# Is a metric tensor field the same thing as $ds² = -dt² + dx²+ dy² + dz²$?

I am having trouble understanding the nature of the metric tensor field on spacetime manifolds.

In particular, a Riemannian manifold $(M,g)$ is defined as a real smooth manifold $M$ equipped with an inner product $g_p$ on the tangent space $T_pM$ at each point $p$ that varies smoothly from point to point in the sense that if $X$ and $Y$ are vector fields on $M$, then $p \mapsto g_p(X(p),Y(p))$ is a smooth function. The family $g_p$ of inner products is called a Riemannian metric tensor.

But in my physics classes, I often hear the equation

$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$

referred to as a "metric."

Is it a Riemannian metric?

Can $ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$ be written as a tensor field?

• Hi Stan Shunpike: The question (v1) asks too many things at the same time. It is preferred to only ask one question per post. I edited out the metric space part. Please feel free to re-ask that at a later time in a separate post. Jan 15 '15 at 19:02
• It's not positive definite, and that's because spacetime is a pseudo-Riemannian manifold; for standard Minkowski the signature is $(1,3)$ rather than Euclidean $(4,0)$. Jan 15 '15 at 19:38

Really this should be written $g = -dt \otimes dt+dx \otimes dx + dy \otimes dy + dz \otimes dz$. Here $g$ is a bilinear form, aka a (0,2) tensor. It eats two vectors and spits out a real number in a way which is linear in each slot separately..

If I have a pair of vectors $v_1 = (t_1,x_1,y_1,z_1)$ and $v_2 = (t_2,x_2,y_2,z_2)$, then applying this tensor to the pair of vectors gives

$g(v_1,v_2) = -t_1t_2+x_1x_2+y_1y_2+z_1z_2$

This looks a lot like the usual inner product between two vectors, except it is not positive definite, since the inner product of a "time vector" with itself is negative (so time is kind of "imaginary" if you want to think of it that way).

This is not quite a Riemannian metric since it is not positive definite, but it is what is called a pseudo-riemannian metric.

• +1: But I'm always confused when users write such nice answers but don't upvote the question for which the answer was written. Jan 15 '15 at 20:37
• @joshphysics I have rectified the situation. Basically I just read the question and immediately wanted to answer, and the resulting time delay caused me to forget to up vote. Jan 15 '15 at 20:48
• Is $g$ a function? I recognize it is a tensor and therefore a multilinear map. But the way it is written above, it looks like it takes $v_1$ and $v_2$ as inputs like a function Jan 16 '15 at 3:49
• @StanShunpike Multilinear maps are functions. They're just linear in each argument. Jan 16 '15 at 4:54

You are right to be confused and no, you're not going crazy so there is no need to reach for the haloperidol yet: this is simply a sloppy, but ultimately well understood (in physics) use of language. As in the other answers, the name "metric" simply refers to its algebraic (in the sense of "shape" of equations) likeness to an inner product. As in the other answers, however, it is a degenerate bilinear form.

One thing that I believe that the other answers haven't mentioned, which I think is important, is that the (pseudo) "distance" function $d:\mathcal{M}\to\mathbb{R};\;\; d(X,\,Y)= \sqrt{|g(X-Y,\,X-Y)|}$ is not subadditive, i.e. it does NOT fulfill the triangle inequality. Thus the induced pseudo norm fails to have two of the fundamental properties of metric space (in the topological sense) i.e. there are nonzero null vectors and we don't have subadditivity. This "failure" is what makes for time dilation and what makes the twin "paradox" so interesting (the zigzag path of the spacefaring twin has a shorter "length" than the vertical path of the homebody twin).

You might like to privately call $g$ a "pseudo metric" or Minkowskian metric - I do. The name "pseudo Riemannian manifold" is also very widely used for a manifold kitted with a degenerate billinear form, but in all other respects like a Riemannian manifold, since many of the theorems of Riemannian geometry are also true for pseudo Riemannian manifolds. In particular, the "fundamental" theorem of Riemannian geometry (see Wiki page of this name), that there is a unique (Levi Civita) connexion (covariant derivative) that absorbs the torsion (sets it to nought), is true.

This is just the inner product of the tensor field on two vectors. Given $$\eta_{00} = -1 \\ \eta_{ij} = \delta_i^j$$ with all other components zero, the inner product $$ds^2 = ds_\mu ds^\mu = \eta_{\mu \nu} ds^\mu ds^\nu$$ is, for $ds^\mu = (dt, dx, dy, dz)$, $$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2.$$

(I think I got my vectors/one-forms straight, but please correct me if I didn't.)