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While solving a derivation in statistical mechanics I came across a result which was derived from expression: $p\propto \exp{( -\frac{C_{V}}{2k T^{2}} \Delta{T^{2}}- \frac{1}{2kT \kappa _{T} V}\Delta{V^{2}}})$

The result is which is written after writing one line which goes as: which shows that the fluctuations in T and V are statistically independent, Gaussian variables!. A quick glance at above equation yields: $\overline{(\Delta{T^{2}})}=\frac{kT^{2}}{C_{V}}, \overline{(\Delta{V^{2}})}=kT \kappa _{T} V$ Where bar over T and V represents average. I don't understand how statistically independent Gaussian variables led to this result and what are statistically independent Gaussian variables. Please help me understand this.

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A Gaussian distribution, often called normal distribution, is a (continuous) probability distribution of the form, $$ f_X\left(x\vert\mu,\,\sigma\right)\sim A\exp\left(-\left(x-\mu\right)^2/2\sigma^2\right) $$ for the normalization constant $A$, mean value $\mu$ and variance $\sigma^2$.

The author assumes that the temperature and volume distributions follow normal distributions (with variances of $k_BT/C_v$ and $k_BT\kappa_TV$) that are not correlated, so the product of these two is straight-forward: $$ f_T\cdot f_V\sim\exp\left(-\Delta T^2/2\sigma_T^2\right)\cdot\exp\left(-\Delta V^2/\sigma_V^2\right) $$ And with simple properties of exponentials, leads to the result given by the author.

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  • $\begingroup$ Sir in $f_T\cdot f_V\sim\exp\left(-\Delta T^2/2\sigma_T^2\right)\cdot\exp\left(-\Delta V^2/\sigma_V^2\right)$ shouldn't the denominator contain $2 sigma_V^2$ also if we compare $ f_X \left(x\vert\mu,\,\sigma\right)\sim A\exp\left(-\left(x-\mu\right)^2/2\sigma^2\right )$ with $p\propto \exp{( -\frac{C_{V}}{2k T^{2}} \Delta{T^{2}}- \frac{1}{2kT \kappa _{T} V}\Delta{V^{2}}})$ then variances should give the result which I required. $\endgroup$ – harshit singh Apr 28 '18 at 15:46
  • $\begingroup$ Sir, are averages of squares of fluctuations equal to variances. If this is the case then simply comparing with probability distribution formula should give the required result. $\endgroup$ – harshit singh Apr 28 '18 at 16:03
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If $X$ and $Y$ are jointly Gaussian (or in this case, $T$ and $V$), and uncorrelated, it can be shown that $$f_{XY}(x,y) = f_X(x)f_Y(y);$$ this assures independence.

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  • $\begingroup$ Pardon me, does $f_{xy}$ means partial derivative with respect to x and y. $\endgroup$ – harshit singh Apr 21 '18 at 4:51
  • $\begingroup$ @harshitsingh no, it's a probability distribution of random variables $X$, $Y$. $\endgroup$ – Kyle Kanos Apr 21 '18 at 11:34
  • $\begingroup$ OK. Please, can you explain how we got to that result in my problem using the hint you have provided. I can't seem to understand this. $\endgroup$ – harshit singh Apr 21 '18 at 12:36

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