Normal Derivative at Boundary Surface S Explained

[ssky]

\frac{\partial }{\partial n} is the normal derivative at the boundary surface S in the outward direction defines as

\frac{\partial }{\partial n}= l_{x}\frac{\partial }{\partial x}+l_{y}\frac{\partial }{\partial y}+l_{z}\frac{\partial }{\partial z}

What is that mean? I have no knowledge about this formula.

can someone explain to me:

1- the meaning of this formula.
2- the meaning of \l_{x},l_{y},l_{z} and is there a defination for them.

 

[Unrest]

1- It's the derivative of some quantity along the direction of the surface normal, rather than the usual way of differentiating along the direction of an axis.

2- l_x, l_y, l_z are the components of the surface normal vector

An example, in the special case where the surface is parallel to the yz plane, and facing +x:
lx=1, ly=0, lz=0
so
<br /> \frac{\partial }{\partial n}= \frac{\partial }{\partial x}<br />

 

[chiro]

\frac{\partial }{\partial n} is the normal derivative at the boundary surface S in the outward direction defines as

\frac{\partial }{\partial n}= l_{x}\frac{\partial }{\partial x}+l_{y}\frac{\partial }{\partial y}+l_{z}\frac{\partial }{\partial z}

What is that mean? I have no knowledge about this formula.

can someone explain to me:

1- the meaning of this formula.
2- the meaning of \l_{x},l_{y},l_{z} and is there a defination for them.

Think about the dot product and its relevance in this context.

 

[ssky]

1- It's the derivative of some quantity along the direction of the surface normal, rather than the usual way of differentiating along the direction of an axis.

2- l_x, l_y, l_z are the components of the surface normal vector

An example, in the special case where the surface is parallel to the yz plane, and facing +x:
lx=1, ly=0, lz=0
so
<br /> \frac{\partial }{\partial n}= \frac{\partial }{\partial x}<br />

Thanks...

but you said that :

l_x, l_y, l_z are the components of the surface normal vector

and i found in a book that l_x, l_y, l_z being the direction cosines.

so, is there a different between the two meanings?

 

[ssky]

Think about the dot product and its relevance in this context.

Yes, but I'm confused about the n

 

[Rasalhague]

Have you heard of "directional derivatives"? Your normal derivative is just the directional derivative in the direction of a vector normal to a given surface. To calculate the value of a directional derivative at some point, in a direction specified by a unit vector, we can take the dot product of that unit vector with the gradient. Suppose we have a normal vector defined at some point on a surface:

\mathbf{n} = l_x \mathbf{i} + l_y \mathbf{j} + l_y \mathbf{k},

with unit length:

\left \| \mathbf{n} \right \| = 1,

and a scalar field f with gradient

\nabla f = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j} + \frac{\partial f }{\partial z} \mathbf{k}.

Then value of the directional derivative is

\nabla f \cdot \mathbf{n} = l_x \frac{\partial f}{\partial x} + l_y \frac{\partial f}{\partial y} + l_z \frac{\partial f }{\partial z},

evaluated at that point. (The direction cosines of a unit vector are its scalar components in the x, y and z directions; you can find the direction cosine in a particular direction by taking the dot product of the vector with the standard basis vector, \mathbf{i}, \mathbf{j} or k corresponding to that direction.)

Sometimes the gradient operator (operator: a function of functions) is expressed like this, on it own, without any scalar field, "plugged in" yet:

\nabla = \frac{\partial }{\partial x} \mathbf{i} + \frac{\partial }{\partial y} \mathbf{j} + \frac{\partial }{\partial z} \mathbf{k}.

Now, another way of expressing this whole idea is to not mention the gradient, as such, but simply to think of the directional derivative operator as acting, in its own right, on a scalar field. In that case, we can write

\frac{\partial }{\partial n} = l_x \frac{\partial }{\partial x} + l_y \frac{\partial }{\partial y} + l_z \frac{\partial }{\partial z}.

When we input a scalar field into this operator, the result is (defined to be) the same as taking the dot product of a unit normal vector with the gradient of the scalar field, i.e. the value of the directional derivative in the direction normal to the surface:

l_x \frac{\partial f}{\partial x} + l_y \frac{\partial f}{\partial y} + l_z \frac{\partial f }{\partial z}.

 

[Rasalhague]

Thanks...

but you said that :

l_x, l_y, l_z are the components of the surface normal vector

and i found in a book that l_x, l_y, l_z being the direction cosines.

so, is there a different between the two meanings?

There's no difference if the vector has unit length. The component of a vector in the direction of a particular coordinate axis is the vector's direction cosine for that direction multiplied by its length. So if the length of the vector is 1, the component is the direction cosine (multiplied by 1), which is just the direction cosine.

 

[chiro]

Yes, but I'm confused about the n

It is like your other derivatives, but in the context of a vector. Your derivative measures the change in reference to a particular vector n which points in some direction. The dot product represents in a way your direction cosines which are just projections.

 

[ssky]

Have you heard of "directional derivatives"? Your normal derivative is just the directional derivative in the direction of a vector normal to a given surface. To calculate the value of a directional derivative at some point, in a direction specified by a unit vector, we can take the dot product of that unit vector with the gradient. Suppose we have a normal vector defined at some point on a surface:

\mathbf{n} = l_x \mathbf{i} + l_y \mathbf{j} + l_y \mathbf{k},

with unit length:

\left \| \mathbf{n} \right \| = 1,

and a scalar field f with gradient

\nabla f = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j} + \frac{\partial f }{\partial z} \mathbf{k}.

Then value of the directional derivative is

\nabla f \cdot \mathbf{n} = l_x \frac{\partial f}{\partial x} + l_y \frac{\partial f}{\partial y} + l_z \frac{\partial f }{\partial z},

evaluated at that point. (The direction cosines of a unit vector are its scalar components in the x, y and z directions; you can find the direction cosine in a particular direction by taking the dot product of the vector with the standard basis vector, \mathbf{i}, \mathbf{j} or k corresponding to that direction.)

Sometimes the gradient operator (operator: a function of functions) is expressed like this, on it own, without any scalar field, "plugged in" yet:

\nabla = \frac{\partial }{\partial x} \mathbf{i} + \frac{\partial }{\partial y} \mathbf{j} + \frac{\partial }{\partial z} \mathbf{k}.

Now, another way of expressing this whole idea is to not mention the gradient, as such, but simply to think of the directional derivative operator as acting, in its own right, on a scalar field. In that case, we can write

\frac{\partial }{\partial n} = l_x \frac{\partial }{\partial x} + l_y \frac{\partial }{\partial y} + l_z \frac{\partial }{\partial z}.

When we input a scalar field into this operator, the result is (defined to be) the same as taking the dot product of a unit normal vector with the gradient of the scalar field, i.e. the value of the directional derivative in the direction normal to the surface:

l_x \frac{\partial f}{\partial x} + l_y \frac{\partial f}{\partial y} + l_z \frac{\partial f }{\partial z}.

Thank you so much... for your explanation

Now, I really understood what do you mean...

 

[ssky]

It is like your other derivatives, but in the context of a vector. Your derivative measures the change in reference to a particular vector n which points in some direction. The dot product represents in a way your direction cosines which are just projections.

Thank you for your interesting...