## Outline

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

- 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

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

- 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

- 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 11.4where $$ \gamma_v \equiv \left( 1-v^2\right)^{-1/2} $$

- 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 $$ x_{\bar{\alpha}} = {\Lambda^{\beta}}_{\bar{\alpha}} x_{\beta} \quad \mathrm{or} \quad x_{\bar{\alpha}} = {\Lambda_{\bar{\alpha}}}^{\beta} x_{\beta} \, ? $$

- 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: $$ {\Lambda^{\bar{\beta}}}_{\alpha}(\mathbf{v}) {\Lambda^{\nu}}_{\bar{\beta}}(-\mathbf{v}) = {\Lambda^{\bar{\beta}}}_{\alpha}(\mathbf{v}) \left[ {\Lambda^{\bar{\beta}}}_{\nu}(\mathbf{v})\right]^{-1} = {\delta^{\nu}}_{\alpha} $$

- 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 bookthus $$ {\Lambda_{\nu^{\prime}}}^{\mu} {\Lambda^{\sigma^{\prime}}}_{\mu} = {\delta_{\nu^{\prime}}}^{\sigma^{\prime}} $$
- You can obtain the inverse of a Lorentz transformation or the inverse of the metric simply by raising or lowering both the indices

- 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

- 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 \( \vec{U} \cdot \vec{a} =0 \) !

- 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

- 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)