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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.

Kuhlambo
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  • You do have some homogeneity assumption there, translating into a symmetry. Use the symmetry to show that functions of two positions can only depend on the difference, and that functions of one position aren't functions at all. – ACuriousMind Mar 04 '15 at 16:12
  • I have no idea how to do that. The problem is i don't even know were to start searching. – Kuhlambo Mar 04 '15 at 16:16
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    The first thing to realize is that in general you can't. Only a few function have this property, so the job is to show that the particular one you have should be one of that select group. And that goes to symmetry, as ACuriousMind says. – dmckee --- ex-moderator kitten Mar 04 '15 at 16:21
  • here i think it is just assumed that the property is fullfilled. Maybe i could write $g(x,0):=g(x)$ but this woul not explain the integration? What am i missing? – Kuhlambo Mar 04 '15 at 16:26
  • Answer (to yourself) the following questions: What does homogeneity mean? Is a non-constant $f(x)$ compatible with that meaning? Is a $f(x_1,x_2)$ that is not $f(x_1 - x_2)$ compatible with that meaning? – ACuriousMind Mar 04 '15 at 16:28
  • I was thinking for a two argument function to depend only on the difference in arguments it must be invariant to rotations of both arguments and translations of both arguments. So $ g(x_1,x_2)=g(x_1-x_2,0)$ will hold. Right? Not sure if homogeneaty must mean a constant function, that would not really make a function would it? But how does the integration come in to play i really don't get that. – Kuhlambo Mar 04 '15 at 16:37
  • Or does the integral have nothing to do with it at all and they just do it at this point to get to a certain forluma? I would really be glad to know this because it is causing me a really painfull headache at the moment. – Kuhlambo Mar 04 '15 at 16:56
  • So that's a yes? – Kuhlambo Mar 04 '15 at 17:19

1 Answers1

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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.

usumdelphini
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  • Thanks a lot for your answer. So that explains the $ 1 \over V$ on the right, since we are looking at the average. Can you give me any Tip on how to show that such a function of homogeneous space must be equal to a spatial average? – Kuhlambo Mar 05 '15 at 08:18