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How can I calculate a vector derivative (i.e. divergence, gradient, curl) of interpolated data? For sample data, you can use:

f[x_, y_, z_] := Exp[I z] {1, 0, 0} testdata=Flatten[Table[N@{x,y,z,f[x,y,z]},{x,0,4 Pi,Pi/10},{y,0,4 Pi,Pi/10},{z,0,4 Pi,Pi/10}],2]; intf = Interpolation[testdata] 

I know that for 1D data, like

dim1 = Table[N@{z, f[0, 0, z]}, {z, 0, 2 Pi, Pi/10}]; int1 = Interpolation@dim1; 

you can do D[int1[x],x]. However, I can't seem to get convince MMA that the interpolated function intf actually returns a "vector" quantity (i.e. Length@intf[x,y,z]->3), so that I can do something like:

curl = {D[#[[3]],y]-D[#[[2]],z],-(D[#[[3]],x]-D[#[[1]],z]),D[#[[2]],x]-D[#[[1]],y]}&; curl@intf[x, y, z] (* Out[] := {0,0,0} *) 

I've posted my best attempt in an answer below, but I'm wondering what other solution approaches there are.

How can I calculate a vector derivative (i.e. divergence, gradient, curl) of interpolated data? For sample data, you can use:

f[x_, y_, z_] := Exp[I z] {1, 0, 0} testdata=Flatten[Table[N@{x,y,z,f[x,y,z]},{x,0,4 Pi,Pi/10},{y,0,4 Pi,Pi/10},{z,0,4 Pi,Pi/10}],2]; intf = Interpolation[testdata] 

I know that for 1D data, like

dim1 = Table[N@{z, f[0, 0, z]}, {z, 0, 2 Pi, Pi/10}]; int1 = Interpolation@dim1; 

you can do D[int1[x],x]. However, I can't seem to get convince MMA that the interpolated function intf actually returns a "vector" quantity (i.e. Length@intf[x,y,z]->3), so that I can do something like:

curl = {D[#[[3]],y]-D[#[[2]],z],-(D[#[[3]],x]-D[#[[1]],z]),D[#[[2]],x]-D[#[[1]],y]}&; curl@intf[x, y, z] (* Out[] := {0,0,0} *) 

Similarly, this should work for gradient:

grad = {D[#[[1]], x], D[#[[2]], y], D[#[[3]], z]} &; 

and divergence:

div = (D[#[[1]], x] + D[#[[2]], y] + D[#[[3]], z])&; 

I've posted my best attempt in an answer below, but I'm wondering what other solution approaches there are.

How can I calculate a vector derivative (i.e. divergence, gradient, curl) of interpolated data? For sample data, you can use:

f[x_, y_, z_] := Exp[I z] {1, 0, 0} testdata=Flatten[Table[N@{x,y,z,f[x,y,z]},{x,0,4 Pi,Pi/10},{y,0,4 Pi,Pi/10},{z,0,4 Pi,Pi/10}],2]; intf = Interpolation[testdata] 

I know that for 1D data, like

dim1 = Table[N@{z, f[0, 0, z]}, {z, 0, 2 Pi, Pi/10}]; int1 = Interpolation@dim1; 

you can do D[int1[x],x]. However, I can't seem to get convince MMA that the interpolated function intf actually returns a "vector" quantity (i.e. Length@intf[x,y,z]->3), so that I can do something like:

curl = {D[#[[3]],y]-D[#[[2]],z],-(D[#[[3]],x]-D[#[[1]],z]),D[#[[2]],x]-D[#[[1]],z]}&; curl@intf[x, y, z] (* Out[] := {0,0,0} *) 

I've posted my best attempt in an answer below, but I'm wondering what other solution approaches there are.

How can I calculate a vector derivative (i.e. divergence, gradient, curl) of interpolated data? For sample data, you can use:

f[x_, y_, z_] := Exp[I z] {1, 0, 0} testdata=Flatten[Table[N@{x,y,z,f[x,y,z]},{x,0,4 Pi,Pi/10},{y,0,4 Pi,Pi/10},{z,0,4 Pi,Pi/10}],2]; intf = Interpolation[testdata] 

I know that for 1D data, like

dim1 = Table[N@{z, f[0, 0, z]}, {z, 0, 2 Pi, Pi/10}]; int1 = Interpolation@dim1; 

you can do D[int1[x],x]. However, I can't seem to get convince MMA that the interpolated function intf actually returns a "vector" quantity (i.e. Length@intf[x,y,z]->3), so that I can do something like:

curl = {D[#[[3]],y]-D[#[[2]],z],-(D[#[[3]],x]-D[#[[1]],z]),D[#[[2]],x]-D[#[[1]],y]}&; curl@intf[x, y, z] (* Out[] := {0,0,0} *) 

I've posted my best attempt in an answer below, but I'm wondering what other solution approaches there are.

How can I calculate vector derivatives (divergence, gradient, curl) of interpolated data? For sample data, you can use:

f[x_, y_, z_] := Exp[I z] {1, 0, 0} testdata=Flatten[Table[N@{x,y,z,f[x,y,z]},{x,0,4 Pi,Pi/10},{y,0,4 Pi,Pi/10},{z,0,4 Pi,Pi/10}],2]; intf = Interpolation[testdata] 

I know that for 1D data, like

dim1 = Table[N@{z, f[0, 0, z]}, {z, 0, 2 Pi, Pi/10}]; int1 = Interpolation@dim1; 

you can do D[int1[x],x]. However, I can't seem to get convince MMA that the interpolated function intf actually returns a "vector" quantity (i.e. Length@intf[x,y,z]->3), so that I can do something like:

curl = {D[#[[3]],y]-D[#[[2]],z],-(D[#[[3]],x]-D[#[[1]],z]),D[#[[2]],x]-D[#[[1]],z]}&; curl@intf[x, y, z] (* Out[] := {0,0,0} *) 

I've posted my best attempt in an answer below, but I'm wondering what other solution approaches there are.

How can I calculate a vector derivative (i.e. divergence, gradient, curl) of interpolated data? For sample data, you can use:

f[x_, y_, z_] := Exp[I z] {1, 0, 0} testdata=Flatten[Table[N@{x,y,z,f[x,y,z]},{x,0,4 Pi,Pi/10},{y,0,4 Pi,Pi/10},{z,0,4 Pi,Pi/10}],2]; intf = Interpolation[testdata] 

I know that for 1D data, like

dim1 = Table[N@{z, f[0, 0, z]}, {z, 0, 2 Pi, Pi/10}]; int1 = Interpolation@dim1; 

you can do D[int1[x],x]. However, I can't seem to get convince MMA that the interpolated function intf actually returns a "vector" quantity (i.e. Length@intf[x,y,z]->3), so that I can do something like:

curl = {D[#[[3]],y]-D[#[[2]],z],-(D[#[[3]],x]-D[#[[1]],z]),D[#[[2]],x]-D[#[[1]],z]}&; curl@intf[x, y, z] (* Out[] := {0,0,0} *) 

I've posted my best attempt in an answer below, but I'm wondering what other solution approaches there are.