If we applied a delta Dirac function as a force, how can we obtain the acceleration of that object? I know that this is called impulse that changes the velocity, but since there is a change in velocity then there is an acceleration. Even though that practically impossible, but I think the math should give an answer for that.

  • $\begingroup$ The Dirac delta distribution is not a function. Treating it as one (physicists can get away with this, usually), the derivative is zero everywhere except at zero, is infinite at zero, but the area under the curve is finite. (Note well: I'm using physics math here. You'll get a different perspective at the mathematics sister site.) $\endgroup$ – David Hammen Feb 4 '18 at 15:41
  • $\begingroup$ Hi @cyberic: In how many space dimensions? $\endgroup$ – Qmechanic Feb 4 '18 at 15:50
  • $\begingroup$ @Qmechanic let's keep it simple and assume it's in a one dimensional (along x-axis) $\endgroup$ – cyberic Feb 4 '18 at 15:56
  • $\begingroup$ You know $F=ma$. If $F=\delta(t-t_0)$, then $a=\frac{1}{m}\delta(t-t_0)$. $\endgroup$ – Jahan Claes Feb 4 '18 at 16:21
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    $\begingroup$ @David Hammen: When I'm at dinner with mathematicians, I don't discuss religion, politics, or the delta function. $\endgroup$ – JEB Feb 4 '18 at 18:09

The acceleration is proportional to the force according to Newton's second law. Therefore, if you have a delta function force, your acceleration is also a delta function. This will then produce an instantaneous change in the velocity. In thinking about the position v time graph, there will be a "corner" in it since the slope will instantly change.


Example 1. A Dirac delta distribution force $$F(x)~=~-V_0\delta(x-x_0)~=~-V^{\prime}(x)\tag{1}$$ in 1D has a step potential $$V(x)~=~V_0\theta(x-x_0).\tag{2}$$ Mechanical energy conservation yields $$E~=~\frac{m}{2}\dot{x}^2+V(x),\tag{3}$$ or $$\dot{x}~=~\pm \sqrt{\frac{2}{m}(E-V(x))},\tag{4}$$ which may be integrated to yield a continuous piecewise linear position profile of the form $$x(t)~=~x_0+v_0t+\frac{v_1}{2}|t-t_0|.\tag{5}$$ This may be interpreted as scattering of the point particle.

Example 2. A Dirac delta distribution acceleration profile $$\ddot{x}(t)~=~v_1\delta(t-t_0) \tag{6}$$ with jerk $$\dddot{x}(t)~=~v_1\delta^{\prime}(t-t_0) \tag{7}$$ can be integrated to yield eq. (5).


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