How to simplify tensor expression with symbolic coeficients?

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up vote
7
down vote

favorite

I can use Vectors to simplify the following expression:

$Assumptions = (a | b | c) ∈ Vectors[3];
TensorReduce[3 b.a[Cross]c - 3 c.b[Cross]a]

which gives me 0, and correct.

However, if I change the value "3" into a symbol "g", it fails:

$Assumptions = (a | b | c) ∈ Vectors[3];
TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

Or

$Assumptions = (a | b | c) ∈ Vectors[3];
$Assumptions = g ∈ Reals;
TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

Mathematica gives me $g b.atimes c-g c.btimes a$, which should be 0 as well.

How can I do this simplification in Mathematica?

vector tensors

share|improve this question

edited Aug 26 at 6:51

kglr

158k8183382

asked Aug 26 at 6:36

ZHANG Juenjie

449210

add a comment | 

up vote
7
down vote

favorite

I can use Vectors to simplify the following expression:

$Assumptions = (a | b | c) ∈ Vectors[3];
TensorReduce[3 b.a[Cross]c - 3 c.b[Cross]a]

which gives me 0, and correct.

However, if I change the value "3" into a symbol "g", it fails:

$Assumptions = (a | b | c) ∈ Vectors[3];
TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

Or

$Assumptions = (a | b | c) ∈ Vectors[3];
$Assumptions = g ∈ Reals;
TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

Mathematica gives me $g b.atimes c-g c.btimes a$, which should be 0 as well.

How can I do this simplification in Mathematica?

vector tensors

share|improve this question

edited Aug 26 at 6:51

kglr

158k8183382

asked Aug 26 at 6:36

ZHANG Juenjie

449210

add a comment | 

up vote
7
down vote

favorite

up vote
7
down vote

favorite

I can use Vectors to simplify the following expression:

$Assumptions = (a | b | c) ∈ Vectors[3];
TensorReduce[3 b.a[Cross]c - 3 c.b[Cross]a]

which gives me 0, and correct.

However, if I change the value "3" into a symbol "g", it fails:

$Assumptions = (a | b | c) ∈ Vectors[3];
TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

Or

$Assumptions = (a | b | c) ∈ Vectors[3];
$Assumptions = g ∈ Reals;
TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

Mathematica gives me $g b.atimes c-g c.btimes a$, which should be 0 as well.

How can I do this simplification in Mathematica?

vector tensors

share|improve this question

edited Aug 26 at 6:51

kglr

158k8183382

asked Aug 26 at 6:36

ZHANG Juenjie

449210

I can use Vectors to simplify the following expression:

$Assumptions = (a | b | c) ∈ Vectors[3];
TensorReduce[3 b.a[Cross]c - 3 c.b[Cross]a]

which gives me 0, and correct.

However, if I change the value "3" into a symbol "g", it fails:

$Assumptions = (a | b | c) ∈ Vectors[3];
TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

Or

$Assumptions = (a | b | c) ∈ Vectors[3];
$Assumptions = g ∈ Reals;
TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

Mathematica gives me $g b.atimes c-g c.btimes a$, which should be 0 as well.

How can I do this simplification in Mathematica?

vector tensors

share|improve this question

edited Aug 26 at 6:51

kglr

158k8183382

asked Aug 26 at 6:36

ZHANG Juenjie

449210

share|improve this question

share|improve this question

edited Aug 26 at 6:51

kglr

158k8183382

edited Aug 26 at 6:51

kglr

158k8183382

edited Aug 26 at 6:51

kglr

158k8183382

158k8183382

asked Aug 26 at 6:36

ZHANG Juenjie

449210

asked Aug 26 at 6:36

ZHANG Juenjie

449210

asked Aug 26 at 6:36

ZHANG Juenjie

449210

449210

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
7
down vote



accepted

$Assumptions = (a 

TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

0

share|improve this answer

answered Aug 26 at 6:49

kglr

158k8183382

add a comment | 

up vote
5
down vote

$Assumptions=(a|b|c) ∈ Vectors[3] && g ∈ Reals

TensorReduce[g b.a[Cross]c-g c.b[Cross]a]
(*0*)

g doesn't have to be Reals. Complexes works too. It just can't be something totally undefined.

share|improve this answer

answered Aug 26 at 6:54

Bill Watts

1,9581414

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

$Assumptions = (a 

TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

0

share|improve this answer

answered Aug 26 at 6:49

kglr

158k8183382

add a comment | 

up vote
7
down vote



accepted

$Assumptions = (a 

TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

0

share|improve this answer

answered Aug 26 at 6:49

kglr

158k8183382

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

$Assumptions = (a 

TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

0

share|improve this answer

answered Aug 26 at 6:49

kglr

158k8183382

$Assumptions = (a 

TensorReduce[g b.a[Cross]c - g c.b[Cross]a]

0

share|improve this answer

answered Aug 26 at 6:49

kglr

158k8183382

share|improve this answer

share|improve this answer

answered Aug 26 at 6:49

kglr

158k8183382

answered Aug 26 at 6:49

kglr

158k8183382

answered Aug 26 at 6:49

kglr

158k8183382

158k8183382

add a comment | 

add a comment | 

up vote
5
down vote

$Assumptions=(a|b|c) ∈ Vectors[3] && g ∈ Reals

TensorReduce[g b.a[Cross]c-g c.b[Cross]a]
(*0*)

g doesn't have to be Reals. Complexes works too. It just can't be something totally undefined.

share|improve this answer

answered Aug 26 at 6:54

Bill Watts

1,9581414

add a comment | 

up vote
5
down vote

$Assumptions=(a|b|c) ∈ Vectors[3] && g ∈ Reals

TensorReduce[g b.a[Cross]c-g c.b[Cross]a]
(*0*)

g doesn't have to be Reals. Complexes works too. It just can't be something totally undefined.

share|improve this answer

answered Aug 26 at 6:54

Bill Watts

1,9581414

add a comment | 

up vote
5
down vote

up vote
5
down vote

$Assumptions=(a|b|c) ∈ Vectors[3] && g ∈ Reals

TensorReduce[g b.a[Cross]c-g c.b[Cross]a]
(*0*)

g doesn't have to be Reals. Complexes works too. It just can't be something totally undefined.

share|improve this answer

answered Aug 26 at 6:54

Bill Watts

1,9581414

$Assumptions=(a|b|c) ∈ Vectors[3] && g ∈ Reals

TensorReduce[g b.a[Cross]c-g c.b[Cross]a]
(*0*)

g doesn't have to be Reals. Complexes works too. It just can't be something totally undefined.

share|improve this answer

answered Aug 26 at 6:54

Bill Watts

1,9581414

share|improve this answer

share|improve this answer

answered Aug 26 at 6:54

Bill Watts

1,9581414

answered Aug 26 at 6:54

Bill Watts

1,9581414

answered Aug 26 at 6:54

Bill Watts

1,9581414

1,9581414

add a comment | 

add a comment | 

 

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