# Do electrons of opposite spin attract each other? [closed]

Electrons of opposite spin create magnetic fields that are opposite to each other. So, is it possible that, 'such electron pairs attract each other' or ' they distract each other less than electron of same spin'? Thanks.

• Your title seems to contradict your question. Are you talking of the two electrons (presumably attracting each other), or of the two electron pairs? Make up your mind. – Ivan Neretin Dec 25 '20 at 22:27
• I would say two electrons in the same orbital avoid each other, rather. – Karl Dec 25 '20 at 23:47
• Quite the opposite; electrons with the same spin repel each other less. This is a feature commonly known as exchange energy, and is discussed in detail at chemistry.stackexchange.com/questions/58625/…, although the full explanation depends on quantum mechanics. You will certainly come across the notion of exchange energy when you do electronic configurations, although understanding the QM may have to wait until a bit later. – orthocresol Dec 26 '20 at 0:21

## 1 Answer

Electrons can sort of attract. Let's review magnetic dipoles. First, suppose we have a current loop $$I$$ rendered in blue at radius $$a$$ around the $$z$$-axis in the $$xy$$-plane and we want to compute the magnetic field at some point on the $$z$$-axis marked in red. See Figure $$1$$. The source point vector is $$\vec{r}_s=\langle a\cos\phi,a\sin\phi,0\rangle$$ So the element of arc length along the current path is $$d\vec\ell=d\vec{r}_s=\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi$$ The field point vector is $$\vec{r}_f=\langle0,0,z\rangle$$ So the vector from the source point to the field point is $$\vec{r}=\vec{r}_f-\vec{r}_s=\langle-a\cos\phi,-a\sin\phi,z\rangle$$ And its magnitude $$r=\left\lVert\vec{r}\right\rVert=\sqrt{a^2+z^2}$$ And a unit vector in its direction $$\hat{r}=\frac{\vec{r}}{r}=\frac{\langle-a\cos\phi,-a\sin\phi,z\rangle}{\sqrt{a^2+z^2}}$$ So we can get the magnetic field at $$\vec{r}_f$$ via the Biot-Savart law \begin{align}\vec B&=\frac{\mu_0}{4\pi}\int\frac{Id\vec\ell\times\hat{r}}{r^2}\\ &=\frac{\mu_0I}{4\pi}\int_0^{2\pi}\frac{\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi\times\langle-a\cos\phi,-a\sin\phi,z\rangle}{(a^2+z^2)^{3/2}}\\ &=\frac{\mu_0I}{4\pi(a^2+z^2)^{3/2}}\int_0^{2\pi}\langle az\cos\phi,az\sin\phi,a^2\rangle\,d\phi\\ &=\frac{\mu_0}{4\pi}\frac{2\pi a^2I}{(a^2+z^2)^{3/2}}\hat k\approx\frac{\mu_0}{4\pi}\frac{2m}{r^3}\hat k\end{align} Where the magnetic dipole moment of the loop is $$m=\pi a^2I$$ and we are looking at the far field of the dipole so $$a^2+z^2\approx r^2$$. The magnetic scalar potential of a dipole, valid away from its source current, is $$\Phi_m=\frac{\mu_0}{4\pi}\frac{mz}{r^3}$$ so \begin{align}\vec B&=-\vec{\nabla}\Phi_m=-\frac{\mu_0m}{4\pi}\left\langle-\frac32\frac{2xz}{r^5},-\frac32\frac{2yz}{r^5},-\frac32\frac{2z^2}{r^5}+\frac{r^2}{r^5}\right\rangle\\ &=\frac{\mu_0m}{4\pi r^5}\langle3xz,3yz,3z^2-r^2\rangle\end{align} Along the $$z$$-axis, $$x=y=0$$, $$r=z$$ so $$\vec B=\frac{\mu_0}{4\pi}\frac{2m}{r^3}\hat k$$ So the scalar potential checks. Along the $$x$$-axis, $$y=z=0$$, $$r=x$$ and $$\vec B=-\frac{\mu_0m}{4\pi x^3}\hat k$$ so $$\frac{\partial B_z}{\partial x}=\frac{3\mu_0m}{4\pi r^4}$$ This is the same as for a scalar potential of $$\Phi_m=-\frac{\partial B_z}{\partial x}\,xz$$ because then $$\vec B=-\vec{\nabla}\Phi_m=\frac{\partial B_z}{\partial x}\langle z,0,x\rangle=\frac{3\mu_0m}{4\pi r^4}\langle0,0,a\cos\phi\rangle$$ For the field at our loop due to an identical loop at $$\langle-r,0,0\rangle$$. So then the force on our loop is \begin{align}\vec F&=\int Id\vec\ell\times\vec B\\ &=\int_0^{2\pi}I\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi\times\frac{3\mu_0m}{4\pi r^4}\langle0,0,a\cos\phi\rangle\\ &=\frac{3\mu_0mI}{4\pi r^4}\int_0^{2\pi}\langle a^2\cos^2\phi,a^2\sin\phi\cos\phi,0\rangle\,d\phi\\ &=\frac{3\mu_0mI\pi a^2}{4\pi r^4}\hat i=\frac{3\mu_0m^2}{4\pi r^4}\hat i\end{align} So in this configuration the force is repulsive if the dipoles are aligned but attractive if pointed opposite one another. You can see this if you have a neocube where oppositely aligned strings of magnet spheres attract at the equators.

For electrons things are more complicated. If we have little bits of charge $$\Delta q$$ and mass $$\Delta m$$ orbiting around an axis at distance $$a$$ and angular frequency $$\omega$$ then the charge passing a point in the orbit per unit time is $$I=f\Delta q=\frac{\omega\Delta q}{2\pi}$$ so the magnetic dipole moment is $$m=I\pi a^2=\frac12\omega a^2\Delta q$$. The angular momentum is $$L=rp=a\Delta ma\omega$$ so we would get $$\frac mL=\frac12\frac{\Delta q}{\Delta m}$$ But life isn't so simple for a couple of reasons: the ratio is actually a factor of $$g_e\approx2.002$$. Also there is a question of whether we want to consider the total angular momentum of the electrons or the component at right angles to the vector from one to another. We choose the latter and this component is $$L_z=\frac12\hbar$$ so we get $$m=\frac{g_eq_e\hbar}{4m_e}$$ So the magnetic force is $$F=\frac{3\mu_0}{4\pi r^4}\left(\frac{g_eq_e\hbar}{4m_e}\right)^2$$ Comparing with the electrostatic force, $$F=\frac{q_e^2}{4\pi\epsilon_0r^2}=\frac{\mu_0c^2q_e^2}{4\pi r^2}$$ We see that the magnetostatic force is greater when $$r\lt\frac{\sqrt3g_e\hbar}{4m_ec}=\frac{\sqrt3(2.002)(1.055\times10^{-34}\text{J}\cdot\text{s})}{4(9.11\times10^{-31}\text{kg})(2.998\times10^8\text{m}/\text{s})}=3.35\times10^{-13}\text{m}$$ This is so close that it kind of negates the whole calculation though: the electrons are going to be moving pretty fast so there will be a magnetic force due to their relative motion and at these distances you really need quantum electrodynamics to analyze the problem. Still, it gives you an idea of how close you have to get so that the net force has at least a chance of being attractive.