Physical picture of Dirac's delta impulses

Question

In linear response theory, the response $A(t)$ is related to the impulse $g(t)$ by

$$A(t) = \int_{-\infty}^{\infty}\chi(t-t^\prime)g(t^\prime)\, dt^\prime $$

A typical example is the case where $g(t)$ is the Dirac's Delta Function. i.e. $g(t) = \delta(t-t_1) $ and

$$A(t) = \int_{-\infty}^{\infty}\chi(t-t^\prime)\delta(t^\prime-t_1)\, dt^\prime $$

From where

$$A(t) = \chi(t-t_1) $$

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I have some doubts about the physical picture here. I hope that someone can give some light.

Q1: About how a real system should behave

The only one intuitive idea I have about $\delta(t-t_1)$, for getting a picture of the real system, is something that is zero except for one single point where it is something like $\infty$. In order to share the same picture, suppose that we are talking about rigid electric dipoles under an spatially homogeneous electric field. How the dipoles will react? I imagine some alternatives:

• The force exerted is infinite and the dipoles turns completely oriented in the direction of the field. For this to be true, the force must be exerted for a time long enough to let the dipoles rotate to the "equilibrium" position. But at the same time, $E$ is not zero only for one point: $t_1$.
• Force is exerted but stops before the dipoles get parallel to the field, so the remain rotating super fast (sounds crazy).
• A little orientation occurs. (I have no argument that supports this alternative)
• No one is valid because an infinite force is unphysical.
• No possible interpretation.

Q2: Apparent contradiction with relaxation experiment

If the dipoles got (totally or partially) oriented at $t_1$, the system will relax and they will return to the equilibrium position. The physical picture looks similar to the following two step process: Step 1: Exert a constant $E(t<0)=a$ and let the system reach the equilibrium. Step 2: Turn off the field abruptly (let say, at $t=0$). In such case, the well known solution is

$$ A(t) = a \int _t ^\infty \chi(t^\prime)\,dt^\prime \neq \chi(t) $$

(notice that $\chi(t)$ is the solution for a $\delta$ signal at $t=0$)

Q3: On the linearity of the relationship impulse-response

If the impulse is infinitely large at $t_1$, how reasonable is to think that the response will be linear?

Q4: Moleling

If one desires to model a system in order to get $\chi$, and one can get $A(t)$ for any system state. How to prepare the system? This question is quite similar to Q1, but makes more emphasis in the state of the system after the Dirac's Delta impulse instead of the processes itself.

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EDIT Considering $\delta$ as an idealization of a "no so fast" and a "no so infinite" input also gives me some problems. For example, suppose a moderate input that give rise to a moderate $A(t)$ in some total time no too short.

Notice that any of such input $g(t)$ satisfices

$$ g(t) = \int_{-\infty}^{\infty}g(t^\prime)\delta(t-t^\prime)\, dt^\prime $$

so it can be though as a infinite linear combination of Dirac's delta input. From this perspective, if $g(t)$ is reasonably smooth, it seems that it is necessary an infinite number of Dirac's deltas in order to $g(t)$ generates a reasonable response. So just one of the infinite terms (like in $g(t) = \delta(t-t_1) $) wouldn't contribute enough to $A(t)$ for make it noticeable.

But of course it must be false, as $A(t) = \chi(t-t_1)$ and $\chi(t-t_1)$ is not null.

If it an experimental approximation of the Dirac's delta, this problem disappear, but I again would not make sense of the application of the math above.

Answer 1

The only one intuitive idea I have about δ(t−t1), for getting a picture of the real system, is something that is zero except for one single point where it is something like ∞.

Most times when you see infinities in physics, you should not think of something that's infinite, but something that's very large, and we're analyzing what would happen if it kept getting larger. What would be the limiting behavior as something keeps getting bigger.

In the case of the Dirac delta function, you're usually thinking about something that keeps getting larger in amplitude, but at the same time getting shorter in duration, in such a way that the total energy (or momentum or whatever is important to the physical system you're studying) remains the same.

If the impulse is infinitely large at $t_1$, how reasonable is to think that the response will be linear?

It's correct that a real system won't be linear with an infinite input amplitude.

But also, a real system will never encounter a true Dirac impulse as an input.

If the system encounters an input with duration much shorter than any time constant associated with its internal state or output response, then its response will (assuming the system is sufficiently linear) be very close to the same as if the input were an ideal impulse. It will mainly depend on the total energy delivered rather than the amplitude of the signal applied or the exact duration of the impulse or the shape of the signal waveform.

You can also think of the impulse response as being only a mathematical tool to investigate the behavior of linear systems (which themselves are only an approximate model for real-world non-linear systems).

Answer 2

1. Even though you can (sometimes) think of the Dirac Delta function as being infinite when its arguments is zero, you still have to keep track of the strength of the impulse; just saying "it's infinite" is not enough. So let's write $$g(t) = \varepsilon \, \delta(t-t_1)$$ where $\varepsilon$ is a constant. The response at time $t$ is then $$ A(t) = \varepsilon \, \chi (t-t_1) $$ Note that the response is linear in the strength of the perturbation $\varepsilon$, which might address your third question.

   Let's say the response function is a decaying exponential for $t>0$ (it has to be zero for $t<0$ due to causality). Then this equation tells you how a system would respond to an impulse perturbation: it would get a quick jolt, and then relax back to its unperturbed equilibrium state.

2. There is no contradiction in your second question, because the experiment you describe is different from an impulse perturbation. Instead you have $$ g(t) = \begin{cases} \varepsilon & \text{if } t < 0 \\ 0 & \text{if } t \geq 0 \end{cases} $$ so that for $t \geq 0$ $$ A(t) = \int_{-\infty}^t \mathrm{d}t' \, \chi (t - t') g(t') = \varepsilon \int_{-\infty}^0 \mathrm{d}t' \, \chi (t - t') = \varepsilon \int_{t}^{\infty} \mathrm{d}t' \, \chi (t') $$

3. To measure the response function you can do the experiment just described (equilibrate in the presence of a constant perturbation, and then turn it off at time zero), and obtain $\chi$ via $$ \chi(t) = - \frac{1}{\varepsilon} \frac{\mathrm{d}A(t)}{\mathrm{d}t} $$ for $t \geq 0$, which is just the derivative of the previous equation. Alternatively you can use the Fluctuation Dissipation theorem and get the response function from the equilibrium correlation function (in absence of any perturbation) as $$ \chi(t) = \begin{cases} - \frac{\displaystyle 1}{\displaystyle k_{\mathrm{B}}T} \frac{\displaystyle \mathrm{d}}{\displaystyle \mathrm{d}t} <A(0)A(t)> & \text{if } t \geq 0 \\ 0 & \text{if } t <0 \end{cases} $$