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Canadian Journal of Physics

Change in the direction of polarization vector and redshift of an incoming light ray as observed from a rotating frame.

Journal: Canadian Journal of Physics

Manuscript ID cjp-2018-1015.R3 Manuscript Type:ForArticle Review Only Date Submitted by the 28-Jun-2019 Author:

Complete List of Authors: Ghosh, T.; Assam University Sen, A.K.; Assam University,

Keyword: methods: analytical, polarization, reference systems, relativity, redshift

Is the invited manuscript for consideration in a Special Not applicable (regular submission) Issue? :

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Change in the direction of polarization vector and redshift of an incoming light ray as observed from a rotating frame.

T. Ghosh and A. K. Sen Department of Physics, Assam University, Silchar-788011, Assam, India (Dated: Received: date / Accepted: date) In the present work, change in the direction of the polarization vector of an incoming light ray is extensively calculated for a rotating observer. The change in the direction of the polarization vector calculated here is only due to the effect of the non-inertial rotating frame, considering that the light source is at a distance and it is emitting a plane polarized light. The metric tensors for a rotating observer have been collected from existing lit- erature. Accordingly electric displacement and magnetic induction values as applicable for the rotating observer have been calculated. These values are used to calculate the change in the orientation of the electric vector of an incoming plane polarized light ray. Earth has been taken as an example of rotating frame, and the calculated amount of change in the direction of the polarization vector has been found to be dependent on the azimuthal as well as polar co-ordinate of the rotating frame. Present work also discusses the redshift as observed by a rotating observer and the value of the redshift has been calculated for an observer sitting on rotating earth.

I. INTRODUCTION measurement by an accelerated observer and focused on the hypothesis of locality which is described as “The presumed equivalence of an accelerated observer with a momentarily Light is an electromagnetic wave,For and it movesReview along the co-moving Only inertial observer-underlies the standard relativis- curvature of the space-time structure. From this very concept, tic formalism by relating the measurements of an accelerated it is a well known fact that light shows a unique property, observer to those of an inertial observer”. Mashhoon [11] known as the gravitational deflection. Following this idea, discussed briefly, the significance and the limitations of it is obvious that the light ray will be deflected, when it the hypothesis of locality in his work. To understand the passes through a gravitational field, commonly known as motion under the influence of accelerated frame one must gravitational deflection. The gravitational deflection of the consider the hyperbolic motion in space-time, which had been light ray gives undeniable proof of the General Theory of elaborately discussed by Rindler [12]. He was the first person Relativity. Since the establishment of the General Theory who studied relativistic motion under an accelerated system of Relativity, many authors [1–8] have carried out inves- and later Born coined it as hyperbolic motion. Rindler [12] tigations for the proper understanding of the gravitational had obtained the differential equation of motion for a particle deflection and the associated polarization phenomena. As in uniform accelerated frame generalising the geometric gravitational deflection is a well-established phenomenon, characteristics of a rectangular hyperbola in Minkowskian it gives an opportunity to study the polarization state of space-time. It is termed as Hyperbolic motion from the the light ray which passes through such space-time. Some fact that, as seen by the inertial frame observer if we plot authors have found in the past [3, 9], that when a light ray the distance against the time on a Minkowski diagram, it passes through gravitational field having axial symmetry, describes a hyperbola. then its polarization vector is rotated by the gravitational field, quite analogous to the phenomenon which is called Faraday Rotation by the magnetic field. More recently Ghosh and Sen [10] investigated the change in the direction of In case of inertial frames, we can transform any four-vector polarization vector in the case of both Schwarzschild and by performing a multiplication operation by a 4x4 Lorentz Kerr field. They showed that although the Schwarzschild transformation matrix. The electric and magnetic field vectors field does not affect the state of polarization of light, but Kerr are components of a field tensor (Fik) of rank two. Thus by field produces a change in the direction of the polarization operating with Lorentz Transformation Matrix twice we can vector of the electromagnetic wave or light ray. Though transform electric and magnetic vectors between any two there are reported works on the change in the direction of the inertial frames [13] [page no. 66]. polarization vector by the space-time geometry introduced by gravitational mass, there is not much investigation conducted in case of accelerated systems. Thus a natural question lies here, whether the space-time transformation generated by an accelerated observer has any effect on the polarization state of However, when we have linear or rotational accelera- the light ray or not? Here in the present work, the space-time tion, we can not identify any Lorentz like Transformation transformation generated by a rotating frame is the subject of Matrix to perform similar co-ordinate transformation. Let investigation. us consider two reference frames K′′(x′′0,x′′1,x′′2,x′′3) and K(x0,x1,x2,x3) representing accelerated (non-inertial) and in- In a discussion on the accelerated system, Mashhoon ertial frame respectively. Under such a situation the transfor- [11] raised the question about the law which specifies the mation relation can be given by Equations (15a) and (15b) of

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Nelson (1987) as reproduced below [14] : Mashhoon [11] showed that an observer rotating in a frame ∫ ∫ with radius r has transnational rΩ2γ2, where γ is ′′i i t × i t γ i x = x + 0 [(W + w) r] dt + 0 V dt the Lorentz factor, and Ω is the rate of rotation of frame per + 1 (γ − )(V xm)V i ∫ V 2 1 m unit time (in other words angular velocity). Thus in continua- + t 1 (γ − )V [(W + w) × r]m V idt (1) 0 V 2 ∫ 1 m tion of the work by him, the rotating observer can be consid- ′′0 t γ 1 γ m x = ∫ 0 c dt + c Vmx ered as an accelerated system, and the metric for the system t 1 γ × m has been given by [11]; + 0 c Vm [(W + w) r] dt 2 Ω2γ2 g = γ2[ − Ω (r + x1)2 − (x2)2] where w is the rotational velocity of the observer’s frame and 00 1 c2 c2 W is related to linear acceleration of the same frame. Others − Ωγ2 ×⃗ (3) g0α = [ c X] symbols have meaning as explained in Nelson (1987). Since gαβ = −δαβ . this type of transformation relation can not be replaced by a As per Mashhoon [11], the rotating observer can be charac- Lorentz like transformation, so it is more convenient to dis- c cuss the transformation of electric and magnetic field vectors, terised by length γΩ , where in the above 1 2 3 using suitable metric tensors ’gik’[14, 15] X = (x ,x ,x ) vector. Very recently Hauck and Mashhoon [16] had approached this Adams [24] first confirmed from the problem of electromagnetism as observed by an observer in measurement of the apparent radial velocity of Sirius B, since a rotating frame. We shall discuss their results and compare then many observations were made to measure the gravita- with our results towards the end of this paper in section II B. tional red-shift [25, 26]. More recently, Krisher et al. [27] To find out the coordinate transformation relation for the lin- measured the gravitational redshift of the Sun in the year early accelerated frame let us consider a, as the proper linear 1993. Recently Dubey and Sen [28] gave the analytic result acceleration of the observer, andFor it is accelerated Review parallel to for Only the gravitational redshift observed from a rotating body the +x1 axis. In , proper acceleration is an and as observed by an asymptotic observer. The authors also acceleration measurable by an rigidly tied to a calculated the numerical value of redshift of some rotating frame and does not occur by . We consider that heavenly bodies. In section IV, of the present work the red- an observer is being accelerated and its coordinate system is shift, Z as observed by an rotating observer for light coming K˜ (x˜0,x˜1,x˜2,x˜3). Then K(x0,x1,x2,x3) defines the coordinate from an inertial frame has been calculated analytically along frame where the inertial source is situated. with their numerical values and for that matter the observer Now for the constant acceleration, (a) the Rindler co-ordinate was considered on the surface of the earth. for the accelerated observer in instantaneous inertial frame K˜ (as defined above) in terms of the co-ordinate of the inertial frame K can be written as [15, 17–20]: II. DISCUSSION ON ROTATING OBSERVER

2 0 Now let us consider the observer is sitting on a rotat- x0 = ( c + x1) ( ax˜ ) a ˜ sinh c2 ing frame, which has the rate of rotation Ω. According to 2 0 (2) x1 = ( c + x1) ( x˜ ). Mashhoon [11] the distinction between accelerated observer a ˜ cosh c2 in Minkowskian space-time and co-moving inertial observer This is the most general form of Rindler coordinates for a uni- is the presence of acceleration scales associated with non- form accelerated observer. Nelson [14] also derived the trans- inertial observer. In this paper, the rotation is considered formation equation for a non-rotational uniform linear accel- around with the x3 or z-axis. The light source is situated in erated frame. He also derived the equation for the motion a space where frame-dragging and other gravitational effect by the time-dependent, non-gravitational acceleration. Later of rotating frame (body) is negligible. Further the light ray is Nelson [14] extensively worked on the transformation relation coming parallel to x1 or x-axis of the Cartesian system. From between the rest frame and the accelerated frame considering Eqn.(3) the metric elements for the rotating frames can be de- both linear acceleration and rotational acceleration. Later his rived, and these have been given by [11]; work had shown that the relation satisfies the familiar Rindler 2 Ω2γ2 g = γ2[ − Ω (r + x1)2 − (x2)2] metric also. A similar transformation relation can be obtained 00 1 c2 c2 from work by Alsing and Milonni [21]. In their work Alsing γ2Ω 2 g01 = c (x ) and Milonni [21] briefly discussed Hawking-Unruh radiation γ2Ω (4) g = − (x1) temperature for an accelerated frame. Also from their work, 02 c g = 0 the transformation relation for the uniform accelerated frame 03 g = −δ . can be derived as in Eqn.(2). Recently the authors Scarr and αβ αβ Friedman [22] had obtained the velocity as well as coordinate As the components of Ω lie only along x3 or z-axis exists, transformation relation for a uniformly accelerated frame. In the other components of Ω don’t exist (i.e. Ωx = Ωy = 0 one of their previous work, they [23] had discussed the four- and Ωz = Ω). Again, if the rate of rotation Ω becomes zero dimensional covariant relativistic equation and the concept of then from the Eqn.(4), the metric elements are given as a maximal acceleration. Castillo and Sanchez´ [15] studied the g00 = 1,g0α = 0,gαβ = −δαβ , which are the metric for flat uniformly accelerated motion, and estimated the red-shift of space-time. an electromagnetic wave.

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A. Change in polarization vector for a general case So from Eqn.(6) with the help from Eqn.(9) it can be written In order to calculate the amount by which the polarization as [10]: √ vector of the light ray will be changed (rotated), we consider 1 + g g χ ξ √ 00 01 the light ray from the source to be plane polarized (as was observer = arctan( − ) (10) done in our earlier work [10]). To determine the position angle 1 g00g01 So the total change in the direction of polarization vector of of the polarization vector of light ray received by the observer received light ray (say χ) can be calculated from Eqn.(10) and from the distant source, one has to consider the initial compo- Eqn.(5), as: nents of the polarization vector of the light ray Ey and Ez along 2 3 the direction of x and x -axis respectively, where one should χ = χobserver − χsource recall the fact that light ray has been travelling along x1-axis. √ 1 + ( g00)g01 Let us assume that the magnitude ratio between Ey and Ez is = arctan(ξ √ ) − arctanξ 1 − ( g00)g01 ξ, in other words Ey = ξEz [10]. So the initial position angle ( √ ) χ 2ξ( g )g of the polarization vector source is given by; √ 00 01 √ = arctan 2 2 (11) 1 + ξ + ξ ( g00)g01 − ( g00)g01 χ Ey ξ source = arctan( ) = arctan (5) It is clear from Eqn.(11) that the total change in the direction Ez of polarization vector of light ray depends on the initial po- Now the position angle of the polarization vector as will be sition angle of the polarization vector of the light ray emitted measured by the observer in the rotating frame will be [10]: from a source. Thus Eqn.(11) gives the change in the direc- For Reviewtion Only of polarization vector of light ray received by a rotating observer (i.e non-inertial frame). χ Dy observer = arctan( ) (6) Now, for the simplicity of the problem let us consider, E = E , Dz y z (which indicates the polarization vector is held at an angle 45◦ Where Dy and Dz are the electric displacement component with respect to the rotation axis). Then the change in the di- along the x2 and x3-axis respectively. To find out the value rection of polarization vector can be given as: of χ one must find out the relationship between elec- √ observer χ = arctan( g g ) (12) tric displacement vector ⃗D and original electric vector ⃗E. The 00 01 relation between electric displacement vector and the electric vector has been given by [13]: B. Change in the direction of polarization vector as seen from rotating frame ⃗E ⃗D = √ + H⃗ ×⃗g (7) g00 To determine the total change in the direction of polariza- tion vector by a rotating observer, one should recall the met- Where H⃗ is the magnetic induction of the light ray received. ric given in Eqn.(4). From subsection II A, Eqn.(12), it is Now from the Eqn.(7) the components Dy and Dz are deduced clear that only the metric components g00, and g01 given in as[10, 13]: Eqn.(4), are necessary to determine the total change in the di- rection of polarization vector as observed by an observer, who √Ey Dy = + Hzg01 is placed on an accelerated non-inertial frame. Now in Eqn.(4) g00 (8) √Ez the Lorentz factor γ has been introduced by[11]; Dz = − Hyg01 g00 √ 1 The different components of the magnetic field of the incom- γ = (13) − Ωr 2 ing light ray can be written in terms of the electric field using 1 ( c ) the relation Hz = |H⃗ z| = |nˆ × E⃗y| [13] [page no. 112], where nˆ is the unit propagation vector parallel to the propagation of Let us assume that the observer is situated at the azimuthal light ray (eg. parallel to x-axis). Now taking the note that position ϕ and polar position θ on the surface of a rotating Ey = ξEz and considering only the magnitude of H⃗ , we can frame of radius r, so the Cartesian y co-ordinate of the ob- 2 write from Eqn.(8): server is given by (x ) = r sin(θ)sin(ϕ) and x co-ordinate is given by (x1) = r sin(θ)cos(ϕ). Now with the help of Eqn.(4), √ξ Eqn.(12) and Eqn.(13) the total change in the direction of po- Dy = Ez( + ξg01) g00 (9) larization vector for a rotating observer can be written as; √1 Dz = Ez( − g01) g00

https://mc06.manuscriptcentral.com/cjp-pubs Canadian Journal of Physics Page 4 of 11 4 √ χ = arctan( g00g01) [√( )( )] Ω2 Ω2γ2 γ2Ω = arctan γ2[1 − (r + x1)2 − (x2)2] (x2) c2 c2 c v u( )( ) u t 1 Ω2 Ω2γ2 1 Ωr sinθ sinϕ = arctan [1 − (r + r sinθ cosϕ)2 − (r sinθ sinϕ)2]  − Ωr 2 c2 c2 − Ωr 2 c 1 ( c ) 1 ( c ) [√ { }( )] c2 Ω2 (Ωr)2 cΩr sinθ sinϕ = arctan [1 − (r + r sinθ cosϕ)2 − sin2 θ sin2 ϕ (14) c2 − (Ωr)2 c2 c2 − (Ωr)2 c2 − (Ωr)2

Eqn.(14) gives the total change in the direction of polarization g01 component in Eqn.(14) becomes zero which in turn makes vector of light ray as observed by a rotating observer situated χ = 0 (i.e. there is no change in the direction of polarization on the surface of a frame of radius r, with azimuthal position vector). In case of earth the pole is represented by 90o lati- ϕ and polar position θ having rate of rotation Ω. These co- tude, where θ = 0 and the relation between earth latitude (say ordinate ϕ and θ can be identified as longitude and ( 90◦ - Θ) is Θ = (90o −θ). So from above discussion, it is clear that latitude) corresponding to earth. there would be no change in the direction of polarization vec- tor if the light ray is observed from any of the two poles of the For Reviewearth. Only 1. Case 1: For θ = 0, polar position of observer

θ π Now, consider a case when the polar position of the ob- 2. Case 2: For = 2 equatorial position of observer server is given by θ = 0 (observer is sitting on the north pole of rotating frame), and with the help of Eqn.(14) it is clear In this case, let us consider the polar position of the rotating θ π that there is no change in the direction of polarization vector observer is given by = 2 , then with the help of Eqn.(14) the of light ray received by a rotating observer. Putting θ = 0 the total rotation of the polarization vector can be written as;

[√{ ( )}( )] c2 r2Ω2 (Ωr)2 cΩr sinϕ χ π − ϕ 2 − 2 ϕ θ= = arctan 1 (1 + cos ) sin (15) 2 c2 − (Ωr)2 c2 c2 − (Ωr)2 c2 − (Ωr)2

π π π χ π ϕ − θ Now from Eqn.(15) one can observe that, θ= (observer on imuthal position = 2 or 2 and polar position = 2 . 2 ϕ π χ the equator) depends on the azimuthal position (ϕ, equiva- Now for the position = 2 the value of is +ve and for ϕ − π χ lent to earth’s longitude) of the observer. In case ϕ = 0 there = 2 is -ve. These two positions represent the pro- will be no change in the position of polarization vector (as grade and retrograde cases respectively. The numerical value ϕ π − π of χ in both cases are same, but they are opposite in sign (i.e. sin0 = 0) and the rotation is maximum when = 2 or 2 . χθ ϕ π = −χθ π ϕ − π ). So the maximum change in the So the maximum change in the direction of polarization vec- = = 2 = 2 , = 2 tor could be observed when the rotating observer is at az- direction of polarization vector is given by;

[√{ ( )}( )] c2 r2Ω2 (Ωr)2 cΩr χ π − − θ=ϕ= = arctan 1 (16) 2 c2 − (Ωr)2 c2 c2 − (Ωr)2 c2 − (Ωr)2

−5 The rate of rotation of earth, Ωearth is 7.27 × 10 rad/sec by an observer on the equatorial position can be calculated 5 and the average radius of earth is rearth is 63.78 × 10 m. as χ θ ϕ π = 0.33 arcsec. Similarly the total change in earth, = = 2 On the equatorial plane the latitude of the earth is 0 ( cor- the direction of polarization vector at longitude 90◦ west is, θ π ϕ π resonding = ) and, for the azimuthal position, = the χ θ π ϕ − π = −0.33 arcsec. Now, let us consider an ex- 2 ◦ 2 earth, = 2 , = 2 longitude is given by 90 . From Eqn.(16) the total change treme case where Ωr ≃ c, the velocity of the light ray. In that in the direction of polarization vector of light ray, received

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TABLE I. change in the direction of polarization vector χ, in arc sec HH ϕ H 0 15 30 45 60 75 90 θ HH 0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 15 0.0000 0.0222 0.0428 0.0606 0.0742 0.0827 0.0856 30 0.0000 0.0428 0.0827 0.1170 0.1433 0.1598 0.1655 45 0.0000 0.0606 0.1170 0.1655 0.2027 0.2260 0.2340 60 0.0000 0.0742 0.1433 0.0202 0.2482 0.2768 0.2866 75 0.0000 0.0827 0.1598 0.2260 0.2768 0.3088 0.3197 90 0.0000 0.0856 0.1655 0.2340 0.2866 0.3197 0.3309

case, Lorentz factor γ becomes undefined, so the metric ele- varies in a similar fashion for both the azimuthal and the po- θ ϕ π ments g00, g01 and g02 also become undefined, as can be seen lar position. It attains a maximum value when = = 2 , from Eqn.(4). Thus the value of χ cannot be calculated. as mentioned before. This value of rotation which is about The total change in the direction of polarization vector, χ 0.33 arc sec, can be measured with a good polarimeter. In as observed by a rotating observer at equator of earth versus principle, by sitting on earth and by making successive mea- ϕ ϕ − π ϕ the azimuthal position ( ) of the observer is shown in Fig.1. surements at dawn ( = 2 ), mid-noon( = 0) and dusk( Again, in the case when ϕ = π and the total change in the ϕ = π ) with respect to the source of light, we can verify our 2 For Review Only2 direction of polarization vector, χ as a function of the polar calculations. The variation of total change in the direction of position of the observer has been shown in the Fig. 2. polarization vector, against azimuthal position (ϕ) and polar From both the figures Fig.1 and Fig.2 one can observe that χ position (θ) is shown, taking earth as an example of rotating frame in Table 1

We have already mentioned in section I that there are recent amount as: works on electromagnetism for an observer sitting at the ori- gin of a rotating frame. The authors Hauck and Mashhoon [16] derived the following transformation equations (their √( ) equation (45)) for the transformation of electric field as ob- β 2 χ = arctan (β 4 − 3β 2 + 1) (19) served by an observer sitting at the origin of a co-ordiante − β 2 4 Ω (1 ) frame rotating around Z or X(3) axis with angular velocity : γ ′ φ ′ φ E(1) = (E1 cos + E2 sin ) ′ ′ (17) E = −E sinφ + E cosφ (2) 1 2 This Eqn.(19) has been obtained from Eqn.(16) simply replac- Ω β where φ = −Ωt+ ’some offset’, as discussed in detail by ing r by c Hauck and Mashhoon [16]. In order to compare their finding One can see from Fig 3 that the result obtained by following with that of ours as expressed in above Eqn.(16), we proceed our procedure gives slightly higher values of rotation for po- as follows. larization vector as compared to the formulations suggested φ − π by Hauck and Mashhoon [16]. In Eqn.(17), we consider = 2 which corresponds to dawn (side of rotaing earth) in our case and then we are able to write But the values are quite close for the two models genrally ly- ing within 30 arc secs. As β value goes to zero (for zero the expression for the amount (χ1) by which polarization vec- tor of light will be rotated according to Hauck and Mashhoon rotation of frame), the rotation of the polarization vector also [16] goes to zero, as expected by the two models. χ =arctan(γ) − arctan(1) 1 ( √ ) The difference in the two values comes from the fact that in 1 − 1 − β 2 (18) Eqn.(17) ( originally taken from Hauck and Mashhoon [16]), = arctan √ the observer is sitting at the origin of the rotating frame. But − β 2 1 + 1 in the present work the observer is sitting somewhere in the rotating frame with non-zero x and y co-ordinate values. A In above we recall γ = √ 1 . But according to our formula- 1−β 2 more direct comparison is not possible within the scope of the tion as expressed in Eqn.(16), we can write the corresponding present work at this stage and this we postpone for future.

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0.33

0.28

0.22 )

0.17

arcsec (

0.11

0.06 For Review Only

0.00

0 10 20 30 40 50 60 70 80 90

(degree)

χ θ π ϕ FIG. 1. Variation of total change in the direction of polarization vector, (in arcsec) at the equator ( = 2 ) vs the azimuthal position, (in degree)

C. A different case: When the rotation axis of the observer is no rotation. Light ray is emitted from a inertial frame given by along the direction of propagation of light frame K(x0,x1,x2,x3). In this case the transformation equa- tion will be given by Rindler [12], in the present Eqn.(2). Now, consider a case when the light is coming parallel to From this equation (2) we can write the line element as: the rotation axis of the observer’s the rotational frame. In 1′ ax ′ ′ ′ ′ this case, the only component of the rotational vector (of the ds2 = (1 + )dx0 − (dx1 )2 − (dx2 )2 − (dx3 )2 (21) frame) exists along x1-axis. To calculate the change in the c2 direction of polarization vector χ, the values of g and g 00 01 From Eqn.(21) it is clear that the metric element g01 is zero, have to be calculated, from Eqn.(12). The value of g01 can be hence with the help of Eqn.(7) one can obtain the displace- calculated from Eqn.(4) which is, ⃗ ment vector ⃗D = √E . Now the two components of displace- g00 2Ω2 ment vector along x2′ and x3′ directions are given as: r · g01 = 2 (0) = 0 (20) c ′ √Ey Dy = g00 (22) So, from Eqn.(12) it is clear that there will be no change in D′ = √Ez the direction of polarization vector if the light ray is coming z g00 parallel to the rotation axis of the frame and we will get, χ = 0. ′ ′ Here Dy and Dz are the displacement vector received by the observer situated on K′ frame. So the position angle of the polarization vector of the received ray would be : III. DISCUSSION ON OBSERVER WHO IS SITTING ON AN UNIFORM RECTILINEAR ACCELERATED FRAME D′ χ′ y Ey χ observer = arctan( ′ ) = arctan( ) = source (23) Dz Ez Now, suppose that the observer’s frame is moving with a 1 χ′ uniform rectilinear acceleration along the positive x direction where observer is the position angle of the polarization vector ′ 0′ 1′ 2′ 3′ and situated at K (x ,x ,x ,x ,) frame. Then there will be as observed by the receiver and χsource is the position angle of

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0.33

0.28

0.22 )

0.17

arcsec (

0.11

0.06 For Review Only

0.00

0 10 20 30 40 50 60 70 80 90

(degree)

FIG. 2. Variation of total change in the direction of polarization vector, χ (in arcsec) versus the polar position, θ (in degree), where ϕ is kept π constant at 2 .

the polarization vector of emitted light ray. χsource is given in From Eqn.(4) the line element for the rotational observer can the earlier as arctan( Ey ). It is clear from Eqn.(22) that there be written as: Ez √ will be no change in the direction of polarization vector for 0 2 0 1 0 2 δ α β this case. ds = g00(dx ) + g01dx dx + g02dx dx + αβ gαβ dx dx (25) The values of g00, g0x, or g01 and gxx or g11 etc. have been calculated earlier Eqn.(4), for the co-ordinate system (x0,x1,x2,x3) = (ct,x,y,z). Now for our convenience in the present geometry, we can rewrite the metric tensors values in terms of (ct,r,θ,ϕ), which are as follows : IV. REDSHIFT CAUSED BY THE ROTATING FRAME ( ) 2 Ω2γ2 g = 1 [ − Ω (r + r θ ϕ)2 − (r θ ϕ)2] 00 − Ωr 2 1 c2 sin cos c2 sin sin ( 1 ( c ) ) Ω θ ϕ To calculate the redshift when the light is emitted by a dis- g = 1 r sin sin 01 −( Ωr )2 c tant inertial source and observed by a rotational ( non-inertial) 1( c ) (26) observer, the four-velocity of the observer has to be deter- − 1 Ωr sinθ cosϕ g02 = − Ωr 2 c mined, and it is obtained as [13, page no. 23]: 1 ( c ) g03 = 0 gαβ = −δαβ . dxi ui = (24) Thus we may write: ds

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For Review Only

FIG. 3. The figure shows the variation of the amount of rotation (of polarization vector of light) according to the present work (χ) and model by Hauck and Mashhoon [16] χ1

( ) 1 Ω2 Ω2γ2 ds2 = [1 − (r + r sinθ cosϕ)2 − (r sinθ sinϕ)2] (cdt)2 1 − ( Ωr )2 c2 c2 ( c ) ( ) Ωr sinθ sinϕ 1 Ωr sinθ cosϕ 1 − r sinθ sinϕdϕdt + r sinθ cosϕdϕdt (27) − Ωr 2 c − Ωr 2 c 1 ( c ) 1 ( c )

For any calculation of redshift introduced by space-time, we be expanded in series to terms of the first order and it is writ- must begin with the invariant. ten as:

i i ∂Ψ ∂Ψ [kiu ]source = [kiu ]observer (28) Ψ = Ψ + r +t (31) 0 ∂r ∂t Now, the components of four velocity can be written as: From this, one can have[13]

0 dx0 cdt ∂Ψ u = ds = ds − 1 ki = (32) u1 = dx ∂xi ds (29) 2 dx2 u = ds Again it can be shown that[13], 3 dx3 u = i ds k ki = 0 (33) Now let Γ be any arbitrary quantity which describes the wave Γ Eqn.(33) is the fundamental equation of geometrical optics field. For a plane monochromatic wave can be written as and called the eikonal equation. Using Eqn.(32) the compo- [13] [page no. 140]: nents of the wave four-vector, ki are given as: i α Γ = aej(kix )+ (30) k = − ∂Ψ = − ∂Ψ = ν 0 ∂x0 ∂ct c √ − ∂Ψ k1 = ∂xr Where j= −1, ki is wave four vector. One can write the ex- ∂Ψ (34) jΨ k = − pression for the field as, Γ = ae . Ψ is the eikonal. Over 2 ∂xθ Ψ k = − ∂Ψ small region of space and time intervals, the eikonal, can 3 ∂xϕ

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ν τ 0 ϕ is the frequency measured in terms of , 0 and Now for the distance source usource = 1 and usource = 0, there- is defined as[13][page no. 268] ν = − ∂Ψ . Now, one must fore ∂τ0 note that frequency ν expressed in terms of proper time τ0, is different at different point of space. ν′ is the frequency as measured by the rotating observer. In the present case of inter- est where the light ray is coming from distance inertial frame and observed by rotating ( non-inertial) observer situated at [ ( )] ϕ the surface of radius of rotating frame, we can write from the [ ] kϕ u k u0 = k u0 1 + (37) Eqn.(33) (please also see [28]): 0 source 0 0 k0 u observer ( ) ϕ i kϕ u [kiu ]source = [k0u0 1 + 0 ]source (35) k0 u and ( ) dϕ Ω ϕ Now, for the observer dt = , ( rotational velocity of ob- i kϕ u 0 [kiu ]observer = [k0u0 1 + 0 ]observer (36) server’s frame) and [u ]observer is given by(from Eqn.(29)) k0 u

0 For Review Only1 [u ]observer = √ (38) [ − Ω2r2 ( + θ ϕ) − Ω2r2 θ ϕ] + Ω2r2 2 θ 1 c2 1 sin cos c2 sin sin 2 c2 sin

ϕ uobserver Ω Noting that 0 = c from Eqn.(37) we can have: uobserver   kϕ 1 + Ω ν′ √ k0 c  = ν (39) [ − Ω2r2 ( + θ ϕ) − Ω2r2 θ ϕ] + Ω2r2 2 θ 1 c2 1 sin cos c2 sin sin 2 c2 sin

At the location of the observer the observed frequency is de- Again, in case of photon the energy E and linear momentum p fined by ν′. To calculate the redshift, one must calculate the are expressed as, E = pc considering the rest mass of photon kϕ is zero. The angular momentum about the rotation axis of value of [ ]observer. For a rotating frame like earth, the rel- k0 θ ϕ ativistic action, S for a particle can be expressed as [13, 28]: observer can be expressed as [28] L = pr sin sin . So the kϕ = − + ϕ + + θ value of is r sinϕ sinθ. So from Eqn.(39) it can be written S Et L Sr S , where E is the conserved energy and k0 L is the components of angular momentum around the axis of as: = − ∂S symmetry. Now the four momentum is pi ∂xi and for the propagation of photon we can replace action S by the eikonal, Ψ . Finally, we can write

k = − ∂Ψ = E 0 ∂x0 c ∂Ψ (40) kϕ = − ∂ϕ = −L

  1 + r sinθ sinϕ Ω ν′ √ c  = ν (41) [ − Ω2r2 ( + θ ϕ) − Ω2r2 θ ϕ] + Ω2r2 2 θ 1 c2 1 sin cos c2 sin sin 2 c2 sin

ν′ The redshift, Z is generally defined by the relation. Z = ν − 1. So the redshift as estimated by a rotating observer will be given by:

https://mc06.manuscriptcentral.com/cjp-pubs Canadian Journal of Physics Page 10 of 11 10 √ ′ − Ω2r2 θ ϕ − Ω2r2 θ ϕ Ω2r2 2 θ ν [1 2 (1 + sin cos ) 2 sin sin ] + 2 2 sin Z = − 1 = c c c − 1 (42) ν θ ϕ Ω 1 + r sin sin c

ν′ ν′ β v Now when ν < 1, then we call it redshift and when ν > 1, where = c , v is the translational velocity of the observer we call it blueshift. From Eqn.(42) one can observe that for parallel to x1-axis. Here as the observer is sitting on the rotat- equatorial plane there will be no shift of wavelength when ing frame with radius r, the polar and azimuthal positions by θ the azimuthal position (ϕ) is zero, but the shift will be ex- and ϕ respectively, so the the velocity will be γ2Ωr sinθ cosϕ. ϕ π ϕ − π tremum at = 2 (dusk side) and = 2 (dawn side). We note that γ is the Lorentz factor and Ω is the angular ve- −5 We know for earth Ωearth is 7.27 × 10 rad/sec and the locity of the rotating frame. So the β can be expressed by × 5 β γ2 Ωr θ ϕ average radius of earth ( rearth) is 63.78 10 m. There- the relation = c sin cos . The redshift related to the fore, the redshift values under different conditions will be : relativistic Doppler effect is: − π π ≃ − × 6 Zearth,ϕ= ,θ= 1.604429 10 , i.e. redshift occurs here 2 2 π λ ′ ϕ − Dop 1 at dusk. Again for the azimuthal position = 2 (i.e. at − − ZDop = 1 = Ωr 1 (44) dawn) on the equatorial plane of earth, the redshift will be λDop γ(1 − γ2 sinθ cosϕ) −6 c Z ϕ − π θ π ≃ 1.604436×10 which shows a blueshift. earth, = 2 , = 2 Now if we vary the polar position of the observer then at both A simple mathematical analysis will show that, the results ob- the poles there will be no shift of the wavelength of the re- tained in Eqn.(41) and Eqn.(43) are essentially the same. ceived light. Again from Eqn.(42)For one can findReview that for both OnlyThe Eqn.(41) was obtained through a rigorous procedure in- cases when either θ or ϕ becomes zero, the value of Z also volving metric for non-inertial accelerated (rotational) frame. becomes zero. From our above discussion it is clear that the These results for the relativistic redshift are same as the redshift attains extremum value when θ = π and ϕ = − π Doppler redshift derived considering the rotational frame. π 2 2 or 2 . Moreover it can be said that redshift is maximum at Thus the redshift effect for the rotational frame is equivalent ϕ π θ π ϕ − π θ π = 2 , = 2 and blushift is maximum at = 2 , = 2 . to the relativistic redshift caused by the translational motion Let us consider a light ray consisting of Lyman − α line of the observer. 15 which has frequency, νLy−α = 2.47 × 10 Hz (λLy−α = ◦ 1215.673123A) and is emitted from a source placed at some ϕ − π V. CONCLUSION distance inertial frame. Now, at dawn ( = 2 ) the re- −α ν′ × 15 ceived frequency of Ly will be − π = 2.470004 10 Hz 2 ◦ In this paper, we studied the rotation of the polarization λ ′ ϕ π ( − π = 1215.671178A), again at dusk ( = ) the received vector of light, coming from a distant inertial source and 2 2 ′ 15 ′ frequency of Ly − α will be ν π = 2.46999 × 10 Hz (λ π = as viewed by a non-inertial observer sitting on a rotating 2 2 ◦ frame. The amount of rotation primarily depends on the four 1215.675068A). But there will be no shift at mid-noon (ϕ = variables, i.e. (r, θ, ϕ) coordinate of the rotating observer, 0). Thus we see the difference in the third place after decimal and the rate of rotation Ω, of the coordinate frame. The when wavelength is expressed in Angstroms. A present day maximum rotation of the polarization vector can be observed, high resolution astronomical spectrometer can resolve spec- if the position of the observer is on the equatorial plane and tral lines with milli-Angstroms resolutions. So in principle  π the azimuthal position is 2 on the rotating frame. As an one can verify the predictions of our analysis on the redshift extension of the present work, the earth has been taken as calculations . an example of a rotating frame, where the observer is sitting on the surface of the earth and the maximum change in the direction of polarization vector, has been found to occurs at latitude 0◦ (θ = 90◦), and longitude 90◦. This maximum A. Relativistic Doppler Shift for rotating frame value is χ θ ϕ π = 0.33 arc-sec. Again if we consider earth, = = 2 that the rotating axis of the frame is along the direction of As in the case for rotating observer, it is clear that the ob- propagation of light, then there will be no change in the server is either moving towards the source or away from the direction of the polarization vector. source, so this causes a relativistic Doppler effect. In this present work let us consider the observer is on the rotating In the present work, we also report the amount of the red- λ ′ frame which receives a light ray with wavelength Dop from shift as observed by an observer sitting on a rotating frame. It a distance source which initially emits a light ray with wave- has been found that there will be a redshift in dusk (prograde) λ length Dop. The two wavelengths are related by the relation case and a blueshift for the dawn (retrograde) case (i.e. max- in case of relativistic Doppler effect [29]: ϕ π imum redshift at azimuthal position = 2 on the equatorial plane and maximum blueshift at azimuthal position ϕ = − π ). λ 2 Dop γ − β We calculate the red shifted and blue shifted values of wave- λ ′ = (1 ) (43) Dop length taking Lyman-alpha line as a test case. It is also shown

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that results from our calculations considering metric tensors Silchar, India for useful discussions and suggestions. T. for an accelerated (rotational) frame, come out to be same as Ghosh is also thankful to Dr. Anuj Kumar Dubey, and Dr. the one which can be obtained considering the expression for Amritaksha Kar Department of Physics and Dr. Hirak Chat- relativistic Doppler shift. terjee Department of Chemistry, Assam University, for pro- From the above discussion, it is clear that both the changes viding help and support in programming and making plots. in the direction of polarization vector (χ) and redshift (Z) are We are thankful to anonymous referee of this paper, for his/her ϕ π maximum at the azimuthal position, = 2 and polar position, valuable comments which we believe, have improved the θ π χ = 2 . Both and Z become zero when either one of the quality of our work. coordinates ϕ and θ becomes zero or 180◦. Again one must χ π notice that the value of and Z in the azimuthal positions 2 − π and 2 have the same modulas value but with opposite sign.

ACKNOWLEDGMENTS

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