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Virtual displacement

  1. Oct 3, 2011 #1
    I have been reading Classical Mechanics by goldstein for over 2 months. I have encountered the concept of Virtual Disolacement in that book. Some of my friends have told me that it means a very small desplacement that you cannot see..
    However, I am not sure if it is right. Then what is the difference between virtual and ordinary displacement? Can anyone please tell me what virtual displacement really means??
  2. jcsd
  3. Oct 3, 2011 #2


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    I understand you. I also read Goldstein now and then, and I asked myself the same question. But I think it gets a little more clear in Chapter 2, about the Hamiltonian formulation.

    Imagine that there is an extra time dimension, call it s, that is independent of the ordinary time t, and imagine that the system can move in s-time as well as in t-time. The constraint equations (holonomic) may contain t explicitly, but not s.

    So, when the system moves in s-time, for a fixed t, the contraints will only depend explicitly upon the space coordinates. We obtain the actual motion in time t for the system if we keep s=0.

    A virtual dispacement [itex]\delta q[/itex] of some generarlized space coordinate q can then be interpreted as a differential displacement of q in s-time: [itex]\delta q = \frac{\partial q}{\partial s}ds[/itex], and actually, in every formula where [itex]\delta q[/itex] occurs, we can use [itex]\frac{\partial q}{\partial s}[/itex], evaluated at s=0, instead. The formulas we obtain in this way shall be valid for all possible motions of the system in s-time, and therfore, for all possible values of the [itex]\frac{\partial q}{\partial s}[/itex], for independent q:s. This will lead to Lagrange's equations just as in Chapter 1 in Goldstein.
    Last edited: Oct 3, 2011
  4. Oct 3, 2011 #3
    I have a few questions regarding your explanations, Erland.
    How can you bring 2 time dimensions independent of each other? What does it mean? I mean time is only 1 dimension..
    So youre saying that virtual displacement is the displacement in s-time, which is different from the normal t- time.. But thats still bothering me..Because I cant properly understand what s-time is.
    Can you tell me what the differnce is between t time and s time??
  5. Oct 4, 2011 #4


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    Well, the s-time doesn't exist in the real world, it's just imagination, that why it's called virtual displacement. And it is not necessary to consider it as a time-coordinate (that's just how I personally think of it), but as any parameter which the generalized coordinates is imagined to depend upon, so that instead of q(t) we have q(t,s). In Chapter 2 in Goldstein, this parameter is called [itex]\alpha[/itex]. We consider the system as moving wrt both ordinary time and s-time, and consider displacements wrt s (it is probably because when I think of motion, I think of it wrt time, that I consider s as a kind of time coordinate). But it doesn't exist in reality, it is just a way to generate possible paths, one path for each fixed value of s, and the actual path will be that one for which the derivative wrt s of the integral of the lagrangian vanishes. Look up Chapters 1 and 2 in Goldstein.
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