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enter image description here

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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2 Answers 2

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The books is talking about hyperbolic angles, which are different from ordinary (circular) angles. The math is similar but uses hyperbolic trig functions. The slope of the line is $\tanh θ$ as you said, but the corresponding hyperbolic angle is $\tanh^{-1} \tanh θ = θ$.

The angle on the piece of paper is $\tan^{-1} \tanh θ$, but that has no geometric meaning in this context. It isn't preserved by Lorentz boosts, while the hyperbolic angle is.

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  • $\begingroup$ Thank you benrg. So on the piece of paper the lines $x, x'$ have an angle of ${\tan}^{-1}(\tanh{\theta})$ but in the Minkowski space the angle is $\theta$ ? Is that right? $\endgroup$
    – Kashmiri
    Sep 24 at 6:30
  • $\begingroup$ @Kashmiri That's right (see the added paragraph). $\endgroup$
    – benrg
    Sep 24 at 6:46
  • $\begingroup$ Great! Thank you from bottom of my heart sir! :) $\endgroup$
    – Kashmiri
    Sep 24 at 6:50
  • $\begingroup$ Is it also that the hyperbolic angles of all points on the line $x'$ are same and equal to $\theta$ ? $\endgroup$
    – Kashmiri
    Sep 24 at 11:42
  • $\begingroup$ Applying a Lorentz boost will slide the (x, t) coordinate on a hyperbola, then won't it's hyperbolic angle change? You said it remains the same. Can you please elaborate on that. Thank you $\endgroup$
    – Kashmiri
    Sep 24 at 15:09
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As @benrg says, the "Minkowskian-angle" (called "rapidity") uses the hyperbolic functions, and not the circular functions. With velocity $(v/c)=\tanh\theta$, we have time-dilation factor $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$ and Doppler factor $k=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=\exp\theta$.

Geometrically, an angle (a circular angle) is the arc-length of a circular arc divided by its radius. For Minkowski spacetime, the Minkowskian-angle is the Minkowski-arc-length of a Minkowski-circular-arc (a "hyperbolic arc") divided by its Minkowski-radius.

Alternatively, we can characterize an angle as the twice the area of a circular sector, divided by the square-of-the-radius. Similarly, we can characterize a Minkowski-angle (the rapidity) as the twice the area of a hyperbolic sector, divided by the square-of-the-Minkowski-radius.

Because of these characterizations, for relativity, one should really draw a hyperbolic-arc for the rapidity, and not a circular-arc, as you have drawn.

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  • $\begingroup$ Thank you it is getting clearer. Is it that all the points on the $x'$ line have the same hyperbolic angle, $\theta$ ? $\endgroup$
    – Kashmiri
    Sep 24 at 13:58
  • $\begingroup$ @Kashmiri Yes, all of the events on the $x'$-axis are on the line with slope $ct=x\tanh\theta$. Personally, I would reserve "rapidity" to refer to the angular-like measure between timelike lines. The indicated quantity between these spacelike-lines which are coplanar with the 4-velocities are numerically equal to the rapidity in Minkowski spacetime (but not in a Galilean spacetime). $\endgroup$
    – robphy
    Sep 24 at 18:47

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