Lorentz Transformation In-Terms Of Hyperbolic Rotation

Lorentz transformation is how an observer sees an event when he's moving on different points of spacetime. I would like to notice right here, in particular (or, general) relativity, our universe and the spacetime are interchangeable terms (For why, see this). But, in reality, the same observer can’t be in several locations at a time to see the identical occasion. Thus, we have to have not less than two observers. One owns his coordinate at $(x,t)$ and one other 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 because to make our proposal easy. (1+1) dimensions imply one time dimension and one house dimension. But, we are able to lengthen to a motion the place we need to deal with all three area coordinates. I’m letting this job to you. Please let me know in the touch upon how you might do it. So, the Lorentz transformation gives 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. You might conscious or not that the matrix is another means of representing equations. So, in matrix form, we now have our Lorentz transformation (say LT for brief) as

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

This implies, if you multiply proper-hand aspect matrices then, you will get the equation 1. If we suppose velocity of gentle $c$ to be 1 then,

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

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

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

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 \occasions \frac1\sqrt1- v^2)^2$

$ = \frac11- v^2 - \fracv^21- v^2$

$ = \frac1- v^21- v^2 = 1$

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

Equally, $L_22^2 - L_21^2 = \gamma^2 - (-v\gamma)^2 = 1$.

Such a id might be present in hyperbolic trigonometric perform which I imply 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

$\beginpmatrixx' \\ t' \endpmatrix = \startpmatrixcosh(\eta) & sinh(\eta) \\ sinh(\eta) & cosh(\eta) \endpmatrix\beginpmatrixx \\ t \endpmatrix$.

This can be a Lorentz transformation by way of hyperbolic rotation.

Now, we may have two questions:

1. Why can we make by way of hyperbolic trigonometric features? The answer is kind of easy. Because we are able to have more mathematical tools (or, flexibility) when we working in hyperbolic trigonometric functions than in algebraic operate in order that we will shortly remedy the problem.

2. Why 相対性理論 崩壊 $L$ 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 of the factors of spacetime which is seen by the primary observer will shift (strictly talking, linearly rework) towards the opposite course of him (i.e. second observer). But, if we make our spacetime with hyperbolic coordinates then, all of the points rotate on the hyperbola alongside by some angle. So, the mathematics for this case is Lorentz transformation in terms of hyperbolic rotation.