Subfield: Modern Physics

4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.
Context question:
(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form

$$
\begin{aligned}
X(x, t, v) & =A(v) x+B(v) t \\
T(x, t, v) & =C(v) x+D(v) t
\end{aligned}
$$

where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.
Context question:
(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)
Context answer:
\boxed{证明题}


Context question:
(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:

$$
\begin{aligned}
A(v)^{2}-B(v) C(v) & =1 \\
D(v)^{2}-B(v) C(v) & =1 \\
C(v)(A(v)-D(v)) & =0 \\
B(v)(A(v)-D(v)) & =0 .
\end{aligned}
$$

(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.
Context answer:
\boxed{证明题}


Question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:

$$
\begin{aligned}
x^{\prime} & =A(v) x-v A(v) t \\
t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t
\end{aligned}
$$

Solution:
The $F^{\prime}$ frame is defined by $x^{\prime}=0$, therefore $A(v) x+B(v) t=A(v) v t+B(v) t$ plugging in the equation $x=v t$. This implies that $A v+B=0$. This let's all the undetermined functions $A, B, C, D$ to be solved in terms of $A$.