# 8.022 (E&amp;M) – Lecture 12 ```8.022 (E&amp;M) – Lecture 12
Topics:
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Continuation of Special Relativity
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Transformation of Electric Fields
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Relativistic Momentum and Energy
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Transformation of Forces
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Prove that E and B are equivalent in different reference frames
z
y
Electric Fields in Motion
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Consider a parallel plate capacitor
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Squared plates of side L in the xy plane
Charge Q distributed on the plates:
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σ=Q/L2
Electric field // z axis:
2
 E = 4πσ = 4πQ/L
x
L
The capacitor is now boosted with velocity v in the x direction
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How does E transform?
L/γ
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E’=4πσ’=4πQ’/A’= 4πQ/LL’= 4πQ/L(L/γ)= γ 4πQ/A= γE
E '⊥ = γ E
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Conclusion:
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Expected result: the field lines get thicker as L shrinks…
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8.022 – Lecture 12
2
1
z
y
Electric Fields in Motion (2)
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Now orient the capacitor with the plates in the yz plane:
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Charge Q distributed on the plates:
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σ=Q/L2
Electric field // z axis:
2
 E = 4πσ = 4πQ/L
L
Boost again the capacitor with velocity v in the x direction
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x
How does E transform?
E’=4πσ’=4πQ’/A’= 4πQ/L’yL’z= 4πQ/A= E
Conclusion:
E ' // = E
L
Expected result: the field lines keep the same distance…
How does the capacitance change?
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8.022 – Lecture 12
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Momentum and Energy
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For a particle of mass m moving with velocity u
Classical definitions:
G
G
• M om entum : p = m u
1
• Kinetic energy: E kin = m u 2
2
Relativistic definition
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G
• Momentum: p = γ u mu
• Energy:
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Why? See handout #2
E = γ u mc 2
where γu is the relativistic γ factor: γu = 1/(1-u2/c2)
For low velocities, the new formulae reproduce old ones (Taylor!)
1 u2
G
G
G
G
• p = γ u mu ~ (1 +
− ...) mu ~ mu
2 c2
1 u2
1
• E kin = γ u mc 2 ~ (1 +
− ...) mc 2 ~ mc 2 + mu 2
2 c2
2
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8.022 – Lecture 12
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2
y
’
y
Transformation of p and E
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z’
x’
x
Consider 2 inertial reference frames: O and O’
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O’
O
z
v
O’ is moving w.r.t. O with velocity v // x axis
How do momentum and energy Lorentz transform?
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One can demonstrate that
⎧ E ' = γ v ( E − β v cpx )
⎪ p ' = γ ( p − β E / c)
⎪ x
v
x
v
⎨
⎪ p 'y = py
⎪p' = p
z
⎩ z
Sorry not today: no time!
Where γu = 1/(1-βv2) and βv2=v2/c2
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8.022 – Lecture 12
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Transformation of Forces
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In an inertial R.F. O a force Fx is acting on a body of mass
m
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Body is initially at rest: p=0 at t=0
Small acceleration &AElig; non relativistic velocities involved in O
dpx
dt
1
1 Fx
• C h a n g e in p o s itio n : ∆ x = a ∆ t 2 =
∆t2
2
2 m
• F o rc e F // to x a x is : F x =
• C h a n g e in E n e rg y :
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∆E =
(∆ p )
2m
2
=
( Fx ∆ t ) 2
2m
How do these quantities look like in the Lab Frame O’?
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NB: O’ is moving with velocity v//x axis wrt the Frame O
G. Sciolla – MIT
8.022 – Lecture 12
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3
Forces // v
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How does Fx look in the Lab Frame O’ ?
dp ' x
dt '
R em em ber Lorentz transform ations: E ' = γ v ( E − β v cp x ) and p ' x = γ v ( p x − β v E / c )
F 'x =
Fx 2 ∆ t
∆ [γ ( p x − β E / c ) ] ∆ p x − β ∆ E / c F x − β / c 2 m
∆ p 'x
F 'x =
=
=
=
v
v 1 Fx
∆t '
v ⎞⎤
⎡ ⎛
∆t − 2 ∆x
∆t
1− 2
∆ ⎢γ ⎜ t − 2 x ⎟ ⎥
c
c
2 m
c
⎠⎦
⎣ ⎝
For ∆ t → 0, this becom es:
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Conclusion:
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Fx 2 ∆ t
2m = F
F ' x = lim F ' x = lim
x
∆t → 0
∆t → 0
v 1 Fx
∆t
1− 2
c 2 m
Fx − β / c
The component of the force // to vRF is constant:
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F 'x = Fx
8.022 – Lecture 12
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Forces perpendicular to v
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How does Fy look in the Lab Frame O’ ?
F 'y =
dp ' y
dt '
R em em ber Lorentz transform ations: E ' = γ v ( E − β v cp x ) and p ' y = p y
Fy
∆p y
=
v
v 1 Fx
∆t '
v ⎞⎤
⎛
⎞
⎛
⎞
⎡ ⎛
γ ⎜ ∆t − 2 ∆x ⎟
∆t ⎟
γ ⎜1 − 2
∆ ⎢γ ⎜ t − 2 x ⎟ ⎥
c
c 2 m
c
⎝
⎠
⎝
⎠
⎠⎦
⎣ ⎝
Fy
Fy
F ' y = lim F ' x = lim
For ∆ t → 0, this becom es:
=
∆t → 0
∆t → 0
F
1
v
γ
⎛
⎞
x
∆t ⎟
γ ⎜1 − 2
c 2 m
⎝
⎠
F 'y =
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∆p 'y
=
∆p y
=
Conclusion:
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Components of force perpendicular to vRF are contracted:
this–consistent
G.IsSciolla
MIT
with what8.022
we–found
Lecture 12
F 'y =
Fy
γ
8
4
Please pay attention: this is difficult!
Force by current on moving charge
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Description in the Lab Frame
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Electrically neutral wire carrying a current
Positive charge density λ+= λ+REST= λ0.
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NB: these charges are at rest in O’
Negative charge density λ−= λ−ΜΟΤ= −λ0
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NB: these charges are moving with velocity u
−λ0 is not the density of the electrons in their reference frame O
A charge Q outside the wire moves to right with velocity v
&AElig;
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Q
v
charge Q
+ charges
- charges
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8.022 – Lecture 12
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λ of negative charges in their RF
We said: λ−= λ−ΜΟΤ= −λ0
 What is charge density in their RF? λ0REST
 First attempt:
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 Charge density λ0 = Q/L where L = length of the wire in Lab frame
 In lab frame: λ−ΜΟΤ = Q/L= -λ0
 In O (in rest with - charges), length of wire appears contracted: L’:L/γ
&AElig; λ−REST = Q/L’= Qγ/L= -γλ0
&AElig; λ−REST &gt; λ −ΜΟΤ WRONG!
Why? There is no such thing as the wire. Just the length of + and – charges which
happen to be the same in the Lab reference frame but not elsewhere.
 Second attempt:
 The electrons will think: our length in our own RF is L’-. In the
reference frame of the lab, boosted wrt us by a velocity –v, this
length will be contracted by a factor γ: L’-= γL
&AElig; λ−REST = Q/L-’= Q/γL= -λ0 /γ
&AElig; λ−REST &lt; λ −ΜΟΤ
G. Sciolla – MIT
8.022 – Lecture 12
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5
Force by current on moving charge
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What forces act on the charge Q? Lab frame:
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Wire is neutral: no electric filed E
Current will generate magnetic filed B:
 Current in the wire: I=dQ/dt=λ dx/dt=λ u
0
0
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Ampere’s law: B = 2I = 2λ0u
cr
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cr
v
c
Magnetic force acting on charge Q: F = Q B = Q
2λ0uv
c2r
Direction?
 Right hand rule: repulsive force
 NB: I opposite to v electrons!
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8.022 – Lecture 12
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Force in charge’s rest frame?
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Let’s now move to the charges rest frame:
charge Q
+ charges
- charges
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Velocities involved:
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Charge Q: at rest by definition
u−v
Negative charges in the wire: velocity u’=(u-v)relativistic sum u ' =
1 − uv / c 2
Positive charges in the wire: velocity –v
Is there any force acting on Q?
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There must be: Relativity Principle!
No magnetic force: the charge is at rest!
G. Sciolla – MIT
8.022 – Lecture 12
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6
Charge densities in Q’s RF
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Are we in trouble? Let’s see what happens to the charges in the wire Positive charges: 
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Charge density in charge’s reference frame: λ+’=Q/L’=γvλ0
Negative charges:
( − ) = -γ 1− β β λ
−λ
λ-'=γ u'λ- REST = γ u' 0 = γ u γ v (1− β u β v )
u v) 0
v(
γu
γu
Goal: calculate γ u' . Let's start calculating 1/γ 2 u' = 1 − β 2 u'
⎛ u−v ⎞
2
⎜
⎟
⎛ βu − β v ⎞ 1 − 2βu β v + β u 2 β v 2 − (β u − β v )2
1− uv / c 2 ⎠
⎛u'⎞
=
= 1− ⎜ ⎟ = 1 − ⎝
=
1−
⎜
⎟ =
2
c2
⎝c⎠
(1− β u β v )
⎝ 1 − βu β v ⎠
2
2
1 − β 2 u'
=
(1 − β v 2 )(1 − β u 2 )
(1 − β u β v )
2
=
1
γ uγ
2
2
v
(1− β u β v )
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⇒
2
γ u' = γ u γ v (1− β u β v )
8.022 – Lecture 12
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Force in charge’s rest frame
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Net charge density in Q’s reference frame:
λ 'NET = λ+ ' + λ-' = γ v λ0 - γ v (1 − β u β v ) λ0 = γ v β u β v λ0 = γ v
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In this RF there is a net charge on the wire! &AElig; Electric field!
E'=
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2λ 'NET
r
= γv
2uvλ0
rc 2
Electric field &AElig; force F’ will act on the charge Q
F ' = QE ' = γ v
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uv
λ0
c2
2Quvλ0
rc 2
(repulsive)
Is there a Magnetic field as well?
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Yes, but it does not excerpts any force on Q because Q is at rest
G. Sciolla – MIT
8.022 – Lecture 12
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Comparison of forces in the 2 RFs
In lab frame:
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Repulsive magnetic force acting on charge Q:
2λ uv
v
F = Q B = Q 20
cr
c
In Q’s rest frame:
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Repulsive electric force acting on charge Q:
F ' = QE ' = γ v Q
2λ0uv
c2r
Are results consistent?
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Yes! We have seen that forces in direction perpendicular to v
transform as
Fy
F 'y =
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γ
8.022 – Lecture 12
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Thoughts on this problem
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Is the comparison fair?
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No, on the contrary!
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In one RF we have a magnetic force, in the other an electric force
Are we comparing apples and oranges?
This results proves that Electricity and Magnetism are intimately
connected!
Physics is consistent!
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Principle of relativity demands that the 2 observers will come to
the same conclusions
The details of the calculation (Electric? Magnetic?) are different in
the different RF, but ultimately irrelevant.
G. Sciolla – MIT
8.022 – Lecture 12
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Summary of Special Relativity
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Speed of light and physics are the same in all RF
Consequences in mechanics
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Time dilation
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Length contraction
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Moving objects appear shorter along direction of motion: ∆L = γ ∆L’
Force transformation
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Moving clocks run slower ∆t’ = γ ∆t
Components // v: constant; perpendicular to v: contracted: F ' y =
Fy
γ
Consequences in E&amp;M
Pure B in one RF looks like E in another And vice versa, pure E in one RF looks like E+B in another 
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Because there is a force even in the particle’s reference frame
G. Sciolla – MIT
8.022 – Lecture 12
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Outlook
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Today:
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Conclusion of Introduction to Special Relativity
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Transformations for momentum, energy and forces
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Proved that E and B are intimately connected
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Two observers, “relativistically” consistent results
Next time:
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Back to Magnetism
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Ampere’s law, Biot-Savart, Vector potential
G. Sciolla – MIT
8.022 – Lecture 12
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