Download 21.3 Finding Scalar Potentials

∇ × a = 0 is usually used a practical test of whether a field is conservative or not (by simply
Again first check that ∇ × a = 0,
checking to see if indeed ∇ × a = 0).
(∇ × a)i = ǫijk ∂j (2(cl xl )xk + xl xl ck )
Note: φ(r) is only determined up to a constant: if ψ = φ + constant then ∇ψ = ∇φ and
ψ can equally well serve as a potential. The freedom in the constant corresponds to the
freedom in choosing r 0 to calculate the potential. Equivalently the absolute value of a scalar
potential has no meaning, only potential differences are significant.
21.3
= ǫijk (2cl δjl xk + 2c · rδjk + 2xl δlj ck )
= 2ǫijk (cj xk + xj ck ) = 0
Thus
φ(r) =
Finding Scalar Potentials
Z r
a(r ′ ) · dr′ =
0
Z 1
Z
1
a(λ r) · (dλ r)
0
2 (c · λ r) λ r + λ2 r 2 c · (dλ r)
Method (1): Integration along a straight line.
=
We have shown that the scalar potential φ(r) for a conservative vector field a(r) can be
constructed from a line integral which is independent of the path of integration between the
Z
= 2 (c · r) r · r + r 2 (c · r)
endpoints. Therefore, a convenient way of evaluating such integrals is to integrate along a
straight line. Depending on the convergence of the integral we have two (obvious) choices
= r 2 (c · r)
0
1
λ2 dλ
0
In general this is an elegant method.
• r 0 = 0 if φ(r0 ) finite (ie an irrelevant constant) so setting r ′ = λr we have
φ(r) = r ·
Z
Method (2): Direct Integration “by inspection” (guessing).
λ=1
a(λr) dλ ,
Sometimes the result can be directly spotted,
λ=0
Example 1 (revisited):
• r 0 = ∞ if φ(r0 ) finite (ie an irrelevant constant) so setting r ′ = λr we have
φ(r) = r ·
Z
λ=∞
a = (2xy + z 3 , x2 , 3xz 2 )
∂ 2
∂
∂
=
(x y + z 3 ), (x2 y + z 3 ), (x2 y + z 3 )
∂x
∂y
∂z
= ∇ (x2 y + z 3 )
a(λr) dλ .
λ=1
Example 1:
Let a = (2xy + z 3 )ex + x2 ey + 3xz 2 ez .
Can be tricky to spot though.
We first find that ∇ × a = 0, so the field is conservative. Thus
φ(r) =
Z
1
Z
1
Method (3): Direct Integration.
a(λr) · rdλ
0
=
As a = ∇φ then
[(2λ2 xy + λ3 x3 )x + λ2 x2 y + λ3 3xz 3 ]dλ
0
2
3
2
= 2/3x y + 1/4xz + 1/3x y + 3/4xz
3
∂φ
= ax (x, y, z)
∂x
∂φ
= ay (x, y, z)
∂y
∂φ
= az (x, y, z)
∂z
and integrate the equations
= x2 y + xz 3
Note: Always check that your φ(r) satisfies a(r) = ∇ φ(r).
2
Example 2: Let a(r) = 2 (c · r) r + r c where a is a constant vector.
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Z
x
φ(x, y, z) =
Z
y
φ(x, y, z) =
z
φ(x, y, z) =
Z
dx′ ax (x′ , y, z) + f (y, z)
dy ′ ay (x, y ′ , z) + g(x, z)
dz ′ az (x, y, z ′ ) + h(x, y)
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and then determine by consistency the ‘integration constants’ f, g, h. A variation is to
differentiate the first equation wrt y and then solve for f , ie
Z x
∂
∂f
∂φ
=
(y, z) = ay (x, y, z)
dx′ ax (x′ , y, z) +
∂y
∂y
∂y
to give a further ‘integration constant’ g(z), which can be found by differentiating the first
equation wrt z.
Example 1 (revisited):
Let a = (2xy + z 3 )ex + x2 ey + 3xz 2 ez . Then
We can simplify the left-hand side of this equation to obtain
Z r
Z tB
Z tB
B
1 d 2
r̈ · ṙ dt = m
m
ṙ dt = 21 m[vB2 − vA2 ],
r̈ · dr = m
2
dt
rA
tA
tA
where vA and vB are the magnitudes of the velocities at points A and B respectively.
The right-hand side simply gives
Z r
Z r
B
B
−
∇V (r) · dr = −
dV = VA − VB
rA
rA
where VA and VB are the values of the potential V at r A and r B , respectively. Therefore
φ = x2 y + xz 3 + f (y, z)
φ = x2 y
φ =
xz 3 + h(x, y)
They agree if we choose f (y, z) = 0, g(x, z) = xz 3 , h(x, y) = x2 y 3.
Straightforward method but can be messy though.
21.4
1 2
1
mv + VA = mvB2 + VB
2 A
2
+ g(y, z)
Conservative Forces: Conservation of Energy
Let us now see how the name conservative field arises. Consider a vector field F (r) corresponding to the only force acting on some test particle of mass m. We will show that for a
conservative force, where we can write
F = −∇V
the total energy is constant in time. (The force is minus the gradient of the (scalar)
potential. The minus sign is conventional and chosen so that potential energy decreases as
and the total energy
E = 21 mv 2 + V
is conserved, i.e. constant in time.
Newtonian gravity and the electrostatic force are both conservative. Frictional forces are not
conservative; energy is dissipated and work is done in traversing a closed path. In general,
time-dependent forces are not conservative.
21.5
Physical Examples of Conservative Forces
Newtonian Gravity and the electrostatic force are both conservative. Frictional forces are not
conservative; energy is dissipated and work is done in traversing a closed path. In general,
time-dependent forces are not conservative.
The foundation of Newtonian Gravity is Newton’s Law of Gravitation. The force F on
a particle of mass m1 at r due to a particle of mass m at the origin is given by
F = −G m m1
the force does work.)
Proof: The particle moves under the influence of Newton’s Second Law:
r
r3
where G = 6.672 59(85) × 10−11 Nm2 kg2 is Newton’s Gravitational Constant.
The gravitational field G(r) (due to the mass at the origin) is
mr̈ = F (r).
F (r) = m1 G(r)
or G(r) = −G m
Consider a small displacement dr along the path taking time dt. Then
mr̈ · dr = F (r) · dr = −∇V (r) · dr.
Integrating this expression along the path from rA at time t = tA to r B at time t = tB yields
m
Z r
B
rA
r̈ · dr = −
Z r
B
rA
∇V (r) · dr.
where the test mass m1 is so small that is gravitational field can be ignored. The gravitational
field is conservative as
(∇ × (r/r 3 ))i = ǫijk ∂j (xk /r 3 ) = ǫijk (δjk /r 3 − 3xj xk /r 5 ) = 0
The gravitational potential defined by
G = −∇φ
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r
r3
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can be obtained by spotting the direct integration (inspection) giving
φ=−
Gm
r
Alternatively, to calculate by a line integral choose r0 = ∞ then
Z 1
Z r
G(λr) · dλr
φ(r) = −
G(r ′ ) · dr′ = −
∞
Z 1∞
Gm (r · r) dλ
Gm
=
=−
r3
λ2
r
∞
NB In this example the vector field G is singular at the origin r = 0. This implies we have
to exclude the origin and it is not possible to obtain the scalar potential at r by integration
along a path from the origin. Instead we integrate from infinity, which in turn means that
the gravitational potential at infinity is zero.
NB Since F = m1 G = −∇(m1 φ) the potential energy of the mass m1 is V = m1 φ. The
distinction (a convention) between potential and potential energy is a common source of
confusion.
Electrostatics: Coulomb’s Law states that the force F on a particle of charge q1 at r in
the electric field E due to a particle of charge q at the origin is given by
F = q1 E =
q1 q r
4πǫ0 r 3
where ǫ0 = 107 /(4πc2 ) = 8.854 187 817 · · · × 10−12 C 2 N −1 m−2 is the Permitivity of Free
Space. Again the test charge q1 is taken as small, so as not to disturb the electric field.
The electrostatic potential is taken as
φ=
q
4πǫ0 r
(obtained by integrating E = −∇φ from infinity to r) and the potential energy of a charge
q1 in the electric field is V = q1 φ.
Note that mathematically electrostatics and gravitation are very similar, the only real difference being that gravity between two masses is always attractive, whereas like charges
repel.
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