Title: ETH_Sem5 / AdvQC / 04_Einsteinian_Mech

Q: What is the relativity principle of Einstein? Why is it necessary?
A: 1. All inertial frames are equivalent.
2. Maxwell's equations have the same form in all inertial frames.
It has been verified in numerous experiments that all observers in all inertial frames, independently of their state of motion, always measure the same speed of light. However, Galilei transformations permit velocities larger than the speed of light in vacuum \(c\).
Another formal deficiency of Newtonian mechanics and therefore of Galilei transformations is the sheer existence of Maxwell's equations, which contain the speed of light \(c\) as a parameter.

Q: What is the space-time interval?
A: The four-dimensional distance or space-time interval \(\spacetime\) between any two events is defined as:
\[\spacetimesq=c^2(t_2-t_1)^2-(\pos_2-\pos_1)^2=c^2(t_2-t_1)^2-\pos_{12}^2\]
For two events connected by a light signal or any 2 events, if we assume homogeneity and isotropy:
\[\spacetimesq = \spacetime^{\prime2} = 0 \qquad\forall\ \mathrm{IS}\strich\]
Furthermore, one calls
\[\spacetimesq = \left\{ \begin{array}{ll} >0 & \text{time-like} \\ =0 & \text{light-like} \\ <0 & \text{space-like} \end{array} \right.\]

Q: What is the Einstein summation convention?
A: All Greek indices occurring twice, once as subscript and once as superscript, are summed from 0 to 3 (for space and time components).
All Latin indices occurring twice are summed from 1 to 3 (space components only).
Example:
\[\begin{aligned} y &= \sum_{i=1}^3 c_ix^i = c_1x^1 + c_2x^2 + c_3x^3 & \\ &= c_ix^i \\ &= \sum_{\mu=0}^3 c_\mu x^\mu = c_0x^0 + c_1x^1 + c_2x^2 + c_3x^3 & \\ &= c_\mu x^\mu \end{aligned}\]
(Mnemonic for co: co is low, \(\coM\)) "If you're co-operating, you're lying low"

Q: What is the metric tensor? What are co- and contravariant space-time 4-vectors?
A: The metric or metric tensor is defined to make the spatial coordinates negative (see space-time interval). Space-time vectors \(x\) equipped with \(g\) is called Minkowski space.
\[g=(\metriccoMV)=(\metriccontraMV)= \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix} \right) = \mathrm{diag}\,(1,-1,-1,-1)\]
contravariant
\[(\contraM)=\begin{pmatrix} x^0 \\ x^1 \\ x^2 \\ x^3\end{pmatrix}=\begin{pmatrix}ct \\ x \\ y \\ z\end{pmatrix}=\begin{pmatrix}ct \\ x^i\end{pmatrix}=\begin{pmatrix}ct \\ \pos\end{pmatrix}\]
covariant
\[(\coM)=\begin{pmatrix}x_0 \\ x_1 \\ x_2 \\ x_3\end{pmatrix}=\begin{pmatrix}ct \\ -x \\ -y \\ -z\end{pmatrix}=\begin{pmatrix}ct \\ x_i\end{pmatrix}=\begin{pmatrix}ct \\ -\pos\end{pmatrix}\]
\[\contraM=\metriccontraMV\coV \means \coM=\metriccoMV\contraV\]
\[\begin{aligned} \spacetimesq &= \sum_{\mu,\nu=0}^3\metriccoMV(\contraM_2-\contraM_1)(\contraV_2-\contraV_1) \\ &= \metriccoMV(\contraM_2-\contraM_1)(\contraV_2-\contraV_1) \\ &= c^2(t_2-t_1)^2-\pos_{12}^2 \\ &\quad\text{(Einstein's summation convention)} \end{aligned}\]

Q: How is the four-dimensional scalar product between any two 4-vectors \(a\) and \(b\) defined?
A: \[a\cdot b = a^Tg\,b = a^\mu b_\mu = a^0b^0-a^ib^i=a^0b^0-\bm{a}\cdot\bm{b}\]
\[\Rightarrow \begin{aligned} \mathrm{d} s^2 &=\mathrm{d} x\cdot \mathrm{d} x \\ &=\mathrm{d} x^T g\, \mathrm{d} x = \metriccoMV\mathrm{d} \contraM\mathrm{d} \contraV = \mathrm{d} \contraM\mathrm{d} \coM \\ &= \mathrm{d} \coM\mathrm{d} \contraM = c^2\mathrm{d} t^2 - \mathrm{d}\pos^2 \end{aligned} \]

Q: What is the Lorentz transformation?
A: Coordinate transformations from \(\mathrm{IS}\) to \(\mathrm{IS}\strich\)
\[x\strich=\Lambda x+b \means x^{\prime\mu}={\Lambda^\mu}_\nu\,\contraV + b^\mu\]
are called Lorentz transformations if they leave the space-time interval \(\spacetimesq\) invariant.
Consequence: Space and time coordinates are no longer independent of each other but may be transformed into each other
\[\begin{aligned} \Lambda &= ({\Lambda^\mu}_\nu) \\ &= \left( \begin{pmatrix} {\Lambda^0}_0 & {\Lambda^0}_1 & {\Lambda^0}_2 & {\Lambda^0}_3 \\ {\Lambda^1}_0 & {\Lambda^1}_1 & {\Lambda^1}_2 & {\Lambda^1}_3 \\ {\Lambda^2}_0 & {\Lambda^2}_1 & {\Lambda^2}_2 & {\Lambda^2}_3 \\ {\Lambda^3}_0 & {\Lambda^3}_1 & {\Lambda^3}_2 & {\Lambda^3}_3 \end{pmatrix} \right) \quad \\ \bar{\Lambda} &= ({\Lambda_\nu}^\mu) \\ &= \left( \begin{pmatrix} \phantom{+}{\Lambda^0}_0 & -{\Lambda^1}_0 & -{\Lambda^2}_0 & -{\Lambda^3}_0 \\ -{\Lambda^0}_1 & \phantom{+}{\Lambda^1}_1 & \phantom{+}{\Lambda^2}_1 & \phantom{+}{\Lambda^3}_1 \\ -{\Lambda^0}_2 & \phantom{+}{\Lambda^1}_2 & \phantom{+}{\Lambda^2}_2 & \phantom{+}{\Lambda^3}_2 \\ -{\Lambda^0}_3 & \phantom{+}{\Lambda^1}_3 & \phantom{+}{\Lambda^2}_3 & \phantom{+}{\Lambda^3}_3 \\ \end{pmatrix} \right) \\ & \Lambda = \pdv{x^{\prime\mu}}{x^\nu} = \partial_\nu x^{\prime\mu} \\ &\bar{\Lambda} = \pdv{x^\mu}{x^{\prime\nu}}= \partial_\nu^\prime x^{\mu} \end{aligned}\]
With
\[\begin{aligned} \mathrm{d} s^{\prime2} & = \mathrm{d} x^{\prime T}g\,\mathrm{d} x\strich = \mathrm{d} x^{T}\Lambda^Tg\,\Lambda\mathrm{d} x \\ & \overset{!}{=}\mathrm{d} s=\mathrm{d} x^Tg\,\mathrm{d} x \end{aligned}\]
we get the fundamental and most important property of Lorentz transformations which completely specifies these transformations:
\[\Lambda^Tg\,\Lambda = g \means {\Lambda^\alpha}_\mu g_{\alpha\beta}{\Lambda^\beta}_\nu = g_{\mu\nu}\]

Q: Write the d'Alembert operator in terms of the 4-gradient!
A: \[\colorboxed{BrickRed}{\dAlembert = \partial_\mu\partial^\mu = \frac{1}{c^2}\pdv{^2}{t^2}-\Laplace}\]
The minus sign in front of the Laplacian is due to the combination of a covariant with a contravariant vector (4-gradient).
\[\partial_\mu = \pdv{}{\coM} = \left(\frac{1}{c}\pdv{}{t},\nabla\right)\]

Q: Compare the position vector, (inverse) transformations, defining property, distance and metric for the 3D with the 4D case!
A: 3D | Quantity | 4D
\(\displaystyle\pos\) | position vector | \(\displaystyle\contraM=(ct,\pos)\)
\(\displaystyle x_i\strich = \sum_{j=1}^3R_{ij}x_j\) | transformations | \(\displaystyle x^{\prime\mu}={\Lambda^\mu}_\nu x^\nu \,, \quad x^{\prime}_\mu={\Lambda_\mu}^\nu x_\nu\)
\(\displaystyle R^{-1} = R^T\) | inverse transformations | \(\displaystyle\Lambda^{-1} = g\,\Lambda^{T}\,g\)
\(\displaystyle R^TR=1\) | defining property | \(\displaystyle\Lambda^T\,g\,\Lambda = g\)
\(\displaystyle \mathrm{d} \pos^2=\sum_{i,j=1}^3\delta_{ij} \mathrm{d} x_i \mathrm{d} x_j\) | distance | \(\displaystyle \mathrm{d} s^2 = \metriccoMV\mathrm{d} \contraM\mathrm{d}\contraV=c^2\mathrm{d} t^2 - \mathrm{d} \pos^2\)
\(\displaystyle\delta_{ij} = \mathrm{diag}(1,1,1)\) | metric | \(\displaystyle\metriccoMV=\mathrm{diag}(1,-1,-1,-1)\)

Q: What is the Lorentz factor?
A: If one looks at two light signals emitted at \(t=t\strich=0\) from \(x=x\strich=0\) in opposite directions, one can derive (using the relativity principle) the Lorentz factor for the 1D:
\[\colorboxed{BrickRed}{\gamma(v) = \gamma = \frac{1}{\sqrt{(1-\frac{v^2}{c^2})}} = {\Lambda^0}_0}\]
where \(v\) is the relative velocity offset between \(\mathrm{IS}\) and \(\mathrm{IS}\strich\).
For \(v\ll c\) we can expand it into the series:
\[\gamma(v)\approx 1 + \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\frac{v^4}{c^4} + \dots \]

Q: What is the general Lorentz boost? Which are the two consequences emerging from the Lorentz boost?
A: Similar to the 1D case, one show:
\[\colorboxed{BrickRed}{\Lambda(\vel)=\left( \begin{matrix} \gamma & -\gamma v_1/c & -\gamma v_2/c & -\gamma v_3/c \\ -\gamma v_1/c & & & \\ -\gamma v_2/c & & \delta_{ij}+\frac{v_iv_j(\gamma-1)}{v^2} & & \\ -\gamma v_3/c & & \\ \end{matrix} \right)}\]
where \(\vel = \sum_i v_i\bm{e}_i\).
\(\Lambda\) mixes spatial and time coordinates!
The Lorentz boost has two special consequences:
Length Contraction: \[l=\frac{l_0}{\gamma}\] \(l_0\) is the proper length (at rest)
Time Dilation: \[t=\gamma\tau \means \mathrm{d} \tau = \frac{\mathrm{d} t}{\gamma}\] \(\tau\) is the proper time (at rest)

Q: What is the general procedure to check if an equation is Lorentz covariant?
A: Transform each component of the equation from \(\mathrm{IS}\) into \(\mathrm{IS}^\prime\) and check if the equation still has the same form.
This means you transform the components using the properties
\[x_\alpha^\prime = {\Lambda_\alpha}^\nu\,x_\nu \qquad x^{\prime\beta} = {\Lambda^\beta}_\nu\,x^\nu \qquad x^{\beta} = g^{\beta\alpha}\,x_\alpha\]
For example, to check if the TDSE is invariant under Lorentz transformation, we check if Hamiltonian and the wavefunction are covariant by themselves. The wavefunction is
\[\begin{aligned} \Psi^{\prime}\left(\bm{r}^{\prime}, t^{\prime}\right) &= \exp \left[\frac{\imag}{\hbar} p_\beta^{\prime} g^{\beta \alpha} x_\alpha^{\prime}\right] \\ & = \exp \left[\frac{\imag}{\hbar} {\Lambda_\beta}^\mu p_\mu g^{\beta \alpha} {\Lambda_\alpha}^\nu x_\nu\right] \\ & = \exp \left[\frac{\imag}{\hbar} {\Lambda_\beta}^\mu g^{\beta \alpha} {\Lambda_\alpha}^\nu p_\mu x_\nu\right] \\ &= \exp \left[\frac{\imag}{\hbar} g^{\mu \nu} p_\mu x_\nu\right] \\ &= \exp \left[\frac{\imag}{\hbar} p_\mu x^\mu\right] \\ &= \exp \left[\frac{\imag}{\hbar} p_\mu g^{\mu \nu} x_\nu\right] = \Psi(\bm{r}, t) \end{aligned}\]
and thus Lorentz covariant. However, the Hamiltonian is not, since
\[\begin{aligned} \hat{H}^\prime &= \imag \hbar c \partial_0^\prime-\frac{\hbar^2}{2 m} \partial_i^\prime \partial^{\prime i} \\ &= \imag \hbar c \partial_0^\prime-\frac{\hbar^2}{2 m} \partial_i^\prime g^{ij} \partial_j^\prime \\ &= \imag \hbar c {\Lambda_0}^\nu \partial_\nu-\frac{\hbar^2}{2 m} {\Lambda_i}^\rho g^{ij}{\Lambda_j}^\sigma \partial_\rho \partial_\sigma \\ &= \imag \hbar c \underbrace{{\Lambda_0}^\nu \partial_\nu}_{\neq \partial_0} - \frac{\hbar^2}{2 m} \left(\underbrace{\partial_k g^{kl} \partial_l}_{\partial_k\partial^k} + \underbrace{\partial_0 g^{00}\partial_0 + \partial_r g^{r 0}\partial_0 + \partial_0 g^{0 s}\partial_s}_{\neq 0} \right) \\ &\neq \hat{H} \end{aligned}\]

Q: What is the 4-velocity and the 4-momentum?
A: 4-velocity \(\bm{\velcontraM}\)
\[\velcontraM = \dv{\contraM}{\tau} = \gamma \begin{pmatrix}c \\ \vel\end{pmatrix}=\gamma\,(c,\vel)\]
Since proper time \(\tau\) is a Lorentz scalar and space-time \(\contraM\) is a Lorentz vector, the 4-velocity \(\velcontraM\) is also a Lorentz 4-vector.
It features the same four-dimensional length in all frames of reference:
\[u\cdot u = u^T\,g\,u = \velcontraM\velcoM=\gamma^2(c^2-\vel^2)=c^2\]
4-momentum \(\bm{p^\mu}\)
\[p^\mu = m\,\velcontraM = \gamma\,m\,(c,\vel) = \left(\frac{E}{c},\bm{p}\right)\]
with the relativistic 3-momentum
\[\bm{p}=\gamma m\vel = \frac{m\vel}{\sqrt{1-\frac{\vel^2}{c^2}}}\]
One can derive the relativistic energy using symmetry properties (Noether):
\[E=\gamma mc^2 = \frac{mc^2}{\sqrt{1-\frac{\vel^2}{c^2}}}\]
\(m\) will always denote the rest mass, independent of the IS, and therefore a fundamental property of the particle.

Q: What is meant by the relativistic energy and momentum are conserved quantities?
A: If an observer in one specific inertial frame of reference IS sees that \(E\) and/or \(\bm{p}\) are conserved throughout a reaction or process, so does any other observer in another frame of reference \(\mathrm{IS\strich}\) related to IS via a Lorentz transformation.

Q: Write down the energy-momentum dispersion relation! What is the non-relativistic limit of the energy expression?
A: One can square the relativistic energy expression
\[E^2=\gamma^2m^2c^4\]
and rewrite it to arrive at the energy-momentum dispersion relation:
\[\colorboxed{BrickRed}{E^2=c^2\mom^2+m^2c^4}\]
The relativistic energy of a free particle is often split into a rest energy and kinetic energy term:
\[E=\gamma mc^2=E=\underbrace{mc^2}_{E_0} + \underbrace{(\gamma-1)mc^2}_{E_\mathrm{kin}}\]
For \(\abs{\vel}\ll c\) the energy can be expressed using the series expansion of the Lorentz factor:
\[E\approx \underbrace{\vphantom{\frac{1}{2}}mc^2}_{E_0} + \underbrace{\frac{1}{2}mv^2}_{\text{non-relativistic } E_\mathrm{kin}} + \underbrace{\frac{3}{8}m\frac{v^4}{c^2}}_{1^{\mathrm{st}} \text{ relativistic approx.}} + \dots\]

Q: Why is \(p^\mu\) also called the energy-momentum vector?
A: Energy and momentum represent the four components of a 4-vector which transform into each other under Lorentz transformations!
In relativistic theories, a clear distinction between energy and momentum is no longer possible.
This peculiarity of special relativity is often reflected by denoting the 4-vector \(p^\mu\) as the energy-momentum vector in order to emphasize the unity of energy and momentum in relativistic theories. (compare: space-time vector \(x^\mu\) combines space and time coordinates into a single 4-vector.)

Q: Write down the covariant four-dimensional Euler-Lagrange equations! What is the 4-potential?
A: \[\pdv{L}{\contraM}-\dv{}{\tau}\pdv{L}{\velcontraM}=0\qquad\forall\mu=0,1,2,3\]
which leads to the Lagrangian for a free particle to be:
\[L(u) = -mc\sqrt{\velcontraM\velcoM}\]
To get the Lagrangian for a particle in an external electromagnetic field, we have to specify the interaction term. This is done with the 4-potential
\[A^\mu = (\phi,\bm{A})\]

Q: Write down the covariant form of the Lagrangian in external electromagnetic fields!
A: Because the electromagnetic force is linear in the field strengths \(\Efield\) and \(\Bfield\), the Lagrangian also has to be a linear function of the 4-potential \(A^\mu\).
Since the Lagrangian is a Lorentz scalar, the yet undetermined interaction term has to be a Lorentz scalar, too!
Since the Lorentz force \(F_L\) has a velocity-dependent component, the simplest choice for the interaction term is proportional to \(A_\mu\velcontraM\), which motivates:
\[\colorboxed{BrickRed}{L_2(x,u)=-mc\sqrt{\velcontraM u_\mu}-\frac{q}{c}A_\mu(x)\velcontraM}\]
As the 3D formulation of the Lagrangian \(L_1 = L_1(\pos, \vel, t)\) has to yield exactly the same action \(S\) as the explicitly covariant Lagrangian \(L_2 = L_2(x, u)\), we can recover the familiar interaction terms:
\[\begin{aligned}-\frac{q}{c}A_\mu(x)\velcontraM\mathrm{d} \tau &= \left(-q\phi+\frac{q}{c}\bm{A}\cdot\vel\right)\mathrm{d} t \\ \thus L_1(\pos, \vel, t) &= -mc^2\sqrt{1-\frac{v^2}{c^2}} - q\phi(\pos,t)+\frac{q}{c}\vel\cdot\bm{A}(\pos,t) \end{aligned} \]

Q: Write down the covariant continuity equation!
A: \[\colorboxed{BrickRed}{\partial_\mu j^\mu=0 \means \pdv{\varrho}{t}+\dop\bm{j}=0}\]
where \(j^\mu\) is the 4-current
\[j(x)=\left(j^\mu(x)\right)=\left(c\varrho(x),\bm{j}(x)\right)=\left(c\varrho(\pos,t),\bm{j}(\pos,t)\right)\]