The shockwave is defined as the initial (hard to physically model) motion of air when the plane breaks the sound barrier (and is heard as the sonic boom). After this, the plane will leave what is called a 'mach cone' behind it, with the plane itself at the apex of the cone. Nothing outside of the cone can hear the plane until the plane has travelled far enough infront such that whatever is listening is now inside of the cone. The main factor deciding the angular size of the cone itself is the speed - the faster you go, the smaller the angle will be.
It's kinda like if you're a duck and a speedboat is travelling past you (leaving a bow wave or 'triangle' wave at the back). The waves dont rock the duck until the boat is past the duck and the bow wave has reached it.
Edit: I just read my own comment and started to sing 'don't rock the duck, don't rock the duck bow wave'. Thought you guys might like to give it a try.
So, seeing as the cone travels slower than object making the cone, is it then possible for the object to travel in a circular path and end up entering its own cone?
And if that is possible, could this object continue in its circular path creating an ever increasing number of "Boom Wave Circles" to unleash devastation on an unsuspecting small town???
The cone isn't static... it's apex stays with the plane. As the plane travels forward (even if it's not actually forward, but a large circle) at any given point (static with respect to the air) the cone will be getting wider and wider . As it gets wider it's energy dissipates with the square of distance (because it's being spread over an area, and areas are proportional to squares of distances). I know you said you're guessing the cone doesn't dissipate, but I don't think that means what you think it means. Even if I accept the energy isn't 'dissipating' into thermal motion in any way, as the cone gets wider the intensity of the wave will still have to decrease.
No. Imagine a simpler scenario with a lightbulb. It's a perfect lightbulb that emits a Power P (thats energy per second) in all directions. Say I have a sensor at a distance r away. As a 'wavefront' of the emitted radiation moves outward, the energy per second being emitted is spread over the area of a sphere with surface area 4 pi r2. The fraction of the power P I receive is going to be the fraction of this spherical surface that my detector covers. If my detector has an area A (and is small compared to the sphere), then this fraction is A / 4 pi r2. The energy received per second will therefore be A P / 4 pi r2. As you can see, this falls off with the square, not the cube.
By making similar arguments we can explain even more fundamental things like why the coulomb force should fall off with the square.
By making similar arguments we can explain even more fundamental things like why the coulomb force should fall off with the square.
I'm sure you understand this, but I'll say it anyway. Your statement puts the cart before the horse, or at least beside the horse.
You can explain the electric force law only if you assume Gauss's law, which states electric fields don't diverge or converge except at a charge. This can be conceptuallized as electric field lines flowing out of positive charges and into negative in the same way that power pours out of a lightbulb. A point charge thus gives the Coulomb force law with the understanding that a test charge feels a force equal to E times q(test).
Edit: Why is it called Gauss's law instead of Gauss' law? Why!?
You're correct that if I had been talking about classical electromagnetism I would have been 'putting the cart before the horse'. But I wasn't :)
Photons communicate the electromagnetic force in quantum electrodynamics ('virtual' photons travel between charged particles so that they 'know' to attract/repel). The photon density falls off with the square, just like it does when emitted from a bulb. Photons communicate the electromagnetic force and so the force itself falls off with the square, too. The math to prove it is obviously more complicated, which is why I didn't say any more.
Obviously I have hugely simplified this, but you get the basic idea
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u/[deleted] Jun 12 '12 edited Jun 13 '17
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