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• 293,778

Playing with complex notation for relativistic applications in a plane

Posted by peeterjoot on April 19, 2011

Motivation.

In the electrodynamics midterm we had a question on circular motion. This screamed for use of complex numbers to describe the spatial parts of the spacetime trajectories.

Let’s play with this a bit.

Our invariant.

Suppose we describe our spacetime point as a paired time and complex number

\begin{aligned}X = (ct, z).\end{aligned} \hspace{\stretch{1}}(2.1)

Our spacetime invariant interval in this form is thus

\begin{aligned}X^2 \equiv (ct)^2 - {\left\lvert{z}\right\rvert}^2.\end{aligned} \hspace{\stretch{1}}(2.2)

Not much different than the usual coordinate representation of the spatial coordinates, except that we have a ${\left\lvert{z}\right\rvert}^2$ replacing the usual $\mathbf{x}^2$.

Taking the spacetime distance between $X$ and another point, say $\tilde{X} = ( c \tilde{t}, \tilde{z})$ motivates the inner product between two points in this representation

\begin{aligned}(X - \tilde{X})^2 &= (ct - c \tilde{t})^2 - {\left\lvert{z - \tilde{z}}\right\rvert}^2 \\ &= (ct - c \tilde{t})^2 - (z - \tilde{z})(z^{*} - \tilde{z}^{*}) \\ &= (ct)^2 - 2 (ct) (c \tilde{t}) + (c \tilde{t})^2 - {\left\lvert{z}\right\rvert}^2 - {\left\lvert{\tilde{z}}\right\rvert}^2 + (z \tilde{z}^{*} + z^{*} \tilde{z}) \\ &= X^2 + \tilde{X}^2 - 2 \left( (ct) (c \tilde{t}) - \frac{1}{{2}}(z \tilde{z}^{*} + z^{*} \tilde{z}) \right) \\ \end{aligned}

It’s clear that it makes sense to define

\begin{aligned}X \cdot \tilde{X} = (ct) (c \tilde{t}) - \text{Real} (z \tilde{z}^{*}),\end{aligned} \hspace{\stretch{1}}(2.3)

consistent with our original starting point

\begin{aligned}X^2 = X \cdot X.\end{aligned} \hspace{\stretch{1}}(2.4)

Let’s also introduce a complex inner product

\begin{aligned}{\langle{{z}}, {{\tilde{z}}}\rangle} \equiv \frac{1}{{2}} \left( z \tilde{z}^{*} + z^{*} \tilde{z}) \right) = \text{Real} (z \tilde{z}^{*}).\end{aligned} \hspace{\stretch{1}}(2.5)

Our dot product can now be written

\begin{aligned}X \cdot \tilde{X} = (ct) (c \tilde{t}) - {\langle{{z}}, {{\tilde{z}}}\rangle}.\end{aligned} \hspace{\stretch{1}}(2.6)

Change of basis.

Our standard basis for our spatial components is $\{1, i\}$, but we are free to pick any other basis should we choose. In particular, if we rotate our basis counterclockwise by $\phi$, our new basis, still orthonormal, is $\{ e^{i\phi}, i e^{i\phi} \}$.

In any orthonormal basis the coordinates of a point with respect to that basis are real, so just as we can write

\begin{aligned}z = {\langle{{1}}, {{z}}\rangle} + i {\langle{{i}}, {{z}}\rangle},\end{aligned} \hspace{\stretch{1}}(3.7)

we can extract the coordinates in the rotated frame, also simply by taking inner products

\begin{aligned}z = e^{i \phi} {\langle{{e^{i \phi}}}, {{z}}\rangle} + i e^{i\phi} {\langle{{i e^{i\phi} }}, {{z}}\rangle}.\end{aligned} \hspace{\stretch{1}}(3.8)

The values ${\langle{{e^{i \phi}}}, {{z}}\rangle}$, and ${\langle{{i e^{i\phi} }}, {{z}}\rangle}$ are the (real) coordinates of the point $z$ in this rotated basis.

This is enough that we can write the Lorentz boost immediately for a velocity $\vec{v} = c \beta e^{i\phi}$ at an arbitrary angle $\phi$ in the plane

\begin{aligned}\begin{bmatrix}ct' \\ {\langle{{e^{i\phi}}}, {{z'}}\rangle} \\ {\langle{{i e^{i\phi}}}, {{z'}}\rangle} \end{bmatrix}=\begin{bmatrix}\gamma & -\gamma \beta & 0 \\ -\gamma \beta & \gamma & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}ct \\ {\langle{{e^{i\phi}}}, {{z}}\rangle} \\ {\langle{{i e^{i\phi}}}, {{z}}\rangle} \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.9)

Let’s translate this to $ct, x, y$ coordinates as a check. For the spatial component parallel to the boost direction we have

\begin{aligned}{\langle{{e^{i\phi}}}, {{x + iy}}\rangle} &= \text{Real} ( e^{-i\phi} (x + i y) ) \\ &= \text{Real} ( (\cos\phi - i \sin\phi)(x + i y) ) \\ &= x \cos\phi + y \sin\phi,\end{aligned}

and the perpendicular components are

\begin{aligned}{\langle{{ i e^{i\phi}}}, {{x + iy}}\rangle} &= \text{Real} ( -i e^{-i\phi} (x + i y) ) \\ &= \text{Real} ( (-i \cos\phi - \sin\phi)(x + i y) ) \\ &= -x \sin\phi + y \cos\phi.\end{aligned}

Grouping the two gives

\begin{aligned}\begin{bmatrix}{\langle{{e^{i\phi}}}, {{x + iy}}\rangle} \\ {\langle{{i e^{i\phi}}}, {{x + iy}}\rangle} \end{bmatrix}=\begin{bmatrix}\cos\phi & \sin\phi \\ -\sin\phi & \cos\phi\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= R_{-\phi}\begin{bmatrix}x \\ y\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.10)

The boost equation in terms of the cartesian coordinates is thus

\begin{aligned}\begin{bmatrix}1 & 0 \\ 0 & R_{-\phi}\end{bmatrix}\begin{bmatrix}c t' \\ x' \\ y'\end{bmatrix}=\begin{bmatrix}\gamma & -\gamma \beta & 0 \\ -\gamma \beta & \gamma & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & R_{-\phi}\end{bmatrix}\begin{bmatrix}c t \\ x \\ y\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.11)

Writing

\begin{aligned}\begin{bmatrix}c t' \\ x' \\ y'\end{bmatrix}={\left\lVert{{\wedge^\mu}_\nu}\right\rVert} \begin{bmatrix}c t \\ x \\ y\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.12)

the boost matrix ${\left\lVert{{\wedge^\mu}_\nu}\right\rVert}$ is found to be (after a bit of work)

\begin{aligned}{\left\lVert{{\wedge^\mu}_\nu}\right\rVert} &=\begin{bmatrix}1 & 0 \\ 0 & R_{\phi}\end{bmatrix}\begin{bmatrix}\gamma & -\gamma \beta & 0 \\ -\gamma \beta & \gamma & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & R_{-\phi}\end{bmatrix} \\ &=\begin{bmatrix}\gamma & - \gamma \beta \cos\phi & -\gamma \beta \sin\phi \\ -\gamma \beta \cos\phi & \gamma \cos^2\phi + \sin^2 \phi & (\gamma -1) \sin\phi \cos\phi \\ -\gamma \beta \sin\phi & (\gamma -1) \sin\phi \cos\phi & \gamma \sin^2\phi + \cos^2\phi \\ \end{bmatrix} \\ \end{aligned}

A final bit of regrouping gives

\begin{aligned}{\left\lVert{{\wedge^\mu}_\nu}\right\rVert} =\begin{bmatrix}\gamma & - \gamma \beta \cos\phi & -\gamma \beta \sin\phi \\ -\gamma \beta \cos\phi & 1 + ( \gamma -1) \cos^2\phi & (\gamma -1) \sin\phi \cos\phi \\ -\gamma \beta \sin\phi & (\gamma -1) \sin\phi \cos\phi & 1 + (\gamma -1) \sin^2\phi \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.13)

This is consistent with the result stated in [1], finishing the game for the day.

References

[1] Wikipedia. Lorentz transformation — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 20-April-2011]. http://en.wikipedia.org/w/index.php?title=Lorentz_transformation&oldid=424507400.