Can Curvature Be Integrated Over Space in General Relativity?

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In summary: The Ricci curvature scalar is integrated against the volume form - a straightforward integration of the curvature. What do you suppose this means? What is the angular deficit per unit volume?In summary, the curvature of GR accumulates, and you can integrate it immediately around various points to find the curvature around a larger area.
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Is the curvature of GR accumulative? Can you integrate the curvature immediately around various points to find the curvature around a larger area? Thanks.
 
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No. You can't go around adding the curvature tensors of different parts of the manifold, they don't exist in the same tangent space. Even if you parallel transported them or Lie dragged them or something, I would have no idea what this "addition" would even mean physically. Even if you just looked at the curvature scalar, which is a scalar function that you can actually integrate, I don't see any meaning in integrating it.

If you draw a circle, then the radius of curvature scalar at each point on the circle is R, the radius of curvature. What does it mean to add them together? You can integrate it and get 2πR2 but this means nothing...
 
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friend said:
Is the curvature of GR accumulative? Can you integrate the curvature immediately around various points to find the curvature around a larger area? Thanks.

If you make this question more specific, it will be possible to answer it. Specify what curvature you mean. Write down the formula.
 
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Matterwave said:
No. You can't go around adding the curvature tensors of different parts of the manifold, they don't exist in the same tangent space. Even if you parallel transported them or Lie dragged them or something, I would have no idea what this "addition" would even mean physically.

Actually...the connection (viewed as a ##GL(n)##-valued 1-form) can be integrated along a path. One needs to use path-ordered integration. The result is a linear operator in ##GL(n)## which tells you how to parallel transport vectors along the path from one endpoint to the other. If you integrate around a closed path, then you end up in the same tangent space you started in, so the result is covariant: it is a ##GL(n)## matrix that tells you about parallel transport around the given loop.

The Riemann tensor (viewed as the curvature 2-form ##\Omega^a{}_b = \frac12 R^a{}_{bcd} \,dx^c \wedge dx^d##) cannot be integrated unless you soak up the extra indices somehow (because ##GL(n)## is non-Abelian and there isn't a way to define an ordering on surfaces). In 2 dimensions, there are no extra indices, and you can integrate the curvature directly; the curvature is proportional to the Euler class, and its integral is the Euler number of your manifold. In higher dimensions, you can write down combinations of products of Riemann with itself which are the Pontrjagin classes, and these can be integrated, giving you other topological invariants.

Even if you just looked at the curvature scalar, which is a scalar function that you can actually integrate, I don't see any meaning in integrating it.

The Ricci scalar measures the angular deficit per unit volume, so actually it's a perfectly sensible thing to integrate.
 
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Ben Niehoff said:
Actually...the connection (viewed as a ##GL(n)##-valued 1-form) can be integrated along a path. One needs to use path-ordered integration. The result is a linear operator in ##GL(n)## which tells you how to parallel transport vectors along the path from one endpoint to the other. If you integrate around a closed path, then you end up in the same tangent space you started in, so the result is covariant: it is a ##GL(n)## matrix that tells you about parallel transport around the given loop.

The Riemann tensor (viewed as the curvature 2-form ##\Omega^a{}_b = \frac12 R^a{}_{bcd} \,dx^c \wedge dx^d##) cannot be integrated unless you soak up the extra indices somehow (because ##GL(n)## is non-Abelian and there isn't a way to define an ordering on surfaces). In 2 dimensions, there are no extra indices, and you can integrate the curvature directly; the curvature is proportional to the Euler class, and its integral is the Euler number of your manifold. In higher dimensions, you can write down combinations of products of Riemann with itself which are the Pontrjagin classes, and these can be integrated, giving you other topological invariants.
The Ricci scalar measures the angular deficit per unit volume, so actually it's a perfectly sensible thing to integrate.

Huh, looks like I should read up more on this. Thanks for the corrections! :D
 
  • #6
Ben Niehoff said:
...
The Ricci scalar measures the angular deficit per unit volume, so actually it's a perfectly sensible thing to integrate.
I find myself thinking about the Einstein-Hilbert action whose equations of motion yield the field equations of GR. See: http://en.wikipedia.org/wiki/Einstein–Hilbert_action

There the Ricci curvature scalar is integrated against the volume form - a straightforward integration of the curvature. What do you suppose this means? What is the angular deficit per unit volume?
 

FAQ: Can Curvature Be Integrated Over Space in General Relativity?

What is curvature accumulation?

Curvature accumulation refers to the process by which curvature, or the measure of how much a surface deviates from being flat, increases over time. This can occur in various systems, such as in physical objects or in mathematical models.

How does curvature accumulation occur?

Curvature accumulation can occur through various mechanisms, such as through repeated bending or deformation of a surface, or through the addition of new material that changes the overall shape of the object. In mathematical models, curvature accumulation can be simulated through the accumulation of small changes over time.

Is curvature accumulation always a bad thing?

Not necessarily. In some cases, curvature accumulation can be desirable, as it can lead to the formation of interesting and complex shapes. However, in other cases, excessive curvature accumulation can result in structural instabilities or deformities that can compromise the function or stability of a system.

Can curvature accumulation be reversed?

In some cases, yes. If the curvature accumulation is due to external forces, it may be possible to reverse the process by applying counteracting forces or by removing the source of the curvature. However, if the curvature accumulation is intrinsic to the system, it may be irreversible.

How is curvature accumulation measured?

The measurement of curvature accumulation depends on the specific system being studied. In physical objects, curvature can be measured using tools such as rulers, protractors, or specialized instruments. In mathematical models, curvature is often calculated using equations and algorithms that take into account the shape and properties of the model.

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