Definition of hypersurface orthogonal

In summary, the term "hypersurface orthogonal" is used in the fields of General Relativity and Differential Geometry to describe a vector field that is irrotational and orthogonal to a family of hypersurfaces. In some cases, this definition may also apply to non-timelike congruences. The existence of hypersurface orthogonal vector fields may depend on the topology of the relevant space, as shown in the example of foliations by codimension one surfaces. Further research is needed to determine the topological conditions for the coexistence of hypersurface orthogonal vector fields and foliations by codimension one surfaces, and if they can be orthogonal to each other.
  • #1
fairy._.queen
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Definition of "hypersurface orthogonal"

Hi all!
I'm not sure if the thread belongs more to General Relativity or Differential Geometry, but I guess the border is labile.

I've come across the term "hypersurface orthogonal" many times, but I still haven't found a clear definition. Apparently, if a flow is geodesic and irrotational, it is also hypersurface orthogonal, whatever this means.

Could you please give me a definition and explain how the hypersurface orthogonality is linked to the geodesy and irrotationality of a flow?

Thanks in advance!
 
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  • #2
https://www.physicsforums.com/showpost.php?p=4483918&postcount=8

Note that a hypersurface orthogonal vector field need not have integral curves that are geodesics; it simply needs to be irrotational. An example of a hypersurface orthogonal vector field that doesn't have geodesic integral curves is the 4-velocity field of the family of Rindler observers in flat space-time.
 
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  • #3
Thank you for your reply!
If I got it correct, I would say the definition is: "A (timelike?) vector [itex]\frac{\partial}{\partial\lambda}[/itex] is said to be hypersurface orthogonal if there exist a foliating of space-time into hypersurfaces of constant [itex]\lambda[/itex] such that [itex]\frac{\partial}{\partial\lambda}[/itex] is everywhere orthogonal to such hypersurfaces". Am I right?

Is it possible to apply the term to non-timelike congruences?
For example, I found this sentence in a paper: "It is proved that the vectors of this eigenframe are hypersurface orthogonal and consequently that a coordinate system exist in which the metric [...] is diagonal"
For the metric to be diagonal, this must mean that you can foliate space-time so that, for instance, the y vector is orthogonal to the surfaces of constant y.
 
  • #4
Yes it can be applied to any vector field although and what you said is fine except that the term only applies to vector fields, not to vectors, At any given event in space-time there always exists a hypersurface that is orthogonal to a given vector so there isn't really anything interesting there.
 
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  • #5
Thanks!
 
  • #6
No problem! By the way, I should probably mention that the simplest definition of hypersurface orthogonality of any vector field ##\xi^a## is simply that ##\xi^a = \alpha \nabla^a \beta## for smooth scalar fields ##\alpha,\beta##. It turns out that locally this is equivalent to ##\xi_{[a}\nabla_{b}\xi_{c]} = 0## (this is a result of Frobenius' theorem for integral subbundles) and this in turn is equivalent to ##\xi^a## being irrotational i.e. ##\epsilon^{abcd}\xi_b \nabla_c \xi_d = 0## which can be shown through a simple calculation.
 
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  • #7
It makes sense. Thanks again!
 
  • #8
Anytime :) cheers!
 
  • #9
WannabeNewton said:
Anytime :) cheers!

WBN: do you know if there are any existence theorem on hypersurface orthogonal vector fields?

I.e. if you have a foliation by codimension one hypersurfaces - what must be satisfied for there to exists an everywhere hypersurface-orthogonal vector field?

I recognize that this would be equivalent the existence of foliations by codimension one surfaces, and one-dimensional curves. The existence of the first does depend on the topology of the manifold. Ex: http://www.map.mpim-bonn.mpg.de/Foliations states a theorem that says that the existence of codimension one foliations depend on the relevant topological space having Euler characteristic 0!

Thus, I wonder what the topological conditions on the first and the latter coexisting. And what if they are orthogonal to each other?
 

1. What is a hypersurface orthogonal?

A hypersurface orthogonal is a mathematical term used to describe a surface that is perpendicular to a given set of curves or surfaces. In simpler terms, it is a surface that is at right angles to another surface or set of surfaces.

2. How is hypersurface orthogonal different from a regular surface?

A regular surface may have varying degrees of curvature and does not necessarily have to be at right angles to another surface. A hypersurface orthogonal, on the other hand, is always perpendicular to a given set of surfaces or curves.

3. What are some real-world examples of hypersurface orthogonal?

Some real-world examples of hypersurface orthogonal can be found in engineering and physics, where it is used to describe surfaces that are perpendicular to the flow of a fluid or electromagnetic field. It is also used in geometry to describe the relationship between curves and surfaces.

4. How is hypersurface orthogonal used in mathematics?

In mathematics, hypersurface orthogonal is used to define the concept of orthogonality in higher dimensions. It is also used in differential geometry to study the relationship between curves and surfaces in multi-dimensional spaces.

5. Can hypersurface orthogonal be visualized?

Yes, hypersurface orthogonal can be visualized in three-dimensional space as a surface that is perpendicular to a given set of surfaces or curves. In higher dimensions, it can be visualized as a hypersurface that is at right angles to a higher-dimensional set of surfaces or curves.

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