Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

D'alembert's Principle

  1. Sep 7, 2013 #1
    /I'm having some doubt with D'alembert's Principle. The principle states that [itex] \sum_{i}(\vec {F}_i - \dot{\vec{p}}_i)\delta\vec{r}_i=0[/itex] but does that mean that each term of the summation must vanish too, or just the sum does? I know that mathematically there's no need that each term shall vanish, but does physical considerations requires them to vanish separately?
  2. jcsd
  3. Sep 7, 2013 #2


    User Avatar
    Science Advisor
    2016 Award

    The point of d'Alembert's principle is to use it for constraint systems. If the constraints are all holonomic, you can introduce independent coordinates [itex]q_k[/itex] writing [itex]\vec{r}_i=\vec{r}_i(q_1,\ldots,q_f)[/itex], where [itex]f[/itex] is the number of independent degrees of freedom. Then you can vary the [itex]q_k[/itex] independently from each other, and this gives you equations of motion in terms of these variables.

    Another method is to keep the Cartesian coordinates [itex]\vec{r}_i[/itex] and implement the constraints with help of Lagrange multipliers. Then you can vary the [itex]\delta \vec{r}_i[/itex] independently. This leads to equations of motion, where additional forces from the constraints are taken into account.
  4. Sep 7, 2013 #3


    User Avatar
    Science Advisor
    Gold Member

    Let's take this in two steps:

    The Principle of Virtual Work tells us that constraints do no work.

    D'Alembert's Principle permits the impressed forces of dynamical systems to be handled in the same way as constraints - the application of this is to find the unknown trajectories caused by the impressed forces via the calculus of variations - thanks to Euler and Lagrange.

    For a system in equilibrium the independence of the coordinates tells us that each of the terms must be zero; for dynamics it is only the sum which must be zero.
  5. Sep 7, 2013 #4
    But why in dynamic situations only the sum must be zero, and not each member of the sum?
  6. Sep 8, 2013 #5


    User Avatar
    Science Advisor
    Gold Member

    In the case of equilibrium each variation has only a single coefficient: the value for the force of that constraint. In linear algebra it is shown that if the variables (coordinates) are linearly independent then the sum of a set of coefficients times the independent variables can only sum to zero when all of the coefficients are identically zero.

    In the dynamical case you have a difference which must be zero ... but now the expression depends upon the trajectories via Newton's second law of motion: so yes, each term must go to zero, and this is the variational principle which is used to find those trajectories.

    Sorry if my earlier statement was misleading and unclear.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook