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Summer Research Fellowship Programme of India's Science Academies

Special theory of relativity

Chandra Kumar Chandravanshi

M.Sc. 2nd semester, Pt. Ravishankar Shukla University, Raipur, Chattisgarh, 492010

Professor Anirban Kundu

University of Calcutta, Kolkata, 700009

Abstract

As I have done my project in the topic Special Theory of Relativity in which I understood that how this theory is applicable to all the relativistic phenomena. The Lorentz Transformation which relates the co-ordinates of two different frames. Using four-vectors I proved the invariance of space-time interval and many other invariants of Lorentz Transformation. In Relativistic kinematics I studied about two-body decays in which I wrote four-momentum for decaying particles and further solving, found that the neutrino was postulated because of the energy distribution of the final state electron and in collider experiments got the result that energy reach for moving-target experiments can be much higher than fixed-target experiments, fixed-target experiment is easy to design. I have done Relativistic formulation of Maxwell's equation, all four Maxwell's equation can be obtained by electromagnetic field tensor and its dual tensor. Further the Electrodynamics of moving bodies tells us that Electric and Magnetic fields are not to be treated separately. Since Lorentz force is an ordinary force and I came to know the four-force called Minkownski force which behaves like Lorentz vector and derived Lorentz force equation. Combined uniform and static electric and magnetic field can be used to get monoenergetic particle beams which are important in the high and low-energy collider experiments and I derived the equation of motion for a field. The generalised momentum equation is the most important equation in the theory of interaction of electromagnetic field with a charged particle. Substitution of the four-differential operator with another new operator and by operating this operator on the electron wave function we can get the entire Quantum electrodynamics. By writing the Langragian for the electromagnetic field I found that the Langrangian density of the electromagnetic field in the presence of an external current is gauge-invariant only because the current is conserved. In short current conservation is equivalent to Gauge invariance. By the experimental tests of Special theory of relativity it proved that there is no any aether medium i.e there is not any absolute frame of reference and the speed of light is constant in all directions.

Abbreviations

Abbreviations
 STRSpecial Theory of Relativity
 IASIndian Academy of Science 

INTRODUCTION

In 1905 Albert Einstein proposed the Special Theory of Relativity in his paper titled ''On the Electrodynamics of Moving Bodies". This theory tells us that how space and time are related for any object moving with consistent speed in a straight path. It is based on two postulates:

1. The physical laws are invariant in all the inertial frame of reference.

2. The speed of light is same for all observers and it does not depend on the speed of observers.

According to this theory many different phenomena occur when any object moves with very high speed, i.e. if any massive object moves with the approximate speed of light then it's mass will start to increase, as well as length contraction and time dilation type phenomena also occurs at this speed. These are the consequences of the Special Theory of Relativity.

LORENTZ TRANSFORMATION

If (t,x,y,z) is the co-ordinate of an inertial frame S and (t',x',y',z') is the co-ordinate of another inertial frame S' which is moving with velocity v along the common x-x' axis, Then the co-ordinates of these two frames are related by (in the natural system of units, i.e., with c = 1)
t' = γ(t - vx), x' = γ(x - vt), y'= y, z'=z, where γ=1v2\gamma=\sqrt{1-v^2} (1)

Here γ is known as Boost factor and considering c=1. For v ≪ 1 we get the Galilean Transformation. Above equation (1) can be represented in the form of matrix as following

Λνμ=(γγv00γvγ0000100001)\Lambda^\mu_{\nu}=\begin{pmatrix} \gamma & -\gamma{v}&0&0 \\-\gamma{v} & \gamma&0&0\\0&0&1&0\\0&0&0&1 \end{pmatrix}\quad

Where Λνμ\Lambda^\mu_{\nu} is known as Lorentz transformation matrix. Note that Λ00>1\Lambda^0_{0}>1 and detΛ=1det\Lambda=1; any transformation that satisfies these two conditions is called proper Lorentz transformation. For a contravariant four-vector { Aμ(A0,A)A^\mu\equiv(A^0,A)} the transformation law is

Aμ=ΛνμAνA'^\mu=\Lambda^\mu_{\nu}A^\nu

Here the repeated indices are summed over.

Similarly for a covariant four-vector { Aμ(A0,A)A_\mu\equiv(A_0,A)} the transformation law is
Aμ=ΛμνAνA'_\mu=\Lambda^\nu_{\mu}A_\nu

Where Λμν\Lambda_\mu^\nu is nothing but the inverse of Λνμ\Lambda_\nu^\mu. Here also the repeated indices are summed over.

VELOCITY ADDITION

Suppose there are three frames S, S' and S". Their origins coincide at t = t' = t" =0. Here S' is moving with a velocity v' with respect to S and S" is moving with velocity v' with respect to S' along the common x-x' axis.

Hence we write Lorentz transformation as

x" = γ'(x' - v't') = γ'[γ(x - vt) - v'γ(t - vx)] (2)

t" = γ'(t' - v'x') = γ'[γ(t - vx) - v'γ(x - vt)]

and the Lorentz Transforation between S and S'' frame

x'' = γ''(x - v''t), t'' = γ''(t - v''x) (3)

Equating the coefficient of x in (2) and (3) and calculating further we get

γ''=γγ'(1+vv') ⇒ v"= 1-(1-v2)(1-v'2)(1-vv')2v"=v+v'1+vv'​​ (4)

This equation (4) is the rule for velocity addition. Hence the velocity of a massive particle will always be less than the speed of light. If v ≪ 1, we get v'' = v+v' the non-relativitic velocity addition rule.

RELATIVISTIC FORMULATION OF MAXWELL'S EQUATION

As we know the four equations of Maxwell for electromagnetism are (in the rationalised Lorentz-Heaviside system)

.B=0,      ×E=Bt\nabla.B=0,           \nabla\times{E}=-\frac{\partial{B}}{\partial{t}}​ (5)

Above equation (5) is known as Faraday's law. Another equation Ampere's law (with Maxwell's correction) is

(.A+ϕt)+(2t22)A=j\nabla(\nabla.A+\frac{\partial\phi}{\partial{t}})+(\frac{\partial^2}{\partial{t^2}}-\nabla^2)A=j (6)​

and Gauss's law gives, after the addition and subtraction of 2ϕt2\frac{\partial^2\phi}{\partial{t^2}}

t(.A+ϕt)+(2t22)ϕ=ρ-\frac{\partial}{\partial{t}}(\nabla.A+\frac{\partial\phi}{\partial t})+(\frac{\partial^2}{\partial t^2}-\nabla^2)\phi=\rho (7)​

Since we know that so we can easily combine equations (6) and (7) as

μμAννμAμ=jν\partial_\mu\partial^\mu A^\nu-\partial^\nu\partial_\mu A^\mu=j^\nu​ (8)

or it can be written in more elegant way-

μFμν=jν\partial_{\mu}F^{\mu\nu}=j^\nu (9)​

where

Fμν=μAννAμF^{\mu\nu}=\partial^{\mu}A^\nu-\partial^{\nu}A^\mu

Equation (13) is known as the Electromagnetic field tensor. It is rank-2 tensor which is antisymmetric by construction and it is most important antisymmetric tensor in physics which can be represented in the form of martrix

Fμν=(0ExEyEzEx0BzByEyBz0BxEzByBx0)F^{\mu\nu}=\begin{pmatrix} 0 & -E_x&-E_y &E_z\\E_x & 0&-B_z&B_y\\E_y&B_z&0&-B_x\\E_z&-B_y&B_x&0 \end{pmatrix}\quad (10)​

and it's dual tensor can be obtained by substituting EB,    BEE\to{B},        B\to-E in the above equation (10) as following

Gμν=(0BxByBzBx0EzEyByEz0ExBzEyEx0)G^{\mu\nu}=\begin{pmatrix} 0 & -B_x&-B_y &-B_z\\B_x & 0&E_z&-E_y\\B_y&-E_z&0&E_x\\B_z&E_y&-E_x&0 \end{pmatrix}\quad (11)​

It is quite straightforward to show that

μGμν=0\partial_{\mu}G^{\mu\nu}=0 (12)​

leads to the second pair of Maxwell's equation.

TRANSFORMATION OF THE FIELDS

Electric and magnetic fields can be transformed under Lorentz transformations. Electricity and magnetism both are relativistc because it depends on the frame of reference. So what appears as an electric phenomena in one frame may appear to be a magnetic phenomena in another frame.

We can derive the laws of field transformation by rank-2 tensor transformation law. Lets take-

F01=Λα0Λβ1FαβF^{01'}=\Lambda_\alpha^0\Lambda_\beta^1 F^{\alpha\beta} (13)​

If the motion is along the common x-axis, so (v,0,0) and only Λ00, Λ10, Λ01\Lambda^0_0, \Lambda^0_1 , \Lambda^1_0 and Λ11\Lambda_1^1 are nonzero but F00=F11=0F^{00}=F^{11}=0 and F10=F01F^{10}=-F^{01}so

Ex=Λ00Λ11F01+  Λ10Λ01F10  -E_x'= \Lambda_0^0 \Lambda_1^1F^{01}+  \Lambda_1^0 \Lambda_0^1F^{10}   (14)​

Substituting the value of Λ from Lorentz transformation matrix in above equation (14) we get

(γ2γ2v2)F01=Ex(\gamma^2-\gamma^2v^2)F^{01}= -E_x (15)​

further we write- F03=Λα0Λβ3FαβF^{03'}=\Lambda_\alpha^0\Lambda_\beta^3F^{\alpha\beta}

Ez=Λ00 Λ33F03+ Λ10 Λ33F13=γ(EzvBy)-E_z= \Lambda^0_0 \Lambda^3_3F^{03}+ \Lambda^0_1 \Lambda^3_3F^{13}= \gamma(-E_z-vB_y)

similarly we get a complete set of transformation of fields

Ex=Ex,         Ey=γ(EyvBz),       Ez=γ(Ez+vBy)E_x'=E_x,                  E_y'=\gamma(E_y-vB_z),             E_z'=\gamma(E_z+vB_y) (16)

Bx=Bx,         By=γ(By+vEz),       Bz=γ(BzvEy)B_x'=B_x,                  B_y'=\gamma(B_y+vE_z),             B_z'=\gamma(B_z-vE_y) (17)​

Above equations (16)and (17) are symmetric under the exchange of EBE\to{B} and BEB\to-E.
Hence we can get the same transformation law from the dual tensor GμνG^{\mu\nu}.

FIELDS DUE TO A UNIFORMLY MOVING PARTICLE

Now we will find the field due to a uniformly moving charged particle which is moving along positive x-axis in the lab frame. For this we will do a very easy approach that is go to the frame where the particle is at rest, after that calculate the fields there and back to the lab frame. For this we need the reverse transformation of equation (16) and (17).

IMG_20190824_115831_449.jpg
    Fields in S due to a uniformly moving particle which is at rest in S'

    Let the particle with chanrge q is moving with a velocity v along the x-axis in S frame and there is a detector at point (0,b,0) in the S frame. Suppose S' be the the frame where the particle is at rest. These two frame coincide at t=t'=0 and let n be the unit vector along the line joining the instantaneous position of the charge (at the origin of S') and the detector (as shown in figure 1).

    Thus n.v=cosψn.v=cos\psi , b=rsinψb=rsin\psi and vt=rcosψvt=-rcos\psi.

    At time t and t' in S and S' respectively, the coordinate of detector in S' is x1=vtx'_1=-vt' , x2=bx'_2=band x3=0x_3'=0. The distance is

    r=(vt)2+b2r'=\sqrt{(vt')^2+b^2} (18)​

    Hence the Electric field component in S' frame are:

    E1=qvt/4πr3E_1'=-qvt'/4{\pi}r'^3

    E2=qb/4πr3E_2'=qb/4{\pi}r'^3

    E3=0E_3'=0

    and the magnetic field component in this frame S' B1, B2, B3B_1',  B_2',  B_3'are all zero.

    Let us now we boost the fields back to the lab frame as it has been stated earlier (replacing v by -v):

    E1=E1=qvtγ/4π(b2+γ2v2t2)3/2E_1=E_1'=-qvt\gamma/4{\pi}(b^2+{\gamma}^2v^2t^2)^{3/2}

    E2=E2=qbγ/4π(b2+γ2v2t2)3/2E_2=E_2'=-qb\gamma/4{\pi}(b^2+{\gamma}^2v^2t^2)^{3/2}

    B3=E2=qvbγ/4π(b2+γ2v2t2)3/2B_3=E_2'=-qvb\gamma/4{\pi}(b^2+{\gamma}^2v^2t^2)^{3/2}

    The above equation can be written in a more compact way. Here E1/E2=vt/bE_1/E_2=-vt/b. So EE is always directed along just as static Coulomb field. Also the denominator (b2+γ2v2t2)3/2(b^2+{\gamma}^2v^2t^2)^{3/2} can be written as r3γ3(1v2sin2ψ)3/2r^3{\gamma}^3(1-v^2sin^2\psi)^{3/2}, so

    E=qrr3γ3(1v2sin2ψ)3/2E=\frac{qr}{r^3{\gamma}^3(1-v^2sin^2\psi)^{3/2}} (19)​

    and the magnetic field is given by

    B=v×EB=v\times{E} (20)

    Hence we conclude that the moving charge produces a magnetic field and if we calculate the Poynting vector E×BE\times{B} from above equation we will get a nonzero result, so the field carries some energy. However it does not radiate. In this way we get that a charge with constant velocity does not radiate, so get radiation one must have an accelerated charge.

    We can say above statement in another way that a charge moving with uniform velocity can be made static in another inertial frame of reference, so that charge does not radiate because physical laws must be invariant in all inertial frames.

    LAGRANGIAN AND EQUATION OF MOTION

    From classical mechanics we know that the Lagrangian L=TVL=T-V is a function of generalised co-ordinate and generalised velovity. Lagrangian can even be the function of time also it let us suppose a system where L=L(q,q˙)L=L(q,\dot{q}). Equation of motion by Euler-Lagrangian can be written as:

    ddt(Lq˙)Lq=0\frac{d}{dt}(\frac{\partial{L}}{\partial{\dot{q}}})-\frac{\partial{L}}{\partial{q}}=0 (21)

    where q is generalised co-ordinates and ​ q˙\dot{q}is generalised velocity.

    The Hamiltonian is given by

    H(q,p)=pq˙L H(q,p)=p\dot{q}-L  where​ p=L/q˙p=\partial{L}/\partial{\dot{q}} (22)

    In classical mechanics if we integrate the total energy of the electromagnetic field over an infinite volume, we get infinity. Thus it is better to talk about a density (i.e. energy density).

    So we can take a Lagrangian density L\mathcal{L}, with

    L dv=L\int{\mathcal{L}}  dv=L (23)​

    Apart from L\mathcal{L} being finite, there is extra advantage of this; the form of action looks better from the relativistic point of view:

    S=L d4xS=\int{\mathcal{L}}  d^4x (24)

    Since here the grneralised coordinate is the field φ\varphi which depends on the coordinates xμx^\mu and μφ\partial^{\mu}\varphi. Now the Euler-Lagrange equation becomes more complicated as following:

    ddt(Lφ˙)+.(L(φ))Lφ=0\frac{d}{dt}(\frac{\partial{\mathcal{L}}}{\partial{\dot{\varphi}}})+\nabla.(\frac{\partial\mathcal{L}}{\partial{(\nabla\varphi)}})-\frac{\partial\mathcal{L}}{\partial\varphi}=0 (25)

    Above equation can be written in more compact notation:

    μ(L(μφ))Lφ=0\partial_\mu(\frac{\partial\mathcal{L}}{\partial{(\partial_\mu\varphi)}})-\frac{\partial\mathcal{L}}{\partial\varphi}=0 (26)

    This equation is known as the equation of motion of a field.

    LAGRANGIAN FOR THE ELECTROMAGNETIC FIELD

    The Lagrangian density must be scalar, because we don't want to get it transformed under Lorentz transformation. Since we know that electromagnetism respects parity. Thus we expect that L\mathcal{L} to be invariant under parity transformation xxx\to-x too. We see that FμνGμνF^{\mu\nu}G_{\mu\nu} which is directly proportional to E.BE.B, is not invariant under parity .

    Thus LFμνFμν\mathcal{L}\propto{F^{\mu\nu}F_{\mu\nu}} now we start with

    L=14FμνFμν\mathcal{L}= -\frac{1}{4}{F^{\mu\nu}F_{\mu\nu}} (27)

    Using the explicit form of FμνF^{\mu\nu} this equation becomes

    L=12(μAνμAνμAννAμ)\mathcal{L}= -\frac{1}{2}(\partial^{\mu}A^\nu\partial_{\mu}A_\nu-\partial^{\mu}A^\nu\partial_{\nu}A_\mu) (28)

    AμA^\muis treated here as the electromagnetic field. Ultimately we are going to quantise AμA^\muand we will find the photon as the exitation quantum of the field. Since above equation has only derivatives of AμA^\mu, not AμA^\mu itself, so the equation of motion of a field can be written as

    μL(μAν)=0\partial_\mu\frac{\partial\mathcal{L}}{\partial{(\partial_{\mu}A_\nu)}}=0 (29)

    Now let us compute L/(μAν){\partial\mathcal{L}}/{\partial{(\partial_{\mu}A_\nu)}}. Second term of equation (28) we write that

    L2(ρAτ) = 12δμρδντηνλημκλAκ+12μAνηνλημκδλρδκτ\frac{\partial\mathcal{L_2}}{\partial{(\partial_{\rho}A_\tau)}}  =  \frac{1}{2}\delta^\rho_\mu\delta^\tau_\nu\eta^{\nu\lambda}\eta^{\mu\kappa}\partial_{\lambda}A_\kappa+\frac{1}{2}\partial_{\mu}A_\nu\eta^{\nu\lambda}\eta^{\mu\kappa}\delta^\rho_\lambda\delta^\tau_\kappa

    =12ηρκητλλAκ+12μAνηνρημτ=\frac{1}{2}\eta^{\rho\kappa}\eta^{\tau\lambda}\partial_{\lambda}A_\kappa+\frac{1}{2}\partial_{\mu}A_\nu\eta^{\nu\rho}\eta^{\mu\tau}

    =τAρ=\partial^{\tau}A^\rho

    so that

    L(μAν) = μAν+νAμ\frac{\partial\mathcal{L}}{\partial{(\partial_{\mu}A_\nu)}}  =  -\partial^{\mu}A^\nu+\partial^{\nu}A^\mu

    =Fμν=-F^{\mu\nu}

    Hence the free field Euler-Lagrange equations become

    μFμν =0\partial_{\mu}F^{\mu\nu}  = 0

    This is nothing but two equations of Maxwell's equations, Gauss' law and 3-component Ampere's law, which is written in the absence of external charge or current densities. It is obvious that if we started with

    L=14GμνGμν\mathcal{L}= -\frac{1}{4}{G^{\mu\nu}G_{\mu\nu}}

    we would have obtained the other two equations of Maxwell.We write the Lagrangian density in terms of FμνF^{\mu\nu} not its dual because we have jμ=(ρ,j)j^\mu=(\rho,j) (four current density) which is nonzero. In such case we can write another term L\mathcal{L} of the form jμAμj^{\mu}A_\mu, and

    L=14FμνFμνjμAμ\mathcal{L}= -\frac{1}{4}{F^{\mu\nu}F_{\mu\nu}}-j^{\mu}A_\mu

    gives the correct equation of motion, namely

    μFμν =jν\partial_{\mu}F^{\mu\nu}  =j^\nu

    We see that there are no such magnetic analogue of jμj^\mu.

    Let us see the Gauge invariance of above equation. Suppose the transformation Aμ Aμ+μλA^\mu{\to}  A^\mu+\partial^\mu\lambda. In equation (25) the term jμAμj^{\mu}A_\mu is apparently not invariant, but gets an extra contribution of jμμλj^\mu\partial_\mu\lambda. However,

    jμμλ =μ(jμλ)(μjμ)λj^\mu\partial_\mu\lambda  = \partial_\mu(j^\mu\lambda)-(\partial_{\mu}j^\mu)\lambda (30)

    In this equation the second term is zero only because the electric four-current is conserved because by the continuity equation. So we conclude that the Lagrangian density of the electromagnetic field in the presence of an external current is gauge-invariant only because the current is conserved. We can say in other words also that: the current is conserved because we demand gauge invariance.

    EXPERIMENTAL TESTS OF SPECIAL THEORY OF RELATIVITY

    There are three major tests of Special Theory of Relativity which are following:

    1. Michelson-Morley Experiment

    2. Kennedy-Thorndike Experiment

    3. Trouton-Noble experiment

    Michelson-Morley Experiment[1]

    The Michelson–Morley experiment was an attempt to detect the existence of aether, a supposed medium permeating space that was thought to be the carrier of light waves. The experiment was performed by Albert A. Michelson and Edward W. Morley. Earth orbits around the sun at a speed of around 30 km/s. The Earth is in motion, so two main possibilities were considered: (1) The aether is stationary and only partially dragged by Earth or (2) The aether is completely dragged by Earth and thus shares its motion at Earth's surface. In addition, Jammes Clerk Maxwell (1865) recognized the electromagnetic nature of light and developed what are now called Maxwell's equation, but these equations were still interpreted as describing the motion of waves through an aether, whose state of motion was unknown.

    Michelson-Morley_experiment_conducted_with_white_light_1.png
      Experimental set-up of Michelson-Morley experiment[2]

      Michelson and Morley had a solution to the problem of how to construct a device sufficiently accurate to detect aether flow. The device they designed, later known as a Michelson-Morley interferometer, sent white light through a half silvered mirror that was used to split it into two beams traveling at right angles to one another. After leaving the splitter, the beams traveled out to the ends of long arms where they were reflected back into the middle by small mirrors. They then recombined on the far side of the splitter in an eyepiece, producing a pattern of constructive and destructive interference whose transverse displacement would depend on the relative time it takes light to transit the longitudinal versus the transverse arms. If the Earth is traveling through an aether medium, a beam reflecting back and forth parallel to the flow of aether would take longer than a beam reflecting perpendicular to the aether because the time gained from traveling downwind is less than that lost traveling upwind.

      800px-Michelson-morley_calculations.svg_1.png
        Expected differential phase shift between light traveling the longitudinal versus the transverse arms of the Michelson–Morley apparatus[3]

        The beam travel time in the longitudinal direction can be derived as follows:

        Light is sent from the source and propagates with the speed of light c in the aether. It passes through the half-silvered mirror at the origin at T = 0. The reflecting mirror is at that moment at distance L (the length of the interferometer arm) and is moving with velocity vv. The beam hits the mirror at time T1T_1 and thus travels the distance cT1cT_1 At this time, the mirror has traveled the distance vT1vT_1. Thus cT1=L+vT1cT_1=L+vT_1 and consequently the travel time T1=L/(cv){\textstyle T_{1}=L/(c-v)}. The same consideration applies to the backward journey, with the sign of v reversed, resulting in cT2=LvT2{\textstyle cT_{2}=L-vT_{2}} and T2=L/(c+v){\textstyle T_{2}=L/(c+v)}. The total travel time T=T1+T2{\textstyle T_{\ell }=T_{1}+T_{2}} is:

        Tl=Lcv+Lc+v=2Lc11v2/c22Lc(1v2/c2)T_l=\frac{L}{c-v}+\frac{L}{c+v}=\frac{2L}{c}\frac{1}{1-v^2/c^2}\approx {\frac{2L}{c}}(1-v^2/c^2)

        The beam is propagating at the speed of light c{\textstyle c} and hits the mirror at time T3T_3, traveling the distance cT3{\textstyle cT_{3}}. At the same time, the mirror has traveled the distance vT3{\textstyle vT_{3}} in the x direction. So in order to hit the mirror, the travel path of the beam is L in the y direction (assuming equal-length arms) and vT3{\textstyle vT_{3}} in the x direction. This inclined travel path follows from the transformation from the interferometer rest frame to the aether rest frame. Therefore, the Pythagorean Theorem gives the actual beam travel distance of L2+(vT3)2{\textstyle {\sqrt {L^{2}+\left(vT_{3}\right)^{2}}}}. Thus cT3=L2+(vT3)2{\textstyle cT_{3}={\sqrt {L^{2}+\left(vT_{3}\right)^{2}}}} and consequently the travel time T3=L/c2v2{\textstyle T_{3}=L/{\sqrt {c^{2}-v^{2}}}} which is the same for the backward journey. The total travel time Tt=2T3{\textstyle T_{t}=2T_{3}} is:

        Tt=2Lc2v2=2Lc11v2c22Lc(1+v22c2){\displaystyle T_{t}={\frac {2L}{\sqrt {c^{2}-v^{2}}}}={\frac {2L}{c}}{\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\approx {\frac {2L}{c}}\left(1+{\frac {v^{2}}{2c^{2}}}\right)}

        The time difference between T and Tt before rotation is given by

        TTt=2c(L1v2c2L1v2c2).{\displaystyle T_{\ell }-T_{t}={\frac {2}{c}}\left({\frac {L}{1-{\frac {v^{2}}{c^{2}}}}}-{\frac {L}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\right).}.

        By multiplying with c, the corresponding length difference before rotation is

        Δ1=2(L1v2c2L1v2c2){\displaystyle \Delta _{1}=2\left({\frac {L}{1-{\frac {v^{2}}{c^{2}}}}}-{\frac {L}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\right)}

        and after rotation

        Δ2=2(L1v2c2L1v2c2){\displaystyle \Delta _{2}=2\left({\frac {L}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}-{\frac {L}{1-{\frac {v^{2}}{c^{2}}}}}\right)}

        Dividing Δ1Δ2{\textstyle \Delta _{1}-\Delta _{2}} by the wavelength λ, the fringe shift n is found:

        n=Δ1Δ2λ2Lv2λc2{\displaystyle n={\frac {\Delta _{1}-\Delta _{2}}{\lambda }}\approx {\frac {2Lv^{2}}{\lambda c^{2}}}}

        Since L ≈ 11 meters and λ≈500 nanometer, the expected fringe shift was n ≈ 0.44. So the result would be a delay in one of the light beams that could be detected when the beams were recombined through interference. Any slight change in the spent time would then be observed as a shift in the positions of the interference fringes. The negative result led Michelson and Morley to the conclusion that there is no measurable aether drift.

        Tt=2Lc,{\displaystyle T_{t}={\frac {2L}{c}},}

        Kennedy Thorndike Experiment[4]

        The Kennedy–Thorndike experiment, first conducted in 1932 by Roy J. Kennedy and Edward M. Thorndike, is a modified form of the Michelson-Morley experimental procedure, testing special relativity. The modification is to make one arm of the classical Michelson–Morley apparatus shorter than the other one. While the Michelson–Morley experiment showed that the speed of light is independent of the orientation of the apparatus, the Kennedy–Thorndike experiment showed that it is also independent of the velocity of the apparatus in different inertial frames.

        1200px-Kennedy-Thorndike_experiment_DE.svg.png
          Experimental set-up for Kennedy-Thorndike Experiment[5]

          Although Lorentz–FitzGerald contraction (Lorentz contraction) by itself is fully able to explain the null results of the Michelson–Morley experiment, it is unable by itself to explain the null results of the Kennedy–Thorndike experiment. Lorentz–FitzGerald contraction is given by the formula

          L=L0 1v2/c2=  L0/γ(v)L=L_{0} \sqrt{1-v^2/c^2}=   L_0/\gamma(v) (31)

          where

          L0L_0 is the proper length (the length of the object in its rest frame),

          LL is the length observed by an observer in relative motion with respect to the object,

          vvis the relative velocity between the observer and the moving object, i.e. between the hypothetical aether and the moving object

          ccis the speed of light,

          and the Lorentz factor is defined as

          γ(v)=1v2/c2\gamma(v)= \sqrt{1-v^2/c^2} (32)

          If the apparatus is motionless with respect to the hypothetical aether, the difference in time that it takes light to traverse the longitudinal and transverse arms is given by:

          TLTT=2(LLLT)cT_L-T_T=\frac{2(L_L-L_T)}{c} (33)

          The time it takes light to traverse back-and-forth along the Lorentz–contracted length of the longitudinal arm is given by:

          TL=T1+T2T_L=T_1+T_2

          =LL/γ(v)cv+LL/γ(v)c+v=\frac{L_L/\gamma(v)}{c-v}+\frac{L_L/\gamma(v)}{c+v}

          =2LL/γ(v)c11v2/c2=\frac{2L_L/\gamma(v)}{c}\frac{1}{1-v^2/c^2}

          =2LLγ(v)c=\frac{2L_L\gamma(v)}c (34)

          where T1 is the travel time in direction of motion, T2 in the opposite direction,viz the velocity component with respect to the luminiferous aether, c is the speed of light, and LL the length of the longitudinal interferometer arm. The time it takes light to go across and back the transverse arm is given by:

          TT=2LTc2v2T_T=\frac{2L_T}{\sqrt{c^2-v^2}}

          =2LTc11v2/c2  =\frac{2L_T}{c}\frac{1}{\sqrt{1-v^2/c^2}}  

          =2LTγ(v)c= \frac{2L_T\gamma(v)}c

          The difference in time that it takes light to traverse the longitudinal and transverse arms is given by:

          TLTT=(2LLLT)γ(v)cT_L-T_T=\frac{(2L_L-L_T)\gamma(v)}c (35)

          Because ΔL=c(TL-TT), the following travel length differences are given (ΔLA being the initial travel length difference and vA the initial velocity of the apparatus, and ΔLB and vB after rotation or velocity change due to Earth's own rotation or its rotation around the Sun)

          ΔLA=2(LLLT)1vA2/c2    \Delta{L_A}=\frac{2(L_L-L_T)}{\sqrt{1-v_A^2/c^2}}      

          ΔLB=2(LLLT)1vB2/c2\Delta{L_B}=\frac{2(L_L-L_T)}{\sqrt{1-v_B^2/c^2}}{} (36)

          In order to obtain a negative result, we should have ΔLA−ΔLB=0. However, it can be seen that both formulas only cancel each other as long as the velocities are the same (vA=vB). But if the velocities are different, then ΔLA and ΔLB are no longer equal (The Michelson–Morley experiment isn't affected by velocity changes since the difference between LL and LTis zero. Therefore, the MM experiment only tests whether the speed of light depends on the orientation of the apparatus.) But in the Kennedy–Thorndike experiment, the lengths LLand LT are different from the outset, so it is also capable of measuring the dependence of the speed of light on the velocity of the apparatus.

          According to the previous formula, the travel length difference ΔLA−ΔLB and consequently the expected fringe shift ΔN are given by (λ being the wavelength):

          ΔN=ΔLAΔLBλ\Delta{N}=\frac{\Delta{L_A}-\Delta{L_B}}\lambda

          =2(LLLT)λ(11vA2/c211vB2/c2)= \frac{2(L_L-L_T)}{\lambda}(\frac{1}{\sqrt{1-v_A^2/c^2}}-\frac{1}{\sqrt{1-v_B^2/c^2}})

          Neglecting magnitudes higher than second order in v/c:

          LLLTλ(vA2vB2c2)\approx\frac{L_L-L_T}{\lambda}(\frac{v_A^2-v_B^2}{c^2}) (37)

          For constant ΔN, i.e. for the fringe shift to be independent of velocity or orientation of the apparatus, it is necessary that the frequency and thus the wavelength λ be modified by the Lorentz factor. This is actually the case when the effect of time dilation on the frequency is considered. Therefore, both length contraction and time dilation are required to explain the negative result of the Kennedy–Thorndike experiment.

          Trouton-Noble Experiment[6]

          The Trouton–Noble experiment was an attempt to detect motion of the Earth through the aether, and was conducted in 1901–1903 by Frederick Thomas Trouton and H.R. Noble. It was based on a suggestion by George FitzGerald that a charged parallel-plate capacitor moving through the aether should orient itself perpendicular to the motion. Like the earlier Michelson-Morley Experiment, Trouton and Noble obtained a null result: no motion relative to the aether could be detected.

          220px-Trouton–Noble_experiment.png
            A circular capacitor B, was fitted into a smooth spherical celluloid ball D that was covered with conductive paint. A mirror attached to the capacitor was viewed through a telescope and allowed fine changes in orientation to be viewed.[7]

            In the experiment, a suspended parallel-plate capacitor is held by a fine torsion fiber and is charged. If the aether theory were correct, the change in Maxwell's Equation due to the Earth's motion through the aether would lead to a torque causing the plates to align perpendicular to the motion. This is given by:

            τ=Ev2c2sin2α\tau=-E'\frac{v^2}{c^2}\sin2\alpha'

            where τ\tauis the torque, EE the energy of the condenser, α\alpha the angle between the normal of the plate and the velocity.

            On the other hand, the assertion of special relativity that Maxwell's equations are invariant for all frames of reference moving at constant velocities would predict no torque (a null result). Thus, unless the aether were somehow fixed relative to the Earth, the experiment is a test of which of these two descriptions is more accurate. Its null result thus confirms Lorentz invariance of special relativity and the absence of any absolute rest frame (or aether).

            REFERENCES

            [1]-https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment

            [2]- https://academy.resonance.is/wp-content/uploads/2014/10/Michelson-Morley_experiment_conducted_with_white_light.png

            [3]-https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment#/media/File:Michelson-morley_calculations.svg

            [4]- https://en.wikipedia.org/wiki/Kennedy%E2%80%93Thorndike_experiment

            [5]- https://upload.wikimedia.org/wikipedia/commons/thumb/d/d4/Kennedy-Thorndike_experiment_DE.svg/1200px-Kennedy-Thorndike_experiment_DE.svg.png

            [6]- https://en.wikipedia.org/wiki/Trouton%E2%80%93Noble_experiment

            [7]- https://upload.wikimedia.org/wikipedia/commons/thumb/5/55/Trouton%E2%80%93Noble_experiment.png/220px-Trouton%E2%80%93Noble_experiment.png

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