Newton's Third Law in the Framework of Special Relativity

0.12 MB·English

by  Miron Tuval

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1 Newton’s Third Law in the Framework of Special
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2 Relativity

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J Miron Tuval and Asher Yahaloma,b

6 a Isaac Newton Institute for Mathematical Sciences,
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20 Clarkson Road, Cambridge CB3 0EH, United Kingdom

] b Ariel University Center of Samaria, Ariel 40700, Israel
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e-mails: [email protected]; [email protected]
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February 12, 2013
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c Abstract
i
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y Newton’sthirdlawstatesthatanyactioniscounteredbyareaction
h ofequalmagnitudebutoppositedirection. Thetotalforceinasystem
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not affected by externalforces is thus zero. However,accordingto the
[
principles of relativity a signal can not propagate at speeds exceeding

1 the speed of light. Hence the action cannot be generated at the same
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timewiththereactionbecausetheinformationabouttheactionhasto
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reach the affected object and the affected object still needs additional
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5 time to react on the source, hence the total force cannot be null at a
2 giventime. Thefollowinganalysisprovidesforabetter understanding
. ofthewaysnaturallawswouldbehavewithintheframeworkofSpecial
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0 Relativity, and on how this understanding may be used for practical
3 purposes.
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1 Introduction
Among the major achievements of Sir Isaac Newton is the formulation of
Newton’s third law stating that any action is countered by a reaction of
equalmagnitudebutoppositedirection[1,2]. Thetotalforceinasystemnot
affected byexternal forces is thuszero. Thislaw has numerousexperimental
verifications and seems to be one of the corner stones of physics. However,
by the middle of the nineteenth century Maxwell has formulated the laws
of electromagnetism in his famous four partial differential equations [3, 4, 5]
which were formulated in their current form by Oliver Heaviside [6]. One of
the consequences of theseequations is that an electromagnetic signalcannot
travel at speeds exceeding that of light. This was later used by Albert
Einstein [7, 4, 5] (among other things) to formulate his special theory of
relativity which postulates that the speed of light is the maximal allowed
velocity in nature. According to the principles of relativity no signal (even
if not electromagnetic) can propagate at superluminal velocities. Hence an
action and its reaction cannot be generated at the same time because the
knowledge abouttheaction has to reach theaffected object and the affected
objectstillneedsadditionaltimetoreactonthesource. Thusthetotalforce
cannot be null at a given time. In consequence, by not holding rigorously
the simultaneity of action and reaction Newton’s third law cannot hold in
exact form butonly as an approximation. Moreover, thetotal force within a
system that is not acted upon by an external force would not be rigorously
null since the action and reactions are not able to balance each other and
the total force on a system which is not affected by an external force in not
null in an exact sense.
Most locomotive systems of today are based on open systems. A rocket
sheds exhaust gas to propel itself, a speeding bullet generates recoil. A car
pushes the road with the same force that is used to accelerate it, the same is
true regarding the interaction of a plane with air and of a ship with water.
However, the above relativistic considerations suggest’s a new type of motor
which is not based on a open system but rather on a closed one.
In this paper we discuss the force between two current carrying coils.
Two mathematical treatments will be given, in one we consider an instan-
taneous action at a distance, here Newton’s third law hold and the total
electromagnetic force will be shown to be equal to zero. Then we consider
thedynamic,electro-magneticconditionwherethereactiontoanactioncan-
not occur before having the action-generated information reach the affected
object, thus bringing about a non-zero resultant.
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Figure 1: Two current loops.
2 The Magneto - Static Condition
Consider a wire segment of length d~l located at ~x carrying a current I
2 2 2
(see figure 1). The magneto-static field at point ~x due to the line segment
1
at ~x is [4]:
2
µ d~l ×~x
dB~(~x )= 0I 2 12. (1)
1 4π 2 |~x |3
12
In the above ~x = ~x − ~x and µ = 4π10−7 is the vacuum magnetic
12 1 2 0
permeability. The magnetic field at point ~x originating from the entire
1
current loop 2 is given by the integral:
µ d~l ×~x
B~(~x )= 0I 2 12. (2)
1 4π 2 |~x |3
I 12
Now if at point ~x there is a wire segment of length d~l carrying a current
1 1
I , then according to the law of Lorentz [4] the wire segment will experience
1
a force:
dF~ = I d~l ×B~(~x ). (3)
12 1 1 1
The entire current loop 1 will experience a total force:
F~ = I d~l ×B~(~x ). (4)
12 1 1 1
I
Inserting equation (2) into equation (4) we obtain the following expression
for the force on the current loop 1:
µ d~l ×(d~l ×~x )
F~ = 0I I 1 2 12 . (5)
12 4π 1 2 |~x |3
I I 12
Using standard vector identities one can show that:
d~l ×(d~l ×~x ) = (d~l ·~x )d~l −(d~l ·d~l )~x , (6)
1 2 12 1 12 2 1 2 12
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hence equation (5) can be written in the form:
µ (d~l ·~x )d~l (d~l ·d~l )~x
F~ = 0I I 1 12 2 − 1 2 12 . (7)
12 4π 1 2"I I |~x12|3 I I |~x12|3 #
However since:
~x 1
12 ~
= −∇ (8)
|~x |3 ~x1|~x |
12 12
in which ∇~ is the gradient operator with respect to the ~x coordinate.
~x1 1
Hence we obtain:
d~l ·~x 1
1 12 = − d~l ·∇~ = 0. (9)
|~x |3 1 ~x1|~x |
I 12 I 12
Thus the first integral in equation (7) does not contribute and we are left
with the expression:
µ (d~l ·d~l )~x
F~ = − 0I I 1 2 12. (10)
12 4π 1 2 |~x |3
I I 12
From the above it is easy to calculate the magneto-static force applied by
current loop 1 on current loop 2 by changing indices 1 ↔ 2:
µ (d~l ·d~l )~x
F~ = − 0I I 2 1 21. (11)
21 4π 2 1 |~x |3
I I 21
However, since:
~x =~x −~x = −(~x −~x ) = −~x , |~x | = |~x | (12)
21 2 1 1 2 12 21 12
We obtain:
F~ = −F~ . (13)
21 12
And the total force on two loop system is null:
F~ = F~ +F~ = 0. (14)
T 12 21
The above result is independent of the geometry of the loops and is in
complete agreement with Newton’s third law.
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3 The Dynamic Electromagnetic Condition
Let us now consider the more general time dependent case. Since for time
dependence one can not separate the magnetic and electric forces which are
connected through Maxwell’s equations. We will consider both the electric
and magnetic parts of the Lorenz force F~ . Let us supposethat the electric
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field E~ and magnetic field B~ are created by current loop 1 and acts upon
current loop 2. Since a conductive loop that carries a neutral charge will
contain both ions and free electrons (in equal amounts) we will have:
F~ = d3x ρ (E~ +~v ×B~)+ d3x ρ (E~ +~v ×B~). (15)
21 2 i2 i2 2 e2 e2
Z Z
In the above we integrate over the entire volume of current loop 2. ρ
i2
and ρ are the ion charge density and electron charge density respectively,
e2
~v and ~v are the ion velocity field and electron velocity field respectively.
i2 e2
We will assume that the magnetization and polarization of the medium are
small and therefore we neglect corrections to the Lorentz force suggested
in [8]. Since the amounts of ions and free electrons are equal in a neutral
conducting loop, we write:
ρ = −ρ . (16)
i2 e2
Thus the electric terms in the above force equation cancel and we are left
with:
F~ = d3x ρ ~v ×B~ + d3x ρ ~v ×B~. (17)
21 2 i2 i2 2 e2 e2
Z Z
In the laboratory frame the ions being at rest we have: ~v = 0. Thus we
i2
arrive at the expression:
F~ = d3x ρ ~v ×B~. (18)
21 2 e2 e2
Z
Introducing the current density: J~ = ρ ~v we obtain the expression:
2 e2 e2
F~ = d3x J~ ×B~. (19)
21 2 2
Z
Now,letusconsiderthecoilthatgeneratesthemagneticfield. Themagnetic
field can be written as follows in terms of its vector potential [4]:
B~ = ∇~ ×A~. (20)
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If the field is generated by a current density J~ in coil 11 we can solve for
1
the vector potential and obtain the result [4]:
µ J~ (~x ,t ) R
A~(~x ) = 0 d3x 1 1 ret , R~ ≡~x , t ≡ t− . (21)
2 1 12 ret
4π R c
Z
In the above t is time and c is the speed of light in vacuum. Combining
equation (21) with equation (20) we arrive at the result:
µ J~ (~x ,t )
B~(~x ) = ∇~ ×A~(~x )= 0 d3x ∇~ × 1 1 ret . (22)
2 ~x2 2 4π 1 ~x2 R !
Z
However, notice that2:
J~ (~x ,t ) J~ (~x ,t )
∇~ × 1 1 ret = ∇~ R×∂ 1 1 ret . (23)
~x2 R ~x2 R R
! !
Since:
R~
∇~ R = − (24)
~x2 R
And:
J~ (~x ,t ) J~ (~x ,t ) ∂ J~ (~x ,t )
1 1 ret 1 1 ret t 1 1 ret
∂ =− − . (25)
R
R R2 Rc
!
Hence:
J~ (~x ,t ) R~ R
∇~ × 1 1 ret = × J~ (~x ,t )+ ∂ J~ (~x ,t ) . (26)
~x2 R R3 1 1 ret c t 1 1 ret
! (cid:18) (cid:18) (cid:19) (cid:19)
Inserting equation (26) into equation (22) we arrive at the result:
µ R~ R
B~(~x ) = 0 d3x × J~ (~x ,t )+ ∂ J~ (~x ,t ) . (27)
2 4π 1R3 1 1 ret c t 1 1 ret
Z (cid:18) (cid:18) (cid:19) (cid:19)
The current density in a thin-conductor loop can be expressed in terms of
the loop’s current as follows:
d3x g(~x )J~ (~x ,t)= dl g(~x ) dA J~ (~x ,t) = d~l g(~x )I (t). (28)
1 1 1 1 1 1 1 1 1 1 1 1
Z Z Z Z
1Itisassumedthatcoil1haszerototalchargedensityasisthecaseforcoil2andany
other neutral coil.
2Weuse thenotation ∂ ≡ ∂ .
y ∂y
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In the above dA is a cross section area element of the loop and g(~x ) is an
1 1
arbitrary function. In terms of the result given in equation (28) one may
write equation (27) as:
µ R~ R
B~(~x ,t) = 0 ×d~l I (t )+ ∂ I (t ) . (29)
2 4π R3 1 1 ret c t 1 ret
I (cid:18) (cid:18) (cid:19) (cid:19)
Inserting equation (29) into equation (19) we arrive at the result:
µ R~ R
F~ = 0 d3x J~ × ×d~l I (t )+ ∂ I (t ) . (30)
21 4π 2 2 R3 1 1 ret c t 1 ret
Z I (cid:18) (cid:18) (cid:19) (cid:19)
Assuming that current loop 2 has also a small cross section area and using
the same argument as in equation (28) we arrive at the result:
µ R~ R
F~ = 0I (t) d~l × ×d~l I (t )+ ∂ I (t ) . (31)
21 4π 2 2 R3 1 1 ret c t 1 ret
I I (cid:18) (cid:18) (cid:19) (cid:19)
3.1 The Quasi-Static Approximation
In the quasi-static approximation we assume that τ = R is small and can
c
be neglected. Under the same approximation t ≃ t. Neglecting all terms
ret
of order τ we arrive at the result:
µ R~
F~ = 0I (t)I (t) d~l × ×d~l . (32)
21 4π 1 2 2 R3 1
I I
However, since R~ =~x we write:
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µ ~x
F~ = 0I (t)I (t) d~l ×( 12 ×d~l ). (33)
21 4π 1 2 2 |~x |3 1
I I 12
Repeating the same type of arguments as in equation (6) we obtain:
µ (d~l ·~x )d~l (d~l ·d~l )~x
F~ = 0I (t)I (t) − 2 12 1 + 1 2 12 . (34)
21 4π 1 2 "I I |~x12|3 I I |~x12|3 #
And using equation (9) with 1 ↔ 2 the first integral vanishes:
µ (d~l ·d~l )~x
F~ = 0I (t)I (t) 1 2 12. (35)
21 4π 1 2 |~x |3
I I 12
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Or:
µ (d~l ·d~l )~x
F~ = − 0I (t)I (t) 1 2 12. (36)
12 4π 1 2 |~x |3
I I 12
This is the quasi-static analogue of equation (10) and is identical to the
static case except for the fact that now the currents are dependent on time.
The same conclusions may be drawn as in the static case. Newton’s third
law holds regardless of the shape of the coils involved and the total force is
null.
This should come as no surprise to the reader since in the quasi-static
approximation we totally neglect the time it takes a signal to propagate
from coil 1 to coil 2 assuming that this time is zero. Different results are
obtained when τ is not neglected as we will see in the next subsection.
3.2 The Case of a Finite τ
Consider the current I(t ) = I(t− R), if R is small but not zero one can
ret c c
write a Taylor series expansion around t in the form:
R ∞ I(n)(t) R
I(t ) =I(t− )= (− )n. (37)
ret
c n! c
n=0
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In the above I(n)(t) is the derivative of order n of I(t). Inserting equation
(37) into equation (21) and taking into account equation (28) we obtain:
∞ (n)
µ I (t ) µ I (t) 1 R
A~(~x ) = 0 d~l 1 ret = 0 1 d~l (− )n (38)
2 1 1
4π R 4π n! R c
I n=0 I
X
Denoting g (R) = 1(−R)n and inserting equation (38) into equation (20)
n R c
we obtain:
∞ (n)
µ I (t)
B~(~x ) = 0 1 ∇~ ×(d~l g (R)) (39)
2 4π n! ~x2 1 n
n=0 I
X
Now writingequation (19) in termsof equation (39) andtaking into account
equation (28) we have:
∞ (n)
µ I (t)
F~ = d~l I (t)×B~ = 0I (t) 1 d~l ×(∇~ g (R)×d~l ). (40)
21 2 2 4π 2 n! 2 ~x2 n 1
I n=0 I I
X
Using a well known vector identity:
d~l ×(∇ g (R)×d~l ) = ∇~ g (R)(d~l ·d~l )−d~l (d~l ·∇~ g (R)) (41)
2 ~x2 n 1 ~x2 n 2 1 1 2 ~x2 n
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We can write:
d~l ×(∇~ g (R)×d~l )= ∇~ g (R)(d~l ·d~l )− d~l d~l ·∇~ g (R)
2 ~x2 n 1 ~x2 n 2 1 1 2 ~x2 n
I I I I I I
(42)
but since d~l ·∇~ g (R) = 0 equation (40) can be written as:
2 ~x2 n
H ∞ (n)
µ I (t)
F~ = 0I (t) 1 ∇~ g (R)(d~l ·d~l ). (43)
21 4π 2 n! ~x2 n 2 1
n=0 I I
X
However, since:
1 1
∇~ g (R) = (− )n∇~ Rn−1 = (− )n(n−1)Rn−2∇~ R
~x2 n c ~x2 c ~x2
1
= (− )n(1−n)Rn−3R~ (44)
c
The force takes the form:
∞ (n)
µ I (t) 1
F~ = 0I (t) 1 (− )n(1−n) Rn−3R~(d~l ·d~l ). (45)
21 2 2 1
4π n! c
n=0 I I
X
We note that there is no first order contribution to the force. Hence the
next contribution to the force after the quasi-static term is second order.
Let us define the dimensionless geometrical factor K~ as:
21n
1
K~ = Rn−3R~(d~l ·d~l ) = −K~ . (46)
21n hn 2 1 12n
I I
in the above h is some characteristic distance between the coils. In terms of
K~ we can write equation (45) as:
21n
∞ (n)
µ I (t) h
F~ = 0I (t) 1 (− )n(1−n)K~ . (47)
21 2 21n
4π n! c
n=0
X
The force due to coil 2 that acts on coil 1 is:
∞ (n)
µ I (t) h
F~ = 0I (t) 2 (− )n(1−n)K~ . (48)
12 1 12n
4π n! c
n=0
X
The total force on the system is thus:
∞
µ (1−n) h
F~ = F~ +F~ = 0 (− )nK~ I (t)I(n)(t)−I (t)I(n)(t) .
T 12 21 4π n! c 12n 1 2 2 1
nX=0 (cid:16) (cid:17)
(49)
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We note that the quasi-static term n = 0 does not contribute to the sum
nor does the n = 1 term. The fact that the retarded field ”corrects” itself
to first order in order to ”mimic” a non retarded field was already noticed
by Feynman [5]. Hence we can write:
∞
µ (1−n) h
F~ = 0 (− )nK~ I (t)I(n)(t)−I (t)I(n)(t) . (50)
T 4π n! c 12n 1 2 2 1
nX=2 (cid:16) (cid:17)
We conclude that in general Newton’s third law is not satisfied, taking the
leading non-vanishing terms in the above sum we obtain:
µ h
F~T ∼= −8π0(c)2K~122 I1(t)I2(2)(t)−I2(t)I1(2)(t)
(cid:16) (cid:17)
(2) (2)
µ h I (t) I (t)
= − 0( )2K~ I (t)I (t) 2 − 1 . (51)
122 1 2
8π c I2(t) I1(t) !
3.3 Two Circular Loops
In the following we will calculate the geometrical factor K~ for a simple
12n
geometry of two circular loops of unit radius and a unit distance located
one atop the other (see figure 2). According to equation (46) we have in the
case h= 13
K~ = − Rn−3R~(d~l ·d~l ). (52)
12n 2 1
I I
Every point on loop 1 and 2 has the coordinates ~x = (cosθ ,sinθ ,1) and
1 1 1
~x = (cosθ ,sinθ ,0), hence:
2 2 2
R~ =~x =~x −~x = (cosθ −cosθ ,sinθ −sinθ ,1) (53)
12 1 2 1 2 1 2
And thus:
R2 = 3−2cos(θ −θ ) (54)
1 2
The line elements can be calculated as follows:
d~l = dθ θˆ = dθ (−sinθ ,cosθ ,0)
1 1 1 1 1 1
d~l = dθ θˆ = dθ (−sinθ ,cosθ ,0) (55)
2 2 2 2 2 2
Hence:
d~l ·d~l = dθ dθ cos(θ −θ ) (56)
1 2 1 2 1 2
3In arbitrary units since K~12n is dimensionless.
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