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Faldo_Elrith

Posted: Wed Jul 02, 2008 9:48 am Post subject: General Relativity

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Hi all. I'm intending to study GR and I need some good tips on tensors and Riemannian geometry. So, any tips? Very Happy

Guitarist

Posted: Wed Jul 02, 2008 1:58 pm Post subject:

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Ya, well, I did attempt to explain tensors a year ago on this very site, I can't find the thread just now, but, as I recall it came at the end of a long tutorial on vector spaces,

I would be willing to start up again in a new thread, but only on the assumption that you knew all there is to know about inner product spaces....

the Riemann geometry would be less easy to explain (not that I know too much about it), but you would certainly need to know all about topological spaces and manifolds, as a minimum.

Are you cool on these "pre-requisites"?

I warn you - at a mathematical level, GR is fiendishly hard. It is not for the ill-prepared!!

Faldo_Elrith

Posted: Thu Jul 03, 2008 6:43 am Post subject:

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Would be really great if you could explain tensors. I have this book, see, and the very first chapter dives straight into tensors without properly saying what these things are supposed to be. Bang Head

Guitarist

Posted: Fri Jul 04, 2008 11:39 am Post subject:

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Yes, and I already said I would be willing to help. But you need to tell us

a) do you have have a working knowledge of vector spaces;

b) do you know what an inner product space is;

c) are you familiar with dual vector spaces;

d) know what a Cartesian product is;

e) know what the tensor product is?

You may, if it is true, answer NO to all the above, then I would still be willing to help, but mainly by pointing to my earlier thread.

But I say again, the mathematics of GR is fiendish.

Faldo_Elrith

Posted: Fri Jul 04, 2008 4:14 pm Post subject:

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a) yes
b) no
c) no
d) yes
e) no

Thanks!

mitchellmckain

Posted: Mon Jul 07, 2008 2:01 am Post subject:

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Joined: 06 Oct 2005
Posts: 2078
Location: Salt Lake City, UTAH, USA

Faldo_Elrith,

I may be able to help a little, but more in the role of a fellow student.

Put quite simply a tensor is a generalization of the idea of vectors and matrices and if you are used to the index notation for vectors and matrices then tensors are a pretty natural extention of these.

It sounds like Guitarist is offering the more mathematically rigorous explanation.

By the way what is the GR text you are talking about.

Guitarist,

I am pretty much a self-tutored novice when it comes to GR and so there may be much that I can learn from you. I worked through the some of the basic calculations in Schutz' book, but not all that recently. I have a masters from the University of Utah, but they really didn't have much to offer in GR. When it comes to playing a teaching role in relativity I pretty much stop at SR.

Quote:

b) do you know what an inner product space is;

Surely Faldo_Elrith you know what an inner product is. Guitarist isn't this pretty much just vector space plus inner product.

Quote:

c) are you familiar with dual vector spaces;

Off the top of my head, I would guess this is basically an extension of the vector space (or rather an inner product space) to include the distinction made between row and collumn vectors (and where the inner product is an operation only between the two different types of vectors).

Quote:

d) know what a Cartesian product is;

I believe this is just the space of ordered pairs of elements of the two spaces you are taking the Cartesian product of.

Quote:

e) know what the tensor product is?

This is just a tensor formed by multiplying the components of two tensors, where none of the indices of any of the tensors are the same. An outer product of a row vector with a collumn vector to produce a matrix is an example of this.

The usual multiplication of matrices and vectors, including the inner product between a row vector with a collumn vector is a contraction of a tensor product.
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Guitarist

Posted: Mon Jul 07, 2008 9:55 am Post subject:

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OK, mitch & Faldo; So we all know what we mean by a vector space. Good.

Let's first review what we mean by an inner product space.

Let

be a vector space, and let $u,\,v \in V$ . Let $V \times V$ be the Cartesian product on this space (strictly, it's the Cartesian product on the underlying set. Don't worry)

Then we we have that $(u,v) \in V \times V$

I define the inner product on

as $g:V \times V \to \mathbb{R},\,g(u,v) = \alpha \in \mathbb{R}$ . In other words, the inner product of 2 vectors is a number (it need not be real, by the way)

An inner product space (IPS) is just a vector space where this construction makes sense.

Let us now find some gadget that takes a vector as argument and returns a number, say $\psi(v) = \beta \in \mathbb{R}$ . This guy is called a "linear functional", and it (and its mates) have the property that

$(\psi + \phi)v = \psi(v) + \phi(v)$ , and

$\alpha(\psi(v)) = \psi(\alpha(v))$

This (with a couple of trivial extras) is sufficient to define $\psi,\,\phi$ as elements in a vector space. It is called the dual space V* to V, and, where V is an IPS, we have following;

$\psi(v) = g(u,v) = \alpha$ for any appropriate choice of $\psi,\,u,\,v$ .

One says that $\psi_u \in V^*$ is the dual vector to $u \in V$ . This is a one-to-one correspondence between V* and V - an isomorphism.

Lemme know if we're all cool with this so far, as it is absolutely fundamental

P.S Oh mitch I believe Schutz uses a slightly different notation/terminology for dual vectors. Doesn't he call them 1-forms? - we can talk about that if you like

mitchellmckain

Posted: Mon Jul 07, 2008 11:46 am Post subject:

Forum Cosmic Wizard

Joined: 06 Oct 2005
Posts: 2078
Location: Salt Lake City, UTAH, USA

Guitarist wrote:

OK, mitch & Faldo; So we all know what we mean by a vector space. Good.

Let's first review what we mean by an inner product space.

Let

as $g:V \times V \to \mathbb{R},\,g(u,v) = \alpha \in \mathbb{R}$ . In other words, the inner product of 2 vectors is a number (it need not be real, by the way)

I am confused by this last parenthetical. Is not your funny looking R the set of real numbers? Are you restricting yourself to inner products that are real but saying that you do not have to?

Guitarist wrote:

An inner product space (IPS) is just a vector space where this construction makes sense.

Let us now find some gadget that takes a vector as argument and returns a number, say $\psi(v) = \beta \in \mathbb{R}$ . This guy is called a "linear functional", and it (and its mates) have the property that

$(\psi + \phi)v = \psi(v) + \phi(v)$ , and

$\alpha(\psi(v)) = \psi(\alpha(v))$

This (with a couple of trivial extras) is sufficient to define $\psi,\,\phi$ as elements in a vector space. It is called the dual space V* to V, and, where V is an IPS, we have following;

$\psi(v) = g(u,v) = \alpha$ for any appropriate choice of $\psi,\,u,\,v$ .

Ok, now you have lost me.

Do you mean that for any $\psi,\,v$ there exists a

such that $\psi(v) = g(u,v)$

Guitarist wrote:

One says that $\psi_u \in V^*$ is the dual vector to $u \in V$ . This is a one-to-one correspondence between V* and V - an isomorphism.

Are you then implying that this existence of a

for every $\psi$ represents a 1-1 correspondence between this vector space of linear functionals $\psi$ and the original vector space V -- i.e. that given any two linear functionals $\psi,\,\phi$ and the corresponding $u,\,v$ (such that for any

, $\psi(w) = g(u,w)$ and $\phi(w) = g(v,w)$ ), you can prove that u=v if and only if $\psi = \phi$

So because of this 1-1 correspondence you are now calling this linear vector space of linear functionals the dual vector space of V (denoted by V*)?

At first this index of u on psi in $\psi_u \in V^*$ confused me but I guess this is just a way representing this 1-1 correspondence and you would read $\psi_u$ as the linear functional corresponding to u.

Guitarist wrote:

P.S Oh mitch I believe Schutz uses a slightly different notation/terminology for dual vectors. Doesn't he call them 1-forms? - we can talk about that if you like

You are correct but I am hardly glued to that one book, I have studied Wald and have struggled with Hawking&Ellis (The Large Scale Structure of Space Time) so I am familiar with their terminology though I do not claim anything approaching a mastery of these texts.
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Guitarist

Posted: Mon Jul 07, 2008 12:29 pm Post subject:

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mitchellmckain wrote:

Are you restricting yourself to inner products that are real but saying that you do not have to?

Yes - for now it is is just for simplicity. If we were in the Math forum, I would simply write the ambiguous field $\mathbb{K}$ . But, since we are in Physics, we require all "measurements" to be real numbers (yes the inner product and its big brother the vector norm are measurements)

Quote:

Do you mean that for any $\psi,\,v$ there exists a

such that $\psi(v) = g(u,v)$

Um, not quite, but close. Turn it around: for any $u,\,v \in V$ such that $g(u,v) =\alpha \in \mathbb{R}$ , there is some $\psi \in V^*$ such that $\psi(v) = g(u,v) = \alpha \in \mathbb{R}$ .

It was merely to emphasize this correspondence that I wrote $\psi_u(v) = g(u,v) = \alpha \in \mathbb{R}$ - it is not standard

Quote:

Are you then implying that this existence of a

for every $\psi$ represents a 1-1 correspondence between this vector space of linear functionals $\psi$ and the original vector space V

I would prefer to think of it as the other way round - the existence of some $\psi$ for each

. But hey - it's an isomorphism, so who cares? You are not wrong!

Quote:

that given any two linear functionals $\psi,\,\phi$ and the corresponding $u,\,v$ (such that for any

, $\psi(w) = g(u,w)$ and $\phi(w) = g(v,w)$ ), you can prove that u=v if and only if $\psi = \phi$

Yep, spot on! Nice work

Quote:

So because of this 1-1 correspondence you are now calling this linear vector space of linear functionals the dual vector space of V (denoted by V*)?

Correct.

mitchellmckain

Posted: Mon Jul 07, 2008 3:36 pm Post subject:

Forum Cosmic Wizard

Joined: 06 Oct 2005
Posts: 2078
Location: Salt Lake City, UTAH, USA

Then we can either wait for Faldo_Elrith's reply or you can continue, because I am with you.
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Faldo_Elrith

Posted: Mon Jul 07, 2008 4:07 pm Post subject:

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Posts: 76

Wow this is great!

Guitarist, can you explain what you mean by "linear functionals"? I'm with you up to that bit there.

Cheers mate.

mitchellmckain

Posted: Mon Jul 07, 2008 8:50 pm Post subject:

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Location: Salt Lake City, UTAH, USA

Faldo_Elrith wrote:

Wow this is great!

Guitarist, can you explain what you mean by "linear functionals"? I'm with you up to that bit there.

Cheers mate.

Well since Guitarist has not answered right away, I will attempt to anwer that. I think it basically means these properties which Guitarist already gave.

Guitarist wrote:

$(\psi + \phi)v = \psi(v) + \phi(v)$ , and

$\alpha(\psi(v)) = \psi(\alpha(v))$

Notice that matrix multiplication has these same properties. That is if A and B are matrices then (A + B) v = Av + B v and for a scalar a, we also have that A (a v) = a (A v).
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Guitarist

Posted: Tue Jul 08, 2008 1:09 am Post subject:

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Faldo_Elrith wrote:

Guitarist, can you explain what you mean by "linear functionals"? I'm with you up to that bit there.

Yeah, I skated over that a bit. mitch was as right as he could be, given the limited information I provided.

First this, though. A "functional" is really just a special sort of function, one that takes a vector (or a function, for that matter) as input and returns a scalar (a number) as output.

I tend to be sloppy, and use the word "map" for anything that can be thought of as an arrow.

Now to show linearity, we really need this. If any mappings whatever, say $f,\,g$ , satisfy

$(\alpha + \beta)f = \alpha(f)+\beta(f)$

$\alpha(f + g) = \alpha(f) + \alpha(g)$ for any scalars $\alpha,\, \beta$ , one say they are linear.

So. When we are handed, say, a vanilla real-valued function we would be inclined to write $f:\mathbb{R} \to \mathbb{R}$ . We would usually take this to mean that this function acts on each and every real number and returns another real number.

In the present case, we have a whole vector space V* of linear functionals, in fact we have one such functional for each input vector. So we may write it like this;

There is a mapping $V^*:V \to \mathbb{R}$ such that, for all $\psi \in V^*,\, v \in V$ , we will have $\psi(v) = \alpha \in \mathbb{R}$ .

Now if an inner product g(u,v)

is defined on

, by the isomorphism $V \cong V^*$ I can always find some $\phi_u$ such that $\phi_u(v) = g(u,v)$ .

So, this is all just a repeat of what I said in an earlier post. For novelty, let me add this:

V^*

is a perfectly respectable vector space. As such it is entitled to its own dual space, which we may write as $(V^*)^* = V^{**}$ . We will expect that the action of V** on V* to be similar to the action of V* on V, that is to return a scalar.

Now, it is a classic result in linear algebra that, whereas the isomorphism $V^* \cong V$ is not "natural" (in the sense that it depends on an arbitrary choice of basis for

), there is a natural isomorphism $V^{**} \cong V$ .

This is going to allow us to do something useful

mitchellmckain

Posted: Tue Jul 08, 2008 3:54 am Post subject:

Forum Cosmic Wizard

Joined: 06 Oct 2005
Posts: 2078
Location: Salt Lake City, UTAH, USA

Guitarist wrote:

So, this is all just a repeat of what I said in an earlier post. For novelty, let me add this:

V^*

), there is a natural isomorphism $V^{**} \cong V$ .

Doesn't this have to do with transformation properties like the difference between covariant and contravariant?

So for a transformation f on V: $u\rightarrow fu$
the linear functionals must transform like $\psi\rightarrow \psi f^{-1}$
so that $\psi(u)\rightarrow \psi(f^{-1} f u) = \psi(u)$

Doesn't this difference in transformation properties result in making the commutivity diagram requirement for natural isomorphisms impossible if it is between linear vector spaces of objects that transform in these two different ways?
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Faldo_Elrith

Posted: Tue Jul 08, 2008 5:05 am Post subject:

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Guitarist, I still don't fully get you. Bang Head

Guitarist wrote:

First this, though. A "functional" is really just a special sort of function, one that takes a vector (or a function, for that matter) as input and returns a scalar (a number) as output.

Well, that bit I understand. But elsewhere it seems that you are also trying to add functionals! How can you do that? You can add vectors, you can add scalars, but you CANNOT add functions. Bugeye

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