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How can I automate this tensor computation?
Why doesn't Mathematica make an obvious simplification?Simplify matrix algebraReduce the output from tuples by including symmetry?How to put the tensor product of two operators onto two variables?Symbolic Tensor AlgebraExpression containing $ttI$ is real — how can I show this is so in my notebook?Working with tensor algebraPython's einsum equivalent in Mathematica?Del as a Differential Operator: (Matrix times Del) cross vectorExplicitly construct tensor quantities with given symmetries
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I am doing this work by hand, but it takes a lot of time and I make several calculation errors, so I was thinking to make Mathematica to calculate this for me, but I am stuck at the very beginning.
I am working with tensors like this:
$XXV_ijk = dfrac16(X_iX_jV_k+X_iX_kV_j+X_jX_iV_k+X_kX_jV_i+X_jX_kV_i+X_kX_iV_j)$
$-dfrac15(delta_ij, (Xcdot X) V_k+delta_ik (Xcdot V) X_j+delta_jk (Xcdot V) X_i)$
where $X_i$ and $V_i$ are the components of the 3-vectors $vecX$ and $vecV$. I have to multiply these tensors, for example
$XXV times XXV = dfrac225V^2+dfrac825 (Xcdot V)^2$
I think I can obtain the first part with Tuples and Total(?) but I don't know how to obtain the part with the Kroeneker deltas; if I can write these tensors correctly I think I can multiply these tensors with . and Transpose.
As @yarchik has pointd out, I have to add that my tensors have unit length
simplifying-expressions tensors code-request
edited Jul 19 at 17:49
Carl Woll
88.4k3 gold badges117 silver badges227 bronze badges
asked Jul 19 at 13:14
mattiav27mattiav27
2,3262 gold badges15 silver badges33 bronze badges
$endgroup$
add a comment |
$begingroup$
I am doing this work by hand, but it takes a lot of time and I make several calculation errors, so I was thinking to make Mathematica to calculate this for me, but I am stuck at the very beginning.
I am working with tensors like this:
$XXV_ijk = dfrac16(X_iX_jV_k+X_iX_kV_j+X_jX_iV_k+X_kX_jV_i+X_jX_kV_i+X_kX_iV_j)$
$-dfrac15(delta_ij, (Xcdot X) V_k+delta_ik (Xcdot V) X_j+delta_jk (Xcdot V) X_i)$
where $X_i$ and $V_i$ are the components of the 3-vectors $vecX$ and $vecV$. I have to multiply these tensors, for example
$XXV times XXV = dfrac225V^2+dfrac825 (Xcdot V)^2$
I think I can obtain the first part with Tuples and Total(?) but I don't know how to obtain the part with the Kroeneker deltas; if I can write these tensors correctly I think I can multiply these tensors with . and Transpose.
As @yarchik has pointd out, I have to add that my tensors have unit length
simplifying-expressions tensors code-request
edited Jul 19 at 17:49
Carl Woll
88.4k3 gold badges117 silver badges227 bronze badges
asked Jul 19 at 13:14
mattiav27mattiav27
2,3262 gold badges15 silver badges33 bronze badges
$endgroup$
$begingroup$
Is there a typo in your last term on the first line, $X_kX_iV_k$? There should be a $j$-index somewhere.
$endgroup$
– Roman
Jul 19 at 14:52
$begingroup$
@Roman yes it is a typo
$endgroup$
– mattiav27
Jul 19 at 15:13
$begingroup$
I think your second term still needs a power correction. The answer should be proportional to $V^2$.
$endgroup$
– yarchik
Jul 19 at 15:40
$begingroup$
@yarchik you are right, but this is a typo not a calculation error...
$endgroup$
– mattiav27
Jul 19 at 15:44
$begingroup$
Note that in your example the the correct result should be $frac825 V^2 + frac225 (Xcdot V)^2$. See all four answers below.
$endgroup$
– Shadowray
Jul 20 at 0:15
add a comment |
$begingroup$
I am doing this work by hand, but it takes a lot of time and I make several calculation errors, so I was thinking to make Mathematica to calculate this for me, but I am stuck at the very beginning.
I am working with tensors like this:
$XXV_ijk = dfrac16(X_iX_jV_k+X_iX_kV_j+X_jX_iV_k+X_kX_jV_i+X_jX_kV_i+X_kX_iV_j)$
$-dfrac15(delta_ij, (Xcdot X) V_k+delta_ik (Xcdot V) X_j+delta_jk (Xcdot V) X_i)$
where $X_i$ and $V_i$ are the components of the 3-vectors $vecX$ and $vecV$. I have to multiply these tensors, for example
$XXV times XXV = dfrac225V^2+dfrac825 (Xcdot V)^2$
I think I can obtain the first part with Tuples and Total(?) but I don't know how to obtain the part with the Kroeneker deltas; if I can write these tensors correctly I think I can multiply these tensors with . and Transpose.
As @yarchik has pointd out, I have to add that my tensors have unit length
simplifying-expressions tensors code-request
edited Jul 19 at 17:49
Carl Woll
88.4k3 gold badges117 silver badges227 bronze badges
asked Jul 19 at 13:14
mattiav27mattiav27
2,3262 gold badges15 silver badges33 bronze badges
$endgroup$
I am doing this work by hand, but it takes a lot of time and I make several calculation errors, so I was thinking to make Mathematica to calculate this for me, but I am stuck at the very beginning.
I am working with tensors like this:
$XXV_ijk = dfrac16(X_iX_jV_k+X_iX_kV_j+X_jX_iV_k+X_kX_jV_i+X_jX_kV_i+X_kX_iV_j)$
$-dfrac15(delta_ij, (Xcdot X) V_k+delta_ik (Xcdot V) X_j+delta_jk (Xcdot V) X_i)$
where $X_i$ and $V_i$ are the components of the 3-vectors $vecX$ and $vecV$. I have to multiply these tensors, for example
$XXV times XXV = dfrac225V^2+dfrac825 (Xcdot V)^2$
I think I can obtain the first part with Tuples and Total(?) but I don't know how to obtain the part with the Kroeneker deltas; if I can write these tensors correctly I think I can multiply these tensors with . and Transpose.
As @yarchik has pointd out, I have to add that my tensors have unit length
simplifying-expressions tensors code-request
simplifying-expressions tensors code-request
edited Jul 19 at 17:49
Carl Woll
88.4k3 gold badges117 silver badges227 bronze badges
asked Jul 19 at 13:14
mattiav27mattiav27
2,3262 gold badges15 silver badges33 bronze badges
edited Jul 19 at 17:49
Carl Woll
88.4k3 gold badges117 silver badges227 bronze badges
asked Jul 19 at 13:14
mattiav27mattiav27
2,3262 gold badges15 silver badges33 bronze badges
edited Jul 19 at 17:49
Carl Woll
88.4k3 gold badges117 silver badges227 bronze badges
edited Jul 19 at 17:49
Carl Woll
88.4k3 gold badges117 silver badges227 bronze badges
edited Jul 19 at 17:49
Carl Woll
88.4k3 gold badges117 silver badges227 bronze badges
88.4k3 gold badges117 silver badges227 bronze badges
asked Jul 19 at 13:14
mattiav27mattiav27
2,3262 gold badges15 silver badges33 bronze badges
asked Jul 19 at 13:14
mattiav27mattiav27
2,3262 gold badges15 silver badges33 bronze badges
asked Jul 19 at 13:14
mattiav27mattiav27
2,3262 gold badges15 silver badges33 bronze badges
2,3262 gold badges15 silver badges33 bronze badges
$begingroup$
Is there a typo in your last term on the first line, $X_kX_iV_k$? There should be a $j$-index somewhere.
$endgroup$
– Roman
Jul 19 at 14:52
$begingroup$
@Roman yes it is a typo
$endgroup$
– mattiav27
Jul 19 at 15:13
$begingroup$
I think your second term still needs a power correction. The answer should be proportional to $V^2$.
$endgroup$
– yarchik
Jul 19 at 15:40
$begingroup$
@yarchik you are right, but this is a typo not a calculation error...
$endgroup$
– mattiav27
Jul 19 at 15:44
$begingroup$
Note that in your example the the correct result should be $frac825 V^2 + frac225 (Xcdot V)^2$. See all four answers below.
$endgroup$
– Shadowray
Jul 20 at 0:15
add a comment |
$begingroup$
Is there a typo in your last term on the first line, $X_kX_iV_k$? There should be a $j$-index somewhere.
$endgroup$
– Roman
Jul 19 at 14:52
$begingroup$
@Roman yes it is a typo
$endgroup$
– mattiav27
Jul 19 at 15:13
$begingroup$
I think your second term still needs a power correction. The answer should be proportional to $V^2$.
$endgroup$
– yarchik
Jul 19 at 15:40
$begingroup$
@yarchik you are right, but this is a typo not a calculation error...
$endgroup$
– mattiav27
Jul 19 at 15:44
$begingroup$
Note that in your example the the correct result should be $frac825 V^2 + frac225 (Xcdot V)^2$. See all four answers below.
$endgroup$
– Shadowray
Jul 20 at 0:15
$begingroup$
Is there a typo in your last term on the first line, $X_kX_iV_k$? There should be a $j$-index somewhere.
$endgroup$
– Roman
Jul 19 at 14:52
$begingroup$
Is there a typo in your last term on the first line, $X_kX_iV_k$? There should be a $j$-index somewhere.
$endgroup$
– Roman
Jul 19 at 14:52
$begingroup$
@Roman yes it is a typo
$endgroup$
– mattiav27
Jul 19 at 15:13
$begingroup$
@Roman yes it is a typo
$endgroup$
– mattiav27
Jul 19 at 15:13
$begingroup$
I think your second term still needs a power correction. The answer should be proportional to $V^2$.
$endgroup$
– yarchik
Jul 19 at 15:40
$begingroup$
I think your second term still needs a power correction. The answer should be proportional to $V^2$.
$endgroup$
– yarchik
Jul 19 at 15:40
$begingroup$
@yarchik you are right, but this is a typo not a calculation error...
$endgroup$
– mattiav27
Jul 19 at 15:44
$begingroup$
@yarchik you are right, but this is a typo not a calculation error...
$endgroup$
– mattiav27
Jul 19 at 15:44
$begingroup$
Note that in your example the the correct result should be $frac825 V^2 + frac225 (Xcdot V)^2$. See all four answers below.
$endgroup$
– Shadowray
Jul 20 at 0:15
$begingroup$
Note that in your example the the correct result should be $frac825 V^2 + frac225 (Xcdot V)^2$. See all four answers below.
$endgroup$
– Shadowray
Jul 20 at 0:15
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can write it directly as you see it
xxv[i_,j_,k_]:= 1/6( x[i]x[j]v[k]+x[i]x[k]v[j]
+x[j]x[i]v[k]+x[k]x[j]v[i]
+x[j]x[k]v[i]+x[k]x[i]v[j] )
-1/5( KroneckerDelta[i,j]Sum[x[l]x[l],l,3]v[k]
+KroneckerDelta[i,k]Sum[x[l]v[l],l,3]x[j]
+KroneckerDelta[j,k]Sum[x[l]v[l],l,3]x[i] )
FullSimplify[ Sum[xxv[i,j,k] xxv[i,j,k],i,3,j,3,k,3],
Assumptions->Sum[x[i]^2,i,3]==1
&&Sum[x[i]v[i],i,3]==xv
&&Sum[v[i]v[i],i,3]==vv]
Out[1]= 2/25 (4 vv + xv^2)
where I assumed that your vector x is normalized
answered Jul 19 at 14:21
yarchikyarchik
4,63610 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
You can cast this as a symbolic tensor question, and make use of my TensorSimplify package. Install the paclet with:
PacletInstall[
"TensorSimplify",
"Site" -> "http://raw.githubusercontent.com/carlwoll/TensorSimplify/master"
]
Once installed, load the package with:
<<TensorSimplify`
Now, define your tensor using TensorProduct:
XXV = 1/3 (TensorProduct[X,X,V] + TensorProduct[X,V,X] + TensorProduct[V,X,X]) -
1/5 (X.X TensorProduct[Inactive[IdentityMatrix][3], V] +
X.V TensorTranspose[TensorProduct[Inactive[IdentityMatrix][3],X],1,3,2] +
X.V TensorProduct[X,Inactive[IdentityMatrix][3]]
);
Note the use of Inactive[IdentityMatrix][3] instead of IdentityMatrix[3]. Then:
TensorSimplify[
TensorContract[TensorProduct[XXV, XXV], 1, 4, 2, 5, 3, 6],
Assumptions -> (X|V) ∈ Vectors[3]
]
2/25 (V.X)^2 X.X + 8/25 V.V (X.X)^2
Using X.X == 1 reproduces your result.
answered Jul 19 at 17:46
Carl WollCarl Woll
88.4k3 gold badges117 silver badges227 bronze badges
$endgroup$
add a comment |
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This is how I'd do it; maybe it's useful for you.
Define $vecX$ and $vecV$ as vectors:
X = Array[x, 3];
V = Array[v, 3];
useful $3times3times3$ tensors for assembling:
a = Outer[Times, X, X, V];
b = (X.X) Outer[Times, IdentityMatrix[3], V];
c = (X.V) Outer[Times, IdentityMatrix[3], X];
assemble $XXV$:
XXV = (a + Transpose[a, 3, 1, 2] + Transpose[a, 2, 3, 1])/3 -
(b + Transpose[c, 3, 1, 2] + Transpose[c, 2, 3, 1])/5;
check a formula:
Total[XXV*XXV, 3] == 2/25 (X.X) ((X.V)^2 + 4 (X.X) (V.V)) // FullSimplify
(* True *)
answered Jul 19 at 15:00
RomanRoman
14.9k1 gold badge20 silver badges51 bronze badges
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$begingroup$
Nice that you verified my answer. I already started to doubt.
$endgroup$
– yarchik
Jul 19 at 15:20
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@yarchik this is one of my calculation errors... I have corrected the formula in my post
$endgroup$
– mattiav27
Jul 19 at 15:35
1
$begingroup$
@mattiav27 it's still wrong, the second term should be 8/25 and you need to specify that you're assuming that $vecX$ has unit length.
$endgroup$
– Roman
Jul 19 at 15:57
add a comment |
$begingroup$
You can implement Einstein's summation convention using, for example, temporary variables as summation indices.
ClearAll[delta, CenterDot, dummyIndexQ, tensorSimplify];
SetAttributes[delta, Orderless];
SetAttributes[CenterDot, Orderless];
dummyIndexQ[x_Symbol] := MemberQ[Attributes[x], Temporary];
tensorSimplificationRules =
delta[a_?dummyIndexQ, a_] :> 3,
delta[a_?dummyIndexQ, b_]^2 :> delta[b, b],
delta[a_, b_?dummyIndexQ] delta[b_, c_] :> delta[a, c],
delta[i_?dummyIndexQ, j_]t_[i_] :> t[j],
(t_[i_?dummyIndexQ])^2 :> (t[CenterDot]t),
t1_[x_?dummyIndexQ] t2_[x_] :> t1[CenterDot]t2
;
tensorSimplify[expr_] := FixedPoint[(Expand[#]//.tensorSimplificationRules)&, expr];
Let's define $XXV_ijk$:
xxv[i_, j_, k_] := (1/3 * (v[k] x[i] x[j] + v[j] x[i] x[k] + v[i] x[j] x[k]) - 1/5 * (delta[i,j] (x[CenterDot]x) v[k] + delta[i,k] (x[CenterDot]v) x[j] + delta[j,k] (x[CenterDot]v) x[i]))
Result for your example $XXV_abc XXV_abc$:
expr = Module[a,b,c, xxv[a,b,c] xxv[a,b,c]];
tensorSimplify[expr]
2/25 (v$cdot$x)^2 (x$cdot$x) + 8/25 (v$cdot$v) (x$cdot$x)^2
Result for more complicated input $XXV_abc XXV_bcd XXV_def XXV_efa$:
expr2 = Module[a,b,c,d,e,f, xxv[a,b,c] xxv[b,c,d] xxv[d,e,f] xxv[e,f,a]];
tensorSimplify[expr2]
$frac524 (xcdot x)^2 (vcdot x)^450625+frac1454 vcdot v (xcdot x)^3 (vcdot
x)^250625+frac1586 (vcdot v)^2 (xcdot x)^450625$
Note that all summation indices must be listed inside the first argument of Module.
answered Jul 19 at 23:48
ShadowrayShadowray
5,8871 gold badge9 silver badges34 bronze badges
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can write it directly as you see it
xxv[i_,j_,k_]:= 1/6( x[i]x[j]v[k]+x[i]x[k]v[j]
+x[j]x[i]v[k]+x[k]x[j]v[i]
+x[j]x[k]v[i]+x[k]x[i]v[j] )
-1/5( KroneckerDelta[i,j]Sum[x[l]x[l],l,3]v[k]
+KroneckerDelta[i,k]Sum[x[l]v[l],l,3]x[j]
+KroneckerDelta[j,k]Sum[x[l]v[l],l,3]x[i] )
FullSimplify[ Sum[xxv[i,j,k] xxv[i,j,k],i,3,j,3,k,3],
Assumptions->Sum[x[i]^2,i,3]==1
&&Sum[x[i]v[i],i,3]==xv
&&Sum[v[i]v[i],i,3]==vv]
Out[1]= 2/25 (4 vv + xv^2)
where I assumed that your vector x is normalized
answered Jul 19 at 14:21
yarchikyarchik
4,63610 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
You can write it directly as you see it
xxv[i_,j_,k_]:= 1/6( x[i]x[j]v[k]+x[i]x[k]v[j]
+x[j]x[i]v[k]+x[k]x[j]v[i]
+x[j]x[k]v[i]+x[k]x[i]v[j] )
-1/5( KroneckerDelta[i,j]Sum[x[l]x[l],l,3]v[k]
+KroneckerDelta[i,k]Sum[x[l]v[l],l,3]x[j]
+KroneckerDelta[j,k]Sum[x[l]v[l],l,3]x[i] )
FullSimplify[ Sum[xxv[i,j,k] xxv[i,j,k],i,3,j,3,k,3],
Assumptions->Sum[x[i]^2,i,3]==1
&&Sum[x[i]v[i],i,3]==xv
&&Sum[v[i]v[i],i,3]==vv]
Out[1]= 2/25 (4 vv + xv^2)
where I assumed that your vector x is normalized
answered Jul 19 at 14:21
yarchikyarchik
4,63610 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
You can write it directly as you see it
xxv[i_,j_,k_]:= 1/6( x[i]x[j]v[k]+x[i]x[k]v[j]
+x[j]x[i]v[k]+x[k]x[j]v[i]
+x[j]x[k]v[i]+x[k]x[i]v[j] )
-1/5( KroneckerDelta[i,j]Sum[x[l]x[l],l,3]v[k]
+KroneckerDelta[i,k]Sum[x[l]v[l],l,3]x[j]
+KroneckerDelta[j,k]Sum[x[l]v[l],l,3]x[i] )
FullSimplify[ Sum[xxv[i,j,k] xxv[i,j,k],i,3,j,3,k,3],
Assumptions->Sum[x[i]^2,i,3]==1
&&Sum[x[i]v[i],i,3]==xv
&&Sum[v[i]v[i],i,3]==vv]
Out[1]= 2/25 (4 vv + xv^2)
where I assumed that your vector x is normalized
answered Jul 19 at 14:21
yarchikyarchik
4,63610 silver badges34 bronze badges
$endgroup$
You can write it directly as you see it
xxv[i_,j_,k_]:= 1/6( x[i]x[j]v[k]+x[i]x[k]v[j]
+x[j]x[i]v[k]+x[k]x[j]v[i]
+x[j]x[k]v[i]+x[k]x[i]v[j] )
-1/5( KroneckerDelta[i,j]Sum[x[l]x[l],l,3]v[k]
+KroneckerDelta[i,k]Sum[x[l]v[l],l,3]x[j]
+KroneckerDelta[j,k]Sum[x[l]v[l],l,3]x[i] )
FullSimplify[ Sum[xxv[i,j,k] xxv[i,j,k],i,3,j,3,k,3],
Assumptions->Sum[x[i]^2,i,3]==1
&&Sum[x[i]v[i],i,3]==xv
&&Sum[v[i]v[i],i,3]==vv]
Out[1]= 2/25 (4 vv + xv^2)
where I assumed that your vector x is normalized
answered Jul 19 at 14:21
yarchikyarchik
4,63610 silver badges34 bronze badges
edited Jul 19 at 14:28
answered Jul 19 at 14:21
yarchikyarchik
4,63610 silver badges34 bronze badges
answered Jul 19 at 14:21
yarchikyarchik
4,63610 silver badges34 bronze badges
answered Jul 19 at 14:21
yarchikyarchik
4,63610 silver badges34 bronze badges
4,63610 silver badges34 bronze badges
add a comment |
add a comment |
$begingroup$
You can cast this as a symbolic tensor question, and make use of my TensorSimplify package. Install the paclet with:
PacletInstall[
"TensorSimplify",
"Site" -> "http://raw.githubusercontent.com/carlwoll/TensorSimplify/master"
]
Once installed, load the package with:
<<TensorSimplify`
Now, define your tensor using TensorProduct:
XXV = 1/3 (TensorProduct[X,X,V] + TensorProduct[X,V,X] + TensorProduct[V,X,X]) -
1/5 (X.X TensorProduct[Inactive[IdentityMatrix][3], V] +
X.V TensorTranspose[TensorProduct[Inactive[IdentityMatrix][3],X],1,3,2] +
X.V TensorProduct[X,Inactive[IdentityMatrix][3]]
);
Note the use of Inactive[IdentityMatrix][3] instead of IdentityMatrix[3]. Then:
TensorSimplify[
TensorContract[TensorProduct[XXV, XXV], 1, 4, 2, 5, 3, 6],
Assumptions -> (X|V) ∈ Vectors[3]
]
2/25 (V.X)^2 X.X + 8/25 V.V (X.X)^2
Using X.X == 1 reproduces your result.
answered Jul 19 at 17:46
Carl WollCarl Woll
88.4k3 gold badges117 silver badges227 bronze badges
$endgroup$
add a comment |
$begingroup$
You can cast this as a symbolic tensor question, and make use of my TensorSimplify package. Install the paclet with:
PacletInstall[
"TensorSimplify",
"Site" -> "http://raw.githubusercontent.com/carlwoll/TensorSimplify/master"
]
Once installed, load the package with:
<<TensorSimplify`
Now, define your tensor using TensorProduct:
XXV = 1/3 (TensorProduct[X,X,V] + TensorProduct[X,V,X] + TensorProduct[V,X,X]) -
1/5 (X.X TensorProduct[Inactive[IdentityMatrix][3], V] +
X.V TensorTranspose[TensorProduct[Inactive[IdentityMatrix][3],X],1,3,2] +
X.V TensorProduct[X,Inactive[IdentityMatrix][3]]
);
Note the use of Inactive[IdentityMatrix][3] instead of IdentityMatrix[3]. Then:
TensorSimplify[
TensorContract[TensorProduct[XXV, XXV], 1, 4, 2, 5, 3, 6],
Assumptions -> (X|V) ∈ Vectors[3]
]
2/25 (V.X)^2 X.X + 8/25 V.V (X.X)^2
Using X.X == 1 reproduces your result.
answered Jul 19 at 17:46
Carl WollCarl Woll
88.4k3 gold badges117 silver badges227 bronze badges
$endgroup$
add a comment |
$begingroup$
You can cast this as a symbolic tensor question, and make use of my TensorSimplify package. Install the paclet with:
PacletInstall[
"TensorSimplify",
"Site" -> "http://raw.githubusercontent.com/carlwoll/TensorSimplify/master"
]
Once installed, load the package with:
<<TensorSimplify`
Now, define your tensor using TensorProduct:
XXV = 1/3 (TensorProduct[X,X,V] + TensorProduct[X,V,X] + TensorProduct[V,X,X]) -
1/5 (X.X TensorProduct[Inactive[IdentityMatrix][3], V] +
X.V TensorTranspose[TensorProduct[Inactive[IdentityMatrix][3],X],1,3,2] +
X.V TensorProduct[X,Inactive[IdentityMatrix][3]]
);
Note the use of Inactive[IdentityMatrix][3] instead of IdentityMatrix[3]. Then:
TensorSimplify[
TensorContract[TensorProduct[XXV, XXV], 1, 4, 2, 5, 3, 6],
Assumptions -> (X|V) ∈ Vectors[3]
]
2/25 (V.X)^2 X.X + 8/25 V.V (X.X)^2
Using X.X == 1 reproduces your result.
answered Jul 19 at 17:46
Carl WollCarl Woll
88.4k3 gold badges117 silver badges227 bronze badges
$endgroup$
You can cast this as a symbolic tensor question, and make use of my TensorSimplify package. Install the paclet with:
PacletInstall[
"TensorSimplify",
"Site" -> "http://raw.githubusercontent.com/carlwoll/TensorSimplify/master"
]
Once installed, load the package with:
<<TensorSimplify`
Now, define your tensor using TensorProduct:
XXV = 1/3 (TensorProduct[X,X,V] + TensorProduct[X,V,X] + TensorProduct[V,X,X]) -
1/5 (X.X TensorProduct[Inactive[IdentityMatrix][3], V] +
X.V TensorTranspose[TensorProduct[Inactive[IdentityMatrix][3],X],1,3,2] +
X.V TensorProduct[X,Inactive[IdentityMatrix][3]]
);
Note the use of Inactive[IdentityMatrix][3] instead of IdentityMatrix[3]. Then:
TensorSimplify[
TensorContract[TensorProduct[XXV, XXV], 1, 4, 2, 5, 3, 6],
Assumptions -> (X|V) ∈ Vectors[3]
]
2/25 (V.X)^2 X.X + 8/25 V.V (X.X)^2
Using X.X == 1 reproduces your result.
answered Jul 19 at 17:46
Carl WollCarl Woll
88.4k3 gold badges117 silver badges227 bronze badges
answered Jul 19 at 17:46
Carl WollCarl Woll
88.4k3 gold badges117 silver badges227 bronze badges
answered Jul 19 at 17:46
Carl WollCarl Woll
88.4k3 gold badges117 silver badges227 bronze badges
answered Jul 19 at 17:46
Carl WollCarl Woll
88.4k3 gold badges117 silver badges227 bronze badges
88.4k3 gold badges117 silver badges227 bronze badges
add a comment |
add a comment |
$begingroup$
This is how I'd do it; maybe it's useful for you.
Define $vecX$ and $vecV$ as vectors:
X = Array[x, 3];
V = Array[v, 3];
useful $3times3times3$ tensors for assembling:
a = Outer[Times, X, X, V];
b = (X.X) Outer[Times, IdentityMatrix[3], V];
c = (X.V) Outer[Times, IdentityMatrix[3], X];
assemble $XXV$:
XXV = (a + Transpose[a, 3, 1, 2] + Transpose[a, 2, 3, 1])/3 -
(b + Transpose[c, 3, 1, 2] + Transpose[c, 2, 3, 1])/5;
check a formula:
Total[XXV*XXV, 3] == 2/25 (X.X) ((X.V)^2 + 4 (X.X) (V.V)) // FullSimplify
(* True *)
answered Jul 19 at 15:00
RomanRoman
14.9k1 gold badge20 silver badges51 bronze badges
$endgroup$
$begingroup$
Nice that you verified my answer. I already started to doubt.
$endgroup$
– yarchik
Jul 19 at 15:20
$begingroup$
@yarchik this is one of my calculation errors... I have corrected the formula in my post
$endgroup$
– mattiav27
Jul 19 at 15:35
1
$begingroup$
@mattiav27 it's still wrong, the second term should be 8/25 and you need to specify that you're assuming that $vecX$ has unit length.
$endgroup$
– Roman
Jul 19 at 15:57
add a comment |
$begingroup$
This is how I'd do it; maybe it's useful for you.
Define $vecX$ and $vecV$ as vectors:
X = Array[x, 3];
V = Array[v, 3];
useful $3times3times3$ tensors for assembling:
a = Outer[Times, X, X, V];
b = (X.X) Outer[Times, IdentityMatrix[3], V];
c = (X.V) Outer[Times, IdentityMatrix[3], X];
assemble $XXV$:
XXV = (a + Transpose[a, 3, 1, 2] + Transpose[a, 2, 3, 1])/3 -
(b + Transpose[c, 3, 1, 2] + Transpose[c, 2, 3, 1])/5;
check a formula:
Total[XXV*XXV, 3] == 2/25 (X.X) ((X.V)^2 + 4 (X.X) (V.V)) // FullSimplify
(* True *)
answered Jul 19 at 15:00
RomanRoman
14.9k1 gold badge20 silver badges51 bronze badges
$endgroup$
$begingroup$
Nice that you verified my answer. I already started to doubt.
$endgroup$
– yarchik
Jul 19 at 15:20
$begingroup$
@yarchik this is one of my calculation errors... I have corrected the formula in my post
$endgroup$
– mattiav27
Jul 19 at 15:35
1
$begingroup$
@mattiav27 it's still wrong, the second term should be 8/25 and you need to specify that you're assuming that $vecX$ has unit length.
$endgroup$
– Roman
Jul 19 at 15:57
add a comment |
$begingroup$
This is how I'd do it; maybe it's useful for you.
Define $vecX$ and $vecV$ as vectors:
X = Array[x, 3];
V = Array[v, 3];
useful $3times3times3$ tensors for assembling:
a = Outer[Times, X, X, V];
b = (X.X) Outer[Times, IdentityMatrix[3], V];
c = (X.V) Outer[Times, IdentityMatrix[3], X];
assemble $XXV$:
XXV = (a + Transpose[a, 3, 1, 2] + Transpose[a, 2, 3, 1])/3 -
(b + Transpose[c, 3, 1, 2] + Transpose[c, 2, 3, 1])/5;
check a formula:
Total[XXV*XXV, 3] == 2/25 (X.X) ((X.V)^2 + 4 (X.X) (V.V)) // FullSimplify
(* True *)
answered Jul 19 at 15:00
RomanRoman
14.9k1 gold badge20 silver badges51 bronze badges
$endgroup$
This is how I'd do it; maybe it's useful for you.
Define $vecX$ and $vecV$ as vectors:
X = Array[x, 3];
V = Array[v, 3];
useful $3times3times3$ tensors for assembling:
a = Outer[Times, X, X, V];
b = (X.X) Outer[Times, IdentityMatrix[3], V];
c = (X.V) Outer[Times, IdentityMatrix[3], X];
assemble $XXV$:
XXV = (a + Transpose[a, 3, 1, 2] + Transpose[a, 2, 3, 1])/3 -
(b + Transpose[c, 3, 1, 2] + Transpose[c, 2, 3, 1])/5;
check a formula:
Total[XXV*XXV, 3] == 2/25 (X.X) ((X.V)^2 + 4 (X.X) (V.V)) // FullSimplify
(* True *)
answered Jul 19 at 15:00
RomanRoman
14.9k1 gold badge20 silver badges51 bronze badges
answered Jul 19 at 15:00
RomanRoman
14.9k1 gold badge20 silver badges51 bronze badges
answered Jul 19 at 15:00
RomanRoman
14.9k1 gold badge20 silver badges51 bronze badges
answered Jul 19 at 15:00
RomanRoman
14.9k1 gold badge20 silver badges51 bronze badges
14.9k1 gold badge20 silver badges51 bronze badges
$begingroup$
Nice that you verified my answer. I already started to doubt.
$endgroup$
– yarchik
Jul 19 at 15:20
$begingroup$
@yarchik this is one of my calculation errors... I have corrected the formula in my post
$endgroup$
– mattiav27
Jul 19 at 15:35
1
$begingroup$
@mattiav27 it's still wrong, the second term should be 8/25 and you need to specify that you're assuming that $vecX$ has unit length.
$endgroup$
– Roman
Jul 19 at 15:57
add a comment |
$begingroup$
Nice that you verified my answer. I already started to doubt.
$endgroup$
– yarchik
Jul 19 at 15:20
$begingroup$
@yarchik this is one of my calculation errors... I have corrected the formula in my post
$endgroup$
– mattiav27
Jul 19 at 15:35
1
$begingroup$
@mattiav27 it's still wrong, the second term should be 8/25 and you need to specify that you're assuming that $vecX$ has unit length.
$endgroup$
– Roman
Jul 19 at 15:57
$begingroup$
Nice that you verified my answer. I already started to doubt.
$endgroup$
– yarchik
Jul 19 at 15:20
$begingroup$
Nice that you verified my answer. I already started to doubt.
$endgroup$
– yarchik
Jul 19 at 15:20
$begingroup$
@yarchik this is one of my calculation errors... I have corrected the formula in my post
$endgroup$
– mattiav27
Jul 19 at 15:35
$begingroup$
@yarchik this is one of my calculation errors... I have corrected the formula in my post
$endgroup$
– mattiav27
Jul 19 at 15:35
1
1
$begingroup$
@mattiav27 it's still wrong, the second term should be 8/25 and you need to specify that you're assuming that $vecX$ has unit length.
$endgroup$
– Roman
Jul 19 at 15:57
$begingroup$
@mattiav27 it's still wrong, the second term should be 8/25 and you need to specify that you're assuming that $vecX$ has unit length.
$endgroup$
– Roman
Jul 19 at 15:57
add a comment |
$begingroup$
You can implement Einstein's summation convention using, for example, temporary variables as summation indices.
ClearAll[delta, CenterDot, dummyIndexQ, tensorSimplify];
SetAttributes[delta, Orderless];
SetAttributes[CenterDot, Orderless];
dummyIndexQ[x_Symbol] := MemberQ[Attributes[x], Temporary];
tensorSimplificationRules =
delta[a_?dummyIndexQ, a_] :> 3,
delta[a_?dummyIndexQ, b_]^2 :> delta[b, b],
delta[a_, b_?dummyIndexQ] delta[b_, c_] :> delta[a, c],
delta[i_?dummyIndexQ, j_]t_[i_] :> t[j],
(t_[i_?dummyIndexQ])^2 :> (t[CenterDot]t),
t1_[x_?dummyIndexQ] t2_[x_] :> t1[CenterDot]t2
;
tensorSimplify[expr_] := FixedPoint[(Expand[#]//.tensorSimplificationRules)&, expr];
Let's define $XXV_ijk$:
xxv[i_, j_, k_] := (1/3 * (v[k] x[i] x[j] + v[j] x[i] x[k] + v[i] x[j] x[k]) - 1/5 * (delta[i,j] (x[CenterDot]x) v[k] + delta[i,k] (x[CenterDot]v) x[j] + delta[j,k] (x[CenterDot]v) x[i]))
Result for your example $XXV_abc XXV_abc$:
expr = Module[a,b,c, xxv[a,b,c] xxv[a,b,c]];
tensorSimplify[expr]
2/25 (v$cdot$x)^2 (x$cdot$x) + 8/25 (v$cdot$v) (x$cdot$x)^2
Result for more complicated input $XXV_abc XXV_bcd XXV_def XXV_efa$:
expr2 = Module[a,b,c,d,e,f, xxv[a,b,c] xxv[b,c,d] xxv[d,e,f] xxv[e,f,a]];
tensorSimplify[expr2]
$frac524 (xcdot x)^2 (vcdot x)^450625+frac1454 vcdot v (xcdot x)^3 (vcdot
x)^250625+frac1586 (vcdot v)^2 (xcdot x)^450625$
Note that all summation indices must be listed inside the first argument of Module.
answered Jul 19 at 23:48
ShadowrayShadowray
5,8871 gold badge9 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
You can implement Einstein's summation convention using, for example, temporary variables as summation indices.
ClearAll[delta, CenterDot, dummyIndexQ, tensorSimplify];
SetAttributes[delta, Orderless];
SetAttributes[CenterDot, Orderless];
dummyIndexQ[x_Symbol] := MemberQ[Attributes[x], Temporary];
tensorSimplificationRules =
delta[a_?dummyIndexQ, a_] :> 3,
delta[a_?dummyIndexQ, b_]^2 :> delta[b, b],
delta[a_, b_?dummyIndexQ] delta[b_, c_] :> delta[a, c],
delta[i_?dummyIndexQ, j_]t_[i_] :> t[j],
(t_[i_?dummyIndexQ])^2 :> (t[CenterDot]t),
t1_[x_?dummyIndexQ] t2_[x_] :> t1[CenterDot]t2
;
tensorSimplify[expr_] := FixedPoint[(Expand[#]//.tensorSimplificationRules)&, expr];
Let's define $XXV_ijk$:
xxv[i_, j_, k_] := (1/3 * (v[k] x[i] x[j] + v[j] x[i] x[k] + v[i] x[j] x[k]) - 1/5 * (delta[i,j] (x[CenterDot]x) v[k] + delta[i,k] (x[CenterDot]v) x[j] + delta[j,k] (x[CenterDot]v) x[i]))
Result for your example $XXV_abc XXV_abc$:
expr = Module[a,b,c, xxv[a,b,c] xxv[a,b,c]];
tensorSimplify[expr]
2/25 (v$cdot$x)^2 (x$cdot$x) + 8/25 (v$cdot$v) (x$cdot$x)^2
Result for more complicated input $XXV_abc XXV_bcd XXV_def XXV_efa$:
expr2 = Module[a,b,c,d,e,f, xxv[a,b,c] xxv[b,c,d] xxv[d,e,f] xxv[e,f,a]];
tensorSimplify[expr2]
$frac524 (xcdot x)^2 (vcdot x)^450625+frac1454 vcdot v (xcdot x)^3 (vcdot
x)^250625+frac1586 (vcdot v)^2 (xcdot x)^450625$
Note that all summation indices must be listed inside the first argument of Module.
answered Jul 19 at 23:48
ShadowrayShadowray
5,8871 gold badge9 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
You can implement Einstein's summation convention using, for example, temporary variables as summation indices.
ClearAll[delta, CenterDot, dummyIndexQ, tensorSimplify];
SetAttributes[delta, Orderless];
SetAttributes[CenterDot, Orderless];
dummyIndexQ[x_Symbol] := MemberQ[Attributes[x], Temporary];
tensorSimplificationRules =
delta[a_?dummyIndexQ, a_] :> 3,
delta[a_?dummyIndexQ, b_]^2 :> delta[b, b],
delta[a_, b_?dummyIndexQ] delta[b_, c_] :> delta[a, c],
delta[i_?dummyIndexQ, j_]t_[i_] :> t[j],
(t_[i_?dummyIndexQ])^2 :> (t[CenterDot]t),
t1_[x_?dummyIndexQ] t2_[x_] :> t1[CenterDot]t2
;
tensorSimplify[expr_] := FixedPoint[(Expand[#]//.tensorSimplificationRules)&, expr];
Let's define $XXV_ijk$:
xxv[i_, j_, k_] := (1/3 * (v[k] x[i] x[j] + v[j] x[i] x[k] + v[i] x[j] x[k]) - 1/5 * (delta[i,j] (x[CenterDot]x) v[k] + delta[i,k] (x[CenterDot]v) x[j] + delta[j,k] (x[CenterDot]v) x[i]))
Result for your example $XXV_abc XXV_abc$:
expr = Module[a,b,c, xxv[a,b,c] xxv[a,b,c]];
tensorSimplify[expr]
2/25 (v$cdot$x)^2 (x$cdot$x) + 8/25 (v$cdot$v) (x$cdot$x)^2
Result for more complicated input $XXV_abc XXV_bcd XXV_def XXV_efa$:
expr2 = Module[a,b,c,d,e,f, xxv[a,b,c] xxv[b,c,d] xxv[d,e,f] xxv[e,f,a]];
tensorSimplify[expr2]
$frac524 (xcdot x)^2 (vcdot x)^450625+frac1454 vcdot v (xcdot x)^3 (vcdot
x)^250625+frac1586 (vcdot v)^2 (xcdot x)^450625$
Note that all summation indices must be listed inside the first argument of Module.
answered Jul 19 at 23:48
ShadowrayShadowray
5,8871 gold badge9 silver badges34 bronze badges
$endgroup$
You can implement Einstein's summation convention using, for example, temporary variables as summation indices.
ClearAll[delta, CenterDot, dummyIndexQ, tensorSimplify];
SetAttributes[delta, Orderless];
SetAttributes[CenterDot, Orderless];
dummyIndexQ[x_Symbol] := MemberQ[Attributes[x], Temporary];
tensorSimplificationRules =
delta[a_?dummyIndexQ, a_] :> 3,
delta[a_?dummyIndexQ, b_]^2 :> delta[b, b],
delta[a_, b_?dummyIndexQ] delta[b_, c_] :> delta[a, c],
delta[i_?dummyIndexQ, j_]t_[i_] :> t[j],
(t_[i_?dummyIndexQ])^2 :> (t[CenterDot]t),
t1_[x_?dummyIndexQ] t2_[x_] :> t1[CenterDot]t2
;
tensorSimplify[expr_] := FixedPoint[(Expand[#]//.tensorSimplificationRules)&, expr];
Let's define $XXV_ijk$:
xxv[i_, j_, k_] := (1/3 * (v[k] x[i] x[j] + v[j] x[i] x[k] + v[i] x[j] x[k]) - 1/5 * (delta[i,j] (x[CenterDot]x) v[k] + delta[i,k] (x[CenterDot]v) x[j] + delta[j,k] (x[CenterDot]v) x[i]))
Result for your example $XXV_abc XXV_abc$:
expr = Module[a,b,c, xxv[a,b,c] xxv[a,b,c]];
tensorSimplify[expr]
2/25 (v$cdot$x)^2 (x$cdot$x) + 8/25 (v$cdot$v) (x$cdot$x)^2
Result for more complicated input $XXV_abc XXV_bcd XXV_def XXV_efa$:
expr2 = Module[a,b,c,d,e,f, xxv[a,b,c] xxv[b,c,d] xxv[d,e,f] xxv[e,f,a]];
tensorSimplify[expr2]
$frac524 (xcdot x)^2 (vcdot x)^450625+frac1454 vcdot v (xcdot x)^3 (vcdot
x)^250625+frac1586 (vcdot v)^2 (xcdot x)^450625$
Note that all summation indices must be listed inside the first argument of Module.
answered Jul 19 at 23:48
ShadowrayShadowray
5,8871 gold badge9 silver badges34 bronze badges
answered Jul 19 at 23:48
ShadowrayShadowray
5,8871 gold badge9 silver badges34 bronze badges
answered Jul 19 at 23:48
ShadowrayShadowray
5,8871 gold badge9 silver badges34 bronze badges
answered Jul 19 at 23:48
ShadowrayShadowray
5,8871 gold badge9 silver badges34 bronze badges
5,8871 gold badge9 silver badges34 bronze badges
add a comment |
add a comment |
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$begingroup$
Is there a typo in your last term on the first line, $X_kX_iV_k$? There should be a $j$-index somewhere.
$endgroup$
– Roman
Jul 19 at 14:52
$begingroup$
@Roman yes it is a typo
$endgroup$
– mattiav27
Jul 19 at 15:13
$begingroup$
I think your second term still needs a power correction. The answer should be proportional to $V^2$.
$endgroup$
– yarchik
Jul 19 at 15:40
$begingroup$
@yarchik you are right, but this is a typo not a calculation error...
$endgroup$
– mattiav27
Jul 19 at 15:44
$begingroup$
Note that in your example the the correct result should be $frac825 V^2 + frac225 (Xcdot V)^2$. See all four answers below.
$endgroup$
– Shadowray
Jul 20 at 0:15