- #1

James MC

- 174

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We define the Dirac delta function of a point mass with mass m located at point x

_{0}as follows:

[tex]\rho(x)=m\delta (x-x_0)[/tex]

If our point mass is represented by a Dirac delta function, our acceleration will need to be too. So we need an acceleration distribution for our point mass that assigns an instantaneous acceleration a to x

_{0}:

[tex]a(x)=a\delta (x-x_0)[/tex]

If that is correct, then is the correct Dirac delta function formulation of f=ma:

[tex] F_i = [\int m\delta (x-x_i)dx] [\int a\delta (x-x_i)dx][/tex]

...for the force of particle i? And, for the total force F

_{T}for N particles, is it:

[tex] F_T = \sum_i [[\int m\delta (x-x_i)dx] [\int a\delta (x-x_i)dx]][/tex]

Okay, so if that's all good, is this the equation for standard mass density distributions (not for point particles, but for mass densities distributed over tiny regions):

[tex] F_T = \sum_i [\int [\rho(x) a(x)]dx][/tex]

Where

*ρ*(x) is now defined at dM/dV and a(x) as da/dV. Notice that the Dirac delta formulation has TWO integrals while the mass density formualtion has ONE integral. I think that's right because you cannot multiply delta functions because, pre-integration, that would be multiplying infinities? Whereas with the densities, you are instead multiplying finite values and so you CAN multiply BEFORE the integration?

Or am I just completely confused here? Any advice would be most welcome!