Divergence and curl - Neil Strickland's home page · div and curl Now suppose we have a vector...

Divergence and curl

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z .

We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h)

=∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h)

=∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h)

=∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h) =

∂f

∂x+∂g

∂y+∂h

∂z

= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h) =

∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h) =

∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h) =

∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h) =

∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field

, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h) =

∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field.

The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h) =

∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u).

Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

div and curl

Now suppose we have a vector field u = (f , g , h), so f , g and h are allfunctions of x , y and z . We can think of ∇ as itself being a strange kind ofvector, in which the entries are differential operators:

∇ =

(∂

∂x,∂

∂y,∂

∂z

).

This means we can make sense of the dot product ∇.u and the cross product∇× u as follows:

∇.u =

(∂

∂x,∂

∂y,∂

∂z

).(f , g , h) =

∂f

∂x+∂g

∂y+∂h

∂z= fx + gy + hz

∇× u = det

i j k∂∂x

∂∂y

∂∂z

f g h

= (hy − gz , fz − hx , gx − fy ).

Note that ∇.u is a scalar field, and ∇× u is a vector field. The scalar field∇.u is called the divergence of u, and is sometimes written as div(u). Thevector field ∇× u is called the curl of u, and is sometimes written curl(u).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2)

= 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x)

= cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x)

= cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x)

= cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x)

= cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x)

= cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x) = cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x) = cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x) = cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x) = cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z)

= 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x) = cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z) = 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x) = cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z) = 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Examples of div and curl

(a) For the vector field u = (x2 + y 2, y 2 + z2, z2 + x2) we have

∇.u =∂

∂x(x2 + y 2) +

∂

∂y(y 2 + z2) +

∂

∂z(z2 + x2) = 2x + 2y + 2z

∇× u = det

i j k∂∂x

∂∂y

∂∂z

x2 + y 2 y 2 + z2 z2 + x2

= (−2z ,−2x ,−2y).

(b) For the vector field u = (sin(x), sin(x), sin(x)) we have

∇.u =∂

∂xsin(x) +

∂

∂ysin(x) +

∂

∂zsin(x) = cos(x) + 0 + 0 = cos(x)

∇× u = det

i j k∂∂x

∂∂y

∂∂z

sin(x) sin(x) sin(x)

= (0,− cos(x), cos(x)).

(c) For the vector field u = (−y , x , z) we have

∇.u =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(z) = 0 + 0 + 1 = 1

∇× u = det

i j k∂∂x

∂∂y

∂∂z

−y x z

= (0, 0, 2).

Div and curl of matrix fields

Consider a vector field

u =

ax + by + czdx + ey + fzgx + hy + iz

=

a b cd e fg h i

xyz

Then

div(u) =∂

∂x(ax + by + cz) +

∂

∂y(dx + ey + fz) +

∂

∂z(gx + hy + iz)

= a + e + i = trace

a b cd e fg h i

curl(u) = det

i j k∂∂x

∂∂y

∂∂z

ax + by + cz dx + ey + fz gx + hy + iz

= (h − f , c − g , d − b)

So curl(u) = 0 if the matrix

a b cd e fg h i

is symmetric.

Div and curl of matrix fields

Consider a vector field

u =

ax + by + czdx + ey + fzgx + hy + iz

=

a b cd e fg h i

xyz

Then

div(u) =∂

∂x(ax + by + cz) +

∂

∂y(dx + ey + fz) +

∂

∂z(gx + hy + iz)

= a + e + i = trace

a b cd e fg h i

curl(u) = det

i j k∂∂x

∂∂y

∂∂z

ax + by + cz dx + ey + fz gx + hy + iz

= (h − f , c − g , d − b)

So curl(u) = 0 if the matrix

a b cd e fg h i

is symmetric.

Div and curl of matrix fields

Consider a vector field

u =

ax + by + czdx + ey + fzgx + hy + iz

=

a b cd e fg h i

xyz

Then

div(u) =∂

∂x(ax + by + cz) +

∂

∂y(dx + ey + fz) +

∂

∂z(gx + hy + iz)

= a + e + i = trace

a b cd e fg h i

curl(u) = det

i j k∂∂x

∂∂y

∂∂z

ax + by + cz dx + ey + fz gx + hy + iz

= (h − f , c − g , d − b)

So curl(u) = 0 if the matrix

a b cd e fg h i

is symmetric.

Div and curl of matrix fields

Consider a vector field

u =

ax + by + czdx + ey + fzgx + hy + iz

=

a b cd e fg h i

xyz

Then

div(u) =∂

∂x(ax + by + cz) +

∂

∂y(dx + ey + fz) +

∂

∂z(gx + hy + iz)

= a + e + i

= trace

a b cd e fg h i

curl(u) = det

i j k∂∂x

∂∂y

∂∂z

ax + by + cz dx + ey + fz gx + hy + iz

= (h − f , c − g , d − b)

So curl(u) = 0 if the matrix

a b cd e fg h i

is symmetric.

Div and curl of matrix fields

Consider a vector field

u =

ax + by + czdx + ey + fzgx + hy + iz

=

a b cd e fg h i

xyz

Then

div(u) =∂

∂x(ax + by + cz) +

∂

∂y(dx + ey + fz) +

∂

∂z(gx + hy + iz)

= a + e + i = trace

a b cd e fg h i

curl(u) = det

i j k∂∂x

∂∂y

∂∂z

ax + by + cz dx + ey + fz gx + hy + iz

= (h − f , c − g , d − b)

So curl(u) = 0 if the matrix

a b cd e fg h i

is symmetric.

Div and curl of matrix fields

Consider a vector field

u =

ax + by + czdx + ey + fzgx + hy + iz

=

a b cd e fg h i

xyz

Then

div(u) =∂

∂x(ax + by + cz) +

∂

∂y(dx + ey + fz) +

∂

∂z(gx + hy + iz)

= a + e + i = trace

a b cd e fg h i

curl(u) = det

i j k∂∂x

∂∂y

∂∂z

ax + by + cz dx + ey + fz gx + hy + iz

= (h − f , c − g , d − b)

So curl(u) = 0 if the matrix

a b cd e fg h i

is symmetric.

Div and curl of matrix fields

Consider a vector field

u =

ax + by + czdx + ey + fzgx + hy + iz

=

a b cd e fg h i

xyz

Then

div(u) =∂

∂x(ax + by + cz) +

∂

∂y(dx + ey + fz) +

∂

∂z(gx + hy + iz)

= a + e + i = trace

a b cd e fg h i

curl(u) = det

i j k∂∂x

∂∂y

∂∂z

ax + by + cz dx + ey + fz gx + hy + iz

= (h − f , c − g , d − b)

= (h − f , c − g , d − b)

So curl(u) = 0 if the matrix

a b cd e fg h i

is symmetric.

Div and curl of matrix fields

Consider a vector field

u =

ax + by + czdx + ey + fzgx + hy + iz

=

a b cd e fg h i

xyz

Then

div(u) =∂

∂x(ax + by + cz) +

∂

∂y(dx + ey + fz) +

∂

∂z(gx + hy + iz)

= a + e + i = trace

a b cd e fg h i

curl(u) = det

i j k∂∂x

∂∂y

∂∂z

ax + by + cz dx + ey + fz gx + hy + iz

= (h − f , c − g , d − b)

So curl(u) = 0 if the matrix

a b cd e fg h i

is symmetric.

grad, div and curl in two dimensions

(a) For a scalar field f in two dimensions, grad(f ) = ∇(f ) = (fx , fy )(a vector field).

(b) For a vector field u = (p, q) in two dimensions, div(u) = ∇.u = px + qy(a scalar field).

(c) For a vector field u = (p, q) in two dimensions,

curl(u) = det

[∂∂x

∂∂y

p q

]= qx − py

(a scalar field, not a vector field as in three dimensions).

grad, div and curl in two dimensions

(a) For a scalar field f in two dimensions, grad(f ) = ∇(f ) = (fx , fy )(a vector field).

(b) For a vector field u = (p, q) in two dimensions, div(u) = ∇.u = px + qy(a scalar field).

(c) For a vector field u = (p, q) in two dimensions,

curl(u) = det

[∂∂x

∂∂y

p q

]= qx − py

(a scalar field, not a vector field as in three dimensions).

grad, div and curl in two dimensions

(a) For a scalar field f in two dimensions, grad(f ) = ∇(f ) = (fx , fy )(a vector field).

(b) For a vector field u = (p, q) in two dimensions, div(u) = ∇.u = px + qy(a scalar field).

(c) For a vector field u = (p, q) in two dimensions,

curl(u) = det

[∂∂x

∂∂y

p q

]= qx − py

(a scalar field, not a vector field as in three dimensions).

Geometric interpretation of div(u)

It works out that the divergence div(u) = ∇.u is positive when the vectors uare spreading out, and negative when they are coming together.

diverging: ∇.u > 0 converging: ∇.u < 0

For the velocity field of an incompressible fluid we will have ∇.u = 0.

Geometric interpretation of curl(u)

In two dimensions, it works out that curl(u) > 0 in regions where the field iscurling anticlockwise, and curl(u) < 0 in regions where it is curling clockwise,and the absolute value of curl(u) is determined by the strength of the curling.

curl(u) > 0, smaller curl(u) < 0, larger

In three dimensions, the field u can curl around any axis. In this context,curl(u) is also a vector field, and it will point along the axis of the curling.

Geometric interpretation of curl(u)

In two dimensions, it works out that curl(u) > 0 in regions where the field iscurling anticlockwise, and curl(u) < 0 in regions where it is curling clockwise,and the absolute value of curl(u) is determined by the strength of the curling.

curl(u) > 0, smaller curl(u) < 0, larger

In three dimensions, the field u can curl around any axis. In this context,curl(u) is also a vector field, and it will point along the axis of the curling.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0

∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0

∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0

∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0

∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0

∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0

∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on.

The various fields are related by the following equations:

∇.E = ρ/ε0

∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0

∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0 ∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0 ∇× E = −B

∇.B = 0

∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0 ∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0 ∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0 ∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0 ∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0 ∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl.

Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Maxwell’s equations

These involve:

I The electric field E, which is a vector field.

I The magnetic field B, which is another vector field.

I The current density J, which is also a vector field.

I The charge density ρ, which is a scalar field.

I Two constants: ε0 ' 8.854× 10−12F/m2 and µ0 ' 1.257× 10−6Hm−1.

The quantities E, B, J and ρ may also depend on time; we write E for ∂E/∂tand so on. The various fields are related by the following equations:

∇.E = ρ/ε0 ∇× E = −B

∇.B = 0 ∇× B = µ0J + µ0ε0E.

This means that:

I The electric field diverges in regions where there is positive charge, andconverges in regions where there is negative charge.

I The magnetic field never diverges or converges.

I Changing magnetic fields cause the electric field to curl.

I Currents cause the magnetic field to curl. Changing electric fields alsocause the magnetic field to curl, but the effect is usually much weaker,because ε0 is small.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows.

Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct))

= 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light)

, andlet α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct))

= 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant.

We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct))

= 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct))

= 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct))

= 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0

= ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0

E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c

= 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0

B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct)))

= −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0)

= E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2

= µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations.

It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Plane wave solution to Maxwell’s equations

One class of solutions to Maxwell’s equations is as follows. Putc = 1/

√µ0ε0 ' 3× 108ms−1 (which turns out to be the speed of light), and

let α be any constant. We can take J = 0 and ρ = 0 and

E = (0, sin(α(x − ct)), 0) B = (0, 0, sin(α(x − ct))/c).

We find that

∇.E =∂

∂ysin(α(x − ct)) = 0 = ρ/ε0 E = (0,−αc cos(α(x − ct)), 0)

∇.B =∂

∂zsin(α(x − ct))/c = 0 B = (0, 0,−α cos(α(x − ct))

∇× E = det

i j k∂∂x

∂∂y

∂∂z

0 sin(α(x − ct)) 0

= (0, 0, α cos(α(x − ct))) = −B

∇× B = det

i j k∂∂x

∂∂y

∂∂z

0 0 sin(α(x − ct))/c

= (0,−α cos(α(x − ct))/c, 0) = E/c2 = µ0ε0 E.

This shows that we do indeed have a solution to the equations. It represents anelectromagnetic wave of wavelength 1/α moving at speed c in the x-direction.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero.

It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0.

Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r

, so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5.

In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5.

Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3

(−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5

(−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3

= 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3

= 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed.

This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.

Stationary charged particle

Another solution to Maxwell’s equations has E = (−xr−3,−yr−3,−zr−3) with

all other fields (B, J and ρ) being zero. It is clear that E = 0 and B = 0, so theonly equations that we need to check are that ∇.E = 0 and ∇× E = 0. Forthis we recall that rx = x/r , so (r−3)x = −3r−4rx = −3xr−5. In the same way,we have (r−3)y = −3yr−5 and (r−3)z = −3zr−5. Using this we find that

(−xr−3)x = 3x2r−5 − r−3 (−xr−3)y = 3xyr−5 (−xr−3)z = 3xzr−5

(−yr−3)x = 3xyr−5 (−yr−3)y = 3y2r−5 − r−3 (−yr−3)z = 3yzr−5

(−zr−3)x = 3xzr−5 (−zr−3)y = 3yzr−5 (−zr−3)z = 3z2r−5 − r−3.

∇.E = (−xr−3)x + (−yr−3)y + (−zr−3)z

= 3x2r−5 − r−3 + 3y 2r−5 − r−3 + 3z2r−5 − r−3

= 3(x2 + y 2 + z2)r−5 − 3r−3 = 3r 2r−5 − 3r−3 = 0.

∇× E =(

(−zr−3)y − (−yr−3)z , (−xr−3)z − (−zr−3)x , (−yr−3)x − (−xr−3)y)

=(

3yzr−3 − 3yzr−3, 3xzr−3 − 3xzr−3, 3xyr−3 − 3xyr−3)

= (0, 0, 0).

This shows that we have a solution to the equations, as claimed. This onerepresents the electric field of a single stationary particle at the origin, with nomagnetic field.