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I am looking at the 4-vector treatment of special relativity, but I have had no formal training in Tensor algebra and thus am having difficulty understanding some of the concepts which appear.

One such concept is the notion of contra- and co-variant vectors. Wikipedia describes a contra-variant vector as varying with the co-ordinates, i.e. the inverse of the reference axes. It gives the example of a change of basis matrix $\mathbf{M}$, and a contra-variant vector $\vec{v}$ and states that in the new set of co-ordinates, we have:

$$\vec{v}'=\mathbf{M}\vec{v}$$

It then states that a covariant vector varies with the co-ordinate axes.

Does this mean that for a covariant vector $\vec{u}$ and change of basis matrix $\mathbf{M}$ we have: $$\vec{u}'=\mathbf{M}^{-1}\vec{u}?$$

I am also having problems understanding the concept of raising and lowering indices. For instance, Wikipedia states that for two four-vectors $\vec{A},\vec{B}$ and using Einstein summation we have:

$$\left\langle \vec{A} \middle|\vec{B} \right\rangle=A^{\mu}B_{\mu}=A_{\nu}B^{\nu}$$

And I am having difficulty understanding how this relates to the other definition of the inner product:

$$\left\langle \vec{A} \middle|\vec{B} \right\rangle =\mathbf{A}^{T}\mathbf{\eta}\mathbf{B}$$

If anyone could help give me a more intuitive understanding of these concepts I would be grateful.

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  • $\begingroup$ "Does this mean that for a covariant vector $\vec{u}$ and change of basis matrix $\mathbf{M}$ we have: $\vec{u}'=\mathbf{M}^{-1}\vec{u}$?" Quick answer: yes, absolutely. The "inner product", i.e. a scalar, between a vector and covector must be independent of co-ordinates, so the covector must transform as you say. Have a look at Kip Thorne's most excellent textbook: read Chapter 1 and I'm sure you'll be much wiser: pma.caltech.edu/Courses/ph136/yr2011/1101.2.K.pdf $\endgroup$ – WetSavannaAnimal Dec 30 '13 at 0:36
  • $\begingroup$ Also have a look at physics.stackexchange.com/q/87775/26076 My favourite way of visualising the tangent space (contravariant or "everday" vectors) is in my answer to that question. The first chapter of Schutz "A First Course on General Relativity" also gives a good description. $\endgroup$ – WetSavannaAnimal Dec 30 '13 at 0:40
  • $\begingroup$ Someone should include in their answer the example that the gradient is actually a covector. I found that example quite helpful when learning about this topic. $\endgroup$ – DanielSank Jul 30 '15 at 17:37
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Contravariant vectors are "standard" vectors. Covariant vectors are linear applications on contavariant vectors producing scalars.

Let us start form the former case. If you fix a couple of bases $\{e_i\}_{i=1,\ldots,n}$ and $\{e'_i\}_{i=1,\ldots,n}$ in the finite dimensional vector space $V$ with dimension $n$, such that $e_i = \sum_j {A^j}_i e'_j$ for set of coefficients ${A^j}_i$ forming a (necessarily) non-singular matrix $A$, you have for a given vector $v \in V$: $$v = \sum_i v^i e_i = \sum_j v'^j e'_j$$ and thuso $$\sum_i v^i \sum_j {A^j}_i e'_j = \sum_j v'^j e'_j$$ so that: $$\sum_j \left( \sum_i {A^j}_i v^i\right) e'_j = \sum_j v'^j e'_j\:.$$ Uniqueness of components of $v$ respect to $\{e'_i\}_{i=1,\ldots,n}$ eventually entails: $$v'^j = \sum_i {A^j}_i v^i\qquad \mbox{where}\quad e_i = \sum_j {A^j}_i e'_j\tag1$$ This is nothing but the standard rule for transforming components of a given contravariant vector when one changes the decomposition basis.

Let us pass to consider covariant vectors. As I said above, a covariant vector is nothing but a linear map $\omega : V \to R$ ($R$ can be replaced by $C$ if dealing with complex vector spaces or the corresponding ring when considering modules). One easily proves that the set of real valued linear applications as above form a vector space, $V^*$, the so-called dual space of $V$. If $\{e_i\}_{i=1,\ldots,n}$ is a basis of $V$, there is an associated basis $$\{e^{*i}\}_{i=1,\ldots,n}$$ of $V^*$, the dual basis, defined by the requirements (in addition to linearity): $$e^{*k}(e_i) = \delta^k_i\tag2$$ Therefore, a covariant vector $\omega \in V^*$ can alway decomposed as follows: $$\omega = \sum_k \omega_k e^{*k}$$ and, using linearity, (2), and $$v = \sum_i v^i e_i$$ one sees that $$\omega(v) = \sum_k \omega_k v^k\:.$$ The RHS doe not depend on the choice of the basis $\{e_i\}_{i=1,\ldots,n}$ and the corresponding $\{e^{*i}\}_{i=1,\ldots,n}$ even if components of covariant and contravariant vectors $\omega$ and $v$ depend on the considered bases. Obviously, changing the basis in $V$ and passing to $\{e'_i\}_{i=1,\ldots,n}$ related to $\{e_i\}_{i=1,\ldots,n}$ through (1), $\{e'_i\}_{i=1,\ldots,n}$ turns out to correspond to a dual basis $\{e'^{*i}\}_{i=1,\ldots,n}$. A straightforward computation based on (2) shows that $$e^{*i} = \sum_j {B_j}^i e'^{*j}$$ where $$B= \left(A^T\right)^{-1}\:.\tag3$$ Consequently, for a covariant vector $$\omega = \sum_i \omega_i e^{*i} = \sum_j \omega'_j e'^{*j}$$ where $$\omega'_j = \sum_j{B_j}^i \omega_i\:.\tag4$$ This relation, together with (3) is nothing but the standard rule for transforming components of a given covariant vector when one changes the decomposition basis.

This structure rarely appears dealing with classical physics, where one usually deals with orthonormal basis. The reason is that when changing basis and passing to another orthonormal basis, the matrix $A$ associating the bases is in the orthogonal group, so that: $$B= \left(A^T\right)^{-1} =A\:.\tag3$$ and one cannot distinguish, working in components, between covariant and contravariant vectors, since the former in (1) and (4) are, in fact, identical. For instance, for a fixed force $F$ applied to a point with velocity $v$, the linear map associating the force with its power as a function of $v$ defines a covariant vector that we could indicate by $"F\cdot"$ $$\pi^{(F)}: v \mapsto F\cdot v$$ where $\cdot$ denotes the standard scalar product in the Euclidean rest space of a reference frame.

If the (real finite dimensional!) vector space $V$ is equipped with a, generally, indefinite, scalar product, that is a non-degenerate symmetric bi-linear map $g : V \times V \to R$, a natural identification of $V$ and $V^*$ arises. It is nothing but the linear and bijective map associating contravariant vectors with covariant vectors: $$V \ni v \mapsto g(v, \:\:)\in V^*$$ Where, obviously $g(v, \:\:) : V \ni u \mapsto g(v, u)\in R$ turns out to be linear and thus define an element of $V^*$ as said. In components, if $u= \sum_i u^i e_i$ and $s= \sum_i s^i e_i$, one has in view of the bilinearity property fulfilled by $g$: $$g(u,s) = \sum_{i,j} g_{ij} u^is^j\qquad \mbox{where}\quad g_{ij} := g(e_i,e_j)\:.$$ The matrix of elements $g_{ij}$ is symmetric and non-singular (as $g$ is symmetric and non-degenerate). With this definition, one easily sees that, if $u\in V$ is a contravariant vector, the associated covariant one $g(u,\:\:)\in V^*$ has components: $$g(u, \:\:\:)_k= \sum_ig_{ki}u^i$$ so that, the scalar product $g(u,v)$ of $u$ and $v$ can also be written: $$g(u,v)= \sum_{ij} g_{ij}u^iv^j = \sum_i v_i u^i\:.$$

Finally, changing basis one has that: $$g(u,s) = \sum_{i,j} g'_{lm} u'^ls'^m\qquad \mbox{where}\quad g'_{lm} := g(e'_l,e'_m)\:,$$ and $$g'_{lm} = \sum_{ij}{B_l}^i {B_m}^j g_{il}\:.$$

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Covariant and contravariant vectors can be thought of as different flavors of vectors in physics. Most of the vectors which occur in the usual classical physics like position, velocity etc. are contravariant, whereas the gradient operator(which is surprisingly vector-like; look at most of the vector identities) is a covariant vector. To be much more mathematically rigorous, they form dual spaces of each other, just like the bra and ket spaces in Quantum Mechanics.

To get a formal understanding of how covariance and contravariance differ, you must see how the corresponding vectors change when subjected to transformations. See, for example : http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors Read the definition section carefully. As a very basic example, the covariant vector of a column vector is a row vector. For a complex number, a similar definition yields its conjugate.

When it comes to inner products, it usually depends upon what space you are working with. If it is matrices, then the inner product is defined in one way. If you are working on a Hilbert Space, the inner product is defined as an integral of the bra and ket vector over all space.

In special relativity, the inner product is usually defined based on the metric. $$\left\langle \vec{A} \mid\vec{B} \right\rangle=\eta_{{\mu\nu}}A^{\mu}B^{\nu}$$

This can be seen as multiplying the matrix $\eta$ with the matrices $A,B$.

The metric is also helpful in raising and lowering the matrices.

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