Physics Asked on May 31, 2021
Any rank 2 tensor would do the same trick, right?
Then, which is the motivation for choosing the metric one?
Also, if you help me to prove that $g^{kp}g_{ip}=delta^k_i$, I would be thankful too ^^
Every vector space $V$ (over the real numbers) comes naturally with it's dual vector space $W$ of linear maps to the real numbers, e.g. $$ varphi in W :Leftrightarrow varphi: V rightarrow mathbb{R} ~ text{ with } ~ varphi(a cdot v + u) = a cdot varphi(v) + varphi(u) ~ ~ forall v,u in V ~ ~ forall a in mathbb R$$
Let $B := {e_1, e_2, dots, e_n }$ be a basis of $V$ then for $x in V$ the expansion $$ x = sum_{i=1}^n X^i e_i $$ yields the so called contravariant coordinates $X^i$ of $x$.
The dual basis to $B$ is defined as the basis $tilde{B} = { varepsilon^1, varepsilon^2, dots, varepsilon^n }$ of $W$ such that $$ varepsilon^i(e_j) = delta^i_j $$ which defines $tilde{B}$ completely because every linear map is already fixed if it is given for all basis vectors of $V$. Then for a $y in W$ the expansion $$ y = sum_{i=1}^n Y_i varepsilon^i $$ yields the so called covariant coordinates $Y_i$ of $y$.
If on the vector space $V$ there is an inner product $(~.~,~.~): V times V rightarrow mathbb{R}$ (for instance the usual dot product) then there is a natural map $tau: V rightarrow W$ between $V$ and $W$:
Let $x in V$ then $$ tau(x) = (x,~.~) in W$$ e.g. for $v in V$ we have $$ tau(x)(v) = (x,v) $$
Let's define the following matrix $g$ $$ g_{i j} := (e_i, e_j) $$ which are the (covariant) components of the twice linear inner product map, also called the metric tensor. In general the basis $B$ is not orthonormal, that is, in general $ g_{i j} ne delta_{ij} $ and therefore $tau(e_i) ne varepsilon^i$
Define the inverse matrix of $g_{ij}$ by $$ g^{ij} := (g^{-1})^{ij} $$ then we have $$ sum_{k=1}^n g^{ik} tau(e_k)(e_j) = sum_{k=1}^n g^{ik} (e_k,e_j) = sum_{k=1}^n g^{ik} g_{kj} = delta^i_j $$
So indeed we have found: $$ varepsilon^i = sum_{k=1}^n g^{ik} tau(e_k) $$
Let's have a quick recap what we have done so far.
Now we have everything at hand to express the covariant (dual) coordinates $X_i$ in terms of the contravariant coordinates $X^i$ of a vector $x in V$. Remember: $$ x = sum_{i=1}^n X^i e_i $$ Now map it to the dual vector $tau(x)$ which is a representation of $x$ in the dual vector space $W$: begin{equation} begin{split} tau(x) & = (x, ~.~) = (sum_{i=1}^n X^i e_i, ~.~) = sum_{i=1}^n X^i (e_i, ~.~) & = sum_{i=1}^n X^i tau(e_i) = sum_{i=1}^n sum_{j=1}^n X^i delta^j_i tau(e_j) & = sum_{i=1}^n sum_{j=1}^n sum_{k=1}^n X^i g_{ik} g^{kj} tau(e_j) & = sum_{k=1}^n left( sum_{i=1}^n X^i g_{ik} right) left( sum_{j=1}^n g^{kj} tau(e_j) right) & = sum_{k=1}^n underbrace{left( sum_{i=1}^n X^i g_{ik} right)}_{=X_k} ~ varepsilon^k end{split} end{equation}
And there you have it $$ X_k = sum_{i=1}^n X^i g_{ik} $$ the covariant coordinates are given by the contraction of $X^i$ and $g_{ik}$. One can now go ahead and show that the natural map $tau$ is invertible and proof the similar result $$ X^i = sum_{k=1}^n g^{ik} X_k $$ but this should be clear since $g_{ij}$ is invertible (if it wasn't, $B$ would be no basis because some $e_i$ were linear combinations of the others).
The crucial thing to observe is that $X^i$ and $X_i$ are not some random numbers but two different things: The former being the coordinates of $x$ in $V$ while the later being the coordinates of $tau(x)$ in $W$.
Correct answer by image on May 31, 2021
To have a good notion of "raising and lowering indices", you need to have a non-degenerate 2-tensor. If it is not non-degenerate, you might send $v^mu$ to $v_mu=0$, in which case you can't say that "lowering the index and then raising it back up" is like doing nothing. Being able to do that is of vital importance. A Riemannian manifold comes, by definition, equipped with such a non-degenerate (symmetric) bilinear form, so there is a canonical choice of notion of raising and lowering indices.
If your manifold is guaranteed to carry other non-degenerate 2-tensors, then you may use those as well to raise and lower indices (though the meaning is different from that of raising and lowering with the metric). This is the case, for instance, if your manifold is symplectic (i.e. carries a closed, non-degenerate 2-form $omega$).
Answered by Danu on May 31, 2021
The metric tensor is defined by its ability to raise and lower indices. Take a finite dimensional vectors space, $V$, with $operatorname{dim}{V} = N$. Given another space, $W$, with dimension $M$ you can construct the space of linear maps between those spaces. We call elements from the space of linear maps matrices, and in this example, they form a vector space that has $operatorname{dim}{L : Vrightarrow W} = Ntimes M$.
If the target space has dimension $1$ (i.e. the map from $V$ to scalars) then the space of maps has dimensions $N$. As we learned from linear algebra, any two finite dimensional spaces with the same dimensionality are isomorphic. We can therefore construct an isomorphic map (i.e. a map that is both 1 to 1 and covers the target space) from the $V$ to $L$. The map from $V$ to $L$ is called the metric.
In other words, when the index is up, the vector is from $V$, when down it's from $L$ (often called the dual-space).
Invertability aside, we also pick the metric to have other properties we desire. In particular, we want the scalars produced by $g(v_1) v_2$ to be invariant under some set of transformations (usually rotations or Lorentz transformations). This is what constrains the metric signature (pattern of signs of eigenvalues).
Also, there is no proof that $g^{mu nu} g_{nu alpha} = delta^mu_{hphantom{mu}alpha}$ because that is the definition of the inverse metric (the map from $L$ to $V$, as opposed to the other way around).
Answered by Sean E. Lake on May 31, 2021
It is a convention$^1$. If a theory has a distinguished invertible rank-2 tensor, why not use it? Examples:
If there are more than one distinguished invertible rank-2 tensor, one would have to make a choice. E.g. bi-metric GR, etc.
--
$^1$More formally, the notion of "raising and lowering indices" reflects the musical isomorphism. See also e.g. this Phys.SE post and links therein.
Answered by Qmechanic on May 31, 2021
Another way of looking at this is with $1$-forms and vectors. Given the vector basis $frac{partial}{partial x^a}$ and the $1$-form ${bfomega}^b$ these are dual if $$ {bfomega}^bleft(frac{partial}{partial x^a}right)~=~delta_a^b. $$ The metric as it changes covariant indices to contravariat indices, to use a somewhat older definition, means that with vector $X_a$ that $$ g_{ab}~=~g(X_a,~X_b), $$ and in a dual meaning that $$ g^{ab}~=~g({bfomega}^aotimes{bfomega}^b), $$ here as a symmetric product. The role of the metric tensor in raising and lowering indices is tied to the dualism between vectors and forms.
Answered by Lawrence B. Crowell on May 31, 2021
Using the metric tensor in this way accounts for the Riesz representation theorem in a non-orthogonal basis.
The Riesz representation theorem establishes a correspondence between linear functionals and vectors by associating a linear functional with the vector that is normal to it's level sets. Contracting with the metric tensor computes this normal vector.
What it means for a vector to be "normal" to a hyperplane depends on the metric, so if you replace the metric tensor with another bilinear form, you will be computing the vector that is "perpendicular" in a different metric.
Answered by Nick Alger on May 31, 2021
We know that we can write a vector as: $$vec{A}=A^ie_i=A_ie^i$$
Then we expect: $$vec{A}cdot e_i=A_i$$ $$vec{A}cdot e^i=A^i$$
Which leads to: $$A_i=vec{A}cdot e_i=(A^ke_k),e_i=A^k(e_k,e_i)$$
But $e_k,e_i$ is our natural definition of $g_{k,i}$.
Therefore: $$A_i=A^kg_{ki}$$
Answered by Stefano on May 31, 2021
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