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Flan
Posted on 03-16-08 01:07 PM (rev. 2 of 03-16-08 01:09 PM) Link | Quote | ID: 80446


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Last view: 4601 days 		A geodesic can mean either:
(a) a point whence curve Q fulfills |Q'| = c, c is nonzero positive.
(b) more commonly, a curve which has property (a) for all points.
For surfaces, we can also state a geodesic curve as one with acceleration perpendicular to the surface at every point, ie, linearly dependent on the unit normal vector.

So a proof that isometries are geodesic-preserving then.

Let's assume a geodesic curve E in surface M, and an isometry f from M to N, which maps E to fE. Then it follows that at every point on E, the tangent vector is defined and constant. For brevity, I'm defining T = E', fT = fE'. And T should read T(p), or "T at point p", but again, brevity.

|T| = c^2 (The square is just for convenience)
T.T = c
D(T.T) = D(c), the Ds are exterior differentiation
2(T'.T) = 0, ie, E'' . E' = 0

In plain English, the acceleration vector is perpendicular to the tangent vector, which lies in the tangent space of M (at point p, obviously). And if we apply the isometry f, then fT'.fT = 0, fT.fT = d, ie, |fT| = d^2. fE is a geodesic.

QED.



Arbe
Posted on 03-16-08 01:14 PM Link | Quote | ID: 80447

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Pac
Posted on 03-16-08 01:27 PM (rev. 2 of 03-16-08 01:27 PM) Link | Quote | ID: 80449


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Flan
Posted on 03-16-08 09:16 PM Link | Quote | ID: 80456


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Posted by Pac
arbe: win

Thank you for your constructive posts.

Arbe
Posted on 03-16-08 09:17 PM Link | Quote | ID: 80457

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Posted by SX
Posted by Pac
arbe: win

Thank you for your constructive posts.


Thank you for using Board2. We're the best of the best for high quality, completely unrelated responses.

NightKev
Posted on 03-16-08 10:17 PM (rev. 2 of 03-16-08 10:17 PM) Link | Quote | ID: 80459


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Posted by SX
In plain English, the acceleration vector is perpendicular to the tangent vector, which lies in the tangent space of M (at point p, obviously). And if we apply the isometry f, then fT'.fT = 0, fT.fT = d, ie, |fT| = d^2. fE is a geodesic.
"Plain English"?

Also, who cares?

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Arbe
Posted on 03-16-08 10:18 PM Link | Quote | ID: 80460

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Posted by NightKev
Posted by SX
In plain English, the acceleration vector is perpendicular to the tangent vector, which lies in the tangent space of M (at point p, obviously). And if we apply the isometry f, then fT'.fT = 0, fT.fT = d, ie, |fT| = d^2. fE is a geodesic.
"Plain English"?

Also, who cares?


Shut up NightKev we dealt with it.

NightKev
Posted on 03-16-08 10:19 PM Link | Quote | ID: 80461


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Posted by ArBeWhat
Posted by NightKev
Posted by SX
In plain English, the acceleration vector is perpendicular to the tangent vector, which lies in the tangent space of M (at point p, obviously). And if we apply the isometry f, then fT'.fT = 0, fT.fT = d, ie, |fT| = d^2. fE is a geodesic.
"Plain English"?

Also, who cares?


Shut up NightKev we dealt with it.
No you didn't.

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Arbe
Posted on 03-16-08 10:19 PM Link | Quote | ID: 80462

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Posted by NightKev
Posted by ArBeWhat
Posted by NightKev
Posted by SX
In plain English, the acceleration vector is perpendicular to the tangent vector, which lies in the tangent space of M (at point p, obviously). And if we apply the isometry f, then fT'.fT = 0, fT.fT = d, ie, |fT| = d^2. fE is a geodesic.
"Plain English"?

Also, who cares?


Shut up NightKev we dealt with it.
No you didn't.


Excuse me? Did you not even read my awesome reply?

NightKev
Posted on 03-16-08 10:23 PM (rev. 2 of 03-16-08 10:23 PM) Link | Quote | ID: 80463


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Posted by ArBeWhat
Posted by NightKev
Posted by ArBeWhat
Posted by NightKev
Posted by SX
In plain English, the acceleration vector is perpendicular to the tangent vector, which lies in the tangent space of M (at point p, obviously). And if we apply the isometry f, then fT'.fT = 0, fT.fT = d, ie, |fT| = d^2. fE is a geodesic.
"Plain English"?

Also, who cares?


Shut up NightKev we dealt with it.
No you didn't.


Excuse me? Did you not even read my awesome reply?
How do you "read" an image?

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Ailure
Posted on 03-16-08 10:53 PM Link | Quote | ID: 80464

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ArBeWhat, not every forum is spatula.

And , I think I just close this.

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