From this link

The Hulse–Taylor system's orbit has decayed since the binary system was initially discovered, in precise agreement with the loss of energy due to gravitational waves. The ratio of observed to predicted rate of orbital decay is calculated to be 0.997±0.002.

But also

The total power of the gravitational radiation (waves) emitted by this system presently, is calculated to be 7.35 × 1024 watts. For comparison, this is 1.9% of the power radiated in light by our own Sun.

So how do we know the Hulse–Taylor system's loss of energy is not (at least partially) due to electromagnetic radiation too? Especially if we don't even see the other pulsar due to its unfavorable inclination?

  • 2
    $\begingroup$ AFAIK, the other star in the Hulse-Taylor system isn't a pulsar, it's a neutron star (though all pulsars are NS, not all NS are pulsars). $\endgroup$
    – Kyle Kanos
    Feb 11 '16 at 13:32
  • $\begingroup$ You are right, but my point holds: Pulses from the companion neutron star have not been detected, but this might only be the result of an unfavorable viewing angle. $\endgroup$ Feb 11 '16 at 13:34
  • $\begingroup$ Probably useful reading: ned.ipac.caltech.edu/level5/ESSAYS/Boughn/boughn.html $\endgroup$
    – Kyle Kanos
    Feb 11 '16 at 13:37
  • $\begingroup$ Kind of amusing that the Wikipedia authors characterize a 1.5 sigma deviation from theory as "in precise agreement." $\endgroup$
    – Rococo
    Feb 16 '16 at 19:59

The emission of gravitational waves causes the separation $r$ between the two binary components to decrease. As they do so, the power emitted in gravitational waves increases as $r^{-5}$. Thus the rate of change of the orbital period is very non-linear, with $dr/dt \propto r^{-3}$.

Now, if the mechanism responsible for the spin-down was somehow pulsar radiation, then you would have to arrange for the pulsar spin to increase with time in order to provide the same accelerating power loss. But that is the opposite of what pulsars do - they all spin down with age and become considerably less powerful. Pulsars are also powered by their spin - that is the rotational kinetic energy is ultimately the source of power. The rotation of the pulsar is not coupled in any straightforward way with the orbital kinetic energy of the binary system. There would be (weak) tidal coupling, but that would tend to slow down the pulsar too.

Aside from this argument, one could also add that the model of the binary system that incorporates the gravitational wave losses fits the data with exquisite precision. Thus any other source of power loss one might hypothesise not only has to provide the right sign for the time derivatives of the power loss, but it also has to get the temporal behaviour exactly right over decades.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.