Canute
08-23-03, 11:02 AM
Is it possible to explain a bit about Dirac's delta function in layman's terms? Probably not. But if it is then can someone please have a shot at it. I'm wondering about the function it fulfils and its general properties.
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View Full Version : Delta Function Canute 08-23-03, 11:02 AM Is it possible to explain a bit about Dirac's delta function in layman's terms? Probably not. But if it is then can someone please have a shot at it. I'm wondering about the function it fulfils and its general properties. lethe 08-23-03, 11:24 AM Originally posted by Canute Is it possible to explain a bit about Dirac's delta function in layman's terms? Probably not. But if it is then can someone please have a shot at it. I'm wondering about the function it fulfils and its general properties. the dirac delta "function" is found in all kinds of applications in physics. you can think of it as the density for a point particle with finite mass. or the charge density of a point particle of finite charge. the density everywhere in space is zero, except at the point where the particle is located, where the density is infinite. if you wanted to find the electric field, for example, from some extended object, you could consider the extended object as an infinite sum of point particles. if you can solve maxwell s equation for the point particle (dirac delta function), then you can get a solution to the more general extended object simply by integrating those solutions (known as green s functions). this method extends quite generally for many types of differential equations, and so you find the dirac delta function in many places. additionaly, the dirac delta function is sometimes used in functional analysis. it is the natural choice to normalize an uncountable basis of an infinite dimensional hilbert space. for this reason it is also seen a lot in quantum mechanics, which usually lives in such a space. this is not really necessary, however, since any hilbert space has a countable basis. the fact that in general the dirac delta function does not live in the hilbert space can also be bothersome to some people. i put the word "function" in quotation marks, because it is not actually a function. Canute 08-23-03, 03:07 PM Thanks. Would is be wrong to say that the delta 'function' is, in a sense, a way of re-normalising the infinite values of points, or is that just gibberish? lethe 08-24-03, 02:39 AM Originally posted by Canute Thanks. Would is be wrong to say that the delta 'function' is, in a sense, a way of re-normalising the infinite values of points, or is that just gibberish? i pretty much have no idea what you re trying to say. what is the "value" of a point? Canute 08-24-03, 07:31 AM Originally posted by lethe i pretty much have no idea what you re trying to say. what is the "value" of a point? Yeah - I thought it might be gibberish. I got the impression that the function of the delta 'function' was to overcome the problem that the calculation of the density of a point in space leads to infinities. I also took your post to say something similar whan you said "It is the natural choice to normalize an uncountable basis of an infinite dimensional hilbert space." However I've probably completely misunderstood it. James R 08-24-03, 08:57 AM In a basic sense, what the Dirac delta function does is to kill integrals. Probably the most important property of the function (technically it's a distribution) is that for any well-behaved function f(x): Integral (-infinity -> infinity) [ f(x) <font face="symbol">d</font>(x - x<sub>0</sub>) ] dx = f(x<sub>0</sub>) lethe 08-24-03, 09:03 AM Originally posted by Canute Yeah - I thought it might be gibberish. well, i m not sure if it s gibberish, because i m not really sure what you mean. certainly it is very sloppy language, which is bad. I got the impression that the function of the delta 'function' was to overcome the problem that the calculation of the density of a point in space leads to infinities. that sounds like a good way to think about it to me. I also took your post to say something similar whan you said "It is the natural choice to normalize an uncountable basis of an infinite dimensional hilbert space." hmm.... Canute 08-24-03, 10:37 AM Originally posted by James R In a basic sense, what the Dirac delta function does is to kill integrals. Probably the most important property of the function (technically it's a distribution) is that for any well-behaved function f(x): Integral (-infinity -> infinity) [ f(x) <font face="symbol">d</font>(x - x<sub>0</sub>) ] dx = f(x<sub>0</sub>) This is like being in France and not speaking French! Does 'kill integrals' mean 'avoid infinite sums of integers'? lethe 08-24-03, 10:54 AM Originally posted by Canute This is like being in France and not speaking French! Does 'kill integrals' mean 'avoid infinite sums of integers'? i think what james said is the mathematical equation that represents the following statement i made: "the delta function shows the density in space of a finite mass point particle" if you want to find the center of gravity of the particle in a nonuniform gravitational field, for example, you have to integrate over space the density times the field. but for a point particle, the integral is trivial. you can simply replace the integration with the value of field at the point where the particle is located times its mass. james integral is those words written in a mathematical equation. Canute 08-24-03, 03:11 PM Originally posted by lethe i think what james said is the mathematical equation that represents the following statement i made: "the delta function shows the density in space of a finite mass point particle" if you want to find the center of gravity of the particle in a nonuniform gravitational field, for example, you have to integrate over space the density times the field. but for a point particle, the integral is trivial. you can simply replace the integration with the value of field at the point where the particle is located times its mass. james integral is those words written in a mathematical equation. Thanks. I'm trying to understand that but have a problem. It seems to say that that to find the centre of gravity of a point particle you take the value of the field at the point and times it by its mass. Am I misreading it? Tom2 08-24-03, 03:17 PM Originally posted by Canute Thanks. I'm trying to understand that but have a problem. It seems to say that that to find the centre of gravity of a point particle you take the value of the field at the point and times it by its mass. Am I misreading it? Let me reproduce lethe's statement: Originally posted by lethe if you want to find the center of gravity of the particle in a nonuniform gravitational field, for example, you have to integrate over space the density times the field. but for a point particle, the integral is trivial. you can simply replace the integration with the value of field at the point where the particle is located times its mass. First, I am almost certain that lethe meant for the first instance of the word "particle" (in blue) to mean "body" (as in an extended object). When the body is taken to be the idealized case of a point particle, then its center of gravity happens to be exactly at the location of the particle. The delta function selects out one particular point on the range of integration. In James' integral, the action of d(x-x<sub>0</sub>) when integrated with f(x) is to return the value f(x<sub>0</sub>). He says that it "kills integrals", but I think it would be more accurate to say that it "kills integrands at all but one point". Canute 08-24-03, 05:44 PM Thanks - I'm getting there. errandir 08-24-03, 06:55 PM I would revise thisOriginally posted by lethe you can think of it as the density for a point particle with finite mass. or the charge density of a point particle of finite charge.to: you can think of it as the unit density function (meaning units of V<sup>-1</sup> where V is the unit of "volume" in the particular space of interest) of a body with infinitesimal volume (point particle of unit value). Originally posted by lethe it is the natural choice to normalize an uncountable basis of an infinite dimensional hilbert space. ...any hilbert space has a countable basis.Say wha? errandir 08-24-03, 07:03 PM I've always understood the dirac delta to be defined using an intergral: The integral of f(x)δ(x-x<sub>0</sub>)dx from x<sub>0</sub> - ε to x<sub>0</sub> + ε (the small neighborhood around x<sub>0</sub>) is equal to f(x<sub>0</sub>). This may then be generalized to higher dimensions. I believe someone posted this earlier on this thread. <i>edit: yes, it was James.</i> To get an intuitive feel for the dirac delta, it has always been explained to me as a limiting case of some pusle (or, in more rigorous cases, some infinite series). The basic idea behind the limiting case of the pulse is that it maintains the same volume in the limiting process, but the width decreases to zero and the height increases to infinity. (The basic idea behind the limiting case of the infinite series is completeness.) Tyger 08-24-03, 08:26 PM One important thing about the delta function that hasn't been mentioned is that it is the continuous analog of the Kroneker delta matrix, which is the n-dimensional unit matrix, and that operations with it are equivalent to operations with matrices. errandir 08-24-03, 09:03 PM Only after a radical redefinition of the inner product. GT_Phys 08-24-03, 09:24 PM Just kinda know that the Dirac delta function is equal to infinity at some point and zero everywhere else, and if you integrate over all the area, the infinity at that one point offsets the zeros and it is equal to one. That's how it crops out integrals and just gives you functions. It's nice and happy and very useful. |