no, because what it's called the "photon momentum" is not connected with its velocity (mv), but with its wave-like behavor, in particularly with the wave vector k:
\vec{p}=\hbar \vec{k}
where k is defined:
\vec{k}=\frac{2\pi}{\lambda} \vec{n}
where n is the wave direction unit vector.
after substitution you obteined:
b_{m}=\frac{1}{L}\int_{-L}^{L}sin(\frac{m\pi x}{L})f(x)d
nothing changes if there is n or m in the final result, you can call the index however you want, so this result transforms into:
b_{n}=\frac{1}{L}\int_{-L}^{L}sin(\frac{n\pi x}{L})f(x)d
\int_{-L}^{L}sin(\frac{m\pi x}{L})dx=0
because the sin function is an odd function and the interval of integration is symmetric. In fact:
\int_{-L}^{L}sin(\frac{m\pi x}{L})dx=\int_{-L}^{0}sin(\frac{m\pi x}{L})dx+\int_{0}^{L}sin(\frac{m\pi x}{L})dx
changing x --> -x in the second...
you must multiply f(x) by (1/L)sin(m pi x/L) and then integrate from -L to L in order to obtein:
\frac{1}{L}\int_{-L}^{L}sin(\frac{m\pi x}{L})f(x)dx=\frac{a_{0}}{2L}\int_{-L}^{L}sin(\frac{m\pi x}{L})dx+\sum a_{n} \frac{1}{L}\int_{-L}^{L}sin(\frac{m\pi x}{L})cos(\frac{n\pi x}{L})dx+\sum b_{n}...
\int_{0}^{p}f(t)dt=\int_{0}^{\frac{p}{2}}f(t)dt+\int_{\frac{p}{2}}^{p}f(t)dt
\int_{\frac{p}{2}}^{p}f(t)dt=\int_{-\frac{p}{2}}^{0}f(t)dt
substituting in the first equation:
\int_{0}^{\frac{p}{2}}f(t)dt+\int_{-\frac{p}{2}}^{0}f(t)=\int_{-\frac{p}{2}}^{\frac{p}{2}}f(t)dt
\int_{0}^{p}f(t)dt=\int_{0}^{\frac{p}{2}}f(t)dt+\int_{\frac{p}{2}}^{p}f(t)dt
then just apply the previous property at the second integral
\int_{\frac{p}{2}}^{p}f(t)dt=\int_{\frac{p}{2}-p}^{p-p}f(t)dt
I hope that it is correct :blushing:
Hi,
this step is connected with the fact that the function f is periodic:
f(t+T)=f(t+T)
so the integral of this function in an interval [a,b] is equal to the integral in [a+T,b+T].
Hi, I'm italian student, excuse me for my english :redface:
The problem is the integration of the two differential equations, in fact:
s1 = ft02/2
s2 = ft0(t-t0) - r(t-t0)2/2
in this case I put "r" positive.
After substitutions and calculations you'll arrive at the solution :smile: