Bohr's model and a statement in my textbook

[photon184739]

In satellite motion around a planet, if the satellite is in orbit, it experiences only the gravitational force and its energy in orbit is constant (see 'https://openstax.org/books/university-physics-volume-1/pages/13-4-satellite-orbits-and-energy'):

It's not radiating energy, is it? Its total energy is constant and a function of 'r' only, which is constant.

 

[PeroK]

The radiation of an accelerating charge is a result from classical EM:

https://en.wikipedia.org/wiki/Larmor_formula

 

[mpresic3]

The earlier answer by PeroK is correct. This is a good question. Your relating quantum with electricity to gravitation and noting the discrepancy illustrates how a physicist thinks by analogy. One could argue the revolving planets also lose energy due to gravitational radiation, but this would be outside the scope of the energy equations you are relating there. Suffice it to say, Bohr's contemporaries in quantum mechanics had to confront the problem that at the rate of radiation, the electron should spiral into the nucleus before a nanosecond. (I think it may be femtosec)

 

[PeterDonis]

In satellite motion around a planet, if the satellite is in orbit, it experiences only the gravitational force and its energy in orbit is constant

Yes. This is a way in which the gravitational force is different from the electromagnetic force; objects that are orbiting other objects due to gravity do not radiate.

 

[mpresic3]

Don't (rotating) binary stars radiate gravitational waves? I keep hearing in the news about gravitational waves. What generates them, if accelerations do not?

 

[vanhees71]

Yes. This is a way in which the gravitational force is different from the electromagnetic force; objects that are orbiting other objects due to gravity do not radiate.

In fact they do. That's what LIGO measures as gravitational wave signals but from much larger objects moving around each other (pairs black holes or neutron stars). The difference between gravitation and electromagnetism in this context is that the coupling constants are largely different. E.g., if you compare the gravitational and the electrostatic force between two protons you find that gravity is by a factor of about $10^{40}$ smaller (though the force laws are the same in the Newtonian approximation of gravity, going with $1/r^2$).

The energy loss of a planet orbiting our Sun due to gravitational waves is thus negligible over the relevant time scales we observe it. That's not the case for an electron classically orbiting an atomic nucleus. The energy loss due to electromagnetic radiation is large enough to make it almost immediately fall into the nucleus. The resolution of this obviously wrong prediction for the electron is quantum mechanics. There are static solutions (energy eigenfunctions) of the Schrödinger equation, i.e., the electron is not moving but just located on an indefinite place around the nucleus. Since nothing is moving or accelerated there's no radiation and you have a stable atom.

The emission of gravitational waves in GR and electromagnetic waves in Maxwellian electromagnetics is quite similar. The main difference is that the multipole expansion of gravitational waves starts with the quadrupole term, such that the irradiated gravitational energy goes with $\omega^6$ (where $\omega$ is the typical frequency of the moving matter), while the em. waves start with the dipole term with the irradiated em. energy going like $\omega^4$. Put in another way the gravitational-wave energy loss goes like the 3rd time derivative of the quadrupole moment of the matter distribution squared, while for em. radiation it's the 2nd time derivative of the dipole moment of the electric charge.

 

[haushofer]

Don't (rotating) binary stars radiate gravitational waves?

Yes. In fact, at far distances the Einstein equations are linearized, becoming similar to the Maxwell equations.

 

[PeterDonis]

Don't (rotating) binary stars radiate gravitational waves? I keep hearing in the news about gravitational waves. What generates them, if accelerations do not?

Yes, binary stars (and other orbiting objects) do radiate gravitational waves, but just "acceleration" is not what generates them, unlike in electromagnetism. EM waves are generated by a time-varying dipole moment, so even a perfectly circular orbit is enough. Gravitational waves are generated by a time-varying quadrupole moment, so a two-body system with perfectly circular orbits does not generate them. You need elliptical orbits; roughly speaking, the larger the objects, the closer the average distance between them, and the more elliptical the orbits, the more gravitational waves are generated. In terms of "acceleration", a constant acceleration is enough to generate EM waves, but you have to have time-varying acceleration to generate gravitational waves.

In fact they do.

Yes, you're right, I was being sloppy. See my response to @mpresic3 above.

 

[PeterDonis]

at far distances the Einstein equations are linearized, becoming similar to the Maxwell equations.

Not quite, since, as noted in my previous post just now, gravitational radiation is quadrupole while EM radiation is dipole.

 

[vanhees71]

Yes, indeed the energy loss goes parametrically like $\ddot{d}^2 \sim \omega^4$ for em. radiation where $d$ is the dipole moment of the charge distribution and like $(\mathrm{d}_t^3 Q) \sim \omega^6$ for gravitational waves where $Q$ is the quadrupol distribution of the mass/energy distribution. See also #6.

 

[haushofer]

Not quite, since, as noted in my previous post just now, gravitational radiation is quadrupole while EM radiation is dipole.

Yes, but isn't this a property of the geodesic equation instead of the Einstein field equations?

 

[vanhees71]

It follows from the Einstein field equations for grav. waves in the same way as it follows from the Maxwell equations for em. waves (at least for the linearized Einstein field equations for weak gravitational waves).

The difference is that for em. waves you have a massless spin-1 field for the grav. waves you have a massless spin-2 field. Both have only 2 polarization degrees of freedom, but the multipole expansions start of course with $J=1$ (i.e., the dipole contribution) for the em. field and with $J=2$ (i.e., the quadrupole contribution) for the grav. field.

This is also clear from the heuristical argument considering the sources. In the em. case it's the charge distribution. Of course there's a monopole component given by the total charge. Since this corresponds to a spherical symmetric situation outside the charge distribution (i.e., in the vacuum) it just gives a Coulomb field, no matter how the charge might move. Of course that's just charge conservation, i.e., outside of the charge distribution, no matter how the charge moves, the total charge (the "monopole moment" of the charge distribution) stays constant in time and gives only rise to a static field and no waves. The next term is usually the dipole moment, given (in terms of Cartesian components) by
$$\vec{P}=\int_V \mathrm{d}^3 x \vec{x} \rho(t,\vec{x}),$$
and in general there's no way to transform this away by any Poincare transformation.

For the gravitational field the sources are the mass distributions (in the here considered quasi-Newtonian limit). Concerning the monopole contribution it's the same as with the em. field and the charge distribution: Outside the mass distribution the solution of this spherically symmetric piece is just the static Schwarzschild solution (Birkhoff's theorem for the gravitational field). The dipole piece is just the center of mass, and you can always go to the center-of-mass frame, so that the dipole term vanishes. So the first term which can give rise to gravitational waves is the quadrupole term.

 

[PeterDonis]

isn't this a property of the geodesic equation instead of the Einstein field equations?

I'm not sure what you mean. The geodesic equation has no content unless you have a solution of the Einstein Field Equations.