K

#### Kashmiri

##### Guest

Kashmiri Asks:

The transformation is given as

To find the angle $x'$ axes makes with $x$ axes I use

$0=(\cosh \theta)(c t)-(\sinh \theta) x$ to get its slope as $Tanh(\theta)$ and hence the angle is ${Tan}^{-1}(Tanh(\theta))$ but the book says the angle is $\theta$ ? What did I do wrong?

*Angle in a space time diagram*FIGURE 4.13 A Lorentz boost as a change of coordinates on a spacetime diagram. The figure shows the grid of $\left(c t^{\prime}, x^{\prime}\right)$ coordinates defined by $(4.18)$ plotted on a $(c t, x)$ spacetime diagram. The $\left(c t^{\prime}, x^{\prime}\right)$ coordinates are not orthogonal to each other in the Euclidean geometry of the printed page. But they are orthogonal in the geometry of spacetime. (Recall the analogies between spacetime diagrams and maps discussed in Example 4.1.) The $\left(c t^{\prime}, x^{\prime}\right)$ axes have to be as orthogonal as the $(c t, x)$ axes because there is no physical distinction between one inertial frame and another. The orthogonality is explicitly verified in Example 5.2. The hyperbolic angle $\theta$ is a measure of the velocity between the two frames.

The transformation is given as

$\begin{aligned} c t^{\prime} &=(\cosh \theta)(c t)-(\sinh \theta) x \\ x^{\prime} &=(-\sinh \theta)(c t)+(\cosh \theta) x \\ y^{\prime} &=y \\ z^{\prime} &=z \end{aligned}$

To find the angle $x'$ axes makes with $x$ axes I use

$0=(\cosh \theta)(c t)-(\sinh \theta) x$ to get its slope as $Tanh(\theta)$ and hence the angle is ${Tan}^{-1}(Tanh(\theta))$ but the book says the angle is $\theta$ ? What did I do wrong?

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