Normal Velocity Gradient Across a Surface: Proving a Result

[Apashanka]

Homework Statement

Unit normal to a surface

Relevant Equations

Unit normal to a surface

I have came across a problem where each point of a surface parallel to the x-y cartesian plane and having it's normal along the z axis is having velocity along the z direction $v_z$ and there exists a velocity gradient across the plane (e.g $v_z(x,y)$ ,
After time $\delta t$ it is written the normal to surface will be $n'=\hat k-(\hat iv_{zx}+\hat j v_{zy})\delta t$ where $v_{ij}=\frac{\partial v_i}{\partial x_j}$ and $\vec r'=\vec r+\vec v \delta t$.
Can anyone please help me out how to proceed to prove this result...

 

[Chestermiller]

What surface is this supposed to be normal to?

 

[Apashanka]

What surface is this supposed to be normal to?

The surface which is initially parallel to x-y cartesian plane having it's unit normal along $\hat k$ and after $\delta t$ having unit normal along n'

 

[Chestermiller]

Imagine a rectangle comprised of material points at time zero, with coordinates along the edges (x,y,z), $(x+\Delta x, y, z)$, $x,y+\Delta y, z)$. What are the coordinates of these same three material points at time $\delta t$?

 

[Apashanka]

Imagine a rectangle comprised of material points at time zero, with coordinates along the edges (x,y,z), $(x+\Delta x, y, z)$, $x,y+\Delta y, z)$. What are the coordinates of these same three material points at time $\delta t$?

Is this rectangle expanding ,if so then in which direction??

 

[Apashanka]

Imagine a rectangle comprised of material points at time zero, with coordinates along the edges (x,y,z), $(x+\Delta x, y, z)$, $x,y+\Delta y, z)$. What are the coordinates of these same three material points at time $\delta t$?

If the rectangular points are having velocity $\vec v(x,y,z)$ then $(x,y,z)$ becomes $(x+v_x(x,y,z)\delta t,y+v_y(x,y,z)\delta t,z+v_z(x,y,z)\delta t)$
Similarly other coordinates also change $x+\Delta x+v(x+\Delta x,y,z)\delta t,y+v_y(x+\Delta x,y,z)\delta t,z+v_z(x+\Delta x,y,z)\delta t$

 

[Chestermiller]

If the rectangular points are having velocity $\vec v(x,y,z)$ then $(x,y,z)$ becomes $(x+v_x(x,y,z)\delta t,y+v_y(x,y,z)\delta t,z+v_z(x,y,z)\delta t)$
Similarly other coordinates also change $x+\Delta x+v(x+\Delta x,y,z)\delta t,y+v_y(x+\Delta x,y,z)\delta t,z+v_z(x+\Delta x,y,z)\delta t$

I thought you said that the velocity is only in the z direction. If that is the case, then these results are not correct.

 

[Apashanka]

I thought you said that the velocity is only in the z direction. If that is the case, then these results are not correct.

If it is so then the three points will become $(x,y,z+v_z(x,y)\delta t) (x+\Delta x,y,z+v_z(x+\Delta x,y)\delta t) (x,y+\Delta y,z+v_z(x,y+\Delta y)\delta t)$
Now they are not in the plane parallel to the x-y cartesian plane
Then sir how to find the unit normal as I asked in the first post??

 

[Chestermiller]

If it is so then the three points will become $(x,y,z+v_z(x,y)\delta t) (x+\Delta x,y,z+v_z(x+\Delta x,y)\delta t) (x,y+\Delta y,z+v_z(x,y+\Delta y)\delta t)$
Now they are not in the plane parallel to the x-y cartesian plane
Then sir how to find the unit normal as I asked in the first post??

Please be patient. Now, what are the coordinates of the three points if we now make the approximations: $$v_z(x+\Delta x,y)=v_z(x,y)+\Delta x\frac{\partial v_z}{\partial x}$$ and $$v_z(x,y+\Delta y)=v_z(x,y)+\Delta y\frac{\partial v_z}{\partial y}$$

 

[Apashanka]

Please be patient. Now, what are the coordinates of the three points if we now make the approximations: $$v_z(x+\Delta x,y)=v_z(x,y)+\Delta x\frac{\partial v_z}{\partial x}$$ and $$v_z(x,y+\Delta y)=v_z(x,y)+\Delta y\frac{\partial v_z}{\partial y}$$

$(x,y,z+v_z(x,y)\delta t) (x+\Delta x,y,z+v_z(x,y)\delta t+\frac{\partial v_z}{\partial x}\Delta x \delta t) (x,y+\Delta y,z+v_z(x,y)\delta t+\frac{\partial v_z}{\partial y}\Delta y\delta t$

 

[Chestermiller]

$(x,y,z+v_z(x,y)\delta t) (x+\Delta x,y,z+v_z(x,y)\delta t+\frac{\partial v_z}{\partial x}\Delta x \delta t) (x,y+\Delta y,z+v_z(x,y)\delta t+\frac{\partial v_z}{\partial y}\Delta y\delta t$

Excellent. Now if you draw a vector at time $\delta t$ between the first material point and the second material point, and another vector between the first point and the third material point, in terms of the unit vectors in the three coordinate directions, what is the equation for each of these two vectors?

 

[Apashanka]

The two vectors are then
$\Delta x \hat i+\frac{\partial v_z}{\partial x}\Delta x\delta t \hat k$
$\Delta y \hat j+\frac{\partial v_z}{\partial y}\Delta y\delta t\hat k$

 

[Chestermiller]

The two vectors are then
$\Delta x \hat i+\frac{\partial v_z}{\partial x}\Delta x\delta t \hat k$
$\Delta y \hat j+\frac{\partial v_z}{\partial y}\Delta y\delta t\hat k$

Excellent. Now, we are trying to find an equation for the normal to the plane in which these two vectors lie. How is the cross product of two vectors related to the plane in which the two vectors lie? So what is the cross product of these two vectors?

 

[Apashanka]

The two vectors are then
$\Delta x \hat i+\frac{\partial v_z}{\partial x}\Delta x\delta t \hat k$
$\Delta y \hat j+\frac{\partial v_z}{\partial y}\Delta y\delta t\hat k$

Yes sir now if we consider a vector $p=a\hat i+b\hat j+c\hat k$ perpendicular to the plane containing this two vectors ,then dot product of $\vec p$ with these two vectors is 0 and using this case $\vec p=c(\hat k-\hat iv_{zx}-\hat j v_{zy})$ for which unit normal ( $\hat p)$ has terms in denominator ,how to get rid of this??

 

[Chestermiller]

Yes sir now if we consider a vector $p=a\hat i+b\hat j+c\hat k$ perpendicular to the plane containing this two vectors ,then dot product of $\vec p$ with these two vectors is 0 and using this case $\vec p=c(\hat k-\hat iv_{zx}-\hat j v_{zy})$ for which unit normal ( $\hat p)$ has terms in denominator ,how to get rid of this??

Don't worry about that for now. Please continue with answers to my questions in post #13.

 

[Mark44]

Yes sir now if we consider a vector $p=a\hat i+b\hat j+c\hat k$ perpendicular to the plane containing this two vectors ,then dot product of $\vec p$ with these two vectors is 0 and using this case $\vec p=c(\hat k-\hat iv_{zx}-\hat j v_{zy})$ for which unit normal ( $\hat p)$ has terms in denominator ,how to get rid of this??

Instead of working with the dot product of your normal vector with each of the vectors in the plane, you could instead calculate the cross product of your two vectors in the plane. The result will be a vector that is perpendicular to each of the two vectors in the plane. It would then be a fairly simple task to normalize the result to get the unit vector you want.

 

[Apashanka]

Excellent. Now, we are trying to find an equation for the normal to the plane in which these two vectors lie. How is the cross product of two vectors related to the plane in which the two vectors lie? So what is the cross product of these two vectors?

Cross product
$-\hat i v_{zx} \Delta x\Delta y \delta t-\hat j v_{zy}\Delta x\Delta y\delta t+\hat k\Delta x\Delta y$

 

[Chestermiller]

Factor out the delta x and delta y. Then normalize to obtain a unit vector. Then linearize with respect to delta t.

 

[Apashanka]

Factor out the delta x and delta y. Then normalize to obtain a unit vector. Then linearize with respect to delta t.

Yes sir then how to normalise it ??there remains term in the denominator...

 

[Chestermiller]

Yes sir then how to normalise it ??there remains term in the denominator...

The denominator has a term proportional to delta t squared. When you linearize, this term is dropped.

 

[Apashanka]

Yes sir got it...
Thanks a lot...
Yours,
@Apashanka