# Error with generators of Lorentz group (basis of Lorentz Lie algebra) [closed]

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 $$\begin{eqnarray*} K_z &=& -i \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} \\ &=& i \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \end{eqnarray*}$$

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## closed as off-topic by ZeroTheHero, Kyle Kanos, Aaron Stevens, Chris♦, user191954 Apr 30 at 5:10

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• 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. – knzhou Apr 27 at 23:12
• 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 – Lopey Tall Apr 28 at 1:16
• It seems to have something to do with using a right-handed coordinate system but I can't place my finger on it – Lopey Tall Apr 28 at 1:21
• 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. – Kyle Kanos Apr 28 at 14:54