- #1

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I have some difficulties of calculating the following integral:

[itex] I=\int _{D}\:\:\:d^{3}q\: d^{3}k\: d^{3}p\:\:F(q^{2}, q.k, q.p, k^{2}, p^{2})[/itex]

where:

D=|k|>1, |k+q|<1 and |p-q|<1

Thanks in advance.

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- Thread starter kassem84
- Start date

- #1

- 13

- 0

I have some difficulties of calculating the following integral:

[itex] I=\int _{D}\:\:\:d^{3}q\: d^{3}k\: d^{3}p\:\:F(q^{2}, q.k, q.p, k^{2}, p^{2})[/itex]

where:

D=|k|>1, |k+q|<1 and |p-q|<1

Thanks in advance.

- #2

- 608

- 1

What is the function F?

- #3

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Hello,What is the function F?

F=e[itex]^{-(q^{2}+q.k+q.p)}[/itex]

The most important thing is how to obtain the boundaries of the integrals. i.e. q,p,k go from where to where?

Thanks.

- #4

- 1,800

- 53

I'm not sure I'm interpreting your notation correctly. Is that dot products in there? Also, the d^3 notation. Is that vectors? Tell you what, if it was just:

[tex]\int\int\int f(q,k,p) dqdkdp[/tex]

then I think we can use Mathematica to obtain the boundaries.

[tex]\int\int\int f(q,k,p) dqdkdp[/tex]

then I think we can use Mathematica to obtain the boundaries.

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- #5

- 13

- 0

I'm not sure I'm interpreting your notation correctly. Is that dot products in there? Also, the d^3 notation. Is that vectors? Tell you what, if it was just:

[tex]\int\int\int f(q,k,p) dqdkdp[/tex]

then I think we can use Mathematica to obtain the boundaries.

Yes, it is the correct notation of the integral I. All the vectors are 3-dimensional in the definition of the function and in the boundary D.

Thanks.

- #6

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- 53

The scalar version is quite interesting. There are two rhomboid regions to integrate over since |k|>1. I believe this is the integral for the region k>1:

[tex]\mathop\iiint\limits_{D} f(p,q,k)dpdqdk=\int_{1}^{\infty}\int_{-1-k}^{1-k}\int_{q-k}^{q+k} f(p,q,k)dpdqdk[/tex]

Perhaps the vector version is similar and you can adapt it to this one.

[tex]\mathop\iiint\limits_{D} f(p,q,k)dpdqdk=\int_{1}^{\infty}\int_{-1-k}^{1-k}\int_{q-k}^{q+k} f(p,q,k)dpdqdk[/tex]

Perhaps the vector version is similar and you can adapt it to this one.

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- #7

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[tex]\int\int\int f(q,k,p) dqdkdp[/tex]

then I think we can use Mathematica to obtain the boundaries.

How can we use mathematica to determine the boundaries- the intersection of the three spheres?

- #8

- 13

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The scalar version is quite interesting. There are two rhomboid regions to integrate over since |k|>1. I believe this is the integral for the region k>1:

[tex]\mathop\iiint\limits_{D} f(p,q,k)dpdqdk=\int_{1}^{\infty}\int_{-1-k}^{1-k}\int_{q-k}^{q+k} f(p,q,k)dpdqdk[/tex]

Perhaps the vector version is similar and you can adapt it to this one.

Thanks. For the vector version, it difficult for me to determine the boundaries on the angles θ and [itex]\phi[/itex].

- #9

- 1,800

- 53

Could you or someone else tell me if I'm interpreting this correctly since I've never worked on one like this before. But first, let's just restrict it to a double integral for now:

[tex]\mathop\iint\limits_{D} f(k,q)d^3q\, d^3 k[/tex]

where each integral is a triple integral in spherical coordinates and:

D={|k|>1, |k+q|<1}

We can compute the outer one easily. Since |k|>1, then for spherical coordinate r, we can write:

[tex]\mathop\int_{r>1}\left( \mathop\int\limits_{S} f(k,q) d^3 q\right)\,d^3 k[/tex]

So what is S? Since |k+q|<1, then that means we need:

[tex]\sqrt{(k_x+q_z)^2+(k_y+q_y)^2+(k_z+q_z)^2}<1[/tex]

for every point in k-space (k_x, k_y, k_z). Now suppose we have for a particular point:

[tex]k=(3,4,7)[/tex]

Then for |k+q|<1, we would have to integrate in q-space over a sphere centered at q=(-3,-4,-7) with radius one. The boundary for that one k-point would be:

[tex](q_x+3)^2+(q_y+4)^2+(q_z+7)^2=1[/tex]

So for just that one k-point, the integral would be:

[tex]\mathop\iiint\limits_{(q_x+3)^2+(q_y+4)^2+(q_z+7)^2\leq 1} f(k,q)d^3q[/tex]

and therefore for all of the k-space, we could then write:

[tex]\mathop\iiint\limits_{k_x^2+k_y^2+k_z^2>1}\iiint_{(q_x-k_x)^2+(q_y-k_y)^2+(q_z-k_z)^2\leq 1} f(k,q)d^3q d^3k[/tex]

Ok, so just for now, can we let p be what ever it has to be to work, say p=(1,1,1) or whatever, can we now compute:

[tex]\mathop\iiint\limits_{k_x^2+k_y^2+k_z^2>1}\iiint_{(q_x-k_x)^2+(q_y-k_y)^2+(q_z-k_z)^2\leq 1} \exp\{-(q^2+q\cdot k+q\cdot(1,1,1))\}d^3q d^3k[/tex]

[tex]\mathop\iint\limits_{D} f(k,q)d^3q\, d^3 k[/tex]

where each integral is a triple integral in spherical coordinates and:

D={|k|>1, |k+q|<1}

We can compute the outer one easily. Since |k|>1, then for spherical coordinate r, we can write:

[tex]\mathop\int_{r>1}\left( \mathop\int\limits_{S} f(k,q) d^3 q\right)\,d^3 k[/tex]

So what is S? Since |k+q|<1, then that means we need:

[tex]\sqrt{(k_x+q_z)^2+(k_y+q_y)^2+(k_z+q_z)^2}<1[/tex]

for every point in k-space (k_x, k_y, k_z). Now suppose we have for a particular point:

[tex]k=(3,4,7)[/tex]

Then for |k+q|<1, we would have to integrate in q-space over a sphere centered at q=(-3,-4,-7) with radius one. The boundary for that one k-point would be:

[tex](q_x+3)^2+(q_y+4)^2+(q_z+7)^2=1[/tex]

So for just that one k-point, the integral would be:

[tex]\mathop\iiint\limits_{(q_x+3)^2+(q_y+4)^2+(q_z+7)^2\leq 1} f(k,q)d^3q[/tex]

and therefore for all of the k-space, we could then write:

[tex]\mathop\iiint\limits_{k_x^2+k_y^2+k_z^2>1}\iiint_{(q_x-k_x)^2+(q_y-k_y)^2+(q_z-k_z)^2\leq 1} f(k,q)d^3q d^3k[/tex]

Ok, so just for now, can we let p be what ever it has to be to work, say p=(1,1,1) or whatever, can we now compute:

[tex]\mathop\iiint\limits_{k_x^2+k_y^2+k_z^2>1}\iiint_{(q_x-k_x)^2+(q_y-k_y)^2+(q_z-k_z)^2\leq 1} \exp\{-(q^2+q\cdot k+q\cdot(1,1,1))\}d^3q d^3k[/tex]

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