Einstein field equation

[Schrödinger's hat]

I'm trying to wrap my head around GR, and one thing I've noticed with every EM or quantum eqn I've encountered can be written in the form:

(curvature+a bit)Potential/vector potential=source

(except dirac, although dirac solutions are a subset of solutions to one like it)

Also reformulating newtonian gravity to be SR invariant gives an equation very much like Maxwell's equation (ie. the weak field approximations up to a couple of constants I couldn't track down).

Ignoring cosmological constant for the moment. Does anyone know if Einstein's field equations can be written in a form a bit like (I apologise for mangling notation)

[math]

[\nabla_\mu,\nabla_nu]_{\alpha\beta} g^{\beta\alpha}+Ag^{\alpha\beta}=kT^{\alpha\beta}

[/math]

where [math] A[/math] is an operator involving two derivatives.

 

Is thinking of a metric as a little bit like a generalisation of the idea of vector potential horribly wrong, or is this concept useful?

[ajb]

Is thinking of a metric as a little bit like a generalisation of the idea of vector potential horribly wrong, or is this concept useful?

 

Maybe you want to think about connections.

[Schrödinger's hat]

Yeah, I'm starting to get a handle on ideas of connections and tangent bundles, but I feel like this is an important step in bridging the way I currently intuitively think of things in terms of potentials, fields and curvature of things embedded in a frame that I have to make before I can get a handle on intrinsic curvature.

 

To clarify, I can handle the idea of [math]g[/math], and what it taking on different values means for spacetime and events within, but I'm having trouble thinking about [math]G[/math].

So far it feels like curvature of [math]g[/math], but in trying to expand things in terms of Christoffel symbols I lose the plot.

In addition there's confusion from it both being used as the metric/operator for mathematical tricks, and as the thing on which we're operating with the Riemann tensor.

I can see curvature (two dervatives of the metric) is one way of thinking how the Ricci tensor comes into it, but I cannot fathom a way for the scalar to come into it other than 'because that's what makes conservation of matter work'.

 

I suppose I could be a bit naive in trying to think of this in terms of 'things' and 'operators' but this is the only way I can see of moving forward.

[Xerxes]

Yeah, I'm starting to get a handle on ideas of connections and tangent bundles

I am sorry to appear rude, but I seriously doubt this. GR is very very hard and requires a deep and working knowledge of:

 

point set topology;

 

differentiable manifolds;

 

vector spaces, in particular tangent spaces;

 

and loads of other mathy stuff which you cannot possibly sidestep (life is sooo hard!!)

 

And if you want to display GR as a theory of connexions on bundles, then you need a deep knowledge of these too. (Though I doubt this can be done - or at least I have never seen anyone attempt it. Have you?)

 

Sorry again, but you have so far failed to demonstrate an understanding of such things. To clarify: There IS no shortcut. This is not a failing on your part, merely an attempt to run before you can walk. For example......

 

I can handle the idea of [math]g[/math], and what it taking on different values means for spacetime and events within, but I'm having trouble thinking about [math]G[/math].

So far it feels like curvature of [math]g[/math],

.....makes almost no sense in differential geometry, which is the subject at hand. Like it or not, an understanding of this subject is a prerequisite for fully understanding E.'s field equations and what they actually mean

 

I suppose I could be a bit naive in trying to think of this in terms of 'things' and 'operators

Like yes, my point exactly

[Schrödinger's hat]

And I apologise if I appeared arrogant. I have no intention of skipping the mathy stuff, but I like to give it as much context as I can before I get started.

 

I was under the impression that I had a vague understanding of the idea of a tangent bundle, but this could easily be mistaken.

Last year I did an algebra course in which I did something with Lie algebras, locally flat spacetime and proving some properties of Casimir invariants and something a lot like the Bianchi identity -- among other things which were all presented extremely abstractly. I managed to do fairly well in the course, but I have absolutely no idea what any of it meant and I remember very little of it.

I was trying to ground some intuition about what [math]G[/math] is in terms of the more familiar wave equations in hopes of bridging some of the gap from the direction of what I am familiar with -- rather than struggling through the abstraction until I realise the gap had been filled.

I also realise that thinking in terms of extrinsic curvature is not very useful and can be quite misleading, but I feel seeing the equation written as (operator)metric=T would help me pin-point the flaws in my reasoning, see why the metric is not like a potential and move forward.

If arranging it in this form is not as simple as I thought then say so and I shall stop annoying you.

[Xerxes]

And I apologise if I appeared arrogant.

I hope I didn't accuse you of arrogance. If it seemed implied, I apologize

 

I was under the impression that I had a vague understanding of the idea of a tangent bundle

And I am sure you do. My doubt was to whether it is relevant to GR.

 

OK fukkit, you asked about the relation between [math]g[/math] and [math]G[/math]. So suppose that [math]V[/math] is an arbitrary vector space. Then of necessity there exists a dual space [math]V^{*}[/math] such that [math]V^{*}: V \to \mathbb{R}[/math]. Let's say [math]\varphi \in V^{*}[/math] such that [math]\varphi(v) = \alpha \in \mathbb{R}[/math]

 

Now suppose an inner product is defined on [math]V[/math]. This may be defined as the mapping from the Cartesian product of vector spaces to the Reals. Since elements in [math]V \times V[/math] are the ordered pairs [math](v,w)[/math] then we will need a mapping, say [math]g:V \times V \to \mathbb{R},\,\,\,g(v.w) = \alpha \in \mathbb{R}[/math].

 

So what is the relation between the [math]\varphi[/math] as defined above and [math]g[/math]? By the definition of an inner product space, one may assign a unique element in [math]V^{*}[/math] say [math]\varphi_w[/math] such that [math]\varphi_w(v)=g(v,w)[/math], which generalizes to [math]\varphi = g(v, \cdot) \in V^{*}[/math]

 

So we may therefore makes the definition that [math]V^{*} \otimes V^{*}: V \times V \to \mathbb{R},\,\,\, g \in V^{*} \otimes V^{*}[/math] which is by definition a type (0,2) tensor.

 

Now it is easily seen that the set of all tensors at a point is a vector space at that point, so using standard notation write [math]g = \sum_{\mu, \nu} g_{\mu \nu}\epsilon^{\mu} \otimes \epsilon^{\nu}[/math] for this element of that vector space, where the "epsilons" are basis vectors and the first term under the sum are called components. But since the power of tensors lies entirely in the fact that any equation using them retains the same form regardless of the choice of coordinates (and hence bases), it is customary to write tensors in component form, hence our tensor is always written as [math]g_{\mu \nu}[/math]

 

It's called "the metric tensor"

 

If arranging it in this form is not as simple as I thought then say so and I shall stop annoying you.

Well is it as simple as you thought? I haven't even started on [math]G[/math], but no, you are not annoying me

[ajb]

Schrödinger's hat: if one wants a very quick introduction on how to preform basic calculations in GR then I recommend Dirac's little book General Theory of Relativity, Princeton University Press; New Ed edition (8 Jan 1996). However, it is very short on details of the mathematics, it just presents for formalism for a very basic understanding.

 

For a better introduction I suggest Sean M. Carroll, Lecture notes on General Relativity, arXiv:gr-qc/9712019v1. This is a far better introduction from a mathematics point of view.

 

Connections are fundamental in general relativity. The two that are central to the theory (not really completely independent) are the Levi-Civita connection on the tangent bundle (or in terms of principles bundles in the frame bundle) and the spin connection on a spin bundle.

 

A spin bundles is a vector bundle that carries a spin representation of SPIN which is the double cover of SO. The spin connection can be interpreted as the gauge field of local Lorentz transformations. If one wants to consider spinors in GR then this is how to do it.

 

I review some of the theory of connections in principle bundles and their relation to gauge theories in a short report than can be found here. I do not discuss Riemannian geometry at all, though the formalism found in the report will cover that situation. In essence, consider the frame bundle of a Reimannian manifold. Anyway, it may help give a wider view of connections.

[Xerxes]

Well, Schroedinger's hat was asking for intuition, so let's see if I can find some BUT BE WARNED I am not a physicist.

 

The bundle approach to GR that I dismissed rather cavalierly seems on reflection to make some sort of sense. Let's start here:

 

Let's take it as read that GR does not actually replace SR but is in fact a generalization of it. The reasons are simple:

 

1) SR assumes "flat" Euclidean spacetime

 

2) GR models spacetime as a possibly non-flat 4-manifold which, by all possible definitions, is locally Euclidean and flat

 

3) Then it may well be that SR applies locally on our non-flat 4-manifold.

 

4) SR depends upon (or defines, your choice) the set of Lorentz transformations on Euclidean space.

 

Picture this: as we "roam" over our spacetime manifold, let's call it [math]M[/math], then at each point [math]m \in M[/math] we may make an arbitrary Lorentz transformation, guaranteeing that SR holds in a "neighbourhood" of [math]m[/math]. The set of all physically realizable Lorentz transformations are the group called [math]SO(1,3)[/math] (assuming a certain metric signature which is not important).

 

Or, to put it another way, at each point [math]m \in M[/math] we may "attach" the Lorentz group. Let's call this group (it's a Lie group BTW) as [math]G[/math] for brevity (though it is almost certainly not the [math]G[/math] that SH was referring to). The (disjoint) set union of this group at all points is called a "principal bundle". It is notated (at least by me) as [math] P(G,M)[/math] and has the following properties.

 

1)[math]P,\,\,G,\,\,M[/math] are all manifolds;

 

2) for each [math]p \in P[/math] there is a projection [math] \pi: P \to M,\,\,\pi(p) = m \in M[/math] whose preimage [math]\pi^{-1}(m) = G[/math] is called a "fibre" over [math]m[/math]. Hold on to this concept

 

So. Although we know exactly how to "travel along" a fibre (simply use the group laws - recall each fibre is a group), there is no assured way to "move between" fibres i.e. different copies of our group - recall they are disjoint (mathematicians say no canonical way). For this we need a connection.

 

Finally, you can think of the 4-manifold spacetime as some sort of "reality" (though it fries my brains), the principle bundle just described is an even higher level of abstraction.

 

Deep-fried brains, maybe?

 

Ummm. Did I promise intuition...........? Duh!

[ajb]

Although we know exactly how to "travel along" a fibre (simply use the group laws - recall each fibre is a group), there is no assured way to "move between" fibres i.e. different copies of our group - recall they are disjoint (mathematicians say no canonical way). For this we need a connection.

 

 

Right, this may be even clearer when thinking about associated vector bundles. A connection allows us to "connect" fibres over different points. Also, via a "covariant derivative" we have a notion of parallel transport of sections. That is we need a connection to define the notion of differentiating sections of a vector bundle in directions along the base manifold.

 

Probably, the notion of a covariant derivative is the most familiar from a physics point of view.

[Xerxes]

A connection allows us to "connect" fibres over different points.

 

Yes. Sticking with connections for now, and remaining true (insofar as this is possible) to the intuitionist nature of this thread, I can offer a coupla of definitions (of a connection on a principal bundle) which may appear wildly different, but are in fact complemtary.

 

My first is easily the most intuitive. So. Our principal bundle, I called it [math]P[/math] is a manifold, and as such is entitled to a set of tangent vectors at each and every point [math]p \in P[/math]. Let's follow convention and call it [math]T_pP[/math]; it's a vector space. Then a connection can be seen as a (smooth) assignment of a "horizontal" subspace [math]H_pP[/math] such that the whole space at this point can be decomposed as the direct sum [math]T_pP = H_pP \oplus V_pP[/math] where the second term is the "vertical part"of this vector space.

 

This is nice. It simply says that element in the vertical subspace "point along" the fibres, while the elements in the horizontal subspace "point between" them. It looks childish, but the math is kosher.

 

My second definition is more challenging, but ultimately more rewarding. So. A connection 1-form is a (smooth) mapping [math]\omega: T_pP \to \mathfrak{g}[/math] the codomain being the Lie algebra of the Lie group [math]G[/math]. Huh? What can this mean?

 

Well it a fact from (fairly) elementary linear algebra that nay vector space that can be decomposed as, say, [math]W = U \oplus V[/math] admits of two projections, [math]p_1:W \to U,\,\,\, p_2: W \to V[/math] with the following rather obvious properties: [math]\ker p_1 = V[/math] and likewise [math] \ker p_2 = U[/math].

 

So. Let's take as a given that any [math]T_pP= V_pP \oplus H_pP[/math] can be decomposed in this way. Note that each fibre in our bundle is a Lie group, which a manifold, whose algebra is simply the vector space tangent to the group identity. So we have that, since the elements in [math]V_pP[/math] "run along" the fibre, we have an isomorphism [math]V_pP \simeq \mathfrak{g}[/math] So now it is easy to see that our connection 1-form (I called it [math]\omega[/math]) is simply the first projection [math]\omega: V_pP \oplus H_pP \to V_pP[/math] up to isomorphism and whose kernel is the subspace [math]H_pP[/math]. Hence my two definitions are complementary!

 

Also, via a "covariant derivative" we have a notion of parallel transport of sections. That is we need a connection to define the notion of differentiating sections of a vector bundle in directions along the base manifold.

Yes, but unfortunately I am being sent overseas for 2 weeks (I call it "being transported"). I am enjoying the rather rambling turn this thread has taken. Do not take my silence as indifference.

[ajb]

The idea of decomposing the tangent bundle of a principle bundle always makes me stop to think about it. Covariant derivatives seem very natural from physics.

 

Of course, these notions are all really the same.

[Schrödinger's hat]

Still reading, just don't have much to say. Thanks for the response (although I can barely follow what Xerxes is saying, last I did that kind of maths it was mostly rubbing objects and their properties together until I got what I wanted)

For a better introduction I suggest Sean M. Carroll, Lecture notes on General Relativity, arXiv:gr-qc/9712019v1. This is a far better introduction from a mathematics point of view.

Yes, I've started reading that recently, working through it as I get the time. Also some Stanford lectures I found, but they are sadly lacking in depth as it was merely an introduction for another course.

What you've been saying agrees with my intuition so far, I guess the word curvature means a very different and/or incorrect thing as far as I was concerned. Perhaps the word I was looking for is, in fact, connection. ie. the way any vectors/tensors/things at one point are transformed by travelling to adjacent points in space-time.

 

 

Random thought: A slight problem with this is the word adjacent -- the concept of adjacency seems predicated on the existence of the metric -- I think this concept of a more generalised and abstracted connection that Xerxes is getting at helps here (if I understand him correctly). Still trying to wrap my head around the idea of a reality based upon a disjoint set. This resonates in some ways with vague notions I have about entanglement, but that discussion is for much, much later.

 

 

 

So. Let's take as a given that any can be decomposed in this way. Note that each fibre in our bundle is a Lie group, which a manifold, whose algebra is simply the vector space tangent to the group identity. So we have that, since the elements in "run along" the fibre, we have an isomorphism So now it is easy to see that our connection 1-form (I called it ) is simply the first projection up to isomorphism and whose kernel is the subspace . Hence my two definitions are complementary!

Having trouble comprehending this bit or possibly the paragraph above.

[ajb]

Having trouble comprehending this bit or possibly the paragraph above.

 

 

Do not let that put you off. In my opinion thinking about principle bundles and connections in them is difficult. Just keep at it.

[Xerxes]

(although I can barely follow what Xerxes is saying,)

As ajb says, don't worry about it, Xerxes was just showing off, rather than offering genuine help. He is a bit like that, you will have to forgive him.

[Schrödinger's hat]

Even if it wasn't the answer I was looking for, it was quite interesting and I'm slightly disappointed that he stopped showing off.

Also why is Xerxes talking about Xerxes in the third person?

[Matt Hunt]

Is thinking of a metric as a little bit like a generalisation of the idea of vector potential horribly wrong, or is this concept useful?

Think of the metric as a form of kinetic energy. You require a bilinear form to define the kinetic energy generally, this is the metric defined on the tangent space.

[Schrödinger's hat]

Think of the metric as a form of kinetic energy. You require a bilinear form to define the kinetic energy generally, this is the metric defined on the tangent space.

 

I don't understand. There's a thought here somewhere about this making kinetic energy non-local.

Also my intuition (hardly a reliable tool in these matters) says that the stress energy tensor being on the other side of the equation (and potentially 0) contradicts this somewhat. Would a (very) generalised v^2 be more appropriate? Is velocity still even a useful concept at this point?

[Xerxes]

I don't understand either. But Schroedinger raises an interesting point. Lemme ramble a bit....

 

A bilinear form is a mapping, say, [math]\langle\cdot \,,\,\cdot \rangle:V \times V \to \mathbb{R}[/math] that is linear in each argument taken separately (though our field does not need to be real, neither do we need always to work with vector spaces - any commutative ring will do).

 

If there is a bilinear form acting on our vector space [math]g:V \times V \to \mathbb{R}[/math] such that [math]g(u,v) \in \mathbb{R}[/math], one declares that we are in a "metric space". The construction [math]g(u,v)[/math] is called an "inner product" - it defines distance length and angle in the loosest possible sense of these terms. By virtue of this mapping this defines the bilinear form [math]g[/math] to be a type (0,2) tensor. Confused? You will be.... wait and I will try to explain notation

 

Nice thing is it easily follows that [math]g(u,\cdot): V \to \mathbb{R}[/math] is a type (0,1) tensor, i.e. a covector, or one-form, or a linear functional aka an element in [math]V^*[/math]. Hold on to this Schroedinger......

 

Writing [math]g(u,\cdot) \equiv g_u[/math] one has that [math]g_u(v) = g(u,v) \in \mathbb{R}[/math] for any [math]v \in V[/math] (though some peeps invert the order here).

 

In the general case that our vector spaces are finite dimensional, whenever [math] V^{**}: V^* \to \mathbb{R}[/math], it is permissible to identify V** with V, so one has [math]v(g_u) = g_u(v) = g(u,v)[/math] so that v is acting as a type (1,0) tensor, a notion that comes as no surprise!

 

caveat - be aware there a lot of hand-waving going on here. I can give chapter and verse if required (though I doubt it would be welcomed).

 

Note also that every type (1,0) tensor is a vector, the converse is not true (likewise tensors of type (0,1))

 

I should point out that in standard notation, it is customary to refer to tensors by their scalar components, so if, say, [math]v =\sum\nolimits_i V^ie_i[/math] in some basis [math]\{e_i\}[/math] then one refers to this type (1,0) tensor as [math]V^i[/math], similarly for type (n,0) and type (0,n) and type (n,m) tensors, but with indices placed in the lower (or both upper and lower) positions accordingly. There is a good reason for this. Anyway, with a slight departure from this notation, the tensor [math]g_{ij}[/math] is called the "metric tensor"

 

So finally. The vector potential is indeed a 1-form, as the notation [math]A_{\mu}[/math] hints, though this doesn't make it (or [math]g_u[/math] for that matter) a metric. Which should be obvious - a metric measures the "distance" or "angle" between 2 entities, not one

 

But I cannot see where kinetic energy comes in. Maybe this claim could justified by the poster.

Edited May 8, 2011 by Xerxes

[ajb]

Think of the metric as a form of kinetic energy. You require a bilinear form to define the kinetic energy generally, this is the metric defined on the tangent space.

 

 

You can think about geodesics and other things in terms of a "Hamiltonian"

 

[math]H = \frac{1}{2} g^{A B}p_{B}p_{A}[/math]

 

on the cotangant bundle of a Riemannian manifold. Here p are the fibre coordinates. This formalism allows one to think of geodesics in terms of minimisation of "energy", you also can formulate this all in terms of a Lagrangian.

 

For me, the interesting thing is this allow Poisson geometry to enter the discussion of general relativity. This is something I must read up on more.

[Xerxes]

Ha! I am going senile! Looking back I see my last post was just a re-hash of an earlier one of mine in this thread. Sorry about that. Anyhoo, to continue my thought train.......

 

Recall I said that for [math]g_u \in V^*,\,\, v \in V[/math] it is permissible to think of the vector [math]v \in V [/math] as an element in V** (via a natural embedding) such that [math]v(g_u) = g_u(v) = g(u,v)[/math] provided only that V a finite-dimensional vector space.

 

First I believe the qualification above is false in all generality - there is a difficult (for me, at least) theorem of Riesz that states that this is also true of any Hilbert space of arbitrary dimension. Ho hum.....

 

Anyway. Look at [math]v(g_u) = g_u(v)=g(u,v)[/math]. This seems mad, right? Vectors and covectors are simultaneously treated as functionals and each as arguments of the other. One restores sanity by defining the bilinear form, sometimes called the natural pairing

 

[math]\langle \, \cdot \,,\, \cdot\rangle: V \times V^* \to \mathbb{R}[/math] where [math]\langle g_u, v\rangle[/math] is called the scalar product of a covector and a vector. Note it cannot possibly be an inner product.

 

So. Legend has it (true or false) that considerations like the above led P.A.M. Dirac to "invent" the so-called bra-ket notation. It goes like this.....

 

Suppose, for now without prejudice (as lawyers say) that [math]\langle u|v \rangle[/math] defines an inner product in the obvious way, then we might just as well regard this as the natural pairing above and say that the object, the bra, [math]\langle u|[/math] is a covector dual to the ket [math]|v\rangle[/math], a vector.

 

This notation is not without its drawbacks, which I don't have time right now to argue, but note that, in general, dual vectors (covectors, 1-forms) always exist and always act on vectors, but do not need an inner product to justify their existence. Now Dirac (as I understand) was trained as a mathematician, but made truly MAJOR contributions to physics, where inner products (or more generally) metrics are pretty much taken for granted, so the ambiguities introduced by this notation (at least as I see them) do not apply.

 

Why am I such a wind-bag?

Edited May 9, 2011 by Xerxes

[Xerxes]

Ya well, I seem to have effectively killed this thread. My apologies to the OPer.

 

When I think I am being "interesting", like here, I am usually on an ego-trip.

 

I cannot help myself - is counselling available?

 

And lurkers beware, I am as often wrong as I am right, but in the absence of posted corrections, how would you know?

[Schrödinger's hat]

Hahah, well. At least you can admit it.

You are forgiven.

Also, I think I'll blame you for my waning interest in GR (I'll get back to it sometime soon, I swear).

[anshul gupta]

Maybe you want to think about connections.

 

Gravity is White.

the realm of God starts from Gravity.

Gravity is now unzipped completely.

[ajb]

Gravity is White.

the realm of God starts from Gravity.

Gravity is now unzipped completely.

 

Ok, so how does this junk help?

 

I am of course talking about affine connections and in particular the Levi-Civita connection.

[Schrödinger's hat]

Ok, so how does this junk help?

 

I am of course talking about affine connections and in particular the Levi-Civita connection.

 

Wiki articles on such things are improving. That diagram for parallel transport is the best representation of the concept I've seen