Curvature as obstruction

Discussion in 'Physics & Math' started by arfa brane, Oct 31, 2017.

1. arfa branecall me arfValued Senior Member

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Leonard Susskind does 8 lectures on general relativity. I think lecture 1 can probably be skipped (that's an opinion mind you).

Anyways, the takeaway from lecture 2 is that curvature is an obstruction to finding a flat metric, or rather to finding a way to define distances between points which requires only Pythagoras. That is, (being able to define) a 'surface' which can have coordinates defined on it which are rectilinear.

The other is that a gravitational source is a spherical body with mass. The obstruction is that tidal forces exist because the field around such a body diverges, so you can't transform (except over small regions) to an accelerated frame which is in free fall (such that gravity vanishes)--the equivalence principle applies in a strictly local sense, trying to make it global encounters the obstruction (tidal forces due to divergence).

Then there is some stuff Gerard t'Hooft wrote in SciAm about what happens when you take a global symmetry and try to make it a local symmetry, which requires the addition of a 'force' to restore the symmetry (the global symmetry recall, is the spherical shape of gravitational sources, locally over small distances and times, this symmetry breaks because it looks flat).

Or something like that.

Another thing, the general theory says nothing else about the structure of bodies with mass (acceleration in a gravitational field is independent of the 'material nature' of a body with mass), that is, it says nothing about the structure of matter fields, treating them only as densities.

The other big theory we have, quantum mechanics, says nothing about what a physical measurement is (in accelerated or inertial frames; frames don't really exist in QM, we have to invent them to do experiments), in fact that theory isn't physical at all; it's about complex amplitudes which define probabilities, right? So that there's a gap between these two theories seems obvious, maybe.
No takeaway there, however.

Last edited: Oct 31, 2017
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3. arfa branecall me arfValued Senior Member

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So, partly I'm kind of saying something to any budding neo-Einstein with a TOE: you will need to explain how the fine-grained structure of matter affects gravity, Einstein's theories simply don't do this. You will also need to explain how a theory of complex probability amplitudes can explain gravity as an interaction between matter particles; Einstein doesn't do that either.

One more thing: entanglement will need to be a part of your theory, and it will need to explain what the connection, if there is one, between entanglement and gravity is. It seems likely that a quantum theory of gravity will necessarily include entanglement and some kind of measurement paradigm; QM says nothing about measurement and gravity is kind of all about measurement.

For instance, your weight on the surface of a gravitating mass constitutes a measurement, so does any weight. If you accelerate under free fall, again that's a measurement.

5. arfa branecall me arfValued Senior Member

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There is a lot to understand about the mathematical treatment of curvature in GR. You can always refer to two dimensional surfaces like the surface of a sphere or of a cone. Curvature can be understood via the notion of parallel transport.

Susskind explains how you can find out if a surface is curved or not by using parallel transport. I found his explanation of doing this on the surface of a cone around a closed curve was fairly intuitive: a cone has zero curvature since you can always slice it open and make it flat; the curvature "lives" at the apex however, since if you parallel transport a vector along a closed curve which goes around the apex it will point in a different direction when you get back to where you started.

This is intuitive because you can start with say, an isosceles triangle that can be rolled up into a cone. If you draw a line from one side of this triangle to the other so its endpoints meet when the triangle is rolled up into a cone, then it's obvious that a vector parallel to one side of the flat triangle (which is really a cone), isn't parallel to the other side. On the other hand any closed curve which doesn't include the apex of the cone is a flat curve, so parallel transport will leave a vector pointing in the same direction it started with.

The next step is that a cone can be made tangent to a sphere at some line of latitude (the cone then is the tangent space at that latitude), so obviously parallel transport will mean a change in the vector's angle, the change depends on the angular excess which is described by the cone (!). This generalises (yay!), so that any curved surface can be said to have a tangent space for any closed curve, including a closed 'curve' which is defined by vectors such as $\partial_x, \partial_y$ where $\partial_x \partial_y$ is proportional to the area inside the curve (which is a rectangle or parallelogram obviously).

This is also why spheres have 1-sphere boundaries--the tangent space at the equator of a sphere is a cylinder, not a cone, although both are flat a cylinder has no singular point (apex).
The mathematics of vector calculus (covariant and contravariant derivatives, gradients, divergence) suffices for curvature in two dimensions (but you get extra terms which are more conveniently defined using what Susskind calls "gammas", actually Christoffel symbols).
You need to upgrade to tensor calculus otherwise, and spacetime is of course four-dimensional, so the rank 1 tensors (vectors) are 4-vectors.

Simple eh?

Last edited: Nov 2, 2017

7. river

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Is not a , sphere , a three dimensional circle ? And a cone a three dimensional shape that has , length , height and breadth .

Just saying .

8. arfa branecall me arfValued Senior Member

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Start with a cylinder. It's a flat surface which is embedded in three dimensions.

A cylinder is the tangent bundle of a circle. It is because a bundle is the disjoint union of sets to form a larger set, the tangents to a circle aren't strictly distinct or disjoint in two dimensions. That is, if you want disjoint sets in the bundle you can't have intersections.
Likewise a cone is the tangent bundle of a circle which is a line of latitude on a sphere.

In each case, the cylinder and cone, you can have them lying on, i.e. tangent to, a sphere.

Watch the first 20 minutes of this lecture from the WE Hereaus Winter School, see what you think.

Last edited: Nov 6, 2017
9. arfa branecall me arfValued Senior Member

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If you have trouble with the concepts of differential calculus, this video tutorial by DrPhysicsA is quite good. He explains in essence, that a tensor is really just a thing that expresses a relation between two vectors (but vectors are just rank 1 tensors anyway, don't forget).

But you're just going from one coordinate system to another (in the above, at least initially), where you have some overlap between 'charts', to find this relation between the coefficients in both systems for the same vector, the tensor is this relation, so the man says.

10. geordiefRegistered Senior Member

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Maybe a different series?

11. arfa branecall me arfValued Senior Member

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Sorry about that. Yes that tutorial is one of the vids. It's probably easier to just do a search at youtube.

Anyway, I'd like to post something about what I think I now understand about gravity.

First of all, it's about differential calculus, not anything necessarily physical--it isn't until you introduce a field with at least a value defined everywhere, i.e. a scalar field, that physics arrives. That is, you need to equip the space with extra structure in the form of vectors. If you have a scalar field already, then you need to add directions, you introduce the idea that a potential changes by increasing or decreasing along a small distance ds. Hence this change will have a direction and differential calculus solves for this.
The spatial derivative of a potential is a force, which is a vector.

12. QuarkHeadRemedial Math StudentValued Senior Member

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Then I do not agree with him. Look.......

Suppose $V$ a Real vector space and that $V^*$ is the set of all mappings $V \to \mathbb{R}$ (this is called the dual of our vector space) so that for any $v \in V,\,\omega \in V^*$ that $\omega(v)= \alpha \in \mathbb{R}$.

It is true that in the case of a finite dimensional vector space ( the proof is tricky) that the dual space to $V^*$ is $V^{**} \equiv V$.

So a tensor of type (0,2) is a bilinear mapping $\omega \otimes \varphi: V \times V \to \mathbb{R}$, where $\omega,\,\varphi \in V^*$

Likewise, a type (2,0) tensor is a bilinear mapping $v \otimes w: V^* \times V^*\to \mathbb{R}$

We may also have, for example, that $\omega \otimes v:V \times V^* \to \mathbb{R}$, this bring a type (1,1)tensor.

This extends, in any finite dimension, to multilinear mappings and thus to type (r,s) tensors

13. arfa branecall me arfValued Senior Member

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Ok, but how do you demonstrate he's wrong and you're right?

I can't see how the rest of your post does this. Further, Einstein in his 1915 paper seems to intimate that tensors are indeed, a way to relate systems of coordinates. The metric tensor preserves distances for instance, but this property is only defined on two different but overlapping charts. So how is a tensor not a thing that relates vectors to vectors?

14. Q-reeusValued Senior Member

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Just diving in and out here to point out something fairly obvious: it depends. A rank 0 tensor field such as that pertaining to (uniform) pressure in a container of gas has - no vectors to compare. A rank 1 tensor field e.g. E field obviously is a vector field. A rank 2 tensor field may map between e.g. vector fields but also can represent e.g. a strain field that is not vectorial in character.
http://mathworld.wolfram.com/TensorRank.html

15. arfa branecall me arfValued Senior Member

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So, rank 1 tensor fields relate scalars to scalars, rank 2 tensor fields relate vectors to vectors.
Perhaps DrPhysicsA is generalising a little.

16. QuarkHeadRemedial Math StudentValued Senior Member

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No.

First you do not say what you mean by "relates" here. I repeat: for any vector space $V$ there exists a dual space $V^*$ such that $V^*:V\to \mathbb{R}$ (which is a mapping, not a "relation") so that for $v \in V$ (a rank 1 tensor) and $\omega \in V^*$ (also a rank 1 tensor) that $\omega(v) = \alpha \in \mathbb{R}$.

Likewise, given the product $V^*\otimes V^*:V \times V \to \mathbb{R}$, then $\omega \otimes \varphi(v,w) \in \mathbb{R}$, or in generalized scalar component form, $\omega_{jk}(v^j,v^k) \in \mathbb{R}$.

So consider the so-called metric tensor at a single point. This has exactly the same form as above, but is equivalent to the so-called inner (or scalar) product of 2 two vectors.

And of course, the distance between any 2 points in any space is the "length" of the vector that connects them. This "length" is also given by the inner product of a vector with itself.

17. arfa branecall me arfValued Senior Member

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Regardless of whether it's accurate to call tensors relations, a thing about the metric I noticed is it seems to be just a way to rewrite Pythagoras' relation between the squares of sides of a right triangle. This rewrite is in the form of differentials and gradients.

There's some hint this doesn't apply when the surface is curved (the metric isn't Euclidean), but not much insight into how the metric 'sees' curvature.

For instance, you can construct a diagram like this:

Which you then think of in terms of general covariance; The $X = x^i$ and $Y = y^i$ can be chosen in any orientation and position relative to each other. So you are constructing it with a common origin for the sake of convenience.

Ok, so you then show that you can write the square of the length OP in terms of differentials, but how does that relate (sorry about that) to curvature? You have to see somehow, that this metric you derive in a flat space extends to non-flat spaces.

Last edited: Nov 11, 2017
18. arfa branecall me arfValued Senior Member

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Suppose I try to draw a small right triangle on a sphere, with one side, $x^1$ on a circle of latitude, then the other side $x^2$ is on a meridian.

If I transport a vector along $x^1$ keeping it parallel to say, $x^2$, then it won't be parallel to this initial direction after some displacement $dx^1$, but parallel to a different circle of latitude at $x^1 + dx^1$, and not to the meridional line there.
I can show, by flattening part of this sphere into part of a triangle, that it's sides are not parallel but meet somewhere.

Is that the point of having a metric tensor, that Pythagoras fails so to say, on a sphere and it can say by how much?

19. arfa branecall me arfValued Senior Member

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Well, DrPhysicsA shows that you can derive a metric on Euclidean 2-space (presumably also on Euclidean n-space), starting with Pythagoras, but using the notion of differential distances.

Because of the rules of calculus, you can do things like: $\frac {\partial y} {\partial x} = \frac {\partial y} {\partial z} \frac {\partial z} {\partial x}$.

Also, if there is only one variable $\frac {dy} {dx} = \frac {\partial y} {\partial x}$, so then $dy = \frac {\partial y} {\partial x} dx$. Now the r.h.s. looks like a linear equation, a gradient times a displacement.

So we can do this with an ordinary triangle like so

Then we have $ds^2 = (dx^1)^2 + (dx^2)^2$.
But this is also $ds^2 = (dx^1)(dx^1) + (dx^2)(dx^2)$.

Using implied summation this is more succinct as $(dx^m)(dx^m)|_{m=1,2}$.

The interesting part is that we can rewrite this as $(dx^m)(dx^n)·\delta_{mn}$.
What this says is you need to relate the use of different indices to a function that "corrects" for unwanted "cross products" that aren't in Pythagoras' formula. It is still the same as the first formula but has indices, and a function acting on the indices.

Last edited: Nov 12, 2017
20. arfa branecall me arfValued Senior Member

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Next, define another indexed coordinate set, $\{ y^1,\; y^2\}$, such that $dy^r dy^s·\delta_{rs}$ is also equal to $(ds)^2$.
This means changing to $\{ y^1,\; y^2\}$ is a transformation of ds that preserves (the square of) its length.

Then we have $dy^r dy^s·\delta_{rs} = dx^m dx^n·\delta_{mn}$

21. arfa branecall me arfValued Senior Member

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Susskind hints that the inner product of a vector space is tied to the metric, since we can say the $x^m$ have unit basis vectors, although we're considering small displacements, $dx^m$, these can still be the coefficients of orthonormal basis vectors ${e^1,\; e^2}$.

Then $dx^m dx^n⋅ \delta_{mn} = dx^m dx^n[e^m\cdot e^n]$.

This is still just two dimensions, because of the dot product between two orthogonal vectors being zero you get the same sum as on the left: $e^1\cdot e^1 + e^2\cdot e^2$, the mixed terms vanish.

So now because $dx^m dx^n = \frac {\partial x^m} {\partial y^r} dy^r \frac {\partial x^n} {\partial y^s} dy^s$ this is a way to transform from the X to the Y system (in flat space). Somehow this object (a rank 2 tensor when the Kronecker delta or inner product is a multiplicand), says something about curvature.

You write: $[e^m \cdot e^n]dx^m dx^n = [e^m \cdot e^n]\frac {\partial x^m} {\partial y^r} \frac {\partial x^n} {\partial y^s} dy^r dy^s = g_{mn}dy^r dy^s$

But parallel transport, defined somehow around a small closed curve (a parallelogram will do), is how curvature is defined. Not being able to keep a vector parallel to itself under transport on a surface is the obstruction to finding a flat space. So how do you do it mathematically? You use a connection, a mathematical object that describes a path through an abstract topological space called a fiber bundle. How do tensors and connections measure curvature, and why do it over very small, even infinitesimally small distances?

One of the things it seems clear you will have to define is changing the direction of transport while keeping the direction of the vector constant. Again, how do you define this?

Last edited: Nov 13, 2017
22. arfa branecall me arfValued Senior Member

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Note the difference between the left and right hand side of the last equation in the above post.

On the left you have $[e^m \cdot e^n]$, on the right you have $[e^m \cdot e^n]\frac {\partial x^m} {\partial y^r} \frac {\partial x^n} {\partial y^s}$ as the tensor.

What kind of object is $\frac {\partial x^m} {\partial y^r} dy^r$? It's an implied sum over r. But, in two dimensions, you write out all the terms thusly: $\frac {\partial x^1} {\partial y^1} dy^1 + \frac {\partial x^1} {\partial y^2} dy^2 + \frac {\partial x^2} {\partial y^1} dy^1 + \frac {\partial x^2} {\partial y^2} dy^2$.

It looks like a thing with coefficients (the differential gradients) and these are what the metric tensor has in it, but it's like you multiply out a set of gradients by the same set (a set product?):
$\{ \frac {\partial x^1} {\partial y^1},\; \frac {\partial x^1} {\partial y^2},\;\frac {\partial x^2} {\partial y^1},\;\frac {\partial x^2} {\partial y^2}\}\cdot\{\frac {\partial x^1} {\partial y^1},\; \frac {\partial x^1} {\partial y^2},\:\frac {\partial x^2} {\partial y^1},\; \frac {\partial x^2} {\partial y^2} \} = \delta_{rm} ⋅ \frac {\partial x^m} {\partial y^s} \frac {\partial x^n} {\partial y^r}$.

Akin to an inner product of a vector with itself.
[Ed. no that's too sloppy, the lhs implies 16 summands, there are only 8, so it should be
$\{ \frac {\partial x^1} {\partial y^1},\; \frac {\partial x^1} {\partial y^2}\} \cdot \{ \frac {\partial x^1} {\partial y^1},\; \frac {\partial x^1} {\partial y^2}\} + \{ \frac {\partial x^2} {\partial y^1},\;\frac {\partial x^2} {\partial y^2}\}\cdot \{ \frac {\partial x^2} {\partial y^1},\;\frac {\partial x^2} {\partial y^2}\} = \delta_{rm} ⋅ \frac {\partial x^m} {\partial y^s} \frac {\partial x^n} {\partial y^r}]$

Last edited: Nov 13, 2017
23. arfa branecall me arfValued Senior Member

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It's occurred to me that on a surface with positive curvature, if you transport in the same direction a pair of vectors which are initially parallel, they will start to converge and not be parallel after some distance, on a surface with negative curvature the pair of vectors will diverge.

I believe transporting a vector in the direction it's pointing is called autoparallel transport; this means the vector is tangent to the direction of "motion"; in physics, the velocity is always tangent to the trajectory of a particle, so must be an example of autoparallel transport. That also means velocity is always in the tangent space of the trajectory.

But you need a mathematical description of this transport. I know that adding a pair of vectors in $\mathbb {R}^n$, means you transport one of them (it doesn't matter which, addition is commutative), so its tail is at the tip of the other--but maybe that doesn't work in curvilinear coordinates.