(Quadrupole tensor; circular binary orbit)

where we factored out the common term 2mr2. The components of the tensor contain the time variation of the quadrupole moment as the masses orbit at frequency m. The strain tensor (9) requires that each term be differentiated with respect to time twice. The result is

—cos 2at —sin 2rnt 0 —sin 2at cos 2at 0 0 0 0_

(Strain tensor for binary, circular orbit)


An additional tensor operation is required to extract the desired observable scalar quantities, the strains h + and hx at different observer positions. For an observer in the equatorial plane, it turns out that h + = hn/2 and hx = 0, where the hn component is the upper left term in the tensor matrix. Thus, for our observer in the equatorial plane, h+ = -

cos 2mt (Strain for equatorial observer; (12.12) circular orbit, m = m1 = m2)

Here, m = 2^/Porb, where Porb is the orbital period, and again m is the mass of each star, m is the orbital angular velocity, r the radius from the center of mass (1/2 the separation), and R the distance to the observer.

The strain h+in (12) will oscillate at twice the orbital frequency. It is apparent in (12) that a large mass, a large orbital radius, and a high orbital frequency all increase the amplitude of the space distortion. At the pole of the orbit (inclination zero), h+ and hx have equal amplitudes, each twice that given for h+ in (12). The result is circular polarization.

Compare (12) to (9) and note that the second time derivative of the tensor Q jk reduces, for this particular situation, to - m2Iz cos 2Mt where Iz = 2 mr2, the moment of inertia about the z axis (Fig. 8a). This is the same result we would have obtained (except for the minus sign) if we had naively used the second time derivative of the moment of inertia about the view (x) axis Ix in (9) instead of Q jk and had also ignored the geometrical factor of 2 introduced in the conversion from h 11 to h+. The moment of inertia about the x axis oscillates as Ix = 2m (r sin Mt )2 = mr 2(1 — cos 2wt). The x-axis observer sees the separation of the two masses oscillating between zero and 2r twice each orbit, i.e., with angular frequency 2m. The second time derivative of Ix substituted into (9) yields the result (12), less the minus sign.

This shows us that mass motions relative to the observer's line of sight give rise to the observed radiation for this particular situation. One can use this result to estimate the magnitudes of the strain from a system with moment of inertia I to be ~ (G/c4)(m2I/R) if the mass distribution does not have a substantial spherical component.

If the mass distribution were purely spherical, such as a sphere with oscillating radius, there would be a large I, but Qjk = 0 for a sphere, so also Qjk = 0. (This notation indicates that each component of the tensor is zero.) Thus, in spite of the large I x, there is no gravitational radiation. If in doubt or in need of a precise result, use the proper quadrupole tensor expressions.

For completeness, we quote here the expressions for h+ and hxfor the more general case where the two masses m1 and m2 are not necessarily equal and the angle between the observer and the pole of the orbit (inclination i; Fig. 8a) need not be 90°. The orbits are again circular and, as before, m is the orbital angular velocity, M = m 1 + m2 and x = m1m2/M.

c R circular

4G M2/3x(GM)2/3cos i hx = —--—---cod 2 m dt inclination i,

circular orbits at (12.13)

c4 R

We have inserted a phase shift of ^ radians (thus removing a minus sign) in each of these expressions (13) to be in accord with convention. The angular position of one of the stars, formerly indicated by Mt, should actually be the integral form shown because m is varying with time due to the inward spiraling of the stars. Finally, these expression make use of Kepler's law, GM = w2a3 to remove the star separation a = r1 + r2 from the final expression (13). The reader can confirm that these expressions reduce to those of (12) for the specified circumstances.

Detection in Virgo cluster

Substitute the values for our system, namely m = 1.4 MQ, r = 1 x 104 m, Porb = 1 x 10—3 s, G/c4 = 0.8 x 10—44 (SI units), into (12) to obtain the result,

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