Orbital decay of PSR B1913+16 and the use of averaged dE/dt vs. instantaneous dE/dt

[Alexrey]

I've been looking at some sections in GR textbooks that deal with the orbital decay of the binary pulsar PSR B1913+16 (Straumann's General Relativity: With Applications to Astrophysics and Padmanabhan's Gravitation have some pretty good sections on this) , and I noticed something strange to do with the way they handled energy loss used in calculating the period change dT/dt of the system.
In both books, they first found the energy loss per period of the system <dE/dt>, and then used this to calculate the period change of the system dT/dt where the relation between the rate of change of energy and the rate of change of period was given by 1/T dT/dt = 2/3E dE/dt (I'm not too sure if this formula is 100% correct as I don't have the books in front of me at the moment). Now, I thought that since we are dealing with energy loss per period, the above formula should have been 1/T dT/dt = 2/3E <dE/dt> , but in both books they seemed to just drop the fact that an energy loss per period was used and treated it as if it was an instantaneous energy loss. Thus their calculation of energy loss "per period" came out with an answer with units of erg per second instead of erg per period as I would have expected, where they then used this answer and the above relation between period change and energy change to find dT/dt with units of seconds per second.

Why was it possible for Straumann and Padmanabhan to treat <dE/dt> (energy loss per period) as dE/dt (instantenous energy loss)?

I apologize for the lack of formulas, but I'm on campus right now and both of my books are at home.

 

[mfb]

I think it is just a sloppy notation.
As the orbits are not closed, an instantaneous "T" is not well-defined, so you have to average over a whole period (or multiple periods) anyway. Afterwards, you can treat it like an instantaneous period change again to integrate it, for example. There, it does not matter whether you averaged over <dE/dt> or not, as just the integrated energy loss within a period matters.

Concerning the relation between energy loss and period:
T^2 = c*E^3 (Kepler, should be a good approximation for the pulsars, too)
derive:
2 T dT/dt = 3c E^2 dE/dt
divide both equations:
1/T dT/dt = 3/2 1/E dE/dt

 

[Alexrey]

Sorry, I'm still a little confused. Do we get a value for the instantaneous energy loss by multiplying the answer of the averaged energy loss per period by the binary system's period to go from erg/period to erg/s?

 

[mfb]

Even if you have perfect knowledge about the system: How do you define "instantaneous energy loss"? Energy in some volume? As the gravitational field is relevant, which volume do you use?
Energy flow at a specific distance? This varies with distance, as the time delay for the gravitational waves changes.

 

[Alexrey]

Wow those are some pretty interesting questions.

 

[Sourabh N]

I am not sure what the definition of <dE/dt> is (I've not read either of those books) but to me <dE/dt> looks like the average energy loss per second, which is different from dE/dT which is the energy loss per period.

 

[pervect]

To my mind, it's rather similar to the problem of compounding interest - i.e. just an accounting issue.

Your energy "account" is losing energy. You can only actually measure the rate at which it is loosing energy compounded over an orbital period. But you can mathematically model it as a continuous function. It's just a book-keeping trick.

 

[PeterDonis]

It's just a book-keeping trick.

Don't let Wall Street know. Next up: credit default swaps on binary pulsars.