The Foundations of Relativity

**Hurkyl** · Jan7-06, 10:46 PM

pervect: I concur with George Jones. The dual basis $LaTeX Code: \\epsilon^i$ to a basis

is a very, well, basic part of linear algebra. Among other things, it is the only basis such that, for any vector v, we have that $LaTeX Code: v = \\sum_i \\epsilon^i(v) e_i$ .

Basically, I'd rather keep $LaTeX Code: \\epsilon^i = g^{ij} e_j$ , something that is true with (1) and not true with (2).

That expression doesn't even make sense. The left hand side is an upper-indexed collection of covectors. The right hand side is an upper-indexed collection of vectors -- the types don't match.

To state it differently, each of the

's is a vector, and thus has an upper index, which I will denote with a dot. Each of the $LaTeX Code: \\epsilon^i$ 's is a covector, and thus has a lower index, which I will denote with another dot. You're claiming that $LaTeX Code: \\epsilon^i_\\cdot = g^{ij} e_j^\\cdot$ , but the indices don't line up.

Another shorter way of saying this - the mapping A from V to V* is actually $LaTeX Code: g^i{}_j$ in tensor notation, a mixed tensor of rank (1,1)

No, it is not! This is an elementary fact about dual spaces: there does not exist a natural map from a vector space to its dual. Loosely speaking, everything transforms wrongly -- if I change coordinates in

by doubling everything, I need to change coordinates in

by halving everything. A tensor would do the wrong thing!

Remember that, in the way people usually do coordinates, the action of a covector $LaTeX Code: \\epsilon^i$ on a vector

is simply given by $LaTeX Code: \\epsilon^i(e_j) = \\epsilon^i_k e^k_j$ -- it is determined by a rank-(1,1) tensor. The only rank-(1,1) tensors whose components are coordinate-independent are the multiples of $LaTeX Code: \\delta^i_j$ , and thus $LaTeX Code: \\epsilon^i(e_j) = c \\delta^i_j$ are the only possible choices if you want the components of this expression to be independent of the coordinates.

**Oxymoron** · Jan8-06, 04:56 AM

Ive been trying to follow this. From what I can gather, the definition of a metric tensor which I wrote about 12 posts ago was the "metric tensor of an inner product". I neglected to write inner product because, well, at that stage it was the only metric tensor I knew.

See I was considering a real inner product space $LaTeX Code: (V,\\cdot)$ with a map

$LaTeX Code: g\\,:\\, V \\times V \\rightarrow \\mathbb{R}$

defined by taking the inner product of two arguments:

$LaTeX Code: g(u,v) = u\\cdot v$

Then one could say that when the tensor $LaTeX Code: g_{ij}$ acts on two elements from

such that

$LaTeX Code: g_{ij}u^iv^j$

then this action is simply taking the inner product. That is

$LaTeX Code: u\\cdot v = g_{ij}u^iv^j$

In my opinion, the Kronecker delta, $LaTeX Code: \\delta^i_j$ should be used with caution in general. For example, in a Euclidean inner product space with a metric tensor you can find an orthonormal basis $LaTeX Code: \\{e_1,\\dots,e_n\\}$ such that

$LaTeX Code: e_i \\cdot e_j = g_{ij} = \\delta_{ij}$

This way, $LaTeX Code: \\delta_{ij}$ is a tensor and we have the special property

$LaTeX Code: g_{ij} = \\delta_{ij} = g^{ij}$

and so every tensor can be written with all its indices raised or lowered, since raising and lowering has no effect on the values of the components of the tensors. But, of course, you can only do this in Cartesian space where there is a good notion of orthonormality.

Im not sure If I have helped or not. I just wanted to clarify what I was talking about.

**robphy** · Jan8-06, 08:14 AM

Originally Posted by pervect

Another shorter way of saying this - the mapping A from V to V* is actually $LaTeX Code: g^i{}_j$ in tensor notation, a mixed tensor of rank (1,1). Defining $LaTeX Code: g^i{}_j$ is as valid a way of defining a metric as defining $LaTeX Code: g_{ij}$ or $LaTeX Code: g^{ij}$

Originally Posted by Oxymoron

In my opinion, the Kronecker delta, $LaTeX Code: \\delta^i_j$ should be used with caution in general.

Just a comment on the Kronecker delta $LaTeX Code: \\delta^i{}_j$ ...
this is sometimes called the "[abstract-]index substitution operator" since $LaTeX Code: p^i=\\delta^i{}_j p^j$ and $LaTeX Code: q_j=\\delta^i{}_j q_i$ . So, it seems to me that, since $LaTeX Code: \\delta^i{}_j=g^i{}_j=g^{ik}g_{kj}=(g^{-1})^{ik}(g)_{kj}$ , specifying $LaTeX Code: g^i{}_j$ cannot uniquely define a metric on V.
[edit]...unless, possibly, you pick out a preferred basis.

**George Jones** · Jan8-06, 08:36 AM

Originally Posted by Oxymoron

In my opinion, the Kronecker delta, $LaTeX Code: \\delta^i_j$ should be used with caution in general.

I cannot emphasize strongly enough that, given a basis for a vector space

, there is no problem with using the Kronecker delta to define an associated dual basis for

, the algebraic dual of

. This is a very useful construction that is independent of the signature of any possible "metric" that is defined on

. As Hurkyl says, this an oft used construction in (multi)linear algebra.

Given a basis $LaTeX Code: \\left\\{ e_{i} \\right\\}$ for

, I prefer to use a different symbol (as mentioned by pervect) for the associated basis of

, i.e., define linear functionals on

by $LaTeX Code: \\omega^{i} \\left( e_{j} \\right) = \\delta^{i}_{j}$ . Then each $LaTeX Code: \\omega^{i}$ lives in

, and $LaTeX Code: \\left\\{ \\omega^{i} \\right\\}$ is a basis for

.

Hurkyl gave a nice property for the basis $LaTeX Code: \\left\\{ \\omega^{i} \\right\\}$ that exists even when there is no metric tensor defined on

. When a (non-degenerate) metric tensor (of any signature!) is defined on

, the $LaTeX Code: \\omega^{i} \\right\\}$ basis for

has another nice property. If the metric is used as a map from

to

, and if the components of a vector

in

with respect to the $LaTeX Code: \\left\\{ e_{i}} \\right}$ basis of

are $LaTeX Code: v^{i}$ , then the components of the components of the covector that the metric maps to are $LaTeX Code: v_{i}$ with respect to the basis $LaTeX Code: \\left\\{ \\omega^{i} \\right\\}$ for

.

The of dual basis

defined using a Kronecker delta is quite important even the inner product on

is not positive definite.

A reason for using different symbols for a basis for

that is dual to a basis for

is as follows.

Let

be an non-degenerate inner product (of any signature) defined on

, and define

$LaTeX Code: e^{i} = g^{ij} e_{j}.$

For each

and

, $LaTeX Code: g^{ij}$ is a real number, and therefore each $LaTeX Code: e^{i}$ is a linear combination of the elements in the basis $LaTeX Code: \\left\\{ e_{i} \\right\\}$ of

. As such, each $LaTeX Code: e^{i}$ is an element of

, i.e., a vector, and not an element of

, i.e., not a covector. This is true in spite of the fact that each $LaTeX Code: e^{i}$ transforms "the wrong way to be a vector".

I gave a short outline of the connection between the abstract multilinear algebra approach to tensors and the transformation approach in this thread.

A metric tensor can be defined without using bases. I am working on a long post about Minkowski spacetime that might bring this thread full circle back to its beginning, and might be amenable to Oxymoron's math(s) background. It starts by defining a Minkowski vector space.

Minkowski spacetime $LaTeX Code: \\left( V,\\mathbf{g}\\right)$ is a 4-dimensional vector space

together with a symmetric, non-degenerate, bilinear mapping $LaTeX Code: g:V\\times V\\rightarrow\\mathbb{R}$ . A vector in

is called a 4-vector, and a 4-vector

is called timelike if $LaTeX Code: g\\left(v,v\\right) >0$ , lightlike if $LaTeX Code: g\\left(v,v\\right) =0$ , and spacelike if $LaTeX Code: g\\left(v,v\\right) <0$ . $LaTeX Code: \\left( V,g\\right)$ is such that: 1) timelike vectors exist; 2)

is spacelike whenever

is timelike and $LaTeX Code: g\\left( u,v\\right)=0$ .

Regards,
George

**Oxymoron** · Jan8-06, 09:23 AM

Before I go on, just a small clarification. So far we have been referring to the metric tensor by $LaTeX Code: g_{ij}$ . Now, if we move to Minkowski spacetime, is the metric tensor given by $LaTeX Code: \\eta_{\\mu\\nu}$ ?

If we consider spacetime, the the action of the metric $LaTeX Code: \\eta_{\\mu\\nu}$ on two arbitrary vectors

is basically an inner product isn't it? Since

$LaTeX Code: \\eta(v,w) = \\eta_{\\mu\\nu}v^{\\mu}w^{\\nu} = v\\cdot w$

Now, since inner products between two vectors always give some scalar, and since a scalar is an index-free entity, they must remain invariant under any sort of Lorentz transformation?

Now lets turn this effect in on itself and take the inner product of a vector with itself. In Euclidean space, such an inner product is, of course, the norm of a vector, and it is always positive. However, in spacetime, this is not the case? Because

$LaTeX Code: \\eta_{\\mu\\nu}v^{\\mu}v^{\\nu} = \\left\\{ \\begin{array}{c} < 0 \\quad \\mbox{timelike}\\\\ = 0 \\quad \\mbox{lightlike}\\\\ > 0 \\quad \\mbox{spacelike} \\end{array}\\right.$

So from such a tensor, we may define the Kronecker delta (which is a tensor of type (1,1)) as

$LaTeX Code: \\delta_{\\mu}^{\\rho} = \\eta_{\\rho\\nu}\\eta^{\\nu\\mu} = \\eta^{\\mu\\nu}\\eta_{\\nu\\rho}$

Is this a sufficient derivation of the Kronecker delta in spacetime? By the way, am I correct in using Greek indices when referring to spacetime coordinates?

If I transform the spacetime metric tensor $LaTeX Code: \\eta$ will its components change IF I consider only flat spacetime? Is the same true for the Kronecker delta in flat spacetime? What will happen if spacetime is not flat?

**George Jones** · Jan8-06, 10:18 AM

Originally Posted by Oxymoron

So far we have been referring to the metric tensor by $LaTeX Code: g_{ij}$ .

Some people use this notation (abstract index notation) for a tensor, while others don't. Some people choose to interpret $LaTeX Code: g_{ij}$ as components of a tensor with respect to a given basis, as I did in post #52. Then each $LaTeX Code: g_{ij}$ is a real number. I have mixed feelings on the subject. Regardless of one's choice of notation, there is an important distinction to be made between a tensor, and the components of a tensor with respect to a basis.

Now, if we move to Minkowski spacetime, is the metric tensor given by $LaTeX Code: \\eta_{\\mu\\nu}$ ?

This notation is often, but not exclusively, used.

the action of the metric $LaTeX Code: \\eta_{\\mu\\nu}$ on two arbitrary vectors is basically an inner product isn't it? Since
$LaTeX Code: \\eta(v,w) = \\eta_{\\mu\\nu}v^{\\mu}w^{\\nu} = v\\cdot w$

Here, you're treating $LaTeX Code: \\eta_{\\mu\\nu}$ , $LaTeX Code: v^{\\mu}$ , and $LaTeX Code: w^{\\nu}$ as real numbers. This can't be done without first choosing a basis. Another important point: the metric exists without choosing a basis. See the bottom of post #68. A interesting and some what challenging exercise is to show that this definition implies the existence of orthonormal bases. Note that I have used (purely as a matter of personal choice) the opposite signature to you. It seems I am in the minority with respect to this on this forum.

Now, since inner products between two vectors always give some scalar, and since a scalar is an index-free entity, they must remain invariant under any sort of Lorentz transformation?

This is the *definition* of a Lorentz transformation. Given

(or if you prefer, $LaTeX Code: \\eta$ ), a Lorentz transformation is a linear mapping $LaTeX Code: L: V \\rightarrow V$ such that

$LaTeX Code: g \\left( Lv, Lv \\right) = g \\left( v, v \\right).$

Exercise: show that this implies $LaTeX Code: g \\left( Lu, Lv \\right) = g \\left( u, v \\right)$ for every

and

in

.

From this defintion it follows that a Lorentz transformation maps an orthonormal basis to another orthomormal basis.

Is this a sufficient derivation of the Kronecker delta in spacetime?

I perfer to think of it this way. The Kronecker delta is a mathematic object defined to be zero of indices are not equal and one if indices are equal. From the definition of $LaTeX Code: \\eta^{\\nu\\mu}$ , it follows that

$LaTeX Code: \\delta_{\\mu}^{\\rho} = \\eta_{\\rho\\nu}\\eta^{\\nu\\mu} = \\eta^{\\mu\\nu}\\eta_{\\nu\\rho}$

This leads to (but does not beg!) the question: How are the $LaTeX Code: \\eta^{\\nu\\mu}$ defined?

By the way, am I correct in using Greek indices when referring to spacetime coordinates?

Again, this notational convention is often, but not exclusively, used. For example, Wald's well-known text on general relativity uses latin indices to think of $LaTeX Code: T_{ij}$ as a tensor, and of $LaTeX Code: T_{\\mu\\nu}$ as the components of a tensor with respect to a given basis. Both sets of indices in this book run over all of spacetime. This part of the abstract index notation to which I referred above.

If I transform the spacetime metric tensor $LaTeX Code: \\eta$ will its components change IF I consider only flat spacetime? ?

In general, yes! Think of the change from inertial to spherical coordinates, etc.

Regards,
George

PS I know you're being overwhelmed by details, and by different points of view, but you're doing great.

**Hurkyl** · Jan8-06, 11:47 AM

Incidentally, just what sort of beast is

supposed to be anyways? Is it just supposed to be a vector-valued one-form, and thus a rank (1,1) tensor? Is it merely an indexed collection of vectors, and that contracting with that index has a radically different significance than with tensors? Or is it something else entirely?

**George Jones** · Jan8-06, 12:29 PM

Originally Posted by Hurkyl

Incidentally, just what sort of beast is supposed to be anyways?

I have been using $LaTeX Code: \\left\\{ e_{1} , \\dots, e_{n} \\right\\}$ as a basis for a n-dimensional vector space

, so the lower index

on $LaTeX Code: e_{i}$ just specifies which element of the basis, which vector in

. Consequently, $LaTeX Code: e_{i}$ is a vector, a (1,0) tensor, an element of

, a linear mapping from

to $LaTeX Code: \\mathbb{R}$ , etc.

Any

in

can be expanded as $LaTeX Code: v = v^{i} e_{i}$ . The upper index on [itex]v^i[itex] specifies which component, i.e., which real number.

These are the conventions that I have been using. The abstract index approach treats indices differently.

Regards,
George

**Hurkyl** · Jan8-06, 12:40 PM

I guess I didn't explain my question well enough:

Each of the individual components of $LaTeX Code: \\delta^i_j$ are real numbers, but taken as a whole, they define a rank (1,1) tensor.

When taken as a whole, are the

supposed to define a rank (1,1) tensor as well?

(Looking again, maybe you did answer my question, saying that as a whole, it's simply supposed to be an indexed collection of vectors, and not form a tensor at all -- but I still feel compelled to restate myself to make sure you're giving the answer I think I'm getting!)

**pervect** · Jan8-06, 01:43 PM

Well, it looks like I'm outvoted. I may have some more comments or questions after I've studies some of the critical responses in more detail.

At the top of the list: if raising an index isn't creating a map from a vector to a co-vector, how should the operation be described? (Maybe it's a non-linear map?).

Meanwhile, this has been a very educational (if rather long) thread.

**George Jones** · Jan8-06, 01:46 PM

Originally Posted by Hurkyl

When taken as a whole, are the supposed to define a rank (1,1) tensor as well?

Ah, I knew there was more to your question than what I saw. I'm not sure, and I have no references here with me. I'd like to look into this later today or tomorrow.

Your question has made me think more about the Kronecker delta. Let $LaTeX Code: \\left\\{e_i\\right\\}$ be a basis for

and $LaTeX Code: \\left\\{\\omega^i\\right\\}$ be the associated dual basis of

. Then the vector-valued on-form $LaTeX Code: \\omega^i \\otimes e_i$ (sum over i) has components $LaTeX Code: \\delta^{i}_{j}$ . Letting this act on

in

gives

$LaTeX Code: \\omega^{i} \\left(v^{j}e_{j} \\right) \\otimes e_i = v^i e_i = v$

Regards,
George

**George Jones** · Jan8-06, 02:00 PM

Originally Posted by pervect

if raising an index isn't creating a map from a vector to a co-vector, how should the operation be described?

.

Raising the indices of elements of basis sets, i.e., going from

to $LaTeX Code: \\omega^i$ is a basis dependent (change the basis and the mapping changes) metric independent linear mapping between

and

.

Rasing the indices of components, i.e., going from

to

is basis independent (in spite of the fact that I've specified it trems of components) metric dependent linear mapping from

to

.

In general, these 2 mappings are not inverses of each other.

Meanwhile, this has been a very educational (if rather long) thread.

Very much so.

Regards,
George

**Hurkyl** · Jan8-06, 03:32 PM

Originally Posted by George Jones

.
Raising the indices of elements of basis sets, i.e., going from to $LaTeX Code: \\omega^i$

Is it accurate to call passing from a basis to the dual basis "raising indices"? I would have said that raising the indices on

produces the collection of vectors

(and thus not a collection of covectors, so that it certainly cannot be the dual basis).

pervect: let me try starting over for this whole discussion!

(And writing it in math-speak instead of physics-speak -- I think any sort of theoretical discussion is more clear in math-speak)

Think back to your introduction to linear algebra. You probably talked about bases, and coordinates with respect to those bases. (You also probably said "Yah, yah" and promptly ignored it, much like I did when I was first introduced.

)

The important thing was that when you selected a basis B for your vector space V, it allows you to write it in terms of coordinates -- it allows you to write the column-vector

.

Continuing on, if you had a linear map $LaTeX Code: T:V \\rightarrow Vsingle-quote$ and you selected bases B and B', it allows you to write the matrix $LaTeX Code: [T]_{B,Bsingle-quote}$ .

This is all important because we have the identity

$LaTeX Code: [T(v)]_{Bsingle-quote} = [T]_{B,Bsingle-quote} [v]_B$

in other words, the column-vector of components of T(v) (with respect to B') are given precisely by multiplying the matrix representation of T (with respect to B and B') by the column-vector of comopnents of v (with repsect to B).

This machinery is exactly what permits us to do elementary linear algebra in terms of matrix arithmetic. Without this machinery in place, we wouldn't be able to talk about things like the components of a vector, or of a matrix.

Matrix arithmetic isn't just good for vectors and linear transformations, though: it is also good for covectors. Just like the vectors of V are naturally modelled as column-vectors and linear transformations $LaTeX Code: V \\rightarrow V$ are naturally modelled as square matrices, we have that covectors in

are naturally modelled as row-vectors.

So once we've chosen a basis B for V, we are very strongly compelled to select a basis

for

that is compatable with matrix arithmetic. In other words, we insist that:

$LaTeX Code: \\omega(v) = [\\omega]_{B^*} [v]_B $

Since each basis vector in B is mapped to a standard-basis column-vector, and each basis covector in

is mapped to a standard-basis row-vector, we must insist that

$LaTeX Code: \\epsilon^i(e_j) = \\delta^i_n $

where

ranges over the basis vectors in B, and $LaTeX Code: \\epsilon^i$ ranges over the basis covectors in

.

It is precisely this choice which allows us to speak about the components of a vector, covector, or in general any tensor, and do our computations in the usual way.

To state this differently, if you do not choose the dual basis in this manner, you absolutely, positively, cannot manipulate tensors in the usual way via their components.

**masudr** · Jan8-06, 04:51 PM

So... if I've gotten this correctly, basis vectors and basis covectors are mapped to each other by the Kronecker delta.

If that is indeed the case, then how can we normally raise and lower components (which is basically transforming vectors into covectors) using the metric, which is not necessarily the Kronecker delta (e.g. in the case of Minkowski space)?

selfAdjoint · Jan8-06, 05:49 PM

Originally Posted by masudr

So... if I've gotten this correctly, basis vectors and basis covectors are mapped to each other by the Kronecker delta.

If that is indeed the case, then how can we normally raise and lower components (which is basically transforming vectors into covectors) using the metric, which is not necessarily the Kronecker delta (e.g. in the case of Minkowski space)?

You use the metric tensor or its inverse. The Kronecker delta is the metric tensor in Euclidean space. In Minkowski space it is $LaTeX Code: \\eta_{\\alpha\\beta} = Diag(1, -1, -1, -1)$ ; in GR it is a general symmetric rank 2 covariant tensor $LaTeX Code: g_{\\alpha\\beta$ .

Thus $LaTeX Code: A_{\\mu} = g_{\\mu\\beta} A^{\\beta}$ , where you must sum over the repeated index beta; this last is called the Einstein convention. He got tired of writing all those sigmas. When you expand there will be one equation for each value of mu.

**masudr** · Jan8-06, 06:36 PM

selfAdjoint,

Yes, but they've just discussed above that the map between vectors (i.e. the space) and covectors (the dual space) is the Kronecker delta, not the metric tensor.

And, you're saying that the map between contravariant vectors (the space) and covariant vectors (the dual space) are the metric tensor, not the Kronecker delta.

So which is correct?