- #1
Peter Pijl
- 1
- 0
Desloge (1989) published the article: ## '##Non-equivalence of a uniformly accelerating reference frame and a frame at rest in a uniform gravitational field## '##. His result, briefly resumed: a uniform gravitational field is not flat, is quite interesting. But the way he proves the result is usable for any gravitational field; I will start with a description of his ‘method’, though in a slightly modified way; the result will be the same.
So, first a definition of a rigid reference system: observers, for example space-ships, provided with an ideal clock, are positioned in an arbitrary gravitational field, in the following way:
- the power of the observers (space-ships) is constant.
- the light-distance, measured by any observer to any other one is constant.
- one of the observers is designed as head observer H; his spatial coordinate will be 0; the spatial coordinate of observer A will be the light-distance, measured by H of A: x##_A##
- the standard clocks of the observers cannot be synchronized; the clock of H will be appointed as coordinate clock. All other observers are supplied with a coordinate clock in the following way: suppose the light-distance of observer A, measured by H, is a; the light-distance of H, measured by A, is b.
Then, b/a determines the ratio between standard clock and coordinate clock; the ratio will be ##\alpha##(x), depending of the spatial distance x of the observer. All coordinate clocks can be synchronized.
We consider the one-direction case.
The frame is rigid; the corresponding coordinate system will be represented by ds##^2 = \alpha(x)^2##(dt##^2## - dx##^2##),(the so-called Weyl -form) where ##\alpha##(t=0) = 1. So local velocity=dx/dt, coordinate velocity. The relationship between local acceleration (this is what every observer really‘feels’, and is able to measure in his space-ship lab, by releasing a free falling particle alongside a rod), and coordinate acceleration, is more complex.
We have to consider the equations for geodesics. We start with the more general form:
$$
\frac{d^2x^a}{du^2}+ \Gamma^a_{bc}\frac{dx^b}{du} \frac{dx^c}{du} =\lambda(u)\frac{dx^a}{du}
$$
In our special case, we have ##\Gamma^a_{bc}##=0, except ##\Gamma^0_{01}=\Gamma^0_{10}=\Gamma^1_{00}=\Gamma^1_{11} = \frac{\alpha '}{\alpha}, \alpha##’ being d##\alpha##(x)/dx. Given that x##_0##=t, x##_1##=x, we find for the first equation:
$$
\frac{d^2t}{du^2}+\frac{2\alpha '}{ \alpha} \frac{dt}{du} \frac{dx}{du} = \lambda(u) \frac{dt}{du}
$$
This holds for any parameter, with corresponding##\lambda##(u). So, also for t(u). Result:
$$
\frac{2 \alpha '}{\alpha} \frac{dx}{dt} = \lambda (t)
$$
The second equation will become:
$$
\frac{d^2x}{dt^2}+\frac{\alpha '}{\alpha} + \frac{\alpha '}{\alpha} \left( \frac{dx}{dt}\right)^2 = \frac{2 \alpha '}{\alpha}\left( \frac{dx}{dt} \right)^2
$$
And therefore
$$
\frac{d^2x}{dt^2} + \frac{\alpha '}{\alpha} - \frac{\alpha '}{\alpha} \left( \frac{dx}{dt}\right)^2 = 0
$$
Suppose, we release a free falling particle at moment t, so dx/dt =0 at that moment: d##^2##x/dt##^2##= ##-\alpha '/\alpha##. In this case, the relation between local and coordinate acceleration is: localacc.=1/##\alpha##. Coordinate acc.= ##\alpha '/\alpha^2##. This holds for every gravitation field. Desloge uses the result for a uniform field; we investigate what comes out for natural gravity. Suppose there is a rigid reference frame, radially positioned with respect to a heavy body. The one-direction form of the Schwarzschild metric is:
$$ds^2 = \left(1-\frac{2m}{x}\right)dt^2 - \left(1-\frac{2m}{x}\right)^{-1}dx^2; \; \; \; x>2m
$$
First, we have to transform the metric in the Weyl-form.
The transformation ##\bar{t}##=t, ##\bar{x}##=x+2m*ln(x - 2m) will do the job.
The result: ds##^2##={1 – 2m/x(##\bar{x}##)(d##\bar{t}^2## - d##\bar{x}^2##).
(There is no mathematical expression for the inverse of ##\bar{x}##(x) ,so we leave it: x(##\bar{x}##).
There is one final task: 1 - 2m/x ##\rightarrow## 1 for ##\bar{x} \rightarrow \infty##. We want the head observer at some finite ##\bar{x}##. We take A, with coordinate ##\bar{x}_A##, and ##\bar{t}##=(1/k). ##\bar{t}##, ##\bar{x}##=(1/k). (x(##\bar{x}##)–x(##\bar{x}_A##)); k = (1 - (2m/x(##\bar{x}_A##)))##^2## will have as result: ##\hat{\alpha}##=(1/k)(1 - 2m/x)##^\frac{1}{2}##; x = x(##\bar{x}(\bar{x}##))and for ##\bar{x}=0##, ##\alpha_A##=1. Now we are able to use the previous result:
\begin{equation}
\frac{d^2\hat{x}}{d\hat{t}^2} = \frac{\alpha '}{\alpha} = \left( \left[1 - \frac{2m}{x}\right]^{-\frac{1}{2}} \frac{2m}{x^2}\frac{dx}{d\bar{x}} \frac{d\bar{x}}{d\hat{x}}\right)\left[1-\frac{2m}{x}\right]^{-\frac{1}{2}} = \frac{2km}{x^2}
\end{equation}
This is the result we expect (k only dependent on the choice of A as head observer), if all observers are measuring with their coordinate clocks. However, an observer, eventually not being member of the reference system, and so measuring with his standard clock, will find:
\begin{equation}
\frac{1}{\alpha}\frac{2km}{x^2} = \left(1-\frac{2m}{x}\right)^{-1}\frac{2km}{x^2}
\end{equation}
The result is classical gravity multiplied with a ##‘##relativistic correction##’##, neglegible in normal circumstances, but considerable in the neighbourhood of the Schwarzschild radius.
P.S. To my opinion, the way Desloge uses the concept of a rigid reference frame, together with the link with a coordinate system is an important contribution to the theory. Still, there are several questions, open to me, which I would like to discuss with those who are interested in the matter.
So, first a definition of a rigid reference system: observers, for example space-ships, provided with an ideal clock, are positioned in an arbitrary gravitational field, in the following way:
- the power of the observers (space-ships) is constant.
- the light-distance, measured by any observer to any other one is constant.
- one of the observers is designed as head observer H; his spatial coordinate will be 0; the spatial coordinate of observer A will be the light-distance, measured by H of A: x##_A##
- the standard clocks of the observers cannot be synchronized; the clock of H will be appointed as coordinate clock. All other observers are supplied with a coordinate clock in the following way: suppose the light-distance of observer A, measured by H, is a; the light-distance of H, measured by A, is b.
Then, b/a determines the ratio between standard clock and coordinate clock; the ratio will be ##\alpha##(x), depending of the spatial distance x of the observer. All coordinate clocks can be synchronized.
We consider the one-direction case.
The frame is rigid; the corresponding coordinate system will be represented by ds##^2 = \alpha(x)^2##(dt##^2## - dx##^2##),(the so-called Weyl -form) where ##\alpha##(t=0) = 1. So local velocity=dx/dt, coordinate velocity. The relationship between local acceleration (this is what every observer really‘feels’, and is able to measure in his space-ship lab, by releasing a free falling particle alongside a rod), and coordinate acceleration, is more complex.
We have to consider the equations for geodesics. We start with the more general form:
$$
\frac{d^2x^a}{du^2}+ \Gamma^a_{bc}\frac{dx^b}{du} \frac{dx^c}{du} =\lambda(u)\frac{dx^a}{du}
$$
In our special case, we have ##\Gamma^a_{bc}##=0, except ##\Gamma^0_{01}=\Gamma^0_{10}=\Gamma^1_{00}=\Gamma^1_{11} = \frac{\alpha '}{\alpha}, \alpha##’ being d##\alpha##(x)/dx. Given that x##_0##=t, x##_1##=x, we find for the first equation:
$$
\frac{d^2t}{du^2}+\frac{2\alpha '}{ \alpha} \frac{dt}{du} \frac{dx}{du} = \lambda(u) \frac{dt}{du}
$$
This holds for any parameter, with corresponding##\lambda##(u). So, also for t(u). Result:
$$
\frac{2 \alpha '}{\alpha} \frac{dx}{dt} = \lambda (t)
$$
The second equation will become:
$$
\frac{d^2x}{dt^2}+\frac{\alpha '}{\alpha} + \frac{\alpha '}{\alpha} \left( \frac{dx}{dt}\right)^2 = \frac{2 \alpha '}{\alpha}\left( \frac{dx}{dt} \right)^2
$$
And therefore
$$
\frac{d^2x}{dt^2} + \frac{\alpha '}{\alpha} - \frac{\alpha '}{\alpha} \left( \frac{dx}{dt}\right)^2 = 0
$$
Suppose, we release a free falling particle at moment t, so dx/dt =0 at that moment: d##^2##x/dt##^2##= ##-\alpha '/\alpha##. In this case, the relation between local and coordinate acceleration is: localacc.=1/##\alpha##. Coordinate acc.= ##\alpha '/\alpha^2##. This holds for every gravitation field. Desloge uses the result for a uniform field; we investigate what comes out for natural gravity. Suppose there is a rigid reference frame, radially positioned with respect to a heavy body. The one-direction form of the Schwarzschild metric is:
$$ds^2 = \left(1-\frac{2m}{x}\right)dt^2 - \left(1-\frac{2m}{x}\right)^{-1}dx^2; \; \; \; x>2m
$$
First, we have to transform the metric in the Weyl-form.
The transformation ##\bar{t}##=t, ##\bar{x}##=x+2m*ln(x - 2m) will do the job.
The result: ds##^2##={1 – 2m/x(##\bar{x}##)(d##\bar{t}^2## - d##\bar{x}^2##).
(There is no mathematical expression for the inverse of ##\bar{x}##(x) ,so we leave it: x(##\bar{x}##).
There is one final task: 1 - 2m/x ##\rightarrow## 1 for ##\bar{x} \rightarrow \infty##. We want the head observer at some finite ##\bar{x}##. We take A, with coordinate ##\bar{x}_A##, and ##\bar{t}##=(1/k). ##\bar{t}##, ##\bar{x}##=(1/k). (x(##\bar{x}##)–x(##\bar{x}_A##)); k = (1 - (2m/x(##\bar{x}_A##)))##^2## will have as result: ##\hat{\alpha}##=(1/k)(1 - 2m/x)##^\frac{1}{2}##; x = x(##\bar{x}(\bar{x}##))and for ##\bar{x}=0##, ##\alpha_A##=1. Now we are able to use the previous result:
\begin{equation}
\frac{d^2\hat{x}}{d\hat{t}^2} = \frac{\alpha '}{\alpha} = \left( \left[1 - \frac{2m}{x}\right]^{-\frac{1}{2}} \frac{2m}{x^2}\frac{dx}{d\bar{x}} \frac{d\bar{x}}{d\hat{x}}\right)\left[1-\frac{2m}{x}\right]^{-\frac{1}{2}} = \frac{2km}{x^2}
\end{equation}
This is the result we expect (k only dependent on the choice of A as head observer), if all observers are measuring with their coordinate clocks. However, an observer, eventually not being member of the reference system, and so measuring with his standard clock, will find:
\begin{equation}
\frac{1}{\alpha}\frac{2km}{x^2} = \left(1-\frac{2m}{x}\right)^{-1}\frac{2km}{x^2}
\end{equation}
The result is classical gravity multiplied with a ##‘##relativistic correction##’##, neglegible in normal circumstances, but considerable in the neighbourhood of the Schwarzschild radius.
P.S. To my opinion, the way Desloge uses the concept of a rigid reference frame, together with the link with a coordinate system is an important contribution to the theory. Still, there are several questions, open to me, which I would like to discuss with those who are interested in the matter.