Rindler Motion in Special Relativity, Part 2: Rindler Coordinates - Comments

  • #26
stevendaryl
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The definition of proper acceleration is ##\dot{u}_a =\nabla_b u_a u^b## and for the metric ##diag(-g_{00}, 1,1,1)## the acceleration of ##u^a=1/\sqrt{g_{00}}\ \partial_t## is just what you got in your post, ##-\partial_x g_{00}/(2\,g_{00})## ( in the x direction ).

I don't know if that helps or hinders.
Yes, so in the particular case in which only ##g_{00}## is different from ##\pm 1##, it works like a potential. Okay. Good to know when it works.
 
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  • #27
Numerical Simulation with Riedler Coordinates:


The Equations of Motions (geodetic Equations) are:
$$\frac{d^2 x^\alpha}{d t^2}=-\Gamma^\alpha_{\beta\gamma}\frac{d x^\beta}{d t}\frac{d x^\gamma}{d t}\quad \quad (1)$$
With c=1 we get for the Riedler Coordinates :
$$\begin{bmatrix}
t_I\\
x_I \\
y_I \\
z_I \\
\end{bmatrix}=
\begin{bmatrix}
\left(x+\frac{1}{a}\right)\sinh(a\,\tau)\\
\left(x+\frac{1}{a}\right)\cosh(a\,\tau)\\
y\\
z
\end{bmatrix} $$
$$\Rightarrow $$

$$\frac{d^2 s^\alpha}{d \tau^2}=- \left( {\frac {d}{dt}}\tau \left( t \right) \right) ^{2}{x(t)}^{2}{a}^
{2}-2\, \left( {\frac {d}{dt}}\tau \left( t \right) \right) ^{2}x(t)a-
\left( {\frac {d}{dt}}\tau \left( t \right) \right) ^{2}+ \left( {
\frac {d}{dt}}x \left( t \right) \right) ^{2}\quad \quad (3)
$$
and the Metric g is:
$$g=\begin{bmatrix}
-\left(x\,a+1\right)^2 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
So the EQM with equation (1) and the metric g are :

Equation (4)
$${\frac {d^{2}}{d{t}^{2}}}\tau \left( t \right) +2\,{\frac {a \left( {
\frac {d}{dt}}x \left( t \right) \right) {\frac {d}{dt}}\tau \left( t
\right) }{x \left( t \right) a+1}}=0\\
{\frac {d^{2}}{d{t}^{2}}}x \left( t \right) + \left( {\frac {d}{dt}}
\tau \left( t \right) \right) ^{2}a \left( x \left( t \right) a+1
\right) =0\\
\frac{d^2}{d t^2}y(t)=0\\
\frac{d^2}{d t^2}z(t)=0\\
$$
Numerical Simulation

To do the simulation we need the initial conditions for the differential equations (4). These must satisfied also the equation (2).
Data :

$$a=10\,, x(0)=2\,,\frac{d x}{d t}\big|_{t=0}=0.1 \,, \frac{d y}{d t}\big|_{t=0}=0.0
\,, \frac{d z}{d t}\big|_{t=0}=0.0 $$
we can calculate with equation (3) and
$$\frac{d^2 s}{d \tau^2}=0\quad \Rightarrow\quad \frac{d \tau}{d t}\big|_{t=0}=\pm 0.004761904762 $$

Result:

can i put hier JPEG Format?
 
Last edited:
  • #28
58
1
Yes, you're right.

[Edit]

I guess there are three different types of acceleration, with components given by:
  1. ##\frac{d^2 x^\mu}{dt^2}##
  2. ##\frac{d^2 x^\mu}{d\tau^2}##
  3. ##\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}##
I guess #3 is "proper acceleration", but I'm not sure whether "coordinate acceleration" means 1 or 2.
For the 2nd one,
$$\frac{d^2x^\mu}{d\tau^2}=\frac{dU^\mu}{d\tau}$$
Generally ##dU^\mu## does not give the relative velocity, so we need something can be directly subtracted, i.e. ##\theta##
For the uniformly accelerating object, the proper acceleration is defined as
$$\alpha=\frac{d\theta}{d\tau}$$
 

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