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The Elementary Lorentz Transformation

To motivate the Lorentz transformation, recall the Galilean transformation between moving coordinate systems:

$\displaystyle x_1'$ $\textstyle =$ $\displaystyle x_1 - vt$ (15.1)
$\displaystyle x_2'$ $\textstyle =$ $\displaystyle x_2$ (15.2)
$\displaystyle x_3'$ $\textstyle =$ $\displaystyle x_3$ (15.3)
$\displaystyle t'$ $\textstyle =$ $\displaystyle t$ (15.4)

(where $K$ is fixed and $K'$ is moving in the 1-direction at speed $v$).


F_j = m \ddot{x}_j = m \ddot{x}_j' = F_j'
\end{displaymath} (15.5)

or Newton's Laws are covariant with respect to the Gallilean transformation.


\frac{\partial }{\partial x_1} = \frac{\partial }{\partial x_1'} + \frac{1}{v}
\frac{\partial }{\partial t'}
\end{displaymath} (15.6)

and so
$\displaystyle \frac{\partial ^2 }{\partial x_1^2}$ $\textstyle =$ $\displaystyle \frac{\partial ^2 }{\partial x_1'^2} + \frac{1}{v^2}
\frac{\partial ^2 }{\partial t'^2} + \frac{2}{v} \frac{\partial ^2 }{\partial x_1'\partial t'}$ (15.7)
$\displaystyle \frac{\partial ^2 }{\partial x_2^2}$ $\textstyle =$ $\displaystyle \frac{\partial ^2 }{\partial x_2'^2}$ (15.8)
$\displaystyle \frac{\partial ^2 }{\partial x_3^2}$ $\textstyle =$ $\displaystyle \frac{\partial ^2 }{\partial x_3'^2}$ (15.9)
$\displaystyle \frac{\partial ^2 }{\partial t^2}$ $\textstyle =$ $\displaystyle \frac{\partial ^2 }{\partial t'^2}.$ (15.10)

Thus if
\left\{ \nabla^2 - \frac{1}{c^2} \frac{\partial ^2 }{\partial t^2} \right\} \psi = 0
\end{displaymath} (15.11)

\left\{ \nabla'^2 - \frac{1}{c^2} \frac{\partial ^2 }{\parti...
...{v} \frac{\partial ^2
\psi}{\partial x_1' \partial t'} \ne 0
\end{displaymath} (15.12)

(!) so that the wave equation, and hence Maxwell's equations which lead directly to the wave equation in free space, are not covariant with respect to the Gallilean transformation! They already determine the permitted velocity of a light wave, and do not allow that velocity to depend on anything but the properties of the medium through which the wave is transmitted.

The simplest linear transformation of coordinates is that preserves the form of the wave equation is easy to determine. It is one that keeps the speed of the (light) wave equal in both the $K$ and the $K'$ frames. Geometrically, if a flash of light is emitted from the (coincident) origins at time $t = t' =
0$, it will appear to expand like a sphere out from both coordinate origins, each in its own frame:

(ct)^2 - (x^2 + y^2 + z^2) = 0
\end{displaymath} (15.13)

(ct')^2 - (x'^2 + y'^2 + z'^2) = 0
\end{displaymath} (15.14)

are simultaneous constraints on the equations. Most generally,
(ct)^2 - (x^2 + y^2 + z^2) = \lambda^2 \left[ (ct')^2 - (x'^2 + y'^2 + z'^2)
\right ]
\end{displaymath} (15.15)

where, $\lambda(v)$ describes a possible change of scale between the frames. If we insist that the coordinate transformation be homogeneous and symmetric between the frames15.1, then
\lambda = 1 .
\end{displaymath} (15.16)

Let us define

$\displaystyle x_0$ $\textstyle =$ $\displaystyle ct$ (15.17)
$\displaystyle x_1$ $\textstyle =$ $\displaystyle x$ (15.18)
$\displaystyle x_2$ $\textstyle =$ $\displaystyle y$ (15.19)
$\displaystyle x_3$ $\textstyle =$ $\displaystyle z$ (15.20)
$\displaystyle (x_4$ $\textstyle =$ $\displaystyle ict \mbox{\vspace{.5in} \rm Minkowski metric})$ (15.21)

Then we need a linear transformation of the coordinates that mixes x and (ct) in the direction of $v$ in such a way that the length

\Delta s^2 = (x_0)^2 - (x_1^2 + x_2^2 + x_3^2)
\end{displaymath} (15.22)

is conserved and that goes into the Gallilean transformation as $v \to
0$. If we continue to assume that $v$ is in the 1 direction, this leads to the Lorentz transformation:
$\displaystyle x_0'$ $\textstyle =$ $\displaystyle \gamma (x_0 - \beta x_1)$ (15.23)
$\displaystyle x_1'$ $\textstyle =$ $\displaystyle \gamma (x_1 - \beta x_0)$ (15.24)
$\displaystyle x_2'$ $\textstyle =$ $\displaystyle x_2$ (15.25)
$\displaystyle x_3'$ $\textstyle =$ $\displaystyle x_3$ (15.26)

where at $x_1' = 0$,
x_1 = vt \to \beta = \frac{v}{c}.
\end{displaymath} (15.27)

\Delta s^2 = \Delta s'^2
\end{displaymath} (15.28)

leads to
x_0^2 - x_1^2 = \gamma^2(x_0^2 - x_1^2) + \gamma^2 \beta^2 (x_1^2 - x_0^2)
\end{displaymath} (15.29)

\gamma^2 (1 - \beta^2) = 1
\end{displaymath} (15.30)

\gamma = \frac{\pm 1}{\sqrt{1 - \beta^2}}
\end{displaymath} (15.31)

where we choose the $+$ sign by convention. This makes $\gamma(0) =
+1$. Finally,
\gamma(v) = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
\end{displaymath} (15.32)

as we all know and love.

Now, let me remind you that when $v « c$,

\gamma(v) = 1 + \frac{1}{2} \frac{v^2}{c^2} + \ldots
\end{displaymath} (15.33)

to lowest surviving order in $\frac{v}{c}$. As we shall see, this is why ``kinetic energy'' in non-relativistic systems (being defined as the total energy minus the potential energy and the rest mass energy) is the usual $\frac{1}{2}mv^2$.

The inverse transformation (from $K'$ to $K$) is also of some interest.

$\displaystyle x_0$ $\textstyle =$ $\displaystyle \gamma (x_0' + \beta x_1')$ (15.34)
$\displaystyle x_1$ $\textstyle =$ $\displaystyle \gamma (x_1' + \beta x_0')$ (15.35)
$\displaystyle x_2$ $\textstyle =$ $\displaystyle x_2'$ (15.36)
$\displaystyle x_3$ $\textstyle =$ $\displaystyle x_3'$ (15.37)

which is perfectly symmetric, with $v \to -v$. It appears that which frame is at ``rest'' and which is moving is mathematically, at least, a matter of perspective.

Finally, if we let

\vec{\beta} = \frac{\bf v}{c}
\end{displaymath} (15.38)

(in an arbitrary direction) then we have but to use dot products to align the vector transformation equations with this direction:
$\displaystyle x_0'$ $\textstyle =$ $\displaystyle \gamma(x_0 - \vec{\beta} \cdot {\bf x})$ (15.39)
$\displaystyle {\bf x}'$ $\textstyle =$ $\displaystyle {\bf x} + \frac{\gamma - 1}{\beta^2} (\vec{\beta} \cdot
{\bf x}) \vec{\beta} - \gamma \vec{\beta} x_0$ (15.40)

I think that you should prove that this is correct as an exercise. Since the direction of $\vec{\beta}$ is arbitrary, it should suffice to show that this reduces the the form above for that direction and an arbitrary transverse direction.

Solution: Note that

{\bf x}_\parallel = \frac{(\vec{\beta} \cdot {\bf x})\vec{\beta}}{\beta^2}
\end{displaymath} (15.41)

Lorentz transform it according to this rule and one gets (by inspection)
{\bf x}_\parallel' = \gamma ({\bf x}_\parallel - \vec{\beta} x_0)
\end{displaymath} (15.42)

as one should. The $x_0'$ transform is obvious. Finally, the other two ($\perp$) components do not get a contribution from $\gamma$. That is,
{\bf x}' = ({\bf x}_\perp + {\bf x}_\parallel) + \gamma {\bf x}_\parallel
- {\bf x}_\parallel - \gamma \vec{\beta} x_0
\end{displaymath} (15.43)

(reconstructing the result above directly) QED.

This is not the most general or convenient way to write the final transform. This is because $\gamma$ and $\beta$ are both related functions; it should not be necessary to use two parameters that are not independent. Also, the limiting structure of the transformation is not at all apparent without considering the functional forms in detail.

It is easy to see from the definition above that

(\gamma^2 - \gamma^2 \beta^2) = 1.
\end{displaymath} (15.44)

The range of $\beta$ is determined by the requirement that the transformation be non-singular and its symmetry:
0 \le \beta < 1 \mbox{\rm    so that    } 1 \le \gamma < \infty.
\end{displaymath} (15.45)

If we think about functions that ``naturally'' parameterize these ranges, they are:
\cosh^2 \xi - \sinh^2 \xi = 1
\end{displaymath} (15.46)

$\displaystyle \beta$ $\textstyle =$ $\displaystyle \tanh \xi \quad = \quad \frac{e^{-\xi} - e^{\xi}}{e^{-\xi} +
e^{\xi}} \quad \in \quad [0,1)$ (15.47)
$\displaystyle \gamma$ $\textstyle =$ $\displaystyle \cosh \xi \quad = \quad \frac{1}{2} (e^{-\xi} + e^{\xi})
\quad \in \quad [1,\infty)$ (15.48)
$\displaystyle \gamma \beta$ $\textstyle =$ $\displaystyle \sinh \xi \quad = \quad \frac{1}{2} (e^{-\xi} -
e^{\xi}) \quad \in \quad [0,\infty) .$ (15.49)

The parameter $\xi$ is called the boost parameter or rapidity. You will see this used frequently in the description of relativistic problems. You will also hear about ``boosting'' between frames, which essentially means performing a Lorentz transformation (a ``boost'') to the new frame. This will become clearer later when we generalize these results further. To give you something to meditate upon, consider in your minds the formal similarity between a ``boost'' and a ``rotation'' between the ${\bf x}_\parallel$ and $x_0$ coordinates where the rotation is through an imaginary angle $i \xi$. Hmmmm.

To elucidate this remark a wee tad more, note that in this parameterization,

$\displaystyle x_0'$ $\textstyle =$ $\displaystyle x_0 \cosh \xi - x_\parallel \sinh \xi$ (15.50)
$\displaystyle x_\parallel'$ $\textstyle =$ $\displaystyle -x_0 \sinh \xi + x_\parallel \cosh \xi$ (15.51)
$\displaystyle {\bf x}_\perp'$ $\textstyle =$ $\displaystyle {\bf x}_\perp .$ (15.52)

What is the $4 \times 4$ transformation matrix (in four dimensions) for this result? Does it look like a ``hyperbolic rotation''15.2 or what?

We have just determined that the (vector) coordinate system transforms a certain way. What, then, of vector fields, or any other vector quantity? How do general vectors transform under a boost? This depends on the nature of the vectors. Many vectors, if not most, transform like the underlying coordinate description of the vectors. This includes the ones of greatest interest in physics. To make this obvious, we will have to generalize a vector quantity to four dimensions.

next up previous contents
Next: 4-Vectors Up: Special Relativity Previous: Einstein's Postulates   Contents
Robert G. Brown 2007-12-28