Correlation funciton of a system with translational symmetry

[Jeremy1986]

Hi guys,
i have been confused by one statement on the spatial correlation funciton in the statistical physics textbook. They say for a spatial correlation function f(x1,x2), where x1 and x2 are the coordinate of particle 1 and 2, if the system has translational symmetry, then f depends only on the distance between 1 and 2, not on the coordinates of the particles. That is, f(x1-x2)=f(|x1-x2|) in this case.
i mean this sounds reasonable, but i just can't tell why. i am wondering if there is a solid demonstration on this. Could anybody give me some explanation on this? Thanks a lot for your help!

 

[vanhees71]

If you have a translationally invariant situation, the correlation function must be translational invariant, because otherwise the system wouldn't be. That means that
$$f(\vec{x}_1,\vec{x}_2)=F(\vec{x}_1-\vec{x}_2).$$
It doesn't tell you that the situation is also isotropic. Only then you can argue further that
$$F(\vec{x}_1-\vec{x}_2)=G(|\vec{x}_1-\vec{x}_2|).$$

 

[Jeremy1986]

If you have a translationally invariant situation, the correlation function must be translational invariant, because otherwise the system wouldn't be. That means that
$$f(\vec{x}_1,\vec{x}_2)=F(\vec{x}_1-\vec{x}_2).$$
It doesn't tell you that the situation is also isotropic. Only then you can argue further that
$$F(\vec{x}_1-\vec{x}_2)=G(|\vec{x}_1-\vec{x}_2|).$$

Thanks vanhees71,
yes, you are right. the second relation holds only if the system has rotational symmetry at the same time. but i don't quite understand the first relation. the textbook gives a denmonstration of the first relation as follows
for f(x1,x2), if we translate the system by x2, then f(x1,x2)=f(x1-x2,0) ,which means that f(x1,x2) depends only on x1-x2.
Here, i think in this case the translational invariant means that the system stay unchanged after translating with any vector. this is different from the translational symmetry we talk about for crystal, because in crytals the system stays unchanged only if you translate it with certain vectors, the basic vertors. I don't know if i am right.

 

[vanhees71]

Sure, translation invariance means that the function must obey
$$f(\vec{x}_1+\vec{a},\vec{x}_2+\vec{a})=f(\vec{x}_1,\vec{x}_2)$$
for all $\vec{a} \in \mathbb{R}^3$. For a crystal that's of course not the case anymore on a scale, where you resolve the microscopic structure. Then it's only translationally invariant for some discrete set of vectors. I'd call this periodic rather than translationally invariant.

 

[Jeremy1986]

Sure, translation invariance means that the function must obey
$$f(\vec{x}_1+\vec{a},\vec{x}_2+\vec{a})=f(\vec{x}_1,\vec{x}_2)$$
for all $\vec{a} \in \mathbb{R}^3$. For a crystal that's of course not the case anymore on a scale, where you resolve the microscopic structure. Then it's only translationally invariant for some discrete set of vectors. I'd call this periodic rather than translationally invariant.

with your kind help, i got better understanding of spatial correlation function. but i have a final puzzle,

for solids or liquids, especially the amorphous solid, they use this kind of spatial correlation function, the so called pair distribution function. in the figure below, there is a definition of such pair distribution function, notice the red-underlined parts

So in this case, of amorphous solids or homogeneous fluid, why the translation invariance holds?

ps:i use an external website to upload the image, i can see the image myself, and i don't know if it works for you.

 

[Jeremy1986]

Sure, translation invariance means that the function must obey
$$f(\vec{x}_1+\vec{a},\vec{x}_2+\vec{a})=f(\vec{x}_1,\vec{x}_2)$$
for all $\vec{a} \in \mathbb{R}^3$. For a crystal that's of course not the case anymore on a scale, where you resolve the microscopic structure. Then it's only translationally invariant for some discrete set of vectors. I'd call this periodic rather than translationally invariant.

so do you mean a solid is macroscopic translational invariant? if we translate it with some distance in macroscopic scale, that solid still stay as it just changed its position. can we call it macroscopic translational invariant? but translational invariant needs both macroscopic or microscopic, so the above relation f(x1-x2)=f(x1-x2) does not hold. am i right?