Hot answers tagged action
16
Could someone please convince me that there is something natural about
the choice of the Lagrangian formulation...
If I ask a high school physics student, "I am swinging a ball on a string around my head in a circle. The string is cut. Which way does the ball go?", they will probably tell me that the ball goes straight out - along the direction the ...
12
The notes from week 1 of John Baez's course in Lagrangian mechanics (http://math.ucr.edu/home/baez/classical/#lagrangian) give some insight into the motivations for action principles.
The idea is that least action might be considered an extension of the principle of virtual work. When an object is in equilibrium, it takes zero work to make an arbitrary ...
10
Dear Ondřeji, a good question but a part of the answer is that your equation for the fluid is underdetermined. It treats $p,\rho$ as independent variables. But the physical system only knows how to behave if you also substitute some equation of state, i.e. a function $p=p(\rho)$ or $p=p(\rho,\vec v)$. Note that your Ansatz for the stress-energy tensor ...
9
In physics, it is often implicitly assumed that the Lagrangian $L=L(q^i,v^i,t)$ depends smoothly on the (generalized) positions $q^i$, velocities $v^i$, and time $t$, i.e. that the Lagrangian $L$ is a differentiable function. Let us now assume that the Lagrangian is of the form
$$L~=~\ell(v^2),\qquad\qquad v~:=~|\vec{v}|,\qquad\qquad(1)$$
where $\ell$ is ...
9
There is also Feynman's approach, i.e. least action is true classically just because it is true quantum mechanically, and classical physics is best considered as an approximation to the underlying quantum approach. See http://www.worldscibooks.com/physics/5852.html or http://www.eftaylor.com/pub/call_action.html .
Basically, the whole thing is ...
8
The intuition for the Lagrangian principle comes specific applications of Newton's laws, especially reversible systems with constraints, like nonspherical particles rolling along complicated surfaces. Newton's formulation of Newton's laws was not the end of the story, because there was more structure in the solutions of these types of problems than that ...
7
We are considering a transformation, which may transform the field variables $\phi^{\alpha}(x)$ and which may transform the space-time points $x^{\mu}$. The transformation in turn apply to
The action $S$.
The Euler-Lagrange equations = the equations of motion (EOM).
A solution of EOM.
If any of the items 1-3 are invariant under the transformation, we ...
7
Building an action: If you know the field content (which I assume means you know the gauge group and reps of all the fields) then:
Write down every term that is Lorentz scalar (so combinations like $\partial_\mu A^\mu$, $\bar{\psi}\gamma^\mu \partial_\mu\psi$ allowed but not things like $\vec{n}\cdot\nabla \phi$ where $\vec{n}$ is some random 3-vector). ...
6
As you can see from the image below, you want the variation of the action integral to be a minimum, therefore $\displaystyle \frac{\delta S}{\delta q}$ must be $0$. Otherwise, you are not taking the true path between $q_{t_{1}}$ and $q_{t_{2}}$ but a slightly longer path. However, even following $\delta S=0$, as you know, you might end up with another ...
6
The action functional and Hamilton's principal function are two different mathematical objects related to the same physical quantity.
The action along a trajectory $\gamma:[t_1,t_2]\rightarrow Q$ is given by
$$
S[\gamma] = \int_{t_1}^{t_2}L(\gamma(t'),\dot\gamma(t'),t')dt'
$$
whereas the pricipal function is the solution of the Hamilton-Jacobi equation
$$
...
6
The Hamiltonian H and Lagrangian L which are rather abstract constructions
in classical mechanics get a very simple interpretation in relativistic quantum mechanics. Both are proportional to the number of phase changes per
unit of time. The Hamiltonian runs over the time axis (the vertical axis in the drawing) while the Lagrangian
runs over the trajectory of ...
5
The dimensions of
the Planck constant $\hbar$,
the action $S$, and
the angular momentum,
are constrained by the following important facts:
A conjugated pair of two observables is quantum mechanically related to the Planck constant $\hbar$ via a Heisenberg uncertainty relation.
A conjugated pair of two variables is classically related to the action ...
5
OP wrote:
As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.
Too often, as a student, one is only shown how to derive Newton's 2nd law from Euler-Lagrange equations by postulating some particular Lagrangian $L$. If one believes that Newton's laws are more natural ...
5
Yes, the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. All the equations of motion may be derived from the condition $\delta S = 0$, the action is stationary (which usually means it has the minimum value on the allowed trajectory/history among all ...
5
Short answer: Lagrangian mechanics only applies to a subset of classical mechanics problems, but when it does, it is mathematically equivalent to Newtonian mechanics (which I take to mean direct application of $\vec{F} = m\vec{a}$).
More detailed rambling: Your class and whatever text you're using ought to cover the equivalence of the Newtonian and ...
5
I) Here we will assume that we ultimately want to consider the full quantum theory, usually written in terms of a gauge-fixed path integral
$$Z~=~\int \!{\cal D}\phi~ \exp\left(\frac{i}{\hbar}S_{\rm gf}[\phi]\right) $$
rather than just the classical action and the corresponding classical equations of motion (with or without gauge-fixing terms). If the ...
5
The symmetry is required to leave the full action $A= \int L dt$ invariant. As can be seen the action is invariant because you get the missing $(1+\epsilon)^2$ factor from the measure $dt$. Now, you can apply the Noether theorem to find the conserved charge which is in this case:
$ Q = 2 E t - Px$
with $E = T+V = \frac{1}{2}m\dot{x}^2 + ...
5
1) In this answer we provide more details for David Bar Moshe's correct answer. The action reads
$$\tag{1}S[q; t_i,t_f]~:=~\int_{t_i}^{t_f} \!dt~ L, \quad L~:=~T-V ,\quad T~:=~\frac{m}{2}\dot{q}^2,\quad V~:=~\frac{\alpha}{q^2}.$$
It is not hard to check that the action has an exact symmetry
$$\tag{2} S[q; t_i,t_f] \quad \longrightarrow \quad S[q^{\prime}; ...
4
I generally tell the story that the action principle is another way of getting at the same differential equations -- so at the level of mechanics, the two are equivalent. However, when it comes to quantum field theory, the description in terms of path integrals over the exponentiated action is essential when considering instanton effects. So eventually one ...
4
We vary the action
$$\delta \int {Ldt} = \delta \int {\int {\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)d^3 xdt = 0} } $$
${\Lambda \left( {A^\nu ,\partial _\mu A^\nu } \right)}$ is the density of lagrangian of the system.\
So:
$$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A^\nu }}\delta A_\nu + \frac{{\partial \Lambda ...
4
Dear amc,
first, write your Lagrangian density as
$$ L = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} (\partial_\mu A_\nu) F^{\mu\nu} $$
Is that fine so far? The $F_{\mu\nu}$ contains two terms that make it antisymmetric in the two indices. However, it's multiplied by another $F^{\mu\nu}$ that is already antisymmetric, so I don't need to antisymmetrize ...
4
1) Firstly, the Lagrangian $L(q(t),v(t),t)$ at some time $t$ is a function of:
the instantaneous position $q(t)$ at the time $t$;
the instantaneous velocity $v(t)$ at the time $t$; and
the time $t$ (also known as explicit time-dependence).
2) Secondly, the (off-shell) action
$$\tag{1} S[q]~:=~ \left. \int_{t_i}^{t_f}\! dt \ ...
4
1) Not all equations of motion (eom) are variational. A famous example is the self-dual five-form in type IIB superstring theory. In classical point mechanics, frictional forces typically lead to non-variational problems. In general, there are consistency requirements that a set of eoms should fulfill in order to be of the variational type.
2) As a simple ...
4
I) At least three different quantities in physics are customary called an action and denoted with the letter $S$.
The (off-shell) action
$$\tag{1}S[q]~:=~ \int_{t_i}^{t_f}\! dt \ L(q(t),\dot{q}(t),t) $$
is a functional of the full position curve/path $q^i:[t_i,t_f] \to \mathbb{R}$ for all times $t$ in the interval $[t_i,t_f]$. See also this question. ...
4
I) Initial value problems and boundary value problems are two different classes of questions that we can ask about Nature.
Example: To be concrete:
an initial value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the initial velocity $v_i$ are given,
while a boundary value problem could be to ask ...
4
The term vanished because we can translate this term to one making a statement about the fields at the boundary and assume that the fields themselves vanish in spatial and temporal infinity.
By Stoke's Theorem, we can translate volume integrals into surface integrals. More specifically Gauss' Theorem states that the integral of a divergence of a field over ...
3
Let us for simplicity consider just classical point mechanics (i.e. a $0+1$ dimensional world volume) with only one variable $q(t)$. (The generalization to classical field theory on an $n+1$ dimensional world volume with several fields is straightforward.)
Let us reformulate the title(v1) as follows:
Why can't the Lagrangian $L$ always be written as a ...
3
Answering the third question, in any mature branch of mathematics or physics, there are always equivalent or near-equivalent formulations of the same structure (it's pretty much a definition of maturity, that some of the relations of a new topic to older topics have been worked out). The Lagrangian can be regarded as a type of functional anti-derivative of a ...
3
This is a standard convexity inequality. If a function has the property that its second derivative is always positive, it has the property that if you replace a,b with $a<b$ with the average $(a+b)/2$, then during the replacement f(a) goes up in value by less than the amount that f(b) goes down (just from the curving-up shape of the function), so that
$$ ...
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