Amnesia's Dream: Per Unit Value Vs Substitution

Monday, December 4, 2017

What if we were to use,

ρ_{n} = \frac{8 π}{λ^{4}}

$\rho_n=\cfrac{8\pi}{\lambda^4}$

why is this different from,

ρ_{n} = \frac{8 π}{c^{3}} f^{2}

$\rho_n=\cfrac{8\pi}{c^3}f^2$

Because in actual,

ρ_{n λ} = \frac{d n_{s}}{d λ} = \frac{8 π}{λ^{4}}

$\rho_{n\,\lambda}=\cfrac{dn_s}{d\lambda}=\cfrac{8\pi}{\lambda^4}$

and

ρ_{n f} = \frac{d n_{s}}{d f} = \frac{8 π}{c^{3}} f^{2}

$\rho_{n\,f}=\cfrac{dn_s}{df}=\cfrac{8\pi}{c^3}f^2$

Because,

f λ = c

$f\lambda=c$

d f λ + f d λ = 0

$df\lambda+fd\lambda=0$

d f = - \frac{f}{λ} d λ

$df=-\cfrac{f}{\lambda}d\lambda$

and

ρ_{n λ} * \frac{d λ}{d f} = \frac{d n_{s}}{d λ} \frac{d λ}{d f} = \frac{d n_{s}}{d f} = ρ_{n f}

$\rho_{n\,\lambda}*\cfrac{d\lambda}{df}=\cfrac{dn_s}{d\lambda}\cfrac{d\lambda}{df}=\cfrac{dn_s}{df}=\rho_{n\,f}$

So,

ρ_{n f} = \frac{8 π}{λ^{4}} * \frac{λ}{f} = \frac{8 π}{λ^{3}} * \frac{1}{f}

$\rho_{n\,f}=\cfrac{8\pi}{\lambda^4}*\cfrac{\lambda}{f}=\cfrac{8\pi}{\lambda^3}*\cfrac{1}{f}$

ρ_{n f} = \frac{8 π}{c^{3}} * \frac{f^{3}}{f} = \frac{8 π}{c^{3}} f^{2}

$\rho_{n\,f}=\cfrac{8\pi}{c^3}*\cfrac{f^3}{f}=\cfrac{8\pi}{c^3}f^2$

It is a mistake to just substitute for

λ = \frac{c}{f}

$\lambda=\cfrac{c}{f}$

ρ_{n} = \frac{8 π}{λ^{4}} = \frac{8 π}{{(\frac{c}{f})}^{4}} = \frac{8 π}{c^{4}} f^{4} \neq ρ_{n f}

$\rho_n=\cfrac{8\pi}{\lambda^4}=\cfrac{8\pi}{\left(\cfrac{c}{f}\right)^4}=\cfrac{8\pi}{c^4}f^4\ne\rho_{n\,f}$

Have a nice day.

Monday, December 4, 2017