All Questions
166
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Deeper explanation for Principle of Stationary Action [duplicate]
The Principle of Stationary Action (sometimes called Principle of Least Action and other names) is successfully applied in a wide variety of fields in physics. It can often can be be used to derive ...
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-2
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1
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108
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Why the choice of Configuration Space in Hamilton's Principle is $(q, \dot{q}, t)$? [closed]
In most physics books I've read, such as Goldstein's Classical Mechanics, the explanation of Hamilton's principle took into consideration the equation (1) known as Action:
$$\displaystyle I = \int_{...
- 113
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votes
3
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Is there a proof that a physical system with a *stationary* action principle cannot always be modelled by a *least* action principle?
I'm aware that with Lagrangian mechanics, the path of the system is one that makes the action stationary. I've also read that its possible to find a choice of Lagrangian such that minimization is ...
- 50.1k
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4
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111
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Directly integrating the Lagrangian for a simple harmonic oscillator
I've just started studying Lagrangian mechanics and am wrestling with the concept of "action". In the case of a simple harmonic oscillator where $x(t)$ is the position of the mass, I ...
- 13
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Is there a Lorentz invariant action for a free multi-particle system?
I want to write down a Lorentz-invariant action of free multi-particle systems.
I know that a Lorentz-invariant action for each particle might be expressed as
$$
S[\vec{r}]=\int dt L(\vec{r}(t),\dot{\...
- 145
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1
answer
59
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Landau/Lifshitz action as a function of coordinates [duplicate]
In Landau/Lifshitz' "Mechanics", $\S43$, 3ed, the authors consider the action of a mechanical system as a function of its final time $t$ and its final position $q$. They consider paths ...
- 277
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1
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Doubts about Noether's theorem derivation
Assume you have an action:
$S[q] = \int L(q, \dot q, t)$ (i.e $q$ is a function of time). (1) Then you do a transformation on $q(t)$ such as $\sigma(q(t), a)$ where $a$ is infinetisemal and this ...
-2
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2
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97
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On the physical meaning of functionals and the interpretation of their output numbers
I am studying about functionals, and while looking for some examples of functionals in physics, I have run into this handout .
Here are two questions of mine.
1- This handout starts as follows (the ...
user386360
2
votes
3
answers
148
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Derivation of Hamiltonian by constraining $L(q, v, t)$ with $v = \dot{q}$
I am trying to reconstruct a derivation that I encountered a while ago somewhere on the internet, in order to build some intuition both for $H$ and $L$ in classical mechanics, and for the operation of ...
- 23
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1
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84
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Hamiltonian analysis of relational $N$-Particle Dynamics
I am following "A Shape Dynamics Tutorial, Flavio Mercati" (https://arxiv.org/abs/1409.0105), and have problems understanding the hamiltonian formulation of $N$-particle dynamics as sketched ...
- 513
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2
answers
153
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What is the most general transformation between Lagrangians which give the same equation of motion?
This question is made up from 5 (including the main titular question) very closely related questions, so I didn't bother to ask them as different/followup questions one after another. On trying to ...
- 785
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712
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Confusion in derivation of Euler-Lagrange equations
Here's a screenshot of derivation of Euler-Lagrange from feynman lecture https://www.feynmanlectures.caltech.edu/II_19.html
My doubt is in the last paragraph. I get that $\eta = 0$ at both ends, but ...
- 301
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1
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148
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How did Landau & Lifshitz (Mechanics) get Equation 2.5?
I understood everything in Landau & Lifshitz's mechanics book until Equation 2.4,but I'm not sure what he means when he says "effecting the variation" and gets Equation 2.5.
Can anyone ...
1
vote
1
answer
51
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Lagrange momentum for position change
After the tremendous help from @hft on my previous question, after thinking, new question popped up.
I want to compare how things behave when we do: $\frac{\partial S}{\partial t_2}$ and $\frac{\...
- 525
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2
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Momentum $p = \nabla S$
My book mentions the following equation:
$$p = \nabla S\tag{1.2}$$ where $S$ is the action integral, nabla operator is gradient, $p$ is momentum.
After discussing it with @hft, on here, it turns out ...
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