Signature of a unique function in energy transfer

Signature of a unique function in energy transfer Subodha Mishra Department of Physics, ITER, Siksha ’O’ Anusandhan University, Bhubaneswar-751030, India (Dated: November 14, 2015) The appearance of a unique function in the energy transfer from one system to the other in different physical situations such as electrical, mechanical, optical and quantum mechanical processes is established in this work.

Energy can neither be created nor be destroyed but can be transformed from one form to another and it can be transferred from one system to another [1]. Though the laws governing the energy transformation and its trasfer from system to system is well known, here we notice a unity in diversity; a unique function appears in all cases of energy trasfer whether it is classical or quantum mechanical process. We consider four examples, well known in elementary physics, from the fields of electricity, mechanics, optics and quantum mechanics. I.

for momentum conservation before and after the collision as P1 + 0 = P 3 + P 4 And for conservation of energy we have, P2 P2 P12 +0= 3 + 4 2m1 2m1 2m2

P3 =



m1 − m 2 m1 + m 2



P1

(6)

P4 =



2m2 m1 + m 2



P1

(7)

In an electrical circuit consisting of an Emf (E) having internal resistance r and an external resiatance ( load) R in series, the current [2] in the circuit given as E . R+r

(1)

The power transferred to the external resistance R is given as

So energy transfered to the second body of mass m2 can be written as Etrans =

Ptrans

  E2 E2 x = I 2R = R = (R + r)2 r (1 + x)2

Ptrans x Etran = 2 = E /r (1 + x)2 tE 2 /r

Since the intial available energy is Ein = take m1 /m2 = x ,we find that

ENERGY TRANSFER IN TWO BODY COLLISION

In this section, we consider an example of elastic collision of two bodies in one dimension [3], Here let us consider a mass (m1 ) coming with momentum P1 colliding with a mass (m2 ) intially at rest (P2 = 0) and then mass m1 moving with momentum P3 and the other mass m2 moving with momentum P4 . So we write the equation

P12 2m1

(8) and we

Etrans m1 m2 x = = 4Ein (m1 + m2 )2 (1 + x)2

(3)

where the right hand side is a function of x only. We will see that in other different physical situations such as mechanical, optical and quantum mechanical, the same x function (1+x) 2 appears in the energy transfer procesess. II.

P2 4m1 m2 P42 = 1 2m2 2m1 (m1 + m2 )2

(2)

where x = R/r. Hence we get the ratio of energy transfered to available energy as

(5)

Solving these two equations, the momenta P3 and P4 after collision are given in terms of P1 as

ELECTRICAL ENERGY TRANSFER TO A RESISTANCE

I=

(4)

III.

(9)

ENERGY TRANSFER IN OPTICAL TRANSMISSION

Here we consider electromagnetic energy transmission [4] when light is incident normally (for simplicity) in the z-direction on a medium having refractive index n1 and traveles to another medium having refactive index n2 . We consider both the medium non-magnetic. The incident, reflected and transmitted electric waves are given as ~ i (z, t) = Ei ei(k1 z−ωt) x E ˆ,

(10)

~ r (z, t) = −Er ei(−k1 z−ωt) x E ˆ, i(k2 z−ωt) ~ Et (z, t) = Et e x ˆ

(11) (12)

and since the magnetic field is related to the electric field ˆ E ~ ~ = k× as B v , corresponding magnetic waves are given as ~ i (z, t) = Bi ei(k1 z−ωt) yˆ, B v1 B r ~ r (z, t) = − ei(−k1 z−ωt) yˆ, B v1 ~ t (z, t) = Bt ei(k2 z−ωt) yˆ B v2

(13)

d2 ψ(x) 2m + 2 (E − V0 )ψ(x) = 0 dx2 h ¯

(14)

(21) 2 2

(15)

Applying the boundary conditions at the boundary (z=0) as E||1 = E||2 and B||1 = B||2 to the wave equations and solving for Er and Et and taking ratio of velocities in the two media as v1 /v2 = n2 /n1 , we get   n1 − n 2 Er = Ei , (16) n1 + n 2   2n1 Ei (17) Et = n1 + n 2 As we know, the average electromagnetic energy flux in the z-direction is given by I=

the potential step [5] defined as V (x) = 0 f or x < 0 and V (x) = V0 f or x > 0. The Schrodinger equation satisfied by the particle is given as

~ × B.ˆ ~ z) (E EB E2 = = µ0 2µ0 2µ0 v

(18)

k The particle has kinetic energy p2 /2m = h¯2m =E− V0 and total energy, E. As we know from elementary quantum mechanics the solution for the wave function for the particle in two regions defined by the potential are given as

ψI = eikx + re−ikx x < 0 ψII = teiqx x > 0.

(22) (23)

where r and t are amplitudes of the reflected and trans√ 2mE/¯ h and q = mitted waves respectively and k = p 2m(E − V0 )/¯ h. The unknown quantities r and t, are determined by using the continuity of ψ and dψ/dx at x = 0. One gets, r = (k − q)/(k + q) and t = 2k/(k + q). Since one can show that the incident current density is jinc = h ¯ k/m and transmitted current density is jtran = (¯ hq/m)|t|2 , the transmission coefficient T (which when multiplied with the incident energy is the fraction of energy transmitted) is given as

So we get coefficient of transmission T as T =

n2 It = Ii n1



Et Ei

2

=

n2 n1



2n1 n1 + n 2

2

T =

(19)

Taking n1 /n2 = x, we can rewrite the above as, T x = 4 (1 + x)2

4kq jtran = |t2 | = jinc (k + q)2

(24)

Taking k/q = x, we have, (20)

T x = 4 (1 + x)2

Maximum energy transfer occurs for x = 1 i.e n1 = n2 . This is obvious as there is no boundary and there is no reflection but only transmission.

V.

(25)

CONCLUSION

Here we consider a quantum particle having mass m and momentum h ¯ k being incident from the left x < 0 on

We notice whether the process is classical or quantum x mechanical, the function (1+x) 2 appears in the energy transfer between two systems. This is a uniqe function which manifests itself in different processes such as electrical, mechanical, optical or quantum mechanical.

1

5

IV. ENERGY TRANSFER IN ONE DIMENSIONAL POTENTIAL STEP

2

3

4

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Mathews P T, 1974, McGraw-Hill Book, UK, 37.