Understanding Differential Geometry in Thermodynamics

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In summary, the exterior product is defined in differential geometry to have an algebraic expression for oriented volume, with the determinant being the simplest expression for oriented volume from vectors.
  • #1
vincentchan
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What is exterior product?
 
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  • #2
vincentchan said:
What is exterior product?

The short version is,where else,than here

NOTE:The ordinary cross product between vectors in E_{3} is a particular case of exterior/wedge product.

Daniel.
 
  • #3
I know its definition, my question is why is it define this way, ie. v^w=-w^v and v^v = 0
 
  • #4
vincentchan said:
I know its definition, my question is why is it define this way, ie. v^w=-w^v and v^v = 0

Nope,that is his definition.It's like asking why is the derivative defined by
[tex] f'(x)=:\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} [/tex]
...That's the definiton and nothing more.Its properties follow as logical consequences from its definition.You can define other form of products,but the definition of exterior product is the same,no matter what happens...

Daniel.
 
  • #5
May I moldify my question the other way, What is exterior product use for and when will we use it (for a real physical application)?
 
  • #6
vincentchan said:
May I moldify my question the other way, What is exterior product use for and when will we use it (for a real physical application)?

AAAAAAAAAAAhhhhhh...Yous should have said that from the beginning.Depends on the physics courses u take and the course material the student is being presented with.Any application into phyiscs of differential geometry will use "manifolds,cotangent spaces,p-forms and wedge products".General relativity,classical mechanics,quantum mechanics,thermodynamics.However,i say again,it's possible that these mathematical notions not be encountered:
For example:Hamiltonian dynamics can be taught without the studying the symplectic manifold called "phase space".
General Relativity can be taught without geometrical description.Take for example,Dirac's course on GR.No manifolds,no p-forms,no Hodge duality,no nothing...
Geometric quantization can be taught only if you're taught classical dynamics and symplectic manifolds,like "phase-space".
Thermodynamics can be taught without making too much mathematical analysis of the differential forms like differential heat and differential work.

Again,it depends on the teacher.If he keeps a mathematically rigurous course,you'll see them.If not,not.

Daniel.
 
  • #7
dextercioby said:
General Relativity can be taught without geometrical description.

Interesting, yet hard to believe...how is this done...please enlighten us...

Geometric quantization can be taught only if you're taught classical dynamics and symplectic manifolds,like "phase-space".

mmm, it's a little bit more then that...Besides it is even better if you do not have any notion on the way classical dynamics work because it must be easier to not get confused...But your list is faaaaaaaaaaar from complete that it is wrong. Also incluse GTR + QFT Thus also QM and special relativity...Ever saw a paper on LQG, check out my journal and you shall be convinced ?

Thermodynamics can be taught without making too much mathematical analysis of the differential forms like differential heat and differential work.

Then how ? No partition functions, no total differentials (how are you going to incorporate reversal processes and therefore hysteresis?) ? What part of thermodynamics are you planning on teaching ?



CONCLUSION : i am glad not to be a student of yours :wink:

marlon
 
  • #8
marlon said:
Interesting, yet hard to believe...how is this done...please enlighten us...

I wrote there.I was referring to the course of Mr.Dirac.Why didn't u read all the way??Complete reading sometimes "enlightens"... :tongue2:


marlon said:
mmm, it's a little bit more then that...Besides it is even better if you do not have any notion on the way classical dynamics work because it must be easier to not get confused...But your list is faaaaaaaaaaar from complete that it is wrong. Also incluse GTR + QFT Thus also QM and special relativity...Ever saw a paper on LQG, check out my journal and you shall be convinced ?

Who said i was making a complete list??Marlon don't read too much into it...You like twisting the words till they come your way...You have developped an awkward and unhealthy habit of imposing your point... :yuck:
Doesn't work with me,though...



marlon said:
Then how ? No partition functions, no total differentials (how are you going to incorporate reversal processes and therefore hysteresis?) ? What part of thermodynamics are you planning on teaching ?

Ever heard of the two formulations of classical reversible processes thermodynamics??Guess not,if u did,u know i was referrng to the neogibbsian formulation...Too bad u'll never be one of my students...You could learn so many...

marlon said:
CONCLUSION : i am glad not to be a student of yours :wink:
marlon

Have it your way...I hope it gives you satisfaction... :tongue2:


Daniel.
 
  • #9
dextercioby said:
I wrote there.I was referring to the course of Mr.Dirac.Why didn't u read all the way??Complete reading sometimes "enlightens"... :tongue2:

No no, that is avoiding an answer...Please explain how this is done...What is the content of this "course" that you obviously followed through completely...I am not asking for your opinion; i am asking for facts: explain how GTR can be derived whithout any geometrical models...

Please for once answer my question...


Who said i was making a complete list??Marlon don't read too much into it...
I certainly didn't...

Twisting words is NOT the same as correcting your mistakes, dear friend. Don't be so easily offended...


Ever heard of the two formulations of classical reversible processes thermodynamics??

Sure, and how do you think the first law of thermodynamics is formulated ? What mathematical concepts are used ? Maybe total differentials ? i don't know ? And what do these total differentials mean and represent physically ?

Guess not,if u did,u know i was referrng to the neogibbsian formulation...

What said anything about Gibbs here ? Please do not bring in topics that have NOTHING to do with this...Let's stick to the facts at hand


Too bad u'll never be one of my students...You could learn so many...

:rofl: :rofl: :rofl: :rofl: :rofl: :rolleyes:
Nothing is more practical then a good theory...ofcourse you first need to understand it... :wink:
marlon
 
  • #10
vincentchan said:
I know its definition, my question is why is it define this way, ie. v^w=-w^v and v^v = 0

With respect to differential geometry, it's defined that way in order to have an algebraic expression for oriented volume. You can easily work backwards from the determinant to find the simplest expression for oriented volume from vectors as the definition of the exterior product.
 
  • #11
marlon said:
No no, that is avoiding an answer...Please explain how this is done...What is the content of this "course" that you obviously followed through completely...I am not asking for your opinion; i am asking for facts: explain how GTR can be derived whithout any geometrical models...
Please for once answer my question...


Section "5" from Mr.Dirac course is called "CURVED SPACE" and is:

Paul Adrien Maurice Dirac said:
One can easily imagine a curved two-dimensional space as a surface immersed in Euclidean three-dimensional space.In the same way,one can have a curved four-dimensional space immersed in a flat space of a larger number of dimensions.Such a curved space is called a Riemann space.A small region of it is approximately flat.
Einsteina assumed that physical space is of this nature and thereby laid the foundation for his theory of gravitation.
For dealing with curved space,one cannot introduce rectilinear system of axes.One has to use curvilinear coordinates such as those dealt with in Section 3.The whole formalism of that section can be applied to curved space,because all the equations are ones which are not disturbed by the curvature.
The invariant distance ds between a point [itex] x^{\mu} [/itex] and a neighboring point [itex] x^{\mu}+dx^{\mu} [/itex] is given by
[tex] ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu} [/tex]
like in (2.1).ds is real for a timelike interval and imaginary for a spacelike interval.
With a new network of curvilinear coordinates,the [itex] g_{\mu\nu} [/itex] ,given as functions of the coordinates,fix all the elements of of distance;so they fix the metric.They determine both the coordinate system and the curvature of space.

Marlon,if that's a way to describe curved space-time in GR,then I'm i going to go to bed... :yuck: I'm not saying it's ballooney,it's that it is outrageously simplistic and mathematically unfounded.



marlon said:
Sure, and how do you think the first law of thermodynamics is formulated ? What mathematical concepts are used ? Maybe total differentials ? i don't know ? And what do these total differentials mean and represent physically ?

Marlon,the differential form of the first principle in CTPCN formulation contains indeed 2 one-forms and one exact differential.The problem was that i assumed (correctly,believe me,personal experience) that students were told:"Hey,this is not a exact total differential."And followed a rather simplistic explanation not once mentioning the words:"manifold,cotangent space,one-form".
I specifically said:
dextercioby said:
Thermodynamics can be taught without making too much mathematical analysis of the differential forms like differential heat and differential work.

By "mathematical analysis" i obviously meant "differential geometry".Telling the students:"[itex]dU [/itex]is a total differential,but [itex] \delta Q [/itex] and [itex] \delta L [/itex] aren't "is not diferential geometry,by no means...



marlon said:
What said anything about Gibbs here ? Please do not bring in topics that have NOTHING to do with this...Let's stick to the facts at hand

I see Gibbs' work provokes you the same feeling of "uncertainty" as do the words "experimental physics" to me... :wink:
For your knowledge,Gibbs rebuilt thermodynamics and took over from where Boltzmann had placed statistical physics.




marlon said:
:rofl: :rofl: :rofl: :rofl: :rofl: :rolleyes:
Nothing is more practical then a good theory...ofcourse you first need to understand it... :wink:
marlon

I see you're pretty tired and took 'than' and 'then' as the same thing... :tongue2:


Daniel.
 
  • #12
back to the original question: a wedge product is merely a coordinjate free way to represent a determinant, or a combination of determinants, hence is a way of measuring lengths, or areas, or volumes, or higher such,..., possibly by integrating over regions of space, possibly curved.
 
  • #13
dextercioby said:
Section "5" from Mr.Dirac course is called "CURVED SPACE" and is:



Marlon,if that's a way to describe curved space-time in GR,then I'm i going to go to bed... :yuck: I'm not saying it's ballooney,it's that it is outrageously simplistic and mathematically unfounded.

Ohh come on, and this is no differential geometry ?

This is nonsense and pure speculation... :rolleyes: :rolleyes:



By "mathematical analysis" i obviously meant "differential geometry".Telling the students:"[itex]dU [/itex]is a total differential,but [itex] \delta Q [/itex] and [itex] \delta L [/itex] aren't "is not diferential geometry,by no means...

Oh yes it really is...



I see Gibbs' work provokes you the same feeling of "uncertainty" as do the words "experimental physics" to me... :wink:

Sorry bit this a wrong conclusion

For your knowledge,Gibbs rebuilt thermodynamics and took over from where Boltzmann had placed statistical physics.
But really, where do you get these false arguments. What you say here is just plain wrong man !




marlon

ps : i suggest we drop this discussion because it has degenerated into speculations on maths and thermodunamics...
 

What is an exterior product?

An exterior product is a type of mathematical operation used in the field of linear algebra. It is used to calculate the area or volume of a geometric shape in a higher-dimensional space.

How is an exterior product calculated?

An exterior product is calculated by taking the cross product of two vectors. This results in a new vector that is perpendicular to both of the original vectors and has a magnitude equal to the area or volume of the shape they form.

What is the difference between an exterior product and a dot product?

The main difference between an exterior product and a dot product is the type of result they produce. An exterior product results in a vector, while a dot product results in a scalar. Additionally, an exterior product is only defined for vector spaces with more than three dimensions, while a dot product is defined for any number of dimensions.

What are some real-world applications of the exterior product?

The exterior product has many practical applications in fields such as physics, engineering, and computer graphics. It is used to calculate the moment of inertia in physics, the torque in engineering, and the surface area in computer graphics.

Are there any limitations to using the exterior product?

One limitation of the exterior product is that it is only defined for vector spaces with an even number of dimensions. Additionally, it can only be used for calculating the area or volume of certain types of shapes, such as polygons and polyhedra.

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