Mixing
Issues evaluating a Fourier transform on a restriction.
The F.T. of the function $R^3rightarrow R -U(x,y,z) $ is:
$$g(k_{1},k_{2},k_{3})=iiint_{R^{3}} U(x,y,z)e^{-i(k_1 x+k_2 y+ k_3 z)}dxdydz$$
By putting $z=0$ in the previous equation, you obtain:
$$g(k_{1},k_{2},k_{3})=iiint_{R^{3}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdydz=$$
$$=iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdyint _{R}dz$$
Now,I don't know how to deal with the last integral $int _{R}dz$. Since previously I have put z=0 then it should be dz=0 and also $int _{R}dz =0 $. However in this case it results $g(k_{1},k_{2},k_{3})$=0 which doesn't make sense. I'd say that the right solution should be:
$$iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$$
That comes if $int _{R}dz=1$.
But I don't know how to proof it, someone knows? Thanks.
fourier-transform
asked Nov 21 '18 at 15:38
Landau
447
|
show 4 more comments
The F.T. of the function $R^3rightarrow R -U(x,y,z) $ is:
$$g(k_{1},k_{2},k_{3})=iiint_{R^{3}} U(x,y,z)e^{-i(k_1 x+k_2 y+ k_3 z)}dxdydz$$
By putting $z=0$ in the previous equation, you obtain:
$$g(k_{1},k_{2},k_{3})=iiint_{R^{3}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdydz=$$
$$=iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdyint _{R}dz$$
Now,I don't know how to deal with the last integral $int _{R}dz$. Since previously I have put z=0 then it should be dz=0 and also $int _{R}dz =0 $. However in this case it results $g(k_{1},k_{2},k_{3})$=0 which doesn't make sense. I'd say that the right solution should be:
$$iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$$
That comes if $int _{R}dz=1$.
But I don't know how to proof it, someone knows? Thanks.
fourier-transform
asked Nov 21 '18 at 15:38
Landau
447
You want to evaluate $iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$ ? Then it is $frac{1}{2pi}int_R g(k_1,k_2,k_3) dk_3$ (write $g(k_1,k_2,k_3)=int_R V(k_1,k_2,z)e^{-i k_3 z}dz$ where $V(k_1,k_2,z) = ldots$)
– reuns
Nov 21 '18 at 16:53
No, I want to proof that $int_{R}dz=1$@reuns
– Landau
Nov 21 '18 at 17:03
$int_a^b dz = b-a, int_{R}dz=infty$
– reuns
Nov 21 '18 at 17:06
Ok, but I have put z=0 so dz=0, so I wuold have said that $int_{R}dz=0$ however it doesn't make sense. It should be results =1.@reuns
– Landau
Nov 21 '18 at 17:12
Nothing of what you wrote makes sense. It seems you don't understand integrals. What are you trying to solve
– reuns
Nov 21 '18 at 17:29
|
show 4 more comments
The F.T. of the function $R^3rightarrow R -U(x,y,z) $ is:
$$g(k_{1},k_{2},k_{3})=iiint_{R^{3}} U(x,y,z)e^{-i(k_1 x+k_2 y+ k_3 z)}dxdydz$$
By putting $z=0$ in the previous equation, you obtain:
$$g(k_{1},k_{2},k_{3})=iiint_{R^{3}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdydz=$$
$$=iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdyint _{R}dz$$
Now,I don't know how to deal with the last integral $int _{R}dz$. Since previously I have put z=0 then it should be dz=0 and also $int _{R}dz =0 $. However in this case it results $g(k_{1},k_{2},k_{3})$=0 which doesn't make sense. I'd say that the right solution should be:
$$iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$$
That comes if $int _{R}dz=1$.
But I don't know how to proof it, someone knows? Thanks.
fourier-transform
asked Nov 21 '18 at 15:38
Landau
447
The F.T. of the function $R^3rightarrow R -U(x,y,z) $ is:
$$g(k_{1},k_{2},k_{3})=iiint_{R^{3}} U(x,y,z)e^{-i(k_1 x+k_2 y+ k_3 z)}dxdydz$$
By putting $z=0$ in the previous equation, you obtain:
$$g(k_{1},k_{2},k_{3})=iiint_{R^{3}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdydz=$$
$$=iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdyint _{R}dz$$
Now,I don't know how to deal with the last integral $int _{R}dz$. Since previously I have put z=0 then it should be dz=0 and also $int _{R}dz =0 $. However in this case it results $g(k_{1},k_{2},k_{3})$=0 which doesn't make sense. I'd say that the right solution should be:
$$iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$$
That comes if $int _{R}dz=1$.
But I don't know how to proof it, someone knows? Thanks.
fourier-transform
fourier-transform
asked Nov 21 '18 at 15:38
Landau
447
asked Nov 21 '18 at 15:38
Landau
447
edited Nov 21 '18 at 17:44
asked Nov 21 '18 at 15:38
Landau
447
asked Nov 21 '18 at 15:38
Landau
447
asked Nov 21 '18 at 15:38
Landau
447
447
You want to evaluate $iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$ ? Then it is $frac{1}{2pi}int_R g(k_1,k_2,k_3) dk_3$ (write $g(k_1,k_2,k_3)=int_R V(k_1,k_2,z)e^{-i k_3 z}dz$ where $V(k_1,k_2,z) = ldots$)
– reuns
Nov 21 '18 at 16:53
No, I want to proof that $int_{R}dz=1$@reuns
– Landau
Nov 21 '18 at 17:03
$int_a^b dz = b-a, int_{R}dz=infty$
– reuns
Nov 21 '18 at 17:06
Ok, but I have put z=0 so dz=0, so I wuold have said that $int_{R}dz=0$ however it doesn't make sense. It should be results =1.@reuns
– Landau
Nov 21 '18 at 17:12
Nothing of what you wrote makes sense. It seems you don't understand integrals. What are you trying to solve
– reuns
Nov 21 '18 at 17:29
|
show 4 more comments
You want to evaluate $iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$ ? Then it is $frac{1}{2pi}int_R g(k_1,k_2,k_3) dk_3$ (write $g(k_1,k_2,k_3)=int_R V(k_1,k_2,z)e^{-i k_3 z}dz$ where $V(k_1,k_2,z) = ldots$)
– reuns
Nov 21 '18 at 16:53
No, I want to proof that $int_{R}dz=1$@reuns
– Landau
Nov 21 '18 at 17:03
$int_a^b dz = b-a, int_{R}dz=infty$
– reuns
Nov 21 '18 at 17:06
Ok, but I have put z=0 so dz=0, so I wuold have said that $int_{R}dz=0$ however it doesn't make sense. It should be results =1.@reuns
– Landau
Nov 21 '18 at 17:12
Nothing of what you wrote makes sense. It seems you don't understand integrals. What are you trying to solve
– reuns
Nov 21 '18 at 17:29
You want to evaluate $iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$ ? Then it is $frac{1}{2pi}int_R g(k_1,k_2,k_3) dk_3$ (write $g(k_1,k_2,k_3)=int_R V(k_1,k_2,z)e^{-i k_3 z}dz$ where $V(k_1,k_2,z) = ldots$)
– reuns
Nov 21 '18 at 16:53
You want to evaluate $iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$ ? Then it is $frac{1}{2pi}int_R g(k_1,k_2,k_3) dk_3$ (write $g(k_1,k_2,k_3)=int_R V(k_1,k_2,z)e^{-i k_3 z}dz$ where $V(k_1,k_2,z) = ldots$)
– reuns
Nov 21 '18 at 16:53
No, I want to proof that $int_{R}dz=1$@reuns
– Landau
Nov 21 '18 at 17:03
No, I want to proof that $int_{R}dz=1$@reuns
– Landau
Nov 21 '18 at 17:03
$int_a^b dz = b-a, int_{R}dz=infty$
– reuns
Nov 21 '18 at 17:06
$int_a^b dz = b-a, int_{R}dz=infty$
– reuns
Nov 21 '18 at 17:06
Ok, but I have put z=0 so dz=0, so I wuold have said that $int_{R}dz=0$ however it doesn't make sense. It should be results =1.@reuns
– Landau
Nov 21 '18 at 17:12
Ok, but I have put z=0 so dz=0, so I wuold have said that $int_{R}dz=0$ however it doesn't make sense. It should be results =1.@reuns
– Landau
Nov 21 '18 at 17:12
Nothing of what you wrote makes sense. It seems you don't understand integrals. What are you trying to solve
– reuns
Nov 21 '18 at 17:29
Nothing of what you wrote makes sense. It seems you don't understand integrals. What are you trying to solve
– reuns
Nov 21 '18 at 17:29
|
show 4 more comments
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You want to evaluate $iint_{R^{2}} U(x,y,0)e^{-i(k_1 x+k_2 y)}dxdy$ ? Then it is $frac{1}{2pi}int_R g(k_1,k_2,k_3) dk_3$ (write $g(k_1,k_2,k_3)=int_R V(k_1,k_2,z)e^{-i k_3 z}dz$ where $V(k_1,k_2,z) = ldots$)
– reuns
Nov 21 '18 at 16:53
No, I want to proof that $int_{R}dz=1$@reuns
– Landau
Nov 21 '18 at 17:03
$int_a^b dz = b-a, int_{R}dz=infty$
– reuns
Nov 21 '18 at 17:06
Ok, but I have put z=0 so dz=0, so I wuold have said that $int_{R}dz=0$ however it doesn't make sense. It should be results =1.@reuns
– Landau
Nov 21 '18 at 17:12
Nothing of what you wrote makes sense. It seems you don't understand integrals. What are you trying to solve
– reuns
Nov 21 '18 at 17:29