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What is the mean curvature vector?

1

0

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What is the mean curvature vector? What is the simplest example of this, maybe in terms of a 2-surface in 3d?

differential-geometry

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asked Mar 22 '17 at 2:30

playplay

343

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As far as I know, the mean curvature is a scalar. It is the trace of the first fundamental form. They call it the mean curvature because that trace is the sum of the principal curvatures, and one can always rescale the metric to change those principal curvatures by a constant factor, and so can arrange to divide out by $2$. In higher dimensions, trace is replaced with tensor contraction, and then you have Ricci curvature and scalar curvature, which play somewhat similar (but also rather different) roles.
  $endgroup$
  – Alfred Yerger
  Mar 22 '17 at 3:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The mean curvature vector is the mean curvature (a scalar) times the normal to the surface (a unit vector), with the sign convention that for a sphere it points inwards.
  $endgroup$
  – Rahul
  Mar 22 '17 at 3:16
  
  

  
  
  
  

add a comment | 

1

0

$begingroup$

What is the mean curvature vector? What is the simplest example of this, maybe in terms of a 2-surface in 3d?

differential-geometry

share|cite|improve this question

asked Mar 22 '17 at 2:30

playplay

343

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As far as I know, the mean curvature is a scalar. It is the trace of the first fundamental form. They call it the mean curvature because that trace is the sum of the principal curvatures, and one can always rescale the metric to change those principal curvatures by a constant factor, and so can arrange to divide out by $2$. In higher dimensions, trace is replaced with tensor contraction, and then you have Ricci curvature and scalar curvature, which play somewhat similar (but also rather different) roles.
  $endgroup$
  – Alfred Yerger
  Mar 22 '17 at 3:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The mean curvature vector is the mean curvature (a scalar) times the normal to the surface (a unit vector), with the sign convention that for a sphere it points inwards.
  $endgroup$
  – Rahul
  Mar 22 '17 at 3:16
  
  

  
  
  
  

add a comment | 

$begingroup$

What is the mean curvature vector? What is the simplest example of this, maybe in terms of a 2-surface in 3d?

differential-geometry

share|cite|improve this question

asked Mar 22 '17 at 2:30

playplay

343

$endgroup$

share|cite|improve this question

asked Mar 22 '17 at 2:30

playplay

343

share|cite|improve this question

asked Mar 22 '17 at 2:30

playplay

343

asked Mar 22 '17 at 2:30

playplay

343

asked Mar 22 '17 at 2:30

playplay

343

1 Answer
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$begingroup$

To get to the mean curvature vector, start with defining surface curvature. We know how to define curvature of curve, but how do you define the curvature of a surface? Simple. Start with a point $p$ on your surface and then take intersections with planes in each direction of $p$.

These coordinate planes give rise to curves which are the result of taking the intersections. Now you can use your theory of curvature for curves! In this case you can chart the nature of a surface by looking at the sign of the curvature locally.

The principal curvatures $k_1,k_s$ are the minimum and maximum values you seen as a result of the computing the curvature for the curves which came from intersecting coordinate planes with $Sigma$. The mean curvature $H = (k_1+k_2)/2$ and the gaussian curvature $G = k_1k_2$.

But what do these values tell you? Let me just take the simple case when $G = 0$. Then that means in some direction your surface has no curvature. For example, take $Sigma = mathbbR^2$. This has no curvature in any direction i.e $G = H = 0$.
I have more stuff here in my notes. (curvature notes)

share|cite|improve this answer

edited Mar 22 '17 at 3:18

answered Mar 22 '17 at 3:05

Faraad ArmwoodFaraad Armwood

7,5812719

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0

$begingroup$

To get to the mean curvature vector, start with defining surface curvature. We know how to define curvature of curve, but how do you define the curvature of a surface? Simple. Start with a point $p$ on your surface and then take intersections with planes in each direction of $p$.

These coordinate planes give rise to curves which are the result of taking the intersections. Now you can use your theory of curvature for curves! In this case you can chart the nature of a surface by looking at the sign of the curvature locally.

The principal curvatures $k_1,k_s$ are the minimum and maximum values you seen as a result of the computing the curvature for the curves which came from intersecting coordinate planes with $Sigma$. The mean curvature $H = (k_1+k_2)/2$ and the gaussian curvature $G = k_1k_2$.

But what do these values tell you? Let me just take the simple case when $G = 0$. Then that means in some direction your surface has no curvature. For example, take $Sigma = mathbbR^2$. This has no curvature in any direction i.e $G = H = 0$.
I have more stuff here in my notes. (curvature notes)

share|cite|improve this answer

edited Mar 22 '17 at 3:18

answered Mar 22 '17 at 3:05

Faraad ArmwoodFaraad Armwood

7,5812719

$endgroup$

add a comment | 

0

$begingroup$

To get to the mean curvature vector, start with defining surface curvature. We know how to define curvature of curve, but how do you define the curvature of a surface? Simple. Start with a point $p$ on your surface and then take intersections with planes in each direction of $p$.

These coordinate planes give rise to curves which are the result of taking the intersections. Now you can use your theory of curvature for curves! In this case you can chart the nature of a surface by looking at the sign of the curvature locally.

The principal curvatures $k_1,k_s$ are the minimum and maximum values you seen as a result of the computing the curvature for the curves which came from intersecting coordinate planes with $Sigma$. The mean curvature $H = (k_1+k_2)/2$ and the gaussian curvature $G = k_1k_2$.

But what do these values tell you? Let me just take the simple case when $G = 0$. Then that means in some direction your surface has no curvature. For example, take $Sigma = mathbbR^2$. This has no curvature in any direction i.e $G = H = 0$.
I have more stuff here in my notes. (curvature notes)

share|cite|improve this answer

edited Mar 22 '17 at 3:18

answered Mar 22 '17 at 3:05

Faraad ArmwoodFaraad Armwood

7,5812719

$endgroup$

add a comment | 

$begingroup$

To get to the mean curvature vector, start with defining surface curvature. We know how to define curvature of curve, but how do you define the curvature of a surface? Simple. Start with a point $p$ on your surface and then take intersections with planes in each direction of $p$.

These coordinate planes give rise to curves which are the result of taking the intersections. Now you can use your theory of curvature for curves! In this case you can chart the nature of a surface by looking at the sign of the curvature locally.

The principal curvatures $k_1,k_s$ are the minimum and maximum values you seen as a result of the computing the curvature for the curves which came from intersecting coordinate planes with $Sigma$. The mean curvature $H = (k_1+k_2)/2$ and the gaussian curvature $G = k_1k_2$.

But what do these values tell you? Let me just take the simple case when $G = 0$. Then that means in some direction your surface has no curvature. For example, take $Sigma = mathbbR^2$. This has no curvature in any direction i.e $G = H = 0$.
I have more stuff here in my notes. (curvature notes)

share|cite|improve this answer

edited Mar 22 '17 at 3:18

answered Mar 22 '17 at 3:05

Faraad ArmwoodFaraad Armwood

7,5812719

$endgroup$

share|cite|improve this answer

edited Mar 22 '17 at 3:18

answered Mar 22 '17 at 3:05

Faraad ArmwoodFaraad Armwood

7,5812719

answered Mar 22 '17 at 3:05

Faraad ArmwoodFaraad Armwood

7,5812719

answered Mar 22 '17 at 3:05

Faraad ArmwoodFaraad Armwood

7,5812719