# LOG#091. Group theory(XI).

Today, we are going to talk about the Lie groups $U(n)$ and $SU(n)$, and their respective Lie algebras, generally denoted by $u(n)$ and $su(n)$ by the physics community. In addition to this, we will see some properties of the orthogonal groups in euclidena “signature” and general quadratic metrics due to their importance in special relativity or its higher dimensional analogues.

Let us remember what kind of groups are $U(n)$ and $U(n)$:

1) The unitary group is defined by:

$U(n)\equiv\left\{ U\in M_{n\times n}(\mathbb{C})/UU^+=U^+U=I\right\}$

2) The special unitary group is defined by:

$SU(n)\equiv\left\{ U\in M_{n\times n}(\mathbb{C})/UU^+=U^+U=I,\det (U)=1\right\}$

The group operation is the usual matrix multiplication. The respective algebras are denoted as we said above by $u(n)$ and $su(n)$. Moreover, if you pick an element $U\in U(n)$, there exists an hermitian (antihermitian if you use the mathematician “approach” to Lie algebras/groups instead the convention used mostly by physicists) $n\times n$ matrix $H$ such that:

$U=\exp (iH)$

Some general properties of unitary and special unitary groups are:

1) $U(n)$ and $SU(n)$ are compact Lie groups. As a consequence, they have unitary, finite dimensional and irreducible representations. $U(n)$ and $SU(n)$ are subgroups of $U(m)$ if $m\geq n$.

2) Generators or parameters of unitary and special unitary groups. As we have already seen, the unitary group has $n^2$ parameters (its “dimension”) and it has rank $n-1$ (its number of Casimir operators). The special unitary group has $n^2-1$ free parameters (its dimension) and it has rank $n-1$ (its number of Casimir operators).

3) Lie algebra generators. The unitary group has a Lie algebra generated by the space of $n^2$ dimensional complex matrices. The special unitary group has a Lie algebra generated by the $n^2-1$ dimensional space of hermitian $n\times n$ traceless matrices.

4) Lie algebra structures. Given a basis of generators $L_i$ for the Lie algebra, we define the constants $C_{ijk}$, $f_{ijk}$, $d_{ijk}$ by the following equations:

$\left[L_i,L_m\right]=C_{ijk}L_k=if_{ijk}L_k$

$L_iL_j+L_jL_i=\dfrac{1}{3}\delta_{ij}I+d_{ijk}L_k$

These structure constants $f_{ijk}$ are totally antisymmetric under the exchange of any two indices while the coefficients $d_{ijk}$ are symmetric under those changes. Moreover, we also have:

$d_{ijk}=2\mbox{Tr}(\left\{L_i,L_j\right\}L_k)$

$f_{ijk}=-2i\mbox{Tr}(\left[L_i,L_j\right]L_k)$

Remark(I):   From $U=e^{iH}$, we get $\det U=e^{i\mbox{Tr} (H)}$, and from here we can prove the statement 3) above.

Remark(II): An arbitrary element of $U(n)$ can be expressed as a product of an element of $U(1)$ and an element of $SU(n)$. That is, we can write $U(n)\cong U(1)\cup SU(n)$, where the symbol $\cong$ means “group isomorphism”.

Example 1. The SU(2) group.

In particular, for $n=2$, we get

$SU(2)=\left\{U\in M_{2\times 2})(\mathbb{C})/UU^+=U^+U=I_{2\times 2},\det U=1\right\}$

This is an important group in physics! It appears in many contexts: angular momentum (both classical and quantum), the rotation group, spinors, quantum information theory, spin networks and black holes, the Standard Model, and many other places. So it is important to know it at depth. The number of parameters of SU(2) is equal to 3 and its rank is equal to one (1). As generators of the Lie algebra associated to this Lie group, called su(2), we can choose for free 3 any independent traceless (trace equal to zero) matrices. As a convenient choice, it is usual to select the so called Pauli matrices $\sigma_i$:

$\sigma_1=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$

$\sigma_2=\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}$

$\sigma_3=\begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}$

In general, these matrices satisfy an important number of mathematical relations. The most important are:

$\left\{\sigma_i,\sigma_j\right\}=2\sigma_i\sigma_j+2i\varepsilon_{ijk}\sigma_k$

and

$\sigma_i\sigma_j=i\varepsilon_{ijk}\sigma_k$

The commutators of Pauli matrices are given by:

$\left[\sigma_i,\sigma_j\right]=2if_{ijk}\sigma_k$

$f_{ijk}=\dfrac{1}{2}\varepsilon_{ijk}$ $d_{ijk}=0$

The Casimir operator/matrix related to the Pauli basis is:

$C(\sigma_i)=\sigma_i^2=\sigma_1^2+\sigma_2^2+\sigma_3^2$

This matrix, by Schur’s lemma, has to be a multiple of the identity matrix (it commutes with each one of the 3 generators of the Pauli algebra, as it can be easily proved). Please, note that using the previous Pauli representation of the Pauli algebra we get:

$\displaystyle{C=\sum_i\sigma_i^2=3I}$

Q.E.D.

A similar relation, with different overall prefactor, must be true for ANY other representation of the Lie group algebra su(2). In fact, it can be proved in Quantum Mechanics that this number is “four times” the $j(j+1)$ quantum number associated to the angular momentum and it characterizes completely the representation. The general theory of the representation of the Lie group SU(2) and its algebra su(2) is known in physics as the general theory of the angular momentum!

Example 2. The SU(3) group.

If n=3, the theory of $SU(3)$ is important for Quantum Chromodynamics (QCD) and the quark theory. It is also useful in Grand Unified Theories (GUTs) and flavor physics.

$SU(3)=\left\{U\in M_{3\times 3})(\mathbb{C})/UU^+=U^+U=I_{3\times 3},\det U=1\right\}$

The number of parameters of SU(3) is 8 (recall that there are 8 different massless gluons in QCD) and the rank of the Lie algebra is equal to two, so there are two Casimir operators.

The analogue generators of SU(3), compared with the Pauli matrices, are the so-called Gell-Mann matrices. They are 8 independent traceless matrices. There are some “different” elections in literature, but a standard choice are the following matrices:

$\lambda_1=\begin{pmatrix}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 &0\end{pmatrix}$

$\lambda_2=\begin{pmatrix}0 & -i & 0\\ i & 0 & 0\\ 0 & 0 &0\end{pmatrix}$

$\lambda_3=\begin{pmatrix}1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 &0\end{pmatrix}$

$\lambda_4=\begin{pmatrix}0 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 &0\end{pmatrix}$

$\lambda_5=\begin{pmatrix}0 & 0 & -i\\ 0 & 0 & 0\\ i & 0 &0\end{pmatrix}$

$\lambda_6=\begin{pmatrix}0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 &0\end{pmatrix}$

$\lambda_7=\begin{pmatrix}0 & 0 & 0\\ 0 & 0 & -i\\ 0 & i &0\end{pmatrix}$

$\lambda_8=\dfrac{1}{\sqrt{3}}\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 &-2\end{pmatrix}$

Gell-Mann matrices above satisfy a normalization condition:

$\mbox{Tr}(\lambda_i\lambda_j)=2\delta_{ij}$

where $\delta_{ij}$ is the Kronecker delta in two indices.

The two Casimir operators for Gell-Mann matrices are:

1) $\displaystyle{C_1(\lambda_i)=\sum_{i=1}^8\lambda_i^2}$

This operator is the natural generalization of the previously seen SU(2) Casimir operator.

2) $\displaystyle{C_2(\lambda_i)=\sum_{ijk}d_{ijk}\lambda_i\lambda_j\lambda_k}$

Here, the values of the structure constans $f_{ijk}$ and $d_{ijk}$ for the su(3) Lie algebra can be tabulated in rows as follows:

1) For $ijk=123,147,156,246,257,345,367,458,678$ we have $f_{ijk}=1,\dfrac{1}{2},-\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},-\dfrac{1}{2},\dfrac{\sqrt{3}}{2},\dfrac{\sqrt{3}}{2}$.

2) If

$ijk=118,146,157,228,247,256,338,344,355,366,377,448,558,668,778,888$

then have

$d_{ijk}=\dfrac{1}{\sqrt{3}},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{\sqrt{3}},-\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{\sqrt{3}},\dfrac{1}{2},\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2\sqrt{3}},-\dfrac{1}{2\sqrt{3}},-\dfrac{1}{2\sqrt{3}},-\dfrac{1}{2\sqrt{3}},-\dfrac{1}{\sqrt{3}}$

Example 3. Euclidean groups, orthogonal groups and the Lorentz group in 4D and general $D=s+t$ dimensional analogues.

In our third example, let us remind usual galilean relativity. In a 3D world, physics is the same for every inertial observer (observers moving with constant speed). Moreover, the fundamental invariant of “motion” in 3D space is given by the length:

$L^2=x^2+y^2+z^2=\delta_{ij}x^ix^j$ $\forall i,j=1,2,3$

In fact, with tensor notation, the above “euclidean” space can be generalized to any space dimension. For a ND space, the fundamental invariant reads:

$\displaystyle{L_N^2=\sum_{i=1}^NX_i^2=x_1^2+x_2^2+\cdots+x_N^2}$

Mathematically speaking, the group leaving the above metrics invariant are, respectively, SO(3) and SO(N). They are Lie groups with dimensions $3$ and $N(N-1)/2$, respectively and their Lie algebra generators are antisymmetric traceless $3\times 3$ and $N\times N$ matrices. Those metrics are special cases of quadratic forms and it can easily proved that orthogonal transformations with metric $\delta_{ij}$ (the euclidean metric given by a Kronecker delta tensor) are invariant in the following sense:

$A^\mu_{\;\;\; i}\delta_{\mu\nu}A^\nu_{\;\;\; j}=\delta_{ij}$

or equivalently

$A\delta A^T=\delta$

using matric notation. In special relativity, the (proper) Lorentz group $L$ is composed by every real $4\times 4$ matrix $\Lambda^\mu_{\;\;\;\nu}$ connected to the identity through infinitesimal transformations, and the Lorentz group leaves invariant the Minkovski metric(we use natural units with $c=1$):

$s^2=X^2+Y^2+Z^2-T^2$ if you use the “mostly plus” 3+1 metric ($\eta=\mbox{diag}(1,1,1,-1)$) or, equivalentaly

$s^2=T^2-X^2-Y^2-Z^2$ if with a “mostly minus” 1+3 metric ($\eta=\mbox{diag}(1,-1,-1,-1)$).

These equations can be also genearlized to arbitrary signature. Suppose there are s-spacelike dimensions and t-time-like dimensions ($D=s+t$). The associated non-degenerated quadratic form is:

$\displaystyle{s^2_D=\sum_{i=1}^sX_i^2-\sum_{j=1}^tX_j^2}$

In matrix notation, the orthogonal rotations leaving the above quadratic metrics are said to belong to the group $SO(3,1)$ (or $SO(1,3)$ is you use the mostly minus convention) real orthogonal group over the corresponding quadratic form. The signature of the quadratic form is said to be $S=2$ or $(3,1)$ (equivalently $\Sigma=3-1=2$ and $(1,3)$ with the alternative convention). We are not considering “degenerated” quadratic forms for simplicity of this example. The Lorentzian or Minkovskian metric are invariant in the same sense that the euclidean example before:

$L^\mu_{\;\;\;\alpha}\eta_{\alpha\beta}L^\mu_{\;\;\;\beta}=\eta_{\alpha\beta}$

$LGL^T=G$

The group $SO(s,t)$ has signature $(s,t)$ or $s-t$ or $s+t$ in non-degenerated quadratic spaces. Obviously, the rotation group $SO(3)$ is a subgroup of $SO(3,1)$ and more generally $SO(s)$ is a subgroup of $SO(s,t)$. We are going to focus now a little bit on the common special relativity group $SO(3,1)$. This group have 6 parameters or equivalently its group dimension is 6. The rank of this special relativity group is equal to 1. We can choose as parameters for the $SO(3,1)$ group 3 spatial rotation angles $\omega_i$ and three additional parameters, that we do know as rapidities $\xi_i$. These group generators have Lie algebra generators $S_i$ and $K_i$ or more precisely, if we define the Lorentz boosts as

$\xi=\dfrac{\beta}{\parallel\beta\parallel}\tanh^{-1}\parallel \beta\parallel$

In the case of $SO(3,1)$, a possible basis for the Lie algebra generators are the next set of matrices:

$iS_1=\begin{pmatrix}0 & 0 & 0& 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\end{pmatrix}$

$iS_2=\begin{pmatrix}0 & 0 & 0& 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\end{pmatrix}$

$iS_3=\begin{pmatrix}0 & 0 & 0& 0\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$

$iK_1=\begin{pmatrix}0 & 1 & 0& 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$

$iK_2=\begin{pmatrix}0 & 0 & 1& 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$

$iK_3=\begin{pmatrix}0 & 0 & 0& 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\end{pmatrix}$

And the commutation rules for these matrices are given by the basic relations:

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

$\left[K_a,K_b\right]=-i\varepsilon_{abc}S_c$

$\left[S_a,K_b\right]=i\varepsilon_{abc}K_c$

Final remark: $SO(s,t)$ are sometimes called isometry groups since they define isometries over quadratic forms, that is, they define transformations leaving invariant the “spacetime length”.

# LOG#039. Relativity: Examples(III).

Example 1. Absorption of a photon by an atom.

In this process, we have from momenergy conservation:

$P^\mu_a P_{a\mu}+2P^\mu_a P_{b\mu}+P^\mu_b P_{b\mu}=P^\mu_c P_{c\mu}$

If the atom is the rest frame, before absorption we get

$P^\mu_a =(m_a c,0,0,0)$

$P^\mu_b=(\dfrac{E_b}{c},p_{bx},0,0)=(\dfrac{E_b}{c},\dfrac{E_b}{c},0,0)=(\dfrac{h\nu}{c},\dfrac{h\nu}{c},0,0)$

Description: an atom “a” at rest with mass $m_a$ absorbs a photon “b” propagating in the x-direction turning itself into an excited atom “c”, moving in the x-axis (suffering a “recoil” after the photon hits it).

The atom after aborption has

$P^\mu_c=(m_c c,0,0,0)$

Therefore, since the photon verifies:

$P^\mu_b P_{b\mu}=\left(\dfrac{h\nu}{c}\right)^2-\left(\dfrac{h\nu}{c}\right)^2=0$

and it is true in every inertial frame. Thus,

$(m_a c)^2+2m_a c\dfrac{h\nu}{c}+0=(m_a c)^2$

Then,

$\boxed{m_c=\sqrt{m_a^2+2m_a\dfrac{h\nu}{c}}=m_a\sqrt{1+2\dfrac{h\nu}{m_a c^2}}}$

Expanding the square root

$m_c\approx m_a\left[ 1+\dfrac{h\nu}{m_a c^2}-\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c}\right)^2+\mathcal{O}(h^3)\right]$

In this case,

$m_c\approx m_a+\dfrac{h\nu}{ c^2}-\dfrac{1}{2}\left(\dfrac{h\nu}{m_a c}\right)^2$

Atom’s rest mass increases by an amount $\dfrac{h\nu}{c^2}$ up to first order in the Planck’s constant, and it decreases up to second order in h a quanity $-\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c^2}\right)^2$ due to motion ( “recoil”). Therefore,

$\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c^2}\right)^2=\dfrac{1}{2}m_a\left( \dfrac{u_a}{c}\right)^2$

In this way, identifying terms: $u_c=\dfrac{h\nu}{m_a c}$

In the laboratory frame, the excited atom velocity is calculated by momenergy conservation. It is simple:

$\dfrac{h\nu}{c}=\gamma_c m_c u_c$

$m_a c^2+h\nu=\gamma_c m_c c^2$

$\dfrac{h\nu}{u_c c}=m_a+\dfrac{h\nu}{c^2}$

Then, we obtain that:

$u_c=\dfrac{h\nu c}{m_a c^2+h\nu}=\dfrac{c}{\left(\dfrac{m_a c^2}{h\nu}\right)+1}\approx \dfrac{h\nu}{m_a}$

where we have used in the last step $m_ac^2>>h\nu$.

Example 2. Emission of a photon by an atom.

An atom c at rest, with $m_c$ the rest mass, emits a photon b with frequency $\dfrac{E_b}{h}=\nu$ in the x-direction, turning itself into a non-excited atom “a”, with $m_a$. What is the energy shift $\Delta E=E_c-E_a$?

$P^\mu_a P_{a\mu}+P^\mu_b P_{b\mu}=P^\mu_c P_{c\mu}$

where

$P^\mu_a$ is $P^\mu_a =(m_a,0,0,0)$ in the rest frame of “a”.

$P^\mu_a$ in the rest from “c” reads $P^\mu_a =\left(\gamma_a m_a c, -\dfrac{E_b}{c},0,0\right)$

$P^\mu_b$ in the rest frame of “c” is $P^\mu_b =\left(\dfrac{E_b}{c},\dfrac{E_b}{c},0,0\right)$

$P^\mu_a$ in the rest frame of “c” is $P^\mu_c=(m_c c,0,0,0)$

In this way, we have

$(m_a c)^2+\gamma _a m_a E_b +\left( \dfrac{E_b}{c}\right)^2=\gamma _am_am_c c^2$

$\gamma_a m_a c^2+E_b=m_c c^2$

$m_a^2c^2+\dfrac{E_b}{c^2}\left(m_c c^2-E_b\right)^2=m_c^2c^2-m_c E_b$

$m_a^2c^2+m_cE_b=m^2_c c^2-m_c E_b$

$m_a^2c^2+m_cE_b=m_c^2c^2-m_cE_b$

From this equations we deduce that

$E_b=\dfrac{m_c c^2-m_a^2c^2}{2m_c}=\dfrac{E_c^2-E_a^2}{2E_c}=(E_c-E_a)\left(1-\dfrac{E_c-E_a}{2E_c}\right)$

And from the definition $E_c-E_a$ we get $E_b=\Delta E\left(1-\dfrac{\Delta E}{2E_c}\right)$

Note: the photon’s energy IS NOT equal to the difference of the atomic rest energies but it is less than that due to the emission process. This fact implies that the atom experieces “recoil”, and it gains kinetic energy at the expense of the photon. There is a good chance for the photon not to be absorbed by an atom of some kind. However, “resonance absorption” becomes problematic. The condition for recoilless resonant absorption to occur nonetheless, e.g., the reabsorption of gamma ray photons by nuclei of the some kind were investigated by Mössbauer. The so-called Mössbauer effect has been important not only to atomic physics but also to verify the theory of general relativity. Furthermore, it is used in materials reseach in present time as well. In 1958, Rudolf L. Mössbauer reported the 1st reoilless gamma emission. It provided him the Nobel Prize in 1961.

Example 3. Decay of two particles at rest.

The process we are going to study is the reaction $C\rightarrow AB$

The particle C decays into A and B. It is the inverse process of the completely inelastic collision we studied in a previous example.

From the conservation of the tetramomentum

$(m_A c)^2+2\gamma_A\gamma_Bm_Am_B(c^2-u_Au_B)+(m_Bc)^2=(m_C c)^2$

Choose the frame in which the following equation holds

$P^\mu_AP_{A\mu}+P^\mu_BP_{A\mu}=P^\mu_{C}P_{A\mu}$

Let $u_A,u_B$ be the laboratory frame velocities in the rest frame of “C”. Then, we deduce

$P^\mu_A=(\gamma_Am_Ac,\gamma_Am_Au_A,0,0)$

$P^\mu_B=(\gamma_Bm_Bc,\gamma_Bm_Bu_B,0,0)$

$P^\mu_C=(m_C c,0,0,0)$

From these equations $(m_Ac)^2+\gamma_A\gamma_Bm_Am_B(c^2-u_Au_B)=\gamma_Am_Am_C c^2$

$-(m_Ac)^2+(m_Bc)^2=(m_Cc)^2-2\gamma_Am_Am_Cc^2$

$2E_Am_C=(m_A^2+m_C^2-m_B^2)c^2$

$\boxed{E_A=\dfrac{(m_A^2+m_C^2-m_B^2)c^2}{2m_C}}$

$\boxed{E_A^{kin}=T_A=\dfrac{(m_A^2+m_C^2-m_B^2)c^2}{2m_C}-m_Ac^2=\dfrac{\left((m_C-m_A)^2-m_B^2\right)c^2}{2m_C}}$

$\boxed{E_B^{kin}=T_B=\dfrac{\left((m_C-m_B)^2-m_A^2)\right)c^2}{2m_C}}$

$\boxed{E_{kin}=T=T_A+T_B=(m_C-m_A-m_B)c^2}$

Therefore, the total kinetic energy of the two particles A,B is equal to the mass defect in the decay of the particle.

Example 4. Pair production by a photon.

Suppose the reaction $\gamma \rightarrow e^+e^-$, in which a single photon ($\gamma$)decays into a positron-electron  pair.

That is, $h\nu\rightarrow e^+e^-$.

$P^\mu_a=P^\mu_c+P^\mu_d$

Squaring the momenergy in both sides:

$P^\mu_a P_{\mu a}=P^\mu_c P_{\mu c}+2P^\mu_c P_{\mu d}+P^\mu_d P_{\mu d}$

In the case of the photon: $P^\mu_a P_{a\mu}=0$

In the case of the electron and the positron: $P^\mu_c P_{\mu c}=P^\mu_d P_{\mu d}=-(m_e c)^2=-(mc)^2$

We calculate the components of momenergy in the center of mass frame:

$P^\mu_c=\left( \dfrac{E_c}{c},p_{cx},p_{cy},p_{cz}\right)$

$P^\mu_d=\left( \dfrac{E_d}{c},-p_{dx},-p_{dy},-p_{dz}\right)$

with $E_c=E_d=mc^2$. Therefore,

$2 P^\mu_c P_{\mu d}=-2\left( \dfrac{E_c^2}{c^2}+p_x^2+p_y^2+p_z^2\right)$

so

$-2(mc)^2-2\left( \dfrac{E_c^2}{c^2}+\mathbf{p}^2\right)=0$

$2(mc)^2+2\left( \dfrac{E_c^2}{c^2}+\mathbf{p}^2\right)=0$

This equation has no solutions for any positive solution of the photon energy! It’s logical. In vacuum, it requires the pressence of other particle. For instance $\gamma \gamma \rightarrow e^+e^-$ is the typical process in a “photon collider”. Other alternative is that the photon were “virtual” (e.g., like QED reactions $e^+e^-\rightarrow \gamma^\star\rightarrow e^+e^-$). Suppose, alternatively, the reaction $AB\rightarrow CDE$. Solving this process is hard and tedious, but we can restrict our attention to the special case of three particles $C,D,E$ staying together in a cluster C. In this way, the real process would be instead $AB\rightarrow C$. In the laboratory frame:

$P^\mu_A+P^\mu_B=P^\mu_C$

we get

$P^\mu_A=\left( \dfrac{E_A}{c},\dfrac{E_A}{c},0,0\right)$

$P^\mu_B=\left( Mc,0,0,0\right)$

and in the cluster reference frame

$P^\mu_C=\left((M+2m)c,0,0,0\right)$

Squaring the momenergy:

$0+2E_AM+(Mc)^2=(M+2m)^2c^2$

$2ME_A=4Mmc^2+4m^2c^2$

$E_A=2mc^2+\dfrac{2m^2c^2}{M}$

$\boxed{E_A=2mc^2\left(1+\dfrac{m}{M}\right)}$

In the absence of an extra mass M, i.e., when $M\rightarrow 0$, the energy $E_A$ would be undefined, and it would become unphysical. The larger M is, the smaller is the additional energy requiere for pair production. If M is the electron mass, and $M=m$, the photon’s energy must be twice the size of the rest energy of the pair, four times the rest energy of the photon. It means that we would obtain

$E=4mc^2=2m_{pair}c^2$

and thus $\gamma =4\rightarrow$

$\beta^2=1-\dfrac{1}{16}=\dfrac{15}{16}$

$\beta=\sqrt{\dfrac{15}{16}}=\dfrac{\sqrt{15}}{4}\approx 0.97$

$v=\dfrac{\sqrt{15}}{4}c\approx 0.97c$

In general, if $m\neq M$ we would deduce:

$\boxed{\beta=\sqrt{1-\dfrac{1}{4\left(1+\frac{m}{M}\right)^2}}}$

Example 5. Pair annihilation of an  electron-positron couple.

Now, the reaction is the annihilation of a positron-electron pair into two photons, turning mass completely into (field) energy of light quanta. $e^+e^-\rightarrow \gamma \gamma$ implies the momenergy conservation

$P^\mu_a+P^\mu_b=P^\mu_c+P^\mu_c$

where “a” is the moving electron and “b” is a postron at rest. Squaring the identity, it yields

$P^\mu_aP_{\mu a}+2P^\mu_aP_{\mu b}+P^\mu_bP_{\mu b}=P^\mu_cP_{\mu c}+2P^\mu_cP_{\mu c}+P^\mu_dP_{\mu d}$

Then, we deduce

$P^\mu_a=\left(\dfrac{E_a}{c},p_{ax},0,0\right)$

$P^\mu_b=\left( m_e c,0,0,0\right)$

while we do know that

$P^\mu_aP_{\mu a}=P^\mu_bP_{\mu b}=-(m_e c)^2$ for the electron/positron and

$P^\mu_cP_{\mu c}=P^\mu dP_{\mu d}=0$ since they are photons. The left hand side is equal to $-2(m_e c)^2+2E_am_e$, and for the momenergy in the right hand side

$P^\mu_c=\left(\dfrac{E_c}{c},p_{cx},0,0\right)$

$P^\mu_d=\left( \dfrac{E_d}{c},p_{dx},0,0\right)$

Combining both sides, we deduce

$(m_a c)^2+E_a m_e=\dfrac{E_cE_d}{c^2}-p_{cx}p_{dx}$

The only solution to the right hand side to be not zero is when we select $p_{cx}=\pm \dfrac{E_c}{c}$ and $p_{dx}=\pm\dfrac{E_d}{c}$ and we plug values with DIFFERENT signs. In that case,

$\boxed{(mc)^2+E_a m=\dfrac{2E_cE_d}{c^2}}$

From previous examples:

$P^\mu_aP_{\mu_b}+P^\mu_bP_{\mu b}=P^\mu_c P_{b\mu}+P^\mu_aP_{\mu b}$

and we evaluate it in the laboratory frame to give

$\boxed{E_a m+(mc)^2=E_c m+E_d m}$

The last two boxed equations allow us to solve for $E_d$

$\boxed{E_d=E_a-E_c+mc^2}$

If we insert this equation into the first boxed equation of this example, we deduce that

$(mc)^2+mE_a=\dfrac{2E_c}{c^2}\left( E_a-E_c+mc^2\right)$

or

$\dfrac{1}{2}mc^2\left(mc^2+E_a\right)=-E_c^2+E_c(E_a+mc^2)$

Solving for $E_c$ this last equation

$\boxed{E_c^{1,2}=E_d^{1,2}=\dfrac{1}{2}\left(E_a+mc^2\right)\pm\sqrt{\dfrac{1}{4}\left(E_a+mc^2\right)^2-\dfrac{1}{2}mc^2\left(mc^2+E_a\right)}}$

$\boxed{E_c^{1,2}=E_d^{1,2}=\dfrac{\left(E_a+mc^2\right)\pm \sqrt{\left(E_a+mc^2\right)\left(E_a-mc^2\right)}}{2}}$

# LOG#037. Relativity: Examples(I)

Problem 1. In the S-frame, 2 events are happening simultaneously at 3 lyrs of distance. In the S’-frame those events happen at 3.5 lyrs. Answer to the following questions: i) What is the relative speed between frames? ii) What is the temporal distance of events in the S’-frame?

Solution. i) $x'=\gamma (x-\beta c t)$

$x'_2-x'_1=\gamma ((x_2-x_1)-\beta c (t_2-t_1))$

And by simultaneity, $t_2=t_1$

Then $\gamma=\dfrac{x'_2-x'_1}{x_2-x_1}=\dfrac{7}{6}$

$\beta=\sqrt{1-\gamma^{-2}}\approx 0.5$

ii) $ct'=\gamma (ct-\beta x)$

$c(t'_2-t'_1)=-\gamma \beta (x_2-x_1)$

since we have simultaneity implies $t_2-t_1=0$. Then,

$c\Delta t'\approx -1.8 lyrs$

Problem 2. In S-frame 2 events occur at the same point separated by a temporal distance of 3yrs. In the S’-frame, $D'=3.5yrs$ is their spatial separation. Answer the next questions: i) What is the relative velocity between the two frames? ii) What is the spatial separation of events in the S’-frame?

Solution. i) $ct'=\gamma (ct-\beta x)$ with $x_1=x_2$

As the events occur in the same point $x_2=x_1$

$c(t'_2-t'_1)=\gamma c (t_2-t_1)$

$\gamma=\dfrac{t'_2-t'_1}{t_2-t_1}=\dfrac{7}{6}$

$\beta=\sqrt{1-\gamma^{-1}}\approx 0.5$

ii) $x'=\gamma (x-\beta c t)$

$x_1=x_2$ implies $x'_2-x'_1=-\gamma \beta c (t_2-t_1)\approx -1.8 lyrs$

Therefore, the second event happens 1.8 lyrs to the “left” of the first event. It’s logical: the S’-frame is moving with relative speed $v\approx c/2$ for $3.5 yrs$.

Problem 3. Two events in the S-frame have the following coordinates in spacetime: $P_1(x_0=ct_1,x_1=x_0)$, i.e., $E_1(ct_1=x_0,x_1=x_0)$ and $P_2(ct_2=0.5x_0, x_2=2x_0)$, i.e., $E_2(ct_2=x_0/2,x_2=2x_0)$. The S’-frame moves with velocity v respect to the S-frame. a) What is the magnitude of v if we want that the events $E_1,E_2$ were simultaneous? b) At what tmes t’ do these events occur in the S’-frame?

Solution. a) $ct'=\gamma (ct-\beta x)$

$t'_2-t'_1=0$ and then $0=\gamma (c(t_2-t_1)-\beta (x_2-x_1))$

$\beta =\dfrac{c(t_2-t_1)}{(x_2-x_1)}=-\dfrac{0.5x_0}{x_0}=-0.5$

b) $t'=\gamma ( 1-\beta x/c)=\gamma ( 1-\beta x/c)$

$t'_1=\dfrac{1}{\sqrt{1-(-0.5)^2}} \left(x_0/c+0.5x_0 /c)\right)\approx 1.7x_0/c$

$t'_2=\dfrac{1}{\sqrt{1-(-0.5)^2}} \left(x_0/c+0.5\cdot 2\cdot x_0/c)\right)\approx 1.7x_0/c$

Problem 4. A spaceship is leaving Earth with $\beta =0.8$. When it is $x_0=6.66\cdot 10^{11}m$ away from our planet, Earth transmits a radio signal towards the spaceship. a) How long does the electromagnetic wave travel in the Earth-frame? b) How long does the electromagnetic wave travel in the space-ship frame?

For the spaceship, $ct=x/\beta$ and for the signal $ct=x+ct_0$. From these equations, we get

$late \beta ct=x$ and $ct=x+ct_0$, and it yields $\beta ct =ct-ct_0$ and thus $t=\dfrac{t_0}{1-\beta}$ for the intersection point. But, $\beta=0.8=8/10=4/5$ and $1-\beta=1/5$. Putting this value in the intersection point, we deduce that the intersection point happens at $t_1=5t_0$. Moreover,

$t_1-t_0=4t_0=4\dfrac{x_0}{0.8c}\approx 11100 s=3.08h=3h 5min$

b) We have to perform a Lorentz transformation from $(ct_0,0)$ to $(ct_1,x_1)$, with $t'=t=0$.

$t_0=x_0/v=2775s$ and $t_1=5t_0=13875s$. Then $x_1=vt_1=5vt_0=5x_0=5\cdot 6.66\cdot 10^{11}m=3.33\cdot 10^{12}m$. And thus, we obtain that $\gamma=5/3$. The Lorentz transformation for the two events read

$(t'_2-t'_1)=\gamma (t_1-t_0)=\gamma (t_1-t_0)-\beta/c(x_1-x_0)=3700 s\approx 1.03h=1h1m40s$

Remarks: a) Note that $t_1-t_0$ and $t'_2-t'_1$ differ by 3 instead of $5/3$. This is due to the fact we haven’t got a time interval elapsing at a certain location but we face with a time interval between two different and spatially separated events.

b)The use of the complete Lorentz transformation (boost) mixing space and time is inevitable.

Problem 5. Two charged particles A and B, with the same charge q, move parallel with $\mathbf{v}=(v,0,0)$. They are separated by a distance d. What is the electric force between them?

$E'=\left(0,\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{d^2},0\right)$

$B'=\left(0,0,\gamma \dfrac{\gamma v q}{4\pi\epsilon_0d^2c^2}\right)$

In the S-frame, we obtain the Lorentz force:

$\mathbf{F}=q\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)=\left(0,\gamma k_C\dfrac{q^2}{d^2}-\gamma \beta^2\dfrac{q^2}{d^2},0\right)=\left(0,\dfrac{k_Cq^2}{\gamma d^2},0\right)$

The same result can be obtained using the power-force (or forpower) tetravector performing an inverse Lorentz transformation.

Problem 6. Calculate the electric and magnetic field for a point particle passing some concrete point.

The electric field for the static charge is: $E=k_C\dfrac{q}{x'^2+y'^2+z'^2}=k_C\dfrac{q}{r'^2}$

with $\mathbf{v}=(v,0,0)$ when the temporal origin coincides, i.e., at the time $t'=t=0$.  Suppose now two points that for the rest observer provide:

$P=(0,a,0)$ and $P'=(-vt',a,0)$. For the electric field we get:

$E'=k_C\dfrac{q}{r'^3}(x',y',z')$ and $E'(t')=k_C\dfrac{q}{(\sqrt{(x'^2+y'^2+z'^2})^3}(x',y',z')$

$E'_(t')=k_C\dfrac{q}{(v^2t'^2+a^2)^{3/2}}(-vt',a,0)$

Then, $E'_p\rightarrow E_p$ implies that $t'=\gamma t=\gamma (t-\dfrac{vx}{c^2})\vert_{x=0}$

$E'_p(t)=k_C\dfrac{q}{(\gamma^2v^2t^2+a^2)^{3/2}}(-\gamma v t,a,0)$

$B'=B'_p(t')=(0,0,0)=B'_p(t)$

$E_p(t)=(E'_{p_x}(t),\gamma E'_{p_y}(t),0)=k_C\dfrac{q}{\gamma^2 v^2t^2+a^2}(-\gamma v t,\gamma a,0)$

$B_p(t)=(B_{p_x},B_{p_y},B_{p_z})=(0,0,\gamma \dfrac{\gamma v E'_p(t)}{c^2})$

$B_p(t)=\dfrac{q}{\gamma^2v^2t^2+a^2}(0,0,\gamma \dfrac{v}{c^2}a)=(0,0,\dfrac{v}{c^2}E_{p_y}(t))$

There are two special cases from the physical viewpoint in the observed electric fields:

a) When P is directly above the charge q. Then $E_p(t=0)=(0,k_C\gamma \dfrac{q}{a^2},0)$

b) When P is directly in front of ( or behind) q. Then, for a=0, $E_p(t)=(-k_C\dfrac{vt}{\gamma^2(v^2t^2)^{3/2}},0,0)$

Note that we have $\dfrac{vt}{(v^2t^2)^{3/2}}\neq \dfrac{1}{v^2t^2}$ if $t<0$.