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Suppose I have a $6N$ dimensional space with points looking like this:

$$(r_x^{(1)},r_y^{(1)},r_z^{(1)}, p_x^{(1)}, p_y^{(1)}, p_z^{(1)},...,r_x^{(N)},r_y^{(N)},r_z^{(N)}, p_x^{(N)}, p_y^{(N)}, p_z^{(N)})$$

This is a phase space for a system of $N$ particles. In this space, I define a set by the following conditions:

$$\sum_{k=1}^{N} \left( \left[ p_{x}^{(k)} \right] ^2 +\left[ p_{y}^{(k)} \right]^2 +\left[ p_{z}^{(k)} \right]^2 \right) =const>0 .$$

$$0\le r_{x}^{(k)} \le l ; \quad 0\le r_{y}^{(k)} \le l ; \quad 0\le r_{z}^{(k)} \le l ; \quad k=1,2,...,N; \quad l=const>0.$$

What is the dimension of this object? How to calculate its "volume"? I was thinking that the first equation specifies a sphere in $3N$ dimensions, which itself is a $3N-1$ dimensional object, but now I'm confused on how to treat the remaining conditions.

PS. Would it be helpful to first consider a similar problem, but with the first condition replaced by

$$0<\sum_{k=1}^{N} \left( \left[ p_{x}^{(k)} \right] ^2 +\left[ p_{y}^{(k)} \right]^2 +\left[ p_{z}^{(k)} \right]^2 \right) \le const?$$

I've seen such procedures done in some textbooks but I don't really understand why not work with the problem directly.

Update: What would be the units of this set be? I'm assuming the $r$'s have units of length and $p$'s of momentum.

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Would it help if I first considered a higher dimensional object, by for instance replacing the equality in the first condition by an inequality? –  DepeHb May 21 '13 at 22:35
    
Do you mean to have the restriction $r_x^{(k)} \geq 0$ on $r_x$ but not $r_y$ and $r_z$? –  Adam Saltz May 21 '13 at 22:40
    
Then you should update the question. It's confusing to have two different versions in the question and the comments, and peope shouldn't have to read through the comments to understand the question. –  joriki May 21 '13 at 23:30
    
Ok, I updated the question. –  DepeHb May 22 '13 at 0:19

2 Answers 2

up vote 2 down vote accepted

Anything specifying a range on a continuous parameter does not typically reduce the dimensionality of the system. If there exist points 'in the middle of' all the ranges, they are locally unaware of the boundary, and hence locally the dimensionality is unaffected.

You have exactly one proper constraint on the motion, which is the total energy condition $\sum \mathbf p^2=E^2=\text{const.}$ There are $6N$ degrees of freedom. Hence your system is $6N-1$ dimensional.

The phase space volume is currently infinite since $r_y,r_z$ can take arbitrarily negative values. If you bound them below by zero then you have a $3N-1$ dimensional sphere of radius $E$ mutiplied by a $3N$ dimensional cube of side length $l$. The total phase space volume is the product of the volumes of these two objects, since they are independent. Note that by the volume of the sphere I mean the surface area in $3N$ dimensions, the intrinsic volume of the sphere as a $3N-1$ ddimensional manifold.

The units of the volume are $(\text{momentum})^{3N-1} (\text{length})^{3N}$

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Thanks for your answer. I corrected the question. The lower bound of zero was meant to be there on all the position variables. I also updated the question with another thing that's puzzling me. –  DepeHb May 22 '13 at 8:29
    
I've edited my answer. Hope this helps. Try not to edit questions by expanding them a lot, it feels like moving the goalposts. Try to ask questions in comments/chat or ask new questions if it's more substantial. –  Sharkos May 22 '13 at 8:49
    
Thanks for your answers and tips. –  DepeHb May 22 '13 at 9:18

I think it'll be helpful to first think of 1 dimensional physical space which for your case of $N$ non interacting particles is actually $2N$ dimensions (position and momentum). In this space, the conditions read: $$p_i = c_i,\ r_i \in[0,l]$$ So the dimension of this space is simply $N$ (since $p$ is constant i.e. zero dimensional). Now let's move to $2D$ physical space ($4N$ phase space). This time we have: $$p_{i1}^2 + p_{i2}^2 = c_i^2,\ r_i \in[0,l]$$ We only have on constraint, so this is a $4N-1$ dimensional space. What is this space? if we consider one particle and ignore one of the $r_i$'s, this is a hollow cylinder in $3D$ space, embedded in a $4D$ space. Now let's return to physical $3D$ space, It should be obvious that the dimension of the space is $6N-1$.

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Thanks for your answer. I also updated the question with another thing that's puzzling me. –  DepeHb May 22 '13 at 8:29

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