Does the Partial Deriv wrt z of VyUz Term Cancel?

[John777]

I am trying to prove two things equal that involves a bunch of dot products, cross products and curls. I can't remember the exact problem, but this demonstrates my question.My question is the left side \delta/\deltax*(U_xV_x + U_yV_y + U_zV_z)

The right side is \delta/\deltax*(V_xU_x + V_yU_y + V_zU_z +
V_yU_z)

does the V_yU_z term cancel since the partial deriv wrt x is 0?

Also do the the other terms become 1 since for example in the partial deriv wrt x of U_xV_x

If my explanation did not make sense here is the original problem grad(U.V) = (U.grad)V + (Vxgrad)U + Ux(gradxV) + Vx(gradxU)

Just looking at the i component The left side yields partial deriv wrt x of (UxVx + UyVy + UzVz)i

The right side however yields terms in the i component that are partials wrt to x, y and z.

 

[Coto]

Assuming u and v to be two vectors denoted by u = (u_x, u_y, u_z), v = (v_x, v_y, v_z) and assuming that the vector components themselves are functions of x,y,z, i.e. that u_i = f_i(x,y,z), v_i=g_i(x,y,z) for i = x,y,z, we cannot say for certain that \partial / \partial x of the cross term v_y u_z is zero unless the functions g_y,f_z do not depend on x (which in general cannot be said.)

Edit: As a simple example, take u_z = x , v_y = x, then what does the partial derivative w.r.t. x of their product equal?

 

[John777]

Assuming u and v to be two vectors denoted by u = (u_x, u_y, u_z), v = (v_x, v_y, v_z) and assuming that the vector components themselves are functions of x,y,z, i.e. that u_i = f_i(x,y,z), v_i=g_i(x,y,z) for i = x,y,z, we cannot say for certain that \partial / \partial x of the cross term v_y u_z is zero unless the functions g_y,f_z do not depend on x (which in general cannot be said.)

Edit: As a simple example, take u_z = x , v_y = x, then what does the partial derivative w.r.t. x of their product equal?

I see your rationale but in trying to make the left side equal the right side how do I cancel out terms such as VzUy d/dx that do not appear on the left side? (This term I am referring to results from the third Ux(gradxV) term.

For example the entire left side i component has three terms, but the entire right side i component has 14 terms 4 of which are negative. So even if those cancel out other terms I am still left with 6 terms whereas I think I need 3 for it to equal the left side.

 

[Coto]

There's some ambiguity in your problem statement that may serve to clarify the problem. For example, you have (v \times \nabla)u, however both those terms are vectors, so is a dot product implied between the two?

 

[Coto]

As a side note: I will say that if the equality is in fact true, that terms should cancel out relatively easily.

 

[John777]

There's some ambiguity in your problem statement that may serve to clarify the problem. For example, you have (v \times \nabla)u, however both those terms are vectors, so is a dot product implied between the two?

Ha! your correct it was a typo my mistake. here is the correct problem:

grad(U.V) = (U.grad)V + (V.grad)U + Ux(gradxV) + Vx(gradxU)

 

[Coto]

My quickest suggestion without having to write out the problem myself would be to expand the left hand side. As an example, carry out the derivation:
\partial / \partial x (u_x v_x + u_y v_y + u_z v_z)

Note: You'll end up with 6 terms.

 

[John777]

Here is just the i components worked out:

left side:

d/dx * (UxVx +UyVy + UzVz)i

right side:

each line will represent one part of right side of given problem

(VxUx * d/dx + VxVy * d/dy + VxUz * d/dz)i

(VxUx *d/dx + VyUx * d/dy + VzUx * d/dz)i

(VxUy * d/dx - VxUy * d/dy + VzUz *d/dx - VxUz *d/dz)i

(VyUz * d/dx - VyUx *d/dy + VzUz *d/dx - VzUx *d/dz)i


Some of these terms cancel and you are left with:

(VxUx * d/dx + VxUx *d/dx + VxUy * d/dx + VzUz *d/dx + VyUz * d/dx + VzUz *d/dx)i

which does not equal the left

 

[John777]

My quickest suggestion without having to write out the problem myself would be to expand the left hand side. As an example, carry out the derivation:
\partial / \partial x (u_x v_x + u_y v_y + u_z v_z)

Note: You'll end up with 6 terms.

I'm not quite sure how you would do that derivation, could you just give me a quick sample?

 

[Coto]

There's a problem with your RHS. You've carried out the operation incorrectly.

Let's consider the i'th term of (u \cdot \nabla)v, it would be:

(u_x \partial / \partial x + u_y \partial / \partial y + u_z \partial / \partial z)v_i

Now, notice that v_i = g_i(x,y,z), so you cannot pull it outside of those partial derivatives like you've done... those partial derivatives are acting on v_i.

In essence, the above statement expanded would be:

u_x \frac{\partial v_i}{\partial x} + u_y \frac{\partial v_i}{\partial y} + u_z \frac{\partial v_i}{\partial z}

As for the derivation of the above (in my previous post), you should be able to use the chain rule and linearity of the partial derivative to solve it.

 

[John777]

There's a problem with your RHS. You've carried out the operation incorrectly.

Let's consider the i'th term of (u \cdot \nabla)v, it would be:

(u_x \partial / \partial x + u_y \partial / \partial y + u_z \partial / \partial z)v_i

Now, notice that v_i = g_i(x,y,z), so you cannot pull it outside of those partial derivatives like you've done... those partial derivatives are acting on v_i.

In essence, the above statement expanded would be:

u_x \frac{\partial v_i}{\partial x} + u_y \frac{\partial v_i}{\partial y} + u_z \frac{\partial v_i}{\partial z}

As for the derivation of the above, you should be able to use the chain rule and linearity of the partial derivative to solve it.

How come it is ok to leave the U outside the partial? I apologize for my ignorance but I am new to this stuff.

 

[Coto]

For all practical purposes, we treat the gradient operator \nabla as if it were a vector with the quantities (\partial / \partial x,\partial / \partial y,\partial / \partial z). This makes for some very appealing shorthand notation for taking curls and divergences, but it can also make it appear as if the vector can just multiply through with other functions.

This is *not* the case! You must be very careful using this notation, and remember that the partial derivatives act on anything that is in front of them.

Now for your question, assuming the vector u to be given as u = (u_x,u_y,u_z) and assuming the gradient *operator* to be given as (\partial / \partial x,\partial / \partial y,\partial / \partial z), what is u \cdot \nabla?

Note: The expression u \cdot \nabla is an example of one of the appealing shorthand forms that we can make if we assume the gradient operator to have that vector like form. Otherwise you would always have to write out the full expression.

 

[John777]

so u.grad = ux d/dx + uy d/dy + uz d/dz

so if u equaled 2xi + 4xyj + 5xyzk

u.grad would equal 2 + 4x + 5xy?

and if it were (u.grad)V where V = Vx + Vy + Vk

it would equal 2Vx + 4xVy + 5xyVk?

 

[Coto]

u.grad would equal 2 + 4x + 5xy?

No, those derivative operators only act on what is in front of them. Would you say the u_x, etc. terms are in front of the derivative operators?

As a second example, what is \nabla \cdot u? What popular operation is this shorthand notation referring to?

 

[John777]

No, those derivative operators only act on what is in front of them. Would you say the u_x, etc. terms are in front of the derivative operators?

As a second example, what is \nabla \cdot u? What popular operation is this shorthand notation referring to?

I am getting more confused. u.grad = ux d/dx + uy d/dy + uz d/dz In this example the U terms seem to be in front of the derivation operators.

I believe you are referring to the dot product? in your example I believe u.grad = ux d/dx + uy d/dy + uz d/dz following the definition where as a simple example a.b = axbx + ayby + azbz

Thanks again for all the help I really appreciate it.

 

[Coto]

Let's start with some examples to lead you straight :) :

1) What is x \frac{d f(x)}{dx} if f(x) = x?

2) What is x \frac{d}{dx} f(x) if f(x) = x?

Now, let's pretend that we don't have any f(x) yet, then we could write:
x \frac{d}{dx}.

However, this is an incomplete object because in order for us to write down an answer, we would need to take that object and act it on some function, f(x), like we did in part 1 and 2.

This is what we have done with this u \cdot \nabla thing. Technically speaking, this is somewhat of an incomplete expression. In order to get something from it, you would need to put a function in front of it.

Notice from part (1) and (2) that when we removed the f(x) from in front of the derivative operator that we ended up with an expression similar to the form u_x \partial / \partial x.

 

[John777]

Let's start with some examples to lead you straight :) :

1) What is x \frac{d f(x)}{dx} if f(x) = x?

2) What is x \frac{d}{dx} f(x) if f(x) = x?

Now, let's pretend that we don't have any f(x) yet, then we could write:
x \frac{d}{dx}.

However, this is an incomplete object because in order for us to write down an answer, we would need to take that object and act it on some function, f(x), like we did in part 1 and 2.

This is what we have done with this u \cdot \nabla thing. Technically speaking, this is somewhat of an incomplete expression. In order to get something from it, you would need to put a function in front of it.

Notice from part (1) and (2) that when we removed the f(x) from in front of the derivative operator that we ended up with an expression similar to the form u_x \partial / \partial x.

So plugging in x for f(x) in the first function will be result in an answer of x. I not totally sure if you are multiplying the second thing by f(x) or not but if you are it would be x^2.

So by in front of the operation you mean to the right as we read so d(x^2)/dx would be 2x?

 

[Coto]

It seems I'm still losing you a bit! The answer to both #1 and #2 is x! They are the same expression written slightly differently.

It's very important that you stop seeing the derivative as something you can multiply. It does not multiply anything, ever. It **acts** on whatever is in front of it, (and yes you are correct, I mean to the right of when I say in front... sorry for that).

So what is #2?

 

[John777]

It seems I'm still losing you a bit! The answer to both #1 and #2 is x! They are the same expression written slightly differently.

It's very important that you stop seeing the derivative as something you can multiply. It does not multiply anything, ever. It **acts** on whatever is in front of it, (and yes you are correct, I mean to the right of when I say in front... sorry for that).

So what is #2?

Ok I see they are both x. I think how the forum was displaying it, I was a little confused as to what you were writing.

 

[Coto]

So can you work backwards from this point and see where the mistake was in your post #13 (https://www.physicsforums.com/showpost.php?p=3027624&postcount=13)

If you see how it works, you should be able to work from this point to prove your original problem, keeping in mind the original posts discussing the problem. This problem is getting you used to dealing with this shorthand notation and it should serve to teach you the importance of avoiding the pitfalls associated with the notation.

 

[John777]

So can you work backwards from this point and see where the mistake was in your post #13 (https://www.physicsforums.com/showpost.php?p=3027624&postcount=13)

If you see how it works, you should be able to work from this point to prove your original problem, keeping in mind the original posts discussing the problem. This problem is getting you used to dealing with this shorthand notation and it should serve to teach you the importance of avoiding the pitfalls associated with the notation.

Great! Thanks for the help I need the basics since I haven't done math in a while and am now going back to school. My professor seems to think everyone has been practicing math everyday and knows a lot, whereas I didn't going into the class.

Its getting pretty late so I will probably try and work this out tomorrow. If I have any questions I'll post them up and hopefully you can respond if you get a chance. Thanks again!

 

[Coto]

No problem John. Good luck!