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What does the term Action means in physics?

  1. Dec 19, 2007 #1
    What does the term "Action" means in physics?

    What does the term "Action" means in physics?
     
  2. jcsd
  3. Dec 19, 2007 #2

    Hootenanny

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    I don't know, but someone usually says it after "lights, camera..." :rolleyes:

    The specific meaning of an 'action' in physics depends on what field one is working in. However, roughly speaking an 'action' is some quantity in a particular [physical] system that can be used to describe how that system behaves.
     
  4. Dec 19, 2007 #3

    f95toli

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    In mechanics (classical and quantum) the usual meaning is the integral of the Lagrangian between two points. It is usually denoted S. There are a few other versions as well, but they all involve the integral of a "path" (in generalized coordinates) between two points.
     
  5. Dec 19, 2007 #4

    robphy

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  6. Dec 19, 2007 #5
    What is meant by varying an action ?

    I've heard about somthing like " varying Einstien-Hilbert Action to get the feild equations".
    What does that mean?
     
  7. Dec 25, 2007 #6

    Mute

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    It means you take a variational derivative of the action. (A variational derivative is essentially the derivative of a functional with respect to a function. A functional is like a function, except that the input is a function and the output is another function). So, if you have an action that is a function of particle trajectory q(t), you can vary the action with respect to q(t): [itex]\delta \mathcal{S}[q(t)]/\delta q[/itex] is the standard notation for a variational derivative.

    The "Principle of Least Action" states that variations in the action should be zero (the name is really a misnomer, as it implies the action is a minimum when this condition is satisfied, but the action could be a maximum or a saddle).

    In Lagrangian Mechanics, the action is defined

    [tex]\mathcal{S} = \int_{t_1}^{t_2}dt L(q,\dot{q},t)[/tex]

    where q(t) is the trajectory of the system. Varying the action with respect to the particle trajectory q(t), after some integration by parts you arrive at

    [tex]\delta \mathcal{S} = \delta\left(\int_{t_1}^{t_2}dt L(q,\dot{q},t)\right) = \int_{t_1}^{t_2}dt\left(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} \right) \delta q[/tex]

    where L is called the Lagrangian, and is assumed to be a function of t (either explicity or implicitly through the q(t)), q(t) and [itex]\dot{q} = dq/dt[/itex]. We want this to be zero, and since [itex]\delta q[/itex] is an arbitrary variation of q, it follows that what's in the brackets must be zero. This relation in the brackets is called the Euler-Lagrange Equations. (Plural because q(t) can really be a vector, and you have an Euler-Lagrange equation for each [itex]q_i[/itex]).

    It turns out that the Euler-Lagrange (EL) equations reproduce Newton's Equations when [itex]L = T - V[/itex], the kinetic energy minus the potential energy (it gets somewhat more complicated for things like General Relativity, which is the example you gave). So, if you can write down the Lagrangian for your system, then given the action defined as above, by varying it you would arrive at the EL equations, and plugging your Lagrangian into that you would derive the equations of motion for your system.

    (As a last note, the EL equations above assume the Lagrangian does not depend on any time derivatives of q higher than [itex]\dot{q}[/itex]. They become modified if it does).
     
    Last edited: Dec 25, 2007
  8. Dec 26, 2007 #7

    rbj

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    i was surprized to see the word "force" so carefully omitted in the explanations. not to equate the two, but they're related.
     
  9. Dec 26, 2007 #8
    Actually, when I studied Hamilitonian Mechanics, it was presented as a way of avoiding explicit enumeration of forces, which gets pretty complicated for complex systems. The point is that by expressing the fundamental principles in terms of the action, you don't have to mentions forces at all if you don't want to.
     
  10. Dec 26, 2007 #9
    actually thats not entirely true for systems with constraints, the lagrange multipliers actually become the forces of constraint, and partial L withrespect to q are the forces related to the potential energy.
     
  11. Dec 26, 2007 #10
    Yes, that's why I said "if you don't want to" - I was sweeping constraint forces under the rug, since I don't think mentioning them really clarifies the topic of this thread, i.e. the use of the action in Physics.
     
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