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I am following a course which contains a part in statistical thermodynamics. One of the questions involves the partition function $Q_N$. I could not figure out the answer of the question myself, so I had a look in the solutions provided by the lecturer. In these solutions he makes use of:

$$\vec{s}=\frac{\vec{r}}{V^\frac{1}{3}} \iff d\vec{s}=\frac{d\vec{r}}{V}$$

I don't understand how these two are equivalent. I would think the following would hold:

$$\vec{s}=\frac{\vec{r}}{V} \iff d\vec{s}=\frac{d\vec{r}}{V}$$

The power $\frac{1}{3}$ is confusing me.

Further information, if required:

$$Q_N=\idotsint\limits_V \exp(-\Phi(\vec{r_1},...,\vec{r_N})/kT)\:d\vec{r_1}\dots d\vec{r_N}$$

$V$ is the volume of the system

$\vec{r_i}$ is an infinitesimal small volume associated with particle $i$

$\vec{s_i}$ is a newly-defined variable which is supposedly handy/needed

Note: I am a master's student in Chemistry, with a bit of a background in physical chemistry, so I would say my mastering of mathematics is (in comparison to mathematicians or physicists) rather poor.

Thomas Russell
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Stijn
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2 Answers2

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If you multiply all the sides of a cube by the same constant $\lambda>0$, the volume will be multiplied by $\lambda^3$. Apply this to an infinitesimal cube.

Siminore
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  • Thank you for your answer. Unfortunately I still do not fully understand... $V$ is simply the volume of the system - a constant, so why would it go from $V^\frac{-1}{3}$ to $V$ when we go from $\vec{s}$ to $d\vec{s}$? – Stijn Oct 30 '14 at 15:29
  • $d\vec{s}=ds_1 , ds_2 , ds_3$. – Siminore Oct 30 '14 at 15:50
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I think that $\mathrm{d}\vec{r}$ isn't actually being used here as the differential of $\vec{r}$, and that's what's confusing you. i.e. the notation really means $\mathrm{d}r_x \, \mathrm{d}r_y \, \mathrm{d}r_z$

  • So, if I understand correctly: $d\vec{s}=(\frac{dr_x}{V^\frac{1}{3}})(\frac{dr_y}{V^\frac{1}{3}})(\frac{dr_z}{V^\frac{1}{3}})=\frac{d\vec{r}}{V}$ (Im not sure why the maths markup is acting up) – Stijn Oct 30 '14 at 15:39