Pull to refresh

Uniform gravity, can it exist?

Level of difficulty Medium
Reading time 3 min
Views 877

3.What's an ugly smiling face?

It's the cat from curved space.

V. Komen, I. Tikhonenkov

In our early published posts we've considered the uniform gravitation field which occupies the whole 3D space for all time. The simple model of it is the space where the strength g of the field has the constant value and direction (Fig.1).

Fig. 1
Fig. 1

We tried to analyse this object as a flat space-time with a standard Minkovski metric ( https://habr.com/en/articles/739714/ [1]) and as a curved space with a model metric ( https://habr.com/en/articles/739700/ [2]). In the flat space the particle moves under an action of a force from the field. The post [1] presents two examples (eqs. (1) and (2)) of motion which differs by force definition. Case (1) is the textbook example of a relativistic motion under a constant force. The case (2) treats the observed inertial mass as a gravitational one so the force is growing with the particle's speed. The dynamics (1) and (2) differ in the character of the motion which is transverse to the field direction, namely it is unbounded for (1) but is bounded for ( 2). In [2] we analysed the motion within a curved space time which metric corresponds to case (2) from [1]. We have reproduced the dynamic qualitatively. For a motion in y-direction (field direction) there was perfect coincidence but for x-direction there were numerical discrepancies. In this article we want to find out the cause this deviation, to find whether it is possible to define a metric tensor corresponding to the textbook case (1) and to give a description of all possible 1D stationary metrics. In addition if there exists the metric tensor which gives the same dynamics as in the case (1) [1] then the concept of the stationary uniform field is correct from the general point of view. So here we are discussing in general the majority of metrics for a stationary gravitation field in one dimension. The only accepted approach so far to apply are field equations (A. Einstein) in an empty space:

R_{\mu\nu}=\Gamma_{\mu\alpha,\nu}^{\alpha}-\Gamma_{\mu\nu,\alpha}^{\alpha}- \Gamma_{\mu\nu}^{\alpha}\Gamma_{\alpha\beta}^{\beta} + \Gamma_{\mu\beta}^{\alpha}\Gamma_{\nu\alpha}^{\beta}=0\hspace{55 mm}(1)

where R_{\mu\nu} is Richie tensor, functions Г are Christoffel's symbols of second kind:

\Gamma_{\mu\nu}^{\alpha}=g^{\alpha\beta}\Gamma_{\beta;\mu\nu}\hspace{120 mm} (2)

Christoffel's symbols of first kind:

\Gamma_{k; ij}=\frac{1}{2}\left( g_{ik,j}+g_{jk,i}-g_{ij,k} \right)\hspace{90 mm}(3)

where g,k designates a partial derivative of g on xk. Covariant components gik of a metric tensor define the square of an interval

ds^2=g_{ik}dx^i dx^k

Contravariant components gpq :


The values xk , k=(0, 1, 2, 3) are coordinates of an event in a space-time of four dimensions. For instance, dx0 = cdt, where t is the world time. We restrict ourselves to the frame where g22=g33 =-1, gjk=0 different indices j and k, g00 and g11 are functions of x1 only. The only nonzero Гkij are:

\Gamma^{0}_{01}=\Gamma^{0}_{10}=\frac{g_{00,1}}{2g_{00}},\hspace{5 mm} \Gamma^{1}_{00}=-\frac{g_{00,1}}{2g_{11}},\hspace{5 mm} \Gamma^{1}_{11}=\frac{g_{11,1}}{2g_{11}}\hspace{60 mm}(4)

Substituting (4) into (1) gives field equations

R_{00}=-\Gamma^{1}_{00,1}+\Gamma^{1}_{00}\Gamma^{0}_{10}-\Gamma^{1}_{00}\Gamma^{1}_{11}=0, \hspace{5 mm} R_{11}=\Gamma^{0}_{01,1}+\Gamma^{0}_{10}\Gamma^{0}_{10}-\Gamma^{0}_{10}\Gamma^{1}_{11}=0 \hspace{5 mm}(5)

or, denoting x1 as y:

\frac{d}{dy}\left( \frac{1}{g_{11}} \frac{dg_{00}}{dy}\right)+\frac{1}{2g_{11}} \frac{dg_{00}}{dy} \left( \frac{1}{g_{11}} \frac{dg_{11}}{dy}-\frac{1}{g_{00}} \frac{dg_{00}}{dy} \right)=0 \hspace{45 mm}\\ \frac{d}{dy}\left( \frac{1}{g_{00}} \frac{dg_{00}}{dy}\right)-\frac{1}{2g_{00}} \frac{dg_{00}}{dy} \left( \frac{1}{g_{11}} \frac{dg_{11}}{dy}-\frac{1}{g_{00}} \frac{dg_{00}}{dy} \right)=0 \hspace{40 mm}(6)

it follows from (6) that

\frac{\frac{d}{dy}\left( \frac{1}{g_{11}} \frac{dg_{00}}{dy}\right)}{\frac{1}{g_{11}} \frac{dg_{00}}{dy}}=- \frac{\frac{d}{dy}\left( \frac{1}{g_{00}} \frac{dg_{00}}{dy}\right)}{\frac{1}{g_{00}} \frac{dg_{00}}{dy}}\hspace{90 mm}(7)


\left( \frac{dg_{00}}{dy} \right)^2\frac{1}{g_{00}g_{11}}=\pm 4k^2\hspace{105 mm}(8)

where k is an arbitrary constant such that k > 0. For instance, the metric ((4), [2])

\tilde{g}_{00}=e^{-2ky}, \hspace{5 mm} \tilde{g}_{11}=-e^{-2ky}

from the preceding post obeys eq. (8)

\left( \frac{d\tilde{g}_{00}}{dy} \right)^2\frac{1}{\tilde{g}_{00}\tilde{g}_{11}}=- 4k^2\hspace{5 mm}, k=g/c^2

So it is allowed by the field equations. And what about the motion induced by a constant force , case (1) from [1]? To clarify the question we'll find the equation which connects t and y when the condition (8) is true. Namely we substitute into Hamilton-Jacobi equation for an action S:

g^{00}\frac{1}{c^2}\left( \frac{\partial S}{\partial t} \right)^2+ g^{11}\left( \frac{\partial S}{\partial y} \right)^2 =  \frac{1}{g^{00}}\frac{1}{c^2}\left( \frac{\partial S}{\partial t} \right)^2+ \frac{1}{g^{11}}\left( \frac{\partial S}{\partial y} \right)^2= m_0^2c^2\hspace{20 mm}(9)

where m0 is the rest mass. The following expression for g11:

g_{11}=-\frac{1}{4k^2}\frac{(dg_{00}/dy)^2}{g_{00}}\hspace{120 mm}(10)

Taking S = -tE+Sy(y) ((6), [2]) , E=m0c2 is the energy of a particle, we obtain

S_y=\int_{0}^{y} \frac{\left| dg_{00}/dy \right|}{2kg_{00}} \sqrt{E^2/c^2-m_0^2c^2g_{00}}dy\hspace{80 mm}(11)

Now since

t=\partial S_y /\partial E\hspace{140 mm} (12)

we have

ct=\int_{0}^{y} \frac{\left| dg_{00}/dy \right|}{2kg_{00}} \frac{dy}{\sqrt{1-g_{00}}}\hspace{110 mm}(13)

Using eqs. (4) from [1]

y=\frac{c^2}{g}\left( \sqrt{1+(gt/c)^2}-1 \right),\hspace{5 mm} \dot{y}=\frac{gt}{\sqrt{1+(gt/c)^2}}\hspace{65 mm}(14)

one can derive

ct=\int_{0}^{y} \frac{(1+gy/c^2)dy}{\sqrt{(1+gy/c^2)^2-1}}\hspace{110 mm}(15)

And now our goal is to find a function g00(y) and k such that being inserted into (13) gives (15). The answer is

g_{00}=\frac{1}{\cosh^2\left( \sqrt{(1+gy/c^2)^2-1} \right)},\hspace{5 mm}  k=\frac{g}{c^2}\hspace{70 mm}(16)

corresponding expression for g11 is

g_{11}=-\frac{\sinh^2\left( \sqrt{(1+gy/c^2)^2-1} \right)}{\cosh^4\left( \sqrt{(1+gy/c^2)^2-1} \right)}\frac{(1+gy/c^2)^2}{(1+gy/c^2)^2-1}\hspace{50 mm}(17)

Thus metric (16)-(17) provides the dynamics which corresponds to miscellaneous example of a motion in the uniform field (see case (1) in [1]). So we can see, that the motion along y direction could be reproduced by some curvature for both cases from [1]. Let us look again what happens if the particle has initial momentum px0 along x direction. Hamilton-Jacobi equation:

 \frac{1}{g^{00}}\frac{1}{c^2}\left( \frac{\partial S}{\partial t} \right)^2+ \frac{1}{g^{11}}\left( \frac{\partial S}{\partial y} \right)^2-\left( \frac{\partial S}{\partial x} \right)^2= m_0^2c^2\hspace{50 mm}(18)

Taking S = -tE+Sy(y)+xpx0 ((12), [2]) we obtain

S_y=\int_{0}^{y} \frac{\left| dg_{00}/dy \right|}{2kg_{00}} \sqrt{E^2/c^2-(m_0^2c^2+p_{x0}^2)g_{00}}dy\hspace{80 mm}(11)

Applying (12) we have again eqs. (13), so time dynamics along y-axis is not affected by nonzero px0. exactly as for case (2) [1] and for [2]. Since E2/c2 = m02c2 + px02 and

\partial S/\partial x = x+\partial S_y/\partial x = 0


x=\frac{p_{x0}}{2k\sqrt{p_{x0}^2+m_0^2c^2}}\int_{0}^{y} \frac{\left| dg_{00}/dy \right|dy}{\sqrt{1-g_{00}}}\hspace{80 mm}(19)

It is seen from (19) that if g00 is a monotonic function of y then x is bounded, namely maximum of x is:

x_{max}=\frac{p_{x0}}{k\sqrt{p_{x0}^2+m_0^2c^2}}\sqrt{1-g_{00}}|_{0}^{\infty} \hspace{80 mm}(20)

For the case of g00 from (16)

x_{max}=\frac{p_{x0}c^2}{g\sqrt{p_{x0}^2+m_0^2c^2}} \hspace{110 mm}(21)

which surprisingly is same as for the case (2) from [2], eq. (15). But for the case (1) [1 ] which the metric (16)-(17) corresponds to the motion along x is unbounded. Thus the dynamics of a relativistic motion under constant force in the flat space cannot be caused by any global curvature allowed by field equations. At this point we have accumulated enough facts to make a conclusion. Suppose in the flat space-time we have some example of motion within the uniform field along y axis in the form

ct=\int_{0}^{y}f(y)dy\hspace{130 mm}(22)

Then a corresponding g00 is defined by the requirement that (13) is equal to (22). The g11 is provided by (17), thus the motion along y in the flat space defines the corresponding metric of a curved space-time, where g00, g11 are function of y only. This procedure being applied to the textbook problem case(1) [] gives qualitatively different motion in the curved space. For the model case (2) the motion in flat and curved spaces are similar but not exactly same ones. So when we are discussing the global uniform field we do not in a strict sense know what we are talking about.

Total votes 4: ↑3 and ↓1 +2
Comments 0
Comments Leave a comment