Physics 616

• Prof. Andrew W. Steiner
(or Andrew or "Dr. Steiner")
• Office hour: 103 South College, Thursday 11am
• Email: awsteiner@utk.edu
• Homework: Electronically as .pdf
• You may work with each other on the homework, but you must write the solution in your own words

Use down and up arrows to proceed to the next or previous slide.

Outline

• Lorentz transformations
• Correction to Einstein summation convention: sum only when indices are up and down, or down and up

Lorentz Transformations

• Boost of velocity $v$ in x direction $$\bar{t} = \frac{t}{\sqrt{1-v^2}} - \frac{v x}{\sqrt{1-v^2}}$$ $$\bar{x} = \frac{-vt}{\sqrt{1-v^2}} + \frac{x}{\sqrt{1-v^2}}$$ $$\bar{y} = y \quad \mathrm{and} \quad \bar{z} = z$$
• As we saw from space-time diagrams, this is a rotation (except replacing cos by cosh)
• Two rotations can be combined
• Rotations have inverses
• Rotations with respect to different axes don't commute
• I.e. the rotations form a group

Lorentz Group

• Homogeneous Lorentz group: boosts and rotations (six generators)
• Poincare (inhomogeneous Lorentz group): Lorentz group and translations in space and time (4 more generators)
• (Homogeneous) Lorentz group is separated into four "parts", the proper orthochoronus Lorentz transformations, and those that change P, T, or both
• Lie group, thus generators create elements in the group using $$\Lambda = \exp \left( i \boldsymbol{\omega} \cdot \mathbf{K} \right) \exp \left( -i \boldsymbol{\theta} \cdot \mathbf{J} \right)$$
• Care must be taken because many authors choose a metric such that $\mathrm{Tr}~\eta = -2$ instead of $+2$ as Schutz has done. With our metric: $$J_1 = i \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ \end{array}\right) \quad K_1 = i \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right) \quad \mathrm{ etc...}$$

• Given a particle traveling with speed $\bar{W}=\Delta \bar{x}/\Delta \bar{t}$ measured in an inertial frame ${\cal \bar{O}}$ traveling with velocity $v$ what is the particle's velocity in frame ${\cal O}$? $$W^{\prime} = \frac{\Delta x}{\Delta t} = \frac{W+v}{1 + W v}$$
• $|W^{\prime}| < 1$ if $|W| < 1$ and $|v| < 1$
• If $|v| < 1$ and $|W| \ll 1$, then $$W^{\prime} = W + v$$

• If the velocity of the observer is not in the direction of motion, then in parallel and perpendicular components: $$\begin{eqnarray} u_{\|}=\left(u^{\prime}_{\|}+v\right) \left( 1 +\mathbf{v} \cdot \mathbf{u}^{\prime} \right)^{-1} \\ \mathbf{u}_{\perp}=\gamma_{v}^{-1} \mathbf{u}^{\prime}_{\perp} \left( 1 +\mathbf{v} \cdot \mathbf{u}^{\prime} \right)^{-1} \end{eqnarray}$$
Jackson 11.4
where $$\gamma_v \equiv \left( 1-v^2\right)^{-1/2}$$

Lorentz Transformations of Vectors

• Consider a contravariant vector, $x^{\alpha}$ as measured in a frame ${\cal O}$
• In a frame ${\cal \bar{O}}$, the new contravariant vector is $$x^{\bar{\alpha}} = {\Lambda^{\bar{\alpha}}}_{\beta} x^{\beta}$$
• ${\Lambda^{\bar{\alpha}}}_{\beta}$ is the Lorentz transformation matrix.
• We say that "x transforms like a vector under Lorentz transformations" or "x is a Lorentz vector"
• The interval $\Delta s^2$ is a "Lorentz scalar" in this sense
• What about $$x_{\bar{\alpha}} = {\Lambda^{\beta}}_{\bar{\alpha}} x_{\beta} \quad \mathrm{or} \quad x_{\bar{\alpha}} = {\Lambda_{\bar{\alpha}}}^{\beta} x_{\beta} \, ?$$

Notation, Basis Vectors, and Inversion

• Where it is not immediately obvious, sometimes an arrow is used to denote a (contravariant) vector $$\vec{A} = A^{\alpha} \vec{e}_{\alpha}$$ for basis vectors $$\left( \vec{e}_{\alpha} \right)^{\beta} = {\delta_{\alpha}}^{\beta}$$
• (I find Schutz' notation on this a bit confusing, but this sort of vector gymnastics is common.)
• What is $\vec{e}_0 \cdot \vec{e}_0$ and $\vec{e}_1 \cdot \vec{e}_1$?
• If a Lorentz transformation is just a "boost" with velocity $\mathbf{v}$, then we can invert the transformation: $${\Lambda^{\bar{\beta}}}_{\alpha}(\mathbf{v}) {\Lambda^{\nu}}_{\bar{\beta}}(-\mathbf{v}) = {\Lambda^{\bar{\beta}}}_{\alpha}(\mathbf{v}) \left[ {\Lambda^{\bar{\beta}}}_{\nu}(\mathbf{v})\right]^{-1} = {\delta^{\nu}}_{\alpha}$$

Inverting with Indices

• Lorentz transformations have an inverse $${\Lambda^{\mu}}_{\alpha}(\mathbf{v}) {\Lambda^{\alpha}}_{\beta}(-\mathbf{v}) = {\delta^{\mu}}_{\beta}$$
• Inversion of a Lorentz transformation can be with indices in the "northeast" and "southwest" corners instead of the typical "northwest" and "southeast" corners. Carroll writes $${\left( \Lambda^{-1} \right)^{\nu^{\prime}}}_{\mu} = {\Lambda_{\nu^{\prime}}}^{\mu}$$
See e.g. 1.28 in Carroll's book
thus $${\Lambda_{\nu^{\prime}}}^{\mu} {\Lambda^{\sigma^{\prime}}}_{\mu} = {\delta_{\nu^{\prime}}}^{\sigma^{\prime}}$$
• You can obtain the inverse of a Lorentz transformation or the inverse of the metric simply by raising or lowering both the indices

Four-velocity of a world line

• Define four velocity $\vec{U}$ to be a vector tangent to a particle's world line and with length one unit of time in the particle's frame
• If a particle is accelerating, there is a momentarily comoving reference frame (MCRF)
• In the MCRF, $\vec{U} = \vec{e}_0$ (where the subscript "0" refers to the particle's frame)
• Then the "four-momentum" is $$\vec{p} = m \vec{U} = \left( E,p^{1},p^{2},p^{3} \right)$$
• If a particle with mass $m$ has velocity $\mathbf{v}$ in the x-direction, then $$\begin{eqnarray} p^{0} &=& m\left(1-v^2\right)^{-1/2} \\ p^{1} &=& m|\mathbf{v}|\left(1-v^2\right)^{-1/2}\\ p^{2} &=& p^3=0 \end{eqnarray}$$
• This is a useful quantity because the four-momentum is conserved
• Sometimes written: $\mathbf{p} = \gamma m \mathbf{v}$ and $E = \gamma m c^2$
E.g. Jackson 11.5

Proper time and Four-velocity

• What is $\vec{U} \cdot \vec{U}$ ?
• The proper time is defined by $$\left( d \tau \right)^2 = - d\vec{x} \cdot d \vec{x} = dt^2 - dx^2 -dy^2 -dz^2$$
• One can show that $\vec{U} = d \vec{x}/d \tau$
• The acceleration four-vector is then $$\vec{a} = \frac{d \vec{U}} {d \tau}$$
• And $\vec{U} \cdot \vec{a} =0$ !

Norms and photons

• The norm (or magnitude) of the four-momentum is $-m^2$
• For a particle moving with velocity $\mathbf{v}$ and an observer moving in frame ${\cal \bar{O}}$ with four-velocity $\vec{U}_{\mathrm{obs}}$ $$- \vec{p} \cdot \vec{U}_{\mathrm{obs}} = \bar{E}$$ where $\bar{E}$ is the energy of the particle relative to the observer
• This expression is frame-invariant
• For photons, $d \tau$ is zero, so it has no four-velocity
• Photons have no rest frame

Group Work

• Using Schutz problems in Chapter 2, do problems 5, 10, and 18 (group 1) 6, 14, and 19 (group 2), 8, 15, and 22 (group 3)