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!


1 Answer 1


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.

  • $\begingroup$ Thanks for your answer! I'm currently on my phone but I'll check the data asap. But what I don't understand is why is the Lorentz Transformation for the time different than the time dilation? I know the time between my two events A and B is t'_B and I know that it has to do with the different place of our event B in our moving Frame. But why is there a difference when we have a clock in x'_A and a clock at x'_B. So when is it ok to use time dilation? $\endgroup$
    – Seen
    Aug 18, 2016 at 7:28
  • $\begingroup$ 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 $(t'_B - t'_C) = (t_B - t_C)/\gamma$. $\endgroup$
    – udrv
    Aug 18, 2016 at 9:34
  • $\begingroup$ In your case the two events take place at different locations in both frames. Otoh the rule for length contraction is that in the observing frame the length is measured at the same time (simultaneous events) between points at different locations and is compared to the rest length. $\endgroup$
    – udrv
    Aug 18, 2016 at 9:35

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