I am trying to understand this page which comes to the following equation:
$$f_{\mathrm{ext}}(t) = mv_0\delta(t).$$
They are attempting to describe the force being applied to a mass as a momentary excitation impulse.
Force = mass * acceleration
Here we have:
Force = mass * initial-impulse-velocity * $\delta(t)$
So what is $\delta(t)$? I am guessing it must be by necessity an inverse of a time increment. ie. 1/the-time-the-impulse-is-imparted-over.
Is that correct? Is there some concept that explains this?
How do I use $\delta(t)$ in other equations?
I mean, I would have thought you could also say:
$$f_{\mathrm{ext}}(t) = m\frac{\delta(v(t))}{\delta(t)}$$
ie. Force is equal to mass times the rate of change of velocity (acceleration).
How does that relate to the equation:
$$f_{mathrm{ext}}(t) = m\cdot v_0\cdot\delta(t)$$
Wouldn't that mean:
$$f_{\mathrm{ext}}(t) = kg \cdot\frac ms\cdot s = kg\cdot m$$
I mean, isn't $\delta(t)$ an increment of time when you say $\dfrac{\delta(v)}{\delta(t)}$ for example?
I'm having trouble understanding this. Hopefully at least my confusion on the terms makes sense.
What exactly is $\delta(t)$ in this equation?
Based on Wikipedia, the Dirac delta can be loosely thought of as a function on the real line
that is defined by
$$ \delta (t)=\begin{cases}+\infty ,&t=0\\0,&t\neq 0\end{cases}$$ and which satisfies
$$ \int _{-\infty }^{\infty }\delta (t)\,dt=1.$$
This is merely a heuristic characterization. The Dirac delta is not a function in the traditional sense, as no function defined on the real numbers has these properties. The Dirac delta function can be rigorously defined either as a distribution or as a measure. See the Wikipedia article for more information.