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The speed composition formulas give: $\theta =\arctan \left(\frac{V'_{y}}{\gamma V}\right)$ If we replace the cylinder with the light coming from a star, i.e. $V' = -c$, we can calculate the aberratio …

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 …

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 relativistic theory, the proper time τ of an object is the time measured in "the" frame of reference of this object, i.e. in a frame of reference where it is motionless. For photons, there is no fr …

The derivative of the momentum with respect to time is the force acting on the particle. If the velocity of the particle varies only in deraction, which means that the force is perpendicular to the ve …

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 …

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'= …

See formula (39) : https://www.scielo.br/j/rbef/a/wh9hgMSRTJzvws8ZxzL8NNp/?format=pdf&lang=en with :$\beta= tanh(\varphi)\;\;,\gamma=cosh(\varphi)\;\;, \beta\gamma = sinh(\varphi)$ https://en.wikipedi …

In the following link https://en.wikipedia.org/wiki/Lorentz_transformation#/media/File:Lorentz_boost_any_direction_standard_configuration.svg , we have: $\ \mathbf{r}'=\mathbf{r}_{\bot} +\gamma( \mat …

From 'reality', if a sound wave is emitted at a point P (shock wave), do we hear the explosion once or twice, even if we are moving with a speed v, change the direction of propagation of the wave, it …

We know that the K-G equation is deduced from the Einstein relation: $E^{2}=m^{2} +\vec{p}^{2} \;\;\;\;$ (with $c=1$) It is known that :$E^{2}=\frac{m^{2}}{1-\beta^{2}}=\left(\frac{m}{1-\beta}\righ …

The direct Lorentz transformations are : $$ \begin{cases} x'=\gamma \left(x-vt\right)\;\;\;\;\;(1)\\ t'=\gamma \left(t-\frac{v}{c^{2}}x\right)\;\;\;\;(2)\end{cases}$$ (1) and (2) gives $$\begin{cases …

we know that: $\;\;E^{2}=(mc^{2})^{2}+p^{2}c^{2}\;\;\;,$ $\;\;\;p=\gamma mv$ $E^{2}=m^{2}c^{4}+\gamma^{2}m^{2}v^{2}c^{2}=m^{2}c^{4}\left(1+\gamma^{2}\frac{v^{2}}{c^{2}}\right)$ $E^{2}=\gamma^{2}m^{2}c …

In the formula$$\Delta t'=\gamma\left(\Delta t-\frac{U\Delta x}{c^{2}}\right)$$ we suppose that two events occur in the same place, i.e.$\;\Delta x=0$ we find in the moving reference frame $\mathcal{R …

The problem comes from the Euclidean representation of the orthogonality of two vectors in Minkowskian space: the scalar product of two vectors x,y with coordinates $x_{i},y_{i}$ (in 2D) is defined in …