Solenoidal and conservative fields

[TrickyDicky]

We seem to be talking at cross purposes here, now I have to refer you to my posts 30, 31, 32 and 34 because when you answer me you seem to ignore the posts after the one you are addressing.

I don't buy it.

What specifically are you not buying? 'cause I ain't selling anything.
Can you answer specifically my questions in the last sentence of my previous post (#34)?

Also, covariant partials DO commute when applied to scalars if your connection is torsion-free. It's when you apply them to vector fields that curvature measures the failure of commutativity.

Well, of course, that's what I've been saying all along. And curl differential operators act on vector fields, right?:uhh:

 

[quasar987]

But I'm not unsatisfied at all, how could I? I agreed with it (just like I agree with what you just posted about exterior derivatives) since I already knew about that generalization. Only problem was that it didn't actually addressed my specific question.
My mind is eased now that I think I found basically what I was looking for.
To summarize this thread I started with some questions about laplacian fields that were answered to my full satisfaction by homeomorphic. Next I brought here a question from another thread that received no answers, about whether there were any circumstances (leaving aside the topological ones, that is I'm considering only oriented, contractible and simply connected smooth 3-manifolds) in which curl of grad was not zero because I had the intuition that it was not the case if the manifold was not flat. To this I received some very informative answers but not satisfactory in the sense of answering directly my doubts.
Finally I found online this very basic result from Riemannian geometry that says that commutativity of the appropriate generalization of mixed partials implies vanishing Riemann tensor, and this I would think that answers my question. My conclusion is that only if the 3-space is Euclidean must we expect the rule "curl of grad=0" to hold. Now do you disagree with this conclusion? If so, where does my logic fail here? Thanks.

As it has been been mentionned previously, if you use the definition of grad, curl, div of Matterwave, then of course, curl o grad = 0 simply because d²=0.

But although you're not unsatisfied with this, this does not address your question! :)

Because, as I gather from post #31, by a "gradient field", you mean a field of covariant 2-tensors of the kind $\nabla F$ (for some 1-form F), and by the "curl" of a covector field A, you mean the the covariant 2-tensor field whose components in local coordinates are $A_{i;j}-A_{j;i}$. Notice that since the Christofel symbol $\Gamma^k_{ij}$ of a levi-civita connection is symmetric in its i,j components, your curl is actually just -2dA in invariant form.

Ok, so, since your gradient is a 2-tensor and your curl operator is defined on 1-forms, how do you take the curl of a gradient field? Do you use the Hodge star for this?

Also, what is your motivation for defining the gradient like this? Usually, we take the gradient of a function. How is your gradient related to functions on M?

 

[TrickyDicky]

As it has been been mentionned previously, if you use the definition of grad, curl, div of Matterwave, then of course, curl o grad = 0 simply because d²=0.

But although you're not unsatisfied with this, this does not address your question! :)

Ok, but do you understand that d²=0 is only valid for a flat connection (flat vector bundle) and therefore valid for vanishing curvature spaces and I'm trying to generalize to spaces with curvature that would have to take into account the curvature form that make D² different than zero?

Because, as I gather from post #31, by a "gradient field", you mean...

I just mean $\nabla f$ like it says in every book.

Also, what is your motivation for defining the gradient like this? Usually, we take the gradient of a function. How is your gradient related to functions on M?

Again I'm taking the gradient of a function like everyone else.

 

[homeomorphic]

We seem to be talking at cross purposes here, now I have to refer you to my posts 30, 31, 32 and 34 because when you answer me you seem to ignore the posts after the one you are addressing.

I don't think those posts adequately address the issues.

Originally Posted by homeomorphic View Post

I don't buy it.

What specifically are you not buying? 'cause I ain't selling anything.
Can you answer specifically my questions in the last sentence of my previous post (#34)?

I told you. I am not sure your definition is independent of coordinate choices, in general. You are going to get a different nabla for each choice of coordinates. Therefore, it's not clear that your definition makes sense.

Originally Posted by homeomorphic View Post

Also, covariant partials DO commute when applied to scalars if your connection is torsion-free. It's when you apply them to vector fields that curvature measures the failure of commutativity.

Well, of course, that's what I've been saying all along. And curl differential operators act on vector fields, right?

The way you applied it, you were applying the partials to the components of the vector field, which are scalar. Covariantly differentiating a vector field is not the same thing as trying to apply your strange curl operator that we haven't even established makes sense.

The non-commutativity that curvature measures results from covariantly differentiating a vector field, not taking the "curl" of it.

 

[quasar987]

Ok, but do you understand that d²=0 is only valid for a flat connection (flat vector bundle) and therefore valid for vanishing curvature spaces and I'm trying to generalize to spaces with curvature that would have to take into account the curvature form that make D² different than zero?

d²=0 only for flat spaces?! The operator "d", called exterior derivative, is defined on any smooth manifold, independantly of any riemannian structure or connection, and always satisfies d²=0. With a connection, there is also the operator $d^{\nabla}:\Omega^p(M,TM)\rightarrow \Omega^{p+1}(M,TM)$ whose square measures curvature, but this is not really relevant here because the "d" appearing in Matterwave's formulae is the exterior derivative d that always square to 0.

I just mean $\nabla f$ like it says in every book.

I assume that here $\nabla$ refers to the levi-civita connection, so that by $\nabla f$ you mean the covariant derivative of f, which is the 1-form $X\mapsto \nabla_Xf$. If so, then notice that $\nabla f = df$, so that according to my remark of post #37 that "your curl = -2d", we then have curl o grad(f) = -2d²f=0.

 

[TrickyDicky]

d²=0 only for flat spaces?! The operator "d", called exterior derivative, is defined on any smooth manifold, independantly of any riemannian structure or connection, and always satisfies d²=0. With a connection, there is also the operator $d^{\nabla}:\Omega^p(M,TM)\rightarrow \Omega^{p+1}(M,TM)$ whose square measures curvature, but this is not really relevant here because the "d" appearing in Matterwave's formulae is the exterior derivative d that always square to 0.


I assume that here $\nabla$ refers to the levi-civita connection, so that by $\nabla f$ you mean the covariant derivative of f, which is the 1-form $X\mapsto \nabla_Xf$. If so, then notice that $\nabla f = df$, so that according to my remark of post #37 that "your curl = -2d", we then have curl o grad(f) = -2d²f=0.

The wikipedia page about Flat vector bundles states that the flatness condition of the connection is equivalent to dd=0.

 

[TrickyDicky]

The non-commutativity that curvature measures results from covariantly differentiating a vector field, not taking the "curl" of it.

Sure, but the curl in curved space involves covariant differentiation or not?

 

[quasar987]

The wikipedia page about Flat vector bundles states that the flatness condition of the connection is equivalent to dd=0.

This is a poor choice of notation imo, because what they mean here by "d" is the operator $d^{\nabla}:\Omega^p(M,E)\rightarrow \Omega^{p+1}(M,E)$, which coincides with d only when the vector bundle E is the trivial line bundle $\mathbb{R}$ over M.

See http://en.wikipedia.org/wiki/Vector-valued_differential_form#Exterior_derivative

 

[Matterwave]

Notice that:

$$A_{i;j}-A_{j;i}=\partial_j A_i - \partial_i A_j +\Gamma^k_{ij}A_k-\Gamma^k_{ji}A_k$$

For any symmetric connection (and a Riemannian connection is a symmetric connection), the last two terms cancel, and you get:

$$A_{i;j}-A_{j;i}=\partial_j A_i - \partial_i A_j$$

that's exactly the components of the exterior derivative (up to choice of normalization):

$$(dA)_{ij}=\partial_j A_i - \partial_i A_j$$

This is of course not 0 for general A! But, what happens when we apply this to a gradient of a function df?

$$(df)_{i;j}-(df)_{j;i}=\partial_j (df)_i - \partial_i (df)_j$$

Which quickly gets us back to:

$$(df)_{i;j}-(df)_{j;i}=\partial_j\partial_i f - \partial_i\partial_j f=0$$

Well, this result should be obvious!

So, even by your own definition, we get 0 always, no matter the curvature.

EDIT: Hmmmm...what's the deal with the Latex not rendering? o.o

 

[tiny-tim]

latex recognises "f}" as a swear-word :yuck: …

$$(df)_{i;j}-(df}_{j;i}=\partial_j (df)_i - \partial_i (df)_j$$

$$(df)_{i;j}-(df}_{j;i}=\partial_j\partial_i f - \partial_i\partial_f=0$$

try …

$$(df)_{i;j}-(df)_{j;i}=\partial_j (df)_i - \partial_i (df)_j$$

$$(df)_{i;j}-(df)_{j;i}=\partial_j\partial_i f - \partial_i\partial_f=0$$

 

[Matterwave]

Thanks XD

I didn't noticed I used the wrong braces.

 

[TrickyDicky]

Notice that:

$$A_{i;j}-A_{j;i}=\partial_j A_i - \partial_i A_j +\Gamma^k_{ij}A_k-\Gamma^k_{ji}A_k$$

Yes, there seems to be a problem with this fromula, I took it from some other physicsforums discussion and it was originally in a different notation , I might have goofed somewhere when changing notation, or maybe the formula was wrong to begin with.

 

[Matterwave]

You should realize from that formula, however, that the exterior derivative itself is fully compatible with the connection since you can express it as either partial derivatives or covariant derivatives.

 

[TrickyDicky]

This is a poor choice of notation imo, because what they mean here by "d" is the operator $d^{\nabla}:\Omega^p(M,E)\rightarrow \Omega^{p+1}(M,E)$, which coincides with d only when the vector bundle E is the trivial line bundle $\mathbb{R}$ over M.

See http://en.wikipedia.org/wiki/Vector-valued_differential_form#Exterior_derivative

Thanks for pointing me to that page. I don't know, there must be something wrong with my reasoning because what I read there seems to confirm what I'm thinking.

 

[Matterwave]

If you want to use the covariant exterior derivative to define a curl, you can (you are free to make whatever definitions you want). But YOU have to come up with the definition.

You seem to just assume that there is a standard definition for curl generalized using the covariant exterior derivative and somehow the definition I gave is not the correct definition. If there is such a standard definition, please find it and share. If not, you are free to come up with it yourself. But either way, we have to agree to a definition first before we can do anything.

 

[TrickyDicky]

You should realize from that formula, however, that the exterior derivative itself is fully compatible with the connection since you can express it as either partial derivatives or covariant derivatives.

Yes, I realize that, the exterior derivative itself is, but I'm concerned about d^2=0 being the case in the presence of a non-flat connection. If the space is curved one must take into account the curvature form and the exterior derivatives turn into exterior covariant derivatives D and D^2≠0, or at least that is my understanding, so we have a curved space with a connection (the Levi-Civita conn.) that being metric compatible measures the curvature of the manifold and according to the pages of wikipedia mentioned by me and quasar987 (or at least what I infer from them) the dd=0 property would be equivalent to the condition of flat vector bundle connection, which if the manifold is curved and we are using the Levi-civita connection wouldn't be the case, right?
Please if you are versed in exterior calculus and curved connections, could you clarify this?

 

[TrickyDicky]

If you want to use the covariant exterior derivative to define a curl, you can (you are free to make whatever definitions you want). But YOU have to come up with the definition.

You seem to just assume that there is a standard definition for curl generalized using the covariant exterior derivative and somehow the definition I gave is not the correct definition. If there is such a standard definition, please find it and share. If not, you are free to come up with it yourself. But either way, we have to agree to a definition first before we can do anything.

I understand your point here, I'll try and find it.

 

[Matterwave]

Yes, I realize that, the exterior derivative itself is, but I'm concerned about d^2=0 being the case in the presence of a non-flat connection. If the space is curved one must take into account the curvature form and the exterior derivatives turn into exterior covariant derivatives D and D^2≠0, or at least that is my understanding, so we have a curved space with a connection (the Levi-Civita conn.) that being metric compatible measures the curvature of the manifold and according to the pages of wikipedia mentioned by me and quasar987 (or at least what I infer from them) the dd=0 property would be equivalent to the condition of flat vector bundle connection, which if the manifold is curved and we are using the Levi-civita connection wouldn't be the case, right?
Please if you are versed in exterior calculus and curved connections, could you clarify this?

dd=0 always. DD=0 only for a flat connection. But D and d are not the same operator. You can have both on a manifold. They act on different objects and give you different objects. d acts on forms (0-forms, 1-forms), or if you want, scalar valued forms. D acts on vector-valued forms. These are vectors whose components are forms (think a column vector, each entry of which is a form). Or, equivalently, forms which take on vector values, i.e. when you act the form on a vector, it gives you back a vector not a scalar

 

[TrickyDicky]

Maybe a basic question here would be: can $\nabla f$ be considered as a covector valued 0-form? If yes then I think the use of the exterior covariant derivative (D) is granted.

 

[TrickyDicky]

See for instance: https://www.physicsforums.com/showthread.php?p=3855127

 

[quasar987]

Maybe a basic question here would be: can $\nabla f$ be considered as a covector valued 0-form? If yes then I think the use of the exterior covariant derivative (D) is granted.

Yes, $\nabla f$ is an element of $\Omega^0(M, T^*M)=\Gamma(T^*M)=\Omega^1(M)$ and you can apply D to it. But on 0-(vector-valued-)forms, D is just $\nabla$, so you get $\nabla^2f\in \Omega^1(M,T^*M)=T_0^2M$, the so-called covariant hessian of f.

Its action on vectors is as follows: $(\nabla^2f)(X,Y)=Y(Xf)-(\nabla_YX)f$.

 

[TrickyDicky]

Yes, $\nabla f$ is an element of $\Omega^0(M, T^*M)=\Gamma(T^*M)=\Omega^1(M)$ and you can apply D to it. But on 0-(vector-valued-)forms, D is just $\nabla$, so you get $\nabla^2f\in \Omega^1(M,T^*M)=T_0^2M$, the so-called covariant hessian of f.

Its action on vectors is as follows: $(\nabla^2f)(X,Y)=Y(Xf)-(\nabla_YX)f$.

Thanks, can we then apply the hodge dual to D($\nabla f$) and get the curl (always considering we are using a Levi-civita connection in a curved 3-manifold to perform the exterior covariant derivative)?

 

[quasar987]

The Hodge dual maps 2-forms to 1-forms. Unfortunately, $\nabla^2f$ is not a 2-form, it is a general covariant 2-tensor. In fact, when the connection is symmetric, as is the case of the levi-civita connection for instance, $\nabla^2f$ is symmetric (as is the hessian in R³ by symmetry of the mixed partials!).

So in the riemannian case, we can't even hope to antisymetrize $\nabla^2f$ and then apply hodge: we'll get 0 all the time... :(

 

[TrickyDicky]

The Hodge dual maps 2-forms to 1-forms. Unfortunately, $\nabla^2f$ is not a 2-form, it is a general covariant 2-tensor. In fact, when the connection is symmetric, as is the case of the levi-civita connection for instance, $\nabla^2f$ is symmetric (as is the hessian in R³ by symmetry of the mixed partials!).

So in the riemannian case, we can't even hope to antisymetrize $\nabla^2f$ and then apply hodge: we'll get 0 all the time... :(

Yes you are right,
Does this have anything to do with the fact that curls refer to infinitesimal rotatations?

 

[quasar987]

I don't know.. it seems like the situation so far is the following:

Given M a manifold with a connection $\nabla$, if we naively define the curl of a 1-form $A\in\Omega^1(M)$ by $\mathrm{curl}(A):=\mathrm{Ant}(\nabla A)\in \Omega^2(M)$, then there are 2 cases:

i) if the connection is symmetric (i.e. torsion free), curl(A) = dA (modulo a multiplicative constant). But this is actually independant of the connection and can be defined on any manifold with or without a connection. And we know that in the levi-civita case, this is indeed the curl as defined by Matterwave, which coincides with the actual curl in the R³ case. In this case, curl o grad = 0. So at this point, we are given the option to revise our definition of curl and we may chose to say it's just "d" after all.

ii) If the connection has torsion, then $\mathrm{curl}(A)=dA - \tau(A,\cdot,\cdot)$, where $\tau(A,X,Y)=A(\nabla_XY-\nabla_YX-[X,Y])$ is the torsion $\left( \begin{array}{c} 2 \\ 1 \end{array}\right)$-tensor. In coordinates, this is

$$\mathrm{curl}(A) = (\partial_iA_j - \partial_jA_i) - (\Gamma_{ij}^k - \Gamma_{ji}^k)A_k$$

(mod constant). So, if we decide to stick with the covariant definition of curl, then curl is only defined on manifolds with connections, and it has the perculiarity that the failure of curl o grad to vanish is a measure not of the curvature of the connection, but rather of its torsion.

 

[TrickyDicky]

I don't know.. it seems like the situation so far is the following:

Given M a manifold with a connection $\nabla$, if we naively define the curl of a 1-form $A\in\Omega^1(M)$ by $\mathrm{curl}(A):=\mathrm{Ant}(\nabla A)\in \Omega^2(M)$, then there are 2 cases:

i) if the connection is symmetric (i.e. torsion free), curl(A) = dA (modulo a multiplicative constant). But this is actually independant of the connection and can be defined on any manifold with or without a connection. And we know that in the levi-civita case, this is indeed the curl as defined by Matterwave, which coincides with the actual curl in the R³ case. In this case, curl o grad = 0. So at this point, we are given the option to revise our definition of curl and we may chose to say it's just "d" after all.

Why just d?, because we are applying it only to one-forms and dA is a two-form, but we want a one-form as result , actually matterwave (and tiny-tim and wikipedia) defined it as *d. if curl is just d, we should admit that for a curved space curl is just D, which I don't think is correct. Btw what is Ant?

ii) If the connection has torsion, then $\mathrm{curl}(A)=dA - \tau(A,\cdot,\cdot)$, where $\tau(A,X,Y)=A(\nabla_XY-\nabla_YX-[X,Y])$ is the torsion $\left( \begin{array}{c} 2 \\ 1 \end{array}\right)$-tensor. In coordinates, this is

$$\mathrm{curl}(A) = (\partial_iA_j - \partial_jA_i) - (\Gamma_{ij}^k - \Gamma_{ji}^k)A_k$$

(mod constant). So, if we decide to stick with the covariant definition of curl, then curl is only defined on manifolds with connections, and it has the perculiarity that the failure of curl o grad to vanish is a measure not of the curvature of the connection, but rather of its torsion.

Hmm, this is a pretty intriguing result, however I was limiting my explorations to Riemannian manifolds with symmetric connection.

 

[quasar987]

Ant, also written Alt, is the antisymetrization operator. It takes a tensor T and spits out an antisymetric one. If T is already antisymetric, Ant(T)=T. If T is symetric, Ant(T)=0. See http://en.wikipedia.org/wiki/Wedge_product#The_alternating_tensor_algebra.

Ok, maybe there is a little too much I neglected to say for my last post for it to be intelligible. Let me try again...

Recall: in post #54-59, we were contemplating to define (in the riemannian context) the curl of a gradient covector field $\nabla f$ by considering $D(\nabla f)=\nabla^2f$ and then going down to 1-forms using hodge duality. But then I said that this doen't quite make sense, because $\nabla^2f$ is not a 2-form. If we want to use hodge on it, we need to anti symetrize it first.

Ok, but this discussion focuses on the curl of a gradient. This restriction is not necessary. More generally, we can define the curl of any 1-form A as $\mathrm{curl}(A):=*\mathrm{Ant}(\nabla A)$ (or $\mathrm{curl}(A)=*\mathrm{Ant}(D A)$ if your prefer). It turns out (after computing) that $\mathrm{Ant}(\nabla A)=dA$, so that our definition is just $\mathrm{curl}(A)=*dA$ after all. So our attempt at a definition of curl via the covariant derivative landed us on the same formula as the one from tiny-tim's post where he defines curl(A) as *dA. In particular, curl o grad = 0 always with our definition.

So that's a little disapointing. But then I noticed that if you back up a little and consider the more general situation of a manifold M with just a connection on it. We may generalize our notion of curl by setting $\mathrm{curl}(A):=\mathrm{Ant}(\nabla A)$. Of course, in this setting we don't have the luxury of a hodge star *, so we must be content with curl being a 2-form.

Then I noticed that if $\nabla$ is symetric, then $\mathrm{curl}(A)=dA$. So modulo a star on the left, this is the same as in the riemannian case. Also, by setting $\mathrm{grad}(f):=\nabla f=df$ in this setting, we have the familiar curl o grad = 0.

However, if $\nabla$ is not symetric, then curl(A)=dA - τ(A, , ) where τ is the torsion tensor. And so, we see that curl o grad ≠ 0 in general. In fact, curl o grad = 0 iff τ=0.

Like you, I found this an interesting and intriguing observation!

 

[TrickyDicky]

Nice recap, if I look back I think I should have realized it when matterwave made clear in a previous post that the formula I posted generalizing the curl was equal to zero with the symmetric connection which implied it didn't if the connection was asymmetric, but as I said I was centering only in the symetric case and was happier thinking there was an error somewhere.

 

[TrickyDicky]

And it is interesting to see that a quick google search shows this result has been used by physicists (even by Einstein in 1928 "in an attempt to match torsion to the electromagnetic field tensor" according to wikipedia) to try to come up with some kind of EM and gravity unified field theory.

 

[quasar987]

Really! Do you know what that means 'unified field theory'? what was Einstein trying to do exactly?

 

[TrickyDicky]

Really! Do you know what that means 'unified field theory'? what was Einstein trying to do exactly?

Are you kidding? Physicists have been trying to geometrically unify electromagnetism and gravitation (more recently QM and GR) for ages.
Einstein was obsessed with this from 1915 to his death without any success, I don't remember about the details (it's been some time since I've read Einstein's biographies), the quote in my last post was from the wikipedia page on Einstein-cartan theory, the first paragraph.

 

[quasar987]

But what was Einstein trying to do exactly. Like, mathematically speaking, do you know what it is that he was trying to construct or find?

 

[Matterwave]

I'm pretty fuzzy on this history, but Einstein was basically trying to find one unifying structure which would describe both Gravity and Electra-magnetism (at the time, these were the only 2 forces known). Einstein thought the best candidate was a unified (classical) field theory (as opposed to a quantum field theory). This means, he was trying to find some field, which permeated all of spacetime, which would describe gravity on the one hand and electromagnetism on the other.

It would be elegant if, like the case for electricity and magnetism, these forces were just the 2 sides of the same coin.

Einstein and Maxwell basically succeeded in unifying electricity and magnetism - showing that they basically combine in 4-D spacetime to form a 2-form field (the Faraday tensor).

I believe Einstein was thinking that one could combine gravity into that picture as well.

In fact, Einstein was quite excited for Kaluza who came up with just such a theory (today called Kaluza-Klein theory). Einstein prompted Kaluza (who was sort of a recluse) to publish his theory. The problem with Kaluza's theory was that it involved an extra spatial dimension (the curvature in which, almost magically, reproduces the Maxwell equations!), but he was unable to explain why this dimension was not accessible/observable to us. Klein later remedied this by postulating that the extra spatial dimension was a tiny compact dimension (obeyed periodic boundary conditions).

There are still other problems with the theory of Kaluza and Klein. Being a classical theory, it does not adequately describe the quantum world.

Einstein was never a fan of quantum mechanics, but, it would seem, it's getting more and more difficult to come up with a picture of the universe which is NOT fundamentally quantum. This may have been the great difficulty which Einstein could not overcome. I believe he was always searching for a classical unified field theory, when, as best as we can tell, any unified field theory should be a quantum one.

 

[homeomorphic]

Einstein and Maxwell basically succeeded in unifying electricity and magnetism - showing that they basically combine in 4-D spacetime to form a 2-form field (the Faraday tensor).

I wouldn't really credit Einstein and Maxwell with that. I forget who was involved at the beginning of the story. Then, Faraday came up with his law which gave a relationship, and Maxwell showed that something similar held for time-varying electric fields. Then, it was really Minkowski (according to Penrose), building on Einstein's work, who came up with the idea of space-time, as well as the Faraday tensor in some form.

 

[Matterwave]

Who to credit with what is always a question in the history of physics. Some people would argue that one should not credit Einstein with SR either, and that it was Poincare and Lorentz who "came up with it". But certainly Einstein was the one to unify the ideas of Length contraction and time-dilation, etc., into a coherent theory.

Maxwell was certainly not the first to see the relationship between electricity and magnetism. Ampere's law, for example already showed some relationship between current and magnetic fields. Faraday's law does the same thing, but in the other direction (changing magnetic fields gives current).

Obviously, physics is a group effort, so to credit anyone or two people with the unification of E&M is not simple. I chose Einstein and Maxwell because they represent to me the people who came up with the basic structure upon which the unification could be done.