Comparing Spacetime and Thermodynamic State Space Manifolds

In summary: The difference between a mathematical model and a physical system is that the mathematical model is a description of the physical system, while the physical system is the thing being described. )In summary, the thermodynamic state space as manifold can only be conceived from an extrinsic point of view.
  • #1
cianfa72
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TL;DR Summary
Spacetime vs thermodynamic state space as manifolds from the intrinsic vs extrinsic point of view
Hi,

I don't know if it is the right place to ask for the following: I was thinking about the difference between the notion of spacetime as 4D Lorentzian manifold and the thermodynamic state space.

To me the spacetime as manifold makes sense from an 'intrinsic' point of view (let me say all the universe lives in such manifold) whereas the thermodynamic state space as manifold actually represents the thermodynamic state of a physical system and as manifold it makes sense only from an 'extrinsic' point of view.

In other words in the thermodynamic state space case we start from an 'ambient' vector space ##\mathbb R^n## in which a physical system is represented as a point in this space. Of course there exist constrains on the possible 'combination' of the state space variables/coordinates (e.g. the amount of Energy) so that the set of possible points as manifold is actually a submanifold of ##\mathbb R^n##.

What do you think about ? Thank you.
 
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  • #2
The entire point of using calculus on manifolds is that you do not require any embedding space. Be it spacetime or the thermodynamic state space.
 
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  • #3
Orodruin said:
The entire point of using calculus on manifolds is that you do not require any embedding space. Be it spacetime or the thermodynamic state space.
Yes, of course: from the calculus point of view there is no difference at all. My point was to understand if there is in principle a difference between the two.

To me the thermodynamic state space cannot be conceived from an intrinsic point of view (it was born in extrinsic view).

##n## in ##\mathbb R^n## is the state space dimension that is the number of thermodynamic variables used (e.g. ##p,V,t,S##). This number is not the number of system's degree of freedom -- i.e. the dimension of the submanifold representing the thermodynamic state of the physical system being described (basically the number of coordinates).
 
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  • #4
cianfa72 said:
To me the thermodynamic state space cannot be conceived from an intrinsic point of view (it was born in extrinsic view).
Why not? It is just a structure telling you about the possible states of a system. That there are some functions on this space that you can measure does not make it any less so.

cianfa72 said:
My point was to understand if there is in principle a difference between the two.
I’d say the big difference is the Lorentzian structure placed on spacetime.
 
  • #5
It seems to me that one should compare
  • the spacetime manifold [itex] M [/itex] with a thermodymamic space [itex] M_{therm} [/itex] (possibly coordinatized by [itex] (U,V,S) [/itex] )
and
  • (say) a spacelike hypersurface in [itex] M [/itex] associated with possible foliation of [itex] M[/itex]
    with a hypersurface [itex] (U,V,S(U,V)) [/itex] in [itex] M_{therm} [/itex] describing the states of a particular substance (like a an ideal gas).
Possibly interesting:
https://physics.stackexchange.com/q...m-of-thermodynamics-based-on-contact-geometry
 
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  • #6
robphy said:
the spacetime manifold [itex] M [/itex] with a thermodymamic space [itex] M_{therm} [/itex] (possibly coordinatized by [itex] (U,V,S) [/itex] )
If we assume a simple thermodynamic state space ##M_{therm}## with only two degree of freedom (i.e. ##M_{therm}## dimension is 2) we can assign to it just two coordinates, for instance ##(V,S)##. In this case the internal energy ##U## is not an indipendent coordinate, it is just a function defined on ##M_{therm}##.
 
  • #7
cianfa72 said:
If we assume a simple thermodynamic state space ##M_{therm}## with only two degree of freedom (i.e. ##M_{therm}## dimension is 2) we can assign to it just two coordinates, for instance ##(V,S)##. In this case the internal energy ##U## is not an indipendent coordinate, it is just a function defined on ##M_{therm}##.
(as I said above) for a particular substance.
Specifying U(S,V) or S(U,V) is essentially defining the equation of state of a particular substance.
 
  • #8
robphy said:
(as I said above) for a particular substance.
Specifying U(S,V) or S(U,V) is essentially defining the equation of state of a particular substance.
Ah ok, basically you are saying that ##M_{therm}## manifold is actually three dimensional (it is coordinatized by ##(U,V,S)## ).

Pick a particular system (substance) amounts to pick a function ##f(U,V,S)## on the manifold and equate it to zero. Then expliciting ##U## we can write ##f(U,V,S)=U - g(V,S)=0## i.e. ##U=g(V,S)##.

This way our system (substance) is defined as the submanifold ##U - g(V,S)=0## of the ##M_{therm}## manifold we started with.

Is that correct ? Thank you.
 
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  • #10
From links at post #6 I was reading about the theory of contact manifold. Pfaffian equations are introduced and a solution of it is actually a submanifold of the manifold we stared with such that the Pfaffian equation's one-form vanish on it.

Take for instance the following one-form field ##\omega = f(p,t)dp + g(p,t)dt## defined on the manifold and consider the Pfaffian equation ##\omega = f(p,t)dp + g(p,t)dt = 0##.

By definition ##h(p,t)=c## is a solution if ##dh## is a non-zero multiple (which is the integrating factor function) of the one-form ##\omega## above. On ##h(p,t)=0## submanifold we have ##dh=0## hence the value ##f(p,t)## and ##g(p,t)## functions take on those points should be zero (i.e. ##\omega## vanishes on that submanifold).

Is that right ? Thank you.
 
  • #11
cianfa72 said:
From links at post #6 I was reading about the theory of contact manifold. Pfaffian equations are introduced and a solution of it is actually a submanifold of the manifold we stared with such that the Pfaffian equation's one-form vanish on it.

Take for instance the following one-form field ##\omega = f(p,t)dp + g(p,t)dt## defined on the manifold and consider the Pfaffian equation ##\omega = f(p,t)dp + g(p,t)dt = 0##.

By definition ##h(p,t)=c## is a solution if ##dh## is a non-zero multiple (which is the integrating factor function) of the one-form ##\omega## above. On ##h(p,t)=0## submanifold we have ##dh=0## hence the value ##f(p,t)## and ##g(p,t)## functions take on those points should be zero (i.e. ##\omega## vanishes on that submanifold).

Is that right ? Thank you.
Why do you think it is right? Take a specific example. Say ##f=1## and ##g=1##, then they are never zero anywhere. The equation becomes ##dp+dt=0##, which integrates to ##p+t=c##, so your ##h(p,t)=p+t##.
 
  • #12
martinbn said:
Take a specific example. Say ##f=1## and ##g=1##, then they are never zero anywhere. The equation becomes ##dp+dt=0##, which integrates to ##p+t=c##, so your ##h(p,t)=p+t##.
Oh Yes, you're right I'm confused. So what does actually mean the one-form ##\omega## vanish ?
 
  • #13
cianfa72 said:
Oh Yes, you're right I'm confused. So what does actually mean the one-form ##\omega## vanish ?
One forms evaluate on vectors. It means that if you restrict the form ##\omega## to points on the surface ##h(p,t)=c## and evaluate on vectors tangent to the surface it will vanish. On other vectors it will not.
 
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  • #14
martinbn said:
One forms evaluate on vectors. It means that if you restrict the form ##\omega## to points on the surface ##h(p,t)=c## and evaluate on vectors tangent to the surface it will vanish. On other vectors it will not.
So the equation ##\omega = f(p,t)dp + g(p,t)dt = 0## actually means: search for ##h(p,t)=c## submanifolds of the given manifold such that the one-form ##\omega## evaluated on the vector space of tangent vectors to that submanifold (i.e on the vector subspace of the tangent space defined at each point of the given manifold) vanish.

Is that correct ? Thanks.
 
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  • #15
cianfa72 said:
So the equation ##\omega = f(p,t)dp + g(p,t)dt = 0## actually means: search for ##h(p,t)=c## submanifolds of the given manifold such that the one-form ##\omega## evaluated on the vector space of tangent vectors to that submanifold (i.e on the vector subspace of the tangent space defined at each point of the given manifold) vanish.

Is that correct ? Thanks.
Yes, but only in dimension 2. In general at each point the form will vanish on a ##n-1## dimensional subspace of the tangent space at that point. This way you get a family of such subspaces one for each point of the manifold. You can then ask if there are submanifolds whose tangent spaces are these. In general it will not be the case. There is a theorem of Frobenious that explains when this happens. In you example of a 2 dimensional manifold, you have a distribution of 1 dimensional subspaces. And those are always integrable. You need to find the integral curves to a vector field.
 
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  • #16
martinbn said:
Yes, but only in dimension 2. In general at each point the form will vanish on a ##n-1## dimensional subspace of the tangent space at that point. This way you get a family of such subspaces one for each point of the manifold. You can then ask if there are submanifolds whose tangent spaces are these. In general it will not be the case. There is a theorem of Frobenious that explains when this happens.
Yes, if ##\omega## is the given one-form field, Frobenious thereom says it is integrable giving you a ##n-1## submanifold if and only if
$$\omega \wedge d \omega = 0$$
My point is that the defintion of contact manifold ##M_{therm}## of odd dimension ##2k+1## requires ##\omega \wedge (d \omega)^k \neq 0## so the Frobenius condition actually does not hold.

As far as I can tell, there exist no submanifolds of dimension ##k+1, k+2...2k## such that ##\omega## evaluated on their associated subspaces of tangent vectors vanishes.

##\omega \wedge (d \omega)^{k+1} = 0## does means there exist a submanifold of dimension ##k## such that the above condition holds.
 
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  • #17
cianfa72 said:
Yes, if ##\omega## is the given one-form field, Frobenious thereom says it is integrable giving you a ##n-1## submanifold if and only if
$$\omega \wedge d \omega = 0$$
My point is that the defintion of contact manifold ##M_{therm}## of odd dimension ##2k+1## requires ##\omega \wedge (d \omega)^k \neq 0## so the Frobenius condition actually does not hold.

As far as I can tell, there exist no submanifolds of dimension ##k+1, k+2...2k## such that ##\omega## evaluated on their associated subspaces of tangent vectors vanishes.
You considered a 2 dimensional manifold in post #11, it cannot be a ##2k+1## dimensional.
 

Related to Comparing Spacetime and Thermodynamic State Space Manifolds

1. How are spacetime and thermodynamic state space manifolds related?

Spacetime and thermodynamic state space manifolds are both mathematical representations of physical systems. Spacetime describes the four dimensions of space and time, while thermodynamic state space represents the possible states of a thermodynamic system. Both manifolds use mathematical coordinates to describe the properties and behavior of these systems.

2. What is the difference between a manifold and a state space?

A manifold is a mathematical concept that describes a space that is locally similar to Euclidean space. It can have any number of dimensions and can represent physical systems such as spacetime or thermodynamic states. A state space, on the other hand, is a specific type of manifold that represents the possible states of a system. It is often used in thermodynamics to describe the properties of a system at different points in time.

3. How do these manifolds help us understand physical systems?

By using mathematical coordinates to describe physical systems, these manifolds allow us to visualize and analyze complex systems in a more simplified manner. They provide a framework for understanding the relationships between different variables and how they change over time. This can help us make predictions and better understand the behavior of these systems.

4. Can these manifolds be applied to other fields besides physics?

Yes, the concept of manifolds can be applied to various fields such as engineering, economics, and computer science. In each field, manifolds are used to represent different types of systems and can aid in understanding and analyzing their behavior. For example, in economics, manifolds can be used to represent the possible states of a market or economy.

5. Are there any limitations to using manifolds to describe physical systems?

While manifolds provide a useful framework for understanding physical systems, they are not without limitations. In some cases, the complexity of a system may make it difficult to accurately represent it using a manifold. Additionally, manifolds are based on mathematical models and may not fully capture the complexities and nuances of a real-world system. Therefore, it is important to use caution when applying manifolds to real-world situations.

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