Full Text
PHYSICS
K. P. STANYUKOVICH
INTERACTION OF TWO BODIES “EMITTING” GAS FLOWS
(Presented by Academician N. N. Bogolyubov, December 7, 1957)
Let there be two stationary spherical bodies, of masses \(M_1\) and \(M_2\), located at a distance \(r_0\) from one another. We shall regard the gas emitted by the bodies as ultrarelativistic; let the equation of state of the gas be
\[ pv=\bar{R}T, \tag{1} \]
where \(\bar{R}=\mathrm{const}\). The adiabatic equation is \(pv^{1/3}=\mathrm{const}\), with \(p=\frac{1}{3}\rho c^2\). It may be assumed approximately that the pressure of the gas flowing out of one body will fall to some pressure of the external medium, determined likewise by the gas flow issuing from the other body (see Fig. 1). Obviously,
\[ r_1^2=r_2^2+r_0^2-2r_0r_2\cos\varphi\cos\lambda, \tag{2} \]
where \(\varphi\) is the “latitude” and \(\lambda\) the “longitude” relative to the second body of a given point located at a distance \(r_1\) from the first body and at a distance \(r_2\) from the second body.
Fig. 1
If \(r_0 \gg R_1\) and \(r_0 \gg R_2\), where \(R_1\) and \(R_2\) are the radii of the bodies \(M_1\) and \(M_2\), then, as is known \((^1)\),
\[ r_2^2=\frac{\bar{m}_2v_{02}}{c}\left(\frac{p_{02}}{p_{\Pi 2}}\right)^{(2-k)/k}, \tag{3} \]
where \(m_2=\bar{m}_2/4\pi\); \(\bar{m}_2\) is the mass of gas emitted by the body \(M_2\); \(p_{02}\), \(v_{02}\) are the pressure and specific volume of the gas in the body \(M_2\). Taking \(k=4/3\), we obtain
\[ \frac{p_{\Pi 2}}{p_{02}}=\left(\frac{\bar{m}_2v_{02}}{cr_2^2}\right)^2. \tag{4} \]
Analogously, for the body \(M_1\) we shall have
\[ \frac{p_{\Pi 1}}{p_{01}}=\left(\frac{\bar{m}_1v_{01}}{cr_1^2}\right)^2. \tag{5} \]
It follows from this that the pressure force, and the proportional reaction force, are \(F\sim m^2/r^2\) (if \(k\ne 4/3\), then this law will no longer hold).
When the gas may be considered as a continuous medium, outflow from both bodies will continue until the pressures equalize, i.e., until the condition
\[ p_{\Pi 1}=p_{\Pi 2}, \tag{6} \]
is satisfied, where \(p_{\Pi 1}\) and \(p_{\Pi 2}\) are the gas pressures for the first and second bodies, respectively. However, if lateral overflow of gas is taken into account, then outflow will continue until, on each elementary area, there is no equa-
the acting forces change as \(\sim \int p\,dr^{2}\). In this case, on the average a condition of the form must be satisfied
\[ p_{\mathrm{p}1} r_1^2 = p_{\mathrm{p}2} r_2^2 . \tag{7} \]
If rarefied gases flow out, then the depth of their mutual penetration must be taken into account. In this case the interaction of the bodies will decrease somewhat.
On the basis of (4) and (5) there is the relation
\(m_1^2 p_{01} v_{01}^{2}/r_1^2 = m_2^2 p_{02} v_{02}^{2}/r_2^2\), or, on the basis of (1), assuming that the temperature \(T_0=\mathrm{const}\),
\(\overline{m}_1\sqrt{v_{01}}/r_1=\overline{m}_2\sqrt{v_{02}}/r_2\).
Eliminating \(r_1^2\) from (2) with the aid of (6), and neglecting the lateral overflow of the gas, which is insignificant, we arrive at the equation of the surface on which \(p_{\mathrm{p}1}r_1^2=p_{\mathrm{p}2}r_1^2\):
\(r_2^2(1-\overline{a})-2r_0r_2\cos\varphi\cos\lambda+r_0^2=0\), where
\(\overline{a}=\overline{m}_1^2 v_{01}/\overline{m}_1^2 v_{02}\), whence
\[ \frac{r_2}{r_0} = \frac{1}{1-\overline{a}} \left[ \cos\varphi\cos\lambda \pm \sqrt{\cos^2\varphi\cos^2\lambda-1+\overline{a}} \right] \tag{8} \]
It is known \({}^{(1)}\) that if \(r_0 \gg R\), then the outflow velocity
\[ \frac{a_{\mathrm{p}}}{c} = 1-\frac{1}{2} \left(\frac{p_{\mathrm{p}}}{p_0}\right)^{2(k-1)/k}; \]
for the body \(M_2\), at \(k=4/3\), this expression has the form
\[ \frac{a_{\mathrm{p}2}}{c} = 1-\frac{1}{2}\sqrt{\frac{p_{\mathrm{p}2}}{p_{02}}} = 1-\frac{m_2 v_{02}}{2cr_2^2}. \tag{9} \]
Since the mass of gas flowing through an elementary area
\(df=r^2\cos\varphi\,d\varphi\,d\lambda\) is
\(d\overline{m}=\overline{m}\,df/4\pi r^2=\overline{m}\cos\varphi\,d\varphi\,d\lambda\), the projection of the force acting on the body \(M_2\) onto the axis \(r_0\) is
\[ -F=j = \int_{0}^{\overline{m}_2} a_{x2} \left( \frac{a_{\mathrm{p}2}}{c} + \frac{c}{3a_{\mathrm{p}2}} \right) \,d\overline{m} = 4m_2c \int_{0}^{\pi/2}\int_{0}^{\pi} \left( \frac{a_{\mathrm{p}2}}{c} + \frac{c}{3a_{\mathrm{p}2}} \right) \cos^2\varphi\cos\lambda\,d\varphi\,d\lambda, \tag{10} \]
where \(a_{x2}=a_2\cos\varphi\cos\lambda\).
Substituting here \(a_{\mathrm{p}2}\) from (9), we arrive at the expression
\(F=\dfrac{{}^{4}/_{3}m_2^2v_{02}}{r_0^2}\,\overline{F}(\overline{a})\), where
\[ \overline{F}(\overline{a}) = \int_{0}^{\pi/2}\int_{0}^{\pi} \left[ \cos\varphi\cos\lambda + \sqrt{\cos^2\varphi\cos^2\lambda-1+\overline{a}} \right]^2 \cos^2\varphi\cos\lambda\,d\varphi\,d\lambda = \]
\[ = 4 \int_{0}^{\pi/2}\int_{0}^{\pi/2} \sqrt{\cos^2\varphi\cos^2\lambda-1+\overline{a}}\, \cos^3\varphi\cos^2\lambda\,d\varphi\,d\lambda . \tag{11} \]
In the general case, it is difficult to compute \(\overline{F}(\overline{a})\). If \(\overline{a}>1\), which may occur when \(M_1/M_2>1\) and, consequently, when \(\overline{m}_1/\overline{m}_2>1\), then
\[ \overline{F}(\overline{a}) \cong \frac{2}{3}\pi\sqrt{\overline{a}} \left(1-\frac{1}{5\overline{a}}\right). \tag{12} \]
In this case
\[ F\cong \frac{m_1m_2\sqrt{v_{01}v_{02}}}{18\pi r_0^2} \left[ 1-\frac{\overline{m}_2^{\,2}v_{02}}{5\overline{m}_1^{\,2}v_{01}} \right]. \tag{13} \]
If \(\overline{a}=1\), then the following relations will hold:
\[ M_1=M_2=M;\qquad \overline{m}_1=\overline{m}_2=\overline{m};\qquad v_{01}=v_{02}=v_0; \]
\[ \overline{F}(\overline{a})=\pi/2;\qquad F=\overline{m}^{\,2}v_0/24\pi r_0^2 . \tag{14} \]
If in (13) we put $\bar a=1$, we obtain
\[ F=\frac{2\bar m^{2}v_{0}}{45\pi r_{0}^{2}}, \tag{15} \]
which differs little from (14). Thus, in calculating $\overline{F}(\bar a)$, it is sufficient to restrict oneself to the first term of the expansion in powers of $\bar a$ and to assume that the investigation can be carried out by means of an interpolation relation that gives correct results for $\bar a\to\infty$ and $\bar a=1$:
\[ F=\frac{\bar m_{1}\bar m_{2}\sqrt{v_{01}v_{02}}}{18\pi r_{0}^{2}} \left[ 1-\frac{\bar m_{2}^{2}v_{02}}{4\bar m_{1}^{2}v_{01}} \right]. \tag{16} \]
For $\bar a\gg 1$
\[ F=\frac{\bar m_{1}\bar m_{2}\sqrt{v_{01}v_{02}}}{18\pi r_{0}^{2}}. \tag{17} \]
It is obvious that the force of interaction between the bodies will be an attractive force, since the gas expands nonuniformly, namely, less when flowing into the region inside the bodies. In this case the difference in the momenta of the outflowing gas will lead to an “attraction” of the bodies. A pressure force $F_p$ also acts on the bodies; however, since $F_p\sim pR^{2}$, while $F\sim pr_{0}^{2}$, we have $F_p/F\sim R^{2}/r_{0}^{2}$, whence it follows that the pressure force may be neglected when $R/r_{0}\gg 1$.
The case $M_{1}=M_{2}=M$ is the most interesting and is treated most correctly, since for $a=1$ the result does not depend on the criteria of equality of pressures or forces. In this case the surface of “interaction” of the flows is a plane perpendicular to $r_{0}$ and bisecting $r_{0}$.
True, some difference in the numerical coefficient in the law of interaction can be explained by lateral overflow of gases for dense flows and, if for not very dense flows one takes into account the depth of their mutual penetration. Both these factors may somewhat reduce the force of interaction.
Having correctly considered the case when $M_{1}=M_{2}$, it is easy, by the usual methods of potential theory (assuming that a superposition of flows takes place, which is valid at large distances from the bodies), to consider the general case when $M_{1}>M_{2}$, assuming that $\bar M_{1}=\bar k M_{2}$, where $\bar k>1$. In this case it will naturally turn out that the influence of body $M_{1}$ on $M_{2}$ is equal to the influence of body $M_{2}$ on $M_{1}$.
Assuming that
\[ \bar m_{1,2}\sqrt{v_{01,2}} \left( 1-\frac{1}{4}\frac{\bar m_{2}^{2}}{\bar m_{1}^{2}}\frac{v_{02}}{v_{01}} \right)^{1/2} =\beta_{1,2}M_{1,2}, \tag{18} \]
we find
\[ F=\frac{\beta_{1}\beta_{2}M_{1}M_{2}}{18\pi r_{0}^{2}} =\frac{GM_{1}M_{2}}{r_{0}^{2}}, \tag{19} \]
where we have taken $\beta_{1}\beta_{2}=18\pi G=\mathrm{const}$.
We have arrived at a law of interaction between bodies of the form of Newton’s or Coulomb’s law.
If we assume that the amount of gas emitted by the bodies per unit time is proportional to the mass of the bodies, then one must put
\[ \overline{m}_1=\overline{\alpha}_1 M_1,\qquad \overline{m}_2=\overline{\alpha}_2 M_2, \tag{20} \]
where
\[ \overline{\alpha}_1=\sqrt{\frac{18\pi G\rho_{01}}{1-\frac14 M_2^2/M_1^2}},\qquad \overline{\alpha}_2=\sqrt{\frac{18\pi G\rho_{02}}{1-\frac14 M_2^2/M_1^2}}, \]
where \(\rho_{01,2}=1/v_{01,2}\). Hence
\[ \rho_{01,2}=\frac{\overline{\alpha}_{1,2}^{\,2}}{18\pi G} \left(1-\frac14\frac{M_2^2}{M_1^2}\right). \tag{21} \]
In the general case one may write
\[ \rho_{01,2}=\frac{\overline{\alpha}_{1,2}^{\,2}}{18\pi G}\,\theta, \tag{22} \]
where \(\theta=\theta_0(1-\frac14 M_2^2/M_1^2)\); for \(M_2/M_1=1\), \(\theta=\frac34\theta_0\), while for \(M_2/M_1\to 0\), \(\theta=1\). The factor \(\theta_0<1\) takes into account the degree of mutual penetration of the gas streams during their interaction.
Let us now consider the case in which the bodies emit not gas particles, but quanta with rest mass equal to zero, and let us suppose that the same equations of state are valid for radiation. Eliminating from the equations
\[ m=\frac{\overline{m}}{4\pi} =\frac{r^2 a}{v\sqrt{1-a^2/c^2}},\qquad \frac{w}{\sqrt{1-a^2/c^2}}=w_0 \tag{23} \]
the quantity \(a\), we find that for radiation the expression
\[ m=\frac{\overline{m}}{4\pi} =\frac{c w_0 r^2\sqrt{1-w^2/w_0^2}}{wv} =\frac{E_0}{4\pi c^2}, \tag{24} \]
will be valid, where \(E_0\) is the total energy of the radiation.
Since
\[ \frac{w}{w_0}=\left(\frac{p}{p_0}\right)^{1/4} =\left(\frac{v_0}{v}\right)^{1/3}, \tag{25} \]
replacing \(w\) by \(p\), we arrive at the equation
\[ \left(\frac{p}{p_0}\right)^{3/2} -\frac{p}{p_0} +\left(\frac{v_0 E_0}{4\pi c^3 r^2}\right)^2=0. \tag{26} \]
At large distances from the bodies
\[ \frac{p}{p_0}\simeq \left(\frac{E_0v_0}{4\pi c^3 r^2}\right)^2,\qquad \frac{w}{w_0}\simeq \frac1r\sqrt{\frac{v_0E_0}{4\pi c^3}}, \tag{27} \]
and for the force of interaction we arrive at the previous result.
As was shown earlier \((^1)\), the solution used has meaning for \(\infty>r\ge r_{\min}=r_{\mathrm{cr}}\), where
\[ r_{\mathrm{cr}}^2=3\sqrt{3}\,m/2\rho_0 c =3\sqrt{3}\,m/8\pi c\rho_0. \]
Replacing \(m=\overline{\alpha}M\), we find that
\[ r_{\mathrm{cr}}^2=3\sqrt{3}\,\overline{\alpha}M/8\pi\rho_0 c. \]
It is natural to assume that \(r_{\mathrm{cr}}=R\), where \(R\) is the radius of the radiating body. Then we shall have
\[ \overline{\alpha}_{1,2} =\frac{8\pi c\rho_{01,2} R_{1,2}^{\,2}} {3\sqrt{3}\,M_{1,2}}. \tag{28} \]
Solving (22) and (28) together, we find
\[ \overline{\alpha}_{1,2} =\frac{27\sqrt{3}\,G M_{1,2}}{4c\theta R_{1,2}^{\,2}}, \qquad \rho_{01,2} =\frac{243G M_{1,2}^{\,2}}{32\pi c^2\theta R_{1,2}^{\,4}}. \tag{29} \]
Thus, we have expressed the quantities \(\overline{\alpha}_{1,2}\) and \(\rho_{01,2}\) in terms of the interaction constant between the bodies, their masses, and their sizes.
In conclusion I express my gratitude to I. G. Aramanovich for valuable discussions.
Received
2 XII 1957
REFERENCES
- K. P. Stanyukovich, DAN, 119, No. 2 (1958).