Asif's equation and Einstien's theory of special relativity

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For history and motivation, see History of special relativity.

For a more accessible and less technical introduction to this topic, see Introduction to asif& § 39;s equation and Einstien's theory of special relativity.

The theory of special relativity plays an important role in the modern theory of classical electromagnetism. Considering deeply the effect of Special relativity in Electromagnetism, when a charge particle moves with high speed as comparable to the speed of light in vacuum tube or in space under influence of electromagnetic field, its mass varies under Lorentz transformation . In physics specially electromagnetism, Asif's equation of charge variation demonstrates the variation of electric charge under Lorentz transformation. The more sophisticated view of electromagnetism expressed by electromagnetic fields in moving inertial frame can be achieved by considering some relativistic effect including charge as well

History

From a historical point of view, it is evident that Maxwell's equations themselves were precursors to the eventual formulation of special relativity by Albert Einstein in 1905 Purcell argued that, the sources which create electric field are at rest with respect to one of the reference frames which is moving with constant velocity. Given the electric field in the frame where the sources are at rest, Purcell asked: what is the electric field in some other frame? considering this, a Pakistani theoritical physicist proposed a simple equation named as “Asif’s equation of charge variation” that is the modern teaching strategy for developing electromagnetic field theory. Asif’s equation solves the mysteries of an electric field, magnetic field, electromagnetic wave and behavior of point charge in inertial and non-inertial frame of references with respect to rest observer. His first demonstration lies under the experiment when he was observing magnetic field around current carrying conductor, he observed that current carrying conductor produces magnetic field around it with some sort of electric field which was perpendicular to the magnetic field. He realized that there should be change in the charge of electrons moving in the conductor that causes net electric field around conductor. Then he started working on the concept of charge variation in moving inertial frame and the corresponding results was published by International Organization of Scientific Research in Journal of Applied Physics

Derivation of Asif’s equation of charge variation

When a charge moves in electric and magnetic field, it experience electromagnetic force. Consider that a charge is moving with the uniform velocity along x-axis with respect to rest observer and electric and magnetic field are applied externally along x-axis and z-axis respectively. The force experienced by charge due to magnetic field is given by.

$F_{b}=q(VB)$

Force due to electric field on charge is given by.

$F_{e}=qE$

Let us suppose that electric and magnetic field are provided in such a way that electric and magnetic force on a charge becomes equal and charge will deviate due to magnetic field and forms circular path if fluorescent screen is placed in front of it.

$F_{b}=F_{e}$ (1)

Since magnetic force experienced by charge provides necessary centripetal force which is given by.

$F_{b}=F_{e}={\frac {mv^{2}}{r}}$

A body experiencing uniform circular motion requires a centripetal force, towards the axis as shown, to maintain its circular path.

Solving for linear momentum of charge, above equation can be written as.

$P=mv=qBr$ (2)

kinetic energy is given by.

$K.E=\int (\mathbf {F} \,ds)$

According to Newton’s second law of motion,:${\mathbf {F}}={\frac {dP}{dt}}$ where "P" is linear momentum. Therefore above equation can be written as.

$K.E=\int ({\frac {dP}{dt}})ds=\int ({\frac {ds}{dt}})dP$

$K.E=\int vds$

From equation (2) putting value of "P" in above equation. Since "B" and "r" are constant term.

$K.E=Br\int vd(q)$

Since charge "q" is variable and depends upon velocity "v". Therefore according to rule of integration we can equate above equation as.

$K.E=Br\int vd(q)=Br\int d(qv)$

Now using product rule of derivative and multiplying and dividing by "v". We obtain

$K.E=\int {\frac {(vdq+qdv)v}{v}}$ (3)

Since charge is moving with the high speed comparable to the speed of light, to achieve desired relation for charge variation, Lorentz factor is applied which is given by.

$\gamma ={\frac {1}{\sqrt {1-v^{2}/c^{2}}}}={\frac {1}{\sqrt {1-\beta ^{2}}}}={\frac {\mathrm {d} t}{\mathrm {d} \tau }}$

Squaring both sides and by cross multiplying Lorentz factor, we have

$\gamma ^{2}(1-{\frac {v^{2}}{c^{2}}})=1$

$\gamma ^{2}c^{2}-\gamma ^{2}v^{2}=c^{2}$

Taking differential

$2\gamma c^{2}d\gamma -2\gamma v^{2}d\gamma -2v\gamma ^{2}dv=0$

$c^{2}d\gamma =v^{2}d\gamma +v\gamma dv$

$c^{2}d\gamma =(vd\gamma +\gamma dv)v$

In special theory of relativity, small change in the Lorentz factor causes due to mass variation, length contraction or time dilation, let us assume that change in the Lorentz factor causes due to charge variation for rest observer in electromagnetic field. Hence replacing "d\gamma" by "dq" and "\gamma" by "q" we get.

$c^{2}dq=(vdq+qdv)v$

Or

$(vdq+qdv)v=c^{2}dq$ (4)

Substituting equation (4) in equation (3), we obtain

$K.E={\frac {Br}{v}}\int c^{2}dq$ (5)

From equation (2) relation for kinetic energy can be derived as.

$K.E={\frac {1}{2}}mv^{2}={\frac {1}{2}}qvBr$

Hence equation (5) can be written as

${\frac {1}{2}}qvBr={\frac {Br}{v}}\int c^{2}dq$

${\frac {1}{2}}qv^{2}=\int c^{2}dq=\int v(vdq+qdv)$

Or

$\int c^{2}dq=\int v(vdq+qdv)$

Multiplying both sides by "2q" then equation becomes

$\int 2qc^{2}dq-\int (2qv^{2}dq+2vq^{2}dv)=0$ (6)

According to product rule of derivative

$\int (2qv^{2}dq+2vq^{2}dv)=\int d(q^{2}v^{2})$

Hence equation (6) becomes

$\int 2qc^{2}dq-\int d(q^{2}v^{2})=0$

Now integrating both sides

$q^{2}c^{2}-q^{2}v^{2}=C$ (7)

where "C" is constant of integration. To find value of C, we consider that at initial state, velocity of charge is zero. Hence equation(7) becomes

$C=q_{0}^{2}c^{2}$

Where "$q_{0}$" is rest charge because velocity of charge particle is zero at initial state. Substituting value of "C" in equation (7), finally we get

$q^{2}c^{2}-q^{2}v^{2}=q_{0}^{2}c^{2}$

${\frac {q^{2}(c^{2}-v^{2})}{c^{2}}}$

$q^{2}(1-{\frac {v^{2}}{c^{2}}})=q_{0}^{2}$

Hence charge of particle in moving inertial frame i-e "q" is given by.

$\mathbf {q} ={\frac {q_{0}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$

It is named as Asif’s Equation of Charge Variation. Where "q" is relativistic charge and "$q_{0}$" is rest charge.

Derivation of E=mc^2 from Asif's equation

From equation(5), as charge starts moving from rest, it varies from "$q_{0}$" to "q" with respect to rest observer.

$K.E={\frac {Br}{v}}\int _{q_{0}}^{q}c^{2}dq$

$K.E={\frac {Brqc^{2}}{v}}-{\frac {Brq_{0}c^{2}}{v}}$

From equation(2), :$qBr=mv$ and $q_{0}=m_{0}v$ , therefore.

$K.E={\frac {mvc^{2}}{v}}-{\frac {m_{0}vc^{2}}{v}}$

$K.E=mc^{2}-m_{0}c^{2}$

The kinetic energy is the difference between relativistic mass energy and rest mass energy Therefore, total relativistic energy is the sum of the kinetic energy and the rest energy as given by.

$E=mc^{2}$

References

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