Do Non-Commuting Derivatives Shape New Physical Theories?

In summary, Has anyone tried to make physical theories where the derivatives do not commute? There is a condition on the derivatives of every function for them to commute which is learned in first year calculus, but people have tried to create theories where this does not hold. Some theories are based on non-commuting differential operators, while others use torsion to break the symmetry of mixed second partial derivatives.
  • #1
MathematicalPhysicist
Gold Member
4,699
371
Has anyone tried to make physical theories where the derivatives do not commute?

I mean there's a condition on the derivatives of every function for them to commute which is learned in first year calculus.

I mean in QM and QFT we grew accustomed to operators that do not commute, so why not also for differential operators?
 
Physics news on Phys.org
  • #3
Sure,in Calculus of Grassmann variables derivatives do anticommute.Basic notion of Supersymmetry in Superspace formalism has it's application.
 
  • #4
From one perspective, all currently known fundamental physics is based on theories where derivatives do not commute.

A simple example: in electromagnetism, we can define
[itex] D_\mu = \partial_\mu - ieA_\mu [/itex] ,
which acts just like (because it is) a derivative operator.

From which we compute
[itex] [D_\mu, D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu \neq 0[/itex]
(at least when there is an electro-magnetic field present). Every force in the standard model is like this (with a minor but important modification). General relativity works similarly as well.
 
  • #5
MathematicalPhysicist said:
Has anyone tried to make physical theories where the derivatives do not commute?

I mean there's a condition on the derivatives of every function for them to commute which is learned in first year calculus.

I mean in QM and QFT we grew accustomed to operators that do not commute, so why not also for differential operators?

Look up Yang-Mills gauge theory and non-abelian SU(N) groups, it's based on non-commuting differential operators. 't Hooft's renormalization of this theory in the early seventies is what basically allows the existence of the SM as we know it. So this is not BTSM at all.
On the other hand, as mentioned in a link above, GR's curvature tensor is also a commutator of covariant derivatives.
 
  • #6
DimReg said:
From one perspective, all currently known fundamental physics is based on theories where derivatives do not commute.

A simple example: in electromagnetism, we can define
[itex] D_\mu = \partial_\mu - ieA_\mu [/itex] ,
which acts just like (because it is) a derivative operator.

From which we compute
[itex] [D_\mu, D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu \neq 0[/itex]
(at least when there is an electro-magnetic field present). Every force in the standard model is like this (with a minor but important modification). General relativity works similarly as well.

What I meant is that when you have something like this:
[tex] \partial_t \partial_x - \partial_x \partial_t [/tex]

This equals zero when certain conditions are met in simple analysis.
 
  • #7
andrien said:
Sure,in Calculus of Grassmann variables derivatives do anticommute.Basic notion of Supersymmetry in Superspace formalism has it's application.

I didn't refer to Grassmann variables, these variables seem strange for me; Integration is the same as derivative?!
 
  • #8
MathematicalPhysicist said:
What I meant is that when you have something like this:
[tex] \partial_t \partial_x - \partial_x \partial_t [/tex]

This equals zero when certain conditions are met in simple analysis.

Oh I see. My first thought is that this would be a lot of work for something we basically already have: a representation of non commuting operators. But my second thought is that this might be a way to create some non standard physical situations, where for example energy and momentum would not be simultaneous observables.

I would guess that you can already do this if you want in Quantum Mechanics, where all you have to do is pick a wavefunction that doesn't meet the conditions for equality of partial derivatives. (On the other hand, maybe being a member of the Hilbert space would rule out the existence of such a function). Looking at this wikipedia article :(http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Requirement_of_continuity)
I am reminded of the dirac delta potential, which seems similar to the example they give under "Requirement of Continuity".
 
  • #9
MathematicalPhysicist said:
What I meant is that when you have something like this:
[tex] \partial_t \partial_x - \partial_x \partial_t [/tex]

This equals zero when certain conditions are met in simple analysis.

Ah, ok, it seems we all thought about covariant derivatives. The only way you can break the symmetry of mixed second partial derivatives besides changing the topological conditions you refer to is by introducing torsion. There are quite a few physical theories that introduce torsion.
 
  • #10
DimReg said:
Oh I see. My first thought is that this would be a lot of work for something we basically already have: a representation of non commuting operators. But my second thought is that this might be a way to create some non standard physical situations, where for example energy and momentum would not be simultaneous observables.

I would guess that you can already do this if you want in Quantum Mechanics, where all you have to do is pick a wavefunction that doesn't meet the conditions for equality of partial derivatives. (On the other hand, maybe being a member of the Hilbert space would rule out the existence of such a function). Looking at this wikipedia article :(http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Requirement_of_continuity)
I am reminded of the dirac delta potential, which seems similar to the example they give under "Requirement of Continuity".

Well we actually look at ringed Hilbert spaces, i.e that the wave function is in Schwartz class. (This is what I remember from reading Atkinson's textbook).

The Schwartz class property doesn't regard continuity of the derivative, but I should recheck it, cause I am talking from my head without really remembering the finer technical details.
 
  • #11
MathematicalPhysicist said:
but I should recheck it, cause I am talking from my head without really remembering the finer technical details.

To be honest, it's clear you know more about this stuff than I do, so I doubt I can help you any further.
 
  • #12
Well what I'd do, would be taking the definition of the derivative, and see when the partial derivatives don't commute and why. I guess they always commute as long as the one's variable which is acting first is not dependent on the others (at least that's what I get by seeing it).
But if you want to make a spacetime variable let us say to explicitly depend on the other, then you reduce the dimension on your problem (getting multipliers for the constraint needed)...
 
  • #13
This sounds somewhat like non-commutative geometry, on which there is a wealth of literature.
 
  • #14
MathematicalPhysicist said:
Has anyone tried to make physical theories where the derivatives do not commute?

I mean there's a condition on the derivatives of every function for them to commute which is learned in first year calculus.

I mean in QM and QFT we grew accustomed to operators that do not commute, so why not also for differential operators?

There are some papers in topology which dealed with this subject. The problem is faced with the appearance of singularities. The well-known example is Diracs Monopole. There, the equation dF=0 is not true anymore and this is equivalent to non-commuting derivatives. Of course it was also resolved using fiber bundles.
This example was the starting point of Harvey and Lawson to develop its theory of singular connections. Then the non-commutativity of the derivatives was used to obtain invariant diferetntial forms (like characterictic classes etc.) Another key example came from complex analysis. Consider
[tex] d\left( \frac{dz}{z}\right) =\delta(z) d\bar{z}\wedge dz [/tex]
with the delta function which can be also written as
[tex] d^2(ln(z))=\delta(z) d\bar{z}\wedge dz [/tex]
and the delta function comes from the residue theorem.
 

1. What is the concept of commutation of derivatives?

The commutation of derivatives is a mathematical technique used to rearrange the order in which derivatives are applied to a function. It involves swapping the order of differentiation, which can result in different outcomes for the function.

2. Why is commutation of derivatives important in scientific research?

Commutation of derivatives is important in scientific research because it allows for the simplification and manipulation of complex equations. This can make it easier to solve problems and find solutions in various fields such as physics, engineering, and economics.

3. Can commutation of derivatives be applied to all types of functions?

No, commutation of derivatives can only be applied to functions that are smooth and continuous. Functions with discontinuities or sharp corners may not follow the usual rules of differentiation and therefore cannot be commuted.

4. What is the difference between commutation of derivatives and the chain rule?

While both commutation of derivatives and the chain rule involve rearranging the order of differentiation, they are different techniques. The chain rule is used when differentiating composite functions, while commutation of derivatives is used to rearrange the order of differentiation for a single function.

5. Are there any limitations to using commutation of derivatives?

Yes, there are limitations to using commutation of derivatives. It may not always be possible to commute derivatives, especially if the function is nonlinear or involves multiple variables. Additionally, commutation of derivatives may not always result in a simplified or more useful expression.

Similar threads

Replies
19
Views
1K
  • Quantum Interpretations and Foundations
Replies
9
Views
2K
  • Quantum Physics
Replies
2
Views
1K
  • Beyond the Standard Models
Replies
1
Views
190
  • Beyond the Standard Models
Replies
3
Views
3K
  • Beyond the Standard Models
Replies
0
Views
505
  • Quantum Interpretations and Foundations
6
Replies
175
Views
6K
  • Beyond the Standard Models
Replies
1
Views
2K
  • STEM Academic Advising
Replies
9
Views
537
  • Quantum Physics
Replies
24
Views
1K
Back
Top