Role of a Time Delay in the Gravitational Two-Body Problem


In the traditional frame of classical electrodynamics, a hydrogen atom would emit electromagnetic waves and thus constantly lose energy (radiative damping), resulting in the fall of the electron on the proton over a finite period of time. The corresponding results were derived under the assumption of the instantaneous interaction between the proton and the electron. In 2004, Raju published a paper where he removed the assumption of the instantaneous interaction and studied the role of a time delay (retardation) in the classical hydrogen atom. He introduced a model of this effect that can be solved analytically. He calculated the radius rsel of a selected orbit, for which this effect (the retardation) and the radiation reaction force would cancel each other with respect to the tangential acceleration. It turned out that rsel is by an order of magnitude smaller than the Bohr radius. In the present paper we use Raju’s model for the corresponding gravitational (rather than electromagnetic) situation. In frames of Newton’s gravity, we calculate analytically the radius and the energy Esel of a selected orbit, for which the retardation effect and the radiation reaction force (caused by the emission of gravitational waves) would cancel each other out with respect to the tangential acceleration. Then we compare our outcome with some results of Kaplan’s solution for the two-body gravitational problem based on the Schwarzschild’s field. We show that Esel is very close to the minimum energy of bound states obtained by Kaplan in Einstein’s gravity. We emphasize that the selected orbit from the present paper represents the second only physical situation (the first being a uniformly accelerated charge) where the radiation reaction force is (effectively) zero, but the radiation-caused energy loss is not zero.

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Appendix A: Comments on the Electrodynamic Two-Body Problem

Appendix A: Comments on the Electrodynamic Two-Body Problem

Raju [2], while applying his model to hydrogen atoms, derived the radius of the selected orbit rsel by balancing two torques: one torque due to the time delay, another—due to the radiation reaction force. Here we check his result through an alternative calculation: by balancing the rate of the energy gain due to the time delay and the rate of the energy loss due to the work done by the radiation reaction force. In the course of this calculation, we also correct some error from Raju paper [2].

For the rate of the energy gain we can start from Eq. (21) and substitute km1m2 by e2:

$$ \left( {{\text{dE}}/{\text{dt}}} \right)_{\text{gain}} = {\text{ e}}^{ 2} {\text{m}}_{ 1} {\text{m}}_{ 2} \omega^{ 2} /\left[ {{\text{c}}\left( {{\text{m}}_{ 1} + {\text{ m}}_{ 2} } \right)^{ 2} } \right]. $$

The square of the Kepler frequency in Eq. (A.1) is

$$ \omega^{ 2} = { 8}\left( {{\text{m}}_{ 1} + {\text{ m}}_{ 2} } \right)\left| {\text{E}} \right|^{ 3} /\left( {{\text{m}}_{ 1} {\text{m}}_{ 2} {\text{e}}^{ 4} } \right), $$

where |E| = − E is the binding energy. After substituting Eq. (A.2) in Eq. (A.1), we get:

$$ \left( {{\text{d}}\left| {\text{E}} \right|/{\text{dt}}} \right)_{\text{gain}} = \, {-}{ 8}\left| {\text{E}} \right|^{ 3} /\left[ {{\text{ce}}^{ 2} \left( {{\text{m}}_{ 1} + {\text{ m}}_{ 2} } \right)} \right]. $$

As for the work done by the radiation reaction force per unit time, we start from the 1st equation in Problem 1 from Sect. 70 of the textbook [1], which for the case of a hydrogen atom can be represented in the form

$$ \left( {{\text{d}}\left| {\text{E}} \right|/{\text{dt}}} \right)_{\text{loss}} = {\text{ 2e}}^{ 6} /\left( { 3 {\text{c}}^{ 3} \mu^{ 2} {\text{R}}^{ 4} } \right), $$

where μ is the reduced mass (defined in Eq. (17)). For circular orbits, the inverse radius 1/R has a simple expression through the binding energy |E|:

$$ 1/{\text{R }} = { 2}\left| {\text{E}} \right|/{\text{e}}^{ 2} . $$

After substituting Eq. (A.5) in Eq. (A.4), we obtain:

$$ \left( {{\text{d}}\left| {\text{E}} \right|/{\text{dt}}} \right)_{\text{loss}} = { 32}\left| {\text{E}} \right|^{ 4} /\left( { 3 {\text{c}}^{ 3} {\text{e}}^{ 2} \mu^{ 2} } \right). $$

By combining Eqs. (A.3) and (A.6), we obtain the final expression for the partial rate of the energy change (while disregarding the loss due to the energy flux through a surface surrounding the volume containing the two-body system):

$$ {\text{d}}\left| {\text{E}} \right|/{\text{dt }} = \, \left( {{\text{d}}\left| {\text{E}} \right|/{\text{dt}}} \right)_{\text{gain}} + \left( {{\text{d}}\left| {\text{E}} \right|/{\text{dt}}} \right)_{\text{loss}} = { 32}\left| {\text{E}} \right|^{ 4} /\left( { 3 {\text{c}}^{ 3} {\text{e}}^{ 2} \mu^{ 2} } \right) \, {-}{ 8}\left| {\text{E}} \right|^{ 3} /\left[ {{\text{ce}}^{ 2} \left( {{\text{m}}_{ 1} + {\text{ m}}_{ 2} } \right)} \right]. $$

From Eq. (A.7) it is easy to find out that d|E|/dt = 0 at the following value of the binding energy

$$ \left| {\text{E}} \right|_{\text{sel}} = {\text{ 3c}}^{ 2} {\text{m}}_{ 1}^{ 2} {\text{m}}_{ 2}^{ 2} /\left[ { 4\left( {{\text{m}}_{ 1} + {\text{ m}}_{ 2} } \right)^{ 3} } \right]. $$

For hydrogen atoms (m1 = me, m2 = mp), Eq. (A.8) yields |E|sel = 208 eV. This is 15 times greater than the actual binding energy of the ground state of hydrogen atoms |E|bind = 13.6 eV.

By substituting |E|sel from Eq. (A.8) in Eq. (A.5), we get the following radius of the selected circular orbit:

$$ {\text{R}}_{\text{sel}} = {\text{ 2e}}^{ 2} \left( {{\text{m}}_{ 1} + {\text{ m}}_{ 2} } \right)^{ 3} /\left( { 3 {\text{c}}^{ 2} {\text{m}}_{ 1}^{ 2} {\text{m}}_{ 2}^{ 2} } \right). $$

For hydrogen atoms Eq. (A.9) yields Rsel = 3.45 × 10−10 cm. This is 15 times smaller than the Bohr radius.

For comparison with the radius of the selected orbit obtained by Raju [2], let us approximate Eq. (A.9) by using m1 ≪ m2:

$$ {\text{R}}_{\text{sel}} \approx \, \left[ { 2 {\text{e}}^{ 2} /\left( { 3 {\text{m}}_{ 1} {\text{c}}^{ 2} } \right)} \right]{\text{m}}_{ 2} /{\text{m}}_{ 1} . $$

The corresponding result from Raju paper [2] can be obtained by equating ω2 ≈ 3m1c2/(2m2R2) from his Eq. (49) to ω2 = e2/R3 from his formula in the 5th line after Eq. (49). This would yield

$$ {\text{R}}_{{{\text{sel}},{\text{Raju}}}} \approx \, \left[ { 2 {\text{e}}^{ 2} /\left( { 3 {\text{c}}^{ 2} } \right)} \right]{\text{m}}_{ 2} /{\text{m}}_{ 1} . $$

From the comparison of Eqs. (A.10) and (A.11), it is seen that the factor m1 is missing in the denominator in Raju’s value of Rsel,Raju. This (typographic?) error of Raju [2] can be traced back to his formula ω2 = e2/R3, which has an incorrect dimension. Namely, the correct formula (which follows from equating the centripetal force to the Coulomb force) should have been ω2 = e2/(m1R3). After correcting this (typographic?) error, the expression (A.11) for Rsel,Raju would coincide with the equilibrium radius Rsel from Eq. (A.10) that we obtained from the energy balance.

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Oks, E. Role of a Time Delay in the Gravitational Two-Body Problem. Found Phys 51, 23 (2021).

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  • Gravitational two-body problem
  • Retardation
  • Radiation reaction force
  • Schwarzschild’s field
  • Kaplan’s solution