No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the angular dependence to the energy flux.

That you need to consider the quadrupole moment is easy to understand: the dipole moment of a mass distribution can always be set to exactly $0$ if the origin of coordinates is chosen in the system center of mass. The monopole moment is the total system mass, which is conserved (in the weak-field limit), hence the monopole field at large distances is always $-GM/r$, a time constant, hence no radiation. The first physically relevant term is thus the quadrupole moment, $D_{ab}$.

Calling $X_{ab}$ the third time derivative of $D_{ab}$, the time-averaged energy flux in an element of solid angle $d\!\Omega$ in the direction $\vec n$, where $\vec n$ is a unit vector, is (Landau & Lifshitz, Field Theory, vol. 2, Ch.13) is:

$$d\!I = \frac{G}{144\pi c^5}\left((X_{ab}n_an_b)^2 + 2X_{ab}X_{ab} -4 X_{ab}X_{ac}n_b n_c \right) d\!\Omega $$

Obviously, $n_a$ are the components of $\vec n$. It should be obvious from the previous argument that GWs cannot be emitted isotropically.

No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the angular dependence to the energy flux.

That you need to consider the quadrupole moment is easy to understand: the dipole moment of a mass distribution can always be set to exactly $0$ if the origin of coordinates is chosen in the system center of mass. The monopole moment is the total system mass, which is conserved (in the weak-field limit), hence the monopole field at large distances is always $-GM/r$, a time constant, hence no radiation. The first physically relevant term is thus the quadrupole moment, $D_{ab}$.

We can use the exact expression to make the above more precise. Calling $X_{ab}$ the third time derivative of $D_{ab}$, the energy flux averaged over one wave period in an element of solid angle $d\!\Omega$ in the direction $\vec n$, where $\vec n$ is a unit vector, is (Landau & Lifshitz, Field Theory, vol. 2, Ch.13) is:

$$d\!I = \frac{G}{144\pi c^5}\left((X_{ab}n_an_b)^2 + 2X_{ab}X_{ab} -4 X_{ab}X_{ac}n_b n_c \right) d\!\Omega $$

Obviously, $n_a$ are the components of $\vec n$. The tensor $D$ is defined as

$$D_{ab} \equiv \int (3x_a x_b -r^2 \delta_{ab}) \rho d\!V$$

which shows $D$ to be symmetric and traceless, $D_{aa} = 0$. Both properties carry over to $X \equiv d^3 D/dt^3$, so that we can always choose a system of axes such that $X$ is (at least instantaneously) diagonal, and traceless:

$$ X =\left(\begin{matrix}X_1 & 0 & 0 \\ 0 & X_2 & 0 \\ 0 & 0 & -(X_1+X_2) \end{matrix}\right) $$

The most symmetric that $X$ can be is by having $X_1 =X_2$, but there is no way that it can have $X_3=X_1=X_2$, because of its being traceless. If we now choose $\vec n$ is the direction identified by $X_1$, we have

$$d\!I = \frac{G}{144\pi c^5}\left(4X_2^2+4X_1X_2 \right) d\!\Omega $$

while if we take $\vec n$ in the direction of $X_3$ we find:

$$d\!I = \frac{G}{144\pi c^5}\left( 2X_1^2 + 2 X_2^2 \right) d\!\Omega $$

which differs from the previous formula whether $X_1 = X_2$ or not.

No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the angular dependence to the energy flux.

That you need to consider the quadrupole moment is easy to understand: the dipole moment of a mass distribution is a constant, and can always be set to exactly $0$ if the origin of coordinates is chosen in the system center of mass. The monopole moment is the total system mass, which is conserved (in the weak-field limit), hence the monopole field at large distances is always $-GM/r$, a time constant, hence no radiation. The first physically relevant term is thus the quadrupole moment, $D_{ab}$.

Calling $X_{ab}$ the third time derivative of $D_{ab}$, the time-averaged energy flux in an element of solid angle $d\!\Omega$ in the direction $\vec n$, where $\vec n$ is a unit vector, is (Landau & Lifshitz, Field Theory, vol. 2, Ch.13) is:

$$d\!I = \frac{G}{144\pi c^5}\left((X_{ab}n_an_b)^2 + 2X_{ab}X_{ab} -4 X_{ab}X_{ac}n_b n_c \right) d\!\Omega $$

Obviously, $n_a$ are the components of $\vec n$. It should be obvious from the previous argument that GWs cannot be emitted isotropically.

No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the angular dependence to the energy flux.

That you need to consider the quadrupole moment is easy to understand: the dipole moment of a mass distribution can always be set to exactly $0$ if the origin of coordinates is chosen in the system center of mass. The monopole moment is the total system mass, which is conserved (in the weak-field limit), hence the monopole field at large distances is always $-GM/r$, a time constant, hence no radiation. The first physically relevant term is thus the quadrupole moment, $D_{ab}$.

Calling $X_{ab}$ the third time derivative of $D_{ab}$, the time-averaged energy flux in an element of solid angle $d\!\Omega$ in the direction $\vec n$, where $\vec n$ is a unit vector, is (Landau & Lifshitz, Field Theory, vol. 2, Ch.13) is:

$$d\!I = \frac{G}{144\pi c^5}\left((X_{ab}n_an_b)^2 + 2X_{ab}X_{ab} -4 X_{ab}X_{ac}n_b n_c \right) d\!\Omega $$

Obviously, $n_a$ are the components of $\vec n$. It should be obvious from the previous argument that GWs cannot be emitted isotropically.