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The Metric Tensor

The section above is still very generic and little of it depends on whether the tensors are three or four or ten dimensional. We now need to make them work for the specific geometry we are interested in, which is one where we will ultimately be seeking transformations that preserve the invariant interval:

(ds)^2 = (dx^0)^2 - (dx^1)^2 - (dx^2)^2 - (dx^3)^2
\end{displaymath} (16.11)

as this is the one that directly encodes an invariant speed of light.

From this point on, we must be careful not to confuse $x
\cdot x = x^2$ and $x^2 = y$, etc. Contravariant indices should be clear from context, as should be powers. To simplify life, algebraically indices are always greek (4-vector) or roman italic (3-vector) while powers are still powers and hence are generally integers.

Let us write this in terms of only contravariant pieces $dx^\mu$. This requires that we introduce a relative minus sign when contracting out the components of the spatial part of the differential only. We can most easily encode this requirement into a special matrix (tensor) called the metric tensor as:

(ds)^2 = g_{\alpha\beta} dx^\alpha dx^\beta
\end{displaymath} (16.12)

The tensor $g$ obviously satisfies the following property:
g_{\alpha\beta} = g_{\beta\alpha}
\end{displaymath} (16.13)

(that is, it is symmetric) because the multiplication in the Einstein summation is ordinary multiplication and hence commutative. It is called the metric tensor because it defines the way length is measured.

At this point if we were going to discuss general relativity we would have to learn what a manifold16.5 s. Technically, a manifold is a coordinate system that may be curved but which is locally flat. By locally flat I mean very specifically that one can cover the entire space with ``patches'' in the neighborhood of points where the coordinate system is locally Euclidean (e.g. Cartesian). An example of a curved space manifold is the surface of a sphere (think the surface of the earth). When we look down at the ground beneath our feet, it looks quite flat and we can draw triangles on it that appear to have interior angles that sum to $\pi$ and we can draw a map of (say) our county that more or less accurately encodes distances on the ground in terms of distances measured on the map. However, if we take too big a patch all of this breaks down. The angles in a triangle sum to strictly more than $\pi$ radians. Maps have to be distorted and chopped into pieces to correctly represent distances on the ground as distances on the flat 2-dimensional map. This is how a manifold works - we can work with it in the local neighborhood of any point as if it is flat, but if we go too far we have to work harder and correct for its curvature, where ``too far'' is obviously defined in terms of the scale of its curvature and some common sense.

General relativity introduces the hypothesis that gravitational fields bend space-time. However, this bending is very, very slight unless one is in a very strong gravitational field, and this bending preserves a local smoothness of space-time so that space-time, although it is no longer strictly Euclidean, is still a manifold and we can do all sorts of transformations in a very general way as long as we restrict the results to a locally flat patch.

In our discussion of special relativity we will assume from the beginning that our space-time is flat and not bent by strong gravitational fields. In this case the metric tensor can be expressed in a very simple form. We will use the Lorentz metric (as opposed to the Minkowski metric that uses $x^4 = ict$ instead of $x^0$). Using our definitions of the $\mu = 0,1,2,3$ coordinates, $g$ in the differentials above is just:

g_{00} = 1, g_{11} = g_{22} = g_{33} = -1
\end{displaymath} (16.14)

and we see that it is not just symmetric, it is diagonal.

The contravariant and mixed metric tensors for flat space-time are the same (this follows by considering the $\frac{\partial x^\alpha}{\partial x^\beta}$ coordinate transformation matrices that define co- and contra-variance):

g_{\alpha\beta} = g_\alpha^\beta = g^{\alpha\beta} .
\end{displaymath} (16.15)

Finally, the contraction of any two metric tensors is the ``identity'' tensor,
g_{\alpha\gamma}g^{\gamma\beta} = \delta_\alpha^\beta =
\delta_{\alpha\beta} = \delta^{\alpha\beta} .
\end{displaymath} (16.16)

Since we want $(ds)^2$ to be (to contract to) a scalar, it is clear that:

$\displaystyle x_\alpha$ $\textstyle =$ $\displaystyle g_{\alpha\beta} x^\beta$ (16.17)
$\displaystyle x^\alpha$ $\textstyle =$ $\displaystyle g^{\alpha\beta} x_\beta$ (16.18)

or the metric tensor can be used to raise or lower arbitrary indices, converting covariant indices to contravariant and vice-versa:
F^{\mu\alpha\nu} = g^{\alpha\beta} F^{\mu\nu}_\beta
\end{displaymath} (16.19)

This is an important trick! Note well that in order to perform a contraction that reduces the rank of the expression by one, the indices being summed must occur as a co/contra pair (in either order). If both are covariant, or both are contravariant, one or the other must be raised or lowered by contracting it with the metric tensor before contracting the overall pair! We use this repeatedly in the algebra in sections below.

Finally we are in a position to see how covariant and contravariant vectors differ (in this metric). We have already seen that ``ordinary'' vectors must linearly transform like contravariant vectors. Given a contravariant vector $(A^0,A^1,A^2,A^3)$ we thus see that

A_0 = A^0, A_1 = -A^1, A_2 = -A^2, A_3 = -A^3
\end{displaymath} (16.20)

A^\alpha = (A^0,{\bf A}), A_\alpha = (A^0,-{\bf A}).
\end{displaymath} (16.21)

Covariant vectors are just spatially inverted contravariant vectors. Note that this definition, together with our definition of the general scalar product, reconstructs the desired invariant:
B \cdot A = B_\alpha A^\alpha = (B^0 A^0 - {\bf B} \cdot {\bf A})
\end{displaymath} (16.22)

This tells us how ordinary quantities transform. However, we are also interested in how tensor differentials transform, since these are involved in the construction of a dynamical system. By considering the chain rule we see that

\frac{\partial }{\partial \bar{x}^{\alpha}} = \frac{\parti...
...partial \bar{x}^{\alpha}}
\frac{\partial }{\partial x^\beta}
\end{displaymath} (16.23)

or, differentiation by a contravariant coordinate transforms like a covariant vector operator. This is more or less the definition of covariant, in fact. Similarly, differentiation with respect to a covariant vector coordinate transforms like a contravariant vector operator. This also follows from the above by using the metric tensor,
\frac{\partial }{\partial x_{\alpha}} = g_{\alpha\beta}
\frac{\partial }{\partial x^{\beta}} .
\end{displaymath} (16.24)

It is tedious to write out all of the pieces of partial derivatives w.r.t. various components, so we (as usual, being the lazy sorts that we are) introduce a ``simplifying'' notation. It does, too, after you get used to it.

$\displaystyle \partial^\alpha$ $\textstyle =$ $\displaystyle \frac{\partial }{\partial x_\alpha} = (\frac{\partial }{\partial x^0}, -
\mbox{\boldmath$\nabla$})$ (16.25)
$\displaystyle \partial_\alpha$ $\textstyle =$ $\displaystyle \frac{\partial }{\partial x^\alpha} = (\frac{\partial }{\partial x^0}, +
\mbox{\boldmath$\nabla$}) .$ (16.26)

Note that we have cleverly indicated the co/contra nature of the vector operators by the placement of the index on the bare partial.

We cannot resist writing down the 4-divergence of a 4-vector:

\partial^\alpha A_\alpha = \partial_\alpha A^\alpha =
...tial A^0}{\partial t} + \mbox{\boldmath$\nabla$}\cdot {\bf A}
\end{displaymath} (16.27)

which looks a lot like a continuity equation or a certain well-known gauge condition. (Medidate on just what $A^\mu$ would need to be for either of these equations to be realized as a four-scalar). Hmmmmmm, I say.

Even more entertaining is the 4-Laplacian, called the D'Lambertian operator:

$\displaystyle \Box$ $\textstyle =$ $\displaystyle \partial_\alpha \partial^\alpha = \frac{\partial ^2 }{\partial x^{02}} -
\nabla^2$ (16.28)
  $\textstyle =$ $\displaystyle \frac{1}{c^2} \frac{\partial ^2 }{\partial t^2} - \nabla^2$ (16.29)

which just happens to be the (negative of the) wave operator! Hmmmmmmmm! By strange coincidence, certain objects of great importance in electrodynamics ``just happen'' to be Lorentz scalars! Remember that I did say above that part of the point of introducing this lovely tensor notation was to make the various transformational symmetries of physical quantities manifest, and this appears to be true with a vengeance!

That was the ``easy'' part. It was all geometry. Now we have to do the messy part and derive the infinitesimal transformations that leave scalars in this metric invariant.

next up previous contents
Next: Generators of the Lorentz Up: The Lorentz Group Previous: Tensors in 4 Dimensions   Contents
Robert G. Brown 2007-12-28