Lorentz Transformation In-Phrases Of Hyperbolic Rotation

Lorentz transformation is how an observer sees an occasion when he's shifting on different factors of spacetime. ローレンツ変換 終焉 would like to note here, in particular (or, normal) relativity, our universe and the spacetime are interchangeable phrases (For why, see this). However, in actuality, the same observer can’t be in several locations at a time to see the same occasion. Thus, we need to have not less than two observers. One owns his coordinate at $(x,t)$ and another at $(x',t')$ is shifting with respect to the primary observer $(x,t)$ with velocity $v$. We're supposing our observers to maneuver in (1+1) dimensions as a result of to make our proposal easy. (1+1) dimensions imply one time dimension and one space dimension. However, we will prolong to a movement the place we have to deal with all three house coordinates. I’m letting this job to you. Please let me know within the comment on how you might do it. So, the Lorentz transformation offers a relation between them as

$x' = \gamma (x - vt)$, and
$t' = \gamma (t - \fracvxc^2)$

(say equation 1) the place, $\gamma = \frac1\sqrt1- \fracv^2c^2$ is a Lorentz factor, and $c$ is the velocity of light. Chances are you'll conscious or not that the matrix is another approach of representing equations. So, in matrix type, we've got our Lorentz transformation (say LT for brief) as

$\startpmatrixx' \\ t'\finishpmatrix = \beginpmatrix\gamma & -v\gamma\\ \frac-v\gammac^2 & \gamma\endpmatrix\startpmatrixx \\ t\finishpmatrix$

This implies, in the event you multiply proper-hand facet matrices then, you can get the equation 1. If we suppose velocity of light $c$ to be 1 then,

$\beginpmatrixx' \\ t'\finishpmatrix = \beginpmatrix\gamma & -v\gamma\\ -v\gamma & \gamma\finishpmatrix\beginpmatrixx \\ t\finishpmatrix$.

If you are curious why I set $c = 1$ then, learn this publish. We wish to see LT when it comes to hyperbolic trigonometric features so, we have an interest in the matrix ($L$ say)

$L = \beginpmatrix\gamma & -v\gamma \\ -v\gamma & \gamma\finishpmatrix = \beginpmatrix L_11 & L_12 \\ L_21 & L_22\endpmatrix$.

So, for those who do square of $L_11$ and $L_12$ (or, $L_22$ and $L_21$) and then subtract them so that you simply get one. i.e.

$L_11^2 - L_12^2$
$ = \gamma^2 - (-v\gamma)^2$

$ = (\frac1\sqrt1- v^2)^2 - (-v \times \frac1\sqrt1- v^2)^2$
$ = \frac11- v^2 - \fracv^21- v^2$

$ = \frac1- v^21- v^2 = 1$
$\therefore L_11^2 - L_12^2 = \gamma^2 - (-v\gamma)^2 = 1$.

Similarly, $L_22^2 - L_21^2 = \gamma^2 - (-v\gamma)^2 = 1$.
This type of identification can be present in hyperbolic trigonometric function which I mean this

$cosh^2(\eta) - sinh^2(\eta) = 1$. So, we set $cosh(\eta) = \gamma$ and $sinh(\eta) = -v\gamma$. Now, we put these in our matrix $L$ and thus

$\startpmatrixx' \\ t' \endpmatrix = \beginpmatrixcosh(\eta) & sinh(\eta) \\ sinh(\eta) & cosh(\eta) \finishpmatrix\startpmatrixx \\ t \finishpmatrix$.

This is a Lorentz transformation when it comes to hyperbolic rotation.
Now, we may have two questions:

1. Why will we make when it comes to hyperbolic trigonometric functions? The answer is kind of easy. As a result of we can have more mathematical instruments (or, flexibility) once we working in hyperbolic trigonometric capabilities than in algebraic perform in order that we can quickly remedy the problem.
2. Why our $L$ is a hyperbolic rotation matrix? Let me make you clear, in the matrix $L$, we only have $v$ which manipulate our transformation. This $v$ is the velocity of second observer $(x',t')$ measured by first observer $(x,t)$. This means if second observer transfer in a velocity $v$ then, he will see all the points of spacetime which is seen by the primary observer will shift (strictly talking, linearly transform) in the direction of the opposite route of him (i.e. second observer). But, if we make our spacetime with hyperbolic coordinates then, all the points rotate on the hyperbola along by some angle. So, the mathematics for this case is Lorentz transformation in terms of hyperbolic rotation.

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