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/sci/ - Science & Math

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File: 9 KB, 602x341, lattice vectors.png [View same] [iqdb] [saucenao] [google]
2864925 No.2864925 [Reply] [Original]

I have a question about the dirac-delta function, in the context of lattices (materials sci/chemistry).

Suppose you have the equation on the left. All vectors are 3 dimensional.
Ignoring the Left hand side (LHS) of the equation for the moment and just looking at the right hand side (RHS) which has the dirac-delta function.

If k vector =/= G vector , then the whole RHS = 0, correct? Because dirac-delta of anything non-zero equals zero.
but what if k vector = G vector?
continued next post.

>> No.2864928
File: 15 KB, 572x509, lattice vectors continued.png [View same] [iqdb] [saucenao] [google]
2864928

I've read that the dirac-delta function is not defined at zero. on the RHS you'd get N times some undefined number.
But now if we give some context to the problem , taking a look at the picture provided in this post, that G vector dot R vector equals an integer multiple of 2 pi.
thus if K vector = G vector then all the exponential terms have a value of 1 and so the sum over all lattice vectors (R vector) of the exponential terms equals should equal the number of lattice vectors, given by N. So, physically speaking, the LHS of the equation should come to N if k vector = G vector
continued next post

>> No.2864942
File: 46 KB, 520x399, 110-zoolander_mugatu_crazy_pills[1].jpg [View same] [iqdb] [saucenao] [google]
2864942

BUT this disagrees with the RHS of the equation. The right hand side of the equation should be N times some undefined number if k vector = G vector, because the dirac-delta function is meant to be undefined at 0 (or in 3 dimensions, 0 vector).
So what the dilly-oh? Can you just fudge it and say that the dirac delta function equals 1 at 0?
Are there any material scientists or chemists or anyone out there who've worked with his identity before and understand what's going on here?
Please help me. The lecture included the question this is relevent to without ever defining the dirac delta function and I have no one I can ask about it in person.

>> No.2864961

Oh boy, this is what happens when non-math students are taught math poorly. First off the Dirac delta is not a function, it's a so called distribution. You can imagine it as a "function" which is everywhere zero outside zero as you said but undefined or infinitely large at zero. Another way of looking at it is as the (distributional) derivative of the Heaviside step function. Now, I have no idea in what context your equation arises and as such I can't tell you if it makes sense or not. It is possible that in this equation a "normalized" variant of the Dirac delta is used which isn't undefined at 0 but rather equal to 1 or 2pi or whatever makes sense.

>> No.2864962

tl;dr is the dirac-delta function always undefined at 0 or in physical sciences is it no problem just to say that it has a value of 1 at zero?

>> No.2864981 [DELETED] 

Are you sure your equation in the OP is right? How do you get G on the RHS when it wasn't there on the left? Shouldn't it just be <span class="math">N \delta(K)[/spoiler]?

>> No.2864976

>>2864961
>It is possible that in this equation a "normalized" variant of the Dirac delta is used which isn't undefined at 0 but rather equal to 1 or 2pi or whatever makes sense.

Okay I'm just going to assume that's the case because otherwise fuck this.
Thank you very much.

>> No.2864978

>>2864961
This guy again, let me add further thoughts. I highly doubt that you can set it equal to 1 or 2pi or whatever at 0 but rather that this equation is meant to realte information as to what happens to the left side when integrated. Another equivalent definition to the previous ones of the Dirac delta is that it is a "function" which is zero outside zero and it's integral over the real numbers is 1. Thus this would indicated that for k=G the LHS integrated over all real numbers is equal to N and if k!=G the LHS integrated is equal to 0. That's the best I can come up with right now.

>> No.2864996

>>2864978
Ok again further thoughts, it seems that this is not supposed to be the Dirac delta but rather the Kronecker delta centered at 0. Then the whole story actually makes sense. Thus delta(0)=1 and delta(x)=0 for x!=0. The equation is at least absolutely correct that way.

>> No.2865018
File: 9 KB, 572x509, wrwerewrewr.png [View same] [iqdb] [saucenao] [google]
2865018

>>2864996
OP here, I don't think so because kronecker delta is used earlier in the question sheet, and it is always written like the above one in this picture, with subscripts, rather than the bottom one of this picture, with brackets indicating it is a "function" of something.

>> No.2865027
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2865027

dirac-delta functions really only have physical meaning inside of an integration. So it's defined as infinite when K = G, but it has an area of 1 (which is why you can integrate over it)

>> No.2865036

One more thing, the right hand side of your equation is actually a periodic delta train. I don't know what the vectors K and G represent, but the left hand side represents the Fourier series representation of a periodic pulse train.

>> No.2865056

OP here. I'm afraid I'll have to leave you gentlemen.

Thank you for all your thoughts.