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LOG#097. Group theory(XVII).

The case of Poincaré symmetry

There is a important symmetry group in (relativistic, quantum) Physics. This is the Poincaré group! What is the Poincaré group definition? There are some different equivalent definitions:

i) The Poincaré group is the isometry group leaving invariant the Minkovski space-time. It includes Lorentz boosts around the 3 planes (X,T) (Y,T) (Z,T) and the rotations around the 3 planes (X,Y) (Y,Z) and (Z,X), but it also includes traslations along any of the 4 coordinates (X,Y,Z,T). Moreover, the Poincaré group in 4D is a 10 dimensional group. In the case of a ND Poincaré group, it has $N(N-1)/2+N$ parameters/dimensions, i.e., the ND Poincaré group is $N(N+1)/2$ dimensional.

ii) The Poincaré group formed when you add traslations to the full Lorentz group. It is sometimes called the inhomogenous Lorentz group and it can be denoted by ISO(3,1). Generally speaking, we will generally have $ISO(d,1)$, a D-dimensional ($D=d+1$) Poincaré group.

The Poincaré group includes as subgroups, the proper Lorentz transformations such as parity symmetry and some other less common symmtries. Note that the time reversal is NOT a proper Lorentz transformation since the determinant is equal to minus one.

Then, the Poincaré group includes: rotations, traslations in space and time, proper Lorentz transformations (boosts). The combined group of rotations, traslations and proper Lorentz transformations of inertial reference frames (those moving with constant relative velocity) IS the Poincaré group. If you give up the traslations in space and time of this list, you get the (proper) Lorentz group.

The full Poincaré group is a NON-COMPACT Lie group with 10 “dimensions”/parameters in 4D spacetime and $N(N+1)/2$ in the ND case.  Note that the boost parameters are “imaginary angles” so some parameters are complex numbers, though. The traslation subgroup of the Poincaré group is an abelian group forming a normal subgroup of the Poincaré group while the Lorentz grou is only a mere subgroup (it is not a normal subgroup of the Poincaré group). The Poincaré group is said, due to these facts, to be a “semidirect” product of traslations in space and time with the group of Lorentz transformations.

The case of Galilean symmetry

We can go back in time to understand some stuff we have already studied with respect to groups. There is a well known example of group in Classical (non-relativistic) Physics.

The Galilean group is the set or family of non-relativistic continuous space-time (yes, there IS space-time in classical physics!) transformations in 3D with an absolute time. This group has some interesting subgroups: 3D rotations, spatial traslations, temporal traslations and proper Galilean transformations ( transformations leaving invariant inertial frames in 3D space with absolute time). Thereforem the number of parameters of the Galilean group is 3+3+1+3=10 parameters. So the Galileo group is 10 dimensional and every parameter is real (unlike Lorentz transformations where there are 3 imaginary rotation angles).

The general Galilean group can be written as follows:

$G\begin{cases} \mathbf{r}\longrightarrow \mathbf{r}'=R\mathbf{r}+\mathbf{x_0}+\mathbf{V}t\\ t\longrightarrow t'=t+t_0\end{cases}$

Any element of the Galileo group can be written as a family of transformations $G=G(R,\mathbf{x_0},\mathbf{v},t_0)$. The parameters are:

i) $R$, an orthogonal (real) matrix with size $3\times 3$. It satisfies $RR^T=R^TR=I$, a real version of the more general unitary matrix $UU^+=U^+U=I$.

ii) $\mathbf{x_0}$ is a 3 component vector, with real entries. It is a 3D traslation.

iii) $\mathbf{V}$ is a 3 component vector, with real entries. It gives a 3D non-relativistic (or galilean) boost for inertial observers.

iv) $t_0$ is a real constant associated to a traslation in time (temporal traslation).

Therefore, we have 10 continuous parameters in general: 3 angles (rotations) defining the matrix $R$, 3 real numbers (traslations $\mathbf{x_0}$), 3 real numbers (galilean boosts denoted by $\mathbf{V}$) and a real number (traslation in time). You can generalize the Galilean group to ND. You would get  $N(N-1)/2+N+N+1$ parameters, i.e, you would obtain a $N(N+3)/2+1$ dimensional group. Note that the total number of parameters of the Poincaré group and the Galilean group is different in general, the fact that in 3D the dimension of the Galilean group matches the dimension of the 4D Poincaré group is a mere “accident”.

The Galilean group is completely determined by its “composition rule” or “multiplication operation”. Suppose that:

$G_3(R_3,\mathbf{z_0},\mathbf{V}_3,t_z)=G_2\cdot G_1$

with

$G_1(R_1,\mathbf{x_0},\mathbf{V}_1,t_x)$

and

$G_2(R_2,\mathbf{y_0},\mathbf{V}_2,t_y)$

Then, $G_3$ gives the composition of two different Galilean transformations $G_1, G_2$ into a new one. The composition rule is provided by the following equations:

$R_3=R_2R_1$

$\mathbf{z_0}=\mathbf{y_0}+R_2\mathbf{x_0}+\mathbf{V}_2 t_x$

$\mathbf{V}_3=\mathbf{V}_2+R_2\mathbf{V}_1$

$t_z=t_x+t_y$

Why is all this important? According to the Wigner theorem, for every continuous space-time transformation $g\in G$ should exist a unitary operator $U(g)$ acting on the space of states and observables.

We have seen that every element in uniparametric groups can be expressed as the exponential of certain hermitian generator. The Galilean group or the Poincaré group depends on 10 parameters (sometimes called the dimension of the group but you should NOT confuse them with the space-time dimension where they are defined). Remarkably, one can see that the Galilean transformations also act on “spacetime” but where the time is “universal” (the same for every inertial observer). Then, we can define

$iK_\alpha=\dfrac{\partial G}{\partial \alpha}\bigg| _{\alpha=0}$

These generators, for every parameter $\alpha$, will be bound to dynamical observables such as: linear momentum, angular momentum, energy and many others. A general group transformation for a 10-parametric (sometimes said 10 dimensional) group can be written as follows:

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10}=\prod_{k=1}^{10}e^{iK_{\alpha_k}\alpha_k}}$

We can apply the Baker-Campbell-Hausdorff (BCH) theorem or simply expand every exponential in order to get

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10})=\prod_{k=1}^{10}e^{iK_{\alpha_k}\alpha_k}=\exp \sum_{k=1}^{10}\omega_k (\alpha_1,\ldots,\alpha_{10})K_{\alpha_k}}$

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10})=I+i\sum_{k=1}^{10}\omega_k(\alpha_1,\ldots,\alpha_{10})K_{\alpha_k}+\ldots}$

The Lie algebra will be given by

$\displaystyle{\left[K_i,K_j\right]=i\sum_{k}c_{ij}^kK_k}$

and where the structure constants will encode the complete group multiplication rules. In the case of the Poincaré group Lie algebra, we can write the commutators as follows:

$\left[X_\mu,X_\nu\right]=\left[P_\mu,P_\nu\right]=0$

$\left[M_{\mu\nu},P_\alpha\right]=\eta_{\mu\alpha}P_\nu-\eta_{\nu\alpha}P_\mu$

$\left[M_{\mu\nu},M_{\alpha\beta}\right]=\eta_{\mu\alpha}M_{\nu\beta}-\eta_{\mu\beta}M_{\nu\alpha}-\eta_{\nu\alpha}M_{\mu\beta}+\eta_{\nu\beta}M_{\mu\alpha}$

Here, we have that:

i) $P$ are the generators of the traslation group in spacetime. Note that as they commute with theirselves, the traslation group is an abelian subgroup of the Lorentz group. The noncommutative geometry (Snyder was a pioneer in that idea) is based on the idea that $P$ and more generally even the coordinates $X$ are promoted to noncommutative operators/variables/numbers, so their own commutator would not vanish like the Poincaré case.

ii) $M$ are the generators of the Lorent group in spacetime.

If we study the Galilean group, there are some interesting commutation relationships fo the corresponding generators (rotations and traslations). There are 6 “interesting” operators:

$K_{i}\equiv \overrightarrow{J}$ if $i=1,2,3$

$K_{i}\equiv \overrightarrow{P}$ if $i=,4,5,6$

These equations provide

$\left[P_\alpha,P_\beta\right]=0$

$\left[J_\alpha,J_\beta\right]=i\varepsilon_{\alpha\beta}^\gamma J_\gamma$

$\left[J_\alpha,P_\beta\right]=i\varepsilon_{\alpha\beta}^\gamma P_\gamma$

$\forall\alpha,\beta=1,2,3$

The case of the traslation group

In Quantum Mechanics, traslations are defined in the space of states in the following sense:

$\vert\vec{r}\rangle\longrightarrow\vert\vec{r}'\rangle =\exp\left(-i\vec{x_0}\cdot \vec{p}\right)\vert \vec{r}\rangle=\vert\vec{r}+\vec{x_0}\rangle$

Let us define two linear operators, $R$ and $R'$ associated, respectively, to initial position and shifted position. Then the transformation defining the traslation over the states are defined by:

$R\longrightarrow R'=\exp\left(-i\vec{x_0}\cdot\vec{p}\right)R\exp \left(i\vec{x_0}\cdot \vec{p}\right)$

where

$R_i\vert\vec{r}'\rangle=\vec{r}_i\vert\vec{r}'\rangle$

Furthermore, we also have

$\left[\vec{x_0}\cdot \vec{p},\vec{y_0}\cdot R\right]=-i\vec{x_0}\cdot\vec{y_0}$

$\left[R_\alpha,p_\beta\right]=i\delta_{\alpha\beta}I$

The case of the rotation group

What about the rotation group? We must remember what a rotation means in the space $\mathbb{R}^n$. A rotation is a transformation group

$\displaystyle{X'=RX\longrightarrow \parallel X'\parallel^2=\parallel X\parallel^2 =\sum_{i=i}^n (x'_i)^2=\sum_{i=1}^n x_i^2}$

The matrix associated with this transformation belongs to the orthogonal group with unit determinant, i.e., it is an element of $SO(N)$. In the case of 3D space, it would be $SO(3)$. Moreover, the ND rotation matrix satisfy:

$\displaystyle{I=X^TX=XX^T\leftrightarrow \sum_{i=1}^N R_{ik}R_{ij}=R_{ik}R_{ij}=\delta_{kj}}$

The rotation matrices in 3D depends on 3 angles, and they are generally called the Euler angles in some texts. $R(\theta_1,\theta_2,\theta_3)=R(\theta)$. Therefore, the associated generators are defined by

$iM_j\equiv\dfrac{\partial R}{\partial \theta_j}\bigg|_{\theta_j=0}$

Any other rotation matric can be decomposed into a producto of 3 uniparametric rotations, rotation along certain 2d planes. Therefore,

$R(\theta_1,\theta_2,\theta_3)=R_1(\theta_1)R_2(\theta_2)R_3(\theta_3)$

where the elementary rotations are defined by

Rotation around the YZ plane: $R_1(\theta_1)=\begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\theta_1 & -\sin\theta_1\\ 0 & \sin\theta_1 & \cos\theta_1\end{pmatrix}$

Rotation around the XZ plane: $R_2(\theta_2)=\begin{pmatrix} \cos\theta_2 & 0 & \sin\theta_2\\ 0 & 1 & 0\\ -\sin\theta_2 & 0 & \cos\theta_2\end{pmatrix}$

Rotation around the XY plane: $R_3(\theta_3)=\begin{pmatrix} \cos\theta_3 & -\sin\theta_3 & 0\\ \sin\theta_3 & \cos\theta_3 & 0\\ 0 & 0 & 1\end{pmatrix}$

Using the above matrices, we can find an explicit representation for every group generator (3D rotation):

$M_1=-i\begin{pmatrix}0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{pmatrix}$

$M_2=-i\begin{pmatrix}0 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{pmatrix}$

$M_3=-i\begin{pmatrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}$

and we also have

$\left[M_j,M_k\right]=i\varepsilon^{m}_{jk}M_{m}$

where the $\varepsilon^m_{jk}=\varepsilon_{mjk}$ is the completely antisymmetry Levi-Civita symbol/tensor with 3 indices. There is a “for all practical purposes” formula that represents a rotation with respect to some axis in certain direction $\vec{n}$. We can make an infinitesimal rotation with angle $d\theta$, due to the fact that rotation are continuous transformations, it commutes with itself and it is unitary, so that:

$R(d\theta)\vec{r}=\vec{r}+d\theta(\vec{n}\times\vec{r}+\mathcal{O}(d\theta^2)=\vec{r}-id\theta M_\alpha\vec{r}+\mathcal{O}(d\theta^2)$

In the space of physical states, with $\vec{k}=\theta\vec{n}$ some arbitrary vector

$\vec{r}'=R\vec{r}\longrightarrow\vert\vec{r}'\rangle=\vert R\vec{r}\rangle=U(R)\vert\vec{r}\rangle=e^{-i\vec{k}\cdot\vec{J}}\vert\vec{r}\rangle=e^{-i(k_xJ_x+k_yJ_y+k_zJ_z)}\vert \vec{r}\rangle$

Here, the operators $J=(J_x,J_y,J_z)$ are the infinitesimal generators in the space of physical states. The next goal is to relate these generators with position operators $Q$ through commutation rules. Let us begin with

$Q\longrightarrow Q'=e^{-i\vec{k}\cdot{J}}Qe^{i\vec{k}\cdot\vec{J}}$

$Q'\vert\vec{r}'\rangle =\vec{r}\vert\vec{r}'\rangle$

Using this last result, we can calculate for any 2 vectors $\vec{k},\vec{n}$:

$\left[\vec{k}\cdot\vec{J},\vec{n}\cdot\vec{Q}\right]=i(\vec{k}\times\vec{n})\cdot\vec{Q}$

or equivalent, in component form,

$\left[J_j,Q_k\right]=i\varepsilon_{jkm}Q_m$

These commutators complement the above commutation rules, and thus, we have in general

$\left[\vec{k}\cdot\vec{J},\vec{n}\cdot\vec{Q}\right]=i(\vec{k}\times\vec{n})\cdot\vec{Q}$

$\left[J_j,J_k\right]=i\varepsilon_{jkm}J_m$

$\left[J_j,Q_k\right]=i\varepsilon_{jkm}Q_m$

In summary: a triplet of rotation operators generates “a vector” somehow.

The case of spinning particles

In fact, these features provide two different cases in the case of a single particle:

i) Particles with no “internal structure” or “scalars”/spinless particles. A good example could it be the Higgs boson.

ii) Particles with “internal” degrees of freedom/structure/particles with spin.

In the case of a particle without spin in 3D we can define the angular momentum operator as we did in classical physics ($L=r\times p$), in such a way that

$J=Q\times P$

Note that the “cross product” or “vector product” in 3D is generally defined if $C=A\times B$ as

$C=A\times B=\begin{vmatrix}i & j & k\\ A_x & A_y & A_z\\ B_x & B_y & B_z\end{vmatrix}$

or by components, using the maginal word XYZZY, we also have

$C_x=A_yB_z-A_zB_y$

$C_y=A_zB_x-A_xB_z$

$C_z=A_xB_y-A_yB_x$

Remember that the usual “dot” or “scalar” product is $A\cdot B=A_xB_x+A_yB_y+A_zB_z$

Therefore, the above operator $J$ defined in terms of the cross product satisfies the Lie algebra of $SO(3)$.

By the other hand, in the case of a spinning particle/particle with spin/internal structure/degrees of freedom, the internal degrees of freedom must be represented by some other operator, independently from $Q,P$. In particular, it must also commute with both operators. Then, by definition, for a particle with spin, the angular momentum will be a sum with two contributions: one contribution due to the “usual” angular momentum (orbital part) and an additional “internal” contribution (spin part). That is, mathematically speaking, we should have a decomposition

$J=Q\times P+S$

with $\left[Q,S\right]=\left[P,S\right]=0$

If $S$, the spin operator, satisfies the above commutation rules (in fact, the same relations than the usual angular momentum), we must impose

$\left[S_j,S_k\right]=i\varepsilon_{jkm}S_m$

The case of Parity P/Spatial inversions

This special transformation naturally arises in some applications. From the pure geometrical viewpoint, this transformation is very simple:

$\vec{r}'=-\vec{r}$

In coordinates and 3D, the spatial inversion or parity is represented by a simple matrix equals to minus the identity matrix

$P=\begin{pmatrix} -1 & 0 & 0\\ 0 & -1 & 0\\ 0& 0 & -1\end{pmatrix}$

This operator correspods, according to the theory we have been studying, to some operator P (please, don’t confuse P with momentum) that satisfies

$PqP^{-1}=-q$

$PpP^{-1}=-p$

and where $q, p$ are the usual position and momentum operators. Then, the operator

$L=q\times p$ is invariant by parity/spatial inversion P, and thus, this feature can be extended to any angular momentum operator like spin S or angular momentum J. That is,

$PJP^{-1}=J$ and $PSP^{-1}=S$

The Wigner’s theorem implies that corresponding to the operator P, a discrete transformation, must exist some unitary or antiunitary operator. In fact, it shows that P is indeed unitary

$P\left[Q_i,P_j\right]P^{-1}=\left[Q_i,P_j\right]=P(i\hbar\delta_{ij})P^{-1}$

If P were antiunitary we should get

$P\left[Q_i,P_j\right]P^{-1}=\left[Q_i,P_j\right]=P(i\hbar\delta_{ij})P^{-1}=-i\hbar\delta_{ij}$

Then, the parity operator P is unitary and $P^{-1}=P$. In fact, this can be easily proved from its own definition.

If we apply two succesive parity transformations we leave the state invariant, so $P^2=I$. We say that the parity operator is idempotent.  The check is quite straightforward

$\vert\Psi\rangle\longrightarrow\vert\Psi'\rangle=PP\vert\Psi\rangle\longrightarrow\vert\Psi\rangle=e^{i\omega}\vert\Psi\rangle$

Therefore, from this viewpoint, there are (in general) only 2 different ways to satisfy this as we have $PP=e^{i\omega}I$:

i) $e^{i\omega}=+1$. The phase is equal to $0$ modulus $2\pi$. We have hermitian operators

$P=P^{-1}=P^+$

Then, the effect on wavefunctions is that $\Psi (P^{-1}(\vec{r}))=\Psi (-\vec{r})$. That is the case of usual particles.

ii) The case $e^{i\omega}=-1$. The phase is equal to $\pi$ modulus $2\pi$. This is the case of an important class of particles. In fact, Steven Weinberg has showed that $P^2=(-1)^F$ where F is the fermion number operator in the SM. The fermionic number operator is defined to be the sum $F=L+B$ where L is now the leptonic number and B is the baryonic number. Moreover, for all particles in the Standard Model and since lepton number and baryon number are charges Q of continuous symmetries $e^{iQ}$  it is possible to redefine the parity operator so that $P^2=I$. However, if there exist Majorana neutrinos, which experimentalists today believe is quite possible or at least it is not forbidden by any experiment, their fermion number would be equal to one because they are neutrinos while their baryon and lepton numbers are zero because they are Majorana fermions, and so $(-1)^F$ would not be embedded in a continuous symmetry group. Thus Majorana neutrinos would have parity equal to $\pm i$. Beautiful and odd, isnt’t it? In fact, if some people are both worried or excited about having Majorana neutrinos is also due to the weird properties a Majorana neutrino would have under parity!

The strange case of time reversal T

In Quantum Mechanics, temporal inversions or more generally the time reversal is defined as the operator that inverts the “flow or direction” of time. We have

$T: t\longrightarrow t'=-t$ $\vec{r}'(-t)=\vec{r}(t)$

And it implies that $\vec{p}(-t)=-\vec{p}(t)$. Therefore, the time reversal operator $T$ satisfies

$TQT^{-1}=Q$

$TPT^{-1}=-P$

In summary: T is by definition the “inversion of time” so it also inverts the linear momentum while it leaves invariant the position operator.

Thus, we also have the following transformation of angular momentum under time reversal:

$TJT^{-1}=-J$

$TST^{-2}=-S$

Time reversal can not be a unitary operator, and it shows that the time reversal T is indeed an antiunitary operator. The check is quite easy:

$T\left[Q,P\right]T^{-1}=\left[TQT^{-1},TPT^{-1}\right]=-\left[Q,P\right]=Ti\hbar T^{-1}$

This equation matches the original definiton if and only if (IFF)

$TiT^{-1}=-i \leftrightarrow TT^{-1}=-1$

Time reversal is as consequence of this fact an antiunitary operator.

LOG#093. Group theory(XIII).

The sixtor or 6D Riemann-Silberstein vector is a complex-valued quantity $\Psi=E+iB$ up to one multiplicative constant and it can be understood as a bivector field in Clifford algebras/geometric calculus/geometric algebra. But we are not going to go so far in this post. We remark only that a bivector field is something different to a normal vector or even, as we saw in the previous post, a bivector field can not be the same thing that a spinor field with $1/2$ spin. Moreover, the electric and magnetic parts of the sixtor transform as vectors under spatial rotations

$(x^0,\mathbf{x})\longrightarrow (x'^0,\mathbf{x'})=(x^0,R\mathbf{x})$

where $C'(x')=RC(x)$ and R being an orientation preserving rotation matrix in the spcial orthogonal group $SO(3,\mathbb{R})$. Remember that

$SO(3)=G=\left\{R\in M_{3\times 3}(\mathbb{R})/RR^T=R^TR=I_3,\det R=+1\right\}$

The group $SO(3,\mathbb{R})$ is bases on the preservation of the scalar (inner) product defined by certain quadratic form acting on 3D vectors:

$(x,y)=x\cdot y=X^TY=(RX,RY)=X^TR^TRY$

and $R^TR=RR^T=I_3$ so $\det R=+1$ for proper rotations. This excludes spatial reflections or parity transformations P, that in fact are important too. Parity transformations act differently to electric and magnetic fields and they have $\det P=-1$. Parity transformations belong to the group of “improper” rotations in 3D space.

However, the electromagnetic field components are NOT related to the spatial components of a 4D vector. That is not true. With respect to the proper Lorentz group:

$SO^+(1,3)=\left\{\Lambda \in M_{4\times 4}(\mathbb{R}/\Lambda^T\eta\Lambda=\eta,\Lambda^0_{\;\;\;0}\geq 1,\det \Lambda=+1\right\}$

and where the metric $\eta=\mbox{diag}(-1,1,1,1)$ is the Minkovski metric. In fact, the explicit representation of

$\Lambda\in SO(1,3)^+$ by its matrix elements

$\Lambda =\Lambda^\mu_{\;\;\; \nu}=\begin{pmatrix}\Lambda^{0}_{\;\;\; 0} & \Lambda^{0}_{\;\;\; 1} & \Lambda^{0}_{\;\;\; 2} & \Lambda^{0}_{\;\;\; 3}\\ \Lambda^{1}_{\;\;\; 0} & \Lambda^{1}_{\;\;\; 1} & \Lambda^{1}_{\;\;\; 2} & \Lambda^{1}_{\;\;\; 3}\\ \Lambda^{2}_{\;\;\; 0} & \Lambda^{2}_{\;\;\; 1} & \Lambda^{2}_{\;\;\; 2} & \Lambda^{2}_{\;\;\; 3}\\ \Lambda^{3}_{\;\;\; 0} & \Lambda^{3}_{\;\;\; 1} & \Lambda^3_{\;\;\; 2} & \Lambda^{3}_{\;\;\; 3}\end{pmatrix}$

Despite this fact, that the electromagnetic field sixtor $\Psi$ transforms as a vector under the COMPLEX special orthogonal group $SO(3,\mathbb{C})$ in 3D space

$SO(3,\mathbb{C})=\left\{Q\in M_{3\times 3}(\mathbb{C})/Q^TQ=QQ^T=I_3,\det Q=1\right\}$

This observation is related to the fact that the proper Lorentz group and the complex rotation group are isomorphic to each other as Lie groups, i.e. $SO(3,\mathbb{C})\cong SO(3,1;\mathbb{R})^+$. This analogy and mathematical result has some deeper consequences in the theory of the so-called Dirac, Weyl and Majorana spinors (quantum fields describing fermions with differnt number of independent “components”) in the massive case.

The puzzle, now,  is to understand why the mass term is forbidden in

$i\Gamma^\mu\partial_\mu \Psi=0$

as it is the case of the electromagnetic (classical) field. Moreover, in this problem, we will see that there is a relation between symmetries and operators of the Lie groups $SO(1,3;\mathbb{R}), SO(3,\mathbb{C}),SO(3,\mathbb{R})$ and the corresponding generators of their respective Lie algebras. Let’s  begin with pure boosts in some space-time plane:

$\Lambda =\begin{pmatrix}\gamma & -\gamma\beta & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$

and where we defined, as usual,

$\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$ and $\beta c=v$, with $\gamma^2-\gamma^2\beta^2=1$.

Then,

$\Lambda=I_4+\beta\begin{pmatrix}0 & -1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}=I_4+\beta L_1$

and where we have defined the boost generator as $L_1$ in the plane $(t,x^1)$, and the boost parameter will be given by the number

$\cosh (\xi_1)=\gamma_1$

Remark: Lorentz transformations/boosts corresponds to rotations with an “imaginary angle”.

Moreover, we also get

$e^{\xi_1L_1}=\begin{pmatrix}\cosh\xi_1 & -\sinh\xi_1 & 0 & 0\\ -\sinh\xi_1 & \cosh\xi_1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 &0 & 1\end{pmatrix}$

Equivalently, by “empathic mimicry” we can introduce the boost generators in the remaining 3 planes $L_2, L_3$ as follows:

$L_2=\begin{pmatrix}0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$

$L_3=\begin{pmatrix}0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\end{pmatrix}$

In addition to these 3 Lorentz boosts, we can introduce and define another 3 generators related to the classicla rotation in 3D space. Their generators would be given by:

$S_1=\begin{pmatrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\end{pmatrix}$

$S_2=\begin{pmatrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\end{pmatrix}$

$S_3=\begin{pmatrix}0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$

Therefore, $S=(S_1,S_2,S_3)$ and $L=(L_1,L_2,L_3)$ span the proper Lorent Lie algebra $so(1,3)^+$ with generators $L_i, S_j$. These generators satisfy the commutators:

$\left[S_a,S_b\right]=-\varepsilon_{abc}S_c$

$\left[L_a,L_b\right]=+\varepsilon_{abc}S_c$

$\left[L_a,S_B\right]=-\varepsilon_{abc}L_c$

with $a,b,c=1,2,3$ and $\varepsilon_{abc}$ the totally antisymmetric tensor. This Levi-Civita symbol is also basic in the $SO(3)$ structure constants. Generally speaking, in the Physics realm, the generators are usually chosen to be hermitian and an additional imaginar $i$ factor should be included in the above calculations to get hermitian generators. If we focus on the group $SO(3)$ over the real numbers, i.e., the usual rotation group, the Lie algebra basis is given by $(S_a)_{bc}=\varepsilon_{abc}$, or equivalently by the matrices

$S_1=\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{pmatrix}$ $S_2=\begin{pmatrix} 0 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{pmatrix}$ $S_3=\begin{pmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}$

and the commutation rules are

$\left[S_a,S_b\right]=-\varepsilon_{abc}S_c$

If the rotation matrix is approximated to:

$R=I_3+\delta R\leftrightarrow R^TR=RR^T=(I_3+\delta R)(I_3+\delta R)^T=I_3+\delta R+\delta R^T+\mathcal{O}(\delta^2)$

then we have that the rotation matrix is antisymmetric, since we should have

$\delta R+\delta R^T=0$ or $\delta R=-\delta R^T$

so the matrix generators are antisymmetric matrices in the $so(3)$ Lie algebra. That is, the generators are real and antisymmetric. The S-matrices span the full $so(3)$ real Lie algebra. In the same way, we could do the same for the complex group $SO(3,\mathbb{C})$ and we could obtain the Lie algebra $so(3)$ over the complex numbers with $\tilde{S_k}=iS_k$ added to the real $S_k$ of the $SO(3)$ real Lie group. This defines a “complexification” of the Lie group and it implies that:

$\left[S_a,S_b\right]=-\varepsilon_{abc}S_c$

$\left[\tilde{S}_a,\tilde{S}_b\right]=+\varepsilon_{abc}S_c$

$\left[\tilde{S}_a,S_b\right]=-\varepsilon_{abc}\tilde{S}_c$

Do you remember this algebra but with another different notation? Yes! This is the same Lie algebra we obtained in the case of the Lorentz group. Therefore, the Lie algebras of $SO(1,3)^+$ and $SO(3,\mathbb{C})$ are isomorphic if we make the identifications:

$S_a\leftrightarrow S_i$ The rotation part of $SO(1,3)^+$ is the rotation part of $SO(3,\mathbb{C})$

$L_a\leftrightarrow \tilde{S}_a$ The boost part of $SO(1,3)^+$ is the complex conjugated (rotation-like) part of $SO(3,\mathbb{C})$

Then for every matrix $\Lambda\in SO(1,3)^+$ we have

$\Lambda=e^{\xi_1L_1+\xi_2L_2+\xi_3L_3+\alpha_1S_1+\alpha_2S_2+\alpha_3S_3}$

and for every matrix $Q\in SO(3,\mathbb{C})$ we have

$Q=e^{\xi_1\tilde{S}_1+\xi_2\tilde{S}_2+\xi_3\tilde{S}_3+\alpha_1S_1+\alpha_2S_2+\alpha_3S_3}$

For instance, a bivector boost in some axis provides:

$e^{\xi_1\tilde{S}_1}=\begin{pmatrix}1 & 0 & 0\\ 0 & \cosh\xi_1 & i\cosh\xi_1\\ 0 & -i\sinh\xi_1 & \cosh\xi_1\end{pmatrix}=\begin{pmatrix}1 & 0 & 0\\ 0 & \gamma_1 & i\gamma_1\beta_1\\ 0 & -i\gamma_1\beta_1 & \gamma_1\end{pmatrix}$

and where in the first matrix, it acts on $\Psi=\dfrac{1}{\sqrt{2}}(E+iB)=\overline{\Psi}_a$, the second matrix (as the first one) acts also on a complex sixtor, and where the rotation around the axis perpendicular to the rotation plane is defined by the matrix operator:

$e^{\alpha_1S_1}=\begin{pmatrix}1 & 0 & 0\\ 0 & \cos\alpha_1 & \sin\alpha_1\\ 0 & -\sin\alpha_1 & \cos\alpha_1\end{pmatrix}$

and this matrix would belong to the real orthogonal group $SO(3,\mathbb{R})$.

Note: Acting $\exp (\xi\tilde{S}_1)$ onto the sixtor $\Psi$ as a bivector field shows that it generates the correct Lorentz transformation of the full electromagnetic field! The check is quite straightforward, since

$\begin{pmatrix}E'_1+iB'_1\\ E'_2+iB'_2\\ E'_3+iB'_3\end{pmatrix}=\begin{pmatrix}1 & 0 & 0\\ 0 & \gamma_1 & i\gamma_1\beta_1\\ 0 & -i\gamma_1\beta_1 & \gamma_1\end{pmatrix}\begin{pmatrix}E_1+iB_1\\ E_2+iB_2\\ E_3+iB_3\end{pmatrix}$

From this complex matrix we easily read off the transformation of electric and magnetic fields:

$E'_1=E_1$

$E'_2=\gamma_1(E_2-\beta_1B_3)$

$E'_3=\gamma_1(E_3+\beta_1B_2)$

$B'_1=B_1$

$B'_2=\gamma_1(B_2+\beta_1E_3)$

$B'_3=\gamma_1(B_3+\beta_1E_2)$

Note the symmetry between electric and magnetic fields hidden in the sixtor/bivector approach!

For the general Lorentz transformation $x'^\mu=\Lambda^\mu_{\;\;\; \nu}x^\nu$ we have the invariants

$\Psi^T\Psi=\Psi^TQ^+Q\Psi=\dfrac{1}{2}(E^2-B^2)+iE\cdot B$

LOG#048. Thomas precession.

LORENTZ TRANSFORMATIONS IN NON-STANDARD FORM

Let me begin this post with an uncommon representation of Lorentz transformations in terms of “uncommon matrices”. A Lorentz transformation can be written symbolically, as we have seen before, as the set of linear transformations leaving invariant

$ds^2=d\mathbf{x}^2-c^2dt^2$

Therefore, the Lorentz transformations are naively $X'=\mathbb{L}X$. Let $\mathbf{A}, \mathbf{B}$ be 3-rowed column matrices and let $M, R, \mathbb{I}$ represent $3\times 3$ matrices and $T$ will be used (unless it is stated the contrary) to denote the matrix transposition ( interchange of rows and columns in the matrix).

The invariance of $ds'^2=ds^2$ implies the following results from the previous definitions:

$\gamma^2-\mathbf{B}^2=1$

$M^TM =\mathbf{A}\mathbf{A}^T+\mathbb{I}$

$M^T\mathbf{B}=\gamma \mathbf{A}\leftrightarrow \mathbf{B}^T M=\gamma \mathbf{A}^T$

Then, we can write the matrix for a Lorent transformation (boost) in the following non-standard manner:

$\boxed{\mathbb{L}=\begin{pmatrix}\gamma & -\mathbf{A}^T\\ -\mathbf{B} & M\end{pmatrix}}$

and the inverse transformation will be

$\boxed{\mathbb{L}^{-1}=\begin{pmatrix}\gamma & \mathbf{B}^T\\ \mathbf{A} & M^T\end{pmatrix}}$

Thus, we have $\mathbb{L}\mathbb{L}^{-1}=\mathbb{I}_{4x4}\equiv \mathbb{E}$, where we also have

$\gamma^2-\mathbf{A}^2=1$

$M\mathbf{A}=\gamma \mathbf{B}$

$MM^T=\mathbf{B}\mathbf{B}^T+\mathbb{I}_{3x3}$

Let us define, in addition to this stuff, the reference frames $S, \overline{S}'$, corresponding to the the coordinates $\mathbf{X}$ and $\overline{\mathbb{X}}'$. Then, the boost matrix will be recasted, if the velocity read $\mathbf{v}=\mathbf{A}/\gamma$, as

$L_{v}=\begin{pmatrix}\gamma & -\gamma \mathbf{v}^T\\ -\gamma \mathbf{v} & \mathbb{I}+\frac{\gamma^2}{1+\gamma}\mathbf{v}\mathbf{v}^T\end{pmatrix}=\begin{pmatrix}\gamma & -\mathbf{A}^T\\ -\mathbf{A} & \mathbb{I}+\frac{\mathbf{A}\mathbf{A}^T}{1+\gamma}\end{pmatrix}$

Remark: a Lorentz transformation will differ from boosts only by rotations in the general case. That is, with these conventions, the most general Lorentz transformations include both boosts and rotations.

For all $\gamma>0$, the above transformation is well-defined, but if $\gamma<0$, then it implies we will face with transformations containing the reversal of time ( the time reversal operation T, please, is a different thing than matrix transposition, do not confuse their same symbols here, please. I will denote it by $\mathbb{T}$ in order to distinguish, althoug there is no danger to that confusion in general). The time reversal can be written indeed as:

$\mathbb{T}=\begin{pmatrix}-1 & \mathbf{0}^T\\ \mathbf{0} & \mathbb{I}\end{pmatrix}$

In that case, ($\gamma<0$), after the boost $L_{v}$, we have to make the changes $\gamma \rightarrow \vert \gamma\vert$ and $\mathbf{A}\rightarrow -\mathbf{A}$. If these shifts are done, the reference frames $\overline{S}$ and $\overline{S}'$ can be easily related

$\overline{X}'=LX=LL^{-1}_{v}\overline{X}$

in such a way that

$LL^{-1}_{v}=\begin{pmatrix}1 & \mathbf{0}\\ \mathbf{0} & R\end{pmatrix}=L_R$

where the rotation matrix is given formally by the next equation:

$R=M-\dfrac{\mathbf{B}\mathbf{A}^T}{1+\gamma}$

R must be an orthogonal matrix, i.e., $R^TR=\mathbb{I}_{3x3}$. Then $(\det (R))^2=1$, or $det R=\pm 1.$. For $\det R=-1$ we have the parity matrix

$\mathbb{P}=\begin{pmatrix}1 & \mathbf{0}^T\\ \mathbf{0} & -\mathbb{I}_{3x3}\end{pmatrix}$

and it will transform right-handed frames to left-handed frames $\overline{S}$ or $\overline{S}'$. The rotation vector $\alpha$ can be defined as well:

$1+2\cos \alpha=Tr (R)\rightarrow \cos\alpha=\dfrac{Tr R-1}{2}$

so $\alpha^\mu=\dfrac{1}{2}\epsilon^{\mu\nu\lambda}R^\nu_{\lambda}\dfrac{\alpha}{\sin\alpha}, \forall 0\leq \alpha<\pi$. The rotation acting on 3-rowed matrices:

$R\mathbf{A}=\mathbf{B}$

implies that $\overline{X}'=R\overline{X}$, and it changes $-\mathbf{A}/\gamma$ of the frame S into $\overline{S}$. Passing from one frame into another, $\overline{S}'$ to $S'$, it implies we can define a boost with $L_{-\mathbf{B}/\gamma}$. In fact,

$L_{-\mathbf{B}/\gamma}L=\begin{pmatrix}1 & \mathbf{0}^T\\ \mathbf{0} & R\end{pmatrix}=L_R$

Q.E.D.

Remark(I): Without the time reversal, we would get $L_{R\mathbf{v}}L_R=L=L_RL_{\mathbf{v}}$

with $\mathbf{v}=\mathbf{A}/\gamma$ and $R=M-\dfrac{\mathbf{BA}^T}{1+\gamma}$.

Remark (II): $L_RL_v\rightarrow L^T=L^T_vL_R^T=L_vL_{R^T}$. If $L^T=L=L_{R\mathbf{v}}L_R$, then the uniqueness of $R\mathbf{v}$ provides that $R=R^T=R^{-1}$, i.e., that R is an orthogonal matrix. If R is an orthogonal matrix and a proper Lorentz transformation ( $det R=+1$), then we would get $\sin\alpha=0$, and thus $\alpha=0$ or $\alpha=\pi$, and so, $R=I$ or $R=2\mathbf{n}\mathbf{n}^T-1$, with the unimodular vector $\mathbf{n}$, i.e., $\vert \mathbf{n}\vert=1$. That would be the case $\forall \mathbf{v}\neq 0$ and $\mathbf{n}=\mathbf{v}/\vert\vert \mathbf{v}\vert\vert$. Otherwise, if $\mathbf{v}=0$, then $\mathbf{n}$ would be an arbitrary vector.

The second step previous to our treatment of Thomas precession is to review ( setting $c=1$) the addition of velocities in the special relativistic realm. Suppose a point particle moves with velocity $\overline{w}$ in the reference frame $\overline{S}$. Respect to the S-frame (in rest) we will write:

$\mathbf{x}=\overline{\mathbf{x}}+\dfrac{\gamma^2}{\gamma+1}(\overline{\mathbf{x}}\mathbf{v})\mathbf{v}+\gamma \mathbf{v}\overline{t}$

and

$t=\gamma \overline{t}+\gamma (\mathbf{v}\overline{\mathbf{x}})$

and with $\overline{x}=\overline{\mathbf{w}}\overline{t}$ we can calculate the ratio $\mathbf{u}=\mathbf{x}/t$:

$\mathbf{u}=\dfrac{\dfrac{\overline{\mathbf{w}}}{\gamma}+\dfrac{\gamma}{1+\gamma}(\mathbf{v}\overline{\mathbf{w}})\mathbf{v}+\mathbf{v}}{1+\mathbf{v}\overline{\mathbf{w}}}$

and thus

$\mathbf{u}\equiv \dfrac{\mathbf{v}+\mathbf{w}_\parallel+(\mathbf{w}_\perp/\gamma)}{1+\mathbf{v}\overline{\mathbf{w}}}$

where we have defined:

$(\mathbf{w}_\perp/\gamma)\equiv\dfrac{\overline{\mathbf{w}}}{\gamma}$

and

$\mathbf{w}_\parallel\equiv \dfrac{\gamma}{1+\gamma}(\mathbf{v}\overline{\mathbf{w}})\mathbf{v}$

Comment: the composition law for 3-velocities is special relativity is both non-linear AND non-associative.

There are two special cases of motion we use to consider in (special) relativity and inertial frames:

1st. The case of parallel motion between frames (or “parallel motion”). In this case $\overline{\mathbf{w}}=\lambda \mathbf{v}$, i.e., $\mathbf{w}\times \mathbf{v}=0$. Therefore,

$\mathbf{u}=\dfrac{\mathbf{v}+\overline{\mathbf{w}}}{1+\mathbf{v}\overline{\mathbf{w}}}$

This is the usual non-linear rule to add velocities in Special Relativity.

2nd. The case of orthogonal motion between frames, where $\mathbf{v}\perp\mathbf{w}$. It means $\mathbf{v}\mathbf{w}=0$. Then,

$\mathbf{u}=\mathbf{v}+\mathbf{w}/\gamma= \mathbf{v}+\overline{\mathbf{w}}\sqrt{1-\mathbf{v}^2}$

This orthogonal motion to the direction of relative speed has an interesting phenomenology, since this inertial motion will be slowed down due to time dilation because the spatial distances that are orthogonal to $\mathbf{v}$ are equal in both reference frames.

Furthermore, we get also:

$\mathbf{u}^2=1-\dfrac{(1-\overline{\mathbf{w}}^2)(1-\mathbf{v}^2)}{(1+\mathbf{v}\overline{\mathbf{w}})}\leq 1$

Indeed, the condition $\mathbf{u}^2=1$ implies that $\overline{\mathbf{w}}^2=1$ or $\mathbf{v}^2=1$, and the latter condition is actually forbidden because of our interpretation of $\mathbf{v}$ as a relative velocity between different frames. Thus, this last equation shows the Lorentz invariance in Special relativity don’t allow for superluminal motion, although, a priori, it could be also used for even superluminal speeds since no restriction apply for them beyond those imposed by the principle of relativity.

THOMAS PRECESSION

We are ready to study the Thomas precession and its meaning. Suppose an inertial frame $\overline{\overline{S}}$ obtained from another inertial frame $\overline{S}$ by boosting the velocity $\overline{w}$. Therefore, $\overline{\overline{S}}$ owns the relative velocity $\mathbf{v}$ given by the addition rule we have seen in the previous section. Moreover, we have:

$\overline{\overline{x}}=L_{\overline{w}}\overline{x}=L_{\overline{w}}L_{\mathbf{v}}x$

Then, we get

$L_{\mathbf{v}}=\begin{pmatrix}\gamma_v & -\gamma_v \overline{\mathbf{v}}^T\\ -\gamma_v \mathbf{v} & \mathbf{1}+\dfrac{\gamma_v^2}{1+\gamma_v}\mathbf{v}\mathbf{v}^T\end{pmatrix}$

$L_{\overline{\mathbf{w}}}=\begin{pmatrix}\gamma_{\overline{\mathbf{w}}} & -\gamma_{\overline{\mathbf{w}}} \overline{\mathbf{w}}^T\\ -\gamma_{\overline{\mathbf{w}}} \overline{\mathbf{w}}^T & \mathbf{1}+\dfrac{\gamma_{\overline{\mathbf{w}}}^2}{1+\gamma_{\overline{\mathbf{w}}}}\overline{\mathbf{w}}\overline{\mathbf{w}}^T\end{pmatrix}$

where

$\gamma_{v}=\dfrac{1}{\sqrt{1-\mathbf{v}^2}}$

$\gamma_{\overline{\mathbf{w}}}=\dfrac{1}{\sqrt{1-\overline{\mathbf{w}}^2}}$

and then

$\boxed{L\equiv=L_{\overline{\mathbf{w}}}L_v=\begin{pmatrix}\gamma & -\mathbf{A}^T\\ -\mathbf{B} & M\end{pmatrix}}$

with

$\gamma (\mathbf{v},\overline{\mathbf{w}})=\gamma_v\gamma_{\overline{w}}(1+\mathbf{v}\overline{\mathbf{w}})\equiv \gamma (\overline{\mathbf{w}},\mathbf{v})$

$\mathbf{A}=\gamma (\mathbf{v},\overline{\mathbf{w}})\overline{\mathbf{w}}o \mathbf{v}$

$\mathbf{B}=\gamma (\overline{\mathbf{w}},\mathbf{v})\mathbf{v}o\overline{\mathbf{w}}$

$M=M(\overline{\mathbf{w}},\mathbf{v})=\mathbf{1}+\dfrac{\gamma_v^2}{1+\gamma_v}\mathbf{v}\mathbf{v}^T+\dfrac{\gamma_{\overline{\mathbf{w}}}^2}{1+\gamma_{\overline{\mathbf{w}}}}\overline{\mathbf{w}}\overline{\mathbf{w}}^T+\gamma_v\gamma_{\overline{\mathbf{w}}}\left( 1+\dfrac{\gamma_v\gamma_{\overline{\mathbf{w}}}}{(1+\gamma_v)(1+\gamma_{\overline{\mathbf{w}}})}\mathbf{v}\overline{\mathbf{w}}\right)\overline{\mathbf{w}}\mathbf{v}$

Here, we have defined:

$\boxed{\overline{\mathbf{w}}o \mathbf{v}\equiv \dfrac{\left( \gamma_{\overline{\mathbf{w}}}\gamma_v\mathbf{v}+\gamma_{\overline{\mathbf{w}}}\overline{\mathbf{w}}+\gamma_{\overline{\mathbf{w}}}\dfrac{\gamma_v^2}{1+\gamma_v}(\overline{\mathbf{w}}\mathbf{v})\right)}{\gamma (\mathbf{v},\overline{\mathbf{w}})}}$

Remark (I): The matrix L given by

$\begin{pmatrix}\gamma & -\mathbf{A}^T\\ -\mathbf{B} & M\end{pmatrix}$

is NOT symmetric as we would expect from a boost. According to our decomposition for the matrix $M$ it can be rewritten in the following way

$\boxed{R=R(\overline{\mathbf{w}},\mathbf{v})=M(\overline{\mathbf{w}},\mathbf{v})-\dfrac{\mathbf{B}\mathbf{A}^T}{1+\gamma}}$

This last equation is called the Thomas precession associated with the tridimensional 3-vectors $\mathbf{v},\overline{\mathbf{w}}$. We observe that R is a proper-orthogonal matrix from the multiplicative property of the determinants and the fact that all boosts have determinant one. Equivalently, from the condition $R=\pm 1$ for all orthogonal matrix R together with the continuous dependence of R on the velocities and the initial condition $R(0,0)=\mathbf{1}$.

Remark (II): From the definitions of M, and the vectors $\mathbf{A},\mathbf{B}$, we deduce that $\mathbf{v}\times \overline{\mathbf{w}}$ is an eigenvector of R with eigenvalue +1 and this gives the axis of rotation. The rotation angle $\alpha$ as calculated from $Tr R=1+2\cos\alpha$ is complicated expression, and only after some clever manipulations or the use of the geometric algebra framework, it simplifies to

$1+\cos\alpha=\dfrac{(1+\gamma_u+\gamma_v+\gamma_{\overline{w}})}{(1+\gamma_u)(1+\gamma_v)(1+\gamma_{\overline{w}})}>0$

In order to understand what this equation means, we have to observe that the components $\mathbf{v}$ and $\overline{\mathbf{w}}$ refer to different reference frames, and then, the scalar product $\mathbf{v}\mathbf{\overline{w}}$ and the cross product $\mathbf{v}\times\overline{\mathbf{w}}$ must be given good analitic expressions before the geometric interpretation can be accomplished. Moreover, if we want to interpret the cross product as an axis in the reference frame $S$, and correspondingly we want to split $L=L_{R\mathbf{v}}L_R$,  by the definition $\overline{\mathbf{w}}o\mathbf{v}$ we deduce that

$\mathbf{v}\times\mathbf{u}=\dfrac{\mathbf{v}\times\overline{\mathbf{w}}}{\gamma_v(1+\mathbf{v}\overline{\mathbf{w}})}$

and thus, the Thomas rotation of the inertial frame S has its axis orhtogonal to the relative velocity vectors $\mathbf{v},\mathbf{u}$ of the reference frame $\overline{\overline{S}}$, $\overline{\overline{S}}$ against S.

By the other hand, if we interpret the above last equation as an axis in the reference frame $\overline{\overline{S}}$, asociated to the split $L=L_RL_\mathbf{u}$, we would deduce that $L_{R\mathbf{u}}L_R$ implies the following consequence. The reference frame $\overline{\overline{S}}$ is got from boosting certain frame S’ obtained itself from a rotation of S by R. Then, $\overline{\overline{S}}$ obtains (compared with S or S’), a velocity whose components are $R\mathbf{u}$ in the inertial frame S’. Reciprocally, the components of the velocity of S or S’ against the frame $\overline{\overline{S}}$ are provided, in $\overline{\overline S}$, by $\overline{\overline{\mathbf{u}}}=-R\mathbf{u}$. Therefore, from the Thomas precession formula for R we observe that $R\mathbf{u}$ differs from $\mathbf{u}$ only by linear combinations of the vectors $\mathbf{v}$ and $\overline{\mathbf{w}}$. With all this results we easily derive:

$\overline{\overline{u}}\times \overline{\overline{\mathbf{w}}}=(-R\mathbf{u})\times (-\overline{\mathbf{w}})\propto \mathbf{v}\times \overline{\mathbf{w}}$

i.e., the axis for the Thomas rotation matrix of $\overline{\overline{S}}$ is orthogonal to the relative velocities $\overline{\overline{\mathbf{u}}}, \overline{\overline{\mathbf{w}}}$ of the inertial frames S, $\overline{S}$ against $\overline{\overline{S}}$. Finally, to find the rotation matrix, it is enough to restrict the problem to the case where $\overline{\mathbf{w}}$ is small so that squares of it may be neglected. In this simple case, R would become into:

$\boxed{R\approx \mathbf{1}+\dfrac{\gamma_v}{1+\gamma_v}\left(\overline{\mathbf{w}}\mathbf{v}^T-\mathbf{v}\overline{\mathbf{w}}^T\right)}$

and where the rotation angle is given by

$\boxed{\alpha\approx -\dfrac{\gamma_v}{1+\gamma_v}\mathbf{v}\times\overline{\mathbf{w}}\approx -\dfrac{\gamma_v^2}{1+\gamma_v}\mathbf{v}\times\mathbf{u}}$

In order to understand the Physics behind the Thomas precession, we will consider one single experiment. Imagine an inertial frame S in accelerated motion with respect to other inertial frame I. The spatial axes of S remain parallel at any time in the sense that the instantaneous reference frame coinciding with S at times $t+\Delta t$ are related by a pure boost in the limit $\Delta t\rightarrow 0$. This may be managed if we orient S with the aid of a very fast spinning torque-free gyroscope. Then, from the inertial frame I, S seems to be rotated at each instant of time and there is a continuous rotation of S against I since the velocity of S varies and changes continuously. This gyroscopic rotation of S relative to I IS the Thomas precession.  We can determine the angular velocity of this motion in a straightforward manner. During the small interval of time $\Delta t$ measured from I, the instantaneous velocity $\mathbf{v}$ of S changes by certain quantity $\Delta \mathbf{v}$, measured from I. In that case,

$\Delta \alpha=-\gamma_v^2\mathbf{v}\times\dfrac{\Delta\mathbf{v}}{(1+\gamma_v)}$

for the rotation vector during a time interval $\Delta t$. Thus, the angular velocity for the Thomas precession will be given by:

$\boxed{\omega_T=-\dfrac{\gamma^2}{1+\gamma_v}\mathbf{v}\times\dfrac{d\mathbf{v}}{dt}}$

or reintroducing the speed of light we get

$\boxed{\omega_T=-\dfrac{\gamma^2}{1+\gamma_v}\mathbf{v}\times\dfrac{1}{c^2}\dfrac{d\mathbf{v}}{dt}=\dfrac{\gamma^2}{1+\gamma_v}\dfrac{1}{c^2}\mathbf{a}\times\mathbf{v}}$

Remark(I): The special relativistic effect given by the Thomas precession was used by Thomas himself to remove a discrepancy and mismatch between the non-relativistic theory of the spinning electron and the experimental value of the fine structure. His observation was, in fact, that the gyromagnetic ratio of the electron calculated from the anomalous Zeeman effect led to a wrong value of the fine structure constant $\alpha$. The Thomas precession introduces a correction to the equation of motion of an electron in an external electromagnetic filed and such a correction induces a correction of the spin-orbit coupling, explaining the correct value of the fine structure.

Remark (II): In the framework of the relativistic quantum theory of the electron, Dirac realized that the effect of Thomas precession was automatically included!

Remark (III): Inside the Thomas paper, we find these interesting words

“(…)It seems that Abraham (1903) was the first to consider in any detail an electron with an axis. Many have since then considered spinning electron, ring electrons, and the like. Compton (1921) in particular suggested a quantized spin for the electron. It remained for Uhlenberg and Goudsmit (1925) to show ho this idea can be used to explain the anomalous Zeeman effect. The asumptions they had to make seemed to lead to optical and relativity doublet separations twice larger than those we observe. The purpose of the following paper, which contains the results mentioned in my recent letter to Nature (1926), is to investigate the kinematics of an electron with an axis on the basis of the restricted theory of relativity. The main fact used is that the combination of two Lorentz transformations without rotation in general is not of the same form(…)”.

From the historical viewpoint it should also be remarked that the precession effect was known by the end of 1912 to the mathematician E.Borel (C.R.Acad.Sci.,156. 215 (1913)). It was described by him (Borel, 1914) as well as by L.Silberstein (1914) in textbooks already 1914. It seems that the effect was even known to A.Sommerfeld in 1909 and before him, perhaps even to H.Poincaré. The importance of Thomas’ work and papers on this subject was thus not only the rediscovery but the relevant application to a virulent problem in that time, as it was the structure of the atomic spectra and the fine structure constant of the electron!

Remark (IV): Not every Lorentz transformation can be written as the product of two boosts due to the Thomas precession!

THE LORENTZ GROUP AS A QUASIDIRECT PRODUCT: QUASIGROUPS, LOOPS AND GYROGROUPS

Even though we have not studied group theory in this blog, I feel the need to explain some group theory stuff related to the Thomas precession here.

The kinematical differences between Galilean and Einsteinian relativity theories is observed at many levels. The essential differences become apparent already on the level of the homogenous groups without reversals (inverses). Let me first consider the Galileo group. It is generated by space rotations $G_R=L_R$ and galilean boosts in any number and order. Using the notation we have developed in this post, we could write $X'=G_\mathbf{v}X$ in this way:

$G_\mathbf{v}=\begin{pmatrix}1 & \mathbf{0}^T\\ -\mathbf{v} & \mathbf{1}\end{pmatrix}$

The following relationships are deduced:

$G_RG_\mathbf{v}=G_{R\mathbf{v}}G_{R}$

$G_{R_1}G_{R_2}=G_{R_1R_2}$

$G_{\mathbf{v}_1}G_{\mathbf{v}_2}=G_{\mathbf{v}_1+\mathbf{v}_2}=G_{\mathbf{v}_2}G_{\mathbf{v}_1}$

In the case of the Lorentz group, these equations are “generalized” into

$L_RL_\mathbf{v}=L_{R\mathbf{v}}L_{R}$

$L_{R_1}L_{R_2}=L_{R_1R_2}$

$L_{\mathbf{v}_1}L_{\mathbf{v}_2}=L_{R(\mathbf{v}_1,\mathbf{v}_2)}L_{\mathbf{v}_1 o \mathbf{v}_2}$

where $R(\mathbf{v}_1,\mathbf{v}_2)$ is the Thomas precession and the circle denotes the nonlinear relativisti velocity addition. Be aware that the domain of velocities in special relativity is $\vert v\vert<1$, in units with c set to unity.

Both groups (Galileo and Lorentz) contain as a subroupt the group of al spatial rotations $G_R\equiv L_R$. The set of galilean or lorentzian boosts $G_v$ and $L_v$ are invariant under conjugation by $G_R=L_R$, since

$G_RG_vG_R^{-1}=G_{Rv}$

$L_RL_vL_R^{-1}=L_{Rv}$

are boosts as well. In the case of the Galileo group, the set of (galilean) boost forms an (abelian) subgroup and then, it provides an invariant group. We can calculate the factor group with respect to it and we will obtain an isomorphic group to the subgroup of space rotations. Using the group law for the Galileo group:

$\underbrace{G_{R_1}G_{v_1}}\underbrace{G_{R_2}G_{v_2}}=G_{R_1R_2}G_{R_2^{-1}v_1+v_2}=G_{R_3}G_{v_3}$

with $R_3=R_2R_1$ and $v_3=R_2^{-1}v_1+v_2$. As a consequence, the homogenous Galileo group (without reversals) is called a semidirect product of the rotation group with the Abelian group $\mathbb{R}^3$ of all boosts given by $\mathbf{v}$.

The case of Lorentz group is more complicated/complex. The reason is the Thomas precession. Indeed, the set of boost does NOT form a subgroup of the Lorentz group! We can define a product in this group:

$\boxed{L_{v_1} oL_{v_2}=L_{v_1 o v_2}}$

but, in the contrary to the result we got with the Galileo group, this condition does NOT define a group structure. In fact, mathematicians call objects with this property groupoids. The domain of velocities of the this lorentzian grupoid becomes a groupoid under the multiplication $v_1 o v_2$. It has dramatic consequences. In particular, the associative does not hold for this multiplication and this groupoid structure! Anyway, a weaker form of it is true, involving the Thomas precession/rotation formula:

$\boxed{(v_1 o v_2) o v_3=(R^{-1}(v_2,v_3)v_1) o (v_2 o v_3)}$

In an analogue way, the multiplication is not commuative in general too, but it satisfies a weaker form of commutativity. While in general groupoids require to distinguish between right and left unit elements (if any), we have indeed $\mathbf{v}=\mathbf{0}$ as a “two-sided” unit element for the velocity groupoid. In the same manner, while in general groupoids right and left inverses may differ (if any), in the case of Lorentz group, the groupoid associated to Thomas precession has a unique two-sided inverse $-\mathbf{v}$ for any $\mathbf{v}$ relative to the groupoid multiplication law. It is NON-trivial ( due to non-associativeness), albeit true, that the equation given by

$v_1 o v_2=v_3$

may be solved uniquely for $v_2$ and, provided we plug $v_2, v_3$, it may be solve uniquely for any $v_1$. A groupoid satisfying this property (i.e., a groupoid that allows such a uniqueness in the solutions of its equation) is called quasi-group.

In conclusion, we can say that the Lorentz group IS, in sharp contrast to the Galileo group, in no way a semidirect product, being what mathematicians and physicists call a simple group, i.e., it is a noncommutative group having no nontrivial invariant subgroup! It is due to the fact that the multiplication rule of the Lorentz group without reversals makes it, in the sense of our previous definitions, the quasidirect product of the rotation group (as a subgroup of the automorphism group of the velocity groupoid)  with the so-called “weakly associative groupoid of velocities”. Here, weakly associative(-commutative) groupoid means the following: a groupoid with a left-sided unit and left-sided inverses with the next properties:

1. Weak associativeness: $R(\mathbf{0},\mathbf{v})=R(-\mathbf{v},\mathbf{v})=\mathbf{1}$

2. Loop property (from Thomas precession formula): $R(v_1,v_2)=R(v_1,v_1 o v_2)$

and where the automorphims group of the velocity groupoid is defined with the next equations

Definition (Automorphism group of the velocity groupoid): $(Sv_1)o(Sv_2)=S(v_1 o v_2)$

Note: an associative groupoid is called semigroup and and a semigroup with two-sided unit element is called a monoid.

This algebraic structure hidden in the Lorentz group has been rediscovered several times along the History of mathematical physics. A groupoid satisfying the loop property has been named in other ways. For instance, in 1988, A. A. Ungar derived the above composition laws and the automorphism group of the Thomas precession R. Independently, A. Nesterov and coworkers in the Soviet Union had studied the same problem and quasigroup since 1986. And we can track this structure even more. 20 years before the Ungar “rediscovery”, H. Karzel had postulated a version of the same abstract object, and it was integrated into a richer one with two compositions (laws). He called it “near-domain”, where the automorphims R (Thomas precessions) were to be realized by the (distributive) left multiplication with suitable elements of the near-domian ( the reference is Abh. Math.Sem.Uni. Hamburg, 1968).

However, Ungar himself developed a more systematic treatment and description for the Thomas precession “groupoid” that is behind all this weird non-associative stuff in the Lorentz-group in 3+1 dimensions. Accorging to his new approach and terminology, the structure is called “gyrocommutative gyrogroup” and it includes the Thomas precession as “Thomas gyration” in this framework. If you want to learn more about gyrogroups and gyrovector spaces, read this article

http://en.wikipedia.org/wiki/Gyrovector_space

Some other authors, like Wefelscheid and coworkers, called K-loops to these gyrogroups. Even more, there are two extra sources from this nontrivial mathematical structure.

Firstly, in Japan, M.Kikkawa had studied certain loops with a compatible differentiable structure called “homegeneous symmetric Lie groups” ( Hiroshima Math. J.5, 141 (1975)). Even though he did not discuss any concrete example, it is natural from his definitions that it was the same structure Karzel found. Being romantic, we can observe certain justice to call K-loops to gyrogroups (since Kikkawa and Karzel discovered them first!). The second source can be tracked in time since the same ideas were already known by L.Sabinin et alii circa 1972 ( Sov. Math. Dokl.13,970(1972)). Their relation to symmetric homogeneous spaces of noncompact type has been discussed some years ago by W. Krammer and H.K.Urbatke, e.g., in Res. Math.33, 310 (1998).

Finally, a purely algebraic loop theory approach (with motivations far way from geometry or physics) was introduced by D. A. Robinson in 1966. In 1995, A. Kreuzer showed thath it was indeed identical to K-loops, again adding some extra nomenclature ( Math.Proc.Camb. Phylos.Soc.123, 53 (1998)).

THOMAS PRECESSION: EASY DEDUCTION

We have seen that the composition of 2 Lorentz boosts, generally with 2 non collinear velocities, results in a Lorentz transformation that IS NOT a pure boost but a composition of a single Lorentz transformation or boost and a single spatial rotation. Indeed, this phenomenon is also called Wigner-Thomas rotation. The final consequence, any body moving on a curvilinear trajectory undergoes and experiences a rotational precession, firstly noted by Thomas in the relativistic theory of the spinning electron.

In this final section, I am going to review the really simple deduction of the Thomas precession formula given in the paper http://arxiv.org/abs/1211.1854

Imagine 3 different inertial observers Anna, Bob and Charles and their respective inertial frames A, B, and C attached to them. We choose A as a non-rotated frame with respect to B, and B as a non-rotated reference frame w.r.t. C. However, surprisingly, C is going to be rotated w.r.t. A and it is inevitable! We are going to understand it better. Let Bob embrace Charles and let them move together with constant velocity $\mathbf{v}$ w.r.t. Anna. In some point, Charles decides to run away from Bob with a tiny velocity $\mathbf{dv'}$ w.r.t. Bob. Then, Bob is moving with relative velocity $-\mathbf{dv'}$ w.r.t. C and Anna is moving with relative velocity $-\mathbf{v}$ w.r.t. B. We can show these events with the following diagram:

Now, we can write Charles’ velocity in the Anna’s frame by the sum $\mathbf{v+dv}$. Since the frame C is rotated with respect to the A frame, his velocity in the C frame will be $\hat{\mathbf{v}}$ will be calculated step to step as follows. Firstly, we remark that

$\hat{\mathbf{v}}\neq -\mathbf{v}-d\mathbf{v}$

Secondly, the angle $d\mathbf{\Omega}$ of an infinitesimal rotation is given by:

$d\mathbf{\Omega}=-\dfrac{\hat{\mathbf{v}}}{\vert \hat{\mathbf{v}}\vert }\times \dfrac{\mathbf{v}+d\mathbf{v}}{\vert \mathbf{v}+d\mathbf{v}\vert}\approx -\dfrac{\hat{\mathbf{v}}}{v^2}\times (\mathbf{v}+ d\mathbf{v})\;\;\; (1)$

The precession rate in the A frame will be provided using the general nonlinear composition rule in SR. If the motion is parallel to the x-axis with velocity $V$, we do know that

$u'_x=\dfrac{u_x-V}{1-\dfrac{u_x V}{c^2}}$

$u'_y=\dfrac{u_y\sqrt{1-\dfrac{V^2}{c^2}}}{1-\dfrac{u_x V}{c^2}}$

$u'_z=\dfrac{u_z\sqrt{1-\dfrac{V^2}{c^2}}}{1-\dfrac{u_x V}{c^2}}$

and where $\mathbf{u}=(u_x,u_y,u_z)$ and $\mathbf{u}'=(u'_x,u'_y,u'_z)$ are the velocities of some object in the rest frame and the moving frame, respectively. For an arbitrary non-collinear, non-orthogonal, i.e., non parallel velocity $\mathbf{V}=(V_x,V_y,V_z)$ we obtain the transformations

$\boxed{\mathbf{u}'=\dfrac{\sqrt{1-\dfrac{V^2}{c^2}}\left(\mathbf{u}-\dfrac{\mathbf{u}\cdot\mathbf{V}}{V^2}\mathbf{V}\right)-\left( \mathbf{V}-\dfrac{\mathbf{u}\cdot\mathbf{V}}{V^2}\mathbf{V}\right)}{1-\dfrac{\mathbf{u}\cdot\mathbf{V}}{c^2}}\;\;\; (2)}$

and where the unprimed and primed frames are mutually non-rotated to each other. Using this last equation, (2), we can easily describe the transition from the frame A to the frame B. It involves the substitutions:

$\mathbf{V}\rightarrow \mathbf{v}$

$\mathbf{u}\rightarrow \mathbf{v}+d\mathbf{v}$

$\mathbf{u}'\rightarrow d\mathbf{v}'$

After leaving the first order terms in $d\mathbf{v}$, we can get the following expansion from eq.(2):

$d\mathbf{v}'\approx \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left(d\mathbf{v}-\dfrac{\mathbf{v}\cdot d\mathbf{v}}{v^2}\mathbf{v}\right)+\dfrac{1}{1-\dfrac{v^2}{c^2}}\dfrac{\mathbf{v}\cdot d\mathbf{v}}{v^2}\mathbf{v}\;\;\; (3)$

Using again eq.(2) to make the transition between the B frame to the C frame, i.e., making the substitutions:

$\mathbf{V}\rightarrow d\mathbf{v}'$

$\mathbf{u}\rightarrow -\mathbf{v}$

$\mathbf{u}'\rightarrow \hat{\mathbf{v}}$

and dropping out higher order differentials in $d\mathbf{v}'$, we obtain the next formula after we neglect those terms

$\boxed{\hat{\mathbf{v}}\approx -\mathbf{v}+\dfrac{\mathbf{v}\cdot d\mathbf{v}'}{c^2}\mathbf{v}-d\mathbf{v}'\;\;\; (4)}$

The final step consists is easy: we plug eq.(3) into eq.(4) and the resulting expression into eq.(1). Then, we divice by the differential $dt$ in the final formula to provide the celebrated Thomas precession formula:

$\boxed{\dot{\Omega}=\dfrac{d\Omega}{dt}=\omega_T=-\dfrac{1}{v^2}\left(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1\right)\mathbf{v}\times \dot{\mathbf{v}}\;\;\; (5)}$

or equivalently

$\boxed{\dot{\Omega}=\dfrac{d\Omega}{dt}=\omega_T=-\dfrac{1}{v^2}\left(\gamma_{\mathbf{v}}-1\right)\mathbf{v}\times \mathbf{a}\;\;\; (6)}$

It can easily shown that these formulae is the same as the given previously above, writing $v^2$ in terms of $\gamma$ and performing some elementary algebraic manipulations.

Aren’t you fascinated by how these wonderful mathematical structures emerge from the physical world? I can say it: Fascinating is not enough for my surprised mind!


text:
# AQA A Level Maths: Pure复习笔记7.1.3 Differentiating Powers of x

### Differentiating Powers of x

#### How do I differentiate expressions involving powers of x?

• Powers of x are differentiated according to the following formulae:

• Differentiating terms with positive integer powers of x is straightforward

• But negative and fractional powers of x can also be differentiated this way

• Don't forget these two special cases:

• These allow you to differentiate constants and linear terms in x

• If the expression involves a sum or difference of terms just differentiate one term at a time

#### Exam Tip

• Take extra care when differentiating negative and fractional powers of x as mishandling negative signs and fractions is a common way to lose marks in these sorts of questions.

format: plain


text:
MathSciNet bibliographic data MR2118865 60K35 (60G17 82B27) Dubédat, Julien ${\rm SLE}(\kappa,\rho)$${\rm SLE}(\kappa,\rho)$ martingales and duality. Ann. Probab. 33 (2005), no. 1, 223–243. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.

 American Mathematical Society 201 Charles Street Providence, RI 02904-6248 USA


text:
G. Armistice Area Apportionment
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

After a drawn-out mooclear arms race, Farmer John and the Mischievous Mess Makers have finally agreed to establish peace. They plan to divide the territory of Bovinia with a line passing through at least two of the n outposts scattered throughout the land. These outposts, remnants of the conflict, are located at the points (x1, y1), (x2, y2), ..., (xn, yn).

In order to find the optimal dividing line, Farmer John and Elsie have plotted a map of Bovinia on the coordinate plane. Farmer John's farm and the Mischievous Mess Makers' base are located at the points P = (a, 0) and Q = ( - a, 0), respectively. Because they seek a lasting peace, Farmer John and Elsie would like to minimize the maximum difference between the distances from any point on the line to P and Q.

Formally, define the difference of a line relative to two points P and Q as the smallest real number d so that for all points X on line , |PX - QX| ≤ d. (It is guaranteed that d exists and is unique.) They wish to find the line passing through two distinct outposts (xi, yi) and (xj, yj) such that the difference of relative to P and Q is minimized.

Input

The first line of the input contains two integers n and a (2 ≤ n ≤ 100 000, 1 ≤ a ≤ 10 000) — the number of outposts and the coordinates of the farm and the base, respectively.

The following n lines describe the locations of the outposts as pairs of integers (xi, yi) (|xi|, |yi| ≤ 10 000). These points are distinct from each other as well as from P and Q.

Output

Print a single real number—the difference of the optimal dividing line. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Examples
Input
2 51 02 1
Output
7.2111025509
Input
3 60 12 50 -3
Output
0.0000000000
Note

In the first sample case, the only possible line is y = x - 1. It can be shown that the point X which maximizes |PX - QX| is (13, 12), with , which is .

In the second sample case, if we pick the points (0, 1) and (0,  - 3), we get as x = 0. Because PX = QX on this line, the minimum possible difference is 0.


text:
# Powershell script to create folders on remote servers

This is pretty much the first Powershell script I've ever done and I am here looking for a quick review of it. Is there something I could have done better? Is the format okay? I don't know much about expected standards, and this was cobbled together as a result of me wanting to learn.

$Company = Read-Host 'Enter Company Name:'$Credentials = Get-Credential

Write-Host 'Thank you! Creating client FTP folder on ServerA.' -ForegroundColor Magenta -BackgroundColor White

Invoke-Command ServerA {mkdir ("e:\ftp\users\" + $args[0] + "\INCOMING")} -ArgumentList$Company -Credential $Credentials Invoke-Command ServerA {mkdir ("e:\ftp\users\" +$args[0] + "\OUTGOING")} -ArgumentList $Company -Credential$Credentials

Write-Host 'Done. Now creating duplicate folder on ServerB.' -ForegroundColor Magenta -BackgroundColor White

Invoke-Command ServerB {mkdir ("e:\ftp\users\" + $args[0] + "\INCOMING")} -ArgumentList$Company -Credential $Credentials Invoke-Command ServerB {mkdir ("e:\ftp\users\" +$args[0] + "\OUTGOING")} -ArgumentList $Company -Credential$Credentials

Write-Host 'Done.' -ForegroundColor Magenta -BackgroundColor White

Whenever we first set up a user for FTP, we have to set up a folder for them. The folder needs to be named after the company, and include "Incoming" and "Outgoing" subfolders. This needs to be created on ServerA, and the exact same thing created on ServerB. This script finds out what the company name is, checks credentials, and then creates the folders in the appropriate places.

Thoughts? Things I could research to improve it?

You're using this string literal in four different places.

"e:\ftp\users\"

I would suggest storing it in a variable. That way, if the path ever needs to change, one change would cascade through all of the places it needs to without any risk of making a mistake in just one of them.

The same could be said of the "\INCOMING" and "\OUTGOING" literals as well.

Congratulations, you managed to get a PS script to run, that's already something!

Jokes aside, you should look into parameterizing the script, so you could invoke it with command-line arguments instead of prompting for them, if that's possible. This would also reduce the maintenance burden on the script, because if the "server A" and "server B" aren't hard-coded in the script, there's nothing to change but how the script is invoked when either server needs to be changed.

Something like this at the start of the script:

param (
$Company = "Contoso Inc."$ServerA = "PathA"
\$ServerB = "PathB"
)

There's no need for Write-Host, if you want to output a string, just output a string:

"Some output"

Will output the string... Some output. Yup. That easy.

In addition to the already great answers it may also be worth your time to modularize the code file so that it's easy to invoke and easy to distribute. that way, instead of having to invoke it as

. ./createcompanyfolders.ps1

you can add it to your modules directory and just type (assuming you named it this way)

New-CompanyFolders

By creating the script as a function in a module, it also makes it easy to add native PowerShell help to your script that describes usage, expectations, etc. All of this is very easy to do without any significant extra work. Remember, documenting your code will help when in 18 months someone has to come along and modify the code, knowing nothing about it. Doubly so because that person is likely to be you!

Here's a great post by Ed Wilson that describes how to modularize an already written PowerShell script and it shows how to include native help as a comment. A little old, but still very relevant.


text:
Skip to Main Content
Chapter 28

After studying this chapter, you should be able to:

• Describe the major functions of the liver with respect to metabolism, detoxification, and excretion of hydrophobic substances.
• Understand the functional anatomy of the liver and the relative arrangements of hepatocytes, cholangiocytes, endothelial cells, and Kupffer cells.
• Define the characteristics of the hepatic circulation and its role in subserving the liver's functions.
• Identify the plasma proteins that are synthesized by the liver.
• Describe the formation of bile, its constituents, and its role in the excretion of cholesterol and bilirubin.
• Outline the mechanisms by which the liver contributes to whole body ammonia homeostasis and the consequences of the failure of these mechanisms, particularly for brain function.
• Identify the mechanisms that permit normal functioning of the gallbladder and the basis of gallstone disease.

The liver is the largest gland in the body. It is essential for life because it conducts a vast array of biochemical and metabolic functions, including ridding the body of substances that would otherwise be injurious if allowed to accumulate, and excreting drug metabolites. It is also the first port of call for most nutrients absorbed across the gut wall, supplies most of the plasma proteins, and synthesizes the bile that optimizes the absorption of fats as well as serving as an excretory fluid. The liver and associated biliary system have therefore evolved an array of structural and physiologic features that underpin this broad range of critical functions.

### Functional Anatomy

An important function of the liver is to serve as a filter between the blood coming from the gastrointestinal tract and the blood in the rest of the body. Blood from the intestines and other viscera reach the liver via the portal vein. This blood percolates in sinusoids between plates of hepatic cells and eventually drains into the hepatic veins, which enter the inferior vena cava. During its passage through the hepatic plates, it is extensively modified chemically. Bile is formed on the other side at each plate. The bile passes to the intestine via the hepatic duct (Figure 28–1).

###### Figure 28–1

Schematic anatomy of the liver. Top: Organization of the liver. CV, central vein. PS, portal space containing branches of bile duct (green), portal vein (blue), and hepatic artery (red). Bottom: Arrangement of plates of liver cells, sinusoids, and bile ducts in a liver lobule, showing centripetal flow of blood in sinusoids to central vein and centrifugal flow of bile in bile canaliculi to bile ducts. (Redrawn and modified from Ham AW: Textbook of Histology, 5th ed. Philadelphia: JB Lippincott Co., 1965.)

In each hepatic lobule, the plates of hepatic cells are usually only one cell thick. Large gaps occur between the endothelial cells, and plasma is in intimate contact with the cells (Figure 28–2). Hepatic artery blood also enters the ...

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text:
# Minimax Expected Measure Confidence Sets for Restricted

December, 1969
Report Number:
617
Authors:
S.N. Evans, B. Hansen and P.B. Stark
Citation:
PDf
Abstract:

We study confidence sets for a parameter $\theta \in \Theta$ that have minimax expected measure among random sets with at least $1-\alpha$ coverage probability. We characterize the minimax sets using duality, which helps to find confidence sets with small expected measure and to bound improvements in expected measure compared with standard confidence sets.

We construct explicit minimax expected length confidence sets for a variety of one-dimensional statistical models, including the bounded normal mean with known and with unknown variance. For the bounded normal mean with unit variance, the minimax expected measure 95% confidence interval has a simple form for $\Theta = [-\tau, \tau]$ with $\tau \le 3.25$. For $\Theta = [-3, 3]$, the maximum expected length of the minimax interval is about 14% less than that of the minimax fixed-length affine confidence interval and about 16% less than that of the truncated conventional interval $[X -1.96, X + 1.96] \cap [-3,3]$.

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Postscript File:


text:
University Calculus: Early Transcendentals (3rd Edition)

$$f^{-1}(x)=b-\sqrt{x+b^2}$$ - Domain: $[-b^2,\infty)$. - Range: $(-\infty, b]$
$$y=f(x)=x^2-2bx\hspace{1cm}b\gt0\hspace{1cm}x\le b$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=x^2-2bx$$ $$y=(x^2-2bx+b^2)-b^2$$ $$y=(x-b)^2-b^2$$ $$(x-b)^2=y+b^2$$ $$|x-b|=\sqrt{y+b^2}$$ As $x\le b$, we take $|x-b|=b-x$ $$b-x=\sqrt{y+b^2}$$ $$x=b-\sqrt{y+b^2}$$ 2) Interchange $x$ and $y$: $$y=b-\sqrt{x+b^2}$$ Therefore, $$f^{-1}(x)=b-\sqrt{x+b^2}$$ - Domain: $x$ is defined where $(x+b^2)\ge0$, or $x\ge -b^2$. So the domain of $f^{-1}$ is $[-b^2,\infty)$. - Range: We have $\sqrt{x+b^2}\ge0$ Therefore, $-\sqrt{x+b^2}\le0$ And $b-\sqrt{x+b^2}\le b$ That means the range of $f^{-1}$ is $(-\infty, b]$ *Check: $$f(f^{-1}(x))=(b-\sqrt{x+b^2})^2-2b(b-\sqrt{x+b^2})=b^2-2b\sqrt{x+b^2}+(x+b^2)-2b^2+2b\sqrt{x+b^2}=2b^2+x-2b^2=x$$ $$f^{-1}(f(x))=b-\sqrt{x^2-2bx+b^2}=b-\sqrt{(x-b)^2}=b-|x-b|$$ We show above that $|x-b|=b-x$: $$f^{-1}(f(x))=b-(b-x)=b-b+x=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$