2. Curved cat's tail

V. Komen, I. Tikhonenkov

In the previous post we've considered a model example of a motion of a free particle within a uniform gravitation field where a coupling to the field is defined by an observed inertion mass (see eq. (2) in https://habr.com/en/articles/739714/). The equation of motion was:

$\frac{d}{dt}\left(\frac{m_0\dot{y}}{\sqrt{1-\dot{y}^2/c^2}}\right)= \frac{m_0g}{\sqrt{1-\dot{y}^2/c^2}} \hspace{100 mm} (1)$

Here m₀ is the rest mass of a particle, g - is the strength of the uniform field. The geometry is shown on Fig.1 below

If

$\dot{y}(0)=0,\hspace{5 mm}y(0)=0,\hspace{5 mm}\dot{x}(0)=0,\hspace{5 mm}x(0)=0\hspace{75 mm}(2)$

then ( https://habr.com/en/articles/739714/ )

$\dot{y}=c\tanh(gt/c),\hspace{5 mm}y=\frac{c^2}{g}\ln(\cosh(gt/c))\hspace{80 mm}(3)$

Now we'll describe the dynamics (1)-(3) by means of a curved space-time. Surely one should use a diagonal metric tensor :

$ds^2 = g_{00}c^2dt^2+g_{11}dx^2+g_{22}dy^2+g_{33}dz^2$

where ds is an interval. A reasonable guess is that g₂₂=g₃₃=-1 and g₀₀,g₁₁ are the functions of y only since the field is uniform and stationary. Suppose that p_x0=p_z0 =0 so dx=dz=0. We try the metric of the form

$ds^2 = g_{00}c^2dt^2+g_{11}dy^2=e^{-2ky}c^2dt^2-e^{-2ky}dy^2\hspace{60 mm}(4)$

where k has to be defined. The dependence of y upon a time can be found using Hamilton-Jacobi equation for the action S:

$g^{00}\frac{1}{c^2}\left(\frac{\partial S}{\partial t}\right)^2+ g^{11}\left(\frac{\partial S}{\partial y}\right)^2 = e^{2ky}\frac{1}{c^2}\left(\frac{\partial S}{\partial t}\right)^2- e^{2ky}\left(\frac{\partial S}{\partial y}\right)^2=m_0^2c^2\hspace{20 mm}(5)$

Its solution we seek in the form

$S=-Et+S_y(y)\hspace{130 mm}(6)$

Here E is the energy of a particle. After a substitution into (5) we obtain:

$S_y=\int_{0}^{y} \sqrt{E^2/c^2-e^{-2ky}m_0^2c^2} dy$

Next the dependence of y upon a time t is determined from

$\frac{\partial S }{\partial E}=-t+\frac{\partial S_y }{\partial E}=0 \hspace{120 mm}(7)$

so

$t=\frac{E}{c^2}\int_{0}^{y}\frac{dy}{\sqrt{E^2/c^2-m_0^2c^2e^{-2ky}}}$

If the particle starts moving from a rest then E/m₀c² =1 . So:

$ct=\int_{0}^{y}\frac{e^{ky}dy}{\sqrt{e^{2ky}-1}}$

and

$\dot{y}=c\tanh(kct),\hspace{5 mm}y=\frac{1}{k}\ln(\cosh(kct))\hspace{80 mm}(8)$

Comparing with (3) we have for k:

$k=g/c^2\hspace{145 mm}(9)$

Seems that we've made a correct guess (4) for the metric tensor. But suppose that the particle has some initial momentum p_x₀ along x-axis while initial velocity along y-axis is still zero. Now the interval is given by

$ds^2 = e^{-2ky}c^2dt^2-e^{-2ky}dy^2-dx^2\hspace{90 mm}(10)$

Hamilton-Jakobi equation reads

$e^{2ky}\frac{1}{c^2}\left(\frac{\partial S}{\partial t}\right)^2- e^{2ky}\left(\frac{\partial S}{\partial y}\right)^2-\left(\frac{\partial S}{\partial x}\right)^2=m_0^2c^2\hspace{60 mm}(11)$

we are looking for a solution in the form:

$S=-Et+S_y(y)+xp_{x0}\hspace{110 mm}(12)$

Next using (6)-(7) we obtain

$S_y=\int_{0}^{y} \sqrt{E^2/c^2-e^{-2ky}(m_0^2c^2+p_{x0}^2)} dy,\hspace{2 mm}t=\frac{E}{c^2}\int_{0}^{y}\frac{dy}{\sqrt{E^2/c^2-(m_0^2c^2+p_{x0}^2)e^{-2ky}}}$

Since that

$E^2/c^2=m_0^2c^2+p_{x0}^2$

we have again the same equation (8) for y(t):

$y=\frac{1}{k}\ln(\cosh(kct))\hspace{125 mm}(13)$

In order to define the motion along x direction we use

$\frac{\partial S}{\partial p_{x0}}=x+\frac{\partial S_y}{\partial p_{x0}}=0\hspace{120 mm}(14)$

so, using that e^ky=cosh(gt/c), we obtain for a curved space-time

$x=\frac{p_{x0}c^2}{g\sqrt{m_0^2c^2+p_{x0}^2}}\tanh(gt/c)\hspace{100 mm}(15)$

it differs from the corresponding equation for a flat space (eq. (9) in https://habr.com/en/articles/739714/ ):

$x=\frac{p_{x0}c^2}{g\sqrt{m_0^2c^2+p_{x0}^2}}\arctan\sinh(gt/c)\hspace{90 mm}(16)$

The motion along x directions is bounded in both cases of a flat and curved space but corresponding maximum values of x differs by a factor \pi/2. The dependence of y upon x for the motion in a curved space is shown on Fig.2. The equation to plot

$\eta = \frac{1}{2}\ln\left(\frac{\chi^2}{\chi^2-\xi^2(1+\chi^2)}\right),\hspace{5 mm}\xi=xg/c^2, \hspace{5 mm}\eta=yg/c^2$

The dependence of y upon x for the motion in a flat space is shown on Fig.3. The equation to plot

$\eta=-\ln\left(\cos\left(\frac{\xi}{\chi} \sqrt{1+\chi^2}\right)\right),\hspace{5 mm}\xi=xg/c^2, \hspace{5 mm}\eta=yg/c^2$

The dynamics shown on Figs 2 and 3 looks similar but with a slight quantitative differences. What could be the cause of it? The wrong metrics? That is the choice is not unique and one can use the better one. Or perhaps it is impossible in general to emulate by means of a global curvature the dynamics in a flat space-time even for a so simple object as the uniform field? The reasonable idea is to find answers using general field equations developed for a stationary 1D gravitation field and we'll consider the issue in the next post.

Uniform gravity, can it exist?