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Degrees of Freedom total confusion

  1. Mar 10, 2010 #1
    What are degrees of Freedom in the Lagrangian/Hamiltonian formulation of classical mechanics?

    I've been getting very confused trying to understand this concept.

    Would six degrees of freedom mean up, down, left, right, back & forward?

    I've seen it said that 3 degrees of freedom mean the x, y & z axes, so my above sentence is really only 3 degrees of freedom, is it?

    Then, I seen the Susskind classical mechanics lecture claim there are 6 degrees of freedom, x, y & z axes & then position, velocity & acceleration.

    But, then I read in a previous physicsforums post the claim that velocity & acceleration are not dimensions ergo they are not degrees of freedom!

    The wikipedia article is not in the least helpful & I've tried to watch the nptelhrd lectures to only get more confused.

    What does it mean to have a degree of freedom constricted also?

    Thanks it would be really helpful for somebody to shed some light on this idea for me.
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Mar 10, 2010 #2

    ideasrule

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    The degrees of freedom are what variables can be changed independently of each other. If you have a set of equations that involve x,y,and z, those equations have 3 degrees of freedom, since x, y, and z can all change independently of the others. If the equations happen to involve vx, vy, and vz, there are now 6 degrees of freedom.
     
  4. Mar 10, 2010 #3
    So the position of each "particle" in space, i.e. it's ability to move freely in any axis represents the 3 degrees of freedom.

    Then, if this particle has a velocity this allows three more degrees of freedom?

    I can't picture how the constraints work in a system like the one in the picture. Where is the constraint, why are there 5 degrees of freedom in this system? The body is still able to move in all axes...

    edit: the weird "r'ed" arrows are position vectors, as are the "v's" velocity vectors...
     

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  5. Mar 10, 2010 #4

    Ben Niehoff

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    A point particle in R^3 has only 3 degrees of freedom: its (x,y,z) position. The velocities are NOT additional DoF, because they are just time derivatives of position.

    An extended body has 3 additional degrees of freedom because it can rotate around its center of mass, and angular momentum has 3 components.

    In your system, I assume you have point particles, else you would have 6 degrees of freedom as just mentioned: position of center of mass, and rotation about center of mass. But since you have point particles, you have one less DoF (i.e., 5), because there can be no angular momentum around the line connecting the particles.

    Another way to think of this is "How many numbers do I have to specify to completely describe the system at a given point in time?"
     
  6. Mar 10, 2010 #5
    That's amazing! So I'll just clarify my thoughts, degrees of freedom are three degrees of freedom represented as positions in space or a "phase space" w/ respect to the bodies center of mass & if this body is extended it has the capability to rotate about it's center of mass so these rotations enable three further degrees of freedom. These rotations are captured in angular momentum.

    Yeah I had "big" point particles :tongue: I suppose I would be better in thinking of this system as classical for now. So, if I am to think clasically this is a macroscopic body & there are 6 degrees of freedom of this system, the body can rotate about a fixed point.

    I just have one more question, I heard in the lecture linked to above that each body considered seperately has it's own no. of degrees of freedom.
    So, in the picture if I consider the two bodies connected to the red bar as a single body I get 6 DoF.
    If I look at one of the masses as a seperate entity, there are 3 degrees of freedom, as the bar that the mass is connected to prevents any angular movements.

    Correct? Even a lil?:redface:
     
  7. Mar 10, 2010 #6
    Well you would count the position and velocity degrees of freedom as separate degrees of freedom. Just in your case the energy doesn't depend on the position degrees of freedom so you ignore them. You will only have the translational kinetic and the rotational kinetic degrees of freedom that Ben worked out to be 5.

    If this diatomic molecule was trapped in a harmonic oscillator trap or the particles interacted through a spring-like force, then you would have to include the position degrees of freedom.
     
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