First, you have to distinguish mathematical proofs from physics "proofs". You can't really rigorously prove statements about the real world at the same level of rigor that one has in maths.
The unimodularity (Jacobian = 1 or -1, to agree with the usual definitions of the adjective) of the Lorentz transformations is trivial to prove in maths, whatever definition of the Lorentz transformations we adopt.
Something else is the "proof" that the transformation responsible for changing the frame in physics is unimodular. Such a "proof" has to accept some physical assumptions that are ultimately justified empirically.
The usual "proof" is simple and you sketched it at the beginning.
In the empty space, the transformation $B$ (boost) from an observer at rest to an observer moving by the speed $v$ in the $z$ direction is given by
$$ (t,x,y,z)\mapsto (t',x',y',z')$$
Because of time- and space-translational symmetry, this map has to be linear (with a possible shift, i.e. an inhomogeneous term, generalizing Lorentz to Poincaré transformations). Because it is linear, its Jacobian is constant, so it is just one number-valued function of $v$.
By using the rotational symmetry, one can write
$$ B(-v) = R_\pi B(v) R_\pi $$
In words, the boost by the opposite speed may be obtained by rotating the system by $\pi$ around e.g. $x$-axis, boosting it by $+v$, and rotating back. The Jacobian is a determinant so for the product transformation above, it's just the product of the determinants and the determinants for the rotations are one (one could have also used parity so that the signs from these two determinants would cancel as well but I chose to avoid negative-determinant transformations).
It follows that
$$\det B(-v) = \det B(v)$$
i.e. the determinant is an even function of $v$. At the same moment, the boost by $-v$ is nothing else than the transformation switching back to the original frame, i.e.
$$B(-v) = B(v)^{-1}$$
which also means (in combination with the previous identity, to eliminate $\det B(-v)$)
$$\det B(v) = \det B(v)^{-1}$$
and the determinant is therefore $\pm 1$. Because it's $1$ for $v=0$ (identity) and it is a continuous function, we must have $\det B(v)=1$ for all $v$.
If you want to apply the Lorentz transforms, make sure to define each event carefully and assign to it the correct coordinates. The Lorentz transform will do the rest.
In your particular case:
- Event A = light emitted from the flashlight. Coordinates in F': say $x'_A = 0$, $t'_A = 0$.
- Event B = light triggers detector. Coordinates in F': $x'_B = x'_A + d = d$, $t'_B = (x'_B-x'_A)/c = d/c$.
- Apply Lorentz transform from F' to F to get coordinates in frame F:
$$x_A = \gamma(x'_A + v t'_A) = 0$$
$$t_A = \gamma(t'_A + vx'_A/c^2) = 0$$
$$x_B = \gamma(x'_B + v t'_B) = \gamma(d + vd/c) = \gamma(1+ v/c) d$$
$$t_B = \gamma(t'_B + vx'_B/c^2) = \gamma(d/c + vd/c^2) = \gamma(1+ v/c) d/c$$
Now look at what this means physically: F sees the light beam chasing after the detector over a distance $x_B - x_A = \gamma(1+ v/c) d$ and reaching it in a time $t_B - t_A = \gamma(1+ v/c) d/c = (x_B - x_A)/c$, at a velocity $(x_B - x_A)/(t_B - t_A) = c$.
Further exercises: 1) Check the specific figures for your data. 2) Check that the space-time interval between events $A$ and $B$ is indeed the same in both frames.
Best Answer
Lorentz boost is simply a Lorentz transformation which doesn't involve rotation. For example, Lorentz boost in the x direction looks like this:
\begin{equation} \left[ \begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \newline -\beta \gamma & \gamma & 0 & 0 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{array} \right] \end{equation}
where coordinates are written as (t, x, y, z) and
\begin{equation} \beta = \frac{v}{c} \end{equation} \begin{equation} \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{equation}
This is a linear transformation which given coordinates of an event in one reference frame allows one to determine the coordinates in a frame of reference moving with respect to the first reference frame at velocity v in the x direction.
The ones on the diagonal mean that the transformation does not change the y and z coordinates (i.e. it only affects time t and distance along the x direction). For comparison, Lorentz boost in the y direction looks like this:
\begin{equation} \left[ \begin{array}{cccc} \gamma & 0 & -\beta \gamma & 0 \newline 0 & 1 & 0 & 0 \newline -\beta \gamma & 0 & \gamma & 0 \newline 0 & 0 & 0 & 1 \end{array} \right] \end{equation}
which means that the transformation does not affect the x and z directions (i.e. it only affects time and the y direction).
In order to calculate Lorentz boost for any direction one starts by determining the following values:
\begin{equation} \gamma = \frac{1}{\sqrt{1 - \frac{v_x^2+v_y^2+v_z^2}{c^2}}} \end{equation} \begin{equation} \beta_x = \frac{v_x}{c}, \beta_y = \frac{v_y}{c}, \beta_z = \frac{v_z}{c} \end{equation}
Then the matrix form of the Lorentz boost for velocity v=(vx, vy, vz) is this:
\begin{equation} \left[ \begin{array}{cccc} L_{tt} & L_{tx} & L_{ty} & L_{tz} \newline L_{xt} & L_{xx} & L_{xy} & L_{xz} \newline L_{yt} & L_{yx} & L_{yy} & L_{yz} \newline L_{zt} & L_{zx} & L_{zy} & L_{zz} \newline \end{array} \right] \end{equation}
where
\begin{equation} L_{tt} = \gamma \end{equation} \begin{equation} L_{ta} = L_{at} = -\beta_a \gamma \end{equation} \begin{equation} L_{ab} = L_{ba} = (\gamma - 1) \frac{\beta_a \beta_b}{\beta_x^2 + \beta_y^2 + \beta_z^2} + \delta_{ab} = (\gamma - 1) \frac{v_a v_b}{v^2} + \delta_{ab} \end{equation}
where a and b are x, y or z and δab is the Kronecker delta.