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If we applied a Dirac delta 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.

Qmechanic
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cyberic
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2 Answers2

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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.

BioPhysicist
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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 interpreted as a piecewise constant velocity profile of the form $$\dot{x}~=~v_0 +\frac{v_1}{2}{\rm sgn}(t-t_0)\tag{5}$$ for some constants $v_0$ and $v_1$ that depend on $m,E,V_0$. Eq. (5) may be integrated to yield a continuous piecewise linear position profile of the form $$x(t)~=~x_0+v_0(t-t_0)+\frac{v_1}{2}|t-t_0|.\tag{6}$$ This may be interpreted as scattering of the point particle.

Example 2. A Dirac delta distribution acceleration profile $$\ddot{x}(t)~=~v_1\delta(t-t_0) \tag{7}$$ with jerk $$\dddot{x}(t)~=~v_1\delta^{\prime}(t-t_0) \tag{8}$$ can be integrated to yield eq. (6).

Qmechanic
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