Timeline for Why do I need the Beltrami identity to solve the brachistochrone problem?
Current License: CC BY-SA 4.0
18 events
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| Jul 5, 2023 at 6:31 | history | edited | Eli | CC BY-SA 4.0 |
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| Jul 4, 2023 at 21:12 | history | edited | Buzz♦ | CC BY-SA 4.0 |
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| Jul 4, 2023 at 20:12 | answer | added | Eli | timeline score: 1 | |
| Jul 3, 2023 at 20:09 | history | edited | Eli | CC BY-SA 4.0 |
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| Jul 3, 2023 at 16:18 | answer | added | Michael Seifert | timeline score: 3 | |
| Jul 3, 2023 at 15:03 | comment | added | Michael Seifert | @mikestone If I'm not mistaken the first integral comes from the fact that $\partial f/\partial x = 0$, implying that $d/dt(\partial f/\partial \dot{x}) = 0$. (b) This sort of "identical" variational principle can still be of use; for example, one can derive the geodesic equation on a manifold by defining an arbitrary curve $\vec{q}(\lambda)$ and then varying the arc length $S = \int ds = \int \sqrt{ g_{ij} \dot{q}^i(\lambda) \dot{q}^j(\lambda)} d \lambda$. | |
| Jul 3, 2023 at 14:46 | history | edited | Eli | CC BY-SA 4.0 |
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| Jul 3, 2023 at 8:47 | comment | added | Frobenius | @Eli : Precisely. It's not wrong and identical to my notes equation (1.13). Note that for your constant we have $\;k=1/D=1/(2R)\;$ where $\;R=D/2\;$ the radius of the circle. | |
| Jul 2, 2023 at 21:08 | comment | added | Frobenius | Related : Another Solution To Brachistochrone Problem. My notes on brachistochrone | |
| Jul 2, 2023 at 18:55 | history | edited | Eli | CC BY-SA 4.0 |
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| Jul 2, 2023 at 18:40 | history | edited | Eli | CC BY-SA 4.0 |
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| Jul 2, 2023 at 18:28 | history | edited | Eli | CC BY-SA 4.0 |
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| Jul 2, 2023 at 18:18 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Jul 2, 2023 at 18:04 | history | edited | Eli | CC BY-SA 4.0 |
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| Jul 2, 2023 at 17:56 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Jul 2, 2023 at 17:40 | comment | converted from answer | mike stone | How do you get your "first integraL". As you have two variables $x(t)$, $y(t)$ the first integral should be $$ f- \dot x \frac{\partial f}{\partial \dot x} -\dot y \frac{\partial f}{\partial \dot y}=c $$ Also you have a constraint $v= \sqrt{\dot x^2+\dot y^2}$ which makes your functional $$ T= \int \frac v v dt= \int dt, $$ which, being identically satisfied, does not seem much use. | |
| Jul 2, 2023 at 17:37 | comment | added | naturallyInconsistent | It seems to be the same answer, just expressed very differently and very much more difficult to get the answer from the method avoiding Beltrami identity. Why should textbook authors waste paper talking about solution methods that are unnecessarily difficult? | |
| Jul 2, 2023 at 17:30 | history | asked | Eli | CC BY-SA 4.0 |