poo_factor Jun 6 2023 at 09:47

Uniform gravity, can it exist?

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3 min

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1. The motion of a free particle-like cat

V. Komen, I. Tikhonenkov

Good morning! You've woke up. Having prepared coffee and toasts of bread you are drifting from a kitchen to a table before a large wall TV. The left hand keeps a small plate with toasts and the right one controls a coffee level within your beloved mug. The life is plotted for ten seconds to come. But Ervin is already worried that you, as usual, made a conspiracy and decided not to share your breakfast with him. So the damned cat thrusts himself across you pass, hits your legs. By the next moment the plate, toasts and the mug are falling. And â€“ yes! All those precious things reached a floor level by the same time. Physics...It's how it shows up, unexpected. And we used to identify ourselves as physicists. The unexpected thing also is that we still do. It's whence our motivation originates. We cannot pass by any falling objects quietly.

The dynamics of the plate, mug, coffee drops are identical, its independent upon the mass of an object. It is said that they undergo the influence of Earths' gravitation field which results in a force equal to the product of a mass and local strength of the field. This force is responsible for change of a momentum so the object moves. Or it is in the state of a free fall. It means that there are no forces but object follows its natural way in the curved space-time, the geodesic line. No gravity â€“ the space-time is flat and the world-line of free the falling cat is a straight one â€“ poor Ervin moves with a constant velocity. If we are again near the Earth, the gravity is here and the world line of a free falling cat transforms to a cyclic closed orbit.

The field near the Earth surface has one peculiar property namely it is close to the uniform field with constant strength within a small region of space and time. The motion of bodies including cats is well described by classical mechanics. We want to consider the dynamic in the uniform field which occupies the whole space for all time. Certainly the velocities must grow up and one has to apply relativistic equations. We consider two examples which differ by the definition of a force inflicted by a uniform field. The geometry is shown in Fig.1

First example is the motion of a particle with the rest mass m₀ under an action of a force which is constant in the inertial laboratory frame:

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

Here c is a speed of light, g - the strength of a uniform field which is oriented along y-axis, \dot{y}- is the velocity of a particle, a 'dot' stands for a derivative upon a world time t. The coupling to the gravitation field is defined by the rest mass m₀ of a particle. Second example is based on the assumption that the coupling to the filed is defined by observed inertial mass

$m_i=\frac{m_0 }{\sqrt{1-\dot{y}^2/c^2}} \hspace{110 mm}$

which, by means of the equivalence principle, is equal to a gravitational one m_g. So (1) becomes

$\frac{d}{dt}\left(\frac{m_0 \dot{y}}{\sqrt{1-\dot{y}^2/c^2}} \right)= m_ig=m_gg=\frac{m_0 g}{\sqrt{1-\dot{y}^2/c^2}} \hspace{60 mm} (2)$

The case (1) is a standard one for an accelerated motion. The case (2) is a model one and it is added for two reasons. First is that looks more corresponding to the equivalence principle and the second we will state at the end of this post. For initial conditions

$y(0)=0,\dot{y}(0)=0,\hspace{5 mm} x(0)=0,\dot{x}(0)=0 \hspace{85 mm} (3)$

the case (1) has a solution :

$\dot{y}=gt/\sqrt{1+(gt/c)^2}, \hspace{5 mm} y=\frac{c^2}{g} \left(\sqrt{1+(gt/c)^2}-1\right)\hspace{55 mm} (4)$

It is instructive also to consider a bit extended example with a particle having some initial momentum p_0x along x-axis with equations :

$\frac{d}{dt}\left(\frac{m_0 \dot{y}}{\sqrt{1-\dot{y}^2/c^2-\dot{x}^2/c^2}} \right)=m_0 g \hspace{10 mm} \frac{m_0 \dot{x}}{\sqrt{1-\dot{y}^2/c^2-\dot{x}^2/c^2}} =p_{0x} \hspace{10mm} (5)$

We have then:

$\dot{y}=\tilde{g}t/\sqrt{1+(\tilde{g}t/c)^2}, \hspace{5 mm} y=\frac{c^2}{\tilde{g}} \left(\sqrt{1+(\tilde{g}t/c)^2}-1\right),\hspace{5 mm}\tilde{g}=\frac{m_0c}{\sqrt{m_0^2c^2+p_{x0}^2}}g \hspace{5 mm}(6)$ $\dot{x}=\frac{p_{x0}c}{\sqrt{m_0^2c^2+p_{x0}^2}}\frac{1}{\sqrt{1+(\tilde{g}t/c)^2}},\hspace{5 mm} x=\frac{p_{x0}c}{m_0g}\ln\left( \tilde{g}t/c+\sqrt{1+(\tilde{g}t/c)^2} \right)\hspace{10 mm}(7)$

The velocity \dot{x} goes to zero while t tends to infinity. It is because of an inertia of the particle growths and since p_x0 is constant. But the motion along x-axis is unbounded. The motion along y has the same pattern but with different effective strength (6) of a uniform field. Looks like the field becomes weaker while initial energy of a particle is growing. The dependence of y upon x is shown on Fig.2. The equation to plot

$\eta=2\sqrt{1+\chi^2}\sinh^2 (\xi/2\chi),\hspace{5 mm}\xi=xg/c^2 , \hspace{5 mm}\eta=yg/c^2$

Case (2) is also readily solved:

$\dot{y}=c\tanh(gt/c),\hspace{5 mm}y=\frac{c^2}{g}\ln(\cosh(gt/c))\hspace{70 mm}(8)$

Suppose again that the particle possesses some initial momentum along x-axis while initial velocity along y-axis is still zero. The solution for y and \dot{y} are given by (8). Solutions for x:

$\dot{x}=\frac{p_{x0}c}{\sqrt{m_0^2c^2+p_{x0}^2}}\frac{1}{\cosh(gt/c)},\hspace{5 mm} x=\frac{p_{x0}c^2}{g\sqrt{m_0^2c^2+p_{x0}^2}}\arctan(\sinh(gt/c)) \hspace{10 mm}(9)$

One can see that like to the case (1) \dot{x} goes to zero while t increases and \dot{y} increases and approaches c. But the total displacement along x direction now is bounded by

$x_{max}=\frac{\pi}{2}\frac{p_{x0}c^2}{g\sqrt{m_0^2c^2+p_{x0}^2}} \hspace{100 mm}(10)$

The dependence of y upon x is shown on Fig.3. The equation to plot

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

Let us define the time it takes for a particle to reach infinity in its own frame. In the case (1) the proper time differential is

$d\tau=ds/c={dt}{\sqrt{1-\dot{y}^2/c^2-\dot{x}^2/c^2}}= \frac{m_{0}c}{\sqrt{m_0^2c^2+p_{x0}^2}} \frac{dt}{\sqrt{1+(\tilde{g}t/c)^2}}\hspace{20 mm}(11)$

and the time of moving to infinity is infinite:

$\tau_{\infty }=\int_{0}^{\infty}d\tau= \frac{m_{0}c^2}{\tilde{g}\sqrt{m_0^2c^2+p_{x0}^2}} \ln\left( \tilde{g}t/c+\sqrt{1+(\tilde{g}t/c)^2} \right)|_{0}^{\infty}\hspace{30 mm}(12)$

For the case (2) the proper time differential is

$d\tau= \frac{m_{0}c}{\sqrt{m_0^2c^2+p_{x0}^2}} \frac{dt}{\cosh(gt/c)}\hspace{90 mm}(13)$

so the proper time of a motion to infinity is finite

$\tau_{\infty }=\int_{0}^{\infty}d\tau= \frac{\pi c}{2 g}\frac{m_{0}c}{\sqrt{m_0^2c^2+p_{x0}^2}} \hspace{30 mm}(14)$

We see that two approaches exhibit different dynamics so why one should be bothered with the model case (2)? The reason is that we intend to describe the uniform field as being resulted from a global space-time curvature and we have a corresponding metrics which emulates the dynamics of a model case (2). This question will be addressed in the next our post.

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