# Lorentz Transformation Example & Inversion

We have two frames of reference: frame $F$ and frame $F'$ such that $F'$ is moving at velocity $v$ in the positive $x$ direction of $F$. I get the overall idea of special relativity (time dilation, length contraction etc.) but I still have problems with working on the equations. Lets say our velocity is $0.7c$ of $F'$. In $F'$ is a flash lamp and in distance $d'$ (in $x'$ direction) is a detector installed. The time between emission and detection of the light pulse for the observer in $F'$ is $t' = 1.5 *{10}^{-8}s$.

I calculated the distance $d'$, which is approx $4.5m$ and the timeinterval between emission and detection of the light-pulse for the observer in the resting reference frame: $t = 2.1 * {10}^{-8}s$ with $t = \gamma * t'$.

Now I have to to calculate the spatial distance $x$ between the emission and the detection for the observer in $F$. I tried 2 separate ways. First with the space-time invariant $$(c*t')^2 - (d')^2 = (c*t)^2-(x)^2$$ and got $6.3m$ iirc. The second way was using the Lorentz Transformation, the obvious solution I guess, but I'm new to special relativity so forgive me. That's the point where I got confused with the equations.

"The Lorentz transform for the $x$ coordinate is given by: $$x'=\gamma (x-vt)$$ Everything on the RHS of this equation is measured in the frame $F$ and every thing on the LHS is measured in frame $F'$."

That was posted in another thread. What I don't get is the reference of the variables. So my $x'$ is my $d'$ of course. My distance in the moving frame. My $x$ is what I'm searching for. The distance the resting observer is seeing (?). My velocity is the velocity of the moving frame (so $0.7c$, right?) and now my time.. I thought I should use the time measured in the resting frame, so $t = 2.1 * {10}^{-8}s$. But that doesn't give me the same solution as the the space-time invariant. If I take $t'$ however, it works. But why? Did I mix something up? If $t'$ is the right time to use, why don't write the Lorentz Transformation differently?

Sorry but it really confused me - everywhere it is somewhat ambiguous and not absolutely clear what the variables are referring to. An example would help me a lot!

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.

• 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.
• The basic rule is that time dilation refers to events that take place at the same location in one frame. For instance if event C takes place in F' at $x'_C = x'_B = d$ and $t'_C = t'_A = 0$, then as observed in F it takes place at $x_C = \gamma d$ and $t_C = \gamma v d/c^2$. Then the durations to event B in the two frames compare as $t'_B - t'_C = d/c$ and $t_B - t_C = \gamma (1 + v/c) d/c - \gamma v d/c^2 = \gamma d/c = \gamma (t'_B - t'_C)$ or $(t'_B - t'_C) = (t_B - t_C)/\gamma$. – udrv Aug 18 '16 at 9:34