On the role of magnetic fields into the dynamics and gravitational wave emission of binary neutron stars

Analyzing a magnetized BNS system in the Newtonian regime, we assume that each object is characterized intrinsically by its mass $M_{1}$ and $M_{2}$ and its magnetic dipole moment $\mathbf{m}_{1}$, $\mathbf{m}_{2}$. We isolate the magnetic effect neglecting other NS properties. The positions of the NS, $\mathbf{r}_{i}(t)$, are determined in a reference frame with origin in the center of mass of the system. In this frame the condition $M_{1}\mathbf{r}_{1}+M_{2}\mathbf{r}_{2}=0$ follows by definition. This reference frame has the advantage that the dynamics reduces to the one-body description with reduced mass $\mu=M_{1}M_{2}/M$, and position $\mathbf{r}\equiv\mathbf{r}_{1}-\mathbf{r}_{2}$. $M=M_{1}+M_{2}$ is the total mass of the system. In this work, we use the magnetostatic approximation for the interior of the stars considering that the net electric charge is zero [38, 39]. We also assume that the external magnetic field of each star is a perfect dipole field ${\bf m}_{i}$ and consequently, the magnitude of the magnetic moment is only related to the radius of the star as described in [14]³³3The radius of each star is related to their mass through an equation of state (EoS) [40]. A realistic EoS of NS matter is not known precisely yet, but for masses from 1 to $2M_{\odot}$ [41, 22] different EoS give radii varying from 8 to 16 km [24, 42]. For simplicity, we use a typical value for the radius of the stars as $12$ km..

In the magnetostatic regime, the magnetic field ${\bf B}_{1}$ due to star 1 at any point $\mathbf{x}$, is given by:

$${\bf B}_{1}({\bf x})=\frac{3\hat{{\bf n}}_{1}\left(\hat{{\bf n}}_{1}\cdot{\bf m}_{1}\right)-{\bf m}_{1}}{|{\bf x}-{\bf r}_{1}|^{3}}\,,$$

(1)

where $\hat{{\bf n}}_{1}=({\bf x}-{\bf r}_{1})/|{\bf x}-{\bf r}_{1}|$.

The magnetic potential energy resulting from the interaction of a magnetic dipole ${\bf m}$ with an external magnetic field ${\bf B}$ is given by the dot product $U_{m}=-{\bf m}\cdot{\bf B}$ [43].

The magnetic potential energy at position ${\bf r}_{2}$, is thus

$$U_{m}(1\rightarrow 2)=-{\bf m}_{2}\cdot{\bf B}_{1}({\bf r}_{2})=-{\bf m}_{2}\cdot\frac{3\hat{{\bf r}}\left(\hat{{\bf r}}\cdot{\bf m}_{1}\right)-{\bf m}_{1}}{|{\bf r}|^{3}}=-\frac{3({\bf m}_{2}\cdot\hat{{\bf r}})\left(\hat{{\bf r}}\cdot{\bf m}_{1}\right)-{\bf m}_{2}\cdot{\bf m}_{1}}{|{\bf r}|^{3}},$$

(2)

where we have used Eq. (1) and the unit vector $\hat{{\bf r}}=({\bf r}_{2}-{\bf r}_{1})/|{\bf r}_{2}-{\bf r}_{1}|$.

Following [37, 44] we assume the magnetic moments remain parallel to the total angular momentum $\mathbf{L}=\mu\,\left(\mathbf{r}\times\dot{\mathbf{r}}\right)$ during the inspiral, thus ${\bf m}_{1}\cdot\hat{\bf r}={\bf m}_{2}\cdot\hat{\bf r}$=0, consequently the magnetic torque between the dipoles $\mathbf{N}=\mathbf{m}_{1}\times\mathbf{B}_{2}$ vanishes [43] and Eq. (2) become

$$U_{m}(1\rightarrow 2)=\frac{{\bf m}_{2}\cdot{\bf m}_{1}}{|{\bf r}|^{3}}\ .$$

(3)

Furthermore we will assume the magnetic moments for each star are of the form $m_{i}=R_{i}^{3}\,B_{i}/2$, thus we introduce a magnetic parameter $b$ as the dot product of the magnetic moments:

$$b\equiv{\bf m}_{1}\cdot{\bf m}_{2}=\pm\frac{(R_{1}R_{2})^{3}B_{1}B_{2}}{4}\,,$$

(4)

where $+$ or $-$ indicates if the dipoles are aligned or anti-aligned. Using Eq. (4) we show that it is possible to encode the magnetic interaction between the dipoles of the binary thorough the magnetic potential energy of the form $U_{m}=b/r^{3}$.

At this point, we can estimate, in orders of magnitude, the fraction of the magnetic potential respect to the gravitational contribution $U_{g}=-GM\mu/r$, as $U_{m}/U_{g}=-b/(GM\mu r^{2})$ or

$$\frac{U_{m}}{U_{g}}=1.0\times 10^{-4}\left(\frac{R_{1}}{12\,\mathrm{km}}\right)^{3}\left(\frac{B_{1}}{10^{16}\,\mathrm{G}}\right)\left(\frac{R_{2}}{12\,\mathrm{km}}\right)^{3}\left(\frac{B_{2}}{10^{16}\,\mathrm{G}}\right)\left(\frac{2.8\,M_{\odot}}{M}\right)\left(\frac{0.7\,M_{\odot}}{\mu}\right)\left(\frac{12\,\mathrm{km}}{r}\right)^{2},$$

(5)

where we have used the Eq. (4) and re-scaled the parameters of BNS in terms of canonical NS with equal masses $\sim 1.4\,M_{\odot}$ and radii $\sim 12$ km. For the magnetic field strength we use $\sim 10^{16}$G because it is the maximum estimated for magnetars (but not theorized).

We can observe from Eq. (5), that the contribution of magnetic over the gravitational potential is ${\cal O}(10^{-4})$. It is in accordance with the post-Newtonian analyzes, which reports that the magnetic dipole-dipole interaction produces a 2PN correction [14, 45, 38].

As described above, the magnetic interaction between the dipoles of the stars in the binary can be described using a central potential. Thus the gravito-magnetic interaction, in the center of mass frame, is given by the sum of the magnetic $U_{m}$ and gravitational $U_{g}$ potential:

$$U(r)=-\frac{GM\mu}{r}+\frac{b}{r^{3}}=-\frac{GM\mu}{r}\left(1-\frac{b}{GM\mu r^{2}}\right)\ ,$$

(6)

where $b$ has been defined in Eq. (4). With this potential we can use the lagrangian formalism to reduce the dynamics of the binary to an equivalent one-body problem for the reduced mass $\mu$ located at ${\bf r}$. The position of each star is recover using the relations ${\bf r}_{1}=(M_{2}/M)\,{\bf r}$ and ${\bf r}_{2}=-(M_{1}/M)\,{\bf r}$. Since the potential $U$ depends only on the position $r$, the total angular momentum is conserved and thus the movement is restricted to a plane. For simplicity, we choose the equatorial plane. The Lagrangian for the BNS system is thus defined as the difference of the kinetic energy $T=\mu\leavevmode\nobreak\ \dot{r}^{2}/2+\mu r^{2}\dot{\varphi}^{2}/2$, and the potential energy given in Eq. (6),

$$\mathcal{L}(r,\varphi)=\frac{\mu}{2}\dot{r}^{2}+\frac{\mu}{2}r^{2}\dot{\varphi}^{2}+\frac{GM\mu}{r}-\frac{b}{r^{3}}\,.$$

(7)

Since the Lagrangian in Eq. (7) is independent of $\varphi$, one obtains the conservation of the angular momentum directly from the Euler-Lagrange equations

$$\frac{d}{dt}(\mu r^{2}\dot{\varphi})=0\hskip 8.5359pt\Rightarrow\hskip 8.5359ptl=\mu r^{2}\dot{\varphi}={\rm const}\ .$$

(8)

The total energy of the system $E=T+U$ can be written as

$$E=\frac{1}{2}\mu\dot{r}^{2}+V_{\rm eff}(r)\ ,$$

(9)

where $V_{\rm eff}(r)\equiv\frac{l^{2}}{2\mu r^{2}}-\frac{GM\mu}{r}+\frac{b}{r^{3}}$ is an effective potential and we have used the relation between $l$ and $\dot{\varphi}$ given in Eq. (8). On the other hand, the Euler-Lagrange equation for the coordinate $r$ is

$$\ddot{r}+\frac{GM}{r^{2}}-\frac{l^{2}}{\mu^{2}r^{3}}-\frac{3b}{\mu r^{4}}=0\ .$$

(10)

Setting a variable change $u=1/r$ we have

$$\dot{r}=-\frac{1}{u^{2}}\dot{u}=-\frac{l}{\mu}\frac{\dot{u}}{\dot{\varphi}}=-\frac{l}{\mu}\frac{du}{d\varphi}\qquad{\rm and}\qquad\ddot{r}=-\frac{l}{\mu}\dot{\varphi}\frac{d^{2}u}{d\varphi^{2}}=-\frac{l^{2}}{\mu^{2}}u^{2}\frac{d^{2}u}{d\varphi^{2}}\ ,$$

(11)

so that, after some simplifications, Eq. (10) becomes

$$\frac{d^{2}u}{d\varphi^{2}}+u-\frac{1}{R}=\delta_{b}u^{2}\ ,$$

(12)

where

$$R\equiv\frac{l^{2}}{GM\mu^{2}}\qquad{\rm and}\qquad\delta_{b}=-\frac{3\mu b}{l^{2}}\ .$$

(13)

The right hand side of Eq. (12) is a nonlinear term induced by the dipole interaction. The solutions with no magnetic field $b=0=\delta_{b}$, are the conic sections $u(\varphi)=\frac{1}{R}\left(1+\epsilon\cos\varphi\right)$, with eccentricity $\epsilon^{2}\equiv 1+2El^{2}/(G^{2}M^{2}\mu^{3})$. Notice that the nonlinear term in Eq. (12) has the same form as the relativistic correction to the Newtonian potential given by the Schwarzschild spacetime in a relativistic treatment. In the following section we show that circular orbits are allowed for a range of values of the magnetic strength.

It is know that the emission of gravitational radiation tends to circularize elliptical orbits to the degree that before merger, the orbits have been circularized [46, 47, 48]. Since the circular motion dominates the BNS dynamics during the inspiral phase we focus our analysis for this type of orbits.

Circular orbits ($r=$ cte.) may be possible if the condition $\ddot{r}=0$ in Eq. (10) is satisfied. In the scenario described in this work, it happens to be the case for a combination of the magnetic field strength and the angular momentum. By setting $\ddot{r}=0$ in Eq. (10) and solving for $r$ we obtain

$$r_{\rm c}=\frac{R}{2}\left(1+\sqrt{1+\frac{12\,b}{GM\mu\,R^{2}}}\right)\ ,$$

(14)

where we have used $R$ defined in Eq. (13). For $b>0$ there is always a circular orbit; for $b<0$ there is a critical value $b_{c}=-\frac{l^{4}}{12GM\mu^{3}}$ below of which circular orbits ceases to exit. For more negative values of $b$ the effective potential has not extreme points. Fig. 1 displays $V_{\rm eff}$ for some representative values of $b$.

The angular momentum $l$ for circular orbits is expressed by:

$$l_{b}=\mu\sqrt{G\,M\,r_{c}\left(1-\frac{3\,b}{G\,M\,\mu\,r_{c}^{2}}\right)}\ ,$$

(15)

where subscript $b$ denotes a dependence of the variable to the magnetic parameter $b$. Moreover, from Eq. (8), $\dot{\varphi}=l/\mu r^{2}$, the orbital frequency in circular orbits is given by

$$\dot{\varphi}_{b}=\sqrt{\frac{G\,M}{r_{c}^{3}}\left(1-\frac{3\,b}{G\,M\,\mu\,r_{c}^{2}}\right)}\ ,$$

(16)

this expression is the analogous to the third Kepler law for circular orbits and will be use in the next section.

Finally, the total energy of circular orbits is

$$E_{b}=-\frac{G\,M\,\mu}{2r_{c}}\,\left(1+\frac{b}{G\,M\,\mu\,r_{c}^{2}}\right)\ ,$$

(17)

which is equal to the minimum of the effective potential $V_{\rm eff}$ in Eq. (9).

A slightly non-circular orbit will oscillate in and out about a central radius. In the absence of magnetic field the allowed bounded orbits are ellipses. The presence of the magnetic term modifies this behaviour; the orbit looks like an ellipse which slowly rotates about the center, this phenomenon is known as the precession in the orbit.

Eq. (11) can be solved numerically quite straightforwardly, however, some useful information can be inferred from the solution in the limit of small magnetic interaction. Assuming $\delta_{b}<<R$, we seek approximate solutions of Eq. (11) of the form

$$u=u_{0}+\beta u_{1}\ ,\qquad{\rm with}\qquad\beta=\frac{\delta_{b}}{R}\ ,$$

(18)

were we have neglected higher powers of $\delta_{b}/R$. Substituting Eq. (18) into Eq. (12) and collecting terms of the same order, one gets

	$\displaystyle\frac{d^{2}(R\,u_{0})}{d\varphi^{2}}+R\,u_{0}$	$\displaystyle=$	$\displaystyle 1,$		(19)
	$\displaystyle\frac{d^{2}(R\,u_{1})}{d\varphi^{2}}+R\,u_{1}$	$\displaystyle=$	$\displaystyle(R\,u_{0})^{2}\ .$		(20)

The solution of the first equation, as mentioned above, is the conic section

$$R\,u_{0}=1+\epsilon\cos\varphi\ ,$$

(21)

where $\epsilon^{2}\equiv 1+\frac{2El^{2}}{G^{2}M^{2}\mu^{3}}$. On the other hand a particular solution of Eq. (20) is

$$R\,u_{1}=\left(1+\frac{\epsilon^{2}}{2}+\epsilon\varphi\sin\varphi-\frac{\epsilon^{2}}{6}\cos(2\varphi)\right)\ .$$

(22)

Notice the third term increases with each orbit and becomes the most relevant. Neglecting the other corrections we can thus write Eq. (18), in the limit $\beta<<1$, as

	$\displaystyle Ru$	$\displaystyle\approx$	$\displaystyle 1+\epsilon\cos\varphi+\beta\epsilon\varphi\sin\varphi$		(23)
		$\displaystyle\approx$	$\displaystyle 1+\epsilon\cos\left(\varphi-\beta\varphi\right)\ .$

Thus, the period of the orbit is no longer $2\pi$ but rather

$$\frac{2\pi}{1-\beta}\approx 2\pi(1+\beta)=2\pi\left(1+\frac{\delta_{b}}{R}\right)\ .$$

(24)

The precession, in radians per orbit, is therefore given by

$$\Delta\varphi=2\pi-2\pi\left(1+\frac{\delta_{b}}{R}\right)=2\pi\frac{\delta_{b}}{R}\ .$$

(25)

Substituting the expressions of $R$ and $\delta_{b}$ given by Eq. (13) and the expression of $b$ in terms of the strength of the magnetic field, we obtain the precession of the orbit $\Delta\,\varphi=\frac{2\pi\delta_{b}}{R}=\mp 6\pi GM\mu^{3}R_{1}^{3}R_{2}^{3}B_{1}B_{2}/(4l^{4})$. For typical values one gets

$\displaystyle\Delta\,\varphi=\mp 2.83\times 10^{-7}\left(\frac{R_{1}}{12\,\mathrm{km}}\right)^{3}\left(\frac{B_{1}}{10^{16}\,\mathrm{G}}\right)\left(\frac{R_{2}}{12\,\mathrm{km}}\right)^{3}\left(\frac{B_{2}}{10^{16}\,\mathrm{G}}\right)\left(\frac{M}{2.8\,M_{\odot}}\right)\left(\frac{\mu}{0.7\,M_{\odot}}\right)^{3}\left(\frac{l_{0}}{l}\right)^{4}.$

(26)

where the angular momentum $l$ is obtained from the Keplerian expression $l^{2}=GM\mu^{2}a\left(1-\epsilon^{2}\right)$, for a binary with $a=10^{3}\,{\rm km}$, $l_{0}=2.69\times 10^{50}\mathrm{g\,cm}^{2}{\rm s}^{-1}=1.35\times 10^{7}\,M_{\odot}\,{\rm km}^{2}\,{\rm s}^{-1}$.

In the following section we shall consider the lost of energy of the binary due to GW emission using the quadrupolar formalism. We focus on circular orbits since as mentioned above, GW emission tend to circularize the orbits during the inspiral.

The decrease in the separation of the neutron stars, going through a succession of quasi circular orbits, is driven by the emission of GW until the stars merge. However, when the stars are very close to each other, the dynamics is dominated by strong field effects and any approximation of GR used to describe the dynamics ceases to be valid.

As the two stars rotate around each other their orbital distances decreases, this causes the frequency of the GWs to increase. However, the frequency is only valid up to a maximum frequency beyond which the inspiral phase ends and $r_{\rm min}$ is reached.

The time in which the binary system reaches a minimal radius $r_{\rm min}$ can be computed as

$$\tau_{b}=\int_{r_{0}}^{r_{\rm min}}(dr/dt)_{b}^{-1}\,dr\ ,$$

(36)

where $r_{0}$ is the separation between the stars at time $t=0$ and the change of $r$ with time $(dr/dt)_{b}$ is given by Eq. (34).

Departing from circular orbits and the second mass moment given in Eq. (32) and using Eq. (29), the polarization amplitudes are

	$\displaystyle h_{+_{b}}(t)$	$\displaystyle=-\frac{4G^{2}M\mu}{c^{4}dr_{b}(t)}\left(1-\frac{3b}{GM\mu r_{b}(t)^{2}}\right)\cos(2\varphi_{b}(t))\ ,$		(37)
	$\displaystyle h_{{\times}_{b}}(t)$	$\displaystyle=-\frac{4G^{2}M\mu}{c^{4}dr_{b}(t)}\left(1-\frac{3b}{GM\mu r_{b}(t)^{2}}\right)\sin(2\varphi_{b}(t))\ ,$

where $r_{b}(t)$ is given by the integration of Eq. (34). The strain of the GW is defined as $h_{b}=\sqrt{h_{+_{b}}^{2}+h_{{\times}_{b}}^{2}}$ and using Eq. (37), the strain is given by

$$h_{b}(t)=\frac{4G^{2}M\mu}{c^{4}\,dr_{b}(t)}\left(1-\frac{3b}{GM\mu r_{b}(t)^{2}}\right).$$

(38)

Notice that setting $b=0$ in Eqs. (34, 35) they reduce to the well known expressions given for instance in reference [49].

For GWs, the term $2\varphi_{b}$ in Eq. (37) can be approximated as $2\varphi_{b}\simeq 2\dot{\varphi}_{b}t$ where the GW frequency, $\omega_{\rm GW}=2\dot{\varphi}_{b}=2\omega_{b}$ can be obtained. The frequency measured in Hertz is simply $\nu_{b}=\frac{\omega_{GW}}{2\pi}$. number of cycles during the inspiral phase can thus be computed as

$$\mathcal{N}_{b}=\int_{0}^{\tau_{b}}\,\nu_{b}(t)\,dt\ .$$

(39)

The derivative of the orbital frequency, $\dot{\omega}_{b}$, can be constructed by the chain’s rule: $\dot{\omega}_{b}=\left(\frac{d\omega_{b}}{dr}\right)\left(\frac{dr}{dt}\right)$, where $\left(\frac{d\omega_{b}}{dr}\right)$ is directly calculated from the Eq. (16), and $\left(\frac{dr}{dt}\right)$ is substituted from Eq. (34). By this way, we obtain

$\displaystyle\dot{\omega}_{b}=\frac{96G^{3}M^{2}\mu}{5c^{5}r^{4}}\sqrt{\frac{G\,M}{r^{3}}\left(1-\frac{3\,b}{G\,M\,\mu\,r^{2}}\right)}\left(1-\frac{3b}{GM\mu r^{2}}\right)^{2}\left(1-\frac{5b}{GM\mu r^{2}}\right)\left(1+\frac{3b}{GM\mu r^{2}}\right)^{-1}.$

(40)

Finally, considering the circular radius of the non-magnetized BNS in Eq. (16), we write the radius as $r=(GM/(\omega_{0}^{2}))^{1/3}$, with $\dot{\varphi}_{0}=\omega_{0}$, where we are denoting $\omega_{0}$ as the angular frequency which corresponds to this circular non magnetic case. After some algebraic manipulation one gets

$$\dot{\omega_{b}}=\frac{96\,G^{5/3}M^{2/3}\mu{\omega}_{0}^{11/3}}{5c^{5}}\sqrt{1-3k\,{\omega}_{0}^{4/3}}\left(1-5k\,{\omega}_{0}^{4/3}\right)\left(1-3k\,{\omega}_{0}^{4/3}\right)^{2}\left(1+3k\,{\omega}_{0}^{4/3}\right)^{-1},$$

(41)

where $k\equiv b/(GM)^{5/3}\mu$.

In this section, we show how the expressions for the gravitational luminosity Eq. (33) and the change in the orbital period Eq. (35) can be used to determine the total mass of the binary. Furthermore, from this total mass, we obtain a measure of the individual masses. Let us assume an scenario in which we define the ratio $Q=\frac{\dot{P}}{P}$, and the gravitational luminosity $L$ are known during the inspiral phase.

From Eqs. (33, 35) and after some algebra we get the next expression for the total mass

$$M=\frac{5Lr^{9}Q^{2}c^{5}}{96G^{2}}\frac{\sqrt{3}A(Lr^{3}+Qb)+f_{1}}{f_{2}+\sqrt{3}Af_{3}}\ ,$$

(42)

and the reduced mass

$$\mu=-\frac{48G}{Lr^{11}Q^{3}c^{5}}\left(\sqrt{3}Af_{4}+f_{5}\right)\ ,$$

(43)

where

	$\displaystyle A$	$\displaystyle=$	$\displaystyle\sqrt{3L^{2}r^{6}+20QbLr^{3}+12Q^{2}b^{2}},$
	$\displaystyle f_{1}$	$\displaystyle=$	$\displaystyle 3L^{2}r^{6}+13QbLr^{3}+6Q^{2}b^{2},$
	$\displaystyle f_{2}$	$\displaystyle=$	$\displaystyle 9L^{4}r^{12}+96L^{3}r^{9}Qb+320L^{2}r^{6}Q^{2}b^{2}+389Lr^{3}Q^{3}b^{3}+144Q^{4}b^{4},$
	$\displaystyle f_{3}$	$\displaystyle=$	$\displaystyle 3L^{3}r^{9}+22QbL^{2}r^{6}+44Q^{2}b^{2}Lr^{3}+Q^{3}b^{3},$
	$\displaystyle f_{4}$	$\displaystyle=$	$\displaystyle 3L^{2}r^{6}+14QbLr^{3}+12Q^{2}b^{2},$
	$\displaystyle f_{5}$	$\displaystyle=$	$\displaystyle 9L^{3}r^{9}+72QbL^{2}r^{6}+144Q^{2}b^{2}Lr^{3}+72Q^{3}b^{3}.$		(44)

For a vanishing magnetic field $b=0$, we recover the known expressions for the total and reduced mass [36],

$$M_{0}=\frac{5Q^{2}c^{5}r^{3}}{288G^{2}L}\qquad{\rm and}\qquad\mu_{0}=-\frac{864GL^{2}}{5Q^{3}c^{5}r^{2}}\ .$$

(45)

In order to better understand, the impact of the magnetic field in the determination of $\mu$ and $M$ we define the variable

$$x\equiv\frac{Qb}{L\,r^{3}}\ .$$

(46)

By writing Eq. (42) in terms of $x$, we get $M(x)=M_{0}f_{M}(x)$ where

$$f_{M}(x)\equiv\frac{3\left(3+13x+6x^{2}+\sqrt{3}A_{x}(1+x)\right)}{\sqrt{3}A_{x}f_{6}+f_{7}},$$

(47)

with the definitions

	$\displaystyle A_{x}$	$\displaystyle=$	$\displaystyle\sqrt{12x^{2}+20x+3},$
	$\displaystyle f_{6}$	$\displaystyle=$	$\displaystyle 3+22x+44x^{2}+24x^{3},$
	$\displaystyle f_{7}$	$\displaystyle=$	$\displaystyle 9+96x+320x^{2}+384x^{3}+144x^{4}.$		(48)

The reduced mass in terms of the parameter $x$ is obtained from Eq. (43) as $\mu(x)=\mu_{0}f_{\mu}(x)$, where

$$f_{\mu}(x)\equiv\frac{1}{2}+\sqrt{3}A_{x}\left(\frac{1}{6}+\frac{7x}{9}+\frac{2x^{2}}{3}\right)+4x+8x^{2}+4x^{3}.$$

(49)

According to Eqs. (42, 43) for given values of the gravitational luminosity $L$, and the logarithmic change of the GW frequency $Q$, one may deduce a value of the mass $M$ and the reduced mass $\mu$. Any deviation from the values in Eqs. (45) (equivalently if $f_{M}\neq 1$ and $f_{\mu}\neq 1$) can thus be associated with a presence of magnetic field.

The determination of the individual masses of the binary can be obtained from the definition of total mass and reduced mass as

	$\displaystyle M_{1}(x)$	$\displaystyle=$	$\displaystyle\frac{1}{2}M_{0}f_{M}(x)\left(1-\sqrt{1-\frac{1}{4}\frac{\mu_{0}}{M_{0}}\frac{f_{\mu}(x)}{f_{M}(x)}}\right)\ ,$
	$\displaystyle M_{2}(x)$	$\displaystyle=$	$\displaystyle\frac{1}{2}M_{0}f_{M}(x)\left(1+\sqrt{1-\frac{1}{4}\frac{\mu_{0}}{M_{0}}\frac{f_{\mu}(x)}{f_{M}(x)}}\right)\ .$		(50)

Notice however, the dependence of the individual mass depends on $M_{0}$ and $\mu_{0}$ and not on the individual mass of the binary with zero magnetic field.

As we describe in the following section a nonzero magnetic field is consistent with the current uncertainly in the estimation of the mass for systems with equal and non-equal masses. Furthermore the maximum and minimum mass estimation for these systems can be used to determine a bound for the maximum and minimum value of the allowed magnetic field. Fig. 2 shows the plot of $f_{M}$ and $f_{\mu}$ in terms of $x$. We focus on a region close to $x=0$ (no magnetic field) since we are interested in the small deviations produced by the magnetic field. Close the origin the slope of $f_{M}$ is negative whereas the slope of $f_{\mu}$ is positive and as we show below, this behavior will cause an over or under estimation in the mass of each component in the presence of magnetic field. Values of $x>0$ correspond to a binary with anti-aligned dipoles and $x<0$ correspond to aligned dipoles. This is consistent with the fact that the anti-aligned configuration storage more potential energy.

In the following section we show the effect of the magnetic field in the gravitational strain and mass estimation using some examples of astrophysical relevance.

In this section we will consider the inspiral stage in circular motion using two BNS systems: one with equal masses of $M_{1}=M_{2}=1.4\,M_{\odot}$ ($M=2.8\,M_{\odot}$, $\mu=0.7\,M_{\odot}$) and other with masses $M_{1}=1.8\,M_{\odot},\;M_{2}=1\,M_{\odot}$ ($M=2.8\,M_{\odot}$, $\mu=0.643\,M_{\odot}$) and an initial separation of $r_{0}=100$ km. We will describe qualitatively the effect of the magnetic field in some relevant variables. We take the radius of both neutron stars as $R=12\,{\rm km}$, and set the minimal radius as $r_{\rm min}=24\,{\rm km}$. At this stage, we take a fixed value of the magnetic field $B_{1}=B_{2}=8\times 10^{16}$ G and determine the effect of the relative alignment of the dipoles on the inspiral and the consequences in the resulting waveform. Using the parameter mentioned before, we calculate the time to reach the minimum radius $\tau_{b}$, for each configuration with anti-aligned $b<0$ and aligned $b>0$ dipoles. For comparison, we also present the case without magnetic fields $b=0$. The results are shown in Table 1. We see that the BNS with equal masses merges before the one with different masses independently of the magnetic fields. In contrast, for both BNS systems, when $b<0$, the time $\tau_{b}<\tau_{0}$ and when $b>0$, then $\tau_{b}>\tau_{0}$.

The separation $r(t)$ is computed from $t=0$ to $\tau_{b}$ for each case by solving numerically the differential equation (34). In Fig. 3 is shown $r(t)$ for cases with equal and unequal masses. The relative alignment, as stated above, is given by the sign of $b$.

Additionally, we determine the GW frequency from $\nu_{b}(t)=\dot{\varphi_{b}}(r(t))/\pi$ and using Eq. (16). The resulting frequency as a function of time is plotted in Fig. 4.

The waveforms $h_{+}$ and $h_{\times}$ are obtained from Eq. (37) and the numerical integration of Eq. (16) and Eq. (34). In Fig. 5, it is plot the polarization $h_{+}$ considering a distance $d=40$ Mpc (the one reported for the detected event GW170817). This distance gives a strain at the minimal radius of $h_{b}^{\rm(min)}=10^{-20}$, as is show in Fig. 5.

The number of cycles computed from Eq. (39) are presented in Table 2. We observe that the BNS with equal masses has less number of cycles ${\cal N}_{b}$ than the BNS with non-equal masses independently of the magnetic dipole alignment. We see that regardless the relation between the individual masses, when $b<0$ then ${\cal N}_{b}<{\cal N}_{0}$ and vise versa, when $b>0$ then ${\cal N}_{b}>{\cal N}_{0}$.

We have included the cases with equal and different individual masses, which has a more noticeable effect on the collision time in the wave frequency and in the emitted waveform.

In the next section we shall consider the case with variable magnetic field.

We proceed further in our analysis. We now vary the magnitude of the magnetic field and describe the corresponding effect in the inspiral variables: the time to reach the minimal radius, the gravitational luminosity, the strain, and the frequency of the GWs. For the analysis in this section, we consider again two BNS systems, one with equal masses of $M_{1}=M_{2}=1.4\,M_{\odot}$ and other with masses $M_{1}=1.8\,M_{\odot},\;M_{2}=1\,M_{\odot}$. We consider magnetic fields with strength between $B=10^{12}$ G and $B=8\times 10^{16}$ G, consistent with the values and predictions reported in [13, 10]. These magnitudes of the magnetic field imply that the parameter $b$ is within the interval $-4.77\,\times 10^{69}\;{\rm g\,cm}^{5}/{\rm s}^{2}<b<4.77\times 10^{69}\;{\rm g\,cm}^{5}/{\rm s}^{2}.$ In Fig. 6 the value of $\tau_{b}$ as a function of $b$ is plotted for both BNS systems. Note that $\tau_{0}:=\tau_{b}(b=0)$.

As it can be seen in Fig. 6, independently of the magnetic field, the system with equal masses reaches the minimum radius in a shorter time $\tau_{b}$ than in the case of different masses. However, it is clear in both cases that if $b<0$, then $\tau_{b}<\tau_{0}$ and if $b>0$, then $\tau_{b}>\tau_{0}$. We can interpret this result as showing that a magnetic configuration with $b>0$ ($b<0$) produces a slight increase (decrease) in the time to reach $r_{min}$. This qualitative statement can be quantified through the magnetic deviation defined as the ratio:

$$\Delta X=\frac{X_{b}-X_{0}}{X_{0}},$$

(51)

where $X_{b}$ is a variable that depends on the magnetic parameter $b$ and $X_{0}=X_{b}(b=0)$. Following our analysis on $\tau_{b}$, $\Delta\tau=\frac{\tau_{b}-\tau_{0}}{\tau_{0}}$. In the cases where $\tau_{b}>\tau_{0}$, then $\Delta\tau>0$, this happens for a configuration with $b<0$. In contrast, when $\tau_{b}<\tau_{0}$, then $\Delta\tau<0$ which happens for $b>0$. The order of magnitude of $\Delta\tau$ is shown in Table 3, that is the same for BNS with equal and unequal masses. The change in the time to reach the minimum radius can amount to $10^{-4}$ with respect to the change in the time without magnetic influence.

Let now consider the strain $h_{b}^{(\rm min)}$ at $r_{\rm min}$ given by Eq. (38). If $b<0$, then $h_{b}^{\rm min}<h_{0}^{\rm min}$; in contrast if $b>0$, then $h^{\rm min)}_{b}>h^{\rm min}_{0}$. In other words, when the magnetic configuration is such that $b<0$, the strain of the magnetized system is smaller than the strain of the non-magnetized system at $r_{\rm min}$. The opposite happens when the are such that $b<0$, see Fig. 7.