Oxymoron said:
Q1) First a question on the most basic special relativity spacetime metric
s^2 = -c^2t^2 + x^2 + y^2 + z^2.
If we take this to be the metric in flat 4D Euclidean space, then this really isn't a Euclidean space at all. One of the metric axioms is not obeyed: The positive-definite one.
The term metric has a different meaning in relativity than it does in a real analysis course.
I woudn't call this flat 4D Euclidean space, I would call it Minkowski spacetime. Any inertial observer can decompose spacetime into space and time, and the 3-dimensional space for any observer does have a metric in the positive-definite sense.
Q2) Is it true that a metric for flat space will have constant entries? As I understand it, if one considers flat space then the basis vectors are not dependent on the choice of coordinates.
Emphatically no. None of this is true, for example, in spherical coordinates for Minkowski spacetime.
Is it true, then, in curved spacetime a metric would be coordinate dependent since entries in the matrix metric would change from one place to another?
Yes. (Non-zero spacetime curvature) implies (there exits no global coordinate system in which the components of the metric tensor are all constant). Contrapositive: (there exits a global coordinate system in which the components of the metric tensor are all constant) implies (spacetime is flat).
You can see that these statements are true by looking at coordinate expressions for the components of the curvature tensor in terms of derivatives of the metric tensor. The converses of these statements are, however, false.
Even the true statements are a little subtle. Think of the 2-dimensional surface formed by a (loooong...) piece of paper roled up into a tube, i.e., S^1 x R. A curved surface, right? Well, no, not in the sense of curvature in relativity. As surface embedded in 3-dimesional space, this cylinder has non-zero (EDIT)
extrinsic curvature. That accounts for all of the curvature we see, i.e., the surface of a cylinder has zero *intrinsic* curvature. Think of the paper as being graph paper. Before it's rolled, the piece of paper is flat, and the lines on it are geodesics. After it's rolled up, the graph paper is still flat, and the lines are still geodesics.
Similarly, we don't don't think of spacetime as being embedded in a higher dimensional space, and the curvature that we calculate is all intrinsic. Just as it's possible to have (intinsically) flat 2-dimesional spaces that aren't R^2, it's possible to have a flat spacetime that isn't Minkowski spacetime.
Suppose you're given a spacetime metric that has non-constant components. In general, there is no way to decide if the metric corresponds to a flat or a curved spaces. Who knows - you may have been given the metric for Minkowski spacetime expressed in an esoteric an unfamiliar set of curvilinear coordinates. To find out, by hook or by crook or by computer, calculate the connection coefficients. Again, look at the coordinate expressions for the components of the curvature tensor in terms of the connection coefficients.
Q3) In Minkowski space, are coordinates generally denoted by x^{\mu} where mu ranges from zero to 3. My question is, I have seen
x^{\mu} = (ct,x,y,z)
and
x^{\mu} = (x,y,z,ct)
Is it simply a matter of opinion where the "ct" bit is written? Is this what is meant by the "signature" of the metric?
Yes, it is purely a matter of convention where the "ct" bit is written. It's pretty standard these days to use the former convention, and I think this what everyone on this forum uses.
No, this not what is meant by the signature of the metric. This refers to whether x^0 y^0 - x^1 y^1 - x^2 y^2 - x^3 y^3 or -x^0 y^0 + x^1 y^1 + x^2 y^2 + x^3 y^3 is taken to the Minkowsk inner product. The generalizes to non-flat spacetimes. Most people on this forum (and most relativists?) use the latter; I (and the high-energy crowd) much prefer the latter.
Q4) The infinitessimal length, dx. Is it a "4-vector"? I mean, in calculus we are taught that this is almost irrelevant, but now it is an actual vector (even a tensor?). Is this ok? Can you think of it as a infinitessimal vector?
Inuitively one thinks of dx as being an infinitesimal vector, but it is actually a one-form, i.e., a covector field.
Q5) \bold{g}_{\mu\nu} is generally considered as the "invariant" metric tensor in Minkowski space, no?
I don't know what you mean, here. The g_{uv} are the components of the metric tensor with respect to a (fields of) basis vectors. As I have said before, these components are *defined* by g_{uv} = g(e_u , e_v). I would never bold the first g (scalars!) but I sometimes bold the second g. This is true in all spacetimes and for all (0,2) tensors, not just in Minkowski spacetime, and not just for the metric tensor.
Are elements of the metric are actually "interior products" of the basis vectors?
I would say components, not elements, and it depends on what you mean by '"interior products" of the basis vectors.'
If the basis is orthonormal then the metric tensor is simpy a diagonal metric matrix?
Look at the definition I gave for the components of the metric tensor. Do this definition answer this question?
Q6) When is g_{\mu\nu} \equiv \eta_{\mu\nu}?
Beats me; this is not a notation that I care to use.
The metric tensor tensor for Minkowki can be defined without using either basis vectors or coordinates. I did this in a post in one of your threads, but I don't remember which one.
eta_{uv} refers to the metric for Minkowski spacetime in a specific coordinate system. Generally, when changing kernel letter is used for a given tensor, e.g., T_{uv} and T_{u'v'}. What happens when you switch from cartesian to spherical coordinates? Do you switch from eta to g? I prefer to use g all the time and let context sort things out.
Regards,
George
