201902210242 Exercise 5.4.1

Similarly, check that $\displaystyle{\frac{\partial x'^\mu}{\partial x^\nu}}=\varLambda^\mu_{\enspace\nu}$ when $\mu =x$ and $\nu =t$ and when $\mu =\nu =y$ .

Roughwork.

Equation (5.12):

$\begin{aligned} t' & = \gamma (t-\beta x) \\ x' & = \gamma (-\beta t+x) \\ y' & = y \\ z' & = z \end{aligned}$

and

$\begin{aligned} t & = \gamma (t'+\beta x') \\ x & = \gamma (\beta t'+x') \\ y & = y' \\ z & = z' \end{aligned}$

Equation (5.13a):

$\displaystyle{\frac{\partial x'^\mu}{\partial x^\nu}}=\varLambda^\mu_{\enspace\nu} =\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

Proof.

$\displaystyle{\frac{\partial t'}{\partial t}}=\gamma$ , $\displaystyle{\frac{\partial t'}{\partial x}}=-\gamma\beta$ , $\displaystyle{\frac{\partial t'}{\partial y}}=0$ , $\displaystyle{\frac{\partial t'}{\partial z}}=0$

$\displaystyle{\frac{\partial x'}{\partial t}}=-\gamma\beta$ , $\displaystyle{\frac{\partial x'}{\partial x}}=\gamma$ , $\displaystyle{\frac{\partial x'}{\partial y}}=0$ , $\displaystyle{\frac{\partial x'}{\partial z}}=0$

$\displaystyle{\frac{\partial y'}{\partial t}}=0$ , $\displaystyle{\frac{\partial y'}{\partial x}}=0$ , $\displaystyle{\frac{\partial y'}{\partial y}}=1$ , $\displaystyle{\frac{\partial y'}{\partial z}}=0$

$\displaystyle{\frac{\partial z'}{\partial t}}=0$ , $\displaystyle{\frac{\partial z'}{\partial x}}=0$ , $\displaystyle{\frac{\partial z'}{\partial y}}=0$ , $\displaystyle{\frac{\partial z'}{\partial z}}=1$

M	T	W	T	F	S	S
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4	5	6	7	8	9	10
11	12	13	14	15	16	17
18	19	20	21	22	23	24
25	26	27	28

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