Theory of Special Relativity

Topics: ity lds 8.022 (E&M) – Lecture 12 Continuation of Special RelativTransformation of Electric FieRelativistic Momentum and Energy Transformation of Forces Prove that E and B are equivalent in different reference frames 2 Electric Fields in Motion iσ=Q/L2 E = 4πσ = 4π2 v in the x direction E’=4πσ’=4πππγ)= γ 4πγE Conclusion: ix z y L L/γ 'E Eγ⊥ = G. Sciolla – MIT 8.022 – Lecture 12 Consider a parallel plate capacitor Squared plates of sde L in the xy plane Charge Q distributed on the plates: Electric field //z axis: Q/LThe capacitor is now boosted with velocity How does E transform? Q’/A’= 4Q/LL’= 4Q/L(L/Q/A= Expected result: the field lnes get thicker as L shrinks… 13 Electric Fields in Motion (2) σ=Q/L2 E = 4πσ = 4π2 v in the x direction E’=4πσ’=4ππyL’z= 4πQ/A= E Conclusion: ix z y //'E E= L L G. Sciolla – MIT 8.022 – Lecture 12 Now orient the capacitor with the plates in the yz plane: Charge Q distributed on the plates: Electric field //z axis: Q/LBoost again the capacitor with velocity How does E transform? Q’/A’= 4Q/L’Expected result: the field lnes keep the same distance… How does the capacitance change? 4 Momentum and Energy u Classical definitions: where γu γ factor: γu = 1/(1-u2/c2) 212 p mu • = • = G G 2u u p mu E mc γ γ • = • = G G 2 2 2 2 2 2 2 2 1 ~ (1 ...) ~ 2 1 1E ~ (1 ...) ~ 2 2 u u u p c u mc mc mc mu c γ γ • = + − • = + − + G G G G G. Sciolla – MIT 8.022 – Lecture 12 For a particle of mass m moving with velocity Relativistic definition is the relativistic For low velocities, the new formulae reproduce old ones (Taylor!) kin M omentum: Kinetic energy: E mu Momentum: Energy: kin mu mu mu Why? See handout #2 25 Transformation of p and E v Sorry not today: no time! γu = 1/(1-βv 2) and βv 2=v2/c2 ' ( ) ' ( /) ' ' v v x x v x v y y z z E E p p p p p p γ β γ β = −⎧ ⎪ = −⎪⎨ = ⎪ ⎪ =⎩ x y z x’ y ’ z’ O O’ v G. Sciolla – MIT 8.022 – Lecture 12 Consider 2 inertial reference frames: O and O’ O’ is moving w.r.t. O with velocity //x axis How do momentum and energy Lorentz transform? One can demonstrate that Where cp E c 6 Transformation of Forces x m Small acceleration Æ v( ) 2 2 2 2 //1 1 2 2 ( ) 2 2 x x x x dpF dt F x at t m p F tE m m • = • ∆ = ∆ = ∆ ∆ ∆ • ∆ = = G. Sciolla – MIT 8.022 – Lecture 12 In an inertial R.F. O a force Fis acting on a body of mass Body is initially at rest: p=0 at t=0 non relativistic velocities involved in O How do these quantities look like in the Lab Frame O’? NB: O’ is moving with velocity //x axis wrt the Frame O Force F to x axis: Change in position: Change in Energy: 3MIT 8.022 – Lecture 12 7 Forces //v How does Fx lConclusion: vRF [ ] 2 22 2 ' ' ' ' ( ) ' ( /) /' /2' 1' 1 2 t ( /) x x v v x x v x v x x xx x x x p E dpF dt E E p p F tF cp p mF v Fvt t x t c c m c vt x c γ β γ β γ β ββ γ = = − = − ∆−∆∆ ∆ = − ⎡ ⎤⎛ ⎞−⎜ ⎟⎢ ⎥ = = = ∆ ∆− ∆ − ∆ → ⎣ ⎦ ∆ ⎝ ⎠ ∆ 2 0 0 2 /2' ' 11 2 x x x x xt t x F tF c mF F FFv t c m β ∆→ ∆→ ∆− = = = − ∆ ' x xF F= G. Sciolla – MIT 8.022 – Lecture 12 ook in the Lab Frame O’ ? The component of the force //to is constant: Remember Lorentz transformations: and For 0, this becomes: cp E c E c − ∆ lim lim G. Sciolla – 8 Forces perpendicular to v How does Fy Conclusion: vRF are contracted: 2 2 2 0 ' ' ' ' ( ) ' ' ' 1' 1 2 t ' ' y y v v x y y y y y y y x y xt t dpF dt E E p p p p p FF v Fvt vt t x t c x c c m F F γ γγ γ β ∆→ ∆ ⎡ ⎤⎛ ⎞−⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ = = − = ∆ ∆ ∆ = = = = ∆ ⎛ ⎞ ⎛ ⎞∆− ∆ − ∆∆ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ∆→ = = 0 2 11 2 y y x F F Fv t c m γ γ→ = ⎛ ⎞− ∆⎜ ⎟⎝ ⎠ ' y y FF γ = look in the Lab Frame O’ ? Components of force perpendicular to Remember Lorentz transformations: and For 0, this becomes: lim lim cp Is this consistent with what we found about E? 49 Force by current on moving charge λ+= λ+ REST= λ0. λ−= λ− ΜΟΤ0 velocity u Æ −λ0 is not ivelocity v v I + charges -charges charge Q Q G. Sciolla – MIT 8.022 – Lecture 12 Description in the Lab Frame Electrically neutral wire carrying a current Positive charge density NB: these charges are at rest in O’ Negative charge density = −λNB: these charges are moving with the density of the electrons in ther reference frame O A charge Q outside the wire moves to right with Please pay attention: this is difficult! λ of negative charges in their RF We said: λ−= λ−ΜΟΤ= −λ0 REST What is charge density in their RF? λ0 First attempt: 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 REST > λ −ΜΟΤÆ λ− REST = Q/L’= Qγ/L= -γλ0 Æ λ− 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 REST < λ −ΜΟΤÆ λ− REST = Q/L-’= Q/γL= -λ0 /γ Æ λ− G. Sciolla –MIT 8.022 – Lecture 12 10 5 Force by current on moving charge What forces act on the charge Q? Lab frame: Wire is neutral: no electric filed E Current will generate magnetic filed B: Current in the wire: I=dQ/dt=λ0dx/dt=λ0u Ampere’s law: B= 2I = 02 uλ cr cr Magnetic force acting on charge Q: 0 2 2 uvvFQ B Q c c r λ = = Direction? Right hand rule: repulsive force NB: I opposite to v electrons! G. Sciolla – MIT 8.022 – Lecture 12 11 12 Force in charge’s rest frame? Let’s now move to the charges rest frame: Charge Q: at rest by definition Is there any force acting on Q? There must be: Relativity Principle! + charges -charges charge Q 2 ' 1 /uv u uv c − = − G. Sciolla – MIT 8.022 – Lecture 12 Velocities involved: Negative charges in the wire: velocity u’=(u-v)relativistic sum Positive charges in the wire: velocity –v No magnetic force: the charge is at rest! 6 Charge densities in Q’s RF Are we in trouble? Let’s see what happens to the charges in the wire Positive charges: Charge density in charge’s reference frame: λ+’=Q/L’=γvλ0 Negative charges: REST '= λγ λ = γu' −λ0= γγ (1−ββ)(−) = -γ(1−ββ)λ-u'-uv uv v uv 0γu γu 21 2Goal: calculate γ . Let's start calculating 1/γ =−β u'u' u' − 2⎛ uv ⎞ 22 − 22 2 1−β2u' = 1−⎜⎛ u' ⎞ ⎜⎝ 1− uv /c2 ⎟⎠ = 1−⎜⎛ βu −βv ⎞ 12ββ+β β − (β −β) ⎝1 −ββv ⎠⎟ uvuv uv ⎝ c ⎠⎟ = 1 − c2 =(1−ββ)2 = u uv 22(1 −β )(1 −β )1 ⇒vu== γu' =γγ (1−ββ)uv uv uv uv uv 22(1 −ββ)2 γ γ (1−ββ)2 G. Sciolla –MIT 8.022 – Lecture 12 13 14 Force in charge’s rest frame Æ Electric field! Electric field Æ ( )NET 0 0 0 02+ -v v v v ' = ' + ' = -1 u vuv uv c λ λ λλ λ λ γ γ γ γ= =− NET 0 2v 22'' uv r E λλ γ== 0 2v 2 ()' ' rc F λγ= = G. Sciolla – MIT 8.022 – Lecture 12 Net charge density in Q’s reference frame: In this RF there is a net charge on the wire! force F’ will act on the charge Q Is there a Magnetic field as well? Yes, but it does not excerpts any force on Q because Q is at rest ββλ ββ rc repulsiveQuv QE 7 15 Comparison of forces in the 2 RFs In Q’s rest frame: Are results consistent? v transform as 0 2 2uvv c λ = = 0 2 2'' v uvF Q cr λγ= = 'y y FF γ = G. Sciolla – MIT 8.022 – Lecture 12 In lab frame: Repulsive magnetic force acting on charge Q: Repulsive electric force acting on charge Q: Yes! We have seen that forces in direction perpendicular to FQ B Q cr QE Thoughts on this problem Is the comparison fair? In one RF we have a magnetic force, in the other an electric force Are we comparing apples and oranges? No, on the contrary! This results proves that Electricity and Magnetism are intimately connected! Physics is consistent! 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 16 8 Summary of Special Relativity Speed of light and physics are the same in all RF Consequences in mechanics Time dilation Moving clocks run slower ∆t’ = γ∆t Length contraction Moving objects appear shorter along direction of motion: ∆L = γ ∆L’ Force transformation Fy Components //v: constant; perpendicular to v: contracted: F' y = γ Consequences in E&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 Because there is a force even in the particle’s reference frame G. Sciolla –MIT 8.022 – Lecture 12 17 18 Outlook Today: Transformations for momentum, energy and forces iviiiNext time: G. Sciolla – MIT 8.022 – Lecture 12 Conclusion of Introduction to Special Relativity Proved that E and B are intimately connected Two observers, “relatstcally” consstent results Back to Magnetism Ampere’s law, Biot-Savart, Vector potential 9