1. Note that
  
  $\begin{displaymath}\gamma = \frac{1}{(1 -\beta^2)^{1/2}} .\end{displaymath}$
  
  Here $\beta^2$ ranges from zero to one, as does $1 -\beta^2$ and its square root. Thus $\gamma$ is 1 divided by a quantity less than or equal to 1, giving
  
  $\begin{displaymath}\gamma \ge 1 .\end{displaymath}$
2. The proper distance between 2 events is invariant; it's always given by
  
  $\begin{displaymath}\Delta s^2 = (c\Delta t)^2 - \Delta x^2 - \Delta y^2- \Delta z^2 .\end{displaymath}$
  
  In the frame where both events are simultaneous, $\Delta t = 0$ . Thus the distance squared between events in this frame is
  
  $\begin{displaymath}\Delta x^2 + \Delta y^2+\Delta z^2 = - \Delta s^2 .\end{displaymath}$
  
  Since the separation is spacelike, the invariant $\Delta s^2$ is negative. The distance between events in this simultaneous frame is thus the magnitude of the (invariant) proper distance,
  
  $\begin{displaymath}\sqrt{ \Delta x^2 + \Delta y^2+\Delta z^2 } = \vert \Delta s \vert .\end{displaymath}$
3. Again , the invariant proper distance is given by
  
  $\begin{displaymath}\Delta s^2 = (c\Delta t)^2 - \Delta x^2 - \Delta y^2- \Delta z^2 .\end{displaymath}$
  
  In the frame where both events occur at the same place, $\Delta x =\Delta y = \Delta z =0$ . Thus the time between events in this frame is given by
  
  $\begin{displaymath}\Delta s^2 = (c\Delta t)^2 .\end{displaymath}$
  
  Since the separation is timelike, the invariant $\Delta s^2 = c^2 \Delta \tau^2$ is positive. The time between events in this colocated frame is thus the proper time,
  
  $\begin{displaymath} \Delta t = \frac{\Delta s}{c} = \Delta \tau .\end{displaymath}$
From the description of A and B as simultaneous events, with a distance to the right of A, we have

$\begin{displaymath}x^\mu_A = \left( \begin{array}{c} 0\ 0\ 0\ 0 \end{array}\... ...\begin{array}{c} 0\ \Delta s\ 0\ 0 \end{array} \right) ,\end{displaymath}$

in the lab frame, taking to occur at the origin. We now consider how Lorentz boosts by velocity along the x-axis can affect these observed spacetime coordinates:

$\begin{eqnarray*} x^\mu_A \ &\rightarrow &\ \left(\ \begin{array}{cccc} \gamma&-... ..., \Delta s\\ \gamma\, \Delta s\\ 0\\ 0 \end{array}\ \right)\ \ , \end{eqnarray*}$

where and are defined as usual, with .
1. From the above Lorentz transformation, observers traveling with velocity $v\hat{x}$ see event at , event B at $t_B=-\gamma\beta\, \Delta s/c$ . $\gamma$ and are automatically positive, and $\Delta s$ is taken positive in the problem statement. If is positive (the observer travels to the right), then $\beta$ is positive implying negative ; B occurs before A. However, if is negative (the observer travels to the left), then $\beta$ is negative implying positive ; B occurs after A. Thus different observers observe as either before or after .
2. Observers traveling with velocity $v\hat{x}$ see A at the origin, and B at $x = \gamma \Delta s$ , a distance $\gamma \Delta s$ to the right of A. This distance is greater than $\Delta s$ , because $\gamma \ge 1$ .
  For observers travelling in some other direction, a more general Lorentz boost must be considered. This is slightly beyond the scope of this class. One could obtain the boost matrix for an arbitrary direction by a rotational similarity transformation of the boost matrix given. It is easily shown that this boost also places B at least $\Delta s$ to the right of A.
From the description of A and B as collocated events, with a time later than A, we have

$\begin{displaymath}x^\mu_A = \left( \begin{array}{c} 0\ 0\ 0\ 0 \end{array}\... ...{array}{c} c\Delta \tau \ 0 \ 0\ 0 \end{array} \right) ,\end{displaymath}$

in the lab frame, taking to occur at the origin. We now consider how Lorentz boosts by velocity along the x-axis can affect these observed spacetime coordinates:

$\begin{eqnarray*} x^\mu_A \ &\rightarrow &\ \left(\ \begin{array}{cccc} \gamma&-... ...u\\ -\gamma\beta\, c\Delta \tau\\ 0\\ 0 \end{array}\ \right)\ \ ,\end{eqnarray*}$

where and are defined as usual, with .
1. From the above Lorentz transformation, observers traveling with velocity $v\hat{x}$ see event at , event B at $x_B=-\gamma\beta\, c\Delta \tau$ . Again, $\gamma$ , and $\Delta \tau$ are all positive. If is positive (the observer travels to the right), then $\beta$ is positive implying negative ; B occurs to the left of A. However, if is negative (the observer travels to the left), then $\beta$ is negative implying positive ; B occurs to the right of A. Thus different observers observe as either to the right, or to the left, of .
2. Observers traveling with speed see A at , and B at $t = \gamma \Delta \tau$ , a time $\gamma \Delta \tau$ after A. This time is greater than $\Delta \tau$ , because $\gamma \ge 1$ .
1. We defined the column vector $x^\mu$ as
  
  $\begin{displaymath}x^{ \mu} = \left( \begin{array}{c}ct\ x\ y\ z\end{array} \right)\ \end{displaymath}$
  
  and the metric $\eta_{\mu\nu}$ as a tensor whose entries we have written as the matrix
  
  $\begin{displaymath}\eta_{\mu\nu} = \left( \begin{array}{rrrr} 1&&&\ &-1&&\ &&-1&\ &&&-1\end{array} \right)\end{displaymath}$
  
  in row-column order ( $\mu$ indexes rows, $\nu$ indexes columns). We then defined the covector $x_\mu$ as the row vector with components
  
  $\begin{displaymath}x_\mu = \eta_{\mu\nu} x^{\nu} ,\end{displaymath}$
  
  with an implicit sum over $\nu$ . The sum over $\nu$ , for fixed $\mu$ , takes the dot product of row $\mu$ of $\eta_{\mu\nu}$ with the column vector $x^\nu$ . Thus the components of $x_\mu$ are given by matrix multiplication
  
  $\begin{displaymath}\left( \begin{array}{rrrr} 1&&&\ &-1&&\ &&-1&\ &&&-1\end{... ... \begin{array}{c}ct\ -x\ -y\ -z\end{array} \right)\quad . \end{displaymath}$
  
  Since $x_\mu$ is a row vector with these components, we have
  
  $\begin{displaymath}x_\mu = \left(\begin{array}{cccc}ct&-x&-y&-z\end{array} \right) .\end{displaymath}$
  
  We also could have simply done the implicit sum without relying on matrix notation: to find a particular component $x_\mu$ , we must sum $\eta_{\mu\nu} x^{\nu}$ over all values of $\nu$ ; however, since $\eta_{\mu\nu}$ is diagonal, we only have one nonzero contribution, when $\nu = \mu$ , thus
  
  $\begin{displaymath}x_\mu = \left\{ \begin{array}{rl} x^\mu& \mbox{when } \mu = 0\\ - x^\mu& \mbox{when } \mu \ne 0 \end{array} \right. \end{displaymath}$
  
  giving the row vector
  
  $\begin{displaymath}x_\mu = \left(\begin{array}{cccc}ct&-x&-y&-z\end{array} \right) .\end{displaymath}$
2. $\begin{displaymath}x_\mu x^\mu = \left(\begin{array}{cccc}ct&-x&-y&-z\end{arr... ...end{array} \right)\quad = \quad (ct)^2 -x^2 - y^2 -z^2 \quad .\end{displaymath}$
  
  Again we could have simply done the sum
  
  $\begin{displaymath}x_\mu x^\mu = x_0 x^0 + x_1 x^1 + x_2 x^2 + x_3 x^3 = (ct)^2 -x^2 - y^2 -z^2 \quad .\end{displaymath}$
1. Evaluating the sum,
  
  $\begin{eqnarray*} 0 = \partial_\mu j^{ \mu} &=& \partial_0 j^0 + \partial_... ...rac{\partial{\rho}}{\partial{t}} + {\nabla} \cdot \vec{j} .\end{eqnarray*}$
2. We write the charge in a fixed volume as
  
  $\begin{displaymath}Q = \int_V \rho dV , \end{displaymath}$
  
  where the charge density $\rho$ is a function of spacetime, described by the independent variables and $\vec{x}$ . The rate
  
  $\begin{displaymath}\dot{Q} = \int_V \frac{\partial \rho}{\partial t} dV , \end{displaymath}$
  
  at which charge increases within the volume must correspond to charge flowing in along the boundary surface:
  
  $\begin{displaymath}\dot{Q} = - \int_S \vec{j} \cdot \vec{\hat{n}} dA , \end{displaymath}$
  
  where $\vec{\hat{n}}$ is an outward pointing normal vector and the sign comes because we want flow in, not out. Note that $\vec{j}=\rho \vec{v}$ is just the local vector flow of charge. The divergence theorem then gives
  
  $\begin{displaymath}\dot{Q} = - \int_V {\nabla} \cdot \vec{j} dV . \end{displaymath}$
  
  If we insist that this equal our original expression for $\dot{Q}$ for arbitrarily small volumes, we obtain the charge conservation law
  
  $\begin{displaymath}\frac{\partial{\rho}}{\partial{t}} + {\nabla} \cdot \vec{j}= 0 ,\end{displaymath}$
  
  which we have just shown to be Lorentz invariant.
Recall the given tensor :

$\begin{displaymath}F^{\mu\nu} = \left( \begin{array}{rrrr} 0&-E_x&-E_y&-E_z\ E... ...-B_z&B_y\ E_y&B_z&0&-B_x\ E_z&-B_y&B_x&0 \end{array} \right)\end{displaymath}$
1. For $\nu = 0$ , we have the equation
  
  $\begin{displaymath}\partial_\mu F^{\mu 0} = j^0/c .\end{displaymath}$
  
  Plugging in the component values
  
  $\begin{displaymath}\begin{array}{rl} j^0 = &c\rho \\ F^{\mu 0} = &\mbox{column ... ...u = \mbox{a spatial index } i \end{array} \right. \end{array}\end{displaymath}$
  
  gives for the sum over $\mu$
  
  $\begin{displaymath}\sum_{i=1}^3 \partial_i E_i = {\nabla} \cdot \vec{E}= \rho ,\end{displaymath}$
  
  Gauss' law. (Note that because $F^{00}$ vanishes, $\mu = 0$ made no contribution to the sum.)
2. For $\nu = 1$ , we have
  
  $\begin{displaymath}\partial_\mu F^{\mu 1} = j_x/c ,\end{displaymath}$
  
  where is the -component of $\vec{j}$ . Reading off the components $F^{\mu 1}$ (column 1 of the given matrix) and performing the sum over $\mu$ gives
  
  $\begin{eqnarray*} j_x/c &=& \partial_0 (-E_x) + \partial_2 (B_z) + \partial_3... ...tial{E_x}}{\partial{t}} + \partial_y B_z - \partial_z B_y . \end{eqnarray*}$
  
  Rearranging gives the x-component of Ampere's law:
  
  $\begin{displaymath} ( {\nabla} {\times} \vec{B} )_x = \frac{1}{c} \left(... ...frac{\partial{\vec{E}}}{\partial{t}} + \vec{j} \right)_x .\end{displaymath}$
  
  By inspection, the $\nu = 2,3$ terms just give the and components, so altogether the spatial $\nu$ components of the equation are equivalent to Ampere's law:
  
  $\begin{displaymath} {\nabla} {\times} \vec{B}= \frac{1}{c} \left(\ \frac{\partial{\vec{E}}}{\partial{t}} + \vec{j} \right) .\end{displaymath}$
1. 1. $a_\mu b^\mu$ -- Lorentz invariant. This is the inner product of a 4-vector and 4-covector, thus invariant. Also note it has no free indices (only matched and summed ones), and is thus invariant.
  2. $a_\mu b^\nu$ -- Has 1 free lower and 1 free upper index, thus transforms like a (1,1) tensor .
  3. $T_{\mu\nu} v^\nu$ -- Has 1 free lower index, thus transforms like a 4-covector.
  4. $T_{\mu\nu}T^{\mu\nu}$ -- Lorentz invariant. This has no free indices (only matched and summed ones), and is thus invariant.
  5. $\vec{E}$ (electric field) -- mixes with $\vec{B}$ under Lorentz transformations, since $\vec{E}$ and $\vec{B}$ are both elements in the rank (2,0) tensor $F_{\mu\nu}$ .
  6. $\vec{j}$ (electric current) -- mixes with $c\rho$ under Lorentz transformations, since $(c\rho, -\vec{j})$ is the 4-covector $j_\mu$ .
2. 1. $\vec{F}= m\vec{a}$ -- NOT Lorentz covariant. It is not clear how $\vec{F}$ and $\vec{a}$ might be generalized, to be part of an object with definite Lorentz tranformation properties. As stated, Lorentz tranformation properties of this equation are unclear.
  2. $\frac{d^2 x^\mu}{dt^2} = F^\mu$ -- NOT Lorentz covariant. The right hand side transforms like a 4-vector, while the left hand side does not. The left hand side's transformation properties are muddled by the appearance of , which transforms in some complicated way since itself is not Lorentz invariant.
  3. $\frac{d^2 x^\mu}{d\tau^2} = F^\mu$ -- Lorentz covariant. The proper time $\tau$ is Lorentz invariant, so both sides here transform as 4-vectors. This turns out to be the right generalization of $\vec{F}= m\vec{a}$ , with proper identification of .
  4. $\partial _\mu\partial^\mu \phi + m^2 \phi = 0$ -- Lorentz covariant. Both sides are Lorentz invariant (transform as scalars). Here $\phi$ is a scalar (non-transforming) field. The rest mass is invariant, as is the d'Alembertian operator $\partial _\mu\partial^\mu$ , the inner product of a covector and vector operator.
  5. $p_\mu p^\mu = m^2$ -- Lorentz covariant. Again, both sides are Lorentz invariant (transform as scalars).
  6. $F_{\mu\nu} = \partial _\mu A_\nu - \partial _\nu A_\mu$ -- Lorentz covariant. Both sides transform as rank (0,2) tensors, with 2 free lower indices.