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What is the meaning of $A. nabla $?
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Suppose you have a vector field $A = A_1 hati + A_2hatj+ A_3hatk $. Then $nabla cdot A $ would represent the divergence. But what does $A cdot nabla$ mean below, and what would it come out to be? The relevant problems are (b) and (d) below. 
vector-analysis vector-fields
edited Jul 21 at 8:40
Chandler Watson
417320
asked Jul 21 at 6:34
user187604
916
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up vote
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Suppose you have a vector field $A = A_1 hati + A_2hatj+ A_3hatk $. Then $nabla cdot A $ would represent the divergence. But what does $A cdot nabla$ mean below, and what would it come out to be? The relevant problems are (b) and (d) below.
vector-analysis vector-fields
edited Jul 21 at 8:40
Chandler Watson
417320
asked Jul 21 at 6:34
user187604
916
Where did you see $A. nabla$ ?
– Lord Shark the Unknown
Jul 21 at 6:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose you have a vector field $A = A_1 hati + A_2hatj+ A_3hatk $. Then $nabla cdot A $ would represent the divergence. But what does $A cdot nabla$ mean below, and what would it come out to be? The relevant problems are (b) and (d) below.
vector-analysis vector-fields
edited Jul 21 at 8:40
Chandler Watson
417320
asked Jul 21 at 6:34
user187604
916
Suppose you have a vector field $A = A_1 hati + A_2hatj+ A_3hatk $. Then $nabla cdot A $ would represent the divergence. But what does $A cdot nabla$ mean below, and what would it come out to be? The relevant problems are (b) and (d) below.
vector-analysis vector-fields
edited Jul 21 at 8:40
Chandler Watson
417320
asked Jul 21 at 6:34
user187604
916
edited Jul 21 at 8:40
Chandler Watson
417320
edited Jul 21 at 8:40
Chandler Watson
417320
edited Jul 21 at 8:40
Chandler Watson
417320
417320
asked Jul 21 at 6:34
user187604
916
asked Jul 21 at 6:34
user187604
916
asked Jul 21 at 6:34
user187604
916
916
Where did you see $A. nabla$ ?
– Lord Shark the Unknown
Jul 21 at 6:39
add a comment |Â
Where did you see $A. nabla$ ?
– Lord Shark the Unknown
Jul 21 at 6:39
Where did you see $A. nabla$ ?
– Lord Shark the Unknown
Jul 21 at 6:39
Where did you see $A. nabla$ ?
– Lord Shark the Unknown
Jul 21 at 6:39
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
If $mathbf A = pmatrixa_x\a_y\a_z$, then
$$(mathbf Acdotnabla)phi = a_xfracpartialpartial xphi + a_yfracpartialpartial yphi + a_zfracpartialpartial zphi$$
Basically, you treat $nabla$ as a vector of derivatives and do vector algebra, except that you are careful not to move terms across the derivative.
answered Jul 21 at 6:59
celtschk
28.1k65495
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If $mathbf A = pmatrixa_x\a_y\a_z$, then
$$(mathbf Acdotnabla)phi = a_xfracpartialpartial xphi + a_yfracpartialpartial yphi + a_zfracpartialpartial zphi$$
Basically, you treat $nabla$ as a vector of derivatives and do vector algebra, except that you are careful not to move terms across the derivative.
answered Jul 21 at 6:59
celtschk
28.1k65495
add a comment |Â
up vote
3
down vote
accepted
If $mathbf A = pmatrixa_x\a_y\a_z$, then
$$(mathbf Acdotnabla)phi = a_xfracpartialpartial xphi + a_yfracpartialpartial yphi + a_zfracpartialpartial zphi$$
Basically, you treat $nabla$ as a vector of derivatives and do vector algebra, except that you are careful not to move terms across the derivative.
answered Jul 21 at 6:59
celtschk
28.1k65495
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If $mathbf A = pmatrixa_x\a_y\a_z$, then
$$(mathbf Acdotnabla)phi = a_xfracpartialpartial xphi + a_yfracpartialpartial yphi + a_zfracpartialpartial zphi$$
Basically, you treat $nabla$ as a vector of derivatives and do vector algebra, except that you are careful not to move terms across the derivative.
answered Jul 21 at 6:59
celtschk
28.1k65495
If $mathbf A = pmatrixa_x\a_y\a_z$, then
$$(mathbf Acdotnabla)phi = a_xfracpartialpartial xphi + a_yfracpartialpartial yphi + a_zfracpartialpartial zphi$$
Basically, you treat $nabla$ as a vector of derivatives and do vector algebra, except that you are careful not to move terms across the derivative.
answered Jul 21 at 6:59
celtschk
28.1k65495
answered Jul 21 at 6:59
celtschk
28.1k65495
answered Jul 21 at 6:59
celtschk
28.1k65495
answered Jul 21 at 6:59
celtschk
28.1k65495
28.1k65495
add a comment |Â
add a comment |Â
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Where did you see $A. nabla$ ?
– Lord Shark the Unknown
Jul 21 at 6:39