# Off-diagonal terms in metric for 4D space-time [closed]

Consider a delta between two events in 4D space-time written as a 4-vector, $x^\mu=(dt, dR)$. The time $dt$ is a scalar difference in time. The 3-vector $dR$ points some direction in space. One calculates the distance this delta represents by contracting it with a metric tensor like so: $$g_{\mu \nu} x^\mu x^\nu = g_{00} x^0 x^0 +g_{ii} x^i x^i + 2 g_{0i} x^0 x^i + 2 g_{ij} x^i x^j$$

For diagonal metric matrices like the Minkowski metric, one has only the diagonal elements. That accounts for a term of dt times dt, and the dot product of the 3-vector. This is reasonable since neither has a direction. A metric with a non-zero $g_{0i}$ term will be the product of a scalar $dt$ and a 3-vector $dR$. A scalar times a 3-vector should result in a 3-vector. A metric with a non-zero $g_{ij}$ term is even worse. It cannot be an axial vector since an axial vector must be anti-symmetric ($g_{ij} = - g_{ji}$). Any such metric would have an interval that is not invariant under a rotation and thus not conserve angular momentum.

A delta between two events in a 4D space-time written as a quaternion puts all the terms in their correct place: $$x^2=(x^0 x^0 - x^i \cdot x^i, 2 x^0 x^i)$$ There are no cross-terms for a square since an event always points in the same direction as itself. Only the first term is invariant under a Lorentz transformation, the other three are Lorentz covariant, transforming like $\gamma^2 \beta$. The contraction of 2 4-vectors using a metric produces one value. Squaring a quaternion produces 4. It is simple enough to map where similar components go.

Tensors and differential geometry can work in arbitrary dimensions. Quaternions are restricted to a 4 dimensional space. Until we discover super-symmetric particles, physics too may be constrained to a 4 dimensional space.

Why are the off-diagonal elements of a symmetric metric tensor an acceptable practice since a scalar times a 3-vector should be a 3-vector?

## closed as unclear what you're asking by ACuriousMind♦, Kyle Kanos, JamalS, Danu, PraharMar 23 '15 at 15:31

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• Is "a delta" a difference? Also, you cannot, unless you are on $\mathbb{R}^4$ with a constant metric, simply take the "difference" of two events in that fashion, because points on a general manifold aren't vectors! The metric tensor acts at every point on the tangent spaces, not on the space itself. You are correct that metrics with off-diagonal element are not rotationally invariant, but where's the problem with that? I don't understand the question. – ACuriousMind Mar 17 '15 at 14:54
• Yes, I did mean a simple difference. With the machinery of differential geometry, I was wrong to do the simple subtraction as if the two events were vectors. Thanks. – sweetser Mar 17 '15 at 16:18
• dt is no scalar since in general it doesn't transform as dt=dt'. Based on the same argument dR is no vector in spacetime. – image Mar 18 '15 at 17:46
• @MarcelKöpke points out an issue of terminology. Scalar, vector, divergence, gradient, cross product and curl were all coined I believe by Hamilton to describe parts of quaternions. The first term of a quaternion can then be called a Hamilton scalar, the next three a Hamilton vector. No one does this. Instead one goes based on context. In modern terms, one often uses a name that characterizes how something transforms. A scalar is invariant under a Lorentz transformation. That is not the case for dt. Similarly for dR. – sweetser Mar 18 '15 at 21:05

## 2 Answers

the quantity $ds^2 = g_{\mu,\nu} \ dx^\mu \ dx^\nu$ is a measure for distance on a manifold. It is indeed invariant under coordinate transformations, since $g_{\mu,\nu}$ transforms like $$\tilde{g}_{\mu,\nu} = \frac{\partial x^\alpha}{\partial \tilde{x}^\mu} \frac{\partial x^\beta}{\partial \tilde{x}^\nu} \ g_{\alpha,\beta}$$ and $dx^\mu$ transforms like $$d\tilde{x}^\mu = \frac{\partial \tilde{x}^\mu}{\partial x^\alpha} \ dx^\alpha$$ so that we have $$d\tilde{s}^2 = \tilde{g}_{\mu,\nu} \ d\tilde{x}^\mu \ d\tilde{x}^\nu = \frac{\partial x^\alpha}{\partial \tilde{x}^\mu} \frac{\partial x^\beta}{\partial \tilde{x}^\nu} \ g_{\alpha,\beta} \ \ \frac{\partial \tilde{x}^\mu}{\partial x^\sigma} \ dx^\sigma \ \ \frac{\partial \tilde{x}^\nu}{\partial x^\delta} \ dx^\delta = g_{\alpha,\beta} \ dx^\alpha \ dx^\beta = ds^2$$

One can get the length $L_\gamma$ of a path $\gamma$ by parametrisation of $ds(\lambda)$ (along the path) and integration: $$L_\gamma = \int_\gamma ds(\lambda) = \int_\gamma \sqrt{ g_{\mu,\nu}(\lambda) \ \frac{dx^\mu(\lambda)}{d \lambda} \ \frac{dx^\nu(\lambda)}{d \lambda}} \ d\lambda$$ This length is also invariant under coordinate transformations since $ds$ is already invariant.

One can now assign a distance $D$ of two points by the minimal path-length which is required to connect those points: $$D = \inf \ \{L_\gamma \ | \ \gamma \ \textrm{ connets the two points} \}$$

This is also invariant under coordinate transformations since every $L_\gamma$ is already invariant. I guess this is what you meant by "delta".

All this works for every metric, even for those with off-diagonal elements. So the distance of two points is invariant under rotations.

• This is a solid technical answer. I only question the last sentence (see @ACuriousMind comment above which I hope I am not misreading). The symmetry of $g_{ij} = g_{ji}$ in particular causes issues under a rotation. I feel confident the transformation law is flawless. – sweetser Mar 18 '15 at 22:36
• Rotations are nothing more than a special type of coordinate transformation. we "proofed" it for general coordinate transformations, so this holds for rotations, too (I assume with rotations you mean ordinary space rotations). – image Mar 18 '15 at 23:34
• @sweetser I just reread the comment from ACuriousMind. He states that the metric tensor is not in general invariant under rotations (or general coordinate transformations), which is true. But it suffices, that it is covariant, to allow that the line element ds is invariant. I guess that's the main trouble you have. – image Mar 18 '15 at 23:46
• Yes, that is it. The metric tensor is covariant under a rotation (we know how it changes). The line element ds is invariant under rotation, it is a basic feature of an interval. This answer ignores the quaternion discussion probably viewed as an irrelevant tangent. There is no metric for a quaternion product, no connection, and thus no link to the tools of differential geometry. – sweetser Mar 19 '15 at 11:16

Based on the comments from @ACuriousMind, the question itself is poorly formed. To calculate an interval between two events on a general manifold, one needs both a metric and a connection. Using the process of parallel transport, one moves one of the points next to the other. Then one can subtract the two points. The metric at that point is then used to calculate the interval.

A quaternion product has nothing to do with either metrics nor connections. It is a perhaps curious observation that the first term of the square of a quaternion is identical to a Minkowski metric contraction.

• No, Marcel explained above how to calculate the interval, and it doesn't involve transporting a point to another. Parallel transport is for vectors, you don't a parallel transport a point on a manifold to another point. My guess is you are incorrectly assuming that because you can specify a point with a coordinate tuple that it is like a four-vector. A coordinate tuple is not a four-vector. – BuddyJohn Mar 7 '17 at 10:12
• @BuddyJohn A point in a manifold such as $dx^\mu$ is a place to attach a tangent (or cotangent) space where a tangent four-vector can live. I am not as clear about how to view the point since it does have a raised index which makes it look like it should be considered to be a rank 1 tensor. – sweetser Mar 7 '17 at 20:37
• The labels like $\square^\mu$ for components need to be interpreted in context. For example the labels for coordinate tuples, vectors, Dirac matrices, or some of the labels in the QCD Lagrangian, do not mean these are all similar types of objects or even the same number of components. – BuddyJohn Mar 8 '17 at 3:19
• $dx^\mu$ is not a point on a manifold. In this notation, $x^\mu$ are the coordinates for a point on the manifold, and are not a tensor. And at a point $p$, $dx^\mu(p)$ is a vector in the tangent space at $p$. So $g_{\mu\nu}(p)\ dx^\mu(p)\ dx^\nu(p)$ is using the metric at $p$ to contract two vectors in the tangent space at $p$, and results in a scalar, the invariant square of an infinitesimal line element. This is sometimes just called the line element or an interval, which may have lead to the misunderstanding that it is a finite difference, but it only infinitesimal. – BuddyJohn Mar 8 '17 at 3:22
• Thanks for the more explicit notation. I would also write out the two summations that are normally dropped due to the Einstein summation convention. One thing that bothers me a little is the statement that $x^\mu(p)$ would not not transform like a tensor. I would think the definition of $dx^\mu(p)$ might be something like: $lim_{\delta \rightarrow 0}x^\mu(p) - x^\mu(p + \delta) = dx^\mu(p)$. Of course $x^\mu(p)$ is not a differential line element, but it still would be a point at the start of the path. – sweetser Mar 9 '17 at 20:13