- #1
AStaunton
- 100
- 1
want to show v_peak=sqrt(2kT/m):
the point here is that have a maximum when df/dv=0
max-boltz equation again is:
f(v)=4\pi(\frac{m}{2\pi mkT})^{\frac{3}{2}}v{}^{2}e^{-\frac{mv^{2}}{2kT}}
and taking derivative and setting=0 and then dividing out any constant expressions gives:
\frac{df}{dv}=(2v)e^{-mv^{2}/2kT}-(2kT)e^{-mv^{2}/2kT}=0\implies e^{-mv^{2}/2kT}(2v-2kT)=0
I then divided out the (2v-2kT) term and took natural log of both sides:
-\frac{mv^{2}}{2kT}=1\implies v=\sqrt{\frac{-2kT}{m}}
clearly the minus sign should not be in there...besides that the answer is correct...can someone please tell me how I ended with an extra minus sign within the squareroot?ie.what mistake did I make do produce a minus sign along the way...
Another question about the above is:
I divided out the (2v-2kT/m) term and then proceeded to take logs and so on...however since this term also contains a v, I expected that I should if I wanted be able to divide out the exponential term instead, leaving me with:
2v-2kT/m=0
and again solving this equation for v should give me another way to find v_peak, however from this equation v=kT/m which is not the v_peak expression I wanted...can someone tell me why this is so?
the point here is that have a maximum when df/dv=0
max-boltz equation again is:
f(v)=4\pi(\frac{m}{2\pi mkT})^{\frac{3}{2}}v{}^{2}e^{-\frac{mv^{2}}{2kT}}
and taking derivative and setting=0 and then dividing out any constant expressions gives:
\frac{df}{dv}=(2v)e^{-mv^{2}/2kT}-(2kT)e^{-mv^{2}/2kT}=0\implies e^{-mv^{2}/2kT}(2v-2kT)=0
I then divided out the (2v-2kT) term and took natural log of both sides:
-\frac{mv^{2}}{2kT}=1\implies v=\sqrt{\frac{-2kT}{m}}
clearly the minus sign should not be in there...besides that the answer is correct...can someone please tell me how I ended with an extra minus sign within the squareroot?ie.what mistake did I make do produce a minus sign along the way...
Another question about the above is:
I divided out the (2v-2kT/m) term and then proceeded to take logs and so on...however since this term also contains a v, I expected that I should if I wanted be able to divide out the exponential term instead, leaving me with:
2v-2kT/m=0
and again solving this equation for v should give me another way to find v_peak, however from this equation v=kT/m which is not the v_peak expression I wanted...can someone tell me why this is so?