Recent content by zeeshahmad

I have a past paper question from statistical physics: By assuming that ##\hbar^2 k^2=p^2##, I arrived at the result: The interparticle spacing, ##a^3 =\frac{V}{N}## is $$ a^3 >> e^{-\frac{p^2}{2mk_B T}} \lambda_{deB}^3$$ Is my assumption correct? and does the result complete the purpose of...

is it 1 when, theta is greater than Phi?

i get $$p(x)=\frac{1}{2\Phi}(\frac{D}{D^2+x^2})$$

oh I get it, $$\int{p(x) dx} = \frac{1}{2\Phi} (arctan(\frac{x+\delta x}{D})-arctan(\frac{x}{D}))$$ if that is right, we would then differentiate it.. but what about the \delta x ? How do we take care of that?

for $$arctan(\frac{x}{D})$$ it is $$\frac{D}{D^2+x^2}$$ but why do we need the derivative? also, for small delta x, the expression is 0, isn't it? (Edit:sorry for messing up the variables again, I typed small d's before)

$$\frac{1}{2\Phi} (arctan(\frac{x+\delta x}{D})-arctan(\frac{x}{D}))$$

the same expression

Thankyou for the detailed explanation :smile:

Yes. $$P(x, x+\delta x)=P(Dtan(\phi), Dtan(\phi+\delta\phi))$$ $$\int_x^{x+\delta x} \!{p(x)} = \int_{Dtan(\phi)}^{Dtan(\phi+\delta\phi)} <-not-sure-what-goes-in-there$$

Nice posting to you again :approve: Actually I have got the lecture notes, in which it tells the meaning, but I don't understand it: "Consider a distribution with average value μ and standard deviation σ from which a sample measurements are taken, i.e. \mu = \left\langle x \right\rangle...

Could someone explain the meaning of "Mean Squared Deviation"? Also, in <x1+x2+..xn> what is the meaning of the pointy brackets <..> ?

\int_{\theta}^{\theta + \delta \theta} p(\phi) d\phi =\int_{\theta}^{\theta + \delta \theta} \frac{1}{2\Phi} d\phi =\frac{d\delta}{2\Phi} ...:rolleyes:

All of that is correct, just that the last two statements are not in the question (not that I disagree with them), and \theta is only mentioned (1) in the hint and (2) in d = D tan(θ) But the hint says exactly as mentioned in OP.

\int_{\theta}^{\theta + \delta \theta} p(\phi) d\phi =\int_{\theta}^{\theta + \delta \theta} \frac{1}{2\Phi} d\phi =\frac{1}{2} \left[ ln(\phi) \right] (i can't put the limits here with latex) =\frac{1}{2} (ln(\theta + \delta \theta)-ln(\theta)) = ln(\sqrt{1+\frac{\delta...