Learn Relativity Knowledge Memo

This article is a memo after study, the level is limited, if there is an error, please understand.
The parameters without ' are the parameters of the static reference frame X, and the parameters with ' are the parameters of the moving reference frame X'.

1. Lorentz transformation

Lorentz transformations can be seen as axioms, so let's start there.
Let $$k=\sqrt{1-u^2/c^2}$$
式1:$$x^{\prime }=\frac{x-ut}{k}$$
式2:$$t^{\prime }=\frac{t-ux/c^2}{k}$$

1. First analyze formula 1

Equation 1 means that the origins of the coordinates of the two reference frames coincide at the beginning, and from the point of view of the villain standing still on the X reference frame, when the X' frame of reference starts to move in a straight line at a uniform velocity u, it is known that a certain point is in the X frame of reference. The position x and the running time t of the reference frame X' (the time viewed by X) can be used to calculate the corresponding position x' of the point in the X' frame of reference at this time (t).

2. Derivation of scale effect by Lorentz transformation

Scale effect means that when a ruler with a length of l moves along the direction of the ruler scale, for a stationary viewer, the length of the ruler will become shorter and become l ∗ kl* $l\ast length\; of\; k$ .

How to use the Lorentz transformation to derive the scaling effect?

In formula 1, when t=0, when the coordinate origins of the two reference systems coincide, the coordinates of the X reference system are $l\ast At\; point\; k$ , the reading in the X' frame of reference is l. That is, the ruler whose length is l in the X' frame of reference looks shortened in the X frame of reference, and becomes$l\ast k$。

3. Reanalysis formula 2

Since the formula 2 does not contain x', it can be considered that the X' frame of reference has no scale, and naturally it does not matter that it is moving (because a moving smooth stick cannot be seen to be moving), but exists everywhere on the X' frame of reference and can be seen The "alarm clocks" who get up and do not exercise are turning. Taking x and t, the alarm clock reading of the corresponding point of point x on the X' frame of reference can be obtained at time t. If the origin of x=0 is taken, the corresponding reading at x'=0 is $t^{\prime }=t/k$。

看$ux/c^2$ items. This item tells us that at any moment, from the perspective of the observer on the X frame of reference, the time at different positions in the X' frame of reference is different. If "alarm clocks" are placed on various points in the X' frame of reference, the time of each alarm clock is different. In the positive direction of u, the farther the position is, the more "past" the alarm clock time is. In the negative direction of u, the farther the position is, the more "future" the alarm clock time will be.

It's a bit counterintuitive here. In the negative direction of u, the farther the position is, the more "future" the alarm clock time will be. This means that things in the future have already happened in the distance at this time. So can we now see future events? This is not acceptable. Look at $ux/c^2$项。
$$\frac{ux}{c^2}=\frac{u}{c}\frac{x}{c}$$
The first term u\c can be seen as a ratio that is always less than 1. The second term x\c is the time required for light to travel x distance. Multiplying the two is a number that is always less than the time it takes light to travel x distance. What happens in the future at a distance of x takes the time required for light movement x distance to reach the origin, and then is captured by the viewer. At this time, for the viewer, it is not the future but the past.

4. Derivation of clock slow effect by Lorentz transformation

The clock slow effect means that for a stationary viewer, a moving alarm clock "turns" slower than a stationary alarm clock.

At first glance, from formula 2, $t^{\prime }=t/k$ ,
just the opposite of the slow clock effect. It is easy to make mistakes here and requires careful analysis.
In formula 2, let$x=0$ , get$t^{\prime }=t/k$ . This formula means that the viewer divides the reading of the static clock by k at the origin of the X frame of reference, which is the reading of the alarm clock at that point of the X' frame of reference. In this scene, for the viewer of the X frame of reference, the alarm clock does not move, which is inconsistent with the scene in the slow clock effect.

The correct way of thinking is to set x=ut, that is, the clock under consideration is moving at speed u. Insert into formula 2, $t^{\prime }=t\ast k$ , the correct clock-slow effect formula is obtained.

Come back and analyze Equation 2 further.
As previously analyzed, at the origin of the X reference system, there is a relationship t'=t/k, and t' is relatively small. And at x=ut, t'=t*k, t' is larger. Then let x=vt, t=t', we can get
$v=\frac{(1-k)c^2}{u}$, when v is equal to this value, t=t', the two are equal.

2. Use the Lorentz transformation to process the scene of the accelerated frame of reference

Refer to an after-school question in Feynman's lecture notes.
It is known that the spaceship at rest in the initial state is subjected to a constant acceleration g', and the spaceship travels for time t for a stationary observer, find the distance x traveled by the spaceship, the speed v of the spaceship, the acceleration g of the spaceship, and the spaceship t'.

1. Find the relationship between acceleration a and a'.

For the spaceship, there is formula 3:
$$g^{\prime }d{t}^{\prime }=d{v}^{\prime }$$
The formula 4 is obtained from the velocity synthesis formula:
$$v+dv=\frac{v+d{v}^{\prime }}{1+\frac{etc\_{\_}^{\prime }}{c^2}}$$
tidy up and omit $\frac{dvd{v}^{\prime }v}{c^2}$After the term, we get formula 5:
$$gdt=dv=g^{\prime }d{t}^{\prime }\ast k^2=g^{\prime }dt\ast k^3$$
The relationship formula 6 of the acceleration can be obtained:
$$g=g^{\prime }\ast k^3$$

2. Find the speed of the spaceship

由式5，$$g^{\prime }dt=\frac{dv}{(1-\frac{v^2}{c^2}{)}^{\frac{3}{2}}}$$
For the integral on the right, considering that v\c is greater than 0 and less than 1, we can set v\c=sina.
$${\int }_{}^{}(1-\frac{v^2}{c^2}{)}^{\frac{-3}{2}}dv=c{\int }_{}^{}(1-\frac{v^2}{c^2}{)}^{\frac{-3}{2}}d\frac{v}{c}=c{\int }_{}^{}(1-si{n}^{2}a{)}^{\frac{-3}{2}}dsina$$
because$dsina=cosada$ , finishing the right side is equal to$$c{\int }_{}^{}se{c}^{2}ada$$
table lookup$se{c}^{}$The integral of${}^2$$a$$d$$a$$tana$ , the original formula after arrangement is formula 7:
$$v=\frac{g^{\prime }t}{\sqrt{1+\frac{(g^{\prime }t{)}^2}{c^2}}}$$

3. Find the flight distance

From formula 7, flight distance:
$${\int }_{}^{}vdt={\int }_{}^{}\frac{g^{\prime }t}{\sqrt{1+\frac{(g^{\prime }t{)}^2}{c^2}}}dt$$
令$\frac{g^{\prime }t}{c}=u$，原式载
$$\frac{c^2}{g^{\prime }}{\int }_{}^{}\frac{u}{\sqrt{1+u^2}}du$$
再令$u=tana$ , the original formula is
$$\frac{c^2}{g^{\prime }}{\int }_{}^{}\frac{sin}{co{s}^{2a}\_}da$$
After looking up the table and integrating, the original formula is$$\frac{c^2}{g^{\prime }cosa}$$
由$\frac{g^{\prime }t}{c}=u$和$u=tana$ , the flight distance is formula 8:
$$x=\frac{c}{g^{\prime }}\sqrt{c^2+(g^{\prime }t{)}^2}{|}_{t1}^{t2}$$

4. Find the flight time t' of the spaceship

由$d^{\prime }t=\sqrt{1-\frac{v^2}{c^2}}dt$和$v=\frac{g^{\prime }t}{\sqrt{1+\frac{(g^{\prime }t{)}^2}{c^2}}}$，
得$$d{t}^{\prime }=\frac{c}{\sqrt{c^2+(g^{\prime }t{)}^2}}dt$$
令$u=\frac{g^{\prime }t}{c}$得$$d{t}^{\prime }=\frac{c}{g^{\prime }}\frac{1}{\sqrt{1+u^2}}du$$
令$u=tana$以及$dtana=se{c}^{2}ada$得
$$d{t}^{\prime }=\frac{c}{g^{\prime }}secada$$
table look-up integral
$$t^{\prime }=\frac{c}{g^{\prime }}ln(seca+tana)$$
由$\frac{g^{\prime }t}{c}=u$和$u=tana$，得
$$t^{\prime }=\frac{c}{g}ln(\frac{g^{\prime }t+\sqrt{c^2+(g^{\prime }t{)}^2}}{c})$$

5. Summary

Under the initial conditions of time and distance of 0, the spacecraft is subjected to acceleration g', then
$$g=g^{\prime }\ast k^3$$
$$v=\frac{g^{\prime }t}{\sqrt{1+\frac{(g^{\prime }t{)}^2}{c^2}}}$$
$$x=\frac{c}{g^{\prime }}\sqrt{c^2+(g^{\prime }t{)}^2}{|}_{t1}^{t2}$$
$$t^{\prime }=\frac{c}{g}ln(\frac{g^{\prime }t+\sqrt{c^2+(g^{\prime }t{)}^2}}{c})$$
take t=5 years ($5\ast 365\ast 24\ast 3600$ seconds), g'=9.8, c=300000000 Calculate it:
the running speed of the spaceship is 0.9817c.
The flight distance is 4.12 light-years.
The flight time on the spacecraft was 2.273 years.