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

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1. Apr 10, 2018

### stevendaryl

Staff Emeritus
2. Apr 11, 2018

### sweet springs

I have questions.
I suppose
$$g(x)=-\frac{c^2}{x}$$
where x is distance from Event Horizon which is at backward of rocket and is shared by every part of the rocket. Gravity potential is by integrating this,
$$\phi(x)=c^2 log\frac{x}{x_0}+\phi_0$$

Any proper acceleration systems of different acceleration values are the same one. Their distance from E.H. ,x, make apparent difference i.e. values of proper acceleration.

Do these make sense?

Last edited: Apr 11, 2018
3. Apr 11, 2018

### stevendaryl

Staff Emeritus
I have to think about the significance of the potential that you are deriving. Normally, the significance of a potential is that the combination of kinetic energy plus potential is constant for an object dropped into that potential. Right off the bat, I don't see how that comes into play.

A freely falling object in the accelerated $R, \theta$ coordinate system will satisfy two conservation equations:

1. $\frac{1}{2} m (\frac{dR}{d\tau})^2 - \frac{1}{2} m R^2 (\frac{d\theta}{d\tau})^2 = -\frac{1}{2} m c^2$
2. $R^2 \frac{d\theta}{d\tau} = \mathcal{E}$

where $\tau$ is proper time for the object. The first equation is really just the equation $\frac{d\tau}{dt} = \sqrt{1 - \frac{1}{c^2}(\frac{dx}{dt})^2}$ squared and multiplied by $\frac{1}{2} mc^2$ to look more like a Newtonian conservation of energy equation. The second is a consequence of the fact that the "gravitational potential" is independent of the coordinate $\theta$. $\mathcal{E}$ is an arbitrary constant. If we assume that the object is dropped from "rest" at position $R_0$, then the relationship between $\theta$ and $\tau$ is initially: $\theta = \frac{c \tau}{R_0}$, so $\frac{d\theta}{d\tau} = \frac{c}{R_0}$. So the constant $\mathcal{E}$ is just $c R_0$. So the constant $\mathcal{E}$ turns out to be $c R_0$.

We can stick the second into the first to get a new conservation equation:

$\frac{1}{2} m (\frac{dR}{d\tau})^2 - \frac{1}{2} m c^2 \frac{(R_0)^2}{R^2} = -\frac{1}{2} m c^2$

This is an exact equation,

So in terms of $\tau$, the path of a freefalling object seems more like an inverse-square potential, instead of a logarithmic potential.

I don't immediately see how the potential you derive can be understood in terms of the accelerated coordinate system. Maybe I need to look at the nonrelativistic limit of the exact equation. I'll have to think about it more.

4. Apr 11, 2018

### stevendaryl

Staff Emeritus
So even though I'm not exactly sure what the significance of your "gravitational potential" is, you're absolutely right that all accelerated rockets look the same. The only difference from one case to another is where the "height" is measured from.

5. Apr 11, 2018

### sweet springs

Metric of Rindler system is

$$ds^2=(\frac{x}{L})^2 c^2 dt^2 - dx^2 -dy^2 -dz^2$$
where x is distance from E.H. , L=c^2/g(L) where g(L) is gravity or acceleration at x=L. proper time at x=L is chosen as world time.

Approximately,
$$\phi(x)=(g_{00}-1)\frac{c^2}{2}=[(\frac{x}{L})^2-1]\frac{c^2}{2}$$
$$g(x)=-\nabla \phi(x)=-x\frac{c^2}{L^2}$$

This is a acceleration in world time. In proper time there we put square of coefficient
$$d\tau(x)=\frac{x}{L}d\tau(L)=\frac{x}{L}dt$$

$$proper \ g(x)=-(\frac{L}{x})^2\nabla \phi(x)=-\frac{c^2}{x}$$

Integration I did seems wrong due to the fact proper acceleration changes according to g_00(x) in different x. Integration of proper acceleration seems meaningless.

Last edited: Apr 11, 2018
6. Apr 11, 2018

### sweet springs

$$proper\ g(x)\ *\ small\ proper\ time\ at\ x= -\frac{c^2dt}{L}$$

It is relative speed to instantaneous IFR after small time. As it does not depend on x, local instantaneous IFRs are integrated to a common global instantaneous IFR. Rindler system holds global instantaneous IFRs and changes them every moment of time. The above formula gives relative speed of Old instantaneous IFR and New instantaneous IFR.

Say we divide rocket into parts according to x and each part move with proper g(x), Bell's paradox threads connecting parts wil not be torn apart. If rocket body is made of Born rigid material, not only engine but all the parts of rocket follow proper g(x).
Stress works on body so that proper g(x) take place everywhere on the body. Born rigidity body can do it with infinite elastic coefficient so no change of proper length.

Last edited: Apr 11, 2018
7. Apr 12, 2018

### Staff: Mentor

There is no such thing as "infinite elastic coefficient". The finite speed of light places finite limits on material properties (basically that the speed of sound in the material cannot exceed the speed of light).

8. Apr 12, 2018

### Staff: Mentor

I'm not sure what you're claiming here, but if you are claiming that you can have all parts of the rocket have exactly the same proper acceleration and not separate as in the Bell spaceship paradox, you are incorrect.

9. Apr 12, 2018

### sweet springs

Thanks. I agree with your saying about ordinary real matters. Born rigid body is theoretical one that cannot be made up of real matters.

I have a question.
I said approximately but I am not sure about it. I should appreciate it if you would teach me this relation of phi and g_00 is strict one or only an approximation in such and such conditions.

Last edited: Apr 12, 2018
10. Apr 12, 2018

### sweet springs

No. I intend to claim the case all parts are tied with threads and will move with their proper acceleration g(x) , function of positions. I apologize my saying "Bell's thread" in post might have misled you.

11. Apr 12, 2018

### sweet springs

Obviously when we say x=L+h h is height from the "ground" x=L this formula becomes
$$proper\ g(x)=proper\ g(L+h)=\frac{g(L)}{1+h\frac{g(L)}{c^2}}$$
as I wrote in other thread. There stevendaryl derived this by approximation. I would like to know how accurate this relation is. This is the background of my question in previous post#9.
So in other words I would like to know how to get proper gravity acceleration g(x), not through phi but directly from g_00. Best.

Last edited: Apr 12, 2018
12. Apr 12, 2018

### sweet springs

According to text of Landau-Lifshitz section 88 problem 1,
for bodies staying still,
$$proper\ g=-c^2 \nabla ln\ \sqrt{g_{00}}$$
thus
$$proper\ g=-\frac{c^2}{x}$$ or
$$proper\ g(x)=proper\ g(L+h)=\frac{g(L)}{1+h\frac{g(L)}{c^2}}$$
holds strictly.

From this I assume the relation
$$proper\ \phi(x)=c^2\ ln\ \sqrt{g_{00}}$$
holds for still object in stationary gravity field at least. My post #2 turns out to be right as for proper phi. But I am not sure of this yet.

Last edited: Apr 12, 2018
13. Apr 12, 2018

### stevendaryl

Staff Emeritus
The issue is exactly what is the significance of that $\phi(x)$. In nonrelativistic physics, the significance of $\phi$ is that there is a total energy that is conserved:$\frac{1}{2} m\dot{x}^2 + \phi(x)$. But there doesn't seem to be anything similar to that in the Rindler problem. There are conserved quantities involved, but not involving that, as far as I can see.

14. Apr 12, 2018

### sweet springs

In Rindler system
$$proper\ \phi(x)=c^2 ln\ \frac{x}{L}=c^2\ ln(1+\frac{h}{L})=c^2\frac{h}{L}+o((\frac{h}{L})^2)$$
so in case |h|<<L only the first term survives and it is familiar "mgh".

15. Apr 12, 2018

### sweet springs

Going back to reality, I have scarce idea of architechtur or structural technology of towers in various shape and size, I am sure that experts have well established skill to estimate and calculate on structural mechanics so that it is practically rigid and does not hardly collapse. If tower is extremely high change of gravity according to height or distance from the Earth with acceleration g(x) might be considered together with time dilation with it. Same thing we will do in construction of rockets with Rindler's g(x) L<x<L+H above mentioned where L corresponds to rear end and L+H corresponds to top end.

Last edited: Apr 12, 2018
16. Apr 12, 2018

### jbriggs444

Born rigidity characterizes the relationship between the positions in parts of an extended object over time. Indeed, there is no physical object that can maintain exact Born rigidity in the face of a momentary stress applied at a single point. However, there are many physical objects that can maintain Born rigidity in the face of a continuous stress applied on a single surface. e.g. the Eiffel tower. It is supported by its base and maintains Born rigidity in the face of that continuously applied force.

You are correct that the Eiffel tower will only maintain approximate rigidity in the face of a minor earthquake or strong and gusting wind.

17. Apr 12, 2018

### Staff: Mentor

The conditions any structure is subjected to on Earth are extremely mild compared to what we consider in relativity problems. 1 g acceleration is a very small acceleration in relativistic terms; calculate $c^2 / g$ for a 1 g acceleration and see what length $L$ it corresponds to.

18. Apr 12, 2018

### Staff: Mentor

Curves of constant $x, y, z$ in Rindler coordinates are orbits of a timelike Killing vector field, so there is a conserved energy associated with that KVF that is a constant of geodesic motion. The problem is that, unlike non-relativistic Newtonian mechanics or Schwarzschild spacetime in GR, the potential associated with this conserved energy does not approach a finite limit at infinity; it increases without bound. That is what prevents it from having a useful physical interpretation globally.

But if you restrict to a finite region, things work just the way you are describing: the potential is constant along each worldline in the Rindler congruence, and the kinetic energy of a freely falling object, relative to observers on each Rindler worldline, varies with the potential on each worldline in just the right way. For example, suppose we drop an object from the top of a Rindler rocket of length $L$. We work in an inertial frame in which the dropped object is at rest; so at the instant the object is dropped, the bottom of the rocket is at $x = x_0$ and the top of the rocket is at $x = x_0 + L$. Then the kinetic energy of the bottom of the rocket, relative to the dropped object, is just its kinetic energy in the frame we have chosen, which is simply $\left( \gamma - 1 \right) m$ (I'm using units where $c = 1$), where $\gamma$ is what you call $V^0$ in your article. This will also turn out to be the same as the change in potential from the top to the bottom of the rocket.

19. Apr 12, 2018

### sweet springs

In my post #15 I said about rear engine case. For front engine, no built on but hanged down take place. No compression but torsion works.
This reminds me that A. Gaudi used a model of church in Spain hanged upside down to investigate structure. No compression but torsion of the same magnitude work on every parts, I think.
If all parts has its own engine producing g(x) and tied with threads, no push and pull work on threads as I noted in my post #11.

20. Apr 12, 2018

### stevendaryl

Staff Emeritus
Okay, I see where a logarithmic "potential" can come into play, in general.

The equations of motion for a particle freefalling in curved spacetime are given by the geodesic equations:

$\ddot{x^\mu} = -\Gamma^\mu_{\nu \lambda} \dot{x^\nu} \dot{x^\lambda}$

where a dot above a variable means a derivative with respect to proper time. I'm going to assume that there is one time coordinate, $x^0$, and three spatial coordinates, $x^j$ ($j$ ranges from 1 to 3). I'm going to assume that the components of the metric tensor are independent of time. I'm also going to assume that the metric tensor is diagonal (so $g_{\mu \nu} = 0$ unless $\mu = \nu$).

Let's figure out the spatial acceleration for a test particle that is initially at rest: $\dot{x^j} \approx 0$. We can compute $\dot{x^0}$ in terms of the metric tensor:

$d\tau^2 = g_{\mu \nu} dx^\mu dx^\nu \Rightarrow 1 = g_{\mu \nu} \dot{x^\mu} \dot{x^\nu}$ Since $\dot{x^0}$ is the only nonzero component of the 4-velocity, we have:

$1 = g_{00} (\dot{x^0})^2$

So $\dot{x^0} = \frac{1}{\sqrt{g_{00}}}$.

So the geodesic equation gives us the proper acceleration for a spatial coordinate:
$\ddot{x^j} = -\Gamma^j_{0} \dot{x^0} \dot{x^0} = -\Gamma^j_{0} \frac{1}{g_{00}}$

Now, we can use a definition of $\Gamma$ in terms of the metric tensor:

$\Gamma^j_{00} = \frac{1}{2} g^{jk} (\partial_0 g_{k0} + \partial_0 g_{0k} - \partial_k g_{00})$

Since the metric components are independent of $x^0$, this simplifies to:
$\Gamma^j_{00} = -\frac{1}{2} g^{jk} \partial_k g_{00}$

Plugging this into the acceleration equation:
$\ddot{x^j} = \frac{1}{2} g^{jk} \partial_k g_{00} \frac{1}{g_{00}} = g^{jk} \partial_k\ ln \ \sqrt{g_{00}} = \partial^j \ ln \ \sqrt{g_{00}}$

So the initial proper acceleration is analogous to Newtonian physics with a potential of $\phi = ln \ \sqrt{g_{00}}$. However, that's only true initially, when you can ignore terms involving $\dot{x^j}$.

So this is the same as the Landau-Lifshitz result in post #12.

Last edited: Apr 12, 2018