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The question may sound silly. If it is I'm sorry for it but I just couldn't find an answer anywhere else.

I have just learned about vector spaces and their properties and on the other hand have also started with Lagrangian mechanics. The author writes: "The configuration of the system of N particles, moving freely in space, may be represented by the position of a single point in 3N dimensional space, which is called the configuration space of the system."

My questions here are how is it possible for us to visualize the position of say 700 particles using just a single point in 2100 dimensional space? I cannot make any sense of it. Since there are no constraints, for every particle we are adding up 3 dimensions; What is the advantage in doing so?

And is this in anyway related to vector spaces?

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  • $\begingroup$ What about e.g. the Wikipedia article is unclear to you? $\endgroup$ – ACuriousMind Aug 27 '15 at 15:16
  • $\begingroup$ I believe the answer is no, because it doesn't make sense to add two different configurations to each other, and addition of elements is necessary for the space to be considered a vector space. $\endgroup$ – march Aug 27 '15 at 16:28
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My questions here are how is it possible for us to visualize the position of say 700 particles using just a single point in 2100 dimensional space?

Think of it as a map. You can read a map and loom around and you can learn how to relate the common points. So of you have 700 particles there are 2100 scalars. So have a giant 2100 dimensional vector space. You can place the first three to be the xyz of the first particle then the next three for the xyz of the second particle and so on.

Just think of it as a map given one you find the other one by finding the correspondence. So once you get hood at moving between the two you are fine. You can start with a simpler example. Three particles on a line. Then you find that the plane x=y is two of them touching, as are the planes x=z and y=z and the line x=y=z is all three touching.

What is the advantage in doing so?

Lagrangian mechanics would be the biggest reason.

And is this in anyway related to vector spaces?

Yes. It doesn't have to be for instance if there are constraints. But unlike one of the comments you can consider the configuration at two times and subtract them yo get a displacement in configuration space and then you can scale that by the time interval and thus get a derivative. Derivatives often are related to vector space structures.

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  • $\begingroup$ You mean to say that configuration do behave as vector spaces? If so what will adding of two configurations give me if subtraction gives displacement? And are inner products associated with them? $\endgroup$ – Weezy Aug 30 '15 at 12:52
  • $\begingroup$ @eoshah Some configuration spaces behave as vector spaces. Adding them isn't as physical as subtracting because when you add them the answer depends on which configuration is the origin configuration If you treat the configuration as a displacement from the origin then the sum is the sum of the displacements. And yes there can be an inner product again that helps for taking the limit to get that derivative. $\endgroup$ – Timaeus Aug 30 '15 at 13:56
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They are not related structurally: Configuration space is a manifold which in general has no vector space structure. For example $\mathbb{R}$, the configuration space of a free particle moving on a line can be viewed as a vector space (you can sensibly "add" two configurations to get a new one and so on), but if you constrain it to move on a circle this structure is lost.

There is an interesting link between the two though which appears in quantum mechanics. One can naturally associate a vector space to a classical configuration space: If your system has configuration space $M$, one can take the vector space to be $L^2(M)$, the space of square integrable functions on $M$. This is the natural setting of the quantum version of this classical system.

(Thanks to ACuriousMind for pointing out the very fundamental flaw in the first version of this answer!)

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  • $\begingroup$ Wrong, the symplectic manifold you are talking about is the phase space, which is the cotangent bundle of the configuration space. Quantization takes place on the phase space, but indeed obtains the wavefunctions, broadly speaking, as the square-integrable functions on the original configuration space. $\endgroup$ – ACuriousMind Aug 27 '15 at 19:46
  • $\begingroup$ Ah I see! I didn't think about this carefully enough. Since the relation between position and momentum plays an important role in quantum mechanics, phase space (the total space of $M$ and $T^*M$) is what's quantized... I'll edit the answer. $\endgroup$ – childofsaturn Aug 27 '15 at 19:51
  • $\begingroup$ Okay so what I could understand from your answer was that configuration spaces can behave like vector spaces but this is not the general case.? Also I couldn't understand the 2nd paragraph and the comments to this answer since I'm relatively new to quantum mechanics. $\endgroup$ – Weezy Aug 30 '15 at 12:57
  • $\begingroup$ @eoshah Right, they can be vector spaces, but they don't have to be (for example when there are constraints). It is probably better to just learn what they are for many examples. It's also like $\mathbb R^2$ if you use polar coordinates of isn't as easy to add vectors though it still is a vector space. Sometimes it is about what is easier rather than what is possible. $\endgroup$ – Timaeus Aug 30 '15 at 14:02

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