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I am wondering if a physical system subjected to an impulse $=A\delta(x-a)$ makes any sense. I reckon that a force could take that form -- thought of as a finite impulse applied over an infinitesimal time interval. But if the impulse takes that form that would imply infinite velocity?? Thanks.

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No. Step change in velocity. Infinite acceleration. – Mike Dunlavey Nov 2 '11 at 18:10

No it would not imply infinite velocity, however, the way you wrote it complicates the matter a great deal. You used the form $F\propto\delta(x-a)$, suggesting that this is a function of position with the use of the variable $x$. Your wording doesn't necessarily imply this, because it is certainly vague enough for one to interpret it more than one way. I will address two functional forms for the force applied to an object to be sure that I answer the question.

Firstly, consider a force that is applied with a Dirac delta as a position.

$$F= A \delta(x-a) $$

Now a force applied with a Dirac delta over time. These are two completely different problems.

$$F=B \delta(t-b)$$

Consider the problem in 1-D, object moving to the right with velocity $v_0$ and after it passes the point $a$ or time passes the point $b$ it obtains a new velocity, $v_f$. The final velocity is found explicitly in the case of the function of time using the concept of impulse.

$$v_f = v_0 + \frac{1}{m}\int_{b_{-}}^{b_{+}} F(t) dt = v_0 + \frac{B}{m}$$

My guess is that this is what you wanted to ask about. Here, $B$ is exactly the impulse delivered. I used notation $b_{\pm}$ to indicate points that are just before or just after the time in question.

For the other case, where the force is a function of position, the new energy can be explicitly calculated in a similar manner.

$$E_f = E_0 + \int_{a_{-}}^{a_{+}} F(x) dx = E_0 + A$$

$$E_0 = \frac{1}{2} m v_0^2$$

$$v_f = \sqrt{ \frac{2 E_f}{m} } = \sqrt{v_0^2+ \frac{ 2 A}{m} } $$

Now the amount it is sped up by depends on the initial velocity before it passes the (spatial) point $a$. This should be correct.

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Since Zassounotsukushi explained the classical delta-in-time problem, I will expand on the delta-in-space problem which was the original question.

Whether a delta-function potential makes sense depends on the dimension, and for higher than one dimension, it only makes sense if you use quantum mechanics. In one dimension you can use a delta-function potential to produce a sensible exactly solvable quantum model.

If you have a d-dimensional delta-like potential, it can be taken to be a constant V on a sphere of radius R. So long as the wavelength of the particle is much longer than R, the internal structure of the potential is unimportant, and there is a scaling limit of small R, which can be thought of as the scattering off a delta-function potential. But this is only strictly a delta-function potential when the scaling law for V in the zero R limit is given by $V(R)\propto 1/R^d$, so that all the scattering amplitudes converge to a nontrivial limit. This is not the right scaling, except in 1d.

You can think of different scalings as a renormalization of the coefficient of the delta function. Already in two dimensions, you get a log-running of the coefficient of the delta-function in order to get a sensible point limit. This was studied briefly in a series of papers in the 1990s, as a toy model for field theory renormalization. In higher dimensions, you need a power-scaling for V as a function of R.

Classically, in two dimensions, a delta-function potential is useless, because you only hit the delta function on a measure zero set of paths. It gets worse in higher dimensions. But quantum mechanically, these are sensible problems in all dimensions--- they give the scaling for S-wave scattering off an arbitrary potential in the long-wavelength limit.

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