Meaning of general Lorentz transformations

Question

According to Wikipedia, the Lorentz transformations for two inertial frames are written:$$\begin{cases} t'=\gamma(t-\frac{\mathbf {r}_{\parallel }.\mathbf{v} }{c^{2}} )\;\;\;(*)\\\mathbf {r'}_{\parallel }=\gamma(\mathbf {r}_{\parallel }-\mathbf{v}t)\\\mathbf{r'}_{\perp } =\mathbf {r} _{\perp } \end{cases}$$

with: $\mathbf{r}=\mathbf {r} _{\perp }+ \mathbf {r} _{\parallel }\;\;, \;\mathbf{r'}=\mathbf {r'} _{\perp }+ \mathbf {r'} _{\parallel }$

If we assume that $\;\mathbf {r'}_{\parallel }=\mathbf{0}\;$, we have $\mathbf {r}_{\parallel }=\mathbf{v}t\;$and we integrate it into (*) we have$$t'=\frac{t}{\gamma}$$

In this case, is it a change of coordinates or notations?

I add a simplified diagram to bring out the physical meaning of the phenomenon that I want to study and for which I applied this transformation,the lines $(\mathbf{c}t'=\mathbf{r}')\perp (\mathbf{u}t'=\mathbf{L'})$,if we apply the Pythagorean theorem for this case, we have $\;\;t'= \frac{t}{\sqrt{1+\frac{u ^{2}}{c^{2}}}}$

the inverse transformations give $$\begin{cases} t=\gamma(t''+\frac{\mathbf {r''}_{\parallel }.\mathbf{v} }{c^{2}} )\;\;\;\\\mathbf {r}_{\parallel }=\gamma(\mathbf {r''}{\parallel }+\mathbf{v}t'')\\\mathbf{r''}_{\perp } =\mathbf {r} _{\perp } \end{cases}$$

with: $\mathbf{r}=\mathbf {r} _{\perp }+ \mathbf {r} _{\parallel }\;\;, \;\mathbf{r''}=\mathbf {r''} _{\perp }+ \mathbf {r''} _{\parallel }$

If we assume that $\;\mathbf {r}_{\parallel }=\mathbf{0}\;$, we have $\mathbf {r''}_{\parallel }=-\mathbf{v}t''\;$ so$$t''=\gamma t$$ what can be deduced directly from the diagram.

we can formalize ( see diagram) by a vector 'representation':$\;\mathbf{r}=\mathbf{v}t''+\mathbf{r''}=\mathbf{r''}+\mathbf{L''}$ or $$\mathbf{r}=(\mathbf{r''},\mathbf{L''})$$ the Minkowskian scalar product gives :$$\mathbf{r}^{2}=\mathbf{r''}^{2}-\mathbf{L''}^{2}$$ for $\mathcal{R}$: $\mathbf{c}=(\mathbf{c},0)$ and for $\mathcal{R'}$ $$\mathbf{r}=\mathbf{c}t=(\mathbf{c}''t'',\mathbf{L''})=\gamma(\mathbf{c}''t,\mathbf{v}t)$$ with $||\mathbf{c}||=||\mathbf{c}''||=c$

we simplify by $t$ : $$\mathbf{c}=\gamma(\mathbf{c}'',\mathbf{v})$$ from which we can deduce the relation :$$1=\gamma^{2}\left (1-\frac{v^{2}}{c^{2}}\right)$$

To obtain the first case of LTs, we must replace $ut'$ by $ut$ in the diagram to get: $\;t'=t/\gamma\;$ it is the frame of reference $\mathcal{R} $ which is moving!

How to formalize the case of the diagram where $\mathcal{R'}$ is in motion? (the observer in $\mathcal{R'}$ encounters the signal or event perponducularly).

Answer 1

Let's look at the bigger picture first. A Lorentz transformation maps a set of coordinates $(t,\mathbf r)$ to another set of coordinates $(t',\mathbf r')$. More specifically, if an observer $\mathcal O$ assigns coordinates $(t,\mathbf r)$ to a point in spacetime, the Lorentz transformation tells you what another observer $\mathcal O\,'$ will assign as coordinates $(t',\mathbf r')$ to that same point in spacetime. The Lorentz transformation assumes that $\mathcal O\,'$ is moving with speed $v$ with respect to $\mathcal O$.

Now that we have that out of the way, let's see what situation you have described. By setting $\mathbf r_{\parallel}$ to zero, you have selected a subset of all possible points in spacetime. The time coordinate and perpendicular position are unspecified.

To make things more tractable, let's also set $\mathbf r_{\perp}=0$ and assume $\mathbf r'$ is aligned with the x'-axis. This gives us the 1D Lorentz transformation:

$$\cases{ t'=\gamma\left(t-x\dfrac{v}{c^2}\right)\\x'=\gamma(x-vt)}$$

Let's parametrize the path in primed space using $X'(\tau)=(t'(\tau),x'(\tau))=(\tau,0)$. What you have shown is that this path $(t',x')=(\tau,0)$ for observer $\mathcal O\,'$ corresponds to $X(\tau)=(t,x)=(\gamma \tau, \gamma v\tau)=(t,vt)$ for observer $\mathcal O$.

Does that make sense? Yes! The path in primed space is that of a stationary observer, i.e. you describe the path that $\mathcal O\,'$ takes. What you have shown is that the time of a stationary observer $\tau$ is related to the time measured by a moving observer by $t=\gamma \tau$. The time of a stationary observer is also called the proper time and is usually denoted by $\tau$. See what I did there?

So to repeat: you have changed coordinates, but only for a subset of all space. The Lorentz transformation can transform all points in spacetime. You have restricted this to points with $\mathbf r_{\parallel}'=0$. If you interpret these points as a worldline, you describe something that is travelling with the observer.

Answer 2

I will call these observers $S$ and $S'$. The condition $\mathbf{r}_{\parallel}=\mathbf{v}t$ implies that $S$ is measuring a point in space that travels with velocity $\mathbf{v}$, with the same direction and speed as $S'$.

Since this point is co-local with $S'$, the time-dilation effects are being experienced for $S$. If we assume there is a clock moving at velocity $\mathbf{v}$, then its tick period will be longer for $S$ than for $S'$. This is what $t'=t/\gamma$ is saying, it's simply $t=\gamma t'$.

Notice that under $\mathbf{v}\to-\mathbf{v}$ and $x^\mu\to x'^{\mu}$ (and vice-versa) you would get the usual $t'=\gamma t$ since now the clock is stationary with respect to the original frame of reference.

Answer 3

We can formalize this case by analogy:

$\;\mathbf{r}=\mathbf{u}t'+\mathbf{r'}=\mathbf{r'}+\mathbf{L'}$ or $$\mathbf{r}=(\mathbf{r'},\mathbf{L'})$$ the Euclidean scalar product gives ( we do not reverse the relationship like...): :$$\mathbf{r}^{2}=\mathbf{r'}^{2}+\mathbf{L'}^{2}$$ for $\mathcal{R'}$ $$\mathbf{r}=\mathbf{c}t=(\mathbf{c}'t',\mathbf{L'})=\kappa(\mathbf{c}'t,\mathbf{u}t)$$ with $||\mathbf{c}||=||\mathbf{c}'||=c$

we simplify by $t$ : $$\mathbf{c}=\kappa(\mathbf{c}',\mathbf{u})$$ from which we can deduce the relation :$$\kappa=\frac{1}{\sqrt{1+\frac{u^{2}}{c^{2}}}}$$