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Simple equations in Dirac Delta function terms

  1. Feb 2, 2013 #1
    Hi there,

    I'm trying to comprehend Dirac Delta functions. Here's something to help me understand them; let's say I want to formulate Newton's second law F=MA (for point masses) in DDF form. Is this correct:

    [tex] F_i = \int [m_i\delta (x-x_i) a_i\delta (x-x_i)]dx[/tex]
    Or is it this:
    [tex] F_i = [\int m_i\delta (x-x_i)dx] [\int a_i\delta (x-x_i)dx][/tex]
    Or is the idea of a Dirac delta function for the instantaneous acceleration of a point mass not well defined? ...or?

    I intially thought the first equation was correct, but then I worried that infinities were being multiplied, and so, I figured that each delta function would need to be integrated first, before the multiplication takes place, hence the second equation.

    Any thoughts would be most welcome!
     
  2. jcsd
  3. Feb 2, 2013 #2

    chiro

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    Hey James MC.

    One suggestion I would have is to think whether the mass appears out of nowhere, whether the acceleration appears out of no-where, or whether both appear out of nowhere.

    This will help you reconcile the formulation mathematically with that of the physical kind.

    I am going to assume that the mass already exists and the acceleration changes as a result of some impulse.
     
  4. Feb 2, 2013 #3
    Hi Chiro,

    That seems like a reasonable assuption. But how does that help? Presumably we just represent the additional entity whose force is inducing the acceleration with an extra subscript, as we would in the simple formalism. So in DDF formalism:
    [tex] F_{ij} = \int [m_i\delta (x-x_i) a_{ij}\delta (x-x_i)]dx[/tex]
    Where, Fij is the force on i due to j and aij is the acceleration of i due to j.

    My question is: is this equation correct, or is it mathematically defective, say, because it multiplies distributions, when distributions ought not be multiplied?
     
  5. Feb 3, 2013 #4

    chiro

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    Well if the mass already exists then the delta function you are modulating with shouldn't appear in the integrand.

    However if the acceleration due to impulse exists as an impulse, then the dirac delta for this should occur at the given point.

    Also the other thing is how the force is distributed: is it done like a very simple rigid body dynamics problem (i.e. a chunk of mass, simple moment of inertia or an even simpler point particle) or is it done as something more complex (different forces applied over the whole body)?

    I'm guessing you have a really simple object that is moving on one axis and you are "hitting" it with an impulse driving it momentarily accelerate in that direction.

    If this is the case, then the first thing is to resolve the mass distribution and force distribution (acceleration) on that mass.

    Do you have a specific example you are working with?
     
  6. Feb 3, 2013 #5
    I guess the specific example I'm working with is the simplest possible. Certainly a point mass in one dimension (which I hope to extend to three once I've understood DDF formalism). Accordingly, it may even be simpler if we treat the accceleration as coming out of nowhere; I guess I was just thinking that if a point mass is accelerating at time t1 then we can use F=ma (in DDF form or otherwise) to define its force at t1, without requiring knowledge of the properties of the thing that exerted the force.
    I thought that xi defines the given point? After all, that's the one point that the delta function assigns infinity to, '0' elsewhere.
    I don't know what it would mean for the mass to not exist, in this context, but it sounds like you're answering my original question by saying no to the first equation? Thus, by "the delta function you are modulating" I take it you just mean the acceleration delta function? I'm curious as to why this is wrong.

    It might be worth noting, a large portion of what I need to know can be answered without worrying much at all about the physical situation. Thus, a related question, is whether this is correct:
    [tex] \int [2\delta (x-x_i) 3\delta (x-x_i)]dx = 2*3 = 6[/tex]
    So here I'm just substituting arbitrary numbers for physical constants. Is there a problem with having two delta functions together in the one integrand? If so, is the solution to simply multiply two integrands in the manner suggested by my first post?
     
    Last edited: Feb 3, 2013
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