Star with quadrupole in binary system violates Newton's 3th law?

Question

Suppose that, in a binary system of two stars, the star A (and only the star A) has a non-zero quadrupole moment $Q_A$. Then, the star B feels the usual gravity force plus an additional force, since the potential star B feels is:

\begin{equation} \Phi_B = - \dfrac{GM_A}{r} - \dfrac{3 G Q_A}{2 r^3} . \end{equation}

On the other hand, star A feels only the usual gravity force, since $Q_{B}=0$ for hypothesis, and hence the second term above is not present for star A:

\begin{equation} \Phi_A = - \dfrac{GM_B}{r} . \end{equation}

However, applying Newton's third law, one would expect that the forces on the two stars are equals (in modulo), but this is not the case, since force on B has an additional term. When $Q_A=0$ everything is ok and Newton's 3th law works. How is it possible?

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EDIT: Based on the answer below, the force experienced by star A is:

\begin{equation}\tag{1} F_A = -M_A \partial_r \Phi_A - \dfrac{1}{6}\partial_r \partial_i\partial_j \Phi_A Q_{A}^{ij} \end{equation}

where $\Phi_A$ is the potential experienced by star A in its CM. On the other hand, the force acting on star B is:

\begin{equation}\tag{2} F_B = -M_B \partial_r \Phi_B \end{equation}

where I used $Q_B=0$ by hypothesis. Using definitions of $\Phi_{A,B}$ above, I find

\begin{equation} F_A = - \dfrac{GM_A M_B}{r^2} - \dfrac{GM_B Q_A}{r^4} \end{equation} \begin{equation} F_B = - \dfrac{GM_A M_B}{r^2} - \dfrac{9}{2}\dfrac{GM_B Q_A}{r^4} \end{equation}

which depend both on $Q_A$ but are still not equal. I suspect an error in the calculation of the term $\partial_i \partial_j \partial_k \Phi_A Q^{ij}$ [ the only important component of $Q$ is that along the line connecting the stars, i.e. $Q=Q^{rr}$ in polar coordinates. So, in Eq. (1) I just did $\partial_r \partial_r \partial_r \Phi_A Q^{rr} = \dfrac{6GM_B}{r^4} Q^{rr}$ ].

Why there is a discrepancy by a factor 9/2?

Answer 1

While the potential $\Phi_A$ seen by star $A$ is as you've given above, the force it experiences is not simply $-M_A \nabla \Phi_A$. Only if an object is spherically symmetric will the gravitational force on it be equal to the force on a point mass at the same position as the object's CM. In general, the force on an extended object will depend not just on its mass $M$ but also on its higher-order multipole moments.

To show this mathematically, let's consider an extended object with a mass density $\rho(\mathbf{r})$ in some external potential $\Phi(\mathbf{r})$ satisfying $\nabla^2 \Phi = 0$. Choose the origin of our coordinate system to be the center of mass of the object. The net force on this object is $$ F_i = -\iiint \rho(\mathbf{r}) \partial_i \Phi \, d^3 r $$ Expanding $\partial_i \Phi$ in a Taylor series about the origin, we have \begin{align} F_i &= -\iiint \rho \left[ \left. \partial_i \Phi \right|_\mathbf{0} + \left. r_j \partial_i \partial_j \Phi \right|_\mathbf{0} + \left. \frac{1}{2} r_j r_k \partial_i \partial_j \partial_k \Phi \right|_\mathbf{0} \right] \, d^3 r \\ &= - \left. \partial_i \Phi \right|_\mathbf{0} \left[ \iiint \rho \, d^3 r \right] - \left. \partial_i \partial_j \Phi \right|_\mathbf{0} \left[ \iiint r_j \rho \, d^3 r \right] - \frac{1}{2} \left. \partial_i \partial_j \partial_k \Phi \right|_\mathbf{0} \left[ \iiint r_j r_k \rho \, d^3 r \right] - \cdots\\ \end{align} The first integral can be seen to simply be the mass $M$ of the object. The second integral is proportional to the dipole moment of the object, which vanishes because we have chosen our origin to be at the center of mass. Finally, the third term can be rewritten in terms of the quadrupole moment of the object: \begin{align} \left. \partial_i \partial_j \partial_k \Phi \right|_\mathbf{0} &\left[ \iiint r_j r_k \rho \, d^3 r \right] \\ &= \left. \partial_i \partial_j \partial_k \Phi \right|_\mathbf{0} \left[ \iiint \left(r_j r_k \rho - \frac{1}{3} r^2 \delta_{jk} \right) \, d^3 r \right] + \frac{1}{3} \left. \partial_i \partial_j \partial_k \Phi \right|_\mathbf{0} \delta_{jk} \left[ \iiint r^2 \rho \, d^3 r \right] \\ &= \frac{1}{3} \left. \partial_i \partial_j \partial_k \Phi \right|_\mathbf{0} Q_{jk} + \frac{1}{3} \left. \partial_i \nabla^2 \Phi \right|_\mathbf{0} \left[ \iiint r^2 \rho \, d^3 r \right] \end{align} and the last term vanishes because $\nabla^2\Phi = 0$ at all points.

So putting this all together, we have $$ F_i = - M \left. \partial_i \Phi \right|_\mathbf{0} - \frac{1}{6} \left. \partial_i \partial_j \partial_k \Phi \right|_\mathbf{0} Q_{jk} - \cdots $$ where the ellipsis stands for higher-order couplings to even higher multipole moments.

In your case, then, the fact that star $A$ has a quadrupole moment means that the force on it is not equal to $-M_A \nabla \Phi_A$; there will be an additional term that depends on the quadrupole moment of $A$. Meanwhile, star $B$ doesn't have its own quadrupole moment; but the potential $\Phi_B$ that is seen by star $B$ has a term proportional to the quadrupole moment of $A$. For both objects, the force experienced depends on the quadrupole moment of $A$.

Answer 2

When you consider an interaction between two things, the potential energy due to interaction is a shared entity between both of them.

Consider a mass connected to another mass via a spring. In the usual form, you have a mass connected via a spring to a wall, and you get the $\dfrac12kx^2$, but when you have two movable masses connected by the same spring, it will be $\dfrac12k(x_B-x_A)^2$, i.e. one spring EPE term shared between both of them, not two separate terms.

Same thing here, you write down the GPE for both particles at once, and so it is $$\text{GPE}=-\frac{GMm}r-\frac{3GMQ}{2r^3}$$

In practice, you just work using reduced mass and things will be less confusing.