# Why use the tetrad formalism?

1. Dec 29, 2008

### pellman

A little background for the beginners who may be reading. As originally formulated by Einstein the dynamical quantify of GR is the spacetime metric $$g_{\mu\nu}(x)$$. One can introduce quantities called a tetrad field $$e^{I}_{\mu}(x)$$ such that

$$e^{I}_{\mu}e^{J}_{\nu}\eta_{IJ}=g_{\mu\nu}$$

everywhere, where $$\eta_{IJ}$$ is the flat Minkowski metric. Then you rewrite all the GR equations in terms of $$e^{I}_{\mu}(x)$$.

But we have introduced extra degrees of freedom. Any two tetrad fields related by a (local?) Lorentz transformation are equivalent. Why would we want to do that?

2. Dec 29, 2008

### Mentz114

The tetrad formalism is coordinate independent, so we are free to choose one to suit a particular problem, if it helps. For instance, in setting up a freely-falling frame in the Schwarzschild geometry.

3. Dec 29, 2008

### George Jones

Staff Emeritus
It's not "instead of the metric," it's taking components of the metric tensor with respect to an orthonormal basis instead of taking components of the same metric tensor with respect to a coordinate basis. Just as the same vector has different sets of components depending on the basis used, so does the metric tensor $g$. $g_{\mu , \nu}$ usually denote the components of $g$ with respect to a coordinate basis. By definition, the components of $g$ with respect to an orthonormal basis are $\eta_{\mu \nu}$.
No, it's just a change of basis (at all events). In any vector space, including a tangent space of the spacetime manifold, there is freedom for choice of basis. Tretrads are particular linear combinations of coordinate basis vectors.
Orthonormal frames give the physical components of vectors and tensors. Even in general relativity, Lorentz transformations relate the physical components of a tensor for two different observers at the same spacetime event.

In the notation $e_{\mu}^{I}$, $I$ denotes which basis vector and $\mu$ denotes the components of vector $e^I$ with respect to a basis of coordinate vectors. I prefer not to use this notation. I prefer to use $e_\mu$, with $\mu$ playing the role of $I$, i.e., each $\mu$ is a basis vector, not the components of a vector.

See

4. Dec 29, 2008

### pellman

Isn't the same true of the metric tensor? It is a tensor after all.

5. Dec 29, 2008

### Mentz114

What's your point ? We aren't free to choose any metric tensor, but given one, reformulating it in tetrads is safe because of the coordinate independence.

I'm not an expert in this. The example I mention is the only one I've seen worked out.

Last edited: Dec 29, 2008
6. Dec 29, 2008

### pellman

Fix a coordinate system (or fix one for each open set in the covering). Now you have an explicit differential equation in the components $$g_{\mu\nu}$$. We solve it and find $$g_{\mu\nu}(x)$$. I presume it is not unique; there is some gauge freedom. But we now have our family of solutions $$\{g_{\mu\nu}(x)\}$$.

Now we do the same thing expressed terms of the tetrad field. But now we find that for each specific solution $$g_{\mu\nu}(x)$$ there is a host of tetrad solutions $$e_{\mu}^{I}(x)$$, $$e'_{\mu}^{I}(x)$$, $$e''_{\mu}^{I}(x)$$, etc., such that

$$g_{\mu\nu}(x)=e_{\mu}^{I}(x)e_{\nu}^{J}(x)\eta_{IJ}=e'_{\mu}^{K}(x)e'_{\nu}^{L}(x)\eta_{KL}=...$$

The various tetrad fields are related to each by Lorentz transformations $$e'_{\mu}^{K}(x)=\Lambda^K_Ie_{\mu}^{I}(x)$$

The equivalence between these different tetrad fields for a given $$g_{\mu\nu}(x)$$ are in addition to the equivalence between two metrics which differ only by a gauge, or the equivalence between component solutions as measured from different coordinate systems (general covariance).

To take a simpler analogy, it seems to me like the following: suppose we have differential equation (real coefficients) of a single real function $$f(x)$$. But then introduce functions $$u(x)$$ and $$v(x)$$ such that $$u^2+v^2=f$$. There may be family of solutions $$\{f_i\}$$ such that (for instance) $$f_i(x)-f_j(x)=constant$$, ie, a gauge freedom. But then for each of these individual functions, there are infinitely many pairs $$(u(x),v(x))$$ subject to the condition above.

Isn't this the same sort of thing? Why would I want to introduce u and v, when all that really mattered was f?

7. Dec 29, 2008

### atyy

MTW says something like suppose you do an experiment in an aeroplane, you can use any coordinates you want, but the x,y,z and clock stuck to the walls of your plane are special. Apart from this, tetrad fields or vierbeins seem to be used to look for local Lorentz violation (Mattingly, http://relativity.livingreviews.org/Articles/lrr-2005-5/index.html) and also to couple gravity to fermions (Bern, http://relativity.livingreviews.org/Articles/lrr-2002-5/index.html). The references are kind of incidental to your question, but maybe you can use them to find more relevant ones.

8. Dec 30, 2008

### George Jones

Staff Emeritus
Pellman, how comfortable are you with tensors as multi-linear mappings and vectors as differential operators?

9. Dec 31, 2008

### schieghoven

One important reason for doing this is that it is the only way to get spin-half particles onto a curved spacetime. In conventional GR without tetrads, there isn't really an intrinsic mechanism which guarantees that the local symmetry group is the Lorentz group, and this is crucial for constructing spinors. Instead, one probably ends up with the diffeomorphism group, which has no spinor representations.

The tetrad formalism solves this: it provides the connection between the metric and a locally orthonormal frame with Minkowski signature, so it highlights the Lorentz structure of spacetime. The set of tetrads is precisely a representation of the Lorentz group, and this has the double covering group which provides Weyl spinors. Electrons (Dirac spinors) are a pair of Weyl spinors.

Hope that helps,

Dave

10. Dec 31, 2008

### pellman

I'm pretty familiar with these concepts.

Ok. I'll buy that. Thanks, Dave

I don't really get it yet. But I suspect that if I keep this in mind as I read, it'll come.

11. Dec 31, 2008

### shoehorn

Bingo. In my opinion this is by far the most important reason for using vierbeins in general relativity and generalizations thereof. The simple fact is that there exists no convincing way to include fermionic degrees of freedom in GR without resorting to the use of $e^I(x)$ over $g_{ab}(x)$.

For what it's worth, I never found Chandresakhar's line of reasoning on the power of tetrads in convincing; I can't think of a single calculation in classical GR that is significantly easier with tetrads than it is using the metric.