Thomas precession in relativity was already known to Ludwik Silberstein in 1914.^[2] But the only knowledge Thomas had of relativistic precession came from de Sitter's paper on the relativistic precession of the moon, first published in a book by Eddington.^[3]

In 1925 Thomas recomputed the relativistic precessional frequency of the doublet separation in the fine structure of the atom. He thus found the missing factor 1/2, which came to be known as the Thomas half.

This discovery of the relativistic precession of the electron spin led to the understanding of the significance of the relativistic effect. The effect was consequently named "Thomas precession".

Consider a physical system moving through Minkowski spacetime. Assume that there is at any moment an inertial system such that in it, the system is at rest. This assumption is sometimes called the third postulate of relativity.^[4] This means that at any instant, the coordinates and state of the system can be Lorentz transformed to the lab system through some Lorentz transformation.

Let the system be subject to external forces that produce no torque with respect to its center of mass in its (instantaneous) rest frame. The condition of "no torque" is necessary to isolate the phenomenon of Thomas precession. As a simplifying assumption one assumes that the external forces bring the system back to its initial velocity after some finite time. Fix a Lorentz frame O such that the initial and final velocities are zero.

The Pauli–Lubanski spin vector S_μ is defined to be (0, S_i) in the system's rest frame, with S_i the angular-momentum three-vector about the center of mass. In the motion from initial to final position, S_μ undergoes a rotation, as recorded in O, from its initial to its final value. This continuous change is the Thomas precession.^[5]

Consider the motion of a particle. Introduce a lab frame Σ in which an observer can measure the relative motion of the particle. At each instant of time the particle has an inertial frame in which it is at rest. Relative to this lab frame, the instantaneous velocity of the particle is v(t) with magnitude |v| = v bounded by the speed of light c, so that 0 ≤ v < c. Here the time t is the coordinate time as measured in the lab frame, not the proper time of the particle.

Apart from the upper limit on magnitude, the velocity of the particle is arbitrary and not necessarily constant; its corresponding vector of acceleration is a = dv(t)/dt. As a result of the Wigner rotation at every instant, the particle's frame precesses with an angular velocity given by the equation^[6][7][8][9]

where × is the cross product and

${\displaystyle \gamma ={\dfrac {1}{\sqrt {1-{\dfrac {|\mathbf {v} (t)|^{2}}{c^{2}}}}}}}$

is the instantaneous Lorentz factor, a function of the particle's instantaneous velocity. Like any angular velocity, ω_T is a pseudovector; its magnitude is the angular speed the particle's frame precesses (in radians per second), and the direction points along the rotation axis. As is usual, the right-hand convention of the cross product is used (see right-hand rule).

The precession depends on accelerated motion, and the non-collinearity of the particle's instantaneous velocity and acceleration. No precession occurs if the particle moves with uniform velocity (constant v so a = 0), or accelerates in a straight line (in which case v and a are parallel or antiparallel so their cross product is zero). The particle has to move in a curve, say an arc, spiral, helix, or a circular orbit or elliptical orbit, for its frame to precess. The angular velocity of the precession is a maximum if the velocity and acceleration vectors are perpendicular throughout the motion (a circular orbit), and is large if their magnitudes are large (the magnitude of v is almost c).

In the non-relativistic limit, v → 0 so γ → 1, and the angular velocity is approximately

${\displaystyle {\boldsymbol {\omega }}_{\text{T}}\approx {\frac {1}{2c^{2}}}\mathbf {a} \times \mathbf {v} }$

The factor of 1/2 turns out to be the critical factor to agree with experimental results. It is informally known as the "Thomas half".

The description of relative motion involves Lorentz transformations, and it is convenient to use them in matrix form; symbolic matrix expressions summarize the transformations and are easy to manipulate, and when required the full matrices can be written explicitly. Also, to prevent extra factors of c cluttering the equations, it is convenient to use the definition β(t) = v(t)/c with magnitude |β| = β such that 0 ≤ β < 1.

The spacetime coordinates of the lab frame are collected into a 4×1 column vector, and the boost is represented as a 4×4 symmetric matrix, respectively

${\displaystyle X={\begin{bmatrix}ct\\x\\y\\z\end{bmatrix}}\,,\quad B({\boldsymbol {\beta }})={\begin{bmatrix}\gamma &-\gamma \beta _{x}&-\gamma \beta _{y}&-\gamma \beta _{z}\\-\gamma \beta _{x}&1+(\gamma -1){\dfrac {\beta _{x}^{2}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{x}\beta _{y}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{x}\beta _{z}}{\beta ^{2}}}\\-\gamma \beta _{y}&(\gamma -1){\dfrac {\beta _{y}\beta _{x}}{\beta ^{2}}}&1+(\gamma -1){\dfrac {\beta _{y}^{2}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{y}\beta _{z}}{\beta ^{2}}}\\-\gamma \beta _{z}&(\gamma -1){\dfrac {\beta _{z}\beta _{x}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{z}\beta _{y}}{\beta ^{2}}}&1+(\gamma -1){\dfrac {\beta _{z}^{2}}{\beta ^{2}}}\\\end{bmatrix}}}$

and turn

${\displaystyle \gamma ={\frac {1}{\sqrt {1-|{\boldsymbol {\beta }}|^{2}}}}}$

is the Lorentz factor of β. In other frames, the corresponding coordinates are also arranged into column vectors. The inverse matrix of the boost corresponds to a boost in the opposite direction, and is given by B(β)⁻¹ = B(−β).

At an instant of lab-recorded time t measured in the lab frame, the transformation of spacetime coordinates from the lab frame Σ to the particle's frame Σ′ is

${\displaystyle X'=B({\boldsymbol {\beta }})X}$

1

and at later lab-recorded time t + Δt we can define a new frame Σ′′ for the particle, which moves with velocity β + Δβ relative to Σ, and the corresponding boost is

${\displaystyle X''=B({\boldsymbol {\beta }}+\Delta {\boldsymbol {\beta }})X}$

2

The vectors β and Δβ are two separate vectors. The latter is a small increment, and can be conveniently split into components parallel (‖) and perpendicular (⊥) to β^{[nb 1]}

${\displaystyle \Delta {\boldsymbol {\beta }}=\Delta {\boldsymbol {\beta }}_{\parallel }+\Delta {\boldsymbol {\beta }}_{\perp }}$

Combining (1) and (2) obtains the Lorentz transformation between Σ′ and Σ′′,

${\displaystyle X''=B({\boldsymbol {\beta }}+\Delta {\boldsymbol {\beta }})B(-{\boldsymbol {\beta }})X'\,,}$

3

and this composition contains all the required information about the motion between these two lab times. Notice B(β + Δβ)B(−β) and B(β + Δβ) are infinitesimal transformations because they involve a small increment in the relative velocity, while B(−β) is not.

The composition of two boosts equates to a single boost combined with a Wigner rotation about an axis perpendicular to the relative velocities;

${\displaystyle \Lambda =B({\boldsymbol {\beta }}+\Delta {\boldsymbol {\beta }})B(-{\boldsymbol {\beta }})=R(\Delta {\boldsymbol {\theta }})B(\Delta \mathbf {b} )}$

4

The rotation is given by is a 4×4 rotation matrix R in the axis–angle representation, and coordinate systems are taken to be right-handed. This matrix rotates 3d vectors anticlockwise about an axis (active transformation), or equivalently rotates coordinate frames clockwise about the same axis (passive transformation). The axis-angle vector Δθ parametrizes the rotation, its magnitude Δθ is the angle Σ′′ has rotated, and direction is parallel to the rotation axis, in this case the axis is parallel to the cross product (−β)×(β + Δβ) = −β×Δβ. If the angles are negative, then the sense of rotation is reversed. The inverse matrix is given by R(Δθ)⁻¹ = R(−Δθ).

Corresponding to the boost is the (small change in the) boost vector Δb, with magnitude and direction of the relative velocity of the boost (divided by c). The boost B(Δb) and rotation R(Δθ) here are infinitesimal transformations because Δb and rotation Δθ are small.

The rotation gives rise to the Thomas precession, but there is a subtlety. To interpret the particle's frame as a co-moving inertial frame relative to the lab frame, and agree with the non-relativistic limit, we expect the transformation between the particle's instantaneous frames at times t and t + Δt to be related by a boost without rotation. Combining (3) and (4) and rearranging gives

${\displaystyle B(\Delta \mathbf {b} )X'=R(-\Delta {\boldsymbol {\theta }})X''=X'''\,,}$

5

where another instantaneous frame Σ′′′ is introduced with coordinates X′′′, to prevent conflation with Σ′′. To summarize the frames of reference: in the lab frame Σ an observer measures the motion of the particle, and three instantaneous inertial frames in which the particle is at rest are Σ′ (at time t), Σ′′ (at time t + Δt), and Σ′′′ (at time t + Δt). The frames Σ′′ and Σ′′′ are at the same location and time, they differ only by a rotation. By contrast Σ′ and Σ′′′ differ by a boost and lab time interval Δt.

Relating the coordinates X′′′ to the lab coordinates X via (5) and (2);

${\displaystyle X'''=R(-\Delta {\boldsymbol {\theta }})X''=R(-\Delta {\boldsymbol {\theta }})B({\boldsymbol {\beta }}+\Delta {\boldsymbol {\beta }})X\,,}$

6

the frame Σ′′′ is rotated in the negative sense.

The rotation is between two instants of lab time. As Δt → 0, the particle's frame rotates at every instant, and the continuous motion of the particle amounts to a continuous rotation with an angular velocity at every instant. Dividing −Δθ by Δt, and taking the limit Δt → 0, the angular velocity is by definition

${\displaystyle {\boldsymbol {\omega }}_{T}=-\lim _{\Delta t\rightarrow 0}{\frac {\Delta {\boldsymbol {\theta }}}{\Delta t}}}$

7

It remains to find what Δθ precisely is.

The composition can be obtained by explicitly calculating the matrix product. The boost matrix of β + Δβ will require the magnitude and Lorentz factor of this vector. Since Δβ is small, terms of "second order" |Δβ|², (Δβ_x)², (Δβ_y)², Δβ_xΔβ_y and higher are negligible. Taking advantage of this fact, the magnitude squared of the vector is

${\displaystyle |{\boldsymbol {\beta }}+\Delta {\boldsymbol {\beta }}|^{2}=|{\boldsymbol {\beta }}|^{2}+2{\boldsymbol {\beta }}\cdot \Delta {\boldsymbol {\beta }}}$

and expanding the Lorentz factor of β + Δβ as a power series gives to first order in Δβ,

${\displaystyle {\begin{aligned}{\frac {1}{\sqrt {1-|{\boldsymbol {\beta }}+\Delta {\boldsymbol {\beta }}|^{2}}}}&=1+{\frac {1}{2}}|{\boldsymbol {\beta }}+\Delta {\boldsymbol {\beta }}|^{2}+{\frac {3}{8}}|{\boldsymbol {\beta }}+\Delta {\boldsymbol {\beta }}|^{4}+\cdots \\&=\left(1+{\frac {|{\boldsymbol {\beta }}|^{2}}{2}}+{\frac {3}{8}}|{\boldsymbol {\beta }}|^{4}+\cdots \right)+\left(1+{\frac {3}{2}}|{\boldsymbol {\beta }}|^{2}+\cdots \right){\boldsymbol {\beta }}\cdot \Delta {\boldsymbol {\beta }}\\&\approx \gamma +\gamma ^{3}{\boldsymbol {\beta }}\cdot \Delta {\boldsymbol {\beta }}\end{aligned}}}$

using the Lorentz factor γ of β as above.

Introducing the boost generators

${\displaystyle K_{x}={\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\\\end{bmatrix}}\,,\quad K_{y}={\begin{bmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0\end{bmatrix}}\,,\quad K_{z}={\begin{bmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{bmatrix}}}$

and rotation generators

${\displaystyle J_{x}={\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\\\end{bmatrix}}\,,\quad J_{y}={\begin{bmatrix}0&0&0&0\\0&0&0&1\\0&0&0&0\\0&-1&0&0\end{bmatrix}}\,,\quad J_{z}={\begin{bmatrix}0&0&0&0\\0&0&-1&0\\0&1&0&0\\0&0&0&0\end{bmatrix}}}$

along with the dot product · facilitates the coordinate independent expression

${\displaystyle \Lambda =I-\left({\frac {\gamma -1}{\beta ^{2}}}\right)({\boldsymbol {\beta }}\times \Delta {\boldsymbol {\beta }})\cdot \mathbf {J} -\gamma (\gamma \Delta {\boldsymbol {\beta }}_{\parallel }+\Delta {\boldsymbol {\beta }}_{\perp })\cdot \mathbf {K} }$

which holds if β and Δβ lie in any plane. This is an infinitesimal Lorentz transformation in the form of a combined boost and rotation^{[nb 2]}

${\displaystyle \Lambda =I-\Delta {\boldsymbol {\theta }}\cdot \mathbf {J} -\Delta \mathbf {b} \cdot \mathbf {K} }$

where

${\displaystyle \Delta {\boldsymbol {\theta }}=\left({\frac {\gamma -1}{\beta ^{2}}}\right){\boldsymbol {\beta }}\times \Delta {\boldsymbol {\beta }}={\frac {1}{c^{2}}}\left({\frac {\gamma ^{2}}{\gamma +1}}\right)\mathbf {v} \times \Delta \mathbf {v} }$

${\displaystyle \Delta \mathbf {b} =\gamma (\gamma \Delta {\boldsymbol {\beta }}_{\parallel }+\Delta {\boldsymbol {\beta }}_{\perp })}$

After dividing Δθ by Δt and taking the limit as in (7), one obtains the instantaneous angular velocity

${\displaystyle {\boldsymbol {\omega }}_{T}={\frac {1}{c^{2}}}\left({\frac {\gamma ^{2}}{\gamma +1}}\right)\mathbf {a} \times \mathbf {v} }$

where a is the acceleration of the particle as observed in the lab frame. No forces were specified or used in the derivation so the precession is a kinematical effect - it arises from the geometric aspects of motion. However, forces cause accelerations, so the Thomas precession is observed if the particle is subject to forces.

Thomas precession can also be derived using the Fermi-Walker transport equation.^[10][11][12] This is most easily done for a particle with spin with CCW circular motion in the x-y plane at constant velocity v. The spin is assumed to have a component in the x-y plane that precesses. As will be shown, the spin 4-vector component ${\displaystyle S^{0}}$ obeys the differential equation $${\displaystyle {\frac {d^{2}}{d\,{\tau }^{2}}}\,\,S^{0}\,=\,-{\gamma }^{2}\,{\omega }^{2\,(proper)}\,S^{0}}$$

${\displaystyle \tau }$ is proper time in the particle's accelerated frame and ${\displaystyle {\omega }^{(proper)}}$ is the angular velocity observed by the particle. This represents sinusoidal motion with angular velocity ${\displaystyle \gamma \,\omega ^{(proper)}}$as observed by the particle. ${\displaystyle S^{0}}$ is most positive when the spin direction in the x-y plane is aligned with the velocity direction and most negative when the directions are oppositely aligned. Since the period between alignments is less than the orbital period, the spin precession is retrograde. The retrograde angular velocity of the precession is $${\displaystyle {\omega }_{T}^{(proper)}\,=\,(\gamma \,-\,1)\,{\omega }^{(proper)}}$$ The ${\displaystyle T}$ subscript denotes Thomas precession. Since the laboratory time ${\displaystyle t}$ is a constant multiple ${\displaystyle \gamma }$ of the proper time ${\displaystyle \tau }$, this same relation holds for these angular velocities measured in the laboratory frame. Their ratio doesn't change.

This is the same formula as obtained earlier since $${\displaystyle \mathbf {a} \,\times \,\mathbf {v} \,=\,-\,{\omega }^{(lab)}\,v^{2}\,\mathbf {\hat {z}} \,\Rightarrow \,{\frac {1}{c^{2}}}\,{\frac {{\gamma }^{2}}{\gamma +1}}\,\mathbf {a} \,\times \,\mathbf {v} \,=\,-(\gamma -1)\,\omega ^{lab}\,\mathbf {\hat {z}} }$$

${\displaystyle \mathbf {\hat {z}} }$ is a unit vector in the +z direction.

Without loss of generality, the unit of time may be chosen so that the orbital angular velocity ${\displaystyle {\omega }^{(proper)}\,=\,1}$, the origin of time chosen so that the initial motion is in the +x direction, and the unit of distance chosen so that the speed of light ${\displaystyle c=1}$.

Use the metric ${\displaystyle g_{00}=-1}$. The 4-velocity is ^[13][14] $${\displaystyle u\,=\,(\gamma ,\,\gamma \,v\,\cos(\tau ),\,\gamma \,v\sin(\tau ),\,0)}$$ The 4-acceleration is^[15] $${\displaystyle a\,=\,(0,\,-\gamma \,v\,\sin(\tau ),\,\gamma \,v\cos(\tau ),\,0)}$$ The time derivative of the 4-acceleration is $${\displaystyle {\frac {d}{d\,\tau }}\,a\,=\,(0,\,-\gamma \,v\,\cos(\tau ),\,-\gamma \,v\sin(\tau ),\,0)\,=\,-u\,+\,(\gamma ,0,0,0)}$$

The Fermi-Walker transport equation for the 4-spin S is^[16] $${\displaystyle {\frac {d}{d\,\tau }}\,S\,=\,u\;a\cdot S\,-\,a\,\;u\cdot S\,=\,u\;a\cdot S}$$ since ${\displaystyle u\cdot S=0}$. If ${\displaystyle S}$ is replaced by ${\displaystyle u}$, then, as expected, ${\displaystyle {\frac {d}{d\tau }}\,u=a}$ since ${\displaystyle u\cdot a=0}$ and ${\displaystyle u\cdot u=-1}$. The Fermi-Walker transport equation for ${\displaystyle S^{0}}$ with this circular motion is $${\displaystyle {\frac {d}{d\,\tau }}\,S^{0}\,=\,\gamma \,a\cdot S}$$ Taking the time derivative of this equation and substituting for ${\displaystyle {\frac {d}{d\,\tau }}\,a}$ and ${\displaystyle {\frac {d}{d\,\tau }}\,S}$ gives the equation completing the derivation $${\displaystyle {\frac {d^{2}}{d\,{\tau }^{2}}}\,S^{0}\,=\,-{\gamma }^{2}\,S^{0}}$$

The unit vectors ${\displaystyle e_{x}=(0,1,0,0)}$ and ${\displaystyle e_{y}=(0,0,1,0)}$ are time-dependent linear combinations of ${\displaystyle u-\gamma \,e_{0}}$ and the 4-acceleration ${\displaystyle a}$. Here ${\displaystyle e_{0}=(1,0,0,0)}$. Then use $${\displaystyle {\begin{aligned}S\cdot (u-\gamma \,e_{0})&=\gamma \,S^{0}\\S\cdot a&=\,\gamma ^{-1}\,{\frac {d\,S^{0}}{d\,\tau }}\end{aligned}}}$$ ${\displaystyle e_{x}}$ and ${\displaystyle e_{y}}$ are the linear combinations $${\displaystyle {\begin{aligned}e_{x}\,&=\,\gamma ^{-1}\,v^{-1}\,(u-\gamma \,e_{0})\,\cos(\tau )\,-\gamma ^{-1}\,v^{-1}\,a\,\sin(\tau )\\e_{y}\,&=\,\gamma ^{-1}\,v^{-1}\,(u-\gamma \,e_{0})\,\sin(\tau )\,+\gamma ^{-1}\,v^{-1}\,a\,\cos(\tau )\end{aligned}}}$$ Substituting $${\displaystyle {\begin{aligned}S_{x}\,=\,S\,\cdot \,e_{x}\,&=\,v^{-1}\,S^{0}\,\cos(\tau )\,-\gamma ^{-2}\,v^{-1}\,{\frac {d\,S^{0}}{d\,\tau }}\,\sin(\tau )\\S_{y}\,=\,S\,\cdot \,e_{y}\,&=\,v^{-1}\,S^{0}\,\sin(\tau )\,+\gamma ^{-2}\,v^{-1}\,{\frac {d\,S^{0}}{d\,\tau }}\,\cos(\tau )\end{aligned}}}$$ Classically, ${\displaystyle S^{0}}$ is given by $${\displaystyle S^{0}\,=\,|\mathbf {S} _{\perp }|\,\gamma \,v\,\cos(\gamma \,\tau +\phi )}$$ where ${\displaystyle |\mathbf {S} _{\perp }|}$ is the magnitude of the spin component in the x-y plane in the rest frame. This gives $${\displaystyle {\begin{aligned}S_{x}\,&=\,|\mathbf {S} _{\perp }|\,(\sin(\tau )\sin(\gamma \,\tau +\phi )+\gamma \,\cos(\tau )\,\cos(\gamma \,\tau +\phi ))\\S_{y}\,&=\,|S_{\perp }|\,(-\cos(\tau )\,\sin(\gamma \,\tau +\phi )\,+\gamma \,\sin(\tau )\,\cos(\gamma \,\tau +\phi ))\end{aligned}}}$$ For ${\displaystyle \tau =0}$ and ${\displaystyle \phi =0}$ the spin 4-vector is, as expected, $${\displaystyle S\,=\,|\mathbf {S} _{\perp }|\,(\gamma \,v,\,\gamma ,\,0,\,0)}$$ ${\displaystyle S^{0}}$ is most positive when ${\displaystyle \gamma \,\tau _{n}+\phi =2\,n\,\pi }$ for integer n. Then $${\displaystyle {\begin{aligned}S_{x}\,&=\,|\mathbf {S} _{\perp }|\,\gamma \,\cos(\tau _{n})\\S_{y}\,&=\,|S_{\perp }|\,\gamma \,\sin(\tau _{n})\end{aligned}}}$$ At these times, the ${\displaystyle u_{x}}$ and ${\displaystyle u_{y}}$ components of the 4-velocity are $${\displaystyle {\begin{aligned}u_{x}\,&=\,\gamma \,v\,\cos(\tau _{n})\\u_{y}\,&=\,\gamma \,v\,\sin(\tau _{n})\end{aligned}}}$$ ${\displaystyle S^{0}}$ is most positive when the spin and velocity directions are aligned.

These equations for ${\displaystyle S^{0},\,S_{x},\,S_{y}}$ agree with the solution for the spin 4-vector given in Exercise 6.9 in Misner, Thorne, and Wheeler^[17] if we let ${\displaystyle |\mathbf {S_{\perp }} |=\hbar /{\sqrt {2}}}$ and ${\displaystyle \phi =\pi /2}$ and if we also let ${\displaystyle \cos(\tau )\rightarrow -\sin(\tau )}$ and ${\displaystyle \sin(\tau )\rightarrow \cos(\tau )\;}$since the particle there is initially moving in the +y direction instead of in the +x direction here.

Biquaternions or complex quaternions make the derivation of Thomas precession simple. See the article Quaternion Lorentz Transformations for details and references about the equations used here.

Assume the particle is initially at rest. Consider an infinitesimal boost ${\displaystyle B_{y}}$ in the +y direction followed by a boost ${\displaystyle B_{x}}$ in the +x direction. To first order, the initial small perpendicular boost does not change the magnitude of the resultant velocity, only its resultant direction. This can be represented as a small rotation followed by a boost of the same magnitude, only in a slightly different direction.

The product of the two boosts is $${\displaystyle {\begin{aligned}\mathbf {Q} \,=\,\mathbf {B} _{x}\,\mathbf {B} _{y}\,&=\,(\cosh(\alpha /2)+i\,\sinh(\alpha /2)\,\mathbf {I} )\;(\cosh(\delta \alpha /2)+i\,\sinh(\delta \alpha /2)\,\mathbf {J} )\\&=\cosh(\alpha /2)\,\cosh(\delta \alpha /2)+i\,\sinh(\alpha /2)\,\cosh(\delta \alpha /2)\,\mathbf {I} \,+i\,\cosh(\alpha /2)\,\sinh(\delta \alpha /2)\,\mathbf {J} \,-\,\sinh(\alpha /2)\,\sinh(\delta \alpha /2)\,\mathbf {K} \end{aligned}}}$$ This transforms ${\displaystyle \mathbf {X} \,=t\,+\,i\,x\,\mathbf {I} \,+\,i\,y\,\mathbf {J} \,+\,i\,z\,\mathbf {K} }$ as ${\displaystyle \mathbf {X} '\,=\,\mathbf {Q} \,\mathbf {X} \,{\overline {\mathbf {Q} ^{*}}}}$ where the combined overline and asterisk on the last ${\displaystyle \mathbf {Q} }$ mean to do ${\displaystyle i\rightarrow -i,\;\mathbf {I} \rightarrow -\mathbf {I} ,\;\;\mathbf {J} \rightarrow -\mathbf {J} ,\;\mathbf {K} \rightarrow -\mathbf {K} }$. These are the two conjugates as defined in the article Biquaternions. It can be verified that the boosts transform ${\displaystyle \mathbf {X} }$ as expected. The boost ${\displaystyle \mathbf {B} _{y}}$ is applied first.

This can also be written as a rotation followed by a boost. Let ${\displaystyle \delta \theta _{rot}}$ be the spatial rotation angle and let ${\displaystyle \delta \theta _{boost}}$ be the angle of the resultant boost direction with respect to the +x axis. It can be verified that the rotation is CCW for positive ${\displaystyle \delta \theta _{rot}}$.

Solve for them to first order in ${\displaystyle \delta \alpha }$ by equating to $${\displaystyle \mathbf {Q} \,=\,(\cosh(\alpha /2)+i\,\sinh(\alpha /2)\,(\mathbf {I} \,+\delta \theta _{boost}\,\mathbf {J} ))\;(1+{\frac {\delta \theta _{rot}}{2}}\,\mathbf {K} )}$$ To first order in ${\displaystyle \delta \alpha }$, this gives the pair of equations for the coefficients of ${\displaystyle \mathbf {K} }$ and ${\displaystyle \mathbf {J} }$, respectively$${\displaystyle {\begin{aligned}\cosh(\alpha /2)\,{\frac {\delta \theta _{rot}}{2}}\,&=\,-\sinh(\alpha /2)\,\delta \alpha /2\\\sinh(\alpha /2)\,(\delta \theta _{boost}-{\frac {\delta \theta _{rot}}{2}})\,&=\,\cosh(\alpha /2)\,\delta \alpha /2\end{aligned}}}$$ The ratio of the second equation to the first equation is $${\displaystyle {\begin{aligned}{\frac {2\,\delta \theta _{boost}-\delta \theta _{rot}}{\delta \theta _{rot}}}\,&=\,-\coth ^{2}(\alpha /2)\\\Rightarrow \;{\frac {\theta _{rot}}{\theta _{boost}}}&=\,-(\gamma -1)\end{aligned}}}$$ The minus sign indicates that the precession is retrograde.

In quantum mechanics Thomas precession is a correction to the spin-orbit interaction, which takes into account the relativistic time dilation between the electron and the nucleus in hydrogenic atoms.

Basically, it states that spinning objects precess when they accelerate in special relativity because Lorentz boosts do not commute with each other.

To calculate the spin of a particle in a magnetic field, one must also take into account Larmor precession.

The rotation of the swing plane of Foucault pendulum can be treated as a result of parallel transport of the pendulum in a 2-dimensional sphere of Euclidean space. The hyperbolic space of velocities in Minkowski spacetime represents a 3-dimensional (pseudo-) sphere with imaginary radius and imaginary timelike coordinate. Parallel transport of a spinning particle in relativistic velocity space leads to Thomas precession, which is similar to the rotation of the swing plane of a Foucault pendulum.^[18] The angle of rotation in both cases is determined by the area integral of curvature in agreement with the Gauss–Bonnet theorem.

Thomas precession gives a correction to the precession of a Foucault pendulum. For a Foucault pendulum located in the city of Nijmegen in the Netherlands the correction is:

${\displaystyle \omega \approx 9.5\cdot 10^{-7}\,\mathrm {arcseconds} /\mathrm {day} .}$

Note that it is more than two orders of magnitude smaller than the precession due to the general-relativistic correction arising from frame-dragging, the Lense–Thirring precession.

• Velocity-addition formula
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