How do i correctly go from a two variable function to a function of difference?

45 Views Asked by At

I would like to know how I can go from a two argument function $g(x_1,x_2)$ formally correct to a function of the difference of the parameters $g(x_1-x_2)=g(x)$ this seems to involve integration over $x_2$ but I have no idea what to look for to find this.

I have this example from Chaikin/Lubensky, but the steps in between and the reasoning are missing. This is about the pair distribution function $g$.

$$\langle n(x_1)\rangle g(x_1,x_2)\langle n(x_2)\rangle= \langle \sum \limits_{\alpha \neq \alpha'} \delta(x_1- x_\alpha) \delta(x_2 - x_{\alpha'})\rangle$$

and then for a homogeneous fluid we have

$$\langle n\rangle^2 g(x_1-x_2)= {1 \over V } \int dx_2 \langle \sum \limits_{\alpha \neq \alpha'} \delta(x_1- x_\alpha) \delta(x_2 - x_{\alpha'})\rangle$$

What happens to the left side and why? Please be detailed. If possible I would really like to know how this can be done formally correct.

1

There are 1 best solutions below

1
On BEST ANSWER

In physics, a homogeneous material or system has the same properties at every point. By translation invariance, one means independence of (absolute) position, especially when referring to a law of physics, or to the evolution of a physical system. If this is the case, then, all the functions which express properties of the material, cannot depend on the point where they are calculated.

In your specific calculation, the pair distribution function is by definition a function of two points $x_1$ and $x_2$. The only way $g(x_1,x_2)$ can depend on the two positions $x_1$, $x_2$ without breaking the assumption of symmetry, is if it depends on their distance.

Indeed, this is true in general: any function of space, in an homogeneous system, is wither a constant or a function of distances between points.

Going back to your calculation:

  • the function $n(x)$ must be the same at $x_1, x_2$ and in general at any $x$, so that $n(x)= \text{constant}:= n$
  • $g(x_1,x_2)$ must be a $g(|x_1-x_2|)$ because of the arguments above.
  • Then, what has happened in the r.h.s.? Because of homogeneity, the function on the right hand side

$$F(x_1,x_2):= \langle \sum \limits_{\alpha \neq \alpha'} \delta(x_1- x_\alpha) \delta(x_2 - x_{\alpha'})\rangle$$

cannot depend on $x_1,x_2$ but must depend on their difference. Moreover, the function will be equal to its average over the space, in one variable, always because of the homogeneity. The object that you wrote in the last step, on the r.h.s., is the spatial average of $f(x_2):=F(x_1,x_2)$ keeping $x_1$ fixed.

In the following steps, I can imagine that Chaikin and Lubensky show how this average can be written as a function of $x_1-x_2$ only.