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The first variation formula

0

$begingroup$

I took this from Da Carmo (page 344), not that is matters. I am going to use his curves and surface formula just to ensure the answer doesn't require any geometry answerers as I really feel this is analysis. If you would like he has the same proof in his Riemannian geometry book on page 196.

A regular parametrized curve by arc length (so it is a geodesic?) $alpha:[0,ell] to S$ is a geodesic $iff$ for every proper variation $h: [0,ell] times (-epsilon, epsilon) to S$ of $alpha$, $L'(0) = 0$

Here $L'(0) = -int_0^ell left < A(s),V(s) right > ds$ with $V(s) = partial h/partial t (s,0)$ and $A(s) = (D/partial s) partial h/partial s (s,0).$

Note that $L'(0)$ is the derivative of the energy function.

Okay my question lies in $leftarrow$. Basically he says suppose $L'(0) = 0$, then we consider the variational field $V(s) = g(s)A(s)$. Why? I thought $V$ is supposed to be generic, why would we consider only forms of $V$ like that?

real-analysis differential-geometry riemannian-geometry calculus-of-variations

share|cite|improve this question

edited Mar 27 at 3:23

Hawk

asked Mar 27 at 3:15

HawkHawk

5,5851140110

$endgroup$

add a comment | 

0

$begingroup$

I took this from Da Carmo (page 344), not that is matters. I am going to use his curves and surface formula just to ensure the answer doesn't require any geometry answerers as I really feel this is analysis. If you would like he has the same proof in his Riemannian geometry book on page 196.

A regular parametrized curve by arc length (so it is a geodesic?) $alpha:[0,ell] to S$ is a geodesic $iff$ for every proper variation $h: [0,ell] times (-epsilon, epsilon) to S$ of $alpha$, $L'(0) = 0$

Here $L'(0) = -int_0^ell left < A(s),V(s) right > ds$ with $V(s) = partial h/partial t (s,0)$ and $A(s) = (D/partial s) partial h/partial s (s,0).$

Note that $L'(0)$ is the derivative of the energy function.

Okay my question lies in $leftarrow$. Basically he says suppose $L'(0) = 0$, then we consider the variational field $V(s) = g(s)A(s)$. Why? I thought $V$ is supposed to be generic, why would we consider only forms of $V$ like that?

real-analysis differential-geometry riemannian-geometry calculus-of-variations

share|cite|improve this question

edited Mar 27 at 3:23

Hawk

asked Mar 27 at 3:15

HawkHawk

5,5851140110

$endgroup$

add a comment | 

$begingroup$

I took this from Da Carmo (page 344), not that is matters. I am going to use his curves and surface formula just to ensure the answer doesn't require any geometry answerers as I really feel this is analysis. If you would like he has the same proof in his Riemannian geometry book on page 196.

A regular parametrized curve by arc length (so it is a geodesic?) $alpha:[0,ell] to S$ is a geodesic $iff$ for every proper variation $h: [0,ell] times (-epsilon, epsilon) to S$ of $alpha$, $L'(0) = 0$

Here $L'(0) = -int_0^ell left < A(s),V(s) right > ds$ with $V(s) = partial h/partial t (s,0)$ and $A(s) = (D/partial s) partial h/partial s (s,0).$

Note that $L'(0)$ is the derivative of the energy function.

Okay my question lies in $leftarrow$. Basically he says suppose $L'(0) = 0$, then we consider the variational field $V(s) = g(s)A(s)$. Why? I thought $V$ is supposed to be generic, why would we consider only forms of $V$ like that?

real-analysis differential-geometry riemannian-geometry calculus-of-variations

share|cite|improve this question

edited Mar 27 at 3:23

Hawk

asked Mar 27 at 3:15

HawkHawk

5,5851140110

$endgroup$

share|cite|improve this question

edited Mar 27 at 3:23

Hawk

asked Mar 27 at 3:15

HawkHawk

5,5851140110

share|cite|improve this question

edited Mar 27 at 3:23

Hawk

asked Mar 27 at 3:15

HawkHawk

5,5851140110

asked Mar 27 at 3:15

HawkHawk

5,5851140110

asked Mar 27 at 3:15

HawkHawk

5,5851140110

1 Answer
1

active

oldest

votes

0

$begingroup$

You want to prove that if for any proper variation $L'=0$ then the curve is a geodesic. Then, you can choose any particular variation and, if that suffices to prove that the curve is a geodesic, you are done.

A similar example: a sequence converges to $L$ iff all of its subsequences converge to $L$. Proof from right to left: Take the sequence itself.

share|cite|improve this answer

answered Mar 27 at 7:47

GReyesGReyes

2,57315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you choose a particular variation, then you haven't shown for every variation. You only show the particular one works. The analogy you gave isn't the same because the statement requires extracting the subsequence from the original one.
  $endgroup$
  – Hawk
  Mar 27 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You do not have to show that $L'=0$ for every variation. You assume that. Then, you can use this information in any way you want to conclude that it is a geodesic. The way this information is used in this case is by using a particular variation. You know that $L'=0$ for that particular variation because you assume this to be the case for any variation.
  $endgroup$
  – GReyes
  Mar 27 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $L'=0$ for any variation $Longrightarrow$ $L'=0$ for some particular variation $Longrightarrow$ the curve is a geodesic.
  $endgroup$
  – GReyes
  Mar 27 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand still, the question asks us to prove $L' = 0$ for every variation of $alpha$. Not every variation vector field takes the form $f(s)A(s)$
  $endgroup$
  – Hawk
  Mar 28 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is not asking that. You should prove that if for every variation you have $L'=0$ then the curve is a geodesic. You assume that $L'(0)=0$, you do not have to prove that. You have to use that knowledge to conclude that the curve is a geodesic.
  $endgroup$
  – GReyes
  Mar 28 at 4:32
  

  
  
  
  

 | 
show 7 more comments

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0

$begingroup$

You want to prove that if for any proper variation $L'=0$ then the curve is a geodesic. Then, you can choose any particular variation and, if that suffices to prove that the curve is a geodesic, you are done.

A similar example: a sequence converges to $L$ iff all of its subsequences converge to $L$. Proof from right to left: Take the sequence itself.

share|cite|improve this answer

answered Mar 27 at 7:47

GReyesGReyes

2,57315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you choose a particular variation, then you haven't shown for every variation. You only show the particular one works. The analogy you gave isn't the same because the statement requires extracting the subsequence from the original one.
  $endgroup$
  – Hawk
  Mar 27 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You do not have to show that $L'=0$ for every variation. You assume that. Then, you can use this information in any way you want to conclude that it is a geodesic. The way this information is used in this case is by using a particular variation. You know that $L'=0$ for that particular variation because you assume this to be the case for any variation.
  $endgroup$
  – GReyes
  Mar 27 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $L'=0$ for any variation $Longrightarrow$ $L'=0$ for some particular variation $Longrightarrow$ the curve is a geodesic.
  $endgroup$
  – GReyes
  Mar 27 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand still, the question asks us to prove $L' = 0$ for every variation of $alpha$. Not every variation vector field takes the form $f(s)A(s)$
  $endgroup$
  – Hawk
  Mar 28 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is not asking that. You should prove that if for every variation you have $L'=0$ then the curve is a geodesic. You assume that $L'(0)=0$, you do not have to prove that. You have to use that knowledge to conclude that the curve is a geodesic.
  $endgroup$
  – GReyes
  Mar 28 at 4:32
  

  
  
  
  

 | 
show 7 more comments

0

$begingroup$

You want to prove that if for any proper variation $L'=0$ then the curve is a geodesic. Then, you can choose any particular variation and, if that suffices to prove that the curve is a geodesic, you are done.

A similar example: a sequence converges to $L$ iff all of its subsequences converge to $L$. Proof from right to left: Take the sequence itself.

share|cite|improve this answer

answered Mar 27 at 7:47

GReyesGReyes

2,57315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you choose a particular variation, then you haven't shown for every variation. You only show the particular one works. The analogy you gave isn't the same because the statement requires extracting the subsequence from the original one.
  $endgroup$
  – Hawk
  Mar 27 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You do not have to show that $L'=0$ for every variation. You assume that. Then, you can use this information in any way you want to conclude that it is a geodesic. The way this information is used in this case is by using a particular variation. You know that $L'=0$ for that particular variation because you assume this to be the case for any variation.
  $endgroup$
  – GReyes
  Mar 27 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $L'=0$ for any variation $Longrightarrow$ $L'=0$ for some particular variation $Longrightarrow$ the curve is a geodesic.
  $endgroup$
  – GReyes
  Mar 27 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand still, the question asks us to prove $L' = 0$ for every variation of $alpha$. Not every variation vector field takes the form $f(s)A(s)$
  $endgroup$
  – Hawk
  Mar 28 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is not asking that. You should prove that if for every variation you have $L'=0$ then the curve is a geodesic. You assume that $L'(0)=0$, you do not have to prove that. You have to use that knowledge to conclude that the curve is a geodesic.
  $endgroup$
  – GReyes
  Mar 28 at 4:32
  

  
  
  
  

 | 
show 7 more comments

$begingroup$

You want to prove that if for any proper variation $L'=0$ then the curve is a geodesic. Then, you can choose any particular variation and, if that suffices to prove that the curve is a geodesic, you are done.

A similar example: a sequence converges to $L$ iff all of its subsequences converge to $L$. Proof from right to left: Take the sequence itself.

share|cite|improve this answer

answered Mar 27 at 7:47

GReyesGReyes

2,57315

$endgroup$

share|cite|improve this answer

answered Mar 27 at 7:47

GReyesGReyes

2,57315

answered Mar 27 at 7:47

GReyesGReyes

2,57315

answered Mar 27 at 7:47

GReyesGReyes

2,57315