Divergence theorem with rank 2 tensor.

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$int_Vpartial_j T_ijdV = int_partial VT_ijdS_j $ is the divergence theorem for a second rank tensor. I need to show that this is true.I tried to mimic the proof for "the normal" divergence theorem but couldn't succed:

$int_Vvec∇cdot vecFdV =int_partial VvecFcdot dvecS$,

$V=[(x,y,z)|(x,y) in D, f(x,y)<z<g(x,y)]$,

$int_V(partial_x F_x+partial_y F_y+partial_z F_z)dxdydz$ this is where it stopps. How does one interpet this to a second rank tensor?

Does any one know a way to tackle this problem? Thank you for answers.

multivariable-calculus tensors

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asked Jul 19 at 14:57

123

515

add a comment | 

up vote
0
down vote

favorite

$int_Vpartial_j T_ijdV = int_partial VT_ijdS_j $ is the divergence theorem for a second rank tensor. I need to show that this is true.I tried to mimic the proof for "the normal" divergence theorem but couldn't succed:

$int_Vvec∇cdot vecFdV =int_partial VvecFcdot dvecS$,

$V=[(x,y,z)|(x,y) in D, f(x,y)<z<g(x,y)]$,

$int_V(partial_x F_x+partial_y F_y+partial_z F_z)dxdydz$ this is where it stopps. How does one interpet this to a second rank tensor?

Does any one know a way to tackle this problem? Thank you for answers.

multivariable-calculus tensors

share|cite|improve this question

asked Jul 19 at 14:57

123

515

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

$int_Vpartial_j T_ijdV = int_partial VT_ijdS_j $ is the divergence theorem for a second rank tensor. I need to show that this is true.I tried to mimic the proof for "the normal" divergence theorem but couldn't succed:

$int_Vvec∇cdot vecFdV =int_partial VvecFcdot dvecS$,

$V=[(x,y,z)|(x,y) in D, f(x,y)<z<g(x,y)]$,

$int_V(partial_x F_x+partial_y F_y+partial_z F_z)dxdydz$ this is where it stopps. How does one interpet this to a second rank tensor?

Does any one know a way to tackle this problem? Thank you for answers.

multivariable-calculus tensors

share|cite|improve this question

asked Jul 19 at 14:57

123

515

$int_Vpartial_j T_ijdV = int_partial VT_ijdS_j $ is the divergence theorem for a second rank tensor. I need to show that this is true.I tried to mimic the proof for "the normal" divergence theorem but couldn't succed:

$int_Vvec∇cdot vecFdV =int_partial VvecFcdot dvecS$,

$V=[(x,y,z)|(x,y) in D, f(x,y)<z<g(x,y)]$,

$int_V(partial_x F_x+partial_y F_y+partial_z F_z)dxdydz$ this is where it stopps. How does one interpet this to a second rank tensor?

Does any one know a way to tackle this problem? Thank you for answers.

multivariable-calculus tensors

share|cite|improve this question

asked Jul 19 at 14:57

123

515

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 14:57

123

515

asked Jul 19 at 14:57

123

515

asked Jul 19 at 14:57

123

515

515

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

First note that $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ can be written as
$int_Omega partial_i F_i , dV = oint_partialOmega F_i dS_i.$

Now let $vec A = (A_i)$ be a constant vector field. Then
$$
A_i int_Omega partial_j T_ij , dV
= int_Omega partial_j (A_i T_ij) , dV
= oint_partial Omega A_i T_ij , dS_j
= A_i oint_partial Omega T_ij , dS_j.
$$
This is valid for all constant $vec A,$ e.g. for $vec A = e_1, e_2, e_3$ which gives that
$$int_Omega partial_j T_ij , dV = oint_partialOmega T_ij , dS_j$$
for $i=1,2,3,$ i.e. the formula is valid for all $i.$

share|cite|improve this answer

answered Jul 19 at 19:38

md2perpe

5,93011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for answer, but I don't really understand the argument. How come you make them $int_Omega partial_j (A_i T_ij) , dV = oint_partial Omega A_i T_ij , dS_j$ in the first place, and why would it matter if it is valid for all constant vectors?
  – 123
  Jul 19 at 20:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  That equality is just $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ for $F_j = A_i T_ij,$ and you know from before that this is true.
  – md2perpe
  Jul 19 at 20:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $A_i u_i = A_i v_i$ for all constant $A_i$ then $u_i = v_i$ (for all $i$). This is because you can take $vec A = e_j,$ the unit vector field in the $j$th coordinate direction. But since $A_i = (e_j)_i = delta_ij$ (Kronecker delta), you then get $u_j = v_j$ for that $j$. But since you can to that for any $j$ you get $u_j = v_j$ for all $j$.
  – md2perpe
  Jul 19 at 20:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, the product $A_i T_ij$ is a rank 1 tensor on which we can apply our already known theorem/formula.
  – md2perpe
  Jul 19 at 20:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Very informally you might say so, but if you want to be mathematically strict you should not use that wording.
  – md2perpe
  Jul 19 at 20:59
  

  
  
  
  

 | 
show 4 more comments

Your Answer

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

First note that $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ can be written as
$int_Omega partial_i F_i , dV = oint_partialOmega F_i dS_i.$

Now let $vec A = (A_i)$ be a constant vector field. Then
$$
A_i int_Omega partial_j T_ij , dV
= int_Omega partial_j (A_i T_ij) , dV
= oint_partial Omega A_i T_ij , dS_j
= A_i oint_partial Omega T_ij , dS_j.
$$
This is valid for all constant $vec A,$ e.g. for $vec A = e_1, e_2, e_3$ which gives that
$$int_Omega partial_j T_ij , dV = oint_partialOmega T_ij , dS_j$$
for $i=1,2,3,$ i.e. the formula is valid for all $i.$

share|cite|improve this answer

answered Jul 19 at 19:38

md2perpe

5,93011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for answer, but I don't really understand the argument. How come you make them $int_Omega partial_j (A_i T_ij) , dV = oint_partial Omega A_i T_ij , dS_j$ in the first place, and why would it matter if it is valid for all constant vectors?
  – 123
  Jul 19 at 20:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  That equality is just $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ for $F_j = A_i T_ij,$ and you know from before that this is true.
  – md2perpe
  Jul 19 at 20:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $A_i u_i = A_i v_i$ for all constant $A_i$ then $u_i = v_i$ (for all $i$). This is because you can take $vec A = e_j,$ the unit vector field in the $j$th coordinate direction. But since $A_i = (e_j)_i = delta_ij$ (Kronecker delta), you then get $u_j = v_j$ for that $j$. But since you can to that for any $j$ you get $u_j = v_j$ for all $j$.
  – md2perpe
  Jul 19 at 20:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, the product $A_i T_ij$ is a rank 1 tensor on which we can apply our already known theorem/formula.
  – md2perpe
  Jul 19 at 20:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Very informally you might say so, but if you want to be mathematically strict you should not use that wording.
  – md2perpe
  Jul 19 at 20:59
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote



accepted

First note that $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ can be written as
$int_Omega partial_i F_i , dV = oint_partialOmega F_i dS_i.$

Now let $vec A = (A_i)$ be a constant vector field. Then
$$
A_i int_Omega partial_j T_ij , dV
= int_Omega partial_j (A_i T_ij) , dV
= oint_partial Omega A_i T_ij , dS_j
= A_i oint_partial Omega T_ij , dS_j.
$$
This is valid for all constant $vec A,$ e.g. for $vec A = e_1, e_2, e_3$ which gives that
$$int_Omega partial_j T_ij , dV = oint_partialOmega T_ij , dS_j$$
for $i=1,2,3,$ i.e. the formula is valid for all $i.$

share|cite|improve this answer

answered Jul 19 at 19:38

md2perpe

5,93011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for answer, but I don't really understand the argument. How come you make them $int_Omega partial_j (A_i T_ij) , dV = oint_partial Omega A_i T_ij , dS_j$ in the first place, and why would it matter if it is valid for all constant vectors?
  – 123
  Jul 19 at 20:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  That equality is just $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ for $F_j = A_i T_ij,$ and you know from before that this is true.
  – md2perpe
  Jul 19 at 20:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $A_i u_i = A_i v_i$ for all constant $A_i$ then $u_i = v_i$ (for all $i$). This is because you can take $vec A = e_j,$ the unit vector field in the $j$th coordinate direction. But since $A_i = (e_j)_i = delta_ij$ (Kronecker delta), you then get $u_j = v_j$ for that $j$. But since you can to that for any $j$ you get $u_j = v_j$ for all $j$.
  – md2perpe
  Jul 19 at 20:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, the product $A_i T_ij$ is a rank 1 tensor on which we can apply our already known theorem/formula.
  – md2perpe
  Jul 19 at 20:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Very informally you might say so, but if you want to be mathematically strict you should not use that wording.
  – md2perpe
  Jul 19 at 20:59
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote



accepted

up vote
1
down vote



accepted

First note that $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ can be written as
$int_Omega partial_i F_i , dV = oint_partialOmega F_i dS_i.$

Now let $vec A = (A_i)$ be a constant vector field. Then
$$
A_i int_Omega partial_j T_ij , dV
= int_Omega partial_j (A_i T_ij) , dV
= oint_partial Omega A_i T_ij , dS_j
= A_i oint_partial Omega T_ij , dS_j.
$$
This is valid for all constant $vec A,$ e.g. for $vec A = e_1, e_2, e_3$ which gives that
$$int_Omega partial_j T_ij , dV = oint_partialOmega T_ij , dS_j$$
for $i=1,2,3,$ i.e. the formula is valid for all $i.$

share|cite|improve this answer

answered Jul 19 at 19:38

md2perpe

5,93011022

First note that $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ can be written as
$int_Omega partial_i F_i , dV = oint_partialOmega F_i dS_i.$

Now let $vec A = (A_i)$ be a constant vector field. Then
$$
A_i int_Omega partial_j T_ij , dV
= int_Omega partial_j (A_i T_ij) , dV
= oint_partial Omega A_i T_ij , dS_j
= A_i oint_partial Omega T_ij , dS_j.
$$
This is valid for all constant $vec A,$ e.g. for $vec A = e_1, e_2, e_3$ which gives that
$$int_Omega partial_j T_ij , dV = oint_partialOmega T_ij , dS_j$$
for $i=1,2,3,$ i.e. the formula is valid for all $i.$

share|cite|improve this answer

answered Jul 19 at 19:38

md2perpe

5,93011022

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 19:38

md2perpe

5,93011022

answered Jul 19 at 19:38

md2perpe

5,93011022

answered Jul 19 at 19:38

md2perpe

5,93011022

5,93011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for answer, but I don't really understand the argument. How come you make them $int_Omega partial_j (A_i T_ij) , dV = oint_partial Omega A_i T_ij , dS_j$ in the first place, and why would it matter if it is valid for all constant vectors?
  – 123
  Jul 19 at 20:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  That equality is just $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ for $F_j = A_i T_ij,$ and you know from before that this is true.
  – md2perpe
  Jul 19 at 20:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $A_i u_i = A_i v_i$ for all constant $A_i$ then $u_i = v_i$ (for all $i$). This is because you can take $vec A = e_j,$ the unit vector field in the $j$th coordinate direction. But since $A_i = (e_j)_i = delta_ij$ (Kronecker delta), you then get $u_j = v_j$ for that $j$. But since you can to that for any $j$ you get $u_j = v_j$ for all $j$.
  – md2perpe
  Jul 19 at 20:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, the product $A_i T_ij$ is a rank 1 tensor on which we can apply our already known theorem/formula.
  – md2perpe
  Jul 19 at 20:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Very informally you might say so, but if you want to be mathematically strict you should not use that wording.
  – md2perpe
  Jul 19 at 20:59
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for answer, but I don't really understand the argument. How come you make them $int_Omega partial_j (A_i T_ij) , dV = oint_partial Omega A_i T_ij , dS_j$ in the first place, and why would it matter if it is valid for all constant vectors?
  – 123
  Jul 19 at 20:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  That equality is just $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ for $F_j = A_i T_ij,$ and you know from before that this is true.
  – md2perpe
  Jul 19 at 20:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $A_i u_i = A_i v_i$ for all constant $A_i$ then $u_i = v_i$ (for all $i$). This is because you can take $vec A = e_j,$ the unit vector field in the $j$th coordinate direction. But since $A_i = (e_j)_i = delta_ij$ (Kronecker delta), you then get $u_j = v_j$ for that $j$. But since you can to that for any $j$ you get $u_j = v_j$ for all $j$.
  – md2perpe
  Jul 19 at 20:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, the product $A_i T_ij$ is a rank 1 tensor on which we can apply our already known theorem/formula.
  – md2perpe
  Jul 19 at 20:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Very informally you might say so, but if you want to be mathematically strict you should not use that wording.
  – md2perpe
  Jul 19 at 20:59
  

  
  
  
  

Thank you for answer, but I don't really understand the argument. How come you make them $int_Omega partial_j (A_i T_ij) , dV = oint_partial Omega A_i T_ij , dS_j$ in the first place, and why would it matter if it is valid for all constant vectors?
– 123
Jul 19 at 20:04

Thank you for answer, but I don't really understand the argument. How come you make them $int_Omega partial_j (A_i T_ij) , dV = oint_partial Omega A_i T_ij , dS_j$ in the first place, and why would it matter if it is valid for all constant vectors?
– 123
Jul 19 at 20:04

1

1

That equality is just $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ for $F_j = A_i T_ij,$ and you know from before that this is true.
– md2perpe
Jul 19 at 20:29

That equality is just $int_Omega vecnabla cdot vecF , dV = oint_partial Omega vecF cdot dvecS$ for $F_j = A_i T_ij,$ and you know from before that this is true.
– md2perpe
Jul 19 at 20:29

1

1

If $A_i u_i = A_i v_i$ for all constant $A_i$ then $u_i = v_i$ (for all $i$). This is because you can take $vec A = e_j,$ the unit vector field in the $j$th coordinate direction. But since $A_i = (e_j)_i = delta_ij$ (Kronecker delta), you then get $u_j = v_j$ for that $j$. But since you can to that for any $j$ you get $u_j = v_j$ for all $j$.
– md2perpe
Jul 19 at 20:36

If $A_i u_i = A_i v_i$ for all constant $A_i$ then $u_i = v_i$ (for all $i$). This is because you can take $vec A = e_j,$ the unit vector field in the $j$th coordinate direction. But since $A_i = (e_j)_i = delta_ij$ (Kronecker delta), you then get $u_j = v_j$ for that $j$. But since you can to that for any $j$ you get $u_j = v_j$ for all $j$.
– md2perpe
Jul 19 at 20:36

1

1

Yes, the product $A_i T_ij$ is a rank 1 tensor on which we can apply our already known theorem/formula.
– md2perpe
Jul 19 at 20:49

Yes, the product $A_i T_ij$ is a rank 1 tensor on which we can apply our already known theorem/formula.
– md2perpe
Jul 19 at 20:49

1

1

Very informally you might say so, but if you want to be mathematically strict you should not use that wording.
– md2perpe
Jul 19 at 20:59

Very informally you might say so, but if you want to be mathematically strict you should not use that wording.
– md2perpe
Jul 19 at 20:59

 | 
show 4 more comments

 

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