# lecture09 ```Force as gradient of potential energy
 
 dU ( x, y, z )  dW  Fdr  Fx dx  Fy dy  Fz dz 
Fx   U x ,
Fy   U y ,
Fz   z.

F   xˆ U x  yˆ U y  zˆ U z   gradU  U
Definition of the
operator “nabla”:
  xˆ
Force is equal to
potential energy



 yˆ  zˆ
x
y
z
f  x, y, z   gradf  x, y, z 
Nabla times a scalar function
is gradient of the function,
and it is a vector
1
Example: Find force if potential energy is α/r.
r x y z
2
2
2
dU dx  
 1 r 
x
  3
x
r

F   xˆ U x  yˆ U y  zˆ U z

F    xxˆ  yyˆ  zzˆ  r 3


F   r r 3   rˆ r 2
Question: Find is sign of the coefficient α in the following situations?
a) Newton’s force of gravity
b) Coulomb force between two positive charges
c) Coulomb force between two negative charges
d) Coulomb force between a positive charge and a negative charge
2
Same properties of nabla
f  xˆ
f
f
f
 yˆ
 zˆ
x
y
z
 f x f y f z
 f 


 divf
x y z
xˆ
 
 f 
x
fx
yˆ

y
fy
  xˆ



 yˆ  zˆ
x
y
z
Nabla times a scalar function
is gradient of the function
Nabla dot product a vector function
is divergence of the function
zˆ
 df df y   df x df z   df y df x 

  yˆ 
  curlf  rotf
 xˆ  z 


  zˆ
z
 dy dz   dz dx   dx dy 
fz
Nabla cross product a vector function
is curl or rotor of the function
Gradient and curl/rotor of a function are vectors
3
Stokes’s theorem
 
 
   F  ds   F  dr
The surface integral of the curl of a vector field over a surface S
is equal to the line integral of the vector field over its boundary
Another definition of conservative force
  

 F r dr  0    F  0
4
Example: Is Coulomb’s force conservative?

 3
2
ˆ
F  r r  r r
r  x2  y2  z 2
Fx   x 
3x r
1 Fy
  3  4
 3xy 5 

y y  r 
r y
r
x



 F  0



Fx Fy
 F z 

0
y
x


Force conservative and
work is path independent
Example: Find potential energy for Coulomb’s force.
  
r
r 
U  W    F r dr    F r dr    2 dr
r0
r0
r0 r

r
U  U r   U r0  

r


r0
If U r0     0 then U r  

r
5
Time dependent potential energy
 
 

 F r , t dr  0    F r , t   0
If a force satisfies these conditions
then potential energy can be defined

r  



U (r , t )  U (r0 , t )    F r , t dr
r0
U
U
U
U

 U
dU (r , t ) 
dx 
dy 
dz 
dt  U  dr 
dt
U
dt
x
y
z
t
t
 dU  dT 

t
 

U  dr   F  dr  dW  dT
dU  dT 
U
dt
t
If potential energy depends on time then
mechanical energy is not conserved it
gets transformed some other forms of
energy or to external systems.
6
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