The Physics Travel Guide

Thomas Precession

Prerequisites: Lorentz Group. Full path (8 pages)

Intuitive

Thomas precession normally occurs when a spinning particle such as an electron orbits around an atomic nucleus (figure 2); this contributes to its effective spin-orbit coupling, which in turn affects spectral lines (see e.g. [41, section 11.8]).https://arxiv.org/abs/1711.05753

Concrete

Consider a particle with spin S⃗\vec S (defined in its instantaneous rest frame) moving with velocity v⃗(t)\vec v(t) and acceleration a⃗=v⃗˙\vec a = \dot{\vec v} in an inertial lab frame, with γ=(1−v2/c2)−1/2\gamma = (1 - v^2/c^2)^{-1/2}. At each instant the rest frame is obtained from the lab frame by a pure boost with velocity v⃗\vec v. A moment later it is obtained by a pure boost with velocity v⃗+a⃗ δt\vec v + \vec a\,\delta t. Two successive pure boosts in different directions do not combine to a pure boost: the product is a pure boost followed by a rotation (the Wigner rotation of the Abstract layer). So even with no torque acting, the sequence of rest frames rotates relative to the lab axes, and the spin, which keeps its direction in the rest frame, precesses as seen from the lab. The angular velocity of this precession is ω⃗T=γ2γ+1 a⃗×v⃗c2  ≈  12c2 a⃗×v⃗(v≪c),\vec\omega_T = \frac{\gamma^2}{\gamma + 1}\,\frac{\vec a \times \vec v}{c^2} \;\approx\; \frac{1}{2c^2}\,\vec a\times\vec v \quad (v \ll c), and in the absence of torques dS⃗dt=ω⃗T×S⃗\frac{d\vec S}{dt} = \vec\omega_T \times \vec S. The effect appears only for a⃗\vec a not parallel to v⃗\vec v, as expected since collinear boosts commute. For v≪cv \ll c, the rotation angle in a time δt\delta t is av δt/(2c2)a v\,\delta t/(2c^2) when a⃗⊥v⃗\vec a \perp \vec v, which is the angle found from the Baker-Campbell-Hausdorff expansion quoted in the Abstract layer.

Circular orbit. For uniform circular motion with angular velocity ω\omega we have a=vωa = v\omega and a⃗⊥v⃗\vec a\perp\vec v, so ∣ω⃗T∣=γ2γ+1 v2c2 ω=γ2−1γ+1 ω=(γ−1) ω,|\vec\omega_T| = \frac{\gamma^2}{\gamma+1}\,\frac{v^2}{c^2}\,\omega = \frac{\gamma^2 - 1}{\gamma + 1}\,\omega = (\gamma - 1)\,\omega , using γ2v2/c2=γ2−1\gamma^2 v^2/c^2 = \gamma^2 - 1. The direction of a⃗×v⃗\vec a\times\vec v is opposite to the orbital angular velocity, so after one revolution the spin has turned by 2π(γ−1)2\pi(\gamma - 1) against the sense of the orbit.

Spin-orbit coupling in atoms. An electron in an atom sees, in its rest frame, a magnetic field from the moving nucleus. With a central potential energy V(r)V(r) and gyromagnetic factor gg, the naive interaction energy is Unaive=g2m2c2 1rdVdr S⃗⋅L⃗.U_{\text{naive}} = \frac{g}{2m^2c^2}\,\frac{1}{r}\frac{dV}{dr}\,\vec S\cdot\vec L . This energy describes precession in the electron rest frame, which itself rotates relative to the lab with ω⃗T\vec\omega_T. The energy in the lab frame is U=Unaive+S⃗⋅ω⃗TU = U_{\text{naive}} + \vec S\cdot\vec\omega_T. Using ma⃗=−r⃗rdVdrm\vec a = -\frac{\vec r}{r}\frac{dV}{dr} and L⃗=m r⃗×v⃗\vec L = m\,\vec r\times\vec v, to lowest order in v/cv/c one finds S⃗⋅ω⃗T=−12m2c21rdVdr S⃗⋅L⃗\vec S\cdot\vec\omega_T = -\frac{1}{2m^2c^2}\frac{1}{r}\frac{dV}{dr}\,\vec S\cdot\vec L, and so U=g−12m2c2 1rdVdr S⃗⋅L⃗.U = \frac{g - 1}{2m^2c^2}\,\frac{1}{r}\frac{dV}{dr}\,\vec S\cdot\vec L . For g≈2g \approx 2 this is half the naive result. The factor 1/21/2 is the Thomas factor, and it is needed to reproduce the observed fine structure. The Dirac equation gives the same result automatically.

Abstract

Abstractly, the Thomas precession is due to an additional rotation the spin of the particle picks up as a result of the orbit movement. In this sense, the phenomenon is due to a geometric phase/Berry phase.

From a group theoretic perspective, Thomas precession occurs because Lorentz boosts do not commute. For example, we have [Ky, Kx]=i Jz .[K_y,\,K_x] = i\,J_z \, . The rotation that we get by combining boosts is known as a Wigner rotation.

[T]he revolution of the particle around the nucleus can be interpreted as a sequence of Lorentz boosts.[…] Abstractly, the effect is due to Wigner rotations [3] that appear in representations of the Poincaré group, so it is purely kinematical—it follows solely from space-time symmetries. https://arxiv.org/abs/1711.05753

To get the Lorentz transformation mapping from frame 1 to 3 directly, we compose boosts:

exp⁡(ayc δt Ky+axc δt Kx)exp⁡(vxc Kx)=exp⁡(vxc Kx+axc δt Kx+ayc δt Ky+ay vx2 c2 [Ky, Kx] δt+⋯ )(1)\exp\left(\frac{a_y}{c} \,\delta t\, K_y + \frac{a_x}{c} \,\delta t\, K_x\right) \exp\left(\frac{v_x}{c}\,K_x\right) = \exp\left(\frac{v_x}{c}\,K_x + \frac{a_x}{c}\,\delta t\,K_x + \frac{a_y}{c}\,\delta t\, K_y + \frac{a_y\,v_x}{2\,c^2}\,[K_y,\,K_x]\,\delta t + \cdots\right)\tag{1}

by the Dynkin formula version of the Baker-Campbell-Hausdorff theorem. From the commutation relationships for the Lorentz group we get [Ky, Kx]=i Jz[K_y,\,K_x] = i\,J_z, so that ay vx2 c2 [Ky, Kx] δt\frac{a_y\,v_x}{2\,c^2}\,[K_y,\,K_x]\,\delta t corresponds to a rotation about the zz axis of ay vx δt/(2 c2)a_y\,v_x\,\delta t / (2\,c^2) radians (here, somewhat obviously Jz, Jy, JzJ_z,\,J_y,\,J_z are "generators of rotations", *i.e.* tangents to rotations about the x, y, zx,\,y,\,z axes at the identity, respectively, whilst Jz, Jy, JzJ_z,\,J_y,\,J_z are "generators" of boosts).https://physics.stackexchange.com/a/99459/37286

Why is it interesting?

[T]he phenomenon of Thomas precession is one of the best established kinematical consequences of Poincaré symmetryhttps://cqgplus.com/2018/02/15/can-you-see-asymptotic-symmetries/

Path to this page

  1. Calculus
  2. Linear Algebra
  3. Group Theory
  4. Newton's Second Law
  5. Vector Calculus
  6. Newtonian Mechanics
  7. Special Relativity
  8. Lorentz Group