Can we say that Galilean Spacetime has the signature $(0,0,0,+)$?

Question

Minkowski Spacetime can be treated as either having the signature $(+,-,-,-)$ or more commonly $(+,+,+,-)$ as in Minkowski Spacetime the spacetime interval between two events is the quantity that is the same for all observers, and with $ct=w$, and $\Delta{s^2}$ giving the spacetime interval between two events $\Delta{s^2}=\Delta{x^2}+\Delta{y^2}+\Delta{z^2}-\Delta{w^2}$. There is also a Euclidean Signature with the signature $(+,+,+,+)$, in which the spacetime interval between any two events is just given by the euclidean distance formula $\Delta{s^2}=\Delta{x^2}+\Delta{y^2}+\Delta{z^2}+\Delta{w^2}$.

In a sense Galilean Spacetime is in between Minkowski and Euclidean Spacetime as in Minkowski Spacetime the spacetime path of a body accelerating at a constant rate is a hyperbole, in Euclidean Spacetime the spacetime path of an object with constant acceleration is a semi circle, and in Galilean Spacetime the path of an object accelerating at a constant rate is a parabola.

In Galilean Spacetime, if two events aren't simultaneous, then the time between the two events events is the same in all reference frames, but the spatial distance between two events is not the same in all reference frames, and the time between two events does not depend on their separation in the x, y, or z dimensions.

So can we label Galilean Spacetime as having the signature $(0,0,0,+)$?

Answer 1

The notion of signature makes sense for any quadratic form, for elements of a vector space. In the case of the metric, $g$, we define $g_{ab}=g(e_a,e_b)$ where $\{e_a\}_a$ forms a basis of the vector space.

In the case of a smooth manifold, the relevant vector space to consider is the tangent space at a point. Then the signature is the number of positive, negative and zero eigenvalues of $g_{ab}$, counting multiplicities.

If you write $(+,-,-,-)$ for example, this means one positive and three negative eigenvalues. In your case, $(0,0,0,+)$ means an eigenvalue of zero with multiplicity three and one positive eigenvalue.

Note that $g$ is defined for elements of the tangent space at a point $p$ on the manifold, so technically your signature can differ point to point. However, this is only a situation to be concerned by if you are dealing with some kind of discontinuity or degeneracy.

Answer 2

See III.A of this arXiv paper (and references stated therein). This goes under the name of "Newton-Cartan-spacetime" or Newton-Cartan theory.

• Absolut time (coordinate-independently) is modeled by a "Clock $1$-form" $\theta$ on the manifold. For a global time coordinate $t$ it can be written as $\theta = dt$.
• Space is modeled as a symmetric, covariant $2$-tensor field which can be thought of as "the tensor product of the basis vectors which span space" and whose kernel is given by $\theta$.

Hence, one can't describe a Galilean spacetime just in terms of a quadratic form (for which the notion of signature would make sense), but one needs two ingredients: A symmetric $2$-tensor marking space (with "signature" $(+1, +1, +1, 0)$ if you want to call it so) and a clock-form with "signature" $(0, 0, 0,+1)$.

Answer 3

We cannot say that Galilean Spacetime has the signature (0,0,0,+) because it cannot have a single metric.

A single metric does not work, because the speed of light is infinite.

See answer https://physics.stackexchange.com/a/368714/97461 from Galilean spacetime interval?.

https://en.wikipedia.org/wiki/Newton%E2%80%93Cartan_theory

one starts with a smooth four-dimensional manifold M and defines two (degenerate) metrics. A temporal metric ${\displaystyle t_{ab}}$ with signature ${\displaystyle (1,0,0,0)}$, used to assign temporal lengths to vectors on M and a spatial metric $h^{ab}$ with signature ${\displaystyle (0,1,1,1)}$. One also requires that these two metrics satisfy a transversality (or "orthogonality") condition, ${\displaystyle h^{ab}t_{bc}=0}$. Thus, one defines a classical spacetime as an ordered quadruple ${\displaystyle (M,t_{ab},h^{ab},\nabla )}$, where ${\displaystyle t_{ab}}$ and $h^{ab}$ are as described, $\nabla$ is a metrics-compatible covariant derivative operator; and the metrics satisfy the orthogonality condition.