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Differentiating a Vector Field

Hi,

Just need a bit of help with differentiating a vector field.

F=(ax2+bxy+cy22x)i+(x2+xyy2+bz)j+(2y+2z)k F= (ax^2 +bxy +cy^2 -2x)\bold{i} +(x^2 +xy-y^2 +bz)\bold{j}+(2y+2z)\bold{k}

I need to find F\nabla F

I am unsure what to differentiate with respect to as i need to have a vector as the outcome. Do I differentiate w.r.t x for i, w.r.t y for j, and w.r.t to z for k?

Any help is appreciated.
Reply 1
If F=(F1F2F3)F = \begin{pmatrix}F_1 \\ F_2 \\ F_3 \end{pmatrix} is a vector then F=(F1F2F3)\nabla F = \begin{pmatrix} \nabla F_1 \\ \nabla F_2 \\ \nabla F_3 \end{pmatrix} is a matrix, not a vector, where (for instance) F1=(xF1,yF1,zF1)\nabla F_1 = (\partial_x F_1, \partial_y F_1, \partial_z F_1).

If you were differentiating the i\mathbf{i}-component w.r.t. xx and the j\mathbf{j}-component w.r.t. yy and so on then what you'd have would be F\nabla \cdot F and not F\nabla F.
(edited 11 years ago)
Reply 2
Original post by nuodai
If F=(F1F2F3)F = \begin{pmatrix}F_1 \\ F_2 \\ F_3 \end{pmatrix} is a vector then F=(F1F2F3)\nabla F = \begin{pmatrix} \nabla F_1 \\ \nabla F_2 \\ \nabla F_3 \end{pmatrix} is a matrix, not a vector, where (for instance) F1=(xF1,yF1,zF1)\nabla F_1 = (\partial_x F_1, \partial_y F_1, \partial_z F_1).

If you were differentiating the i\mathbf{i}-component w.r.t. xx and the j\mathbf{j}-component w.r.t. yy and so on then what you'd have would be F\nabla \cdot F and not F\nabla F.


Okay, thank you. Thats a big help because the question is to find the values a,b,c such that F is soleneidal. i.e. F=0\nabla \cdot F=0
Reply 3
Original post by HappyHammer15
Okay, thank you. Thats a big help because the question is to find the values a,b,c such that F is soleneidal. i.e. F=0\nabla \cdot F=0


Right. Bear in mind that F\nabla \cdot F is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)
Reply 4
Original post by nuodai
Right. Bear in mind that F\nabla \cdot F is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)


Okay thank you, however, in this example i do not get a variable c in the F\nabla \cdot F. Thank you for your help thus far.
Reply 5
Original post by HappyHammer15
Okay thank you, however, in this example i do not get a variable c in the F\nabla \cdot F. Thank you for your help thus far.


So?
Reply 6
Original post by around
So?


I get a=-0.5 and b = 2. Is c an arbitrary constant then?
Reply 7
Original post by nuodai
Right. Bear in mind that F\nabla \cdot F is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)


To add to this: remember that the notation is suggestive and F\nabla \cdot F is formally like a dot product. i.e. think of =(x,y,z)\nabla = (\frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}) and and F=(Fi,Fj,Fk)F=(F_{\textbf{i}},F_{\textbf{j}},F_{\textbf{k}}) where in your example Fk=2y+2zF_{\textbf{k}} = 2y+2z and so on.

Then, you form the 'dot product'

F=(x,y,z)(Fi,Fj,Fk)=(xFi+yFj+zFk) \displaystyle \nabla \cdot F = \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \cdot (F_{\textbf{i}},F_{\textbf{j}},F_{\textbf{k}}) = \left( \frac{\partial}{\partial x}F_{\textbf{i}}+\frac{\partial}{\partial y}F_{\textbf{j}}+\frac{\partial}{\partial z}F_{\textbf{k}} \right)
where you interpret the 'product' of, say x\frac{\partial}{\partial x} with FiF_{\textbf{i}} in the obvious way i.e. you take the partial derivative of FiF_{\textbf{i}} w.r.t. xx
(edited 11 years ago)
Reply 8
Original post by HappyHammer15
I get a=-0.5 and b = 2. Is c an arbitrary constant then?


Exactly, if it doesn't contribute anything then you can take any value.

e.g. Find all pairs (a,b) such that a = 0.
Reply 9
Original post by Jake22
To add to this: remember that the notation is suggestive and F\nabla \cdot F is formally like a dot product. i.e. think of =(x,y,z)\nabla = (\frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}) and and F=(Fi,Fj,Fk)F=(F_{\textbf{i}},F_{\textbf{j}},F_{\textbf{k}}) where in your example Fk=2y+2zF_{\textbf{k}} = 2y+2z and so on.

Then, you form the 'dot product'

F=(x,y,z)(Fi,Fj,Fk)=(xFi,yFj,zFk) \displaystyle \nabla \cdot F = \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \cdot (F_{\textbf{i}},F_{\textbf{j}},F_{\textbf{k}}) = \left( \frac{\partial}{\partial x}F_{\textbf{i}},\frac{\partial}{\partial y}F_{\textbf{j}},\frac{\partial}{\partial z}F_{\textbf{k}} \right)
where you interpret the 'product' of, say x\frac{\partial}{\partial x} with FiF_{\textbf{i}} in the obvious way i.e. you take the partial derivative of FiF_{\textbf{i}} w.r.t. xx



If i do it in this way i get a very complicated solution that is quite difficult to reduce. Will the way in which i first did it suffice?, i.e FiF_{\textbf{i}} differentiating w.r.t x,Fj F_{\textbf{j}} differentiating w.r.t y and so on.
Reply 10
Original post by HappyHammer15
If i do it in this way i get a very complicated solution that is quite difficult to reduce. Will the way in which i first did it suffice?, i.e FiF_{\textbf{i}} differentiating w.r.t x,Fj F_{\textbf{j}} differentiating w.r.t y and so on.

I never wrote that you should calculate it in any other way... I said to do exactly that i.e. differentiate the i component wrt to x, the j component wrt y etc. and then sum them up. All I wrote was just a convinient way of remembering what the operation F\nabla \cdot F gives you. You shouldn't get anything different at all.

The point is to think of \nabla as a vector and F\nabla \cdot F as a kind of dot product of vectors whose formula in components you are (or at least should be) very familiar with. The reason that it isn't a 'true' dot product is that \nabla is a vector in a different vector space to FF i.e. the dual space.

Note: there was a mistake in my initial post which I have now corrected - I put commas instead of + signs on the right hand side but this shouldn't have affected how you would calculate.
(edited 11 years ago)
Reply 11
Original post by Jake22
I never wrote that you should calculate it in any other way... I said to do exactly that i.e. differentiate the i component wrt to x, the j component wrt y etc. and then sum them up. All I wrote was just a convinient way of remembering what the operation F\nabla \cdot F gives you. You shouldn't get anything different at all.

Note: there was a mistake in my initial post which I have now corrected - I put commas instead of + signs on the right hand side.


Okay, i see now. I was reading =F \nabla =\nabla F and i was trying to compute FF\nabla F \cdot F
This seems like such a silly mistake to make now. Thanks for all of your help though.

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