# Coordinate-free relativity

1. Jun 18, 2010

### Quchen

Hey there,

Does anyone know a book that consequently uses coordinate-free expressions to develop general relativity? I've been looking for something for some time now, but everything I could find just briefly introduced the reader to concepts like exterior algebra, only to (almost) never use the concept again in the rest of the book (example: Einstein's general theory of relativity by Grøn and Hervik).
(I'd really love to see Einstein's field equations compressed to something small yet powerful like it's been done with Maxwell's equations, $\mathrm dF=0\,;$ $\mathrm d*F=4 \pi S$)

David

2. Jun 20, 2010

### Daverz

I would suggest Gravitational Curvature by Frankel.

3. Jun 20, 2010

### Quchen

Sounds good, especially because I've already got his other book, the geometry of physics. I've heard that Straumann wrote an excellent book too, but I cannot find it anywhere to have a look at the contents (at least not enough to spend 80 bucks on it).

4. Jun 21, 2010

### Goldbeetle

Last edited by a moderator: May 4, 2017
5. Sep 5, 2011

### JDoolin

I've heard that concept of "coordinate free relativity" used here and there, but I'm confused on what they mean by that.

Can you describe things without any reference point?

Can you actually picture anything, or draw anything, without it having a position and size? No, I don't think so.

Here, picture a box. Imagine it in your mind. Imagine it in the vacuum of space without any other particle around it.

No, because the moment you pictured the box, there is something else in the scenario: Your eyes. The very act of picturing something invokes an origin. There is no way you can picture anything, without an origin (your eyes), and without picturning it at a certain distance away from you.

And if it is a certain distance away from you, you can define a coordinate system based on that.

Coordinate free relativity, if it means what it sounds like it means, then it is nonsense. So it probably means something else.

6. Sep 5, 2011

### WannabeNewton

There is a level of abstraction involved, yes. Instead of defining tensors in the coordinate dependent way as quantities that transform according to $$T^{\alpha ...\beta }_{\mu ..\nu } = \frac{\partial x^{\alpha }}{\partial x^{\gamma }}...\frac{\partial x^{\beta }}{\partial x^{\delta }}\frac{\partial x^{\sigma }}{\partial x^{\mu }}...\frac{\partial x^{\lambda }}{\partial x^{\nu }}T^{\gamma... \delta }_{\sigma... \lambda }$$ you define an (m, n) tensor as a multi - linear mapping of one - forms and vectors: $$\mathbf{T}:\underbrace{V^{*}\times ...\times V^{*}}_{n}\times \underbrace{V\times ...\times V}_{m} \mapsto \mathbb{R}$$ Instead of giving things in terms of components you give things abstractly in terms of the tensor object itself. Differential forms are very useful in this context because they are formulated in a coordinate - free way.

Last edited: Sep 5, 2011
7. Sep 5, 2011

### JDoolin

I'm not entirely clear on the notation.

$$T^{\alpha ...\beta }_{\mu ..\nu } = \frac{\partial x^{\alpha }}{\partial x^{\gamma }}...\frac{\partial x^{\beta }}{\partial x^{\delta }}\frac{\partial x^{\sigma }}{\partial x^{\mu }}...\frac{\partial x^{\lambda }}{\partial x^{\nu }}T^{\gamma... \delta }_{\sigma... \lambda }$$

... defines a whole lot of different functions, right? The thing on the left hand side
$$T^{\alpha ...\beta }_{\mu ..\nu }$$ represents a function of \alpha, ..., \beta, \mu, ..., \nu, and produces a function. Am I right in assuming that you can replace \alpha, ..., \beta, \mu, ..., \nu with varibles? For nstance, \alpha,...,\beta might be (t,x,y,z) or (t,r,θ,φ) while \mu, ..., \nu would be (t',x',y',z') or (τ,ρ,Θ,Φ) or something along those lines...

(I should be taking into account the fact that there are coordinates on the right-side, too.)

But in the next step, you're taking

$$\mathbf{T}:\underbrace{V^{*}\times ...\times V^{*}}_{n}\times \underbrace{V\times ...\times V}_{m} \mapsto \mathbb{R}$$

Am I correct that each of the V's and V*'s represent some kind of class of vectors? So they are basically the inputs? The V*'s correspond to \alpha,...,\beta, while the V's correspond to \mu,...,\nu?

Then what is actually going on, is you're just notationally describing the math without referring to the coordinates. So the coordinates are actually still there; but they are just hidden, right?

I also notice that you have a bold-faced T over on the left (usually indicating a multi-dimensional quantity), and mapping to a real number on the right. Is that a typo?

In any case, I'm more interested in expanding out the first equation and understanding what it means. The second equation seems to be designed to shovel a bunch of complexity under the rug. I don't mind short-cuts, but I don't like learning the short-cut to an answer, when I don't even know what the question is.

What advantage does this coordinate free notation give you? Could it be described in terms, maybe an Object Oriented Programming expert might understand? Like, maybe V*, V*, V*, V, V, V are the high-level objects, but the end user programmer doesn't need to care or worry about the exact nature of the functions and subroutines the computer uses to render the output. However, without those functions being in place, your computer is not actually going to do anything, right?

So effectively, isn't the coordinate free representation, kind of like saying

Code (Text):
Show[AmazingRenderedOutput]
If someone else has gone through the work of creating the amazing rendered output, and set up the program with a syntax that will show it when you type that command, then you've really done something by typing that command. But it doesn't imply any understanding of what is really going on.

8. Sep 5, 2011

### WannabeNewton

Yes they are essentially just the components. The "..." is just to indicate there can be an arbitrary number of upper and lower indices.
The n $V^{*}$'s correspond to n copies of a dual vector space and the m $V$'s correspond to m copies of the associated vector space. For some point on a manifold, these would be the n copies of the cotangent space and the m copies of the tangent space respectively, and the tensor value of the tensor field at this point maps these copies of the cotangent and tangent space to the reals.
I feel like we might end up misinterpreting each other's words here but no the coordinates are not "hidden". You only need coordinates to obtain the components of a tensor relative to that coordinate's basis. The geometric object itself, which is the tensor, exists regardless of any coordinate system.
I made it bold just to indicate that we are talking about the tensor object and not its components. But it does map to the reals.
You don't rely on coordinates. It is more elegant and highlights the fact that the geometric object exists regardless of a coordinate system. Nature does not supply us with one right?
I am not a programmer by any means so I am sorry if I cannot help you much here but I can try in terms of Java: let's say you write an interface. You cannot really perform anything with the interface alone. What you can do is write classes that all implement this interface and these classes can now give that interface some functionality. Since these classes all implement this interface, they are all related by polymorphism. Even though the interface only has functionality when implemented by a class (or a bunch of classes) it still exists regardless of implementation, it has some abstract definition. Tensors are geometric objects that can be expressed in terms of components relative to a coordinate basis so that you can actually use them for calculations, and you can have a tensor expressed in many different coordinate systems that can be continuously mapped from one to the other, but the tensor itself exists regardless. I hope that helps but again, I'm sorry I couldn't help completely because I am not much of a programmer.

9. Sep 5, 2011

### JDoolin

I think you've basically got the gist of my object oriented programming comparison.

Now, if I understand right, we have a general method that will deal with all manner of coordinate systems. (Cartesian, Cylindrical, Spherical, Parabolic, Paraboloidal, Oblate spheroidal, Prolate spheroidal, Ellipsoidal, Elliptic cylindrical, Toroidal, Bispherical, Bipolar cylindrical, Conical). The methods of General Relativity can operate without error on any well defined coordinate system, and that is fine.

I have no problem with the idea, but just with the words used; calling it "coordinate free." Then it makes it sound like we could actually possibly use General relativity on an undefined coordinate system...

I may be forcing an analogy here, but isn't it sort of like taking a word-processor, which can deal with hundreds of different fonts, and calling it "font-free." Certaninly, you can take away all the fonts and still have the word-processor, but it won't work. It's going to either fail to compile or have a runtime error where it doesn't show you anything you're typing. The word-processor shouldn't be called "font-free" but more "font-ready."

So is the implementation really "coordinate-free" in that it still works, even when there is NO coordinate systems, or is it a "coordinate-ready" methodology that can work on any given coordinates.

10. Sep 5, 2011

### WannabeNewton

It is coordinate - ready in a sense. But you only need coordinates to do calculations. You can express the field equations in terms of the abstract tensor entities and not refer to a coordinate system at all. You are just using the abstract definitions of the tensors instead of how the tensor behaves in a certain coordinate system. It depends on what you mean by "works" because the coordinate - free approach does describe everything its just we can't really do any calculations unless we specify components relative to a basis.

11. Sep 5, 2011

### twofish-quant

It turns out that mathematically a lot of the complexity is unnecessary, and the reason people like coordinate-free descriptions is that it removes a lot of the unnecessary complexity.

Except that what you do is to give a broad description of your function, and once you know about the characteristics of that function, you do general stuff without knowing the details of what is in the inside of the function. If I know that the function is "const" then I can do things that I couldn't do if I didn't know.

This turns out to be important for compiler design.

If you really want your mind blown. Take a look at this wiki page.

http://en.wikipedia.org/wiki/Covariance_and_contravariance_(computer_science)

It turns out that you can use the same language to talk about C++ classes and vector spaces, and there is a branch of mathematics called category theory

But the cool thing is that you can make some statements about the without knowing what is going on. A lot of mathematics involves trying to figure out what the minimum description you need to say something useful.

To use an example. If I know that "dump as html" dumps out html, then I can take that output and pump it into another function that reads in html. I don't know what or how it dumps as html, and I don't care.

12. Sep 5, 2011

### twofish-quant

Yup, and it goes a bit further. I draw a coordinate system on an apple. I change the coordinate system. Nothing really changed. Now I draw a coordinate system on a flat plane. Things really are different. So there is a lot of mathematics that tells you when you are really changing things and when you aren't.

To use a programming analogy. You can write different programs to get the same output, so how do you know if you have two programs that give the same output, if you don't. This gets at the very deep connection between general relativity and compiler design, because when you are writing a compiler, you are trying to translate one set of instruction to another set of instructions that does exactly the same thing, only faster.

One way of thinking about this is imagine a 1GB dimension space in which each axis can take on the value 0 or 1. A computer program is a point in this space. Compiling and running a program is a set of instructions that describe how the point moves through space.

You don't need to define a coordinate system. You can say that space as a certain set of characteristics, and once you list the characteristics of that space, you end up with a set of coordinate systems.

Not really. Another analogy is that you could write a word processor with different languages, and you end up with the same output. The problem with using coordinate systems is that you can have different coordinate systems that describe the same space, and it's not obvious whether two spaces are the "same" or "different."

It works when there are no coordinate systems. You have to describe the space in some way, but you can do it without coordinate systems.

For example, if I say that I have a flat 2-d surface, I've just uniquely described the space. No coordinates. If I want to describe a cube, I describe a flat surface, I describe the way of taping the surfaces together. Again no coordinates.

13. Sep 5, 2011

### twofish-quant

It turns out that you can.

It turns out also that you can.

You can. But you don't have to. People have come up with a set of mathematical definitions that you can talk and think about spaces without using coordinates.

One way of thinking about how this works is imagine you are blind, and try to think about how you would describe an apple, and tell the difference between an apple and a grape to someone else that is blind. It's hard, but it can be done. If someone gives me an apple and a plate, they feel different, and an apple feels more like an orange than a plate does.

Last edited: Sep 5, 2011
14. Sep 6, 2011

### JDoolin

I cannot picture anything without a reference point. Tell me how you do it. How do you "picture" something without referencing your eyes?

Doesn't even a blind person have a sense of where the apple is? Doesn't the blind person have a sense of direction, that they can reach forward an take the apple? That the apple has a size and shape and location. Nature supplies to them a distance scale of "arm's lenghth" or an area scale of "about the size of my palm, from which they should naturally invoke a coordinate system.

Are you saying that if you are blind, all the world is just an amorphous reality with no direction, position, or scale? That's not what I have seen. When I see blind people, though, they seem highly aware of (and quite concerned about) the positions of things relative to themselves.

15. Sep 6, 2011

### redrzewski

16. Sep 6, 2011

### Staff: Mentor

Relativity is fundamentally a geometric theory. You can do a lot of geometry without using coordinates. Even notions of angles and distances are not based on coordinates but are geometrical. In fact, generally you define coordinates based on the underlying geometry, not vice versa.

17. Sep 6, 2011

### JDoolin

What do you mean?

18. Sep 6, 2011

### Ben Niehoff

These are all orthogonal coordinate systems on flat space. Such coordinate systems do not exist on curved spaces! You need something even more general.

We can and we do. I solve Einstein's equations in d dimensions without using coordinates all the time.

It is coordinate-free. It is entirely possible to do computations in coordinate-free notation without ever making reference to any coordinate system.

19. Sep 6, 2011

### Staff: Mentor

I mean basically all of Euclidean geometry can be done without coordinates. For example you can prove that the sum of the interior angles of a triangle is 180 deg, and many other things, all without ever using any coordinates.

20. Sep 7, 2011

### JDoolin

All of the proofs I see on that topic rely on geometric constructions on paper. But if you have drawn it on a piece of paper, (or even if you're just visualizing it as if it were drawn on a piece of paper), then there is a coordinate system.

Can you show me your proof? (or tell me where I can see a similar proof.)