Understanding the Dirac delta function for impulses

Question

I am looking at a very basic impulse system of a ball bouncing on the ground at time $T$. It experiences force for a short time when it hits the ground. Using Newton's 2nd Law:

$ m \ddot{x} = F(t) - mg$

$\displaystyle \int_{T - \varepsilon}^{T + \varepsilon} \! m \frac{\mathrm{d}^2x}{\mathrm{d}t^2} \, \mathrm{d}t = \int_{T - \varepsilon}^{T + \varepsilon} \! F(t) \, \mathrm{d}t \; - \int_{T - \varepsilon} ^{T + \varepsilon} \! mg \, \mathrm{d}t$

$\bigg[ m \frac{\mathrm{d}x}{\mathrm{d}t} \bigg]_{T - \varepsilon}^{T + \varepsilon} = I - 2\varepsilon mg $

Where impulse $I$ is the area under the force curve.

I can see that this can be rewritten as something of the form

$\Delta p = I - O(\varepsilon)$

And as $\varepsilon$ goes to $0$ we get

$\Delta p = I$

However, I have seen the result that the differential equation can be rewritten using the Delta function as

$ m \ddot{x} = -mg +I \delta (t - T)$

I do not understand how this result came about using the Delta function. My question is how can we rewrite the above differential equation ($ m \ddot{x} = F(t) - mg$) using the Dirac Delta function and the impulse $I$?

Answer 1

Consider the 1D Newton's equation \begin{align} m\ddot x = F(t). \end{align} This says that the force experienced by the particle varies as a function of time (in general, the force could also depend on the position $x$).

Suppose $F = 0$, then the particle is experiencing no external forces. Hence, it's either traveling at a constant velocity or it's not moving.

Now, suppose you apply a constant force on the particle over a time interval $[T, T+\Delta t]$, say \begin{align} F(t) = \begin{cases} C & \text{ it } T\le t \le T+\Delta t\\ 0 & \text{ otherwise} \end{cases} \end{align} i.e. you are pushing the particle with constant force from time $T$ to $T+\Delta t$. Then physics says that the particle experiences an impulse of size $I=C\Delta t$.

But you could have applied the following force \begin{align} F(t) = \begin{cases} 2C & \text{ it } T\le t \le T+\frac{1}{2}\Delta t\\ 0 & \text{ otherwise} \end{cases} \end{align} which is twice the magnitude but half the time of the original force. The impulse experienced by the particle is the same. What this is modeling is the fact that you are pushing harder but for less time.

In theory, you could increase the magnitude of the force that you are applying but maintain the same impulse by applying force for a shorter period of time. Furthermore, if you consider the limit of this procedure, i.e. you smack the particle at time $T$ with magnitude $I=C\Delta t$. This is modeled by the delta function, i.e. \begin{align} m\ddot x = I\delta(t-T). \end{align}

Answer 2

I think it's important to understand the full implications of what it means when you write $$ \lim_{\epsilon \to 0} \int_{T - \epsilon}^{T + \epsilon} F(t) dt = I,$$ where $I \ne 0$.

When you write this, you are basically assuming that $F(t)$ has a Dirac delta function in it. If not, then the integral would be $O(\epsilon)$ and $I$ would be $0$.

(As an aside, you might think that you can get a nonzero result by allowing a singularity in the force at time $T$, or something like that. This may be possible, but it would probably be physically indistinguishable from a Dirac delta. The obvious approach, saying $F(t)=(t-T)^\alpha$ for some power $\alpha$, does not work. $I$ will turn out to be 0, $\infty$, or ill-defined.)

So far, I have argued that $I \ne 0$ is more or less equivalent to saying $$ F(t) = I \delta(t - T) + \dots $$ where $\dots$ could represent any other terms that do not contribute a finite impulse at time $T$. Now try plugging this $F(t)$ into $$ m \ddot{x} = F(t) - mg $$ You get the desired result, $$ m \ddot{x} = -mg + I \delta(t - T) $$