2
$\begingroup$

In math books, I saw the metric tensor is defined with the use of the Jacobian matrix as

$$g_{\mu \nu}=J_{\mu}^a \: J_{\nu}^b \: \eta_{ab}\tag{1}$$

where $J_{\mu}^a=\frac{\partial \bar{x}^a}{\partial x^{\mu}}$ (Added: where barred symbols denote Minkowskian coordinates and unbarred ones stand for curvilinear coordinates). And with the matrix notation $\mathrm{g}= \mathrm{J^T} \cdot \eta \cdot \mathrm{J} .$

In 1928 Einstein introduced the $n$-Bein which was further developed and it is known as "tetrad formalism" of GR. The metric tensor in terms of the vierbein (tetrad) field is

$$g_{\mu \nu}={e_{\mu}}^a \: {e_{\nu}}^b \: \eta_{ab}.\tag{2}$$

They both satisfy the orthonormality condition $${e^{\mu}}_a \: {e_{\nu}}^a=\delta^{\mu}_{\nu} \quad {e_{\mu}}^a \: {e^{\mu}}_b=\delta^{b}_{a}.$$

As the Jacobian matrix is bijective iff $\mathrm{J} \neq 0$ so $\bar{\mathrm{J}}=\mathrm{J}^{-1}$ and we also have the same as above.

As (1) and (2) look identical, the question is: what is the difference between the Jacobian matrix and the vielbein matrix? Do they represent the same math objects in the application to 4-dimensional space? My guess is that it is just a matter of terminology and that the Jacobian matrix is used for a broader range of coordinate transformations, though the "vierbein" is the term from the GR that applies to the 4-dimensional case.

References:

  1. Taha Sochi, "Tensor Calculus", https://arxiv.org/abs/1610.04347.
$\endgroup$
  • $\begingroup$ In regards to the last question, vierbein is the 4D case, and vielbein is the term for the formalism in any dimension. $\endgroup$ – JamalS Aug 9 at 16:39
  • $\begingroup$ @JamalS thanks. Obviously Vielbein =n-Bein. $\endgroup$ – Eddward Aug 9 at 18:45
  • $\begingroup$ Comment to the post (v10): Barred & unbarred coordinates are the other way around. $\endgroup$ – Qmechanic Aug 9 at 19:05
  • $\begingroup$ If defined in such was then I also changed in matrix notation to $\mathrm{g}= \mathrm{J^T} \cdot \eta \cdot \mathrm{J} .$ The Jacobian definition of barred via unbarred symbol is arbitrary (as per cited book). But then it is the matter of the order of Jacobian indexes in the definition of $g_{ab}$ $\endgroup$ – Eddward Aug 9 at 19:32
3
$\begingroup$

In a nutshell, vielbeins $e^a_{\mu}$ work more generally for abstract manifolds (up to topological obstructions), and generalize the Jacobian $J^a_{\mu}=\partial y^a/\partial x^{\mu}$, which only works for affine spaces. Unlike the vielbeins, the Jacobian always satisfies an integrability condition $\partial J^a_{\mu}/\partial x^{\nu}=(\mu\leftrightarrow \nu)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. Good answer with good references. So, it basically confirms my initial guess that these are the same objects for Riemannian geometry of 4-dimensional space-time where the GR rules. $\endgroup$ – Eddward Aug 9 at 20:17
  • $\begingroup$ No, a pseudo-Riemannian manifold $(M,g)$ [which is the standard setting for GR] is not necessarily an affine space [which is the standard setting for SR]. $\endgroup$ – Qmechanic Aug 9 at 20:19
  • $\begingroup$ Thank you! It clarifies. $\endgroup$ – Eddward Aug 9 at 23:14
  • $\begingroup$ but leaves me with the question: how the metric tensor is defined then for a pseudo-Riemannian manifold using the expression (1).. $\endgroup$ – Eddward Aug 9 at 23:25
  • 1
    $\begingroup$ @Eddward It simply is not. However, one can define a metric using (2). $\endgroup$ – mmeent Aug 10 at 8:26
1
$\begingroup$

A choice of coordinates $x^\mu$ for some patch of spacetime automatically defines a corresponding basis for the tangent space at each point, with basis vectors $\frac{\partial}{\partial x^\mu}$. This is referred to as a coordinate basis, or sometimes as a holonomic basis.

Of course, a choice of basis is in principle independent of a choice of coordinates. The fact that there is a natural coordinate-induced basis available doesn’t mean we have to use it.

This might lead one to wonder if there are choices of basis which cannot be induced by a coordinate chart, and the answer is a resounding yes. As an example, one can show that the familiar orthonormal polar unit vectors $\hat r$ and $\hat \theta $ are such a choice.

When we go from one coordinates chart to another, the Jacobian matrix provides the corresponding transformation between coordinate-induced bases. However, if a non-holonomic basis is involved then there’s obviously no corresponding Jacobian because the non-holonomic basis doesn’t correspond to a choice of coordinates. Therefore, the change of basis needs to be described by a more general object. This is the vielbein matrix $e_\mu^{\ \ \nu}$.


Consider the following example for the standard Euclidean plane with Cartesian coordinates $(x,y)$. This choice of coordinates corresponds to the (holonomic) basis $\left\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right\}$.

If we shift to polar coordinates $(r,\theta)$, we can find a corresponding polar basis $\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$. Since we have

$$x = r\cos(\theta) \qquad y = r\sin(\theta)$$ it follows that

$$\frac{\partial}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial}{\partial y} = \cos(\theta)\frac{\partial}{\partial x}+\sin(\theta)\frac{\partial}{\partial y}$$ and similarly for $\frac{\partial}{\partial \theta}$. Letting $y\equiv (r,\theta)$, this can be written compactly as

$$\frac{\partial}{\partial y^\mu} = \frac{\partial x^\nu}{\partial y^\mu} \frac{\partial}{\partial x^\nu} \equiv J^\nu_{\ \mu} \frac{\partial}{\partial x^\nu}$$

with $J$ the Jacobian. In this basis, the metric takes the form

$$g = \pmatrix{1& 0 \\ 0 & r^2}$$

which means that this polar basis is orthogonal but not orthonormal. In contrast, consider the basis

$$\hat r \equiv \cos(\theta)\frac{\partial}{\partial x} + \sin(\theta)\frac{\partial}{\partial y}$$ $$\hat \theta \equiv -\sin(\theta)\frac{\partial}{\partial x} + \cos(\theta)\frac{\partial}{\partial y}$$

One can show without much effort that these basis vectors are orthonormal. They are not holonomic, however; one can see this by noting that for a smooth function $f$, $\hat r(\hat \theta f) \neq \hat \theta(\hat r f)$, which means that they cannot be expressed as

$$\hat r = \frac{\partial}{\partial u} \qquad \hat \theta = \frac{\partial}{\partial v}$$ for some coordinates $(u,v)$. Therefore, we cannot write a Jacobian for this coordinate transformation. Instead, writing $(\hat r,\hat\theta) \equiv (\hat e_r, \hat e_\theta)$, the change of basis is provided by

$$e_\mu^{\ \ \nu} = \pmatrix{\cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta)}$$ $$\hat e_\mu = e_{\mu}^{\ \ \nu} \frac{\partial}{\partial x^\nu}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @Eddward Like the Jacobian, the vierbein matrix is a change-of-basis matrix which you use when going from one basis for the tangent space to another. The difference is that a Jacobian only makes sense when the two bases in question are induced by coordinate charts. If one (or both) are non-holonomic bases, then there's nothing to differentiate, so the Jacobian is inappropriate. In this case, you must use a vierbein matrix. I've added an example to my answer which may help. $\endgroup$ – J. Murray Aug 9 at 20:38
  • $\begingroup$ Thank you so much! I think it answer my question in full. I presume also that for a given metric tensor $g_{\alpha \beta}$ the vierbein is not uniquely defined. But maybe this is a subject for another question or my own thoughts.. $\endgroup$ – Eddward Aug 9 at 20:47
  • $\begingroup$ @Eddward Yes, that's right. $e_{\mu}^{\ \ \ \nu}$ is just the matrix which relates one basis to another; it is defined for any two valid bases. It's just like the Jacobian in this way, and has nothing to do with the metric at all. $\endgroup$ – J. Murray Aug 9 at 20:52
  • $\begingroup$ @J. Murray A vielbein is not just a matrix that relates one basis to another, it is a basis. In practice we describe a vielbein relative to some other basis, but this is not an integral part of the concept. (One could introduce a coordinate independent characterization.) Moreover, a vielbein contains exactly the same information as the metric. It is therefor incorrect to say that it has nothing to do with the metric at all. $\endgroup$ – mmeent Aug 10 at 8:25
  • $\begingroup$ @mmeent I am here making the distinction between the vielbein itself and the matrix which relates one basis to another. My point was that the latter object depends on which bases are being related to each other, and is not determined by the metric. The comment is now deleted, but my comment was in response to the question "if the metric is the Schwarzschild metric, what is $e_\mu^{\ \ \nu}$?" $\endgroup$ – J. Murray Aug 10 at 12:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.