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# Posts Tagged ‘metric tensor’

## PHY450H1S. Relativistic Electrodynamics Lecture 11 (Taught by Prof. Erich Poppitz). Unpacking Lorentz force equation. Lorentz transformations of the strength tensor, Lorentz field invariants, Bianchi identity, and first half of Maxwell’s.

Posted by peeterjoot on February 24, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Covering chapter 3 material from the text [1].

Covering lecture notes pp. 74-83: Lorentz transformation of the strength tensor (82) [Tuesday, Feb. 8] [extra reading for the mathematically minded: gauge field, strength tensor, and gauge transformations in differential form language, not to be covered in class (83)]

Covering lecture notes pp. 84-102: Lorentz invariants of the electromagnetic field (84-86); Bianchi identity and the first half of Maxwell’s equations (87-90)

# Chewing on the four vector form of the Lorentz force equation.

After much effort, we arrived at

\begin{aligned}\frac{d{{(m c u_l) }}}{ds} = \frac{e}{c} \left( \partial_l A_i - \partial_i A_l \right) u^i\end{aligned} \hspace{\stretch{1}}(2.1)

or

\begin{aligned}\frac{d{{ p_l }}}{ds} = \frac{e}{c} F_{l i} u^i\end{aligned} \hspace{\stretch{1}}(2.2)

## Elements of the strength tensor

\paragraph{Claim}: there are only 6 independent elements of this matrix (tensor)

\begin{aligned}\begin{bmatrix}0 & . & . & . \\ & 0 & . & . \\ & & 0 & . \\ & & & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)

This is a no-brainer, for we just have to mechanically plug in the elements of the field strength tensor

Recall

\begin{aligned}A^i &= (\phi, \mathbf{A}) \\ A_i &= (\phi, -\mathbf{A})\end{aligned} \hspace{\stretch{1}}(2.4)

\begin{aligned}F_{0\alpha} &= \partial_0 A_\alpha - \partial_\alpha A_0 \\ &= -\partial_0 (\mathbf{A})_\alpha - \partial_\alpha \phi \\ \end{aligned}

\begin{aligned}F_{0\alpha} = E_\alpha\end{aligned} \hspace{\stretch{1}}(2.6)

For the purely spatial index combinations we have

\begin{aligned}F_{\alpha\beta} &= \partial_\alpha A_\beta - \partial_\beta A_\alpha \\ &= -\partial_\alpha (\mathbf{A})_\beta + \partial_\beta (\mathbf{A})_\alpha \\ \end{aligned}

Written out explicitly, these are

\begin{aligned}F_{1 2} &= \partial_2 (\mathbf{A})_1 -\partial_1 (\mathbf{A})_2 \\ F_{2 3} &= \partial_3 (\mathbf{A})_2 -\partial_2 (\mathbf{A})_3 \\ F_{3 1} &= \partial_1 (\mathbf{A})_3 -\partial_3 (\mathbf{A})_1 .\end{aligned} \hspace{\stretch{1}}(2.7)

We can compare this to the elements of $\mathbf{B}$

\begin{aligned}\mathbf{B} = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \partial_1 & \partial_2 & \partial_3 \\ A_x & A_y & A_z\end{vmatrix}\end{aligned} \hspace{\stretch{1}}(2.10)

We see that

\begin{aligned}(\mathbf{B})_z &= \partial_1 A_y - \partial_2 A_x \\ (\mathbf{B})_x &= \partial_2 A_z - \partial_3 A_y \\ (\mathbf{B})_y &= \partial_3 A_x - \partial_1 A_z\end{aligned} \hspace{\stretch{1}}(2.11)

So we have

\begin{aligned}F_{1 2} &= - (\mathbf{B})_3 \\ F_{2 3} &= - (\mathbf{B})_1 \\ F_{3 1} &= - (\mathbf{B})_2.\end{aligned} \hspace{\stretch{1}}(2.14)

These can be summarized as simply

\begin{aligned}F_{\alpha\beta} = - \epsilon_{\alpha\beta\gamma} B_\gamma.\end{aligned} \hspace{\stretch{1}}(2.17)

This provides all the info needed to fill in the matrix above

\begin{aligned}{\left\lVert{ F_{i j} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.18)

## Index raising of rank 2 tensor

To raise indexes we compute

\begin{aligned}F^{i j} = g^{i l} g^{j k} F_{l k}.\end{aligned} \hspace{\stretch{1}}(2.19)

### Justifying the raising operation.

To justify this consider raising one index at a time by applying the metric tensor to our definition of $F_{l k}$. That is

\begin{aligned}g^{a l} F_{l k} &=g^{a l} (\partial_l A_k - \partial_k A_l) \\ &=\partial^a A_k - \partial_k A^a.\end{aligned}

Now apply the metric tensor once more

\begin{aligned}g^{b k} g^{a l} F_{l k} &=g^{b k} (\partial^a A_k - \partial_k A^a) \\ &=\partial^a A^b - \partial^b A^a.\end{aligned}

This is, by definition $F^{a b}$. Since a rank 2 tensor has been defined as an object that transforms like the product of two pairs of coordinates, it makes sense that this particular tensor raises in the same fashion as would a product of two vector coordinates (in this case, it happens to be an antisymmetric product of two vectors, and one of which is an operator, but we have the same idea).

### Consider the components of the raised $F_{i j}$ tensor.

\begin{aligned}F^{0\alpha} &= -F_{0\alpha} \\ F^{\alpha\beta} &= F_{\alpha\beta}.\end{aligned} \hspace{\stretch{1}}(2.20)

\begin{aligned}{\left\lVert{ F^{i j} }\right\rVert} = \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.22)

## Back to chewing on the Lorentz force equation.

\begin{aligned}m c \frac{d{{ u_i }}}{ds} = \frac{e}{c} F_{i j} u^j\end{aligned} \hspace{\stretch{1}}(2.23)

\begin{aligned}u^i &= \gamma \left( 1, \frac{\mathbf{v}}{c} \right) \\ u_i &= \gamma \left( 1, -\frac{\mathbf{v}}{c} \right)\end{aligned} \hspace{\stretch{1}}(2.24)

For the spatial components of the Lorentz force equation we have

\begin{aligned}m c \frac{d{{ u_\alpha }}}{ds} &= \frac{e}{c} F_{\alpha j} u^j \\ &= \frac{e}{c} F_{\alpha 0} u^0+ \frac{e}{c} F_{\alpha \beta} u^\beta \\ &= \frac{e}{c} (-E_{\alpha}) \gamma+ \frac{e}{c} (- \epsilon_{\alpha\beta\gamma} B_\gamma ) \frac{v^\beta}{c} \gamma \end{aligned}

But

\begin{aligned}m c \frac{d{{ u_\alpha }}}{ds} &= -m \frac{d{{(\gamma \mathbf{v}_\alpha)}}}{ds} \\ &= -m \frac{d(\gamma \mathbf{v}_\alpha)}{c \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} dt} \\ &= -\gamma \frac{d(m \gamma \mathbf{v}_\alpha)}{c dt}.\end{aligned}

Canceling the common $-\gamma/c$ terms, and switching to vector notation, we are left with

\begin{aligned}\frac{d( m \gamma \mathbf{v}_\alpha)}{dt} = e \left( E_\alpha + \frac{1}{{c}} (\mathbf{v} \times \mathbf{B})_\alpha \right).\end{aligned} \hspace{\stretch{1}}(2.26)

Now for the energy term. We have

\begin{aligned}m c \frac{d{{u_0}}}{ds} &= \frac{e}{c} F_{0\alpha} u^\alpha \\ &= \frac{e}{c} E_{\alpha} \gamma \frac{v^\alpha}{c} \\ \frac{d{{ m c \gamma }}}{ds} &=\end{aligned}

Putting the final two lines into vector form we have

\begin{aligned}\frac{d{{ (m c^2 \gamma)}}}{dt} = e \mathbf{E} \cdot \mathbf{v},\end{aligned} \hspace{\stretch{1}}(2.27)

or

\begin{aligned}\frac{d{{ \mathcal{E} }}}{dt} = e \mathbf{E} \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.28)

# Transformation of rank two tensors in matrix and index form.

## Transformation of the metric tensor, and some identities.

With

\begin{aligned}\hat{G} = {\left\lVert{ g_{i j} }\right\rVert} = {\left\lVert{ g^{i j} }\right\rVert}\end{aligned} \hspace{\stretch{1}}(3.29)

\paragraph{We claim:}
The rank two tensor $\hat{G}$ transforms in the following sort of sandwich operation, and this leaves it invariant

\begin{aligned}\hat{G} \rightarrow \hat{O} \hat{G} \hat{O}^\text{T} = \hat{G}.\end{aligned} \hspace{\stretch{1}}(3.30)

To demonstrate this let’s consider a transformed vector in coordinate form as follows

\begin{aligned}{x'}^i &= O^{i j} x_j = {O^i}_j x^j \\ {x'}_i &= O_{i j} x^j = {O_i}^j x_j.\end{aligned} \hspace{\stretch{1}}(3.31)

We can thus write the equation in matrix form with

\begin{aligned}X &= {\left\lVert{x^i}\right\rVert} \\ X' &= {\left\lVert{{x'}^i}\right\rVert} \\ \hat{O} &= {\left\lVert{{O^i}_j}\right\rVert} \\ X' &= \hat{O} X\end{aligned} \hspace{\stretch{1}}(3.33)

Our invariant for the vector square, which is required to remain unchanged is

\begin{aligned}{x'}^i {x'}_i &= (O^{i j} x_j)(O_{i k} x^k) \\ &= x^k (O^{i j} O_{i k}) x_j.\end{aligned}

This shows that we have a delta function relationship for the Lorentz transform matrix, when we sum over the first index

\begin{aligned}O^{a i} O_{a j} = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(3.37)

It appears we can put 3.37 into matrix form as

\begin{aligned}\hat{G} \hat{O}^\text{T} \hat{G} \hat{O} = I\end{aligned} \hspace{\stretch{1}}(3.38)

Now, if one considers that the transpose of a rotation is an inverse rotation, and the transpose of a boost leaves it unchanged, the transpose of a general Lorentz transformation, a composition of an arbitrary sequence of boosts and rotations, must also be a Lorentz transformation, and must then also leave the norm unchanged. For the transpose of our Lorentz transformation $\hat{O}$ lets write

\begin{aligned}\hat{P} = \hat{O}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.39)

For the action of this on our position vector let’s write

\begin{aligned}{x''}^i &= P^{i j} x_j = O^{j i} x_j \\ {x''}_i &= P_{i j} x^j = O_{j i} x^j\end{aligned} \hspace{\stretch{1}}(3.40)

so that our norm is

\begin{aligned}{x''}^a {x''}_a &= (O_{k a} x^k)(O^{j a} x_j) \\ &= x^k (O_{k a} O^{j a} ) x_j \\ &= x^j x_j \\ \end{aligned}

We must then also have an identity when summing over the second index

\begin{aligned}{\delta_{k}}^j = O_{k a} O^{j a} \end{aligned} \hspace{\stretch{1}}(3.42)

Armed with these facts on the products of $O_{i j}$ and $O^{i j}$ we can now consider the transformation of the metric tensor.

The rule (definition) supplied to us for the transformation of an arbitrary rank two tensor, is that this transforms as its indexes transform individually. Very much as if it was the product of two coordinate vectors and we transform those coordinates separately. Doing so for the metric tensor we have

\begin{aligned}g^{i j} &\rightarrow {O^i}_k g^{k m} {O^j}_m \\ &= ({O^i}_k g^{k m}) {O^j}_m \\ &= O^{i m} {O^j}_m \\ &= O^{i m} (O_{a m} g^{a j}) \\ &= (O^{i m} O_{a m}) g^{a j}\end{aligned}

However, by 3.42, we have $O_{a m} O^{i m} = {\delta_a}^i$, and we prove that

\begin{aligned}g^{i j} \rightarrow g^{i j}.\end{aligned} \hspace{\stretch{1}}(3.43)

Finally, we wish to put the above transformation in matrix form, look more carefully at the very first line

\begin{aligned}g^{i j}&\rightarrow {O^i}_k g^{k m} {O^j}_m \\ \end{aligned}

which is

\begin{aligned}\hat{G} \rightarrow \hat{O} \hat{G} \hat{O}^\text{T} = \hat{G}\end{aligned} \hspace{\stretch{1}}(3.44)

We see that this particular form of transformation, a sandwich between $\hat{O}$ and $\hat{O}^\text{T}$, leaves the metric tensor invariant.

## Lorentz transformation of the electrodynamic tensor

Having identified a composition of Lorentz transformation matrices, when acting on the metric tensor, leaves it invariant, it is a reasonable question to ask how this form of transformation acts on our electrodynamic tensor $F^{i j}$?

\paragraph{Claim:} A transformation of the following form is required to maintain the norm of the Lorentz force equation

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T} ,\end{aligned} \hspace{\stretch{1}}(3.45)

where $\hat{F} = {\left\lVert{F^{i j}}\right\rVert}$. Observe that our Lorentz force equation can be written exclusively in upper index quantities as

\begin{aligned}m c \frac{d{{u^i}}}{ds} = \frac{e}{c} F^{i j} g_{j l} u^l\end{aligned} \hspace{\stretch{1}}(3.46)

Because we have a vector on one side of the equation, and it transforms by multiplication with by a Lorentz matrix in SO(1,3)

\begin{aligned}\frac{du^i}{ds} \rightarrow \hat{O} \frac{du^i}{ds} \end{aligned} \hspace{\stretch{1}}(3.47)

The LHS of the Lorentz force equation provides us with one invariant

\begin{aligned}(m c)^2 \frac{d{{u^i}}}{ds} \frac{d{{u_i}}}{ds}\end{aligned} \hspace{\stretch{1}}(3.48)

so the RHS must also provide one

\begin{aligned}\frac{e^2}{c^2} F^{i j} g_{j l} u^lF_{i k} g^{k m} u_m=\frac{e^2}{c^2} F^{i j} u_jF_{i k} u^k.\end{aligned} \hspace{\stretch{1}}(3.49)

Let’s look at the RHS in matrix form. Writing

\begin{aligned}U = {\left\lVert{u^i}\right\rVert},\end{aligned} \hspace{\stretch{1}}(3.50)

we can rewrite the Lorentz force equation as

\begin{aligned}m c \dot{U} = \frac{e}{c} \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.51)

In this matrix formalism our invariant 3.49 is

\begin{aligned}\frac{e^2}{c^2} (\hat{F} \hat{G} U)^\text{T} G \hat{F} \hat{G} U=\frac{e^2}{c^2} U^\text{T} \hat{G} \hat{F}^\text{T} G \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.52)

If we compare this to the transformed Lorentz force equation we have

\begin{aligned}m c \hat{O} \dot{U} = \frac{e}{c} \hat{F'} \hat{G} \hat{O} U.\end{aligned} \hspace{\stretch{1}}(3.53)

Our invariant for the transformed equation is

\begin{aligned}\frac{e^2}{c^2} (\hat{F'} \hat{G} \hat{O} U)^\text{T} G \hat{F'} \hat{G} \hat{O} U&=\frac{e^2}{c^2} U^\text{T} \hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} G \hat{F'} \hat{G} \hat{O} U \\ \end{aligned}

Thus the transformed electrodynamic tensor $\hat{F}'$ must satisfy the identity

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} G \hat{F'} \hat{G} \hat{O} = \hat{G} \hat{F}^\text{T} G \hat{F} \hat{G} \end{aligned} \hspace{\stretch{1}}(3.54)

With the substitution $\hat{F}' = \hat{O} \hat{F} \hat{O}^\text{T}$ the LHS is

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} \hat{G} \hat{F'} \hat{G} \hat{O} &= \hat{O}^\text{T} \hat{G} ( \hat{O} \hat{F} \hat{O}^\text{T})^\T \hat{G} (\hat{O} \hat{F} \hat{O}^\text{T}) \hat{G} \hat{O} \\ &= (\hat{O}^\text{T} \hat{G} \hat{O}) \hat{F}^\text{T} (\hat{O}^\text{T} \hat{G} \hat{O}) \hat{F} (\hat{O}^\text{T} \hat{G} \hat{O}) \\ \end{aligned}

We’ve argued that $\hat{P} = \hat{O}^\text{T}$ is also a Lorentz transformation, thus

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{O}&=\hat{P} \hat{G} \hat{O}^\text{T} \\ &=\hat{G}\end{aligned}

This is enough to make both sides of 3.54 match, verifying that this transformation does provide the invariant properties desired.

## Direct computation of the Lorentz transformation of the electrodynamic tensor.

We can construct the transformed field tensor more directly, by simply transforming the coordinates of the four gradient and the four potential directly. That is

\begin{aligned}F^{i j} = \partial^i A^j - \partial^j A^i&\rightarrow {O^i}_a {O^j}_b \left( \partial^a A^b - \partial^b A^a \right) \\ &={O^i}_a F^{a b} {O^j}_b \end{aligned}

By inspection we can see that this can be represented in matrix form as

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.55)

# Four vector invariants

For three vectors $\mathbf{A}$ and $\mathbf{B}$ invariants are

\begin{aligned}\mathbf{A} \cdot \mathbf{B} = A^\alpha B_\alpha\end{aligned} \hspace{\stretch{1}}(4.56)

For four vectors $A^i$ and $B^i$ invariants are

\begin{aligned}A^i B_i = A^i g_{i j} B^j \end{aligned} \hspace{\stretch{1}}(4.57)

For $F_{i j}$ what are the invariants? One invariant is

\begin{aligned}g^{i j} F_{i j} = 0,\end{aligned} \hspace{\stretch{1}}(4.58)

but this isn’t interesting since it is uniformly zero (product of symmetric and antisymmetric).

The two invariants are

\begin{aligned}F_{i j}F^{i j}\end{aligned} \hspace{\stretch{1}}(4.59)

and

\begin{aligned}\epsilon^{i j k l} F_{i j}F_{k l}\end{aligned} \hspace{\stretch{1}}(4.60)

where

\begin{aligned}\epsilon^{i j k l} =\left\{\begin{array}{l l}0 & \quad \mbox{if any two indexes coincide} \\ 1 & \quad \mbox{for even permutations ofi j k l=0123$} \\ -1 & \quad \mbox{for odd permutations of$i j k l=0123} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(4.61)

We can show (homework) that

\begin{aligned}F_{i j}F^{i j} \propto \mathbf{E}^2 - \mathbf{B}^2\end{aligned} \hspace{\stretch{1}}(4.62)

\begin{aligned}\epsilon^{i j k l} F_{i j}F_{k l} \propto \mathbf{E} \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(4.63)

This first invariant serves as the action density for the Maxwell field equations.

There’s some useful properties of these invariants. One is that if the fields are perpendicular in one frame, then will be in any other.

From the first, note that if ${\left\lvert{\mathbf{E}}\right\rvert} > {\left\lvert{\mathbf{B}}\right\rvert}$, the invariant is positive, and must be positive in all frames, or if ${\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}$ in one frame, we can transform to a frame with only $\mathbf{E}'$ component, solve that, and then transform back. Similarly if ${\left\lvert{\mathbf{E}}\right\rvert} < {\left\lvert{\mathbf{B}}\right\rvert}$ in one frame, we can transform to a frame with only $\mathbf{B}'$ component, solve that, and then transform back.

# The first half of Maxwell’s equations.

\paragraph{Claim: } The source free portions of Maxwell’s equations are a consequence of the definition of the field tensor alone.

Given

\begin{aligned}F_{i j} = \partial_i A_j - \partial_j A_i,\end{aligned} \hspace{\stretch{1}}(5.64)

where

\begin{aligned}\partial_i = \frac{\partial {}}{\partial {x^i}}\end{aligned} \hspace{\stretch{1}}(5.65)

This alone implies half of Maxwell’s equations. To show this we consider

\begin{aligned}e^{m k i j} \partial_k F_{i j} = 0.\end{aligned} \hspace{\stretch{1}}(5.66)

This is the Bianchi identity. To demonstrate this identity, we’ll have to swap indexes, employ derivative commutation, and then swap indexes once more

\begin{aligned}e^{m k i j} \partial_k F_{i j} &= e^{m k i j} \partial_k (\partial_i A_j - \partial_j A_i) \\ &= 2 e^{m k i j} \partial_k \partial_i A_j \\ &= 2 e^{m k i j} \frac{1}{{2}} \left( \partial_k \partial_i A_j + \partial_i \partial_k A_j \right) \\ &= e^{m k i j} \partial_k \partial_i A_j e^{m i k j} \partial_k \partial_i A_j \\ &= (e^{m k i j} - e^{m k i j}) \partial_k \partial_i A_j \\ &= 0 \qquad \square\end{aligned}

This is the 4D analogue of

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} f) = 0\end{aligned} \hspace{\stretch{1}}(5.67)

i.e.

\begin{aligned}e^{\alpha\beta\gamma} \partial_\beta \partial_\gamma f = 0\end{aligned} \hspace{\stretch{1}}(5.68)

Let’s do this explicitly, starting with

\begin{aligned}{\left\lVert{ F_{i j} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.69)

For the $m= 0$ case we have

\begin{aligned}\epsilon^{0 k i j} \partial_k F_{i j}&=\epsilon^{\alpha \beta \gamma} \partial_\alpha F_{\beta \gamma} \\ &= \epsilon^{\alpha \beta \gamma} \partial_\alpha (-\epsilon_{\beta \gamma \delta} B_\delta) \\ &= -\epsilon^{\alpha \beta \gamma} \epsilon_{\delta \beta \gamma }\partial_\alpha B_\delta \\ &= - 2 {\delta^\alpha}_\delta \partial_\alpha B_\delta \\ &= - 2 \partial_\alpha B_\alpha \end{aligned}

We must then have

\begin{aligned}\partial_\alpha B_\alpha = 0.\end{aligned} \hspace{\stretch{1}}(5.70)

This is just Gauss’s law for magnetism

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} = 0.\end{aligned} \hspace{\stretch{1}}(5.71)

Let’s do the spatial portion, for which we have three equations, one for each $\alpha$ of

\begin{aligned}e^{\alpha j k l} \partial_j F_{k l}&=e^{\alpha 0 \beta \gamma} \partial_0 F_{\beta \gamma}+e^{\alpha 0 \gamma \beta} \partial_0 F_{\gamma \beta}+e^{\alpha \beta 0 \gamma} \partial_\beta F_{0 \gamma}+e^{\alpha \beta \gamma 0} \partial_\beta F_{\gamma 0}+e^{\alpha \gamma 0 \beta} \partial_\gamma F_{0 \beta}+e^{\alpha \gamma \beta 0} \partial_\gamma F_{\beta 0} \\ &=2 \left( e^{\alpha 0 \beta \gamma} \partial_0 F_{\beta \gamma}+e^{\alpha \beta 0 \gamma} \partial_\beta F_{0 \gamma}+e^{\alpha \gamma 0 \beta} \partial_\gamma F_{0 \beta}\right) \\ &=2 e^{0 \alpha \beta \gamma} \left(-\partial_0 F_{\beta \gamma}+\partial_\beta F_{0 \gamma}- \partial_\gamma F_{0 \beta}\right)\end{aligned}

This implies

\begin{aligned}0 =-\partial_0 F_{\beta \gamma}+\partial_\beta F_{0 \gamma}- \partial_\gamma F_{0 \beta}\end{aligned} \hspace{\stretch{1}}(5.72)

Referring back to the previous expansions of 2.6 and 2.17, we have

\begin{aligned}0 =\partial_0 \epsilon_{\beta\gamma\mu} B_\mu+\partial_\beta E_\gamma- \partial_\gamma E_{\beta},\end{aligned} \hspace{\stretch{1}}(5.73)

or

\begin{aligned}\frac{1}{{c}} \frac{\partial {B_\alpha}}{\partial {t}} + (\boldsymbol{\nabla} \times \mathbf{E})_\alpha = 0.\end{aligned} \hspace{\stretch{1}}(5.74)

These are just the components of the Maxwell-Faraday equation

\begin{aligned}0 = \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} + \boldsymbol{\nabla} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(5.75)

# Appendix. Some additional index gymnastics.

## Transposition of mixed index tensor.

Is the transpose of a mixed index object just a substitution of the free indexes? This wasn’t obvious to me that it would be the case, especially since I’d made an error in some index gymnastics that had me temporarily convinced differently. However, working some examples clears the fog. For example let’s take the transpose of 3.37.

\begin{aligned}{\left\lVert{ {\delta^i}_j }\right\rVert}^\text{T} &= {\left\lVert{ O^{a i} O_{a j} }\right\rVert}^\text{T} \\ &= \left( {\left\lVert{ O^{j i} }\right\rVert} {\left\lVert{ O_{i j} }\right\rVert} \right)^\text{T} \\ &={\left\lVert{ O_{i j} }\right\rVert}^\text{T}{\left\lVert{ O^{j i} }\right\rVert}^\text{T} \\ &={\left\lVert{ O_{j i} }\right\rVert}{\left\lVert{ O^{i j} }\right\rVert} \\ &={\left\lVert{ O_{a i} O^{a j} }\right\rVert} \\ \end{aligned}

If the transpose of a mixed index tensor just swapped the indexes we would have

\begin{aligned}{\left\lVert{ {\delta^i}_j }\right\rVert}^\text{T} = {\left\lVert{ O_{a i} O^{a j} }\right\rVert} \end{aligned} \hspace{\stretch{1}}(6.76)

From this it does appear that all we have to do is switch the indexes and we will write

\begin{aligned}{\delta^j}_i = O_{a i} O^{a j} \end{aligned} \hspace{\stretch{1}}(6.77)

We can consider a more general operation

\begin{aligned}{\left\lVert{{A^i}_j}\right\rVert}^\text{T}&={\left\lVert{ A^{i m} g_{m j} }\right\rVert}^\text{T} \\ &={\left\lVert{ g_{i j} }\right\rVert}^\text{T}{\left\lVert{ A^{i j} }\right\rVert}^\text{T} \\ &={\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ A^{j i} }\right\rVert} \\ &={\left\lVert{ g_{i m} A^{j m} }\right\rVert} \\ &={\left\lVert{ {A^{j}}_i }\right\rVert}\end{aligned}

So we see that we do just have to swap indexes.

## Transposition of lower index tensor.

We’ve saw above that we had

\begin{aligned}{\left\lVert{ {A^{i}}_j }\right\rVert}^\text{T} &= {\left\lVert{ {A_{j}}^i }\right\rVert} \\ {\left\lVert{ {A_{i}}^j }\right\rVert}^\text{T} &= {\left\lVert{ {A^{j}}_i }\right\rVert} \end{aligned} \hspace{\stretch{1}}(6.78)

which followed by careful treatment of the transposition in terms of $A^{i j}$ for which we defined a transpose operation. We assumed as well that

\begin{aligned}{\left\lVert{ A_{i j} }\right\rVert}^\text{T} = {\left\lVert{ A_{j i} }\right\rVert}.\end{aligned} \hspace{\stretch{1}}(6.80)

However, this does not have to be assumed, provided that $g^{i j} = g_{i j}$, and $(AB)^\text{T} = B^\text{T} A^\text{T}$. We see this by expanding this transposition in products of $A^{i j}$ and $\hat{G}$

\begin{aligned}{\left\lVert{ A_{i j} }\right\rVert}^\text{T}&= \left( {\left\lVert{g_{i j}}\right\rVert} {\left\lVert{ A^{i j} }\right\rVert} {\left\lVert{g_{i j}}\right\rVert} \right)^\text{T} \\ &= \left( {\left\lVert{g^{i j}}\right\rVert} {\left\lVert{ A^{i j} }\right\rVert} {\left\lVert{g^{i j}}\right\rVert} \right)^\text{T} \\ &= {\left\lVert{g^{i j}}\right\rVert}^\text{T} {\left\lVert{ A^{i j}}\right\rVert}^\text{T} {\left\lVert{g^{i j}}\right\rVert}^\text{T} \\ &= {\left\lVert{g^{i j}}\right\rVert} {\left\lVert{ A^{j i}}\right\rVert} {\left\lVert{g^{i j}}\right\rVert} \\ &= {\left\lVert{g_{i j}}\right\rVert} {\left\lVert{ A^{i j}}\right\rVert} {\left\lVert{g_{i j}}\right\rVert} \\ &= {\left\lVert{ A_{j i}}\right\rVert} \end{aligned}

It would be worthwhile to go through all of this index manipulation stuff and lay it out in a structured axiomatic form. What is the minimal set of assumptions, and how does all of this generalize to non-diagonal metric tensors (even in Euclidean spaces).

## Translating the index expression of identity from Lorentz products to matrix form

A verification that the matrix expression 3.38, matches the index expression 3.37 as claimed is worthwhile. It would be easy to guess something similar like $\hat{O}^\text{T} \hat{G} \hat{O} \hat{G}$ is instead the matrix representation. That was in fact my first erroneous attempt to form the matrix equivalent, but is the transpose of 3.38. Either way you get an identity, but the indexes didn’t match.

Since we have $g^{i j} = g_{i j}$ which do we pick to do this verification? This appears to be dictated by requirements to match lower and upper indexes on the summed over index. This is probably clearest by example, so let’s expand the products on the LHS explicitly

\begin{aligned}{\left\lVert{ g^{i j} }\right\rVert} {\left\lVert{ {O^{i}}_j }\right\rVert} ^\text{T}{\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ {O^{i}}_j }\right\rVert} &=\left( {\left\lVert{ {O^{i}}_j }\right\rVert} {\left\lVert{ g^{i j} }\right\rVert} \right) ^\text{T}{\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ {O^{i}}_j }\right\rVert} \\ &=\left( {\left\lVert{ {O^{i}}_k g^{k j} }\right\rVert} \right) ^\text{T}{\left\lVert{ g_{i m} {O^{m}}_j }\right\rVert} \\ &={\left\lVert{ O^{i j} }\right\rVert} ^\text{T}{\left\lVert{ O_{i j} }\right\rVert} \\ &={\left\lVert{ O^{j i} }\right\rVert} {\left\lVert{ O_{i j} }\right\rVert} \\ &={\left\lVert{ O^{k i} O_{k j} }\right\rVert} \\ \end{aligned}

This matches the ${\left\lVert{{\delta^i}_j}\right\rVert}$ that we have on the RHS, and all is well.

# References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

## PHY450H1S. Relativistic Electrodynamics Tutorial 1 (TA: Simon Freedman).

Posted by peeterjoot on January 21, 2011

# Worked question.

The TA blasted through a problem from Hartle [1], section 5.17 (all the while apologizing for going so slow). I’m going to have to look these notes over carefully to figure out what on Earth he was doing.

At one point he asked if anybody was completely lost. Nobody said yes, but given the class title, I had the urge to say “No, just relatively lost”.

\paragraph{Q:}
In a source’s rest frame $S$ emits radiation isotropically with a frequency $\omega$ with number flux $f(\text{photons}/\text{cm}^2 s)$. Moves along x’-axis with speed $V$ in an observer frame ($O$). What does the energy flux in $O$ look like?

## A brief intro with four vectors

A 3-vector:

\begin{aligned}\mathbf{a} &= (a_x, a_y, a_z) = (a^1, a^2, a^3) \\ \mathbf{b} &= (b_x, b_y, b_z) = (b^1, b^2, b^3)\end{aligned} \hspace{\stretch{1}}(1.1)

For this we have the dot product

\begin{aligned}\mathbf{a} \cdot \mathbf{b} = \sum_{\alpha=1}^3 a^\alpha b^\alpha\end{aligned} \hspace{\stretch{1}}(1.3)

Greek letters in this course (opposite to everybody else in the world, because of Landau and Lifshitz) run from 1 to 3, whereas roman letters run through the set $\{0,1,2,3\}$.

We want to put space and time on an equal footing and form the composite quantity (four vector)

\begin{aligned}x^i = (ct, \mathbf{r}) = (x^0, x^1, x^2, x^3),\end{aligned} \hspace{\stretch{1}}(1.4)

where

\begin{aligned}x^0 &= ct \\ x^1 &= x \\ x^2 &= y \\ x^3 &= z.\end{aligned} \hspace{\stretch{1}}(1.5)

It will also be convenient to drop indexes when referring to all the components of a four vector and we will use lower or upper case non-bold letters to represent such four vectors. For example

\begin{aligned}X = (ct, \mathbf{r}),\end{aligned} \hspace{\stretch{1}}(1.9)

or

\begin{aligned}v = \gamma \left(c, \mathbf{v} \right).\end{aligned} \hspace{\stretch{1}}(1.10)

Three vectors will be represented as letters with over arrows $\vec{a}$ or (in text) bold face $\mathbf{a}$.

Recall that the squared spacetime interval between two events $X_1$ and $X_2$ is defined as

\begin{aligned}{S_{X_1, X_2}}^2 = (ct_1 - c t_2)^2 - (\mathbf{x}_1 - \mathbf{x}_2)^2.\end{aligned} \hspace{\stretch{1}}(1.11)

In particular, with one of these zero, we have an operator which takes a single four vector and spits out a scalar, measuring a “distance” from the origin

\begin{aligned}s^2 = (ct)^2 - \mathbf{r}^2.\end{aligned} \hspace{\stretch{1}}(1.12)

This motivates the introduction of a dot product for our four vector space.

\begin{aligned}X \cdot X = (ct)^2 - \mathbf{r}^2 = (x^0)^2 - \sum_{\alpha=1}^3 (x^\alpha)^2\end{aligned} \hspace{\stretch{1}}(1.13)

Utilizing the spacetime dot product of 1.13 we have for the dot product of the difference between two events

\begin{aligned}(X - Y) \cdot (X - Y)&=(x^0 - y^0)^2 - \sum_{\alpha =1}^3 (x^\alpha - y^\alpha)^2 \\ &=X \cdot X + Y \cdot Y - 2 x^0 y^0 + 2 \sum_{\alpha =1}^3 x^\alpha y^\alpha.\end{aligned}

From this, assuming our dot product 1.13 is both linear and symmetric, we have for any pair of spacetime events

\begin{aligned}X \cdot Y = x^0 y^0 - \sum_{\alpha =1}^3 x^\alpha y^\alpha.\end{aligned} \hspace{\stretch{1}}(1.14)

How do our four vectors transform? This is really just a notational issue, since this has already been discussed. In this new notation we have

\begin{aligned}{x^0}' &= ct' = \gamma ( ct - \beta x) = \gamma ( x^0 - \beta x^1 ) \\ {x^1}' &= x' = \gamma ( x - \beta ct ) = \gamma ( x^1 - \beta x^0 ) \\ {x^2}' &= x^2 \\ {x^3}' &= x^3\end{aligned} \hspace{\stretch{1}}(1.15)

where $\beta = V/c$, and $\gamma^{-2} = 1 - \beta^2$.

In order to put some structure to this, it can be helpful to express this dot product as a quadratic form. We write

\begin{aligned}A \cdot B = \begin{bmatrix}a^0 & \mathbf{a}^\text{T} \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}\begin{bmatrix}b^0 \\ \mathbf{b}\end{bmatrix}= A^\text{T} G B.\end{aligned} \hspace{\stretch{1}}(1.19)

We can write our Lorentz boost as a matrix

\begin{aligned}\begin{bmatrix}\gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.20)

so that the dot product between two transformed four vectors takes the form

\begin{aligned}A' \cdot B' = A^\text{T} O^\text{T} G O B\end{aligned} \hspace{\stretch{1}}(1.21)

## Back to the problem.

We will work in momentum space, where we have

\begin{aligned}p^i &= (p^0, \mathbf{p}) = \left( \frac{E}{c}, \mathbf{p}\right) \\ p^2 &= \frac{E^2}{c^2} -\mathbf{p}^2 \\ \mathbf{p} &= \hbar \mathbf{k} \\ E &= \hbar \omega \\ p^i &= \hbar k^i \\ k^i &= \left(\frac{\omega}{c}, \mathbf{k}\right)\end{aligned} \hspace{\stretch{1}}(1.22)

### Justifying this.

Now, the TA blurted all this out. We know some of it from the QM context, and if we’ve been reading ahead know a bit of this from our text [2] (the energy momentum four vector relationships). Let’s go back to the classical electromagnetism and recall what we know about the relation of frequency and wave numbers for continuous fields. We want solutions to Maxwell’s equation in vacuum and can show that such solution also implies that our fields obey a wave equation

\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 \Psi}{\partial t^2} - \boldsymbol{\nabla}^2 \Psi = 0,\end{aligned} \hspace{\stretch{1}}(1.28)

where $\Psi$ is one of $\mathbf{E}$ or $\mathbf{B}$. We have other constraints imposed on the solutions by Maxwell’s equations, but require that they at least obey 1.28 in addition to these constraints.

With application of a spatial Fourier transformation of the wave equation, we find that our solution takes the form

\begin{aligned}\Psi = (2 \pi)^{-3/2} \int \tilde{\Psi}(\mathbf{k}, 0) e^{i (\omega t \pm \mathbf{k} \cdot \mathbf{x}) } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(1.29)

If one takes this as a given and applies the wave equation operator to this as a test solution, one finds without doing the Fourier transform work that we also have a constraint. That is

\begin{aligned}\frac{1}{{c^2}} (i \omega)^2 \Psi - (\pm i \mathbf{k})^2 \Psi = 0.\end{aligned} \hspace{\stretch{1}}(1.30)

So even in the continuous field domain, we have a relationship between frequency and wave number. We see that this also happens to have the form of a lightlike spacetime interval

\begin{aligned}\frac{\omega^2}{c^2} - \mathbf{k}^2 = 0.\end{aligned} \hspace{\stretch{1}}(1.31)

Also recall that the photoelectric effect imposes an experimental constraint on photon energy, where we have

\begin{aligned}E = h \nu = \frac{h}{2\pi} 2 \pi \nu = \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.32)

Therefore if we impose a mechanics like $P = (E/c, \mathbf{p})$ relativistic energy-momentum relationship on light, it then makes sense to form a nilpotent (lightlike) four vector for our photon energy. This combines our special relativistic expectations, with the constraints on the fields imposed by classical electromagnetism. We can then write for the photon four momentum

\begin{aligned}P = \left( \frac{\hbar \omega}{c}, \hbar k \right)\end{aligned} \hspace{\stretch{1}}(1.33)

### Back to the TA’s formula blitz.

Utilizing spherical polar coordinates in momentum (wave number) space, measuring the polar angle from the $k^1$ (x-like) axis, we can compute this polar angle in both pairs of frames,

\begin{aligned} \cos \alpha &= \frac{k^1}{{\left\lvert{\mathbf{k}}\right\rvert}} = \frac{k^1}{\omega/c} \\ \cos \alpha' &= \frac{{k^1}'}{\omega'/c} = \frac{\gamma (k^1 + \beta \omega/c)}{\gamma(\omega/c + \beta k^1)}\end{aligned} \hspace{\stretch{1}}(1.34)

Note that this requires us to assume that wave number four vectors transform in the same fashion as regular mechanical position and momentum four vectors. Also note that we have the primed frame moving negatively along the x-axis, instead of the usual positive origin shift. The question is vague enough to allow this since it only requires motion.

\paragraph{check 1}

as $\beta \rightarrow 1$ (ie: our primed frame velocity approaches the speed of light relative to the rest frame), $\cos \alpha' \rightarrow 1$, $\alpha' = 0$. The surface gets more and more compressed.

In the original reference frame the radiation was isotropic. In the new frame how does it change with respect to the angle? This is really a question to find this number flux rate

\begin{aligned}f'(\alpha') = ?\end{aligned} \hspace{\stretch{1}}(1.36)

In our rest frame the total number of photons traveling through the surface in a given interval of time is

\begin{aligned}N &= \int d\Omega dt f(\alpha) = \int d \phi \sin \alpha d\alpha = -2 \pi \int d(\cos\alpha) dt f(\alpha) \\ \end{aligned} \hspace{\stretch{1}}(1.37)

Here we utilize the spherical solid angle $d\Omega = \sin \alpha d\alpha d\phi = - d(\cos\alpha) d\phi$, and integrate $\phi$ over the $[0, 2\pi]$ interval. We also have to assume that our number flux density is not a function of horizontal angle $\phi$ in the rest frame.

In the moving frame we similarly have

\begin{aligned}N' &= -2 \pi \int d(\cos\alpha') dt' f'(\alpha'),\end{aligned} \hspace{\stretch{1}}(1.39)

and we again have had to assume that our transformed number flux density is not a function of the horizontal angle $\phi$. This seems like a reasonable move since ${k^2}' = k^2$ and ${k^3}' = k^3$ as they are perpendicular to the boost direction.

\begin{aligned}f'(\alpha') = \frac{d(\cos\alpha)}{d(\cos\alpha')} \left( \frac{dt}{dt'} \right) f(\alpha)\end{aligned} \hspace{\stretch{1}}(1.40)

Now, utilizing a conservation of mass argument, we can argue that $N = N'$. Regardless of the motion of the frame, the same number of particles move through the surface. Taking ratios, and examining an infinitesimal time interval, and the associated flux through a small patch, we have

\begin{aligned}\left( \frac{d(\cos\alpha)}{d(\cos\alpha')} \right) = \left( \frac{d(\cos\alpha')}{d(\cos\alpha)} \right)^{-1} = \gamma^2 ( 1 + \beta \cos\alpha)^2\end{aligned} \hspace{\stretch{1}}(1.41)

Part of the statement above was a do-it-yourself. First recall that $c t' = \gamma ( c t + \beta x )$, so $dt/dt'$ evaluated at $x=0$ is $1/\gamma$.

The rest is messier. We can calculate the $d(\cos)$ values in the ratio above using 1.34. For example, for $d(\cos(\alpha))$ we have

\begin{aligned}d(\cos\alpha) &= d \left( \frac{k^1}{\omega/c} \right) \\ &= dk^1 \frac{1}{{\omega/c}} - c \frac{1}{{\omega^2}} d\omega.\end{aligned}

If one does the same thing for $d(\cos\alpha')$, after a whole whack of messy algebra one finds that the differential terms and a whole lot more mystically cancels, leaving just

\begin{aligned}\frac{d\cos\alpha'}{d\cos\alpha} = \frac{\omega^2/c^2}{(\omega/c + \beta k^1)^2} (1 - \beta^2)\end{aligned} \hspace{\stretch{1}}(1.42)

A bit more reduction with reference back to 1.34 verifies 1.41.

Also note that again from 1.34 we have

\begin{aligned}\cos\alpha' = \frac{\cos\alpha + \beta}{1 + \beta \cos\alpha}\end{aligned} \hspace{\stretch{1}}(1.43)

and rearranging this for $\cos\alpha'$ gives us

\begin{aligned}\cos\alpha = \frac{\cos\alpha' - \beta}{1 - \beta \cos\alpha'},\end{aligned} \hspace{\stretch{1}}(1.44)

which we can sum to find that

\begin{aligned}1 + \beta \cos\alpha = \frac{1}{{\gamma^2 (1 - \beta \cos \alpha')^2 }},\end{aligned} \hspace{\stretch{1}}(1.45)

so putting all the pieces together we have

\begin{aligned}f'(\alpha') = \frac{1}{{\gamma}} \frac{f(\alpha)}{(\gamma (1-\beta \cos\alpha'))^2}\end{aligned} \hspace{\stretch{1}}(1.46)

The question asks for the energy flux density. We get this by multiplying the number density by the frequency of the light in question. This is, as a function of the polar angle, in each of the frames.

\begin{aligned}L(\alpha) &= \hbar \omega(\alpha) f(\alpha) = \hbar \omega f \\ L'(\alpha') &= \hbar \omega'(\alpha') f'(\alpha') = \hbar \omega' f'\end{aligned} \hspace{\stretch{1}}(1.47)

But we have

\begin{aligned}\omega'(\alpha')/c = \gamma( \omega/c + \beta k^1 ) = \gamma \omega/c ( 1 + \beta \cos\alpha )\end{aligned} \hspace{\stretch{1}}(1.49)

Aside, $\beta << 1$,

\begin{aligned}\omega' = \omega ( 1 + \beta \cos\alpha) + O(\beta^2) = \omega + \delta \omega\end{aligned} \hspace{\stretch{1}}(1.50)

\begin{aligned}\delta \omega &= \beta, \alpha = 0 \qquad \text{blue shift} \\ \delta \omega &= -\beta, \alpha = \pi \qquad \text{red shift}\end{aligned} \hspace{\stretch{1}}(1.51)

The TA then writes

\begin{aligned}L'(\alpha') = \frac{L/\gamma}{(\gamma (1 - \beta \cos\alpha'))^3}\end{aligned} \hspace{\stretch{1}}(1.53)

although, I calculate

\begin{aligned}L'(\alpha') = \frac{L}{\gamma^4 (\gamma (1 - \beta \cos\alpha'))^4}\end{aligned} \hspace{\stretch{1}}(1.54)

He then says, the forward backward ratio is

\begin{aligned}L'(0)/L'(\pi) = {\left( \frac{ 1 + \beta }{1-\beta} \right)}^3\end{aligned} \hspace{\stretch{1}}(1.55)

For this I get:

\begin{aligned}L'(0)/L'(\pi) = {\left( \frac{ 1 + \beta }{1-\beta} \right)}^4\end{aligned} \hspace{\stretch{1}}(1.56)

It is still bigger for $\beta$ positive, which I think is the point.

If I can somehow manage to keep my signs right as I do this course I may survive. Why did he pick a positive sign way back in 1.34?

# References

[1] J.B. Hartle and T. Dray. Gravity: an introduction to Einsteins general relativity, volume 71. 2003.

[2] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

## Lorentz transformation of the metric tensors.

Posted by peeterjoot on January 16, 2011

Following up on the previous thought, it is not hard to come up with an example of a symmetric tensor a whole lot simpler than the electrodynamic stress tensor. The metric tensor is probably the simplest symmetric tensor, and we get that by considering the dot product of two vectors. Taking the dot product of vectors $a$ and $b$ for example we have

\begin{aligned}a \cdot b = a^\mu b^\nu \gamma_\mu \cdot \gamma_\nu\end{aligned} \hspace{\stretch{1}}(4.17)

From this, the metric tensors are defined as

\begin{aligned}\eta_{\mu\nu} &= \gamma_\mu \cdot \gamma_\nu \\ \eta^{\mu\nu} &= \gamma^\mu \cdot \gamma^\nu\end{aligned} \hspace{\stretch{1}}(4.18)

These are both symmetric and diagonal, and in fact equal (regardless of whether one picks a $+,-,-,-$ or $-,+,+,+$ signature for the space).

Let’s look at the transformation of the dot product, utilizing the transformation of the four vectors being dotted to do so. By definition, when both vectors are equal, we have the (squared) spacetime interval, which based on the speed of light being constant, has been found to be an invariant under transformation.

\begin{aligned}a' \cdot b'= a^\mu b^\nu L(\gamma_\mu) \cdot L(\gamma_\nu)\end{aligned} \hspace{\stretch{1}}(4.20)

We note that, like any other vector, the image $L(\gamma_\mu)$ of the Lorentz transform of the vector $\gamma_\mu$ can be written as

\begin{aligned}L(\gamma_\mu) = \left( L(\gamma_\mu) \cdot \gamma^\nu \right) \gamma_\nu\end{aligned} \hspace{\stretch{1}}(4.21)

Similarily we can write any vector in terms of the reciprocal frame

\begin{aligned}\gamma_\nu = (\gamma_\nu \cdot \gamma_\mu) \gamma^\mu.\end{aligned} \hspace{\stretch{1}}(4.22)

The dot product factor is a component of the metric tensor

\begin{aligned}\eta_{\nu \mu} = \gamma_\nu \cdot \gamma_\mu,\end{aligned} \hspace{\stretch{1}}(4.23)

so we see that the dot product transforms as

\begin{aligned}a' \cdot b' = a^\mu b^\nu ( L(\gamma_\mu) \cdot \gamma^\alpha ) ( L(\gamma_\nu) \cdot \gamma^\beta ) \gamma_\alpha\cdot\gamma_\beta= a^\mu b^\nu {L_\mu}^\alpha{L_\nu}^\beta\eta_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(4.24)

In particular, for $a = b$ where we have the invariant interval defined by the condition $a^2 = {a'}^2$, we must have

\begin{aligned}a^\mu a^\nu \eta_{\mu \nu}= a^\mu a^\nu {L_\mu}^\alpha{L_\nu}^\beta\eta_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(4.25)

This implies that the symmetric metric tensor transforms as

\begin{aligned}\eta_{\mu\nu}={L_\mu}^\alpha{L_\nu}^\beta\eta_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(4.26)

Recall from 3.16 that the coordinates representation of a bivector, an antisymmetric quantity transformed as

\begin{aligned}T^{\mu \nu} \rightarrow T^{\sigma \pi} {L_\sigma}^\mu {L_\pi}^\nu.\end{aligned} \hspace{\stretch{1}}(4.27)

This is a very similar transformation, but differs from the bivector case where our free indexes were upper indexes. Suppose that we define an alternate set of coordinates for the Lorentz transformation. Let

\begin{aligned}{L^\mu}_\nu = L(\gamma^\mu) \cdot \gamma_\nu.\end{aligned} \hspace{\stretch{1}}(4.28)

This can be related to the previous coordinate matrix by

\begin{aligned}{L^\mu}_\nu = \eta^{\mu \alpha } \eta_{\nu \beta } {L_\alpha}^\beta. \end{aligned} \hspace{\stretch{1}}(4.29)

If we examine how the coordinates of $x^2$ transform in thier lower index representation we find

\begin{aligned}{x'}^2 = x_\mu x_\nu {L^\mu}_\alpha {L^\nu}_\beta \eta^{\alpha \beta} = x^2 = x_\mu x_\nu \eta^{\mu \nu},\end{aligned} \hspace{\stretch{1}}(4.30)

and therefore find that the (upper index) metric tensor transforms as

\begin{aligned}\eta^{\mu \nu} \rightarrow\eta^{\alpha \beta}{L^\mu}_\alpha {L^\nu}_\beta .\end{aligned} \hspace{\stretch{1}}(4.31)

Compared to $4.27$ we have almost the same structure of transformation. Are these the same? Does the notation I picked here introduce an apparent difference that does not actually exist? We really want to know if we have the identity

\begin{aligned}L(\gamma_\mu) \cdot \gamma^\nu\stackrel{?}{=}L(\gamma^\nu) \cdot \gamma_\mu,\end{aligned} \hspace{\stretch{1}}(4.32)

which given the notation selected would mean that ${L_\mu}^\nu = {L^\nu}_\mu$, and justify a notational simplification ${L_\mu}^\nu = {L^\nu}_\mu = L^\nu_\mu$.

# The inverse Lorentz transformation

To answer this question, let’s consider a specific example, an x-axis boost of rapidity $\alpha$. For that our Lorentz transformation takes the following form

\begin{aligned}L(x) = e^{-\sigma_1 \alpha/2} x e^{\sigma_1 \alpha/2},\end{aligned} \hspace{\stretch{1}}(5.33)

where $\sigma_k = \gamma_k \gamma_0$. Since $\sigma_1$ anticommutes with $\gamma_0$ and $\gamma_1$, but commutes with $\gamma_2$ and $\gamma_3$, we have

\begin{aligned}L(x) = (x^0 \gamma_0 + x^1 \gamma_1) e^{\sigma_1 \alpha} + x^2 \gamma_2 + x^3 \gamma_3,\end{aligned} \hspace{\stretch{1}}(5.34)

and after expansion this is

\begin{aligned}L(x) = \gamma_0 ( x^0 \cosh \alpha - x^1 \sinh \alpha ) +\gamma_1 ( x^1 \cosh \alpha - x^0 \sinh \alpha )+\gamma_2+\gamma_3\end{aligned} \hspace{\stretch{1}}(5.35)

In particular for the basis vectors themselves we have

\begin{aligned}\begin{bmatrix}L(\gamma_0) \\ L(\gamma_1) \\ L(\gamma_2) \\ L(\gamma_3)\end{bmatrix}=\begin{bmatrix}\gamma_0 \cosh \alpha - \gamma_1 \sinh \alpha \\ -\gamma_0 \sinh \alpha + \gamma_1 \cosh \alpha \\ \gamma_2 \\ \gamma_3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.36)

Forming a matrix with $\mu$ indexing over rows and $\nu$ indexing over columns we have

\begin{aligned}{L_\mu}^\nu =\begin{bmatrix}\cosh \alpha &- \sinh \alpha & 0 & 0 \\ -\sinh \alpha & \cosh \alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.37)

Performing the same expansion for ${L^\nu}_\mu$, again with $\mu$ indexing over rows, we have

\begin{aligned}{L^\nu}_\mu =\begin{bmatrix}\cosh \alpha & \sinh \alpha & 0 & 0 \\ \sinh \alpha & \cosh \alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(5.38)

This answers the question. We cannot assume that ${L_\mu}^\nu = {L^\nu}_\mu$. In fact, in this particular case, we have ${L^\nu}_\mu = ({L_\mu}^\nu)^{-1}$. Is that a general condition? Note that for the general case, we have to consider compounded transformations, where each can be a boost or rotation.

# References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.