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enter image description here

This and similar depictions are used when deriving the equation for the Mach cone angle $\theta$.

It is usually stated that using algebra we can conclude that $$\sin\theta=\frac{ct}{vt},$$ and thus $$θ=\arcsin \left(\frac{c}{v}\right).$$ However, I noticed that that assumes a right angle between $ct$ and the wavefront, which is never mathematically proven.

So, can someone explain why there is a right angle between $ct$ and the wavefront?

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    $\begingroup$ Because a tangent to the circle is at right angles to a radius intersecting at the tangent point. $\endgroup$
    – Jon Custer
    Jul 3 at 14:14
  • $\begingroup$ Ok that makes sense. Is there a mathematical proof for "Because a tangent to the circle is at right angles to a radius intersecting at the tangent point"? $\endgroup$ Jul 3 at 14:26
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    $\begingroup$ You can find some proofs by searching for "why is radius perpendicular to tangent". $\endgroup$ Jul 3 at 14:31
  • $\begingroup$ Just to make sure the angle I get from the equation is only the angle for half a cone right? $\endgroup$ Jul 3 at 14:41
  • $\begingroup$ yes, only half ...... $\endgroup$ Jul 3 at 14:42

1 Answer 1

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This is a standard theorem in Euclidean geometry: the tangent to a circle is perpendicular to the radius at the point of tangency. The proof is a simple proof by contradiction.

Given: a circle $C$, a line tangent to the circle $T$, and a radius, $R$, from the center of the circle to the point of tangency.

Assume: $R$ is NOT perpendicular to $T$

Then there is some other line segment, $S$, from the center of $C$ which is perpendicular to $T$. Since $S$ is perpendicular to $T$ that makes $R$ the hypotenuse of a right triangle. Therefore the length of $R$ is greater than the length of $S$, so $|R|>|S|$.

Now, except for the point of tangency all other points on $T$ are outside of $C$. Since $C$ is defined as all points a distance of $|R|$ away from the center, all points outside of $C$ are a distance $x>|R|$ away from the center. Since $S$ connects the center to such a point we have $|S|>|R|$.

But $|S|>|R|$ contradicts $|R|>|S|$ so our assumption is false. Therefore $R$ is perpendicular to $T$. QED

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