T-force Delta Rgb Ram Software Download __TOP__
Everardo Marquez
where $C$ is a normalization factor. To find $C$, we can simply integrate $P$ over $z$ and $p$ and set the result equal to $1$, using the delta functions to do the $p$ integral and integrating $z$ from $0$ to $h$, and we get the correct answer. My question is, why does this strategy not work if we write$$ P(z,p) = \fracD \delta(z-h+p^2/(2m^2g))? $$In this case, when we attempt to find the normalization $D$, if we try getting rid of the delta function by integrating over $z$ and then we integrate over the appropriate range of $p$, we fail because the $1/p$ integral diverges, whereas the $1/\sqrt2g(h-z)$ integral converges. What am I missing here?
A Dirac delta has units. It has units of one over the argument so since $\int \delta (x) dx=1$ then $\delta (x)$ has units of inverse distance and $\int \delta (p) dp=1$ then $\delta (p)$ has units of inverse momentum.
So a density is just like the Dirac delta it is something you integrate to get the probability but it needs to be a per interval of distance or a per interval of momentum for it to be something you can integrate.
f5d0e4f075