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To demonstrate Lorentz transformations mathematically, we assume that time is a dimension (via linear transformations, etc.), what physical argument requires us to do this? Details for the reason for …

According to Lorentz transformations (two cases: the moving observer approaches or moves away from the body ): $$\Delta x'=L'=\gamma(\Delta x\pm v\Delta t)=\gamma \Delta x \left(1\pm\frac{v}{c}\right) …

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 revers …

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 …

See: wikipedia:.... $\mathbf{r}=\mathbf {r} _{\perp }+ \mathbf {r} _{\parallel }\;\;, \;\mathbf{r'}=\mathbf {r'} _{\perp }+ \mathbf {r'} _{\parallel }$ then the transformations are:$$\begin{cases} t'= …

There is another way of looking at the problem: we replace the coordinates by the speeds in the Lorentz transformations (LT), i.e. $v'=\gamma(v-at)\;\;,\;\;t'=\gamma(t-\frac{ac}{a_{l}^{2}})$ we assume …

The metric for rotating reference frame is:$$ ds^{2}=\left(c^{2}-\omega^{2}r^{2}\right)dt^{2}-2\omega r^{2} d\phi dt-dr ^{2}-r ^{2}d\phi ^{2}\;\;\;\; \,(*)$$ with $\;r\omega=v$ As in any stationary fi …

By definition, the four-vector is:$$X=(ct,\vec{r})$$ a four-velocity by $$V=\frac{dX}{d\tau}=\left(c\frac{dt}{d\tau},\frac{d\vec{r}}{d\tau}\right)=\gamma(c,\vec{v})$$ The Minkowskian scalar product gi …

Suppose we have two stars A and B which explode into supernovas at the same time (~taps), all observers in motion or at rest closer to A see the explosion of A into supernovas before B which is still …

In a right triangle (x,y,r), we have : $r^{2}=x^{2}+y^{2}\;\;\;\;(1)$ $$x^{2}=r^{2}-y^{2}=(r-y)(r+y)=r^{2}\left(1-\frac{y}{r}\right)\left(1+\frac{y}{r}\right)$$ $$\frac{r^{2}}{x^{2}}=\frac{1}{(1-\frac …

First $P'=\gamma'mv'$ instead of $F'=...$ afterwards, we apply the Lorentz transformations for a 4-vector force $\vec{\mathbf{F}}[\gamma\mathbf{F},\frac{\gamma}{c}(\mathbf{F}\mathbf{v})]$: $$F'_{x}=\f …

Geometric representation of an event that happens in M, $x_{M}=0$ and $ct=5s$​​ (explosion of a firecracker) : the wave only arrives at the observer at rest 5 seconds after the meeting of the two fra …

we have $-x_{A}=x_{B}=x\;$ and $\;\;t'=\gamma\left(t-\frac{vx}{c^{2}}\right)$, the difference is $$\Delta t' =\gamma \,\frac{v}{c^{2}}\,\left(-x_{A}+x_{B}\right)=\gamma \,\frac{2vx}{c^{2}}$$ If we re …

I will try with the diagram below, we suppose that the container ijfg is filled with water, the light crosses this container of the face $f$ towards the face $g$ with a speed $v$ and put a time $t$ to …

A simple calculation gives: $$p'c+mc^{2}=E'=\gamma\, m'c^{2}=\sqrt{(p'c)^{2}+(m'c^{2})^{2}}$$ Which give:$$(p'c)^{2}+(mc^{2})^{2}+2p'c\;mc^{2}=(p'c)^{2}+(m'c^{2})^{2}$$ $$(m'c^{2})^{2}=(mc^{2})^{2}+2h …