I try to derive the Lorentz transformation of a Lorenzt transformation frame an inertial frame $O$ to the frame $O'$ of a moving particle at constant speed v. We have four vectors $\textbf{x}'=\Lambda \textbf{x}+\rho$. Without loss of generality, we assume $\rho=0$. We have $$d\textbf{x}'=\Lambda_{j}^id\textbf{x}^j=\Lambda_{0}^idt$$ since the spatial $ds=0$ in $O$. We just write the equation above explicitly separately in space and time. Then my text book claims


I want to ask why and how the velocity $v$ is related to the space and time with prime? For me, it makes more sense to say $v_i=\frac{ds_i'}{dt}$, i.e. the time in $O$. Personally, I feel unclear about two different times. And what does the differentials represent here?

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  • $\begingroup$ Which textbook are you using? $\endgroup$ – Luthien Dec 30 '18 at 19:42
  • $\begingroup$ And what does the differentials represent here? They're simply small changes in the variable. You can read them as "a little bit of..." For example, dt is a little bit of time. In the context of special relativity, you can replace all of the d's with $\Delta$'s, and it makes no difference. $\endgroup$ – user4552 Dec 31 '18 at 1:44
  • $\begingroup$ @BenCrowell Thanks, but why $dx'\neq0$? $dx'$ is also still when measuring inside that primed system? $\endgroup$ – CO2 Dec 31 '18 at 9:31

You have to use four vectors to properly apply lorentz transformations (since these are the objects that transform under the vector representation of the Lorentz group). Namely, the Lorentz transformation is really given by

$$ x'^\mu = \Lambda^\mu_{\ \nu} \ x^\nu $$

(I'm not going to use differential formalism at first--see edit). This is how coordinates transform component wise. If you want to know the velocity in the primed frame you take $\frac{d}{dt'}$ so that

$$ \frac{d x'^\mu}{dt'} = \frac{d}{dt'}\Lambda^\mu_{\ \nu} \ x^\nu $$

Next recall that if we take the primed observer to $\frac{dt'}{dt} = \frac{1}{\gamma(\beta)}$ for observers in standard configuration. It follows that

$$ \frac{d x'^\mu}{dt'} = \gamma(\beta)\frac{d}{dt}\Lambda^\mu_{\ \nu} \ x^\nu $$

And since the $\Lambda$ only depends on $\beta$ implicitly we can pass the time derivative through just yielding what we expected. Namely,

$$ \frac{d x'^\mu}{dt'} = \Lambda^\mu_{\ \nu} \ \gamma(\beta)\frac{d x^\nu}{dt}.$$

Edit: I will address your confusion you mentioned in the final sentence. Everytime you get confused about what the primed and unprimed coordinates just think about it like this. The primed coordinates of an event are *what you would say they were if you were $S'$ carrying your $S'$ axes with you. (Similarly for the unprimed)

Example: Take a clock at rest in $S'$ (say (s)he is wearing a watch). Between the two events &second hand strikes 1 second and second hand strikes 2 seconds. What would $\Delta t'$ be? Well, it would be $\Delta t' = t_2' - t_1' = 2 -1 = 1$ second.

What would $\Delta x'$ be? Remember, $S'$ is carrying their axes with them and don't know that they are moving. So, does the watch move in $S'$? no it's been on his or her wrist the whole time. Therefore, $\Delta x' = 0$ between the two ticks.

To address your concern about the differentials the exact same logic applies, if the time interval arbitrarily small then we write it as a differential $dt'$. (likewise for $dx').

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  • $\begingroup$ You said $\Delta x'=0$ but in the derivation $dx'/dt'\neq0$?? $\endgroup$ – CO2 Dec 31 '18 at 9:34

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