# A Change of Signature in Classical Relativity

#### Gordon Jump

I'm a bit confused about the idea of "Change of Signature in Classical Relativity". As I see it, a metric is just a scalar function that I make up. For example, in the x,y plane I can define the functions x^2+y^2 and x^2-y^2 simultaneously. What, then, is the significance of "changing" the signature of spacetime? Why would there be one metric in one location and another metric elsewhere? I can define whatever metric I want, anywhere.

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#### Dale

Mentor
The metric is not just an arbitrary scalar function. It is the one that represents measurements of geometry including distances, durations, angles, and relative velocities.

#### PeterDonis

Mentor
I'm a bit confused about the idea of "Change of Signature in Classical Relativity"
Where have you seen this idea? Do you have a reference to a textbook or peer-reviewed paper?

#### Gordon Jump

The metric is not just an arbitrary scalar function. It is the one that represents measurements of geometry including distances, durations, angles, and relative velocities.
Thank you, but I don't think this is the correct answer. I measure coordinates, and their infinitesimal changes. I am free to plug those coordinates into any scalar function I choose.

#### Dale

Mentor
Thank you, but I don't think this is the correct answer. I measure coordinates, and their infinitesimal changes. I am free to plug those coordinates into any scalar function I choose.
You have this completely backwards. You define coordinates, you do not measure them. You measure distances and angles and so forth.

Consider a meter stick. The meter stick tells you that the distance between the two points at the end of the stick is 1 m. It does not tell you the coordinate of either point. To get the coordinates you must arbitrarily define an origin and a set of axes and a shape for your coordinates and so forth.

#### Ibix

I am free to plug those coordinates into any scalar function I choose.
In addition to Dale's point, yes, you may plug coordinates into any function you like. But if you want to do physics, the output of that function should be something physically interesting. The metric generates consistent notions of angle and distance, whatever coordinates you use. And the metric that does this for spacetime is Lorentzian, not Euclidean.

You can try using a Euclidean metric if you like, but you will find that a Lorentzian metric produces a consistent "distance" between two events, while a Euclidean one does not.

#### Gordon Jump

You have this completely backwards. You define coordinates, you do not measure them. You measure distances and angles and so forth.

Consider a meter stick. The meter stick tells you that the distance between the two points at the end of the stick is 1 m. It does not tell you the coordinate of either point. To get the coordinates you must arbitrarily define an origin and a set of axes and a shape for your coordinates and so forth.
Yes, and after I measure the coordinates of an object, I am free to enter those numbers into any metric function I choose.

#### Gordon Jump

In addition to Dale's point, yes, you may plug coordinates into any function you like. But if you want to do physics, the output of that function should be something physically interesting. The metric generates consistent notions of angle and distance, whatever coordinates you use. And the metric that does this for spacetime is Lorentzian, not Euclidean.

You can try using a Euclidean metric if you like, but you will find that a Lorentzian metric produces a consistent "distance" between two events, while a Euclidean one does not.
You are agreeing with me. We choose the Lorentzian metric because it is physically interesting, but we could equally well choose the Euclidean metric - it just wouldn't be as interesting. My actual question is "What is the significance of a change of signature in general relativity"

#### Gordon Jump

Where have you seen this idea? Do you have a reference to a textbook or peer-reviewed paper?

#### Dale

Mentor
Yes, and after I measure the coordinates of an object, I am free to enter those numbers into any metric function I choose.
Again, you don’t measure coordinates. All measurements must be invariants and coordinates are not invariant.

You can enter the coordinates into any function that you like, but the one unique (for a given spacetime) function that we call the metric is the one that gives you the measurements of geometry in the spacetime. Any other function besides that specific one is not called the metric.

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#### Gordon Jump

OK - I think I have an answer. Tell me what you think. I have two metrics that I can define:
dx^2+dy^2
and
dx^2 + sin^2( x) dy^2.

These are intrinsically different in some way in that I don't think there is a coordinate transformation that can take one to the other.

#### PeterDonis

Mentor
Change of signature in classical relativity - IOPscience
I can't find this paper on arxiv (and the IOPScience link is behind a paywall), but here is another paper from 1994 by George Ellis that talks about signature change:

#### PeterDonis

Mentor
We choose the Lorentzian metric because it is physically interesting, but we could equally well choose the Euclidean metric - it just wouldn't be as interesting.
Not physically interesting, physically correct. A Euclidean metric does not correctly describe the actual physical geometric relationships between events in the spacetime we inhabit.

My actual question is "What is the significance of a change of signature in general relativity"
But answering that question requires understanding what a metric means, physically--it describes the actual physical geometric relationships between events. You can't change physics by changing math; so you can't just arbitrarily pick a metric and expect it to correctly describe actual physical geometric relationships between events. You have to measure those relationships first, and then find a metric that correctly describes them mathematically.

As far as I can gather from the 1994 arxiv paper I linked to, there is not a good consensus answer to what a Euclidean metric signature on a portion of a spacetime would mean physically. But geometrically it seems pretty straightforward: you have a 4-dimensional manifold which is separated into two regions by a spacelike 3-surface. On one side of that spacelike 3-surface, the manifold has a Lorentzian metric (i.e., a metric with a Lorentzian signature correctly desribes the geometry); on the other side of that spacelike 3-surface, the manifold has a Euclidean metric (i.e., a metric with a Euclidean signature correctly describes the geometry).

As long as both regions of the manifold give the same 3-metric (i.e., the same 3-dimensional geometry) for the spacelike 3-surface, the overall model works fine mathematically. But, as above, what it means physically seems to be an open question; it seems like the reason for considering such models is more to investigate what they might possibly mean physically, by constructing them and then trying to deduce testable consequences.

#### Nugatory

Mentor
OK - I think I have an answer. Tell me what you think. I have two metrics that I can define:
dx^2+dy^2
and
dx^2 + sin^2( x) dy^2.

These are intrinsically different in some way in that I don't think there is a coordinate transformation that can take one to the other.
If they are in fact different (it’s often difficult to determine by inspection that no coordinate transformation will take one expression for a given metric tensor into another expression for the same metric tensor) then they represent different metric tensors for different manifolds with different physical properties. That’s the “intrinsically different” that you’re looking for.

Any given manifold only has one metric tensor, although you can use coordinate transforms to write its components in many different ways. Thus, it’s not an arbitrary choice - whatever manifold you’re working with has a particular metric tensor, and that’s the one you must use if you want results that apply to that manifold. You may transform from one coordinate system to another, but the components of the metric tensor in the new coordinate system are completely determined by applying the tensor transformation rule to their values in the old coordinate system.

#### Ibix

You are agreeing with me. We choose the Lorentzian metric because it is physically interesting, but we could equally well choose the Euclidean metric - it just wouldn't be as interesting. My actual question is "What is the significance of a change of signature in general relativity"
The significance is that the numbers you get don't relate to physical measurements. So I don't think that I am agreeing with you. There's a distinction between maths, where you are free to define anything you like as long as it isn't contradictory, and physics where what you define needs to relate to experimental science. GR is physics, not maths.

"Change of Signature in Classical Relativity"

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