The Mach cone

I’m sure many of you will hear that image and you’ll even know what Mach’s cone is. You will know that it is formed when an airplane passes the speed of sound although you will not understand that the Mach cone, really, is and is called an envelope to an object that travels at speeds superior to those of sound.

That is to say, Mach’s cone, mathematically, is simply to solve and find the envelope to an object that moves at a certain speed.

Let’s go to the mathematical “nougat” (this expresion means that we go deep into the math needed for resolving it).

Suppose that sound travels at a speed c and that an object (an airplane, to be exact) travels at a speed v greater than c (v>c) in one direction. Let’s take that direction (since we presuppose that it does not vary) as one of the axes (to simplify the calculations), for example, the x-axis although it can be any of the other axes or even a definite line. But, I repeat, moving the axes will make the plane move on the x axis.

Since we are, suppose that, as it travels at speed v, at a certain moment the plane will be in the next position (vt, 0, 0).

Let’s move at a constant speed and, easily, we can know its position.

The airplane will generate sound (obviously) that expands spherically, like any wave. Here there is little to scratch.

In an instant T < t this will expand with velocity c (speed that we have assumed of the sound). This wave front will move according to (vT, 0, 0) obviously. Now, at time t, that wave front will be at:

$latex c(t-T)$

With what your equation, or its function to be more exact, will be:

$latex f(x, y, z, T) = (x-vT)^{2}+y^{2}+z^{2}-c^{2}(t-T)^{2}=0$

Obviously, the waves move as spheres of radius c(t-T) as we have said and therefore, to better understand it, we have parameterized these spheres from:

$latex (x-vT)^{2}+y^{2}+z^{2}=c^{2}(t-T)^{2}$

We have taken everything aside and created a function. Simple.

Now a little theory and that is that the equation of an envelope can be found very easily geometrically. And is that if we assume f (x, y, z, c) where c is the parameter, if we eliminate c from the equations:

$latex f(x, y, z, c)=0$
$latex \frac{\partial }{\partial c}f(x, y, z, c)=0$

We get the envelope.

In what happens to us, since we already have f (x, y, z, c), we only need its partial according to the parameter, which in our case is T instead of c. A very simple differentiation:

$latex \frac{\partial }{\partial T}f(x, y, z, T)=-2v(x-vT)+2c^{2}(t-T)=0$

And now it’s time to give the math machine where, it’s best to clear T:

$latex T=\frac{vx-c^{2}t}{v^{2}-c^{2}}$

With what, working with the math machine more and, now that we have T, removing the parenthesis zone from the equation of the function without differentiating to have it clear:

$latex x-vT=x-\frac{v^2x-vc^{2}t}{v^{2}-c^{2}}=\frac{c^{2}}{v^{2}-c^{2}}(vt-x)$

$latex t-T=t-\frac{v^2x-vc^{2}t}{v^{2}}=\frac{v}{v^{2}-c^{2}}(vt-x)$

And if we replace in f (x, y, z, T) = 0

$latex \frac{c^{4}}{(v^{2}-c^{2})^{2}}(vt-x)^{2}+y^{2}+z^{2}-\frac{(cv)^{2}}{(v^{2}-c^{2})^{2}}(vt-x)^{2}$

And, well, a little more working with the math machine we can see and find that there’s a know function, an sphere:

$latex y^{2}+z^{2}=\frac{c^{2}}{(v^{2}-c^{2})^{2}}(v^{2}-c^{2})(vt-x)^{2}=\frac{c^{2}}{v^{2}-c^{2}}(vt-x)^{2}$

Arriving, finally, to the equation called the Mach cone on the x-axis from (vt, 0, 0) at any instant t at any constant velocity v, of course:

$latex x=vt-\frac{\sqrt{v^{2}-c^{2}}}{c}\sqrt{y^{2}+z^{2}}$

That, graphically (and taken from Wikipedia because drawing is not mine work).

Because physics and mathematics are very beautiful.

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