# Configuration space vs physical space

1. Jan 15, 2009

### Demystifier

The question (or puzzle) that I want to pose essentially belongs to classical (not quantum) physics. Nevertheless, there is a reason why I post it here on the forum for quantum physics, as I will explain at the end of this post.

As a simple example, consider the following Hamiltonian:
$$H=\frac{p_1^2}{2m}+\frac{p_2^2}{2m}$$
What does this Hamiltonian describe? Is it one free particle moving in two dimensions, or two free particles moving in one dimension? Clearly, this Hamiltonian describes both. But then, how to distinguish between these two physically different cases? Is there a FORMAL (not purely verbal!) way to distinguish the configuration space from the "physical" space?

This is essentially a classical question, but there are two reasons why I ask this question here:
First, people here are much more clever than people on the forum for Classical Physics.
The second, more important reason is that, although essentially classical, the motivation behind this question is actually quantum. Namely, the idea is that nonlocality of quantum mechanics could be avoided by noting that, ultimately, QM is nonlocal because it is formulated in the configuration space rather than in the "physical" space. For if the configuration space is reinterpreted as a "true physical" space (whatever that means), then QM becomes local in that "true physical" 3n-dimensional space, where n is the number of particles. But then the problem is to explain why the world looks to us as if it was only 3-dimensional (for simplicity, I ignore relativity). To understand that one needs first to understand what exactly makes the standard 3-dimensional physical space more "physical" than the 3n-dimensional configuration space, which is my motivation to ask the question above.

2. Jan 15, 2009

### malawi_glenn

I don't even know what "physical" space is, can you definie it FORMALLY?

3. Jan 15, 2009

### Demystifier

If I could, I would not pose the question above in the first place. :tongue2:
Indeed, the problem can be reduced to the problem of finding the appropriate formal definition of the "physical" space, given that we already understand intuitively what the physical space should be. (You know, the 3-dimensional space on the top of which everything else seem to exist ...)

4. Jan 16, 2009

### ThomasT

3D Euclidian space is the space that corresponds to our sensory apprehension of reality. I don't know if that qualifies as a formal definition of physical space, but sensory data are the criteria by which we evaluate claims about reality. Anyway, it seems reasonable to assume that the reality that we can't directly sense is also 3D Euclidian. Maybe there isn't any deep explanation for this, any more than there is an explanation for the universal scale expansion that is deeper than the expansion itself.

5. Jan 16, 2009

### nughret

This is a question of biology not physics.

6. Jan 16, 2009

### Tac-Tics

When you have a 3n-D configuration space, you're really working with a product of n spaces joined together. It simple describes the number of variables required to completely specify the state of your system.

Similarly, in statistics, if you take the height of a thousand people in a city, your resulting data will be a 1000-D space. You can see the mathematics treating it this way when you look at the standard deviation, which is the shortest distance, given by the Euclidean norm (ie: root of the sum of squares), from the point in 1000-D space describing your sample to the "average" line described by {(t, t, t, ..., t) in R^1000 | t in R}. Does this mean that you can "create" extra dimensions by polling more people. Not really.... It's just a mathematical model.

Similarly, a (classical) single particle in space must be specified with three reals. If you have n particles, you need three variables each. That doesn't change the playing field they are in, though, since each particle lives in a 3D world, not a 3n-D world.

7. Jan 19, 2009

### strangerep

I've resisted the temptation to attempt an answer in case this was one of your
rhetorical questions/puzzles for which you already have an answer. But the
thread has become idle without any resolution so I'll risk making a fool of myself...

Consider an idealization where an elementary system corresponds to a unirrep
of some dynamical Lie algebra. For brevity, let's say it's some sort of symplectic
algebra with Hamiltonian, etc, etc. Some of the algebra's basis elements
correspond to "position" or "configuration". Let's say there's 3 linearly independent
of these (i.e., considering the non-relativistic case).

Depending on the details of the algebra there'll be some Casimirs, and these together
with one other generator classify the possible representations and hence quantum
numbers. In a general dynamical algebra, the position generators and Hamiltonian
are likely participants in (some of) these Casimirs.

A single elementary system is a bit boring. We can find a canonical transformation
that puts it at rest (or some other canonical state, depending on the details of the
algebra). So let's consider a tensor product of 2 such systems and demand that it
also be a unirrep of the same dynamical algebra. The basic generators for each
system commute, but when we examine the quadratic and higher Casimirs for
the combined system we find various constraints on how two systems can
tensor together to get another valid unirrep. (This is reverse-analogous to the
way we get non-trivial Clebsh-Gordan coefficients when we analyze coupling
between two sets of angular momentum generators. The $J^2$ Casimir
makes the decomposition quite non-trivial.)

Now consider a tensor product of 3 elementary systems, #1,#2,#3, that we
require to be a unirrep of the same dynamical algebra. Things get very messy.
But in this case, system #1 can have a "physical space" (i.e., a subset of generators)
in which the interaction and behaviour of the #2 $\otimes$ #3 cluster can be described.
Similarly, each of the other two can have their own "physical spaces". But in
general the three physical spaces do not coincide (cf. the Unruh effect and Rindler
wedges, etc, in accelerating situations).

But you wanted a more rigorous way to distinguish physical space from
configuration space. So I suggest the generators of physical space
corresponds to the sum of all the position generators of the subsystems,
and the "configuration" aspect of the rest of the dynamical behaviour corresponds
to differences between generators of all the different clusters one can construct
that decompose the whole system. E.g., for the 3-subsystem case the (canonical)
physical space corresponds to
$$X_{PHYS} ~:=~ X_{(1)} + X_{(2)} + X_{(3)}$$
where the X's represent vector quantities.

The various other combinations, e.g., $$X_{(1)} - (X_{(2)} + X_{(3)})$$,
then describe configuration aspect(s) of the relative dynamics.

Such a description is not unique, of course. In general, an ideal observer is one
of the subsystems and defines an observer-centric "physical space" via interactions
with other subsystems (e.g., radar method). But I presume you wanted something
more akin to the spacetime background used in relativity.

So, (now that I've possibly exposed myself to a public spanking), what is
your answer to the puzzle?

Last edited: Jan 19, 2009
8. Jan 20, 2009

### Demystifier

Strangerep, this time I really do not have my answer.

Concerning your attempt, it is certainly the most serious one so far. Yet, I feel that it is not really satisfying to me, though I need more time to understand why. :tongue2:

9. Jan 20, 2009

### dx

Can't you rule out the case of two particles in one dimension by noticing that there is no interaction term in the Hamiltonian?

10. Jan 20, 2009

### Demystifier

No. Particles may be able to travel through each other without a recoil or any other interaction.

11. Jan 20, 2009

### dx

Maybe classical relativistic mechanics can give some insight into this. Influences travel through "physical space" at the speed of light, and not necessarily through configuration space. If we have two configuration space coordinates, and the local dynamical evolution of one coordinate depends on the retarded dynamical evolution of the other, then I think we can tell that these are coordinates of two different particles rather than two coordinates of a single particle. So I'm guessing we can see the physical space in the configuration space by analyzing how the coordinates influence each other and the general causal structure generated by the Hamiltonian.

12. Jan 20, 2009

### hellfire

May be it is meaningful to ask first about the relativistic theory. Shouldn't be a constraint between the components of the linear momentum in case that you are talking about one single particle?

13. Jan 20, 2009

### jostpuur

IMO the key is in the interaction terms, and in some kind of resulting "effective dimension". For example, suppose system is described by a following Lagrange's function

$$L:\mathbb{R}^{3N}\times\mathbb{R}^{3N}\to\mathbb{R}$$

$$L(x,\dot{x}) = \sum_{k=0}^{N-1} \frac{1}{2}m_{3k}\big(\dot{x}_{3k}^2 + \dot{x}_{3k+1}^2 + \dot{x}_{3k+2}^2\big) -\underset{k<l}{\sum_{k,l=0}^{N-1}} K_{k,l}\big((x_{3k}-x_{3l})^2 + (x_{3k+1}-x_{3l+1})^2 + (x_{3k+2}-x_{3l+2})^2\big)^{\alpha_{k,l}}$$

In the end, there is no precise way of telling if this should be a one particle in a 3N-dimensional space, N particles in a 3-dimensional space, or 3N particles in one dimension. However, from the form of the Lagrangian one sees that clearly the interpretation of 3 dimensions is somehow favored.

IMO the same effect occurs in the reality. There is no fundamental answer to a question whether our universe contains extremely large number of particles in 3 dimension, or some small number of particles in an extremely large dimensional space. It is form of the interactions which make the universe appear as if 3 dimensional.

14. Jan 20, 2009

### dx

A more precise way of saying what I said above would be: The configuration space coordinates $q_1$ and $q_2$ are coordinates of the same particle if they have the same causal past in the sense of special relativity.

15. Jan 20, 2009

### jostpuur

I just realized that entanglement actually seems to favor the interpretation of extremely large amount of dimensions, in a sense. (edit: hmhmh... but was this what Demystifier already explained in the opening post.... it mentions locality, not entanglement, but perhaps this is the same stuff...)

Consider a wave function $\psi:\mathbb{R}^2\to\mathbb{C}$ describing single particle in a two dimensions. It is not mysterious at all, that coordinates $x_1$ and $x_2$ are correlated in the amplitudes. However, consider a wave function $\psi:\mathbb{R}^2\to\mathbb{C}$ describing two non-interacting particles in one dimension. If $\psi$ does not separate into product of $x_1$ and $x_2$ two depending parts, we get mysterious entanglement, where measurement of the position of one particle affects the position of the other one.

The entanglement start appearing less mysterious when dimensions are increased and number of particles decreased.

Last edited: Jan 20, 2009
16. Jan 21, 2009

### Demystifier

Yes, that was my original motivation too.

17. Jan 21, 2009

### atyy

I agree.

18. Jan 21, 2009

### Demystifier

I don't think that relativity is essential, because we see in our everyday lives what the "physical" space is, without being aware of relativistic effects. Still, your idea leads me to another idea: That the difference between two spaces cannot be seen on the level of point-particles, but only on the level of fields. For example, the field configuration describing one point-particle in two dimensions is
$$\delta(x_1-y_1(t))\delta(x_2-y_2(t))$$
while that describing two point-particles in one dimension is
$$\delta(x-y_1(t)) + \delta(x-y_2(t))$$
Both are defined by two functions $$y_1(t)$$ and $$y_2(t)$$, and yet they look mathematically different. One is the product of two delta functions, while the other is a sum of two delta functions.

But there is also a problem with that. The many-particle wave function is mathematically a field in 3n dimensions, which would imply that physical space of QM is indeed 3n-dimensional, making QM local in the physical space. That is good, but why then we still see the universe as if only 3 dimensions were really physical? This may be related to the fact that quantum effects are not visible on the macroscopic level, which suggests that it is the phenomenon of decoherence that is responsible for the illusion of 3 dimensions in our everyday lives. But then again, how can we see from our fundamental equations that on the macroscopic level the universe will appear as if it had precisely 3 dimensions? And what exactly these fundamental equations are?

Last edited: Jan 21, 2009
19. Jan 21, 2009

### Demystifier

So, are you saying that the known physical laws would allow the existence of living beings that would think that they live in, e.g., 5 "physical" dimensions? I don't think so.

20. Jan 21, 2009

### hellfire

Consider the lagrangian of a free relativistic particle $$\mathcal L = m \sqrt{- \dot q^{\mu} \dot q_{\mu}}$$. How does the problem arise in such a case?

21. Jan 21, 2009

### pellman

I don't quite see the problem. Isn't this simply a case of noting that math is not reality, rather it models reality? and that the same math can model different physical systems? Hence you can't expect to find the physical system identifiable in the math.

How does this question differ from the questions,

Does the Lagrangian $$L=\frac{1}{2}m\dot{q}^2-\frac{1}{2}kq^2$$ describe a bouncing spring or an electric circuit?

Does $$3+4=7$$ describe what happens when you put 3 apples with 4 apples or what happens when you put 3 oranges with 4 oranges?

22. Jan 21, 2009

### Demystifier

Let me see!
The action is
$$m \int d\tau \sqrt{- \dot q^{\mu} \dot q_{\mu}}$$
If this action is all we know about the system, then we do NOT know that the action is also equal to
$$m \int d\tau$$
In other words, we do NOT know that $$\tau$$ is equal to the proper time and that solutions of the equations of motion should satisfy
$$- \dot q^{\mu} \dot q_{\mu}=1$$
Instead, $$\tau$$ is just an auxiliary affine parameter without an explicit physical interpretation. This, by itself, is not yet a problem.
However, now consider a particular solution of the equation of motion
$$q^0=\tau, \; q^1=\tau, \; q^2=\tau, \; q^3=\tau$$
This solution may well be interpreted as 4 NON-relativistic particles moving in 1 spatial dimension, and $$\tau$$ may well be interpreted as an independent scalar parameter, such as the Newton time.

Perhaps you want to add an additional constraint that removes such unphysical solutions. Depending on how exactly you enforce such a constraint, I may discuss what, if any, problem will remain.

23. Jan 21, 2009

### Demystifier

This is an excellent point!
Still, if you see the quantum motivation that I explained in the first post, you might see that it is not that trivial. The quantum formalism may also be interpreted as a local theory in 3n dimensions. Such an interpretation may be a solution of the nonlocality problem related to the entanglement in QM, which suggests that such an interpretation could be more than just a play with mathematical symbols. There could be something PHYSICAL about such an interpretation.
So basically, I want to solve the problem which, as you say, is not really a problem at all, because this solution may help me to solve another, more serious problem (the problem of non-locality in QM).

24. Jan 21, 2009

### pellman

I can't help answer it yet myself.

But I can't shake the feeling that is related to this question I posted a couple of days ago https://www.physicsforums.com/showthread.php?t=285051

Here I was posing a question about how a relativistic N-body problem need 4N coordinates (or does it? that's part of the question) with clock for each particle, rather than 3N+1. DaleSpam's answer referred to how one slices up the manifold.

The problem, in my mind, was that the dynamical equations --in terms of coordinates-- doesn't refer to the manifold. There is no information about the manifold at all in the equations.

I'll be watching this thread intently.

25. Jan 21, 2009

### Demystifier

That's interesting too. I think you need 4n coordinates in order to maintain the manifest Lorentz covariance. Indeed, it is just a special case of the general rule that if you want to maintain a manifest symmetry, then you need to introduce some redundant degrees of freedom. Gauge fixing removes the redundancy, but also the manifest symmetry.

By the way, you may find interesting to see that recently I have used a 4n-formalism to show that nonlocality of QM is compatible with Lorentz invariance, even if explicitly nonlocal Bohmian hidden variables are involved:
http://xxx.lanl.gov/abs/0811.1905 [accepted for publication in Int. J. Quantum Inf.]

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook