UDC 530-12

PHYSICS

K. P. STANYUKOVICH

MOTION OF A MEDIUM WITH ULTRARELATIVISTIC VELOCITIES IN THE GENERAL THEORY OF RELATIVITY

(Presented by Academician Ya. B. Zel’dovich, 9 X 1968)

The basic equations describing the motion of a medium in a centrally symmetric field in the general theory of relativity, and the field itself, will be written in the form1

\[ \frac{1}{c^2\vartheta^2}\,[Au_t+uu_r] -\frac{\omega^2}{c^2}\left[(\ln v)_r-\frac{Au}{c^2}(\ln v)_t\right] = \frac{1}{2u}\,[A\lambda_t+u\lambda_r] +\frac{\theta^2 T^0\sigma_r}{W}; \tag{1} \]

\[ -[A(\ln v)_t+u(\ln v)_r] +\frac{1}{\theta^2}\left[u_r+\frac{Auu_t}{c^2}\right] +\frac{2u}{r} = \frac{u}{2}\left[\lambda_r+\frac{Au\lambda_t}{c^2}\right]; \tag{2} \]

\[ A\sigma_t+u\sigma_r=0. \tag{3} \]

Here \(A=e^{(\lambda-\nu)/2}\); \(u=A\,dr/dt\); \(\theta^2=1-u^2/c^2\); \(W=(p+\varepsilon)v\); \(u\) is the 3-velocity; \(p\) is the pressure; \(v\) is the specific volume; \(\varepsilon=\rho c^2\) is the energy density; \(W\) is the heat content, \(\omega^2/c^2=-(\partial\ln W/\partial\ln v)_\sigma\); \(\omega\) is the speed of sound; \(\sigma\) is the entropy; \(T^0\) is the temperature. In addition one must know the equation of state of the medium \(p=p(\sigma;v)\) and use the identity \(\partial(p;v)/\partial(T;\sigma)=1\).

We shall write the two independent field equations in the form:

\[ (re^{-\lambda})_r = 1-\frac{\chi r^2}{\theta^2}\,[\varepsilon+pu^2/c^2]; \tag{4} \]

\[ A(re^{-\lambda})_t = \frac{\chi ur^2}{\theta^2}\,[\varepsilon+p]. \tag{5} \]

Let us consider the motion when \(u/c=1-2\Delta\), where \(\Delta\ll 1\),

\[ p=\sigma v^{-k}=(k-1)\varepsilon;\qquad \omega^2/c^2=(k-1); \tag{6} \]

neglecting terms of order \(\Delta^2\), we arrive at the system of equations

\[ Ax_\tau+x_r=0, \tag{7} \]

where \(x=\ln[p\,r^{2k/(2-k)}]\), \(\tau=ct\);

\[ Ay_\tau+y_r=0, \tag{8} \]

where \(y=\ln[\Delta e^\lambda r^{4(k-1)/(2-k)}]\);

\[ A\sigma_\tau+\sigma_r=0; \tag{9} \]

\[ \lambda_r=\frac{1}{r} +e^\lambda\left[ \frac{\chi r}{4\Delta}(\varepsilon+p) -\left(\frac{1}{r}+\chi rp\right) \right]; \tag{10} \]

\[ A\lambda_\tau = -\frac{\chi r^2}{4\Delta}e^\lambda(\varepsilon+p). \tag{11} \]

We introduce a new independent variable \(m\) by means of the relation

\[ \chi\left(\frac{\partial m}{\partial r}\right)_t = \frac{\chi r^2}{\theta^2}\left(\varepsilon+\frac{pu^2}{c^2}\right) = 1-(re^{-\lambda})_r, \]

whence

\[ m=\int_0^r \frac{r^2}{\theta^2}\left(\varepsilon+p\frac{u^2}{c^2}\right)\,dr = \frac{r}{\chi}(1-e^{-\lambda}); \]

for \(\Delta \ll 1\) these relations take the form

\[ \varkappa\left(\frac{\partial m}{\partial t}\right)_{r} = \frac{\varkappa r^{2}}{4\Delta}(\varepsilon+p)-\varkappa r^{2}p = 1-(re^{-\lambda})_{r}, \tag{12} \]

\[ m=\int_{0}^{r}\left[\frac{r^{2}}{4\Delta}(\varepsilon+p)-r^{2}p\right]\,dr = \frac{\varkappa r}{\varkappa}(1-e^{-\lambda}). \]

Let us now pass to the independent variables \((m;r)\). The system of equations (10), (11) then takes the form

\[ r\frac{\partial\lambda}{\partial m}\frac{\partial\tau}{\partial r} + \frac{\partial\tau}{\partial m} \left[ 1-e^{-\lambda}(1+\varkappa r^{2}p) + \frac{\varkappa r^{2}}{4\Delta}e^{\lambda}(\varepsilon+p) - r\frac{\partial\lambda}{\partial r} \right]=0, \]

\[ A\frac{\partial\lambda}{\partial m} + \frac{\varkappa r e^{\lambda}}{4\Delta}(\varepsilon+p)\frac{\partial\tau}{\partial m}=0. \]

Since \(\lambda=-\ln(1-\varkappa m/r)\), \(\partial\lambda/\partial m=\varkappa e^{\lambda}/r\); \(\partial\lambda/\partial r=-\varkappa m e^{\lambda}/r^{2}\), from these relations we find

\[ \partial\tau/\partial m=-4\Delta A/r^{2}(p+\varepsilon), \qquad \partial\tau/\partial r=A[1-4\Delta p/(p+\varepsilon)]. \tag{13} \]

Equations (7), (8), (9) in the variables \((m;r)\) may be written in the form

\[ r^{2}p x_{m}=x_{r}; \tag{14} \]

\[ r^{2}p y_{m}=y_{r}; \tag{15} \]

\[ r^{2}p \sigma_{m}=\sigma_{r}. \tag{16} \]

Write equation (14) in the form

\[ e^{x}r^{4(1-k)/(2-k)}\partial x/\partial m=\partial x/\partial r. \tag{17} \]

Its general solution has the form

\[ m=\frac{(2-k)}{(5k-6)}e^{x}r^{(6-5k)/(2-k)}+F_{1}(x) = \frac{2-k}{5k-6}pr^{3}+F_{1}(pr^{2k/(2-k)}). \tag{18} \]

Writing the system (14), (15), and (16) in the independent variables \((x;r)\), we find that \(y=y(x)\), \(\sigma=\sigma(x)\), and, consequently,

\[ m=\frac{2-k}{5k-6}pr^{3}+F_{2}(y) = \frac{2-k}{5k-6}pr^{3}+F_{3}(\sigma). \tag{19} \]

At the same time

\[ \Delta=r^{-4(k-1)/(2-k)}(1-\varkappa m/r)f_{1}(pr^{2k/(2-k)}); \tag{20} \]

\[ \sigma=f_{3}(pr^{2k/(2-k)}). \tag{21} \]

On the basis of equation (12) we have

\[ \left(\frac{\partial m}{\partial r}\right)_{\tau} = \frac{r^{2}(\varepsilon+p)}{4\Delta}-pr^{2} = \frac{k}{4(k-1)}\frac{pr^{2}}{\Delta}-pr^{2} = \frac{k}{4(k-1)}e^{x-y+\lambda}-pr^{2} \tag{22} \]

or

\[ (\partial m/\partial r)_{\tau}=f(x)/(1-\varkappa m)r-pr^{2}, \tag{23} \]

where

\[ f(x)=ke^{x-y}/4(k+1); \qquad r^{2}(\varepsilon+p)/4\Delta=f(x)e^{\lambda}=f(x)/(1-\varkappa m/r). \]

From (18) we find that

\[ \left(\frac{\partial m}{\partial r}\right)_{\tau} = -pr^{2} + \left(\frac{\partial x}{\partial r}\right)_{\tau} \left[ \frac{2-k}{5k-6}e^{x}r^{(6-5k)/(2-k)} + \frac{dF_{1}(x)}{dx} \right]. \tag{24} \]

Comparing (23) and (24), we arrive at the result:

\[ \left(\frac{\partial x}{\partial r}\right)_{\tau} = \frac{f(x)} { \left(1-\frac{\varkappa m}{r}\right) \left( \frac{2-k}{5k-6}e^{x}r^{(6-5k)/(2-k)} + \frac{dF_{1}(x)}{dx} \right) }, \tag{25} \]

where

\[ \frac{m}{r} = \frac{2-k}{5k-6}e^{x}r^{4(1-k)/(2-k)} + \frac{F_{1}(x)}{r}. \]

Formally solving this equation of the form

\[ (\partial x/\partial r)_{\tau}=\theta(x;r), \tag{26} \]

we find

\[ \tau=\bar{\tau}[\xi(x;r)], \tag{27} \]

where \(\bar{\tau}(\xi)\) is an arbitrary function.

It is now easy to determine the quantity

\[ e^{(\lambda-\nu)/2}=A=-\frac{\partial\tau}{\partial m}\frac{r^{2}(p+\varepsilon)}{4\Delta} =-\frac{\partial\tau}{\partial m}\frac{kpr^{2}}{4(k-1)\Delta} =\frac{\partial\tau}{\partial m}\frac{f(x)}{1-\varkappa m/r}. \tag{28} \]

Since \(e^\lambda=1/(1-\varkappa m/r)\), we find

\[ e^{-\nu/2}=-\frac{\partial\tau}{\partial m}\frac{f(x)}{\sqrt{1-\varkappa m/r}}, \tag{29} \]

and thereby completely solve the problem posed.

Let us now turn to simplifications. For \(\varkappa=0\), i.e., in the absence of a gravitational field, we have the solution \((^2)\)

\[ x=x(r-\tau),\qquad y=y(r-\tau),\qquad \sigma=\sigma(r-\tau), \tag{30} \]

which follows immediately from (7), (8), (9) for \(A=1\).

Neglecting in the second equation (13) the quantity \(\Delta\ll 1\), we find that

\[ \partial\tau/\partial r=A. \tag{31} \]

In this case equations (14), (15), (16) take the form

\[ x=x(m),\qquad y=y(m),\qquad \sigma=\sigma(m), \tag{32} \]

\[ (\partial m/\partial r)_\tau=f(x)/(1-\varkappa m/r)=f(m)/(1-\varkappa m/r). \tag{33} \]

For \(\varkappa=0\)

\[ (\partial m/\partial r)_\tau=f(m). \tag{34} \]

Integrating, we find that

\[ r-\varphi(m)=T(\tau)=\tau, \]

whence \(m=\psi(r-\tau)\) and \(x=x(r-\tau)\), which gives the limiting results (30).

For \(\varkappa\ne0\) it is necessary to integrate equation (33):

\[ \frac{dr}{1}=\frac{d\tau}{0}\frac{dm}{f(m)} \left(1-\frac{\varkappa m}{r}\right), \tag{35} \]

whence

\[ \frac{dr}{dm}f(m)=1-\frac{\varkappa m}{r}. \]

Putting \(r=1/\xi\), we shall have

\[ \frac{d\xi}{dm}f(m)=\varkappa m\xi^{3}-\xi^{2} \tag{36} \]

or

\[ d\xi/d\eta=B(\eta)\xi^{3}-\xi^{2}, \tag{37} \]

where \(d\eta=dm/f(m)\); \(B(\eta)=\varkappa m(\eta)\).

This equation is easily solved by numerical methods (for example, by the Runge–Kutta method); as a result, returning to the old variables, we find

\[ D(r;m)=f(\tau), \tag{38} \]

which then permits one formally to express \(x,y,\sigma\) through \(r\) and \(\tau\).

In the case when \(f=f_0=\mathrm{const}\), and this is an important case, equation (35) is integrated at once:

\[ m^{2}\left[\frac{r^{2}}{m^{2}}-\frac{r}{mf_0}+\frac{\varkappa}{f_0}\right] +F(\tau)\left[ \frac{r/m-1/2f_0+(1/2f_0)\sqrt{1-4\varkappa f_0}} {-r/m+1/2f_0+(1/2f_0)\sqrt{1-4\varkappa f_0}} \right]^{1/\sqrt{1-4\varkappa f_0}}. \tag{39} \]

For \(\chi=0\)

\[ \frac{r}{m} f_0=\frac{\psi(\tau)}{m}+1; \tag{40} \]

further, from (39) we have

\[ -\frac{m^2}{f_0^2}\left[\frac{\psi^2}{m^2}+\frac{\psi}{m}\right] -F(\tau)\left[\frac{\psi/m+1}{\psi/m}\right]=0; \]

whence

\[ \frac{m^2}{f_0^2}\frac{\psi^2}{m^2}=F(\tau)=\frac{\psi^2(\tau)}{f_0^2}, \]

which verifies the calculations we have carried out. From (20) it follows that in the case

\[ f_1=r_0^{4(k-1)/(2-k)}=\mathrm{const} \]

with

\[ \chi m/r=1,\qquad r=r_g(m),\qquad \partial r/\partial m=0\quad (f(\chi)=f(m)\ne0). \]

It also follows from this that here \(r=r_{\min}=2Gm_0/c^2=r_g\), where \(m_0=4\pi m/c^2\), \(\Delta=0\), i.e., at the gravitational radius \(u=c\). As \(r\to\infty\), also \(\Delta=0\), \(u=c\). The maximum value \(\Delta=\Delta_{\max}\), i.e., \(u=u_{\min}\), is attained at

\[ r=\frac{3k-2}{4(k-1)}\chi m =\frac{3k-2}{4(k-1)}\frac{2Gm_0}{c^2} =\frac{3k-2}{4(k-1)}r_g . \]

Thus, in the ultrarelativistic limiting case there is a radius \(r=r^*\) at which the velocity is minimal; this, however, still does not mean that the contraction of the medium stops.

Received
4 X 1968

CITED LITERATURE

1. K. P. Stanyukovich, DAN, 182, No. 2 (1968). 2. K. P. Stanyukovich, ZhETF, 36, No. 6 (1959).