for which $f^0 = u$ is the time like direction. By Gram-Schmidt, our space like basis vector is

\begin{aligned}f^1 &\propto e^1 - (e^1 \cdot f^0) f_0 \\ &= (0,1) - (0, 1) \cdot \gamma (1,\beta) \gamma (1, \beta) \\ &= (0,1) - \gamma^2 (-\beta) (1, \beta) \\ &= (\gamma^2 \beta, 1 + \beta^2 \gamma^2 ) \\ &= \left(\gamma^2 \beta, \frac{1}{{1-\beta^2}}(1 -\beta^2 + \beta^2 \right) \\ &= \left(\gamma^2 \beta, \gamma^2 \right) \\ &\propto - \gamma (\beta, 1) \end{aligned}

The negative sign here is a bit of sneaky move and chosen only after calculating the coordinates of the vector in this frame, so that at speed $\beta = 0$, the coordinates in frames $\{e^i\}$ and $\{f^i\}$ are the same. Our basis is then

\begin{aligned}f^0 &= \gamma(1, \beta) \\ f^1 &= -\gamma(\beta, 1) \end{aligned} \hspace{\stretch{1}}(3.80)

One can quickly verify that $f^0 \cdot f^0 = 1, f^1 \cdot f^1 = -1$, and $f^0 \cdot f^1 = 0$. Our reciprocal frame, defined so that $f^i \cdot f_j = {\delta^i}_j$ is

\begin{aligned}f_0 &= \gamma(1, \beta) \\ f_1 &= \gamma(\beta, 1) \end{aligned} \hspace{\stretch{1}}(3.82)

With this basis our coordinate representation is

\begin{aligned}X = \underbrace{(X \cdot f^0)}_{x^0} f_0 + \underbrace{(X \cdot f^1)}_{x^1} f_1,\end{aligned} \hspace{\stretch{1}}(3.84)

and we calculate our coordinates to be

\begin{aligned}x^0 &= c \tau - \gamma p_0 \beta \\ x^1 &= \gamma p_0 \end{aligned} \hspace{\stretch{1}}(3.85)

As a check one can verify that $X = x^0 f_0 + x^1 f_1$ as expected. So we see that in a frame for which the proper velocity is the time like basis vector, our particle is at rest (moving only in time).

Some interesting information can be extracted after making the coordinate calculation. It is interesting to note that the position $x^1 = \gamma p_0$ equals $p_0$ when $\beta = 0$. When the particle is observed at rest in one frame, it remains at rest in the frame for which its proper velocity is the time like direction (the particle’s rest frame). Furthermore, when the particle is observed moving, the position in the particles rest frame is always greater than the observed position $x_0 \gamma \ge x_0$. In other words, the particle’s position appears closer to the origin in the observer’s frame than it is in the rest frame (it’s position is contracted).

Also see that the rest frame time matches the observer frame time when the particle is observed at rest ($\beta = 0$). The time in the rest frame is always less than the time in the observer frame and by increasing $beta$ we can shift the initial time position of the particle in its rest frame as far backwards as we like. Similarly, if the particle is observed moving backwards in the observer frame, the initial time position of the particle in the rest frame can be pushed as far forward in time as we like.

An initially confusing aspect of the given non-inertial worldline.

For the worldline

\begin{aligned}X = \frac{1}{{a}} ( \sinh( a c \tau ), \cosh( a c \tau) ),\end{aligned} \hspace{\stretch{1}}(3.87)

we calculated

\begin{aligned}u &= c ( \cosh( a c \tau ), \sinh( a c \tau) ) \\ f^0 = u/c &= ( \cosh( a c \tau ), \sinh( a c \tau) ) \\ f^1 &= ( \sinh( a c \tau ), \cosh( a c \tau) ).\end{aligned} \hspace{\stretch{1}}(3.88)

\begin{aligned}x^0 &= X \cdot f^0 = 0 \\ x^1 &= X \cdot f^1 = -\frac{1}{{a}},\end{aligned} \hspace{\stretch{1}}(3.91)

the timelike coordinate is zero uniformly? We can verify easily that the position four vector is recovered as expected from $X = x^0 f_0 + x^1 f_1$, but it still seems irregular that we have no timelike coordinate?

Oh! I see. This is a spacelike four vector. Look at the length

\begin{aligned}X^2 = \frac{1}{{a^2}} ( \sinh^2 ( a c \tau) - \cosh^2 (a c \tau) ) = -\frac{1}{{a^2}} < 0.\end{aligned} \hspace{\stretch{1}}(3.93)

Because it is spacelike in one frame, it can only be (just) spacelike in its rest frame.

By calculating this coordinate, we also see that a choice of

\begin{aligned}f^1 = -( \sinh( a c \tau ), \cosh( a c \tau) ),\end{aligned} \hspace{\stretch{1}}(3.94)

would have been a better one. Then our particle’s coordinate in the rest frame would be $1/a$ at $t=0$. With the initial choice of the basis vector $f^1$, it’s coordinate ends up being the inverse of its position at $t=0$.

Split of energy and momentum (VERY ROUGH NOTES).

\paragraph{Disclaimer.} At the very end of the tutorial Simon jotted some very quick notes, and I’ve included what I got of those without editing below. I have yet to go through these and make something coherent of them.

In a coordinate representation, the timelike component of our momentum was obtained by extracting the first coordinate

\begin{aligned}p^0 = (p^0, p^1, p^2, p^3) \cdot (1, 0, 0, 0).\end{aligned} \hspace{\stretch{1}}(3.95)

This was (after scaling) was our energy term $E = c p^0$, and we can extract this in the observer frame by dotting with our observer frame timelike basis vector $e^0$

\begin{aligned}E_\text{observer} = c p \cdot e^0 \equiv c p^0\end{aligned} \hspace{\stretch{1}}(3.96)

In the observers reference frame, where $u^i = c ( 1, 0, 0, 0)$, and $p^i = m u^i$, we have

\begin{aligned}{p}^i = (m c, 0, 0, 0)\end{aligned} \hspace{\stretch{1}}(3.97)

\begin{aligned}{u'}^i_\text{observed} = c \gamma (( 1, v/c , 0, 0)\end{aligned} \hspace{\stretch{1}}(3.98)

\begin{aligned}u^i_\text{observer} c (1, 0, 0, 0)\end{aligned} \hspace{\stretch{1}}(3.99)

\begin{aligned}{u'}^i_\text{observer} =\begin{bmatrix}\gamma & \gamma v/c & 0 & 0 \\ \gamma v/c & \gamma & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.100)

\begin{aligned}p^0 = \gamma m c\end{aligned} \hspace{\stretch{1}}(3.101)

Frequency of light from a distant star (AGAIN VERY ROUGH NOTES).

Suppose we have a star far away. What is the frequency of the light emitted

\begin{aligned}\hat{\omega} = \omega e^{- a c \tau }\end{aligned} \hspace{\stretch{1}}(3.102)

FIXME: derive.

where $\omega$ is the emitted frequency.

FIXME: This implied an elapsed time before the star would no longer be visible?

Lasagna, Take II (regular, plus gluten free)

Posted by peeterjoot on January 27, 2011

Last year, I posted notes on my first completely independent Lasagna attempt:

https://peeterjoot.wordpress.com/2010/02/08/lasagne-feb-7th-attempt/

I’ve tried it again, but this time I have some pictures to go with it.  I also prepared one regular Lasagna, and made an attempt at a second gluten free variation, using the same batch of filling material.  Since I couldn’t find Lasagna style gluten free noodles, I used rice noodles for the gluten free variant, and layered them, in a pseudo-lasagna like fashion

I used way too many rice noodles for that little Lasagna (a one pound package would have done).  For the regular lasagna, I used about 3/4 of a package of spinach noodles, which was also too much, but I’ll be able to use the leftovers for something else.

Ingredients for this attempt included:

• 1 package (500g / 1lb) lean ground beef.
• 1 jar spaghetti sauce.
• 1 container of cottage cheese (don’t have to bother straining).
• Bag of shredded mozzarella on hand.
• Parmesan to grate on top when done.
• A bunch of mushrooms (~12), sliced.
• 3 small zucchinis, sliced.
• 1 yellow pepper.
• A bag of spinach, fried to make it easier to handle when layering.
• A bit of regular and some super old cheddar (only because I ran out of mozzarella actually)
• 4 cloves of garlic, rasped.

I tried doing the prep work more concurrently, and had at any given time, two frying pans going with one of the fry-em-up ingredients (ground beef, spinach, mushrooms, peppers, zukes).  A bit of liquid seems to help expedite the cooking times for some of the vegies.  The two batches of noodles were also prepared in these times.

In the end, combine the meat, peppers, zukes, mushrooms, and the sauce in a big pan, and mix em and cook em a bit all together.

This is also when I used the kitchen rasp on the garlic.

The layers can go in many ways.  Some of them end up with a very nice colour

Both of the Lasagnas were made in advance of the actual meals, so Ive got them both in the freezer right now, not yet tried.  Before covering and freezing them my two creations looked like:

375 F is the temp I have in my notes from last time, but I dont recall for how long.  A half hour with a few minutes on broil at the end will likely do the trick.  Ill be interested to see how these ones turn out (especially the gluten free variation).

Posted in reciepes | Tagged: , | 1 Comment »

PHY450H1S. Relativistic Electrodynamics Lecture 7 (Taught by Prof. Erich Poppitz). Action and relativistic dynamics

Posted by peeterjoot on January 26, 2011

Will now be covering chapter 2 material from the text [1].

Covering Professor Poppitz’s lecture notes: equation of motion, symmetries, and conserved quantities (energy-momentum 4 vector) from relativistic particle action [Wednesday, Jan. 26, Tuesday, Feb. 1]

These notes are also augmented by some additional notes completing an argument on page 53.

The relativity principle

The relativity principle implies that the EOM should be expressed in 4-vector form, just like Newton’s EOM are expressed in 3-vector form

\begin{aligned}m \ddot{\mathbf{r}} = \mathbf{f}\end{aligned} \hspace{\stretch{1}}(2.1)

Observe that in coordinate form this is

\begin{aligned}m \ddot{r}^i = f^i, \qquad i = 1,2,3\end{aligned} \hspace{\stretch{1}}(2.2)

or for a rotated frame $O'$

\begin{aligned}m \ddot{r'}^i = {f'}^i, \qquad i = 1,2,3\end{aligned} \hspace{\stretch{1}}(2.3)

Need to generalize to 4 vectors, so we need 4-velocity and 4-acceleration.

Later we will study action and Lagrangian, and then relativity will require that the action be a Lorentz scalar. The analogy for a Newtonian point particle is a scalar under rotations.

Four vector velocity

\paragraph{Definition:} Velocity s the rate of change of position in $(ct, \mathbf{x})$-space. Position means specifying both $ct$ and $\mathbf{x}$ for a point in spacetime.

PICTURE: $x^0 = ct$ axis up, and $x^1, x^2, x^3$ axis over, with worldline $x = x(\tau)$. Here $\tau$ is a parameter for the worldline, and provides a mapping for the curve in spacetime.

PICTURE: 3D vectors, $\mathbf{r}(t)$, $\mathbf{r}(t + \Delta t)$, and the difference vector $\mathbf{r}(t + \Delta t) - \mathbf{r}(t)$.

We write

\begin{aligned}\mathbf{v}(t) \equiv \lim_{\Delta t \rightarrow zero} \frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{ \Delta t}\end{aligned} \hspace{\stretch{1}}(2.4)

For four vectors we will parameterize the worldline by its “length”, with $O$ taken from some arbitrary point on it. We can also take $\tau$ to be the proper time, and the only difference will be the factor of $c$ (which becomes especially easy with the choice $c=1$ that is avoided in this class).

\begin{aligned}\frac{x^i(\tau + \Delta \tau) - x^i(\tau)}{\Delta \tau}\end{aligned} \hspace{\stretch{1}}(2.5)

We’ll take

\begin{aligned}\Delta \tau = ds = \frac{dx^i}{ds}\end{aligned} \hspace{\stretch{1}}(2.6)

This is a nice quantity, we are dividing a vector by a scalar, and thus get a four vector as a result (i.e. the result transforms as a four vector).

PICTURE: small fragment of a worldline with constant slope over the infinitesimal interval. $dx^0$ up and $dx^1$ over.

\begin{aligned}u^i \equiv \frac{dx^i}{ds}\end{aligned} \hspace{\stretch{1}}(2.7)

\begin{aligned}ds^2 &= (dx^0)^2 - (dx^1)^2 \\ &= c^2 \left( (dt)^2 - \frac{1}{{c^2}} (dx^1)^2 \right) \\ &= c^2 (dt)^2 \left( 1 - \frac{1}{{c^2}} \frac{dx^1}{dt^2} \right) \end{aligned}

Or

\begin{aligned}ds = c dt \sqrt{1 - \frac{1}{{c^2}} \frac{dx^1}{dt^2} }\end{aligned} \hspace{\stretch{1}}(2.8)

NOTE: Prof admits pulling a fast one, since he has aligned the worldline along the $x^1$ axis, however this is always possible by rotating the coordinate system.

\begin{aligned}u^0 &= \frac{dx^0}{ds} \\ &= \frac{c dt}{ c dt \sqrt{ 1 - \mathbf{v}^2/c^2} } \\ &= \frac{1}{ \sqrt{ 1 - \mathbf{v}^2/c^2} } \\ &= \gamma\end{aligned}

\begin{aligned}u^1 &= \frac{dx^1}{ds} \\ &= \frac{dx^1 }{ c dt \sqrt{ 1 - \mathbf{v}^2/c^2} } \\ &= \frac{v^1/c}{ \sqrt{ 1 - \mathbf{v}^2/c^2} } \\ &= \gamma \frac{v^1}{c}\end{aligned}

Similarily

\begin{aligned}u^2 &= \gamma \frac{v^2}{c} \\ u^3 &= \gamma \frac{v^2}{c}\end{aligned}

We’ve now unpacked the four velocity, and have

\begin{aligned}u^i = \left( \gamma, \frac{\mathbf{v}}{c} \gamma \right)\end{aligned} \hspace{\stretch{1}}(2.9)

Length of the four velocity vector

Recall that this length is

\begin{aligned}u^i g_{ij} u^j &= u^i u_i \\ &= u_i u^i \\ &= (u^0)^2 - (u_i)^2 \\ &= \gamma^2 - \gamma^2 \frac{\mathbf{v}}{c} \cdot \frac{\mathbf{v}}{c} \\ &= \gamma^2 \left(1 - \frac{\mathbf{v}^2}{c^2} \right)\end{aligned}

The four velocity in physics is

\begin{aligned}u^i = \left( \gamma, \frac{\mathbf{v}}{c} \gamma \right)\end{aligned} \hspace{\stretch{1}}(2.10)

but in mathematics the meaning of $u^i u_i = 1$ means that this quantity is the unit tangent vector to the worldline.

Four acceleration

In Newtonian physics we have

\begin{aligned}\mathbf{a} = \frac{\mathbf{v}}{dt}\end{aligned} \hspace{\stretch{1}}(2.11)

Our relativistic mapping of this, with $v \rightarrow u^i$ and $t \rightarrow s$, gives

\begin{aligned}w^i = \frac{d u^i}{ds}\end{aligned} \hspace{\stretch{1}}(2.12)

Geometrically $w^i$ is the normal to the worldline. This follows from $u^i g_{ij} u^j = 1$, so

\begin{aligned}\frac{d}{ds} \left( u^i g_{ij} u^j \right) &=\frac{d u^i}{ds} g_{ij} u^j +u^i g_{ij} \frac{d u^j}{ds} \\ &=\frac{d u^i}{ds} g_{ij} u^j +u^j \underbrace{g_{ji}}_{= g_{ij}} \frac{d u^i}{ds} \\ &=\frac{d u^i}{ds} g_{ij} u^j +u^j g_{ji} \frac{d u^i}{ds} \\ &=2 \frac{d u^i}{ds} g_{ij} u^j \end{aligned}

Note that we’ve utilized the fact above that the dummy summation indexes can be swapped (or changed to anything else we feel inclined to use).

The conclusion is that the dot product of the acceleration and the velocity is zero

\begin{aligned}w_i u^i = 0.\end{aligned} \hspace{\stretch{1}}(2.13)

Relativistic action.

\begin{aligned}S_{ab} = ?\end{aligned} \hspace{\stretch{1}}(3.14)

What is the action for a worldline from $a \rightarrow b$.

We want something that has velocity dependence ($u^i$ not $\mathbf{v}$), but that is Lorentz invariant and has only first derivatives.

The relativisitic length is the simplest so we could form

\begin{aligned}\int ds u^i u_i\end{aligned} \hspace{\stretch{1}}(3.15)

but that’s not interesting since $u^i u_i = 1$. We could form

\begin{aligned}\int ds u^i \frac{u_i}{ds} = \int ds w^i u_i\end{aligned} \hspace{\stretch{1}}(3.16)

but then this is just zero.

We could form something like

\begin{aligned}\int ds \frac{w^i}{ds} u_i\end{aligned} \hspace{\stretch{1}}(3.17)

This is non zero and non-constant, but evaluating the EOM for such an action would produce a result that has higher than second order derivatives.

We are left with

\begin{aligned}S_{ab} = \text{constant} \int_a^b ds \end{aligned} \hspace{\stretch{1}}(3.18)

To fix this constant we note that if we want to minimize the action over the infinitesimal interval, then we need a minus sign. Since the Lagrangian has dimensions of energy, and the dimensions of energy times time are momentum, our action must then have dimensions of momentum. So one possible constant that fixes up our dimensions is $mc$. Construct an action with the following form

\begin{aligned}S_{ab} = - m c\int_a^b ds,\end{aligned} \hspace{\stretch{1}}(3.19)

does the job we want. Here “m” is a characteristic of the particle, which \underline{is a Lorentz scalar}. It also happens to have dimensions of mass. With $ds = c dt \sqrt{1 - \mathbf{v}^2/c^2}$, we have

\begin{aligned}S_{ab} = - m c^2 \int_{t_a}^{t_b} dt \sqrt{ 1 - \frac{1}{{c^2}} \left( \frac{d \mathbf{x}(t) }{dt} \right)^2 }\end{aligned} \hspace{\stretch{1}}(3.20)

Now everything looks like it was in classical mechanics.

\begin{aligned}S_{ab} = \int_{t_a}^{t_b} \mathcal{L}(\dot{\mathbf{x}}(t)) dt\end{aligned} \hspace{\stretch{1}}(3.21)

\begin{aligned}\mathcal{L}(\dot{\mathbf{x}}(t)) = -m c^2 \end{aligned} \hspace{\stretch{1}}(3.22)

Now find the extremum of $S$. That problem is really to compute the variation in the action that results from varying the coordinates around the stationary point, and equate that variation to zero to find the extremum

\begin{aligned}\delta S = S[\mathbf{x}(t) + \delta \mathbf{x}(t)] - S[ \mathbf{x}(t) ] = 0\end{aligned} \hspace{\stretch{1}}(3.23)

The usual condition is imposed where we have zero variation of the coordinates at the boundies of the action integral

\begin{aligned}0 = \delta \mathbf{x}(t_a) = \delta \mathbf{x}(t_b) \end{aligned} \hspace{\stretch{1}}(3.24)

Returning to our action we have

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{\mathbf{x}}}} = \frac{\partial {\mathcal{L}}}{\partial {\mathbf{x}}} = 0\end{aligned} \hspace{\stretch{1}}(3.25)

This last is zero because it’s a free particle with no position dependence.

\begin{aligned} 0 &= -m c^2 \frac{d}{dt} \frac{\partial}{\partial \dot{\mathbf{x}}} \sqrt{ 1 - \dot{\mathbf{x}}^2 } \\ &= -m c^2 \frac{d}{dt} \frac{- \dot{\mathbf{x}}}{\sqrt{ 1 - \dot{\mathbf{x}}^2 } } \\ &= m c^2 \frac{d}{dt} \gamma \dot\mathbf{x} \end{aligned}

So we have

\begin{aligned}\frac{d}{dt} (\gamma \dot{\mathbf{x}}) = 0\end{aligned} \hspace{\stretch{1}}(3.26)

By evaluating this, we can eventually show that we can construct a four vector equation. Doing this we have

\begin{aligned}\frac{d}{dt} (\gamma \mathbf{v}) &=\frac{d}{dt} \left( \left(1 - \mathbf{v}^2/c^2\right)^{-1/2} \mathbf{v} \right) \\ &=-2 (-1/2) \mathbf{v} (\mathbf{v} \cdot \dot{\mathbf{v}})/c^2 \left(1 - \mathbf{v}^2/c^2\right)^{-3/2} + \left(1 - \mathbf{v}^2/c^2\right)^{-1/2} \dot{\mathbf{v}} \\ &=\gamma \left( \frac{\mathbf{v} (\mathbf{v} \cdot \dot{\mathbf{v}}) }{ c^2 - \mathbf{v}^2 } + \dot{\mathbf{v}} \right)\end{aligned}

Or

\begin{aligned}\frac{\mathbf{v} (\mathbf{v} \cdot \dot{\mathbf{v}}) }{ c^2 - \mathbf{v}^2 } + \dot{\mathbf{v}} = 0\end{aligned} \hspace{\stretch{1}}(3.27)

Clearly $\dot{\mathbf{v}} = 0$ is a solution, but is it the only solution?

By dotting this with $\mathbf{v}$ we have

\begin{aligned}0 &= \frac{\mathbf{v}^2 (\mathbf{v} \cdot \dot{\mathbf{v}}) }{ c^2 - \mathbf{v}^2 } + \dot{\mathbf{v}} \cdot \mathbf{v} \\ &= (\mathbf{v} \cdot \dot{\mathbf{v}}) \left( 1 + \frac{\mathbf{v}^2}{c^2 - \mathbf{v}^2} \right) \\ &= (\mathbf{v} \cdot \dot{\mathbf{v}}) \frac{c^2}{c^2 - \mathbf{v}^2} \end{aligned}

This implies that $\dot{\mathbf{v}} = 0$ (a contraction) or that $\mathbf{v} \cdot \dot{\mathbf{v}} = 0$. To examine the perpendicularity question, let’s take cross products. This gives

\begin{aligned}0 =\frac{(\mathbf{v} \times \mathbf{v}) (\mathbf{v} \cdot \dot{\mathbf{v}}) }{ c^2 - \mathbf{v}^2 } + \dot{\mathbf{v}} \times \mathbf{v}\end{aligned} \hspace{\stretch{1}}(3.28)

We have found that $\mathbf{v} \cdot \dot{\mathbf{v}} = 0$ and $\mathbf{v} \times \dot{\mathbf{v}} = 0$. This can only mean that $\dot{\mathbf{v}} = 0$, contradicting the assumption that is non-zero. We conclude that $\dot{\mathbf{v}} = 0$ is the only solution to 3.27.

Next time

We want to finish up and show how this results in a four velocity equation. We have

\begin{aligned}\frac{d}{dt} ( \gamma \mathbf{v}) = 0\end{aligned} \hspace{\stretch{1}}(4.29)

which is

\begin{aligned}\frac{d}{dt} ( u^\alpha ) = 0, \qquad \text{for} u^\alpha = u^1, u^2, u^3\end{aligned} \hspace{\stretch{1}}(4.30)

eventually, we will show that we also have

\begin{aligned}\frac{d}{dt} ( u^i ) = 0\end{aligned} \hspace{\stretch{1}}(4.31)

References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

PHY450H1S. Relativistic Electrodynamics Lecture 6 (Taught by Prof. Erich Poppitz). Four vectors and tensors.

Posted by peeterjoot on January 25, 2011

Still covering chapter 1 material from the text [1].

Covering Professor Poppitz’s lecture notes: nonrelativistic limit of boosts (33); number of parameters of Lorentz transformations (34-35); introducing four-vectors, the metric tensor, the invariant “dot-product and SO(1,3) (36-40); the Poincare group (41); the convenience of “upper” and “lower” indices (42-43); tensors (44)

The Special Orthogonal group (for Euclidean space).

Lorentz transformations are like “rotations” for $(t, x, y, z)$ that preserve $(ct)^2 - x^2 - y^2 - z^2$. There are 6 continuous parameters:

\begin{itemize}
\item 3 rotations in $x,y,z$ space
\item 3 “boosts” in $x$ or $y$ or $z$.
\end{itemize}

For rotations of space we talk about a group of transformations of 3D Euclidean space, and call this the $S0(3)$ group. Here $S$ is for Special, $O$ for Orthogonal, and $3$ for the dimensions.

For a transformed vector in 3D space we write

\begin{aligned}\begin{bmatrix}x \\ y \\ z\end{bmatrix} \rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix}' = O \begin{bmatrix}x \\ y \\ z\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.1)

Here $O$ is an orthogonal $3 \times 3$ matrix, and has the property

\begin{aligned}O^T O = \mathbf{1}.\end{aligned} \hspace{\stretch{1}}(2.2)

Taking determinants, we have

\begin{aligned}\det{ O^T } \det{ O} = 1,\end{aligned} \hspace{\stretch{1}}(2.3)

and since $\det{O^\text{T}} = \det{ O }$, we have

\begin{aligned}(\det{O})^2 = 1,\end{aligned} \hspace{\stretch{1}}(2.4)

so our determinant must be

\begin{aligned}\det O = \pm 1.\end{aligned} \hspace{\stretch{1}}(2.5)

We work with the positive case only, avoiding the transformations that include reflections.

The Unitary condition $O^\text{T} O = 1$ is an indication that the inner product is preserved. Observe that in matrix form we can write the inner product

\begin{aligned}\mathbf{r}_1 \cdot \mathbf{r}_2 = \begin{bmatrix}x_1 & y_1 & z_1\end{bmatrix}\begin{bmatrix}x_1 \\ y_2 \\ x_3 \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.6)

For a transformed vector $X' = O X$, we have ${X'}^\text{T} = X^\text{T} O^\text{T}$, and

\begin{aligned}X' \cdot X' = (X^\text{T} O^\text{T}) (O X) = X^\text{T} (O^\text{T} O) X = X^T X = X \cdot X\end{aligned} \hspace{\stretch{1}}(2.7)

The Special Orthogonal group (for spacetime).

This generalizes to Lorentz boosts! There are two differences

\begin{enumerate}
\item Lorentz transforms should be $4 \times 4$ not $3 \times 3$ and act in $(ct, x, y, z)$, and NOT $(x,y,z)$.
\item They should leave invariant NOT $\mathbf{r}_1 \cdot \mathbf{r}_2$, but $c2 t_2 t_1 - \mathbf{r}_2 \cdot \mathbf{r}_1$.
\end{enumerate}

Don’t get confused that I demanded $c^2 t_2 t_1 - \mathbf{r}_2 \cdot \mathbf{r}_1 = \text{invariant}$ rather than $c^2 (t_2 - t_1)^2 - (\mathbf{r}_2 - \mathbf{r}_1)^2 = \text{invariant}$. Expansion of this (squared) interval, provides just this four vector dot product and its invariance condition

\begin{aligned}\text{invariant} &=c^2 (t_2 - t_1)^2 - (\mathbf{r}_2 - \mathbf{r}_1)^2 \\ &=(c^2 t_2^2 - \mathbf{r}_2^2) + (c^2 t_2^2 - \mathbf{r}_2^2)- 2 c^2 t_2 t_1 + 2 \mathbf{r}_1 \cdot \mathbf{r}_2.\end{aligned}

Observe that we have the sum of two invariants plus our new cross term, so this cross term, (-2 times our dot product to be defined), must also be an invariant.

Introduce the four vector

\begin{aligned}x^0 &= ct \\ x^1 &= x \\ x^2 &= y \\ x^3 &= z \end{aligned}

Or $(x^0, x^1, x^2, x^3) = \{ x^i, i = 0,1,2,3 \}$.

We will also write

\begin{aligned}x^i &= (ct, \mathbf{r}) \\ \tilde{x}^i &= (c\tilde{t}, \tilde{\mathbf{r}})\end{aligned}

Our inner product is

\begin{aligned}c^2 t \tilde{t} - \mathbf{r} \cdot \tilde{\mathbf{r}}\end{aligned} \hspace{\stretch{1}}(3.8)

Introduce the $4 \times 4$ matrix

\begin{aligned} \left\lVert{g_{ij}}\right\rVert = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.9)

This is called the Minkowski spacetime metric.

Then

\begin{aligned}c^2 t \tilde{t} - \mathbf{r} \cdot \tilde{\mathbf{r}}&\equiv \sum_{i, j = 0}^3 \tilde{x}^i g_{ij} x^j \\ &= \sum_{i, j = 0}^3 \tilde{x}^i g_{ij} x^j \\ & \tilde{x}^0 x^0 -\tilde{x}^1 x^1 -\tilde{x}^2 x^2 -\tilde{x}^3 x^3 \end{aligned}

\paragraph{Einstein summation convention}. Whenever indexes are repeated that are assumed to be summed over.

We also write

\begin{aligned}X = \begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.10)

\begin{aligned}\tilde{X} = \begin{bmatrix}\tilde{x}^0 \\ \tilde{x}^1 \\ \tilde{x}^2 \\ \tilde{x}^3 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.11)

\begin{aligned}G = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.12)

Our inner product

\begin{aligned}c^2 t \tilde{t} - \tilde{\mathbf{r}} \cdot \mathbf{r} = \tilde{X}^\text{T} G X &=\begin{bmatrix}\tilde{x}^0 & \tilde{x}^1 & \tilde{x}^2 & \tilde{x}^3 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}\begin{bmatrix}\tilde{x}^0 \\ \tilde{x}^1 \\ \tilde{x}^2 \\ \tilde{x}^3 \\ \end{bmatrix}\end{aligned}

Under Lorentz boosts, we have

\begin{aligned}X = \hat{O} X',\end{aligned} \hspace{\stretch{1}}(3.13)

where

\begin{aligned}\hat{O} =\begin{bmatrix}\gamma & - \gamma v_x/c & 0 & 0 \\ - \gamma v_x/c & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.14)

(for x-direction boosts)

$\tilde{X} = \hat{O} \tilde{X}' \tilde{X}^\text{T} = \tilde{X'}^\text{T} \hat{O}^\text{T} \hspace{\stretch{1}}(3.15)$

But $\hat{O}$ must be such that $\tilde{X}^\text{T} G X$ is invariant. i.e.

\begin{aligned}\tilde{X} G X = {\tilde{X'}}^\text{T} (\hat{O}^\text{T} G \hat{O}) X' = {X'}^\text{T} (G) X' \qquad \forall X' \text{and} \tilde{X}' \end{aligned} \hspace{\stretch{1}}(3.16)

This implies

\begin{aligned}\boxed{\hat{O}^\text{T} G \hat{O} = G}\end{aligned} \hspace{\stretch{1}}(3.17)

Such $\hat{O}$‘s are called “pseudo-orthogonal”.

Lorentz transformations are represented by the set of all $4 \times 4$ pseudo-orthogonal matrices.

In symbols

\begin{aligned}\hat{O}^T G \hat{O} = G\end{aligned} \hspace{\stretch{1}}(3.18)

Just as before we can take the determinant of both sides. Doing so we have

\begin{aligned}\det(\hat{O}^T G \hat{O}) = \det(\hat{O}^T) \det(G) \det(\hat{O}) = \det(G)\end{aligned} \hspace{\stretch{1}}(3.19)

The $\det(G)$ terms cancel, and since $\det(\hat{O}^T) = \det(\hat{O})$, this leaves us with $(\det(\hat{O}))^2 = 1$, or

\begin{aligned}\det(\hat{O}) = \pm 1\end{aligned} \hspace{\stretch{1}}(3.20)

We take the $\det 0 = +1$ case only, so that the transformations do not change orientation (no reflection in space or time). This set of transformation forms the group

\begin{aligned}SO(1,3)\end{aligned}

Special orthogonal, one time, 3 space dimensions.

Einstein relativity can be defined as the “laws of physics that leave four vectors invariant in the

\begin{aligned}SO(1,3) \times T^4\end{aligned}

symmetry group.

Here $T^4$ is the group of translations in spacetime with 4 continuous parameters. The complete group of transformations that form the group of relativistic physics has $10 = 3 + 3 + 4$ continuous parameters.

This group is called the Poincare group of symmetry transforms.

More notation

Our inner product is written

\begin{aligned}\tilde{x}^i g_{ij} x^j\end{aligned} \hspace{\stretch{1}}(4.21)

but this is very cumbersome. The convenient way to write this is instead

\begin{aligned}\tilde{x}^i g_{ij} x^j = \tilde{x}_j x^j = \tilde{x}^i x_i\end{aligned} \hspace{\stretch{1}}(4.22)

where

\begin{aligned}x_i = g_{ij} x^j = g_{ji} x^j\end{aligned} \hspace{\stretch{1}}(4.23)

Note: A check that we should always be able to make. Indexes that are not summed over should be conserved. So in the above we have a free $i$ on the LHS, and should have a non-summed $i$ index on the RHS too (also lower matching lower, or upper matching upper).

Non-matched indexes are bad in the same sort of sense that an expression like

\begin{aligned}\mathbf{r} = 1\end{aligned} \hspace{\stretch{1}}(4.24)

isn’t well defined (assuming a vector space $\mathbf{r}$ and not a multivector Clifford algebra that is;)

Example explicitly:

\begin{aligned}x_0 &= g_{0 0} x^0 = ct \\ x_1 &= g_{1 j} x^j = g_{11} x^1 = -x^1 \\ x_2 &= g_{2 j} x^j = g_{22} x^2 = -x^2 \\ x_3 &= g_{3 j} x^j = g_{33} x^3 = -x^3\end{aligned}

We would not have objects of the form

\begin{aligned}x^i x^i = (ct)^2 + \mathbf{r}^2\end{aligned} \hspace{\stretch{1}}(4.25)

for example. This is not a Lorentz invariant quantity.

\paragraph{Lorentz scalar example:} $\tilde{x}^i x_i$
\paragraph{Lorentz vector example:} $x^i$

This last is also called a rank-1 tensor.

Lorentz rank-2 tensors: ex: $g_{ij}$

or other 2-index objects.

Why in the world would we ever want to consider two index objects. We aren’t just trying to be hard on ourselves. Recall from classical mechanics that we have a two index object, the inertial tensor.

In mechanics, for a rigid body we had the energy

\begin{aligned}T = \sum_{ij = 1}^3 \Omega_i I_{ij} \Omega_j\end{aligned} \hspace{\stretch{1}}(4.26)

\begin{aligned}I_{ij} = \sum_{a = 1}^N m_a \left(\delta_{ij} \mathbf{r}_a^2 - r_{a_i} r_{a_j} \right)\end{aligned} \hspace{\stretch{1}}(4.27)

or for a continuous body

\begin{aligned}I_{ij} = \int \rho(\mathbf{r}) \left(\delta_{ij} \mathbf{r}^2 - r_{i} r_{j} \right)\end{aligned} \hspace{\stretch{1}}(4.28)

In electrostatics we have the quadrupole tensor, … and we have other such objects all over physics.

Note that the energy $T$ of the body above cannot depend on the coordinate system in use. This is a general property of tensors. These are object that transform as products of vectors, as $I_{ij}$ does.

We call $I_{ij}$ a rank-2 3-tensor. rank-2 because there are two indexes, and 3 because the indexes range from $1$ to $3$.

The point is that tensors have the property that the transformed tensors transform as

\begin{aligned}I_{ij}' = \sum_{l, m = 1,2,3} O_{il} O_{jm} I_{lm}\end{aligned} \hspace{\stretch{1}}(4.29)

Another example: the completely antisymmetric rank 3, 3-tensor

\begin{aligned}\epsilon_{ijk}\end{aligned} \hspace{\stretch{1}}(4.30)

Dynamics

In Newtonian dynamics we have

\begin{aligned}m \dot{d}{\mathbf{r}} = \mathbf{f}\end{aligned} \hspace{\stretch{1}}(5.31)

An equation of motion should be expressed in terms of vectors. This equation is written in a way that shows that the law of physics is independent of the choice of coordinates. We can do this in the context of tensor algebra as well. Ironically, this will require us to explicitly work with the coordinate representation, but this work will be augmented by the fact that we require our tensors to transform in specific ways.

In Newtonian mechanics we can look to symmetries and the invariance of the action with respect to those symmetries to express the equations of motion. Our symmetries in Newtonian mechanics leave the action invariant with respect to spatial translation and with respect to rotation.

We want to express relativistic dynamics in a similar way, and will have to express the action as a Lorentz scalar. We are going to impose the symmetries of the Poincare group to determine the relativistic laws of dynamics, and the next task will be to consider the possibilities for our relativistic action, and see what that action implies for dynamics in a relativistic context.

References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

Use of __LINE__ in perl.

Posted by peeterjoot on January 24, 2011

I’d wondered a couple times how to do this, and had littered scripts occasionally with various manually created unique text markers for debugging purposes. Here’s an easier way:

print " Line: ", __LINE__, "\n";
`

(works for __FILE__ too).

PHY450H1S. Relativistic Electrodynamics Tutorial 1 (TA: Simon Freedman).

Posted by peeterjoot on January 21, 2011

Worked question.

The TA blasted through a problem from Hartle [1], section 5.17 (all the while apologizing for going so slow). I’m going to have to look these notes over carefully to figure out what on Earth he was doing.

At one point he asked if anybody was completely lost. Nobody said yes, but given the class title, I had the urge to say “No, just relatively lost”.

\paragraph{Q:}
In a source’s rest frame $S$ emits radiation isotropically with a frequency $\omega$ with number flux $f(\text{photons}/\text{cm}^2 s)$. Moves along x’-axis with speed $V$ in an observer frame ($O$). What does the energy flux in $O$ look like?

A brief intro with four vectors

A 3-vector:

\begin{aligned}\mathbf{a} &= (a_x, a_y, a_z) = (a^1, a^2, a^3) \\ \mathbf{b} &= (b_x, b_y, b_z) = (b^1, b^2, b^3)\end{aligned} \hspace{\stretch{1}}(1.1)

For this we have the dot product

\begin{aligned}\mathbf{a} \cdot \mathbf{b} = \sum_{\alpha=1}^3 a^\alpha b^\alpha\end{aligned} \hspace{\stretch{1}}(1.3)

Greek letters in this course (opposite to everybody else in the world, because of Landau and Lifshitz) run from 1 to 3, whereas roman letters run through the set $\{0,1,2,3\}$.

We want to put space and time on an equal footing and form the composite quantity (four vector)

\begin{aligned}x^i = (ct, \mathbf{r}) = (x^0, x^1, x^2, x^3),\end{aligned} \hspace{\stretch{1}}(1.4)

where

\begin{aligned}x^0 &= ct \\ x^1 &= x \\ x^2 &= y \\ x^3 &= z.\end{aligned} \hspace{\stretch{1}}(1.5)

It will also be convenient to drop indexes when referring to all the components of a four vector and we will use lower or upper case non-bold letters to represent such four vectors. For example

\begin{aligned}X = (ct, \mathbf{r}),\end{aligned} \hspace{\stretch{1}}(1.9)

or

\begin{aligned}v = \gamma \left(c, \mathbf{v} \right).\end{aligned} \hspace{\stretch{1}}(1.10)

Three vectors will be represented as letters with over arrows $\vec{a}$ or (in text) bold face $\mathbf{a}$.

Recall that the squared spacetime interval between two events $X_1$ and $X_2$ is defined as

\begin{aligned}{S_{X_1, X_2}}^2 = (ct_1 - c t_2)^2 - (\mathbf{x}_1 - \mathbf{x}_2)^2.\end{aligned} \hspace{\stretch{1}}(1.11)

In particular, with one of these zero, we have an operator which takes a single four vector and spits out a scalar, measuring a “distance” from the origin

\begin{aligned}s^2 = (ct)^2 - \mathbf{r}^2.\end{aligned} \hspace{\stretch{1}}(1.12)

This motivates the introduction of a dot product for our four vector space.

\begin{aligned}X \cdot X = (ct)^2 - \mathbf{r}^2 = (x^0)^2 - \sum_{\alpha=1}^3 (x^\alpha)^2\end{aligned} \hspace{\stretch{1}}(1.13)

Utilizing the spacetime dot product of 1.13 we have for the dot product of the difference between two events

\begin{aligned}(X - Y) \cdot (X - Y)&=(x^0 - y^0)^2 - \sum_{\alpha =1}^3 (x^\alpha - y^\alpha)^2 \\ &=X \cdot X + Y \cdot Y - 2 x^0 y^0 + 2 \sum_{\alpha =1}^3 x^\alpha y^\alpha.\end{aligned}

From this, assuming our dot product 1.13 is both linear and symmetric, we have for any pair of spacetime events

\begin{aligned}X \cdot Y = x^0 y^0 - \sum_{\alpha =1}^3 x^\alpha y^\alpha.\end{aligned} \hspace{\stretch{1}}(1.14)

How do our four vectors transform? This is really just a notational issue, since this has already been discussed. In this new notation we have

\begin{aligned}{x^0}' &= ct' = \gamma ( ct - \beta x) = \gamma ( x^0 - \beta x^1 ) \\ {x^1}' &= x' = \gamma ( x - \beta ct ) = \gamma ( x^1 - \beta x^0 ) \\ {x^2}' &= x^2 \\ {x^3}' &= x^3\end{aligned} \hspace{\stretch{1}}(1.15)

where $\beta = V/c$, and $\gamma^{-2} = 1 - \beta^2$.

In order to put some structure to this, it can be helpful to express this dot product as a quadratic form. We write

\begin{aligned}A \cdot B = \begin{bmatrix}a^0 & \mathbf{a}^\text{T} \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}\begin{bmatrix}b^0 \\ \mathbf{b}\end{bmatrix}= A^\text{T} G B.\end{aligned} \hspace{\stretch{1}}(1.19)

We can write our Lorentz boost as a matrix

\begin{aligned}\begin{bmatrix}\gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.20)

so that the dot product between two transformed four vectors takes the form

\begin{aligned}A' \cdot B' = A^\text{T} O^\text{T} G O B\end{aligned} \hspace{\stretch{1}}(1.21)

Back to the problem.

We will work in momentum space, where we have

\begin{aligned}p^i &= (p^0, \mathbf{p}) = \left( \frac{E}{c}, \mathbf{p}\right) \\ p^2 &= \frac{E^2}{c^2} -\mathbf{p}^2 \\ \mathbf{p} &= \hbar \mathbf{k} \\ E &= \hbar \omega \\ p^i &= \hbar k^i \\ k^i &= \left(\frac{\omega}{c}, \mathbf{k}\right)\end{aligned} \hspace{\stretch{1}}(1.22)

Justifying this.

Now, the TA blurted all this out. We know some of it from the QM context, and if we’ve been reading ahead know a bit of this from our text [2] (the energy momentum four vector relationships). Let’s go back to the classical electromagnetism and recall what we know about the relation of frequency and wave numbers for continuous fields. We want solutions to Maxwell’s equation in vacuum and can show that such solution also implies that our fields obey a wave equation

\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 \Psi}{\partial t^2} - \boldsymbol{\nabla}^2 \Psi = 0,\end{aligned} \hspace{\stretch{1}}(1.28)

where $\Psi$ is one of $\mathbf{E}$ or $\mathbf{B}$. We have other constraints imposed on the solutions by Maxwell’s equations, but require that they at least obey 1.28 in addition to these constraints.

With application of a spatial Fourier transformation of the wave equation, we find that our solution takes the form

\begin{aligned}\Psi = (2 \pi)^{-3/2} \int \tilde{\Psi}(\mathbf{k}, 0) e^{i (\omega t \pm \mathbf{k} \cdot \mathbf{x}) } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(1.29)

If one takes this as a given and applies the wave equation operator to this as a test solution, one finds without doing the Fourier transform work that we also have a constraint. That is

\begin{aligned}\frac{1}{{c^2}} (i \omega)^2 \Psi - (\pm i \mathbf{k})^2 \Psi = 0.\end{aligned} \hspace{\stretch{1}}(1.30)

So even in the continuous field domain, we have a relationship between frequency and wave number. We see that this also happens to have the form of a lightlike spacetime interval

\begin{aligned}\frac{\omega^2}{c^2} - \mathbf{k}^2 = 0.\end{aligned} \hspace{\stretch{1}}(1.31)

Also recall that the photoelectric effect imposes an experimental constraint on photon energy, where we have

\begin{aligned}E = h \nu = \frac{h}{2\pi} 2 \pi \nu = \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.32)

Therefore if we impose a mechanics like $P = (E/c, \mathbf{p})$ relativistic energy-momentum relationship on light, it then makes sense to form a nilpotent (lightlike) four vector for our photon energy. This combines our special relativistic expectations, with the constraints on the fields imposed by classical electromagnetism. We can then write for the photon four momentum

\begin{aligned}P = \left( \frac{\hbar \omega}{c}, \hbar k \right)\end{aligned} \hspace{\stretch{1}}(1.33)

Back to the TA’s formula blitz.

Utilizing spherical polar coordinates in momentum (wave number) space, measuring the polar angle from the $k^1$ (x-like) axis, we can compute this polar angle in both pairs of frames,

\begin{aligned} \cos \alpha &= \frac{k^1}{{\left\lvert{\mathbf{k}}\right\rvert}} = \frac{k^1}{\omega/c} \\ \cos \alpha' &= \frac{{k^1}'}{\omega'/c} = \frac{\gamma (k^1 + \beta \omega/c)}{\gamma(\omega/c + \beta k^1)}\end{aligned} \hspace{\stretch{1}}(1.34)

Note that this requires us to assume that wave number four vectors transform in the same fashion as regular mechanical position and momentum four vectors. Also note that we have the primed frame moving negatively along the x-axis, instead of the usual positive origin shift. The question is vague enough to allow this since it only requires motion.

\paragraph{check 1}

as $\beta \rightarrow 1$ (ie: our primed frame velocity approaches the speed of light relative to the rest frame), $\cos \alpha' \rightarrow 1$, $\alpha' = 0$. The surface gets more and more compressed.

In the original reference frame the radiation was isotropic. In the new frame how does it change with respect to the angle? This is really a question to find this number flux rate

\begin{aligned}f'(\alpha') = ?\end{aligned} \hspace{\stretch{1}}(1.36)

In our rest frame the total number of photons traveling through the surface in a given interval of time is

\begin{aligned}N &= \int d\Omega dt f(\alpha) = \int d \phi \sin \alpha d\alpha = -2 \pi \int d(\cos\alpha) dt f(\alpha) \\ \end{aligned} \hspace{\stretch{1}}(1.37)

Here we utilize the spherical solid angle $d\Omega = \sin \alpha d\alpha d\phi = - d(\cos\alpha) d\phi$, and integrate $\phi$ over the $[0, 2\pi]$ interval. We also have to assume that our number flux density is not a function of horizontal angle $\phi$ in the rest frame.

In the moving frame we similarly have

\begin{aligned}N' &= -2 \pi \int d(\cos\alpha') dt' f'(\alpha'),\end{aligned} \hspace{\stretch{1}}(1.39)

and we again have had to assume that our transformed number flux density is not a function of the horizontal angle $\phi$. This seems like a reasonable move since ${k^2}' = k^2$ and ${k^3}' = k^3$ as they are perpendicular to the boost direction.

\begin{aligned}f'(\alpha') = \frac{d(\cos\alpha)}{d(\cos\alpha')} \left( \frac{dt}{dt'} \right) f(\alpha)\end{aligned} \hspace{\stretch{1}}(1.40)

Now, utilizing a conservation of mass argument, we can argue that $N = N'$. Regardless of the motion of the frame, the same number of particles move through the surface. Taking ratios, and examining an infinitesimal time interval, and the associated flux through a small patch, we have

\begin{aligned}\left( \frac{d(\cos\alpha)}{d(\cos\alpha')} \right) = \left( \frac{d(\cos\alpha')}{d(\cos\alpha)} \right)^{-1} = \gamma^2 ( 1 + \beta \cos\alpha)^2\end{aligned} \hspace{\stretch{1}}(1.41)

Part of the statement above was a do-it-yourself. First recall that $c t' = \gamma ( c t + \beta x )$, so $dt/dt'$ evaluated at $x=0$ is $1/\gamma$.

The rest is messier. We can calculate the $d(\cos)$ values in the ratio above using 1.34. For example, for $d(\cos(\alpha))$ we have

\begin{aligned}d(\cos\alpha) &= d \left( \frac{k^1}{\omega/c} \right) \\ &= dk^1 \frac{1}{{\omega/c}} - c \frac{1}{{\omega^2}} d\omega.\end{aligned}

If one does the same thing for $d(\cos\alpha')$, after a whole whack of messy algebra one finds that the differential terms and a whole lot more mystically cancels, leaving just

\begin{aligned}\frac{d\cos\alpha'}{d\cos\alpha} = \frac{\omega^2/c^2}{(\omega/c + \beta k^1)^2} (1 - \beta^2)\end{aligned} \hspace{\stretch{1}}(1.42)

A bit more reduction with reference back to 1.34 verifies 1.41.

Also note that again from 1.34 we have

\begin{aligned}\cos\alpha' = \frac{\cos\alpha + \beta}{1 + \beta \cos\alpha}\end{aligned} \hspace{\stretch{1}}(1.43)

and rearranging this for $\cos\alpha'$ gives us

\begin{aligned}\cos\alpha = \frac{\cos\alpha' - \beta}{1 - \beta \cos\alpha'},\end{aligned} \hspace{\stretch{1}}(1.44)

which we can sum to find that

\begin{aligned}1 + \beta \cos\alpha = \frac{1}{{\gamma^2 (1 - \beta \cos \alpha')^2 }},\end{aligned} \hspace{\stretch{1}}(1.45)

so putting all the pieces together we have

\begin{aligned}f'(\alpha') = \frac{1}{{\gamma}} \frac{f(\alpha)}{(\gamma (1-\beta \cos\alpha'))^2}\end{aligned} \hspace{\stretch{1}}(1.46)

The question asks for the energy flux density. We get this by multiplying the number density by the frequency of the light in question. This is, as a function of the polar angle, in each of the frames.

\begin{aligned}L(\alpha) &= \hbar \omega(\alpha) f(\alpha) = \hbar \omega f \\ L'(\alpha') &= \hbar \omega'(\alpha') f'(\alpha') = \hbar \omega' f'\end{aligned} \hspace{\stretch{1}}(1.47)

But we have

\begin{aligned}\omega'(\alpha')/c = \gamma( \omega/c + \beta k^1 ) = \gamma \omega/c ( 1 + \beta \cos\alpha )\end{aligned} \hspace{\stretch{1}}(1.49)

Aside, $\beta << 1$,

\begin{aligned}\omega' = \omega ( 1 + \beta \cos\alpha) + O(\beta^2) = \omega + \delta \omega\end{aligned} \hspace{\stretch{1}}(1.50)

\begin{aligned}\delta \omega &= \beta, \alpha = 0 \qquad \text{blue shift} \\ \delta \omega &= -\beta, \alpha = \pi \qquad \text{red shift}\end{aligned} \hspace{\stretch{1}}(1.51)

The TA then writes

\begin{aligned}L'(\alpha') = \frac{L/\gamma}{(\gamma (1 - \beta \cos\alpha'))^3}\end{aligned} \hspace{\stretch{1}}(1.53)

although, I calculate

\begin{aligned}L'(\alpha') = \frac{L}{\gamma^4 (\gamma (1 - \beta \cos\alpha'))^4}\end{aligned} \hspace{\stretch{1}}(1.54)

He then says, the forward backward ratio is

\begin{aligned}L'(0)/L'(\pi) = {\left( \frac{ 1 + \beta }{1-\beta} \right)}^3\end{aligned} \hspace{\stretch{1}}(1.55)

For this I get:

\begin{aligned}L'(0)/L'(\pi) = {\left( \frac{ 1 + \beta }{1-\beta} \right)}^4\end{aligned} \hspace{\stretch{1}}(1.56)

It is still bigger for $\beta$ positive, which I think is the point.

If I can somehow manage to keep my signs right as I do this course I may survive. Why did he pick a positive sign way back in 1.34?

References

[1] J.B. Hartle and T. Dray. Gravity: an introduction to Einsteins general relativity, volume 71. 2003.

[2] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

PHY450H1S. Relativistic Electrodynamics Lecture 5 (Taught by Prof. Erich Poppitz). Spacetime, events, worldlines, spacetime intervals, and invariance.

Posted by peeterjoot on January 19, 2011

Still covering chapter 1 material from the text [1]?

Covering Professor Poppitz’s lecture notes: Using Minkowski diagram to see the perils of superluminal propagation (32.3); nonrelativistic limit of boosts (33); number of parameters of Lorentz transformations (34-35); introducing four-vectors, the metric tensor, the invariant “dot-product and SO(1,3) (36-40); the Poincare group (41); the convenience of “upper” and “lower”indices (42-43); tensors (44)

More on proper time.

PICTURE:1: worldline with small interval.

Considering a small interval somewhere on the worldline trajectory, we have

\begin{aligned}ds^2 = c^2 dt^2 - dx^2 = c^2 {dt'}^2,\end{aligned} \hspace{\stretch{1}}(2.1)

where $dt'$ is the proper time elapsed in a frame moving with velocity $v$, and $dt$ is the time elapsed in a stationary frame.

We have

\begin{aligned}dt' = dt \sqrt{ 1 - (dx/dt)^2/c^2 } = dt \sqrt{ 1 - v^2/c^2 }.\end{aligned} \hspace{\stretch{1}}(2.2)

PICTURE:2: particle at rest.

For the particle at rest

\begin{aligned}c \tau_{21}^\text{stationary} = c ( t_2 - t_1 ) = \int_1^2 ds = \int_1^2 c dt\end{aligned} \hspace{\stretch{1}}(2.3)

PICTURE:3: particle with motion.

“length” of 1-2 “curved” worldline

\begin{aligned}\int_1^2 ds' &= \int_1^2 c dt' \\ &= \int_1^2 c dt \sqrt{ 1 - (d\mathbf{v}/dt)^2 },\end{aligned}

where in this case $[1,2]$ denotes the range of a line integral over the worldline. We see that the multiplier of dt for any point along the curve is smaller than $1$, so that the length along a straight line is longest (i.e. for the particle at rest).

We’ve argued that if 1,2 occur at the same place, the spacetime length of a straight line between them is the longest. This remains the time for all 1,2 timelike separated.

LOTS OF DISCUSSION. See new posted notes for details.

Back to page 18 of the notes.

We’ve argued that $ds_{12} = {ds'}_{12} \implies s_{12} = {s'}_{12}$ for infinitesimal 1,2 even if not infinitesimal.

The idea is to represent the interval between two not close 1,2 as a sum over small $ds$‘s.

P6: $x = x_2 t /t_2$ straight line through origin, with $t \in [0, t_2]$.

P7: zoomed on part of this line.

\begin{aligned}ds^2 &= c^2 dt^2 - dx^2 \\ &= c^2 dt^2 - \left(\frac{x_2}{t_2}\right)^2 dt^2 \\ &= c^2 dt^2 \left( 1 - \frac{1}{{c^2}} \left(\frac{x_2}{t_2}\right)^2 \right) \\ \end{aligned}

or

\begin{aligned}\int_0^1 ds = c \int_0^{t_2} dt \sqrt{ 1 - \frac{1}{{c^2}} \left(\frac{x_2}{t_2}\right)^2 } \end{aligned} \hspace{\stretch{1}}(2.4)

In another frame just replace $t \rightarrow t'$ and $x_2 \rightarrow x_2'$

\begin{aligned}\int_0^1 ds = c \int_0^{t_2'} dt \sqrt{1 - \frac{1}{{c^2}} \left(\frac{x_2'}{t_2'}\right)^2 } \\ \end{aligned} \hspace{\stretch{1}}(2.5)

Length contraction.

Consider $O$ and $O'$ with $O'$ moving in $x$ with speed $v_x > 0$. Here we have

\begin{aligned}x' &= \gamma \left( x - \frac{v_x}{c} ct \right) \\ c t' &= \gamma \left( ct - \frac{v_x}{c} x \right) \end{aligned} \hspace{\stretch{1}}(3.6)

PICTURE: spacetime diagram with $ct'$ at angle $\alpha$, where $\tan \alpha = v_x/c$.

Two points $(x_A,0)$, $(x_B,0)$, with rest length measured as $L = x_B - x_A$. From the diagram $c(t_B - t_A) = \tan\alpha L$, and from 3.6 we have

\begin{aligned}x_A' &= \gamma \left( x_A - \frac{v_x}{c} c t_A \right) \\ x_B' &= \gamma \left( x_B - \frac{v_x}{c} c t_B \right),\end{aligned} \hspace{\stretch{1}}(3.8)

so that

\begin{aligned}L' &= x_B' - x_A' \\ &= \gamma \left( (x_B - x_A) - \frac{v_x}{c} c (t_B -t_A) \right) \\ &= \gamma \left( L - \frac{v_x}{c} \tan \alpha L \right) \\ &= \gamma \left( L - \frac{v_x^2}{c^2} L \right) \\ &= \gamma L \left( 1 - \frac{v_x^2}{c^2} \right) \\ &= L \sqrt{ 1 - \frac{v_x^2}{c^2} } \end{aligned}

Superluminal speed and causality.

If Einstein’s relativity holds, superliminal motion is a “no-no”. Imagine that some “tachyons” exist that can instantaneously transmit stuff between observers.

PICTURE9: two guys with resting worldlines showing.

Can send info back to $A$ before $A$ sends to $B$. Superluminal propagation allows sending information not yet available. Can show this for finite superluminal velocities (but hard) as well as infinite velocity superluminal speeds. We see that time ordering can not be changed for events separated by time like separation. Events separated by spacelike separation cannot be ca usually connected.

References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

PHY450H1S. Relativistic Electrodynamics Lecture 4 (Taught by Prof. Erich Poppitz). Spacetime geometry, Lorentz transformations, Minkowski diagrams.

Posted by peeterjoot on January 18, 2011

Still covering chapter 1 material from the text [1].

Finished covering Professor Poppitz’s lecture notes: invariance of finite intervals (25-26).

Started covering Professor Poppitz’s lecture notes: analogy with rotations and derivation of Lorentz transformations (27-32); Minkowski space diagram of boosted frame (32.1); using the diagram to find length contraction (32.2) ; nonrelativistic limit of boosts (33).

More spacetime geometry.

PICTURE: ct,x curvy worldline with tangent vector $\mathbf{v}$.

In an inertial frame moving with $\mathbf{v}$, whose origin coincides with momentary position of this moving observer $ds^2 = c^2 {dt'}^2 = c^2 dt^2 - \mathbf{r}^2$

“proper time” is

\begin{aligned}dt' = dt \sqrt{ 1 - \frac{1}{{c^2}} \left( \frac{d\mathbf{r}}{dt} \right)^2 } = dt \sqrt{ 1 - \frac{\mathbf{v}^2}{c^2}} \end{aligned} \hspace{\stretch{1}}(2.1)

We see that $latex dt’ 0$, so that $\sqrt{1-\mathbf{v}^2/c^2} < 1$.

In a manifestly invariant way we define the proper time as

\begin{aligned}d\tau \equiv \frac{ds}{c}\end{aligned} \hspace{\stretch{1}}(2.2)

So that between worldpoints $a$ and $b$ the proper time is a line integral over the worldline

\begin{aligned}d\tau \equiv \frac{1}{{c}} \int_a^b ds.\end{aligned} \hspace{\stretch{1}}(2.3)

PICTURE: We are splitting up the worldline into many small pieces and summing them up.

HOLE IN LECTURE NOTES: ON PROPER TIME for “length” of straight vs. curved worldlines: TO BE REVISITED. Prof. Poppitz promised to revisit this again next time … his notes are confusing him, and he’d like to move on.

Finite interval invariance.

Tomorrow we are going to complete the proof about invariance. We’ve shown that light like intervals are invariant, and that infinitesimal intervals are invariant. We need to put these pieces together for finite intervals.

Deriving the Lorentz transformation.

Let’s find the coordinate transforms that leave $s_{12}^2$ invariant. This generalizes Galileo’s transformations.

We’d like to generalize rotations, which leave spatial distance invariant. Such a transformation also leaves the spacetime interval invariant.

In Euclidean space we can generate an arbitrary rotation by composition of rotation around any of the $xy, yz, zx$ axis.

For 4D Euclidean space we would form any rotation by composition of any of the 6 independent rotations for the 6 available planes. For example with $x,y,z,w$ axis we can rotate in any of the $xy, xz, xw, yz, yw, zw$ planes.

For spacetime we can “rotate” in $x,t$, $y,t$, $z,t$ “planes”. Physically this is motion space (boosting a position).

Consider a $x,t$ transformation.

The trick (that is in the notes) is to rewrite the time as an analytical continuation of the time coordinate, as follows

\begin{aligned}ds^2 = c^2 dt^2 - dx^2\end{aligned} \hspace{\stretch{1}}(4.4)

and write

\begin{aligned}t \rightarrow i \tau,\end{aligned} \hspace{\stretch{1}}(4.5)

so that the interval becomes

\begin{aligned}ds^2 = - (c^2 d\tau^2 + dx^2)\end{aligned} \hspace{\stretch{1}}(4.6)

Now we have a structure that is familiar, and we can rotate as we normally do. Prof does not want to go through the details of this “trickery” in class, but says to see the notes. The end result is that we can transform as follows

\begin{aligned}x' &= x \cosh \psi + ct \sinh \psi \\ ct' &= x \sinh \psi + ct \cosh \psi \end{aligned} \hspace{\stretch{1}}(4.7)

which is analogous to a spatial rotation

\begin{aligned}x' &= x \cos \alpha + y \sin \alpha \\ y' &= -x \sin \alpha + y \cos \alpha \end{aligned} \hspace{\stretch{1}}(4.9)

There are some differences in sign as well, but the important feature to recall is that $\cosh^2 x - \sinh^2 x = (1/4)( e^{2x} + e^{-2x} + 2 - e^{2x} - e^{-2x} + 2 ) = 1$. We call these hyperbolic rotations, something that is simply a mathematical transformation. Now we want to relate this to something physical.

\paragraph{Q: What is $\psi$?}

The origin of $O$ has coordinates $(t, \mathbf{O})$ in the $O$ frame.

PICTURE (pg 32): $O'$ frame translating along $x$ axis with speed $v_x$. We have

\begin{aligned}\frac{x'}{c t'} = \frac{v_x}{c}\end{aligned} \hspace{\stretch{1}}(4.11)

However, using 4.7 we have for the origin

\begin{aligned}x' &= ct \sinh \psi \\ ct' &= ct \cosh \psi\end{aligned} \hspace{\stretch{1}}(4.12)

so that

\begin{aligned}\frac{x'}{c t'} = \tanh \psi = \frac{v_x}{c}\end{aligned} \hspace{\stretch{1}}(4.14)

Using

\begin{aligned}\cosh \psi &= \frac{1}{{\sqrt{1 - \tanh^2 \psi}}} \\ \sinh \psi &= \frac{\tanh \psi}{\sqrt{1 - \tanh^2 \psi}}\end{aligned} \hspace{\stretch{1}}(4.15)

Performing all the gory substitutions one gets

\begin{aligned}x' &= \frac{1}{{\sqrt{1 - v_x^2/c^2}}} x+\frac{v_x/c}{\sqrt{1 - v_x^2/c^2}} c t \\ y' &= y \\ z' &= z \\ ct' &= \frac{v_x/c}{\sqrt{1 - v_x^2/c^2}} x+\frac{1}{{\sqrt{1 - v_x^2/c^2}}} c t\end{aligned} \hspace{\stretch{1}}(4.17)

PICTURE: Let us go to the more conventional case, where $O$ is at rest and $O'$ is moving with velocity $v_x$.

We achieve this by simply changing the sign of $v_x$ in 4.17 above. This gives us

\begin{aligned}x' &= \frac{1}{{\sqrt{1 - v_x^2/c^2}}} x-\frac{v_x/c}{\sqrt{1 - v_x^2/c^2}} c t \\ y' &= y \\ z' &= z \\ ct' &= -\frac{v_x/c}{\sqrt{1 - v_x^2/c^2}} x+\frac{1}{{\sqrt{1 - v_x^2/c^2}}} c t\end{aligned} \hspace{\stretch{1}}(4.21)

We want some shorthand to make this easier to write and introduce

\begin{aligned}\gamma = \frac{1}{{\sqrt{1 - v_x^2/c^2}}},\end{aligned} \hspace{\stretch{1}}(4.25)

so that 4.21 becomes

\begin{aligned}x' &= \gamma \left( x - \frac{v_x}{c} ct \right) \\ ct' &= \gamma \left( ct - \frac{v_x}{c} x \right)\end{aligned} \hspace{\stretch{1}}(4.26)

We started the class by saying these would generalize the Galilean transformations. Observe that if we take $c \rightarrow \infty$, we have $\gamma \rightarrow 1$ and

\begin{aligned}x' &= x - v_x t + O((v_x/c)^2)t' &= t + O(v_x/c)\end{aligned} \hspace{\stretch{1}}(4.28)

This is how to remember the signs. We want things to match up with the non-relativistic limit.

\paragraph{Q: How do lines of constant $x'$ and $ct'$ look like on the $x,ct$ spacetime diagram?}

Our starting point (again) is

\begin{aligned}x' &= \gamma \left( x - \frac{v_x}{c} ct \right) \\ ct' &= \gamma \left( ct - \frac{v_x}{c} x \right).\end{aligned} \hspace{\stretch{1}}(4.29)

What are the points with $x' = 0$. Those are the points where $x = (v_x/c) c t$. This is the $ct' axis$. That’s the straight worldline

PICTURE: worldline of $O'$ origin.

What are the points with $ct' = 0$. Those are the points where $c t = x v_x/c$. This is the $x' axis$.

Lines that are parallel to the $x'$ axis are lines of constant $x'$, and lines parallel to $ct'$ axis are lines of constant $t'$, but the light cone is the same for both.

\paragraph{What is this good for?}

We have time to pick from either length contraction or non-causality (how to kill your grandfather). How about length contraction. We can use the diagram to read the $x$ or $ct$ coordinates, or examine causality, but it is hard to read off $t'$ or $x'$ coordinates.

References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

Stokes Theorem for antisymmetric tensors.

Posted by peeterjoot on January 18, 2011

Obsolete with potential errors.

This post may be in error.  I wrote this before understanding that the gradient used in Stokes Theorem must be projected onto the tangent space of the parameterized surface, as detailed in Alan MacDonald’s Vector and Geometric Calculus.

See the post ‘stokes theorem in geometric algebra‘ [PDF], where this topic has been revisited with this in mind.

Original Post:

In [3] I worked through the Geometric Algebra expression for Stokes Theorem. For a $k-1$ grade blade, the final result of that work was

\begin{aligned}\int( \nabla \wedge F ) \cdot d^k x =\frac{1}{{(k-1)!}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {F}}{\partial {a_{u}}} \cdot (dx_r \wedge dx_s \wedge \cdots \wedge dx_t)\end{aligned} \hspace{\stretch{1}}(7.44)

Let’s expand this in coordinates to attempt to get the equivalent expression for an antisymmetric tensor of rank $k-1$.

Starting with the RHS of 7.44 we have

\begin{aligned}F &= \frac{1}{{(k-1)!}}F_{\mu_1 \mu_2 \cdots \mu_{k-1} }\gamma^{\mu_1} \wedge \gamma^{ \mu_2 } \wedge \cdots \wedge \gamma^{\mu_{k-1}} \\ dx_r \wedge dx_s \wedge \cdots \wedge dx_t &=\frac{\partial {x^{\nu_1}}}{\partial {a_r}}\frac{\partial {x^{\nu_2}}}{\partial {a_s}}\cdots\frac{\partial {x^{\nu_{k-1}}}}{\partial {a_t}}\gamma_{\nu_1} \wedge \gamma_{ \nu_2 } \wedge \cdots \wedge \gamma_{\nu_{k-1}}da_r da_s \cdots da_t\end{aligned} \hspace{\stretch{1}}(7.45)

We need to expand the dot product of the wedges, for which we have

\begin{aligned}\left( \gamma^{\mu_1} \wedge \gamma^{ \mu_2 } \wedge \cdots \wedge \gamma^{\mu_{k-1}} \right) \cdot\left( \gamma_{\nu_1} \wedge \gamma_{ \nu_2 } \wedge \cdots \wedge \gamma_{\nu_{k-1}}\right) ={\delta^{\mu_{k-1}}}_{\nu_1} {\delta^{ \mu_{k-2} }}_{\nu_2} \cdots {\delta^{\mu_{1}} }_{\nu_{k-1}}\epsilon^{\nu_1 \nu_2 \cdots \nu_{k-1}}\end{aligned} \hspace{\stretch{1}}(7.47)

Putting all the LHS bits together we have

\begin{aligned}&\frac{1}{{((k-1)!)^2}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {}}{\partial {a_{u}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1} }{\delta^{\mu_{k-1}}}_{\nu_1} {\delta^{ \mu_{k-2} }}_{\nu_2} \cdots {\delta^{\mu_{1}} }_{\nu_{k-1}}\epsilon^{\nu_1 \nu_2 \cdots \nu_{k-1}}\frac{\partial {x^{\nu_1}}}{\partial {a_r}}\frac{\partial {x^{\nu_2}}}{\partial {a_s}}\cdots\frac{\partial {x^{\nu_{k-1}}}}{\partial {a_t}}da_r da_s \cdots da_t \\ &=\frac{1}{{((k-1)!)^2}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {}}{\partial {a_{u}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1} }\epsilon^{\mu_{k-1} \mu_{k-2} \cdots \mu_{1}}\frac{\partial {x^{\mu_{k-1}}}}{\partial {a_r}}\frac{\partial {x^{\mu_{k-2}}}}{\partial {a_s}}\cdots\frac{\partial {x^{\mu_1}}}{\partial {a_t}}da_r da_s \cdots da_t \\ &=\frac{1}{{((k-1)!)^2}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {}}{\partial {a_{u}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1} }{\left\lvert{\frac{\partial(x^{\mu_{k-1}},x^{\mu_{k-2}},\cdots,x^{\mu_1})}{\partial(a_r, a_s, \cdots, a_t)}}\right\rvert}da_r da_s \cdots da_t \\ \end{aligned}

Now, for the LHS of 7.44 we have

\begin{aligned}\nabla \wedge F &=\gamma^\mu \wedge \partial_\mu F \\ &=\frac{1}{{(k-1)!}}\frac{\partial {}}{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}\gamma^{\mu_k} \wedge\gamma^{\mu_1} \wedge \gamma^{ \mu_2 } \wedge \cdots \wedge \gamma^{\mu_{k-1}} \end{aligned}

and the volume element of

\begin{aligned}d^k x &=\frac{\partial {x^{\nu_1}}}{\partial {a_1}}\frac{\partial {x^{\nu_2}}}{\partial {a_2}}\cdots\frac{\partial {x^{\nu_{k}}}}{\partial {a_k}}\gamma_{\nu_1} \wedge \gamma_{ \nu_2 } \wedge \cdots \wedge \gamma_{\nu_k}da_1 da_2 \cdots da_k\end{aligned}

Our dot product is

\begin{aligned}\left(\gamma^{\mu_k} \wedge\gamma^{\mu_1} \wedge \gamma^{ \mu_2 } \wedge \cdots \wedge \gamma^{\mu_{k-1}} \right) \cdot\left( \gamma_{\nu_1} \wedge \gamma_{ \nu_2 } \wedge \cdots \wedge \gamma_{\nu_k} \right)={\delta^{\mu_{k-1}}}_{\nu_1} {\delta^{ \mu_{k-2} }}_{\nu_2} \cdots {\delta^{\mu_{1}} }_{\nu_{k-1}}{\delta^{\mu_{k}} }_{\nu_{k}}\epsilon^{\nu_1 \nu_2 \cdots \nu_{k}}\end{aligned} \hspace{\stretch{1}}(7.48)

The LHS of our k-form now evaluates to

\begin{aligned}(\gamma^\mu \wedge \partial_\mu F) \cdot d^k x &= \frac{1}{(k-1)!}\frac{\partial }{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}{\delta^{\mu_{k-1}}}_{\nu_1} {\delta^{ \mu_{k-2} }}_{\nu_2} \cdots {\delta^{\mu_1} }_{\nu_{k-1}}{\delta^{\mu_k} }_{\nu_k}\epsilon^{\nu_1 \nu_2 \cdots \nu_k}\frac{\partial {x^{\nu_1}}}{\partial {a_1}}\frac{\partial {x^{\nu_2}}}{\partial {a_2}} \cdots \frac{\partial {x^{\nu_k}}}{\partial {a_k}} da_1 da_2 \cdots da_k \\ &= \frac{1}{(k-1)!}\frac{\partial }{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}\epsilon^{\mu_{k-1} \mu_{k-2} \cdots \mu_1 \mu_k}\frac{\partial {x^{\mu_{k-1}}}}{\partial {a_1}}\frac{\partial {x^{\mu_{k-2}}}}{\partial {a_2}} \cdots \frac{\partial {x^{\mu_1}}}{\partial {a_{k-1}}}\frac{\partial {x^{\mu_k}}}{\partial {a_k}} da_1 da_2 \cdots da_k \\ &= \frac{1}{(k-1)!}\frac{\partial }{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}{\left\lvert{\frac{\partial(x^{\mu_{k-1}},x^{\mu_{k-2}},\cdots x^{\mu_1},x^{\mu_k})}{\partial(a_1, a_2, \cdots, a_{k-1}, a_k)}}\right\rvert} da_1 da_2 \cdots da_k \\ \end{aligned}

Presuming no mistakes were made anywhere along the way (including in the original Geometric Algebra expression), we have arrived at Stokes Theorem for rank $k-1$ antisymmetric tensors $F$

\boxed{ \begin{aligned}&\int\frac{\partial }{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}{\left\lvert{\frac{\partial(x^{\mu_{k-1}},x^{\mu_{k-2}},\cdots x^{\mu_1},x^{\mu_k})}{\partial(a_1, a_2, \cdots, a_{k-1}, a_k)}}\right\rvert} da_1 da_2 \cdots da_k \\ &= \frac{1}{(k-1)!} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial }{\partial {a_u}} F_{\nu_1 \nu_2 \cdots \nu_{k-1} }{\left\lvert{\frac{\partial(x^{\nu_{k-1}},x^{\nu_{k-2}}, \cdots ,x^{\nu_1})}{\partial(a_r, a_s, \cdots, a_t)}}\right\rvert} da_r da_s \cdots da_t \end{aligned} } \hspace{\stretch{1}}(7.49)

The next task is to validate this, expanding it out for some specific ranks and hypervolume element types, and to compare the results with the familiar 3d expressions.

References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[3] Peeter Joot. Stokes theorem derivation without tensor expansion of the blade [online]. http://sites.google.com/site/peeterjoot/math2009/stokesNoTensor.pdf.

refreshed older blog post on Stokes Theorem in Geometric Algebra

Posted by peeterjoot on January 17, 2011

In preparation for understanding the tensor Stokes Theorem content from Landau and Lifshitz “Classical Theory of Fields” (the course text for PHY450H1S, relativistic electrodynamics), I went back and reviewed my old notes where I blundered through a derivation of Stokes Theorem in Geometric Algebra. I spotted some errors in the original, as well as the fact that a summary of the final result was completely missing. The old blog post, and the associated pdf file are both now refreshed.