*Notes to a video lecture on http://www.unizor.com*

__Curl in 3D__

This lecture continues analyzing a concept of

*curl*of a vector field, this time in three-dimensional space.

We will extensively use the material of a previous lecture on

*curl*in two-dimensional space.

Consider a three-dimensional vector field - the one that really exists in our three-dimensional world

*with components*

**V(x,y,z)**{

*}*

**V**_{x}(x,y,z),V_{y}(x,y,z),V_{z}(x,y,z)As a working model, we will use velocities of air molecules as a

*vector field*, sometimes calling it a "wind", and we assume that time does not participate in our analysis, so everything is related either to one particular moment in time or the wind does not change with time.

Wind might steadily blow in one direction, which would be an example of unidirectional vector field or might form a tornado, which is a perfect example of a vector field with a non-zero curl.

As in the two-dimensional case, let's consider a tiny pedal wheel that can be placed at any point in the air without fixing its axis of rotation, as we did in the two-dimensional case, where this axis was always parallel to Z-axis.

After some turning around with the wind, provided the wind is not changing, our pedal wheel will establish some position and will or will not spin in some direction with some angular speed.

Its axis of rotation then will point to a direction of the three-dimensional

*curl*of our vector field at that point and the speed of rotation gives the magnitude. These direction and magnitude define a

*curl*as a vector in three-dimensional vector field.

The complication in three-dimensional case is that our tiny pedal wheel can rotate around any axis directed anywhere in space.

Moreover, if it rotates around some axis at some point in space because a vector field curls around that axis, a slight change in the direction of this pedal wheel might still result in its rotation, maybe with lower velocity.

It helps to simplify the problem by breaking it in three already solved ones.

Consider a vector field

*with components*

**V(x,y,z)**{

*}*

**V**_{x}(x,y,z),V_{y}(x,y,z),V_{z}(x,y,z)Let

*curl*be a vector

**V(x,y,z)***with components*

**K(x,y,z)**{

*}*

**K**_{x}(x,y,z),K_{y}(x,y,z),K_{z}(x,y,z)Intuitively, the X-component

*of the vector*

**K**_{x}(x,y,z)*should represent the curl of the projection of the vector field*

**K(x,y,z)***onto the YZ-plane, that is a curl of a two-dimensional vector field with fixed X-component and two other components equal to*

**V(x,y,z)**{

*}*

**V**_{y}(x,y,z),V_{z}(x,y,z)Therefore,

*equals to*

**K**_{x}(x,y,z)*∂*

**V**∂y − ∂_{z}(x,y,z)/**V**∂z_{y}(x,y,z)/Analogously, the Y-component

*of the vector*

**K**_{y}(x,y,z)*should represent the curl of the projection of the vector field*

**K(x,y,z)***onto the ZX-plane, that is a curl of a two-dimensional vector field with fixed Y-component and two other components equal to*

**V(x,y,z)**{

*}*

**V**_{z}(x,y,z),V_{x}(x,y,z)Therefore,

*equals to*

**K**_{y}(x,y,z)*∂*

**V**∂z − ∂_{x}(x,y,z)/**V**∂x_{z}(x,y,z)/Similarly, the Z-component

*of the vector*

**K**_{z}(x,y,z)*should represent the curl of the projection of the vector field*

**K(x,y,z)***onto the XY-plane, that is a curl of a two-dimensional vector field with fixed Z-component and two other components equal to*

**V(x,y,z)**{

*}*

**V**_{x}(x,y,z),V_{y}(x,y,z)Therefore,

*equals to*

**K**_{z}(x,y,z)*∂*

**V**∂x − ∂_{y}(x,y,z)/**V**∂y_{x}(x,y,z)/Now we know all three components of the vector

*that represents*

**K(x,y,z)***curl*.

**V(x,y,z)**This vector can be represented as a sum of its components using the unit vectors along the coordinate axis

*along*

**i***axis,*

**OX***along*

**j***axis and*

**OY***along*

**k***axis:*

**OZ**

**K(x,y,z) = K**

+ K_{x}(x,y,z)·i ++ K

_{y}(x,y,z)·j + K_{z}(x,y,z)·kwhere the components of the curl vector were calculated above:

**K**

=∂_{x}(x,y,z) ==

**V**∂y − ∂_{z}(x,y,z)/**V**∂z_{y}(x,y,z)/

**K**

=∂_{y}(x,y,z) ==

**V**∂z − ∂_{x}(x,y,z)/**V**∂x_{z}(x,y,z)/

**K**

=∂_{z}(x,y,z) ==

**V**∂x − ∂_{y}(x,y,z)/**V**∂y_{x}(x,y,z)/*Curl as a Vector Product*

All we need now to make all the formulae more beautiful is to apply some vector algebra.

Let's take any two vectors in their coordinates form:

**P = P**_{x}·i + P_{y}·j + P_{z}·k

**Q = Q**_{x}·i + Q_{y}·j + Q_{z}·kNow let's form a vector product of these two vectors and express it also in coordinate form, using trivial identities

*⨯*

**i***;*

**i=0***⨯*

**j***;*

**j=0***⨯*

**k***;*

**k=0***⨯*

**i***;*

**j=k***⨯*

**j***;*

**i=−k***⨯*

**j***;*

**k=i***⨯*

**k***;*

**j=−i***⨯*

**k***;*

**i=j***⨯*

**i***.*

**k=−j**Skipping the terms equaled to zero, the vector (cross) product of

*by*

**P***equals to*

**Q***⨯*

**P***⨯*

**Q =**

= P= P

_{x}·Q_{y}·i*⨯*

**j + P**_{x}·Q_{z}·i*⨯*

**k +**

+ P+ P

_{y}·Q_{x}·j*⨯*

**i + P**_{y}·Q_{z}·j*⨯*

**k +**

+ P+ P

_{z}·Q_{x}·k*⨯*

**i + P**_{z}·Q_{y}·k

**j =**

= P

+ P

+ P

= (P

+ (P

+ (P= P

_{x}·Q_{y}·k + P_{x}·Q_{z}·(−j) ++ P

_{y}·Q_{x}·(−k) + P_{y}·Q_{z}·i ++ P

_{z}·Q_{x}·j + P_{z}·Q_{y}·(−i) == (P

_{y}Q_{z}−P_{z}Q_{y})·i ++ (P

_{z}Q_{x}−P_{x}Q_{z})·j ++ (P

_{x}Q_{y}−P_{y}Q_{x})·kNow consider a pseudo-vector

**∇**with components

{

*∂*}

**/**∂**x,**∂**/**∂**y,**∂**/**∂**z**and a vector field

*with components*

**V(x,y,z)**{

*}*

**V**_{x}(x,y,z),V_{y}(x,y,z),V_{z}(x,y,z)Use pseudo-vector

**∇**as vector

*in the above expression for a vector (cross) product and vector*

**P***as vector*

**V(x,y,z)***.*

**Q**Then, omitting

*in vector field and its components for brevity, their vector product would look like*

**(x,y,z)****∇**⨯

**V =**

= (∂= (

**V**∂_{z}/**y −**∂**V**∂_{y}/**z)·i +**

+ (∂+ (

**V**∂_{x}/**z −**∂**V**∂_{z}/**x)·j +**

+ (∂+ (

**V**∂_{y}/**x −**∂**V**∂_{x}/**y)·k**which exactly the same as an expression for

*curl*derived earlier in this lecture.

**V(x,y,z)**We have come to a very short form of a vector

*representing a curl of a vector field*

**K(x,y,z)***at point {*

**V(x,y,z)***}:*

**x,y,z**
*curl V(x,y,z) = *

**∇**⨯

**V(x,y,z)**
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