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Can someone help me figure out why my $J_y$ is incorrect? :/ It's supposed to be \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}

My work:

To calculate the basis of Lorentz Lie algebra (or equivalently the generators of the Lorentz Lie group), $J_x,J_y,J_z,K_x,K_y,K_z$, we take the matrices $R(\hat{i},\theta)$ and $\beta(\hat{i},\delta)$ (that are respectively matrix representations for counter-clockwise rotations about the i-axis by an angle $\theta$, as well as matrix representations for Lorentz boosts in the i-axis direction with rapidity $\delta$) and respectively apply the following formulas, $J_i=i\frac{\partial R(\hat{i},\theta)}{\partial\theta}\mid_{\theta=0}$ and $K_i=-i\frac{\partial \beta(\hat{i},\delta)}{\partial\delta}\mid_{\delta=0}$.

To get $J_x$ we begin

$R(\hat{x},\theta)$ = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & cos(\theta) & -sin(\theta) \\ 0 & 0 & sin(\theta) & cos(\theta) \end{pmatrix}

Differentiating with respect to $\theta$ we have

\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -sin(\theta) & -cos(\theta) \\ 0 & 0 & cos(\theta) & -sin(\theta) \end{pmatrix}

and then taking the limit as $\theta$ goes to zero, we get

$J_x$ = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}

To get $J_y$ we begin

$R(\hat{y},\theta)$ = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & cos(\theta) & 0 & -sin(\theta) \\ 0 & 0 & 1 & 0 \\ 0 & sin(\theta) & 0 & cos(\theta) \end{pmatrix}

Differentiating with respect to $\theta$ we have

\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -sin(\theta) & 0 & -cos(\theta) \\ 0 & 0 & 0 & 0 \\ 0 & cos(\theta) & 0 & -sin(\theta) \end{pmatrix}

and then taking the limit as $\theta$ goes to zero, we get

$J_y$ = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}

To get $J_z$ we begin

$R(\hat{z},\theta)$ = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & cos(\theta) & -sin(\theta) & 0 \\ 0 & sin(\theta) & cos(\theta) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

Differentiating with respect to $\theta$ we have

\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -sin(\theta) & -cos(\theta) & 0 \\ 0 & cos(\theta) & -sin(\theta) & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}

and then taking the limit as $\theta$ goes to zero, we get

$J_z$ = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}

Now recalling the limits of hyperbolic trig functions,

\begin{eqnarray*} \frac{\partial}{\partial x} [sinh(x)] &=& cosh(x) \\ \frac{\partial}{\partial x} [cosh(x)] &=& sinh(x) \end{eqnarray*}

as well as the value of the hyperbolic trig functions at 0,

\begin{eqnarray*} cosh(0) &=& 1\\ sinh(0) &=& 1 \end{eqnarray*}

To get $K_x$ we begin

$\beta(\hat{x},\delta)$ = \begin{pmatrix} cosh(\delta) & -sinh(\delta) & 0 & 0 \\ -sinh(\delta) & cosh(\delta) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

Simultaneously differentiating with respect to $\delta$ and then taking the limit as $\delta$ goes to 0, we have \begin{eqnarray*} K_x &=& -i \begin{pmatrix} 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\ &=& i \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{eqnarray*}

To get $K_y$ we begin

$\beta(\hat{y},\delta)$ = \begin{pmatrix} cosh(\delta) & 0 & -sinh(\delta) & 0 \\ 0 & 1 & 0 & 0 \\ -sinh(\delta) & 0 & cosh(\delta) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

Simultaneously differentiating with respect to $\delta$ and then taking the limit as $\delta$ goes to 0, we have \begin{eqnarray*} K_y &=& -i \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\ &=& i \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{eqnarray*}

To get $K_z$ we begin

$\beta(\hat{z},\delta)$ = \begin{pmatrix} cosh(\delta) & 0 & 0 & -sinh(\delta) \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -sinh(\delta) & 0 & 0 & cosh(\delta) \end{pmatrix}

Simultaneously differentiating with respect to $\delta$ and then taking the limit as $\delta$ goes to 0, we have \begin{eqnarray*} K_z &=& -i \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} \\ &=& i \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \end{eqnarray*}

Linked question from Math Stack

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  • 2
    $\begingroup$ Well, your $R_y$ is off by a sign. If you want to see, just check what it does to a test vector like $\hat{\mathbf{x}}$. It rotates in the opposite sense to all the other ones. $\endgroup$ – knzhou Apr 27 '19 at 23:12
  • $\begingroup$ Holy crap you're right, that $R_y$ rotates clockwise. Can you help explain why that is? I was under the impress that a counterclockwise rotation by an angle $\theta$ along the i-axis took precisely the form I listed. But you are absolutely correct, en.wikipedia.org/wiki/Rotation_matrix#Basic_rotations $\endgroup$ – Lopey Tall Apr 28 '19 at 1:16
  • $\begingroup$ It seems to have something to do with using a right-handed coordinate system but I can't place my finger on it $\endgroup$ – Lopey Tall Apr 28 '19 at 1:21
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    $\begingroup$ Note that $sin$ is interpreted as the product of the variables $s$, $i$ and $n$, rather than the operator $\sin$, which is obtained by writing \sin. $\endgroup$ – Kyle Kanos Apr 28 '19 at 14:54