Understanding the Structure and Transformation of Tensors in Spacetime

[Oxymoron]

Posted by Careful:
It seems nobody answered yet a particular part of your original question which is wether the geometry of spacetime can also be given a distance formulation. It can (in the achronal case), and here are the axioms:

Thanks Careful. I was hoping there was some sort of metric structure on spacetime.

Posted by George Jones:

George Jones So, isn't an inner product on a vector space V a (0,2) tensor, i.e., an element of V^* \otimes V^*?

Well, after all that we have discussed, I suppose it is.

Posted by Masudr:

The metric tensor helps us "lower" or "raise" indices, thus allowing us to make scalars out of tensors.

So what you are saying is, I can turn a covariant tensor into a contravariant tensor (or a vector into a dual vector) by multiplying it by metric tensor.

 

[George Jones]

TI can turn a covariant tensor into a contravariant tensor (or a vector into a dual vector) by multiplying it by metric tensor.

Let v be a 4-vector. Use the metric g to define the covector \tilde{v} associated with v: for every 4-vector w
\tilde{v} \left( w \right) := g \left( v , w \right).
This is the abstract, index-free version of index lowering. To see this, let \left\{ e_{1}, \dots , e_{n} \right\} be a basis for V, and let \left\{ e^{1}, \dots , e^{n} \right\} be the associated basis for V*. Write g_{ij} = g \left( e_{i} , e_{j} \right).

Write v in terms of the basis for V,
v = v^{i} e_{i},
and \tilde{v} in terms of the basis for V*,
\tilde{v} = v_{i} e^{i},.
Then,
\tilde{v} \left( e_{j} \right) = v_{i} e^{i} \left( e_{j} \right) = v_{i} \delta_{j}^{i} = v_{j}.
But, by definition,
\tilde{v} \left( e_{j} \right) = g \left( v , e_{j} \right) = g \left( v^{i} e_{i} , e_{j} \right) = v^{i} g \left( e_{i} , e_{j} \right) = v^{i} g_{ij}.
Combining these results gives
v_{j} = v^{i} g_{ij}.
Much more to follow.

Regards,
George

 

[Oxymoron]

SUMMARY SO FAR...

A tensor is a multilinear map. In fact, it is just a generalization of a linear functional that was in linear algebra. Therefore, a tensor, in it barest form, is

T\,:\,V_1 \times V_2 \times \dots \times V_N \times V^*_1 \times V^*_2 \times \dots \times V^*_M \rightarrow \mathbb{R}

where V_1,V_2,\dots,V_N are vector spaces over a field \mathbb{R} and V^*_1,V^*_2,\dots,V^*_M are the corresponding dual vector spaces over the same field.

The tensor written above is a tensor of type (M,N). As a result, a tensor of type (0,1) is simply a linear functional

T\,:\, V \rightarrow \mathbb{R}

For every vector space V there exists a dual vector space V^* consisting of all linear functionals on V. From now on we will refer to linear functionals on V as covectors, or 1-forms.

Let V be a finite dimensional vector space. Its dimension, n, is the number of basis vectors on V which are needed to uniquely define a linear functional. Thus,

\dim(V) = n

and there are n basis vectors, \{e_1,e_2,\dots,e_n\}. Likewise for the corresponding dual vector space V, whose dimension is \dim(V^*) = n. The basis of the dual vector space consists of n basis vectors \{\epsilon^1,\epsilon^2,\dots,\epsilon^n\} satisfying

\epsilon_i(e_j) = \delta^i_j

where d^i_j is the Kronecker delta.

Now let's add more structure to our vector space by defining an inner product on it. As a result, V becomes an inner product space with

g\,:\, V \times V \rightarrow \mathbb{R}

defined as

g(u,v) = u\cdot v

This bilinear functional is actually a covariant tensor of degree 2 or simply, the metric tensor. Covariant tensors are of the form:

T\,:\,V \times V \times \dots \times V \rightarrow \mathbb{R}

and contravariant tensors are of the form

T\,:\,V^*\times V^* \times \dots \times V^* \rightarrow \mathbb{R}.

A tensor can also be symmetric as in the case of the metric tensor. Such tensors have the following property

g(u,v) = g(v,u)

that is, if we reverse which way we operate on the elements the sign of the tensor does not change.

One of the fundamental concepts of tensors is the ability to raise and lower the indices. If we have a contravariant tensor, T^i and g_{ij}, as we have seen, is a covariant tensor of degree 2. Then

g_{ij}T^i = T_j

and we say that taking the inner product with the metric tensor has lowered a contravariant index to a covariant index. It can be shown that g is invertible, so

g^{ij}T_j = T^i

where g^{ij} = (g_{ij})^{-1}.

 

[pervect]

SUMMARY SO FAR...
A tensor is a multilinear map. In fact, it is just a generalization of a linear functional that was in linear algebra. Therefore, a tensor, in it barest form, is
T\,:\,V_1 \times V_2 \times \dots \times V_N \times V^*_1 \times V^*_2 \times \dots \times V^*_M \rightarrow \mathbb{R}
where V_1,V_2,\dots,V_N are vector spaces over a field \mathbb{R} and V^*_1,V^*_2,\dots,V^*_M are the corresponding dual vector spaces over the same field.
The tensor written above is a tensor of type (M,N). As a result, a tensor of type (0,1) is simply a linear functional
T\,:\, V \rightarrow \mathbb{R}
For every vector space V there exists a dual vector space V^* consisting of all linear functionals on V. From now on we will refer to linear functionals on V as covectors, or 1-forms.
Let V be a finite dimensional vector space. Its dimension, n, is the number of basis vectors on V which are needed to uniquely define a linear functional. Thus,
\dim(V) = n
and there are n basis vectors, \{e_1,e_2,\dots,e_n\}. Likewise for the corresponding dual vector space V, whose dimension is \dim(V^*) = n. The basis of the dual vector space consists of n basis vectors \{\epsilon^1,\epsilon^2,\dots,\epsilon^n\} satisfying
\epsilon_i(e_j) = \delta^i_j
where d^i_j is the Kronecker delta.

This is very good, and extremely well written, except for a minor but potentially important detail

A basis is , in my understanding, is defined as a set of vectors that are linearly independent and span the space - a basis does not have to be orthonormal, e.g.

http://mathworld.wolfram.com/VectorSpaceBasis.html

Thus it is not necessarily true that e_i \cdot e_j = \delta^i_j.

IF e_i \cdot e_j = \delta^i_j THEN \epsilon^i(e_j) = \delta^i_j

So the above statement is not necessarily true, all we need to say at this point is that the \epsilon^i span the dual space and are linearly independent, any other claim is imposing more structure on them than is necessarily true.

Now let's add more structure to our vector space by defining an inner product on it.

After we've done this, we can talk about the e_i being orthonormal, and we can also make the remark that this is equivalent to \epsilon^i (e_j) = \delta^i_j

 

[Oxymoron]

After we've done this, we can talk about the e_i being orthonormal, and we can also make the remark that this is equivalent to \epsilon^i (e_j) = \delta^i_j

You're right. I should have written this first. Thankyou for the compliments though.

Before I go on, I think I left something out. We can form a tensor space by collecting all tensors of a fixed type (r,s). This space is actually a vector space, and the tensors can be added and subtracted together by real numbers. The problem for our tensor space is in defining a basis for it. This is where we need a new operation called the tensor product, denoted as \otimes.

If T is a (k,l) tensor and S is a (m,n) tensor we define T\otimes S as a (k+m,l+n) tensor, by

T\otimes S (\omega^1,\dots,\omega^k,\dots,\omega^{k+m},V^1,\dots,V^l,\dots,V^{l+n})
= T(\omega^1,\dots,\omega^k,V^1,\dots,V^l)S(\omega^{k+1},\dots,\omega^{k+m},V^{l+1},\dots,V^{l+n})

where \omega^i and V^i are distinct dual vectors and vectors. That is, we define the tensor product of T and S by first acting T on the appropriate set of dual vectors and vectors, and then act S on the remainder, and multiply everything together. But this operation is not commutative.

We can now construct a basis for the space of all fixed type tensors, by taking the tensor product of basis vectors and dual vectors. The resulting basis will consist of all the tensors of the following form:

e_{i_1}\otimes \dots \otimes e_{i_k} \otimes \epsilon^{j_1} \otimes \dots \otimes \epsilon^{j_l}

Therefore, every tensor T of the fixed type (k,l) has a unique expansion:

T = T^{i_1\dots i_k}_{j_1\dots j_l} e_{i_1} \otimes \dots \otimes e_{i_k} \otimes \epsilon^{j_1} \otimes \dots \otimes \epsilon^{j_l}

where T^{i_1\dots i_k}_{j_1\dots j_l} = T(\epsilon^{i_1},\dots,\epsilon^{i_r}, e_{j_1}, \dots, e_{j_s}), which are simply the components of the tensor with respect to the basis of V.

But expressing a tensor T as T^{i_1\dots i_k}_{j_1\dots j_l} e_{i_1} is just like expressing a vector by its components - merely a shortcut.

 

[masudr]

Yes that looks good and fairly complete, so far. Some of my understanding might also help.

A nice thing to remember is that components of a vector transform like a vector, but the basis vectors transform like components of a covector; since the vector is actually the components multiplied by the basis vector, our vector is invariant! Similar considerations apply to covector components and basis covectors (although, the opposite, of course). And that's why, after all, tensors are objects which do not change under co-ordinate transformations (don't get confused: the components transform, but the basis transforms in the opposite way).

Another thing to note is that often the simplest basis for a vector space at some point is the co-ordinate basis, namely the vectors pointing along the co-ordinate lines at any point. And a covector basis can be defined as those linear functionals, when acting on the corresponding co-ordinate basis, gives us the Kronecker delta.

PS. I should formalise this in LaTeX, but it's pretty late here.

 

[George Jones]

You're right. I should have written this first.

Actually, in the spite of the presence of a Kronecker delta, the relation \epsilon^i (e_j) = \delta^i_j has nothing to do either with orthogonality, or with a metric tensor. Given a basis for a vector space V, this relation can always be used to define a basis for the vector space V*, the algebraic dual of V. No metric tensor is needed. Also, the relation does not define a metric tensor.

Orthogonality is a condition between vectors in the same vector space, and the e_j and the \epsilon^i live in different vector spaces. The initial basis for V need not be an orthonormal one in order to use the above relation to define a basis for V*.

The construction that I outlined towards the end of post #40 made no use of a metric tensor.

Regards,
George

 

[pervect]

Actually, in the spite of the presence of a Kronecker delta, the relation \epsilon^i (e_j) = \delta^i_j has nothing to do either with orthogonality, or with a metric tensor. Given a basis for a vector space V, this relation can always be used to define a basis for the vector space V*, the algebraic dual of V. No metric tensor is needed. Also, the relation does not define a metric tensor.

Why do you say that?

It seems to me that the relation does define a dot product, and hence the metric tensor, in a very natural way.

If \epsilon^j(e_i) = \delta^i_j, you have defined a particular mapping from basis vectors to basis co-vectors. e_i is associated with \epsilon^i, the co-vector with the same index.

Now it is possible that you do not want to make use of this relationship, but if you don't want to use it, why specify it? I.e. \epsilon^j(e_i) = \delta^i_j has the purpose of singling out a unique co-vector that is associated with a vector with the same index. If there is some other purpose for writing this relationship down, please enlighten me, because I'm missing the point :-(.

Given that we actually make use of this association, we now have a map from the basis vectors to the basis co-vectors - for every e_i, we have singled out a unique \epsilon^i with the same index, thus we have assigned a unique co-vector to every basis vector.

Because an arbitrary vector can be defined by a weighted linear sum of basis vectors, we have also defined a map from every vector u to a unique dual vector, found by substituting all the e_i with the \epsilon^i and keeping the same linear weights.

Given a mapping between vectors and duals, we have defined a dot product.
Given two vectors u and v, we use the above mapping to find u*. u* is a map from a vector to a scalar. Applying this map u* to the vector v, we then have a scalar. Thus we have a mapping - a linear mapping, though I've skipped over proving this - from two vectors (u,v) to a scalar. This linear mapping between two vectors and a scalar defines the dot product, and a metric tensor.

[add]
Something I should add - in this dot product we have defined above, we can now ask - what is e_i \cdot e_j? A little work shows the answer is \delta^i_j. Thus our basis vectors are orthonormal.

The way I view the end result is that if we have a vector space, from any set of basis vectors we can form a consistent dot product in which those basis vectors are orthonormal. However, when dealing with physics, we have a physical notion of distance which imposes a particular dot product, one that arises from the physics (our ability to measure distances). Thus we restrict the mathematical notions of possible dot product to those that are physically meaningful.

 

[George Jones]

Why do you say that?

Let me start with an example. Let (V,g) be Minkowski vector space, and let {e_0, e_1, e_2, e_3} be an orthonormal(i.e, 1 = g(e_0 , e_0) = -g(e_1 , e_1) = -g(e_2 , e_2) = -g(e_3 , e_3)) basis. There is nothing that prohibits taking u = v in an inner product. Let u = v = e_0 + e_1. Using the bilinearity and symmetry of g gives

g(u,u) = g( e_0 + e_1 , e_0 + e_1) = g(e_0 , e_0) +2g(e_0 , e_0) + g(e_1 , e_1) = 1 + 0 + (-1) = 0.

u is a lightlike vector as expected.

Now calculate the inner product using your construction. This gives

(e^0 + e^1) (e_0 + e_1) = e^0(e_0 + e_1) + e^1(e_0 + e_1) = 1 + 0 + 0 + 1 = 2

Regards,
George

 

[George Jones]

I had a similar http://groups.google.ca/group/sci.physics.research/msg/5405b4c4304b8304?dmode=source&hl=en" sci.phyics.research as well.

Regards
George

 

[pervect]

My approach works for Minkowski vector spaces if one takes (1)

\epsilon^0(e_0) = -1 \hspace{.5 in} \epsilon^1(e_1) = 1 \hspace{.5 in} \epsilon^2(e_2) = 1 \hspace{.5 in} \epsilon^3(e_3) = 1

Therfore this is what I always do, and thus I do not assume that (2) \epsilon^i(e_j) = \delta^i_j.

After a bit of head scratching, I think I can see where one _could_ assume that \epsilon^i(e_j) = \delta^i_j, but it strikes me as being very awkward and unnatural.

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

 

[masudr]

I think the problem here (and please correct me if I'm wrong) is that one of you (pervect) is choosing to specifically use the co-ordinate basis (which requires the notion of calculus on the manifolds), which then gives you the Minkowski metric, and the other is using an approach before co-ordinates and calculus have been defined on the manifold.

Of course, any vector space has an infinite number of valid bases, so both are, of course, correct.

 

[robphy]

My approach works for Minkowski vector spaces if one takes (1)
\epsilon^0(e_0) = -1 \hspace{.5 in} \epsilon^1(e_1) = 1 \hspace{.5 in} \epsilon^2(e_2) = 1 \hspace{.5 in} \epsilon^3(e_3) = 1

I'm trying to understand your approach. Are you saying that (1), a relationship between a basis in V and a basis in the dual-space V^*, defines the Minkowski metric [on V]?

 

[pervect]

I'm trying to understand your approach. Are you saying that (1), a relationship between a basis in V and a basis in the dual-space V^*, defines the Minkowski metric [on V]?

Exactly. First we note that a mapping of basis vectors between V and V* defines a general linear map from V to V*.

I.e. in a 2d space {e_0, e_1} we can represent an arbitrary vector v as \alpha e_0 + \beta e_1. Now, since we have a map from the basis vectors e_i of V to the basis vectors \epsilon^i of V*, it's perfectly natural to define \alpha \epsilon^0 + \beta \epsilon^1 as a linear map from an arbitrary element v \in V to some element of V*.

Let A be such a general mapping A : V -> V*. Then for any vector u in V, we have A(u) in V*. A(u) is a map from V -> \mathbb{R} by the defintion of the dual space. Thus if we have two vectors, u and v in V, we can find (A(u))(v) which is a scalar, and bilinear in u and v. This defines a metric.

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

Furthermore, I'm of the opinion that if you have one set of mappings g^i{}_j from V to V*, the ones defined by the tensor "raising/lowering" index conventions, it's very confusing to have a *different* set of mappings "hanging around" that don't follow the tensor "index" conventions, and that the set of mappings that make \epsilon^i(e_j) = \delta^i_j is just such a set of different mappings. It will tend to confuse people, IMO, and I can't see any particular use for it.

I suppose I'm also requiring that g_ij be invertible, but the basis vectors are linearly independent by defintition, and that should be good enough to insure that an inverse exists.

 

[Hurkyl]

pervect: I concur with George Jones. The dual basis \epsilon^i to a basis e_i 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 v = \sum_i \epsilon^i(v) e_i.

Basically, I'd rather keep \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 e_j's is a vector, and thus has an upper index, which I will denote with a dot. Each of the \epsilon^i's is a covector, and thus has a lower index, which I will denote with another dot. You're claiming that \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 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 V by doubling everything, I need to change coordinates in V^* by halving everything. A tensor would do the wrong thing!


Remember that, in the way people usually do coordinates, the action of a covector \epsilon^i on a vector e_j is simply given by \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 \delta^i_j, and thus \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]

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 (V,\cdot) with a map

g\,:\, V \times V \rightarrow \mathbb{R}

defined by taking the inner product of two arguments:

g(u,v) = u\cdot v

Then one could say that when the tensor g_{ij} acts on two elements from V such that

g_{ij}u^iv^j

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

u\cdot v = g_{ij}u^iv^j

In my opinion, the Kronecker delta, \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 \{e_1,\dots,e_n\} such that

e_i \cdot e_j = g_{ij} = \delta_{ij}

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

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]

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

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

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

 

[George Jones]

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

I cannot emphasize strongly enough that, given a basis for a vector space V, there is no problem with using the Kronecker delta to define an associated dual basis for V*, the algebraic dual of V. This is a very useful construction that is independent of the signature of any possible "metric" that is defined on V. As Hurkyl says, this an oft used construction in (multi)linear algebra.

Given a basis \left\{ e_{i} \right\} for V, I prefer to use a different symbol (as mentioned by pervect) for the associated basis of V*, i.e., define linear functionals on V by \omega^{i} \left( e_{j} \right) = \delta^{i}_{j}. Then each \omega^{i} lives in V*, and \left\{ \omega^{i} \right\} is a basis for V*.

Hurkyl gave a nice property for the basis \left\{ \omega^{i} \right\} that exists even when there is no metric tensor defined on V. When a (non-degenerate) metric tensor (of any signature!) is defined on V, the \left\{\omega^{i} \right\} basis for V* has another nice property. If the metric is used as a map from V to V*, and if the components of a vector v in V with respect to the \left\{ e_{i} \right\} basis of V are v^{i}, then the components of the components of the covector that the metric maps to are v_{i} with respect to the basis \left\{ \omega^{i} \right\} for V*.

The of dual basis V* defined using a Kronecker delta is quite important even the inner product on V is not positive definite.

A reason for using different symbols for a basis for V* that is dual to a basis for V is as follows.

Let g be an non-degenerate inner product (of any signature) defined on V, and define

e^{i} = g^{ij} e_{j}.

For each i and j, g^{ij} is a real number, and therefore each e^{i} is a linear combination of the elements in the basis \left\{ e_{i} \right\} of V. As such, each e^{i} is an element of V, i.e., a vector, and not an element of V*, i.e., not a covector. This is true in spite of the fact that each 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 https://www.physicsforums.com/showthread.php?t=105868".

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 \left( V,\mathbf{g}\right) is a 4-dimensional vector space V together with a symmetric, non-degenerate, bilinear mapping g:V\times V\rightarrow\mathbb{R}. A vector in V is called a 4-vector, and a 4-vector v is called timelike if g\left(v,v\right) >0, lightlike if g\left(v,v\right) =0, and spacelike if g\left(v,v\right) <0. \left( V,g\right) is such that: 1) timelike vectors exist; 2) v is spacelike whenever u is timelike and g\left( u,v\right)=0.

Regards,
George

 

[Oxymoron]

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

If we consider spacetime, the the action of the metric \eta_{\mu\nu} on two arbitrary vectors v,w is basically an inner product isn't it? Since

\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 let's 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

\eta_{\mu\nu}v^{\mu}v^{\nu} = \left\{ \begin{array}{c}<br /> &lt; 0 \quad \mbox{timelike}\\<br /> = 0 \quad \mbox{lightlike}\\<br /> &gt; 0 \quad \mbox{spacelike}<br /> \end{array}\right.

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

\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 \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]

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

Some people use this notation (abstract index notation) for a tensor, while others don't. Some people choose to interpret g_{ij} as components of a tensor with respect to a given basis, as I did in post #52. Then each 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 \eta_{\mu\nu}?

This notation is often, but not exclusively, used.

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

Here, you're treating \eta_{\mu\nu}, v^{\mu}, and 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 g (or if you prefer, \eta), a Lorentz transformation is a linear mapping L: V \rightarrow V such that

g \left( Lv, Lv \right) = g \left( v, v \right).

Exercise: show that this implies g \left( Lu, Lv \right) = g \left( u, v \right) for every u and v in V.

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 \eta^{\nu\mu}, it follows that

\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 \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 T_{ij} as a tensor, and of 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 \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]

Incidentally, just what sort of beast is e_i 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]

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

I have been using \left\{ e_{1} , \dots, e_{n} \right\} as a basis for a n-dimensional vector space V, so the lower index i on e_{i} just specifies which element of the basis, which vector in V. Consequently, e_{i} is a vector, a (1,0) tensor, an element of V**, a linear mapping from V* to \mathbb{R}, etc.

Any v in V can be expanded as v = v^{i} e_{i}. The upper index on v^ispecifies which component, i.e., which real number.&lt;br /&gt; &lt;br /&gt; These are the conventions that I have been using. The abstract index approach treats indices differently.&lt;br /&gt; &lt;br /&gt; Regards,&lt;br /&gt; George

 

[Hurkyl]

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

Each of the individual components of \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 e_i 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]

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]

When taken as a whole, are the e_i 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 \left\{e_i\right\} be a basis for V and \left\{\omega^i\right\} be the associated dual basis of V*. Then the vector-valued on-form \omega^i \otimes e_i (sum over i) has components \delta^{i}_{j}. Letting this act on v in V gives

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

Regards,
George

 

[George Jones]

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 e_i to \omega^i is a basis dependent (change the basis and the mapping changes) metric independent linear mapping between V and V*.

Rasing the indices of components, i.e., going from v_i to v^i is basis independent (in spite of the fact that I've specified it trems of components) metric dependent linear mapping from V* to V.

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]

.
Raising the indices of elements of basis sets, i.e., going from e_i to \omega^i[/itex]

<br /> Is it accurate to call passing from a basis to the dual basis &quot;raising indices&quot;? I would have said that raising the indices on e_i produces the collection of vectors e^i (and thus not a collection of covectors, so that it certainly cannot be the dual basis).<br /> <br /> <br /> pervect: let me try starting over for this whole discussion! <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /> (And writing it in math-speak instead of physics-speak -- I think any sort of theoretical discussion is more clear in math-speak)<br /> <br /> Think back to your introduction to linear algebra. You probably talked about bases, and coordinates with respect to those bases. (You also probably said &quot;Yah, yah&quot; and promptly ignored it, much like I did when I was first introduced. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />)<br /> <br /> The important thing was that when you selected a basis <i>B</i> for your vector space <i>V</i>, it allows you to write it in terms of coordinates -- it allows you to write the column-vector [v]_B.<br /> <br /> Continuing on, if you had a linear map T:V \rightarrow V&amp;#039; and you selected bases <i>B</i> and <i>B&#039;</i>, it allows you to write the matrix [T]_{B,B&amp;#039;}.<br /> <br /> This is all important because we have the identity<br /> <br /> [T(v)]_{B&amp;#039;} = [T]_{B,B&amp;#039;} [v]_B<br /> <br /> in other words, the column-vector of components of <i>T(v)</i> (with respect to B&#039;) are given precisely by multiplying the matrix representation of <i>T</i> (with respect to <i>B</i> and <i>B&#039;</i>) by the column-vector of comopnents of <i>v</i> (with repsect to <i>B</i>).<br /> <br /> This machinery is exactly what permits us to do elementary linear algebra in terms of matrix arithmetic. Without this machinery in place, we wouldn&#039;t be able to talk about things like the components of a vector, or of a matrix.<br /> <br /> <br /> <br /> Matrix arithmetic isn&#039;t just good for vectors and linear transformations, though: it is also good for covectors. Just like the vectors of <i>V</i> are naturally modeled as column-vectors and linear transformations V \rightarrow V are naturally modeled as square matrices, we have that covectors in V^* are naturally modeled as row-vectors.<br /> <br /> <br /> So once we&#039;ve chosen a basis <i>B</i> for <i>V</i>, we are <i>very strongly compelled</i> to select a basis B^* for V^* that is compatable with matrix arithmetic. In other words, we insist that:<br /> <br /> &lt;br /&gt; \omega(v) = [\omega]_{B^*} [v]_B&lt;br /&gt;<br /> <br /> Since each basis vector in <i>B</i> is mapped to a standard-basis column-vector, and each basis covector in B^* is mapped to a standard-basis row-vector, we must insist that<br /> <br /> &lt;br /&gt; \epsilon^i(e_j) = \delta^i_n&lt;br /&gt;<br /> <br /> where e_j ranges over the basis vectors in <i>B</i>, and \epsilon^i ranges over the basis covectors in B^*.<br /> <br /> 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.<br /> <br /> To state this differently, if you do not choose the dual basis in this manner, you <i>absolutely, positively, cannot</i> manipulate tensors in the usual way via their components.

 

[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)?

 

[selfAdjoint]

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 \eta_{\alpha\beta} = Diag(1, -1, -1, -1); in GR it is a general symmetric rank 2 covariant tensor g_{\alpha\beta.

Thus 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]

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?

 

[Hurkyl]

I've also been saying that the map from vectors to covectors is not the Kronecker delta.

 

[masudr]

I've also been saying that the map from vectors to covectors is not the Kronecker delta.

Yeah. I've just had a look at Carroll, and I'm now fairly certain about this. We have a set of basis vectors \hat{e}_\mu for our tangent space. We then use this basis to define a basis for our cotangent space, \hat{\theta}^\mu for our cotangent space, so that

\hat{\theta}^\mu\left(\hat{e}_\nu\right)=\delta^\mu_\nu.

We can use the metric, however, to lower and raise componets of various tensors (by contracting over various indices). The crucial point to bear in mind, is that once we've messed around with some tensor by raising/lowering indices, this new tensor doesn't necessarily have a proper geometrical meaning.

The gradient, and it's action on vectors, is perfectly well defined regardless of any metric, whereas the "gradient with upper indices" is not.

EDIT: Furthermore, when lowering or raising indices with the metric, we are talking about equations with actual numbers -- i.e. we have already (and usually implicitly) chosen some convenient basis (usually the co-ordinate basis). This operation is fundamentally different from using the Kronecker delta to define a basis for the cotangent space given a basis for the tangent space (or vice versa).

 

[George Jones]

Is it accurate to call passing from a basis to the dual basis "raising indices"? I would have said that raising the indices on e_i produces the collection of vectors e^i (and thus not a collection of covectors, so that it certainly cannot be the dual basis).

You're right - it's probably not a good idea to call this operation "raising indices".

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)?

The isomorphism between V and V* induced by e_{i} \mapsto \omega^{i} is basis dependent. An arbitary vector v gets mapped to different covectors, depending on the choice of basis.

The isomorphism between V and V* induced by a metric tensor is natural in the sense that the mapping is completely independent of choice of basis. An arbitary vector v gets mapped to the same covector, call it \tilde{v}, even if different bases, say \left\{ e_{i} \right\} and \left\{ e&#039;_{i} \right\}.

If components of v are defined by introducing a basis \left\{ e_{i} \right\}, then \omega^{i} is the basis of V* that makes all the component stuff work out correctly. I tried to show this in post #52. Note also Hurkyl's #77.

In #52 I have made an unfortuante choice of notation, but it is far too late to edit this post. As both Hurkyl and I have pointed out, the notation e^i is best reserved for other vectors that live in V. Whenever you see an e^i in #52, mentally replace it by \omega^i.

In #52, I define \tilde{v} before introducing any bases. I then introduce bases \left\{ e_{i} \right\} and \left\{ \omega^{i} \right\} for V and V* respectively. The vector v and the covector \tilde{v} are then expanded in terms of these bases. At this point, the expansion coefficients are seeming unrelated, but a little use of linearity properties reveals the standard connection.

Regards,
George

 

[Oxymoron]

Suppose I have a vector space V and that it has two possible bases: \{e_i\} and \{e_j&#039;\}. Now suppose that every basis element belonging to the first set of basis vectors is related somehow to every basis element belonging to the second. Say the relationship is related by some sort of matrix multiplication:

e_i = A_i^je&#039;_j
e&#039;_j = A&#039;_j^ke_k

So we can in effect take a basis element e_i and 'transform' it to another basis element e_j&#039; simply by multiplying it by A_i^j. And similarly we can go backwards. So in this case would A_i^j and A&#039;_j^k be related via

A_i^j = [A&#039;_j^k]^{-1}

That is, are they necessarily inverses of each other?

The question then is, do we have the following result?

A&#039;_j^kA_i^j = A_j^kA&#039;_i^j = \delta_i^k

assuming I have all the indices right.


Similarly, it should be possible for the dual basis to undergo a similar transformation:

\epsilon&#039;^j = A_k^j\epsilon^k


After all that, let's see what happens when we take some arbitrary vector, v \in V and covector \omega \in V^*. If we multiply v by the same matrix we multiplied our basis element with, A_i^j we should have:

A^j_iv^i = v&#039;^j

Note that v^i = v^ie_i. And similarly with the covector (\omega = \omega_j\epsilon^j) should transform under a corresponding dual transformation:

\omega_jA&#039;_i^j = \omega&#039;_i

This all looks remarkably similar to the idea of change of basis by matrix transformations that I did in linear algebra. Is what I have done here simply a generalization of it? I mean, from the looks of it I have merely transformed a vector into a new vector with respect to a new basis via a matrix multiplication - the matrix being that very matrix which defines the basis elements in the new basis.

 

[pervect]

Suppose I have a vector space V and that it has two possible bases: \{e_i\} and \{e_j&#039;\}. Now suppose that every basis element belonging to the first set of basis vectors is related somehow to every basis element belonging to the second. Say the relationship is related by some sort of matrix multiplication:
e_i = A_i^je&#039;_j
e&#039;_j = A&#039;_j^ke_k
So we can in effect take a basis element e_i and 'transform' it to another basis element e_j&#039; simply by multiplying it by A_i^j. And similarly we can go backwards. So in this case would A_i^j and A&#039;_j^k be related via
A_i^j = [A&#039;_j^k]^{-1}
That is, are they necessarily inverses of each other?
The question then is, do we have the following result?
A&#039;_j^kA_i^j = A_j^kA&#039;_i^j = \delta_i^k
assuming I have all the indices right.

Yes. Just use linearity, and substitute the second the e_i from the first equation into the e_k into the second equation

<br /> e&#039;_j = A&#039;^k{}_j A^i{}_k e&#039;_i<br />

Since e_j = \delta^i{}_j e_i we know that

A&#039;^k{}_j A^i{}_k = \delta^i{}_j

This all looks remarkably similar to the idea of change of basis by matrix transformations that I did in linear algebra. Is what I have done here simply a generalization of it? I mean, from the looks of it I have merely transformed a vector into a new vector with respect to a new basis via a matrix multiplication - the matrix being that very matrix which defines the basis elements in the new basis.

It's exactly the same thing that you're used to from linear algebra. There is a standard that transform matrices have indexes that run from northwest-southeast. To see how to keep the indexes from displaying right below one another, study the following plaintex/latex pair

x^a' = \Lambda^{a'}{}_{a} x^a

<br /> x^a&#039; = \Lambda^{a&#039;}{}_{a} x^a<br />

the empty pair of brackets are what's needed to insure the northwest-southeast lineup on \Lambda.

(I think there is a better way to do the primes in latex, though, than what I did).

 

[Oxymoron]

Posted By Pervect

There is a standard that transform matrices have indexes that run from northwest-southeast.

Really!? I always wondered why textbooks never seems to be able to line up the indices on their matrices. But I suppose it looks neater.

Posted by Pervect

(I think there is a better way to do the primes in latex, though, than what I did).

Yeah, you can type "\prime". But to me, there is no difference. So I prefer " ' ".

"\prime" A^{\prime}
" ' " A&#039;

 

[Oxymoron]

Consider the following quote:

"Consider a contravariant tensor T = T^i(\bold{x}(t)) defined on the curve C."

could someone explain what this means. I don't see how a tensor can be defined on a curve, it doesn't make sense to me.

The reason for this is, I want to begin to differentiate tensors.


But first I have some questions on transformations of tensors, which I believe will help me understand differentiation.

Let C be a curve given parametrically by x^i = x^i(t) in an (x^i) coordinate system. Consider the tangent vector field T^i defined by

T^i = \frac{dx^i}{dt}

Under a change of coordinates, the same curve is given by \bar{x}^i = \bar{x}^I(t). and the tangent vector field by

\bar{T}^i = \frac{d\bar{x}^i}{dt}

Now we have, via the chain rule

\frac{d\bar{x}^i}{dt} = \frac{d\bar{x}^i}{dx^r}\frac{dx^r}{dt}

But this is nothing more than

\frac{d\bar{x}^i}{dt} = \frac{d\bar{x}^i}{dx^r}T^r

where r replaces the index i.

But what can we conclude from this? Well, in my opinion, the tangent vector field as defined above is a contravariant tensor of order 1, that is, given any curve and any tangent vector to it, under a change of coordinates, the new tangent vector is related to the old by a contravariant tensor of order one. I am not sure if this is correct though.

The problem comes when I try to differentiate w.r.t. t the transformation law of contravariant tensors:

\bar{T}^i = T^r\frac{\partial\bar{x}^i}{\partial x^r}

I get something like (from Schuam)

\frac{d\bart{T}^i}{dt} = \frac{dT^r}{dt}\frac{\partial\bar{x}^i}{\partial x^r} + T^r\frac{\partial^2\bar{x}^i}{\partial x^s\partial x^r}\frac{dx^s}{dt}

From Schuam's Tensor Calculus

...which shows that the ordinary derivative of T along a curve C is a contravariant tensor if and only if the \bar{x}^i are linear functions of the x^r.

Which leads to the following theorem:

The derivative of a tensor is a tensor if and only if coordinate changes are restricted to linear transformations.

Could anyway explain this to me? Apparently, Schaum uses the fact that ordinary differentiation fails in most cases of coordinate transformations, and introduces Christoffel symbols (which I don't understand anyway). Any explanation of this would be helpful (or if you have your own way of introducing this material).