Another wave equation question.

Fallen Seraph

I was going back through the derivation of the mathematical description of a transverse wave and I've come across something which I don't understand

The argument goes
As a function of time y=ACos(wt).
So some time later:

x=v(dt) => dt=x/v

Then, displacement, y, is the same as at the earlier time t-x/v so we get y=ACos(w(t-x/v)).

And from this it becomes our wonderfully useful equation y =ACos(kx-wt), which is fine.

What I don't understand is:
y will be the same at t-x/v, t and also at t+x/v, so why do we choose t-x/v instead of t+x/v?

Son Goku

There's no reason not to choose t+x/v over t-x/v in this particular example. However there are waves in more detailed environments with more complicated equations whose analogue of t+x/v is forbidden. So to keep everything treated consistently we choose t-x/v as the convention, because its analogues in other situations are what always holds up.

You can only pick t+x/v in this case because the future and past are perfectly symmetric.

As another side note, when you go to Quantum Mechanics "t+x/v"-type waves give rise to anti-particles.

Fallen Seraph

Thanks a lot for that.