# Meaning of general Lorentz transformations

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).

• I'm not sure I understand completely your question. The list of equations you've written is a coordinate transformation, by assuming a certain configuration of the event, i.e., $\mathbf{r}'_{||} = 0$, you arrive at $t' = t/\gamma$. Of course, for general $\mathbf{r}'_{||}$, the list of equations should be the coordinate transformations you need. The expression $t' = t/\gamma$ is a coordinate transformation, not a change of notation. Commented Jul 18 at 11:59
• I have never read that there is a contraction of time, which signifies the relationship in question. Commented Jul 18 at 12:06
• It would be instructive for you to find these transformed coordinates on a Minkowski spacetime diagram.
– rob
Commented Jul 18 at 12:13
• @TheTiler You never heard of time dilation? It's the same phenomenon as relativistic length contraction. They're both just ordinary foreshortening due to hyper (4 dimensional) rotation. Commented Jul 18 at 12:13
• I think it is important to point out that in one case the OP "looks like" he got "time contraction", but is actually just doing a inverse Lorentz transform, whereas we usually present a Lorentz transform of time dilation. That is, the OP likely does not understand that both cases are time dilation, just seen in one frame or the other. Commented Jul 18 at 18:34

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.

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.

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 :$$\kappa=\frac{1}{\sqrt{1+\frac{u^{2}}{c^{2}}}}$$