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3530 J. Opt. Soc. Am. A/Vol. 24, No. 11 /November 2007 Neil Ashby

Relativistic effects in earth-orbiting Doppler lidarreturn signals

Neil Ashby1,2,*1Department of Physics, University of Colorado, Boulder, Colorado 80309-0390, USA

2Present address: Time and Frequency Division, Mail Stop 847, National Institute of Standards and Technology,Boulder, Colorado 80302, USA

*Corresponding author: [email protected]

Received February 2, 2007; revised June 27, 2007; accepted July 11, 2007;posted August 20, 2007 (Doc. ID 79581); published October 18, 2007

Frequency shifts of side-ranging lidar signals are calculated to high order in the small quantities �v /c�, wherev is the velocity of a spacecraft carrying a lidar laser or of an aerosol particle that scatters the radiation backinto a detector (c is the speed of light). Frequency shift measurements determine horizontal components ofground velocity of the scattering particle, but measured fractional frequency shifts are large because of thelarge velocities of the spacecraft and of the rotating earth. Subtractions of large terms cause a loss of signifi-cant digits and magnify the effect of relativistic corrections in determination of wind velocity. Spacecraft ac-celeration is also considered. Calculations are performed in an earth-centered inertial frame, and appropriatetransformations are applied giving the velocities of scatterers relative to the ground. © 2007 Optical Society ofAmerica

OCIS codes: 010.1100, 010.3640, 350.5720.

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. INTRODUCTIONide-ranging laser imaging detection and ranging (lidar)easurements of wind from space will probably use a la-

er transmitter in a spacecraft to illuminate aerosol par-icles or molecules that scatter the radiation almost ex-ctly back toward the transmitter. The measuredrequency shift can be processed to yield the velocity ofhe scattering particles relative to the ground. This paperiscusses relativistic corrections to the determination ofhe wind velocity, arising from large spacecraft velocities,arth rotation, and acceleration of the spacecraft towardhe earth. Three different reference frames need to be dis-ussed in this connection: these are the instantaneousest frame of the lidar apparatus, called the lidar restrame (LRF); the earth-centered locally inertial frame,alled the earth-centered inertial (ECI) frame; and a ref-rence frame at rest relative to the surface of the earth athe point where the wind is measured, called the eartheference frame (ERF).

We assume that the position, velocity, and accelerationf the spacecraft are known in the ECI frame because iteems most likely that the spacecraft would be tracked,nd its orbit determined, in such a frame. To simplify theotation, we shall denote quantities measured in the ECIrame without primes. We shall regard the ECI frame, aocally inertial frame whose origin falls along with thearth’s center, as the fundamental frame of reference. Fig-re 1 illustrates the ECI coordinates that have spatialxes in fixed directions in space and origin at earth’senter.

Figure 2 illustrates the LRF, with origin at the lidarransmitter and axes fixed in space when observed fromhe ECI frame. Quantities measured in the LRF will beoubly primed.

1084-7529/07/113530-17/$15.00 © 2

Since the object of the measurement is to determine theind velocity relative to the ground, a reference frameith origin at the ground point below the scatterer iseeded; this point is moving with velocity VE relative tohe ECI frame; its axes do not rotate. This frame is calledhe ERF; quantities measured with respect to the ERFill be denoted with primes. Thus the wind velocity wille denoted by w�, w�, or w in the LRF, ERF, or ECIrame, respectively. A diagram of the ERF is shown in Fig.. Table 1 summarizes the differences among the variouseference systems.

We shall discuss transformations that enable one to de-ermine relations among the radiation paths in the vari-us reference frames. The LRF is the frame in which theeasurement is made, while the ERF is of interest since

he wind velocity in the ERF is the desired goal of theeasurement. We assume that in the ECI frame, light

ravels with speed c in all directions; gravitational slow-ng of the signal pulses amounts to no more than a fewicoseconds and is neglected. No atmospheric refractionffects are included; the derivations given here focusurely on relativistic effects.The outline of the paper is as follows. In Section 2 we

erive the fractional frequency shift observed in the LRF,ssuming the velocity of the apparatus is constant duringhe propagation time of the radiation. First-order relativ-stic corrections are discussed. Section 3 develops Lorentzransformations in vector form, needed in Section 4 to de-ive higher-order relativistic corrections. We use Lorentzransformations in vector form as a compact way of ex-ressing relativistic transformations among the variouseference frames [1–3]. This is useful when the relativeelocities of the reference frames are in arbitrary direc-ions. The vector notation is explained in Section 3. Nu-

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Neil Ashby Vol. 24, No. 11 /November 2007 /J. Opt. Soc. Am. A 3531

erical simulations of measurement sequences and mea-urement processing with relativistic corrections toxtract the horizontal component of the wind are dis-ussed in Section 5. Effects due to spacecraft accelerationre derived in Section 6 from the point of view of the ECIrame, and these and similar higher-order relativistic cor-ections are discussed in Section 7. In Section 8 an alter-ative derivation of the frequency shift is given neglectinghe relative rotation of the ERF and LRF axes. Relativis-ic corrections of order �v /c� and �v /c�2 are discussed.ere v�7000 m s−1, so v /c�3�10−5.Notation. The existence of three or more reference

rames in this problem (there is also an instantaneousest frame for the wind, but such a frame is not used in

ig. 1. Earth-centered inertial coordinates. The origin is atarth’s center, and the xyz axes have fixed directions in space.or example, the x axis can point toward the vernal equinox withhe z axis parallel to earth’s angular velocity vector. The posi-ions of the scattering particle and the spacecraft are respectivelyenoted by r and rL in these coordinates.

ig. 2. Lidar rest frame. This is defined with origin at the lidarransmitter. Quantities measured in this frame such as the windelocity w� and the velocity VE� of the ground point are doublyrimed.

Table 1. Description

eference Frame Origin Axis

CI Earth’s center FixeRF Lidar transmitter FixeRF Ground point FixeCEF Earth’s center Fixe

his paper) results in a notation that is cumbersome inome respects. We denote the velocity of the ground pointelative to the ECI (see Figs. 1 and 2) by VE, and the ve-ocity of the lidar relative to the ECI by VL. We assumehat in a single measurement the radiation is incident onhe aerosol particle for such a short time that w� is con-tant. However, during the few milliseconds (ms) ofropagation time from the lidar to the scatterer and backo the detector, the velocity of the detector relative tohe ECI may have changed. This is fully discussed inection 6.In the calculation given in the following section, we re-

ard the lidar apparatus as being at rest. The aerosol par-icle is assumed to have velocity w� relative to the lidarpparatus.

. FREQUENCY SHIFT IN LIDAR RESTRAME

n Figure 4, the scattering particle is represented as anrregular object that moves with velocity w� in the x� di-ection. The wave vector k� of the incident wave has x�omponent k� cos ��, so that w�k� cos ��=w� ·k�. Twoavefronts for the incident wave are shown, the first la-eled #1 incident that strikes the particle at some initialime t�, and another one labeled #2 incident, a distance

0�=2� /k� away, that will hit the moving scatterer a timet� later. Figure 5 shows first scattered wavefront #1 re-ected as though it were plane, at time �t� later. Duringhe time �t� the scatterer moves a distance w��t� in theositive x� direction. At this instant the second wavefrontmpinges on the particle; this event would generate a sec-nd backscattered wavefront. Meanwhile the first back-

ig. 3. Earth reference frame. This is defined with origin at theround point below the scatterer but fixed on earth’s surface. Theind velocity relative to the surface is the object of the lidar mea-

urement and is denoted by w�. Quantities such as the velocity

L� of the spacecraft in this frame are denoted with primes.

rames of Reference

ions Velocity in ECI Rotation

ace 0 Noneace VL Noneace VE Nonerth 0 Rotating with earth

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cattered wave propagates exactly back toward the trans-itter a distance c�t�. The two backscattered wavefronts

re separated by the wavelength ��.The component of the distance moved by the scattering

article in the direction of the incident ray during timet� is w� cos ��t�. The distance between the two incidentaves is the wavelength �0� of the incident wave, and fromigure 5 it is seen that

�0� = c�t� − w� cos ��t�. �1�

or the ray scattered back toward the detector, the wave-ength is the distance between reflected wave #1 and in-ident wave #2. Denoting the wavelength of the scatteredhotons by ��, we have

�� = c�t� + w� cos ��t�. �2�

ombining the two equations for the incident and scat-ered wavelength,

�0�

��=

c/f 0�

c/f �=

c�t� − w��t� cos � �

c�t� + w��t� cos � �. �3�

herefore the ratio of the backscattered frequency to thencident frequency is

ig. 4. Two wavefronts are shown traveling toward the scatter-ng particle with speed c relative to the LRF. The particle has ve-ocity w�. The incident wavefronts are separated in space by theavelength �0�. The figure is drawn in the LRF at the instant thatavefront #1 strikes the particle; the scatterer is sufficiently far

rom the transmitter that the arriving wavefronts are plane.

ig. 5. The first wavefront generates a spherical backscatteredave, shown at the instant the second wavefront strikes the scat-

erer. The backscattered wave becomes plane as it travels towardhe detector. The scatterer moves with velocity w�. Successiveavefronts striking the scatterer generate a series of backscat-

ered waves separated in space by wavelength � .
�
f �

f 0�=

1 −w�

ccos ��

1 +w�

ccos ��

=

1 −w� · k�

ck�

1 +w� · k�

ck�

. �4�

his result applies when the lidar velocity is constant; apecial case has been discussed by Gudimetla and Kavaya4]. Introducing the measured fractional frequency shiftf � / f 0� by means of

f �

f 0�= 1 +

�f �

f 0�, �5�

e may solve for the quantity involving the wind velocity� to obtain

w� · k�

ck�= −

1

2

�f �

f 0�

1 +1

2

�f �

f 0�

. �6�

hus on the right-hand side of the above equation, onlyhe measured fractional frequency shift is needed in ordero determine the wind velocity. If the measurements cane processed in such a way that only the fractional fre-uency shift is involved, then no error in the wind velocityill occur resulting from a lack of knowledge of the actual

ransmitter frequency.If we denote the unit vector pointing from transmitter

o scatterer by nt�=k� /k�, this result can be written in theorm

w� · nt� =

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

. �7�

he term in the denominator is a relativistic correctionince it is c−1 times the numerator. Equation (7) could bexpanded to second order in the fractional frequencyhift, giving

w� · nt� = −c

2��f �

f 0�� + c��f �

2f 0��2

+ ¯ �8�

or a spacecraft velocity of 7 km/s, the correction repre-ented by the second term in Eq. (8) can contribute morehan 0.15 m/s to the wind velocity.

The right-hand side of Eq. (7) is determined by theeasurement itself. Thus the component of the wind ve-

ocity in the direction of the beam is determined in theRF. The goal, however, is to obtain the velocity relative

o the ground, that is, to find w� in the ERF. Simple Gal-lean transformations lead to appreciable error; insteade need the Lorentz velocity transformations.Equation (7) gives the measured fractional frequency

hift in the LRF, and an observable quantity having theractional frequency shift as a relativistic correction inhe denominator has been derived. Since the quantity ofnterest is the wind or aerosol velocity in the earth-fixed

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rame, the results must be transformed into the ERF. Ad-itional relativistic effects enter when these transforma-ions are applied.

. LORENTZ TRANSFORMATIONS INECTOR FORMe now prepare to derive a general expression for theind velocity, not restricted to leading-order relativistic

orrections. In this section we introduce the vector form ofhe Lorentz transformations, which is a compact way ofriting the transformations. This form of the Lorentz

ransformations can be found in many textbooks [1–3].A general Lorentz transformation of position can be

eparated into a part parallel to the relative velocity,hich suffers Lorentz contraction, and a part perpendicu-

ar to the relative velocity, which is unchanged. Since co-rdinate differences are of interest in the derivations thatollow, constant translations of origins are not significant.

e shall discuss the transformation from ECI to ERF co-rdinates with parallel axes. Let VE be the velocity of theround point in the ERF on earth’s surface, relative to theCI frame. We assume this will be known accurately in

erms of the position of the ground point and the time. Letbe a position vector in the ECI frame, and denote posi-

ion vectors and time measurements in the ERF withrimes. Then the component of r� parallel to VE is

VEVE · r�

VE2 , �9�

nd this part of the vector suffers Lorentz contraction, ex-ressed in the transformation equations by multiplyingy the factor

�E =1

�1 − VE2 /c2

. �10�

he component of r� perpendicular to the velocity VE is

r� −VEVE · r�

VE2 . �11�

hen the Lorentz transformations in vector form are

r = r� −VEVE · r�

VE2 + �E�VEVE · r�

VE2 +

VE

cct�� , �12�

ct = �E�ct� +VE · r�

c � , �13�

ince only the component of r� parallel to VE suffers Lor-ntz contraction, and only the component of r� parallel tohe velocity enters the time transformation. Dotting VEnto Eq. (12),

VE · r = �E�VE · r� +VE

2

cct�� . �14�

he inverses of these transformation equations are

r� = r + ��E − 1�VEVE · r

VE2 − �E

VE

cct,

ct� = �E�ct −VE · r

c � ,

VE · r� = �E�VE · r −VE

2

cct� . �15�

We shall also need velocity transformations, obtainedy taking differentials of both sides of the position andime transformations. Thus, for example, from Eqs. (15),

dr� = dr + ��E − 1�VEVE · dr

VE2 − �E

VE

ccdt; �16�

cdt� = �E�cdt −VE · dr

c � , �17�

nd taking the ratio of corresponding sides of the abovewo equations, we obtain the vector form of the velocityransformation equations for v� in terms of v and VE:

dr�

dt�= v� =

dr + ��E − 1�VEVE · dr

VE2 − �EVEdt

�E�dt −VE · dr

c2 �

=

�1 − VE2 /c2v + �1 − �1 − VE

2 /c2�VEVE · v

VE2 − VE

1 −VE · v

c2

,

�18�

here the velocity of an object measured in the ECI frames v=dr /dt. We also need the inverse of Eq. (18):

v =

�1 − VE2 /c2v� + �1 − �1 − VE

2 /c2�VEVE · v�

VE2 + VE

1 +VE · v�

c2

,

�19�

iving the velocity v in the ECI frame in terms of the ve-ocity v� in the ERF.

. CONSTANT SPACECRAFT VELOCITYn this section we give a relativistic treatment of the lidareasurement, assuming that the spacecraft velocity doesot change during the time interval between transmis-ion and detection of the signal. We assume for simplicityhat the scattering event occurs at a fixed position rs� andime ts� in the LRF; that is, that no significant time delayccurs during the interaction between pulse and scat-erer.

Also, we assume that the detector is effectively at theame point as the lidar transmitter in the LRF, so thathe path of the reflected signal lies exactly on top of theransmitted signal path. Then if d� is the displacement
t

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rom transmitter to scatterer and dd� is the displacementrom scattering particle to detector, we have

dt� + dd� = 0. �20�

he task is to transform Eq. (7) into the ERF, to obtain aseful expression for the wind velocity w�. The process

nvolves transforming first from LRF to ECI and thenrom ECI to ERF. When two parallel-axis Lorentz trans-ormations in different directions are compounded, thexes of the final frame are no longer necessarily parallelo the axes of the initial frame. This issue is discussed inection 9.Suppose that we specify the line of sight (LOS) in the

CI. This is an important case because the spacecraft or-it is most likely to be specified in the ECI, and theround point position and velocity will be known or can bealculated if the ECI time of the lidar shot is known.

Our starting point is therefore Eq. (7). The vector ntrom the lidar transmitter to the aerosol particle is pre-umed known in the ECI system. To obtain an expressionor nt� in terms of nt, use Eq. (15) with VL in place of VEnd r� in place of r�. Then, for the vector displacement be-ween transmitting and scattering events, we have

dt� = rs� − rt� = rs − rt + ��L − 1�VLVL · �rs − rt�

VL2

− �L

VL

c�cts − ctt�, �21�

here

�L =1

�1 −VL

2

c2

. �22�

e define the vector from transmitter to scatter by

rs − rt = dt. �23�

hen by the constancy of the speed of light, �cts−ctt�=dt,nd we have

dt� = dt + ��L − 1�VLVL · dt

VL2 − �L

VL

cdt. �24�

he magnitude dt� of the displacement vector is obtainedy taking the dot product of dt� with itself. Since there arehree terms in Eq. (24), there are basically six terms inhe square:

�dt��2 = dt� · dt� = dt

2 + 2��L − 1��VL · dt�2

VL2 − 2�L

VL · dt

cdt + ��L

2

− 2�L + 1�VL

2�VL · dt�2

VL4 − 2��L

2 − �L�VL

2VL · dt

cdt

+ �L2

VL2

c2 dt2. �25�

here is considerable cancellation. Then using the identi-ies

�L2 − 1

VL2 =

�L2

c2 , 1 + �L2

VL2

c2 = �L2 , �26�

he expression becomes a perfect square, giving the result

dt� = �Ldt�1 −VL · dt

cdt� . �27�

hen we may form an expression for the unit vector nt� inhe LRF in terms of quantities observed or measured inhe ECI:

nt� =dt�

dt�=

nt + ��L − 1�VL�VL · nt�

VL2 − �L

VL

c

�L�1 −VL · nt

c � , �28�

here nt=dt /dt. Also, it then follows that

VL · nt� =VL · nt − VL

2/c

1 −VL · nt

c

. �29�

n the following derivation, the left-hand side of Eq. (28)ill be used as an abbreviation for the right-hand side, inhich the quantities are expressed in the ECI frame.We want to express w� in terms of w� in the ERF. We

hall do this in two stages, first transforming with a par-llel axis transformation to the ECI frame, and then by aarallel axis transformation to the ERF. The velocityransformation to the ECI is

w� =

w + ��L − 1�VLVL · w

VL2 − �LVL

�L�1 −VL · w

c2 � . �30�

herefore, dotting nt� into this expression, we have fromq. (7)

w · nt� + ��L − 1�VL · nt�VL · w

VL2 − �LVL · nt�

�L�1 −VL · w

c2 � =

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

.

�31�

We manipulate this equation so that all terms involv-ng w are isolated as linear terms on the left side. As arst step,

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w · �nt� + ��L − 1�VLVL · nt�

VL2

�L�

= VL · nt� + �1 −VL · w

c2 � −c

2

�f �

f 0�

1 +1

2

�f �

f 0�

, �32�

nd then moving the term on the right that has w in it tohe left,

w · �nt� + ��L − 1�VLVL · nt�

VL2

�L+

VL

c2

−c

2

�f �

f 0�

1 +1

2

�f �

f 0��

= VL · nt� +

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

. �33�

o save writing, we introduce the following abbreviation:

A = nt�/�L + �1 − 1/�L�VLVL · nt�

VL2 +

VL

c2

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

. �34�

hen we obtain

w · A = VL · nt� +

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

. �35�

n the right side of Eq. (35), the quantities are presumednown from the lidar measurement or from computationserformed in the ECI frame. Also, on the left, the vector Aay be presumed known. A is almost but not quite a unit

ector. Also, although A appears to depend on doublyrimed quantities, the appearance of nt� is to be regardeds shorthand for the substitution represented in Eq. (28).nserting Eq. (28) into the definition of A, after some sim-lification we obtain the following expression for A:

A =nt

�L2�1 −

VL · nt

c � −VL

c+

VL

c2

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

. �36�

The next step is to extract that part of w that dependsn w�, the desired wind velocity in the ERF. The velocityransformation from Eq. (19) is

w =

w� + ��E − 1�VEVE · w�

VE2 + �EVE

�E�1 +VE · w�

c2 � . �37�

otting A into this equation, and using Eq. (35), we have

w� · A + ��E − 1�w� · VEVE · A

VE2 + �EVE · A

�E�1 +VE · w�

c2 �

= VL · nt� +

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

. �38�

Again, a term in w� occurs in the denominator. Multi-lying by the factor in the denominator that has w� in it,hen rearranging so that all terms involving w� are on theeft, the result is

w� · �A + ��E − 1�VEVE · A

VE2

�E−

VE

c2 −c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+ VL · nt��=

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+ VL · nt� − VE · A. �39�

e introduce an abbreviation for the quantity in squarerackets:

B = �A + ��E − 1�VEVE · A

VE2

�E−

VE

c2 −c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+ VL · nt�� .

�40�

hen the result is

w� · B =

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+ VL · nt� − VE · A. �41�

The main result, Eq. (41), or appropriate approxima-ions to it, should be used in analyzing Doppler lidar data.e neglected spacecraft acceleration in deriving Eq. (41).bviously numerous relativistic corrections enter the re-

ult. As a first check, let us pass to the nonrelativisticimit, in which all terms involving 1/c are neglected. Then

Wu

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nt� = nt, A = nt, B = A = nt.

e also neglect the term �f � / f � in the denominator andse ��=c / f 0�. We obtain

w� · nt = −��

2�f � + �VL − VE� · nt. �42�

hus the theoretical expressions tell us how to remove theelocities VL and VE from the measurement so that theomponent of wind velocity w� along the LOS is deter-ined. [In Eq. (42) the LOS is specified in the ECI frame.]ince the spacecraft velocity has by far the largest mag-itude of all velocities in the problem, such subtractionsill typically involve the loss of two or three digits of ac-

uracy. For example, if the wind velocity w� were zero,here could still be a large measured Doppler shift due tohe large velocity of the spacecraft relative to the ground.hen Eq. (42) shows that most of the observed Dopplerhift will be canceled by the term VL ·nt, and the remain-ng shift will be canceled by the term −VE ·nt. Equation42) could be tested by observing the backscatteredround return signal.

Keeping terms of O�VL /c� only, we may expand Eq.36):

A � nt�1 +VL · nt

c � −VL

c, �43�

B � A. �44�

he result of substituting these approximations into Eq.41) is

w� · �nt −VL − ntnt · VL

c � =

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+ �VL − VE� · �nt

−VL − ntnt · VL

c � . �45�

he additional terms added to nt inside the parenthesesre an effect of relativistic aberration—the change of ap-arent direction of a light ray when a transformation be-ween relatively moving reference frames is introduced.ere, the terms VL−ntnt ·VL are perpendicular to nt.hus the first-order relativistic corrections serve to adderms that slightly change the direction of the LOS. Therere several relativistic corrections in Eq. (45); most ofhem are of comparable orders of magnitude. On the rightide of this equation, there are corrections of orderVL /c�*VL to the main term, which is VL ·nt. A typical cor-ection contributes about 0.2 ms−1 to the right side, buthe net effect depends on the geometry.

Expanding to second order in powers of �VL /c� andV /c� gives
E
w� · �nt�1 +VL · nt

c � −VL

c+

VL − VE

c2

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+nt

c2 ��VL · nt�2 − VL2 − VE

2 /2�� +VE

c2 �VL · nt

+ VE · nt/2�� = �1 +VE · VL

c2 � −c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+ �VE

− VL� · �nt�1 +VL · nt

c � −VL

c+

nt

c2 ��VL · nt�2 − VL2�� .

�46�

he largest second-order terms contribute corrections tohe wind speed of order �VL /c�2*VL�4�10−6 m/s, or 510−10 rad to the direction of the line of sight. Such small

orrections are probably negligible for near-earth applica-ions. In any case, expansion to second or higher orderives expressions that are so cumbersome that it is prob-bly better to use exact expressions in numerical calcula-ions.

Figure 6 shows an example in which there exists a uni-orm horizontal wind field and a horizontal relative veloc-ty VL−VE of the spacecraft relative to the ground. Thecanning angle of the lidar is at 30° relative to the nadir

ig. 6. Possible situation with VL−VE parallel to the ground, atngle � from the uniform wind field. The scanning angle is 30°rom the nadir.

ig. 7. Error in wind velocity resulting from neglect of denomi-ator in last term of Eq. (41). The three curves are labeled by thecanning angle relative to the nadir.

itdva

ctda

5IEatelt=coifi

R

AGtr�f

Tjas

rmmt

wtm

T

Tf

tstr

Tp

wtsufttt

T

(Tf


n the diagram. Omission of the relativistic corrections inhe denominator in Eq. (45) results in systematic errors inetermining the wind speed that are plotted in Fig. 7 forarious scanning angles. In this example VE=300 ms−1,nd the altitude of the lidar was 400 km.To obtain a better idea of the importance of relativistic

orrections, we have developed a numerical simulation ofhe measurement process involving an earth-orbiting li-ar apparatus. This will be explained in the next sectionnd some examples will be discussed.

. SIMULATIONS: EXAMPLESn this section we describe numerical simulations usingq. (41). These examples illustrate how the theory is to bepplied and also indicate how large the relativistic correc-ions can be. The scattering particle is assumed to be onarth’s surface, at the chosen observation point ofatitude/longitude �40.00°N,−105°W�. At this pointhe wind is assumed to be constant, w��50.000 m/s E,47.000 m/s N�. We assume the space-raft is in a Keplerian orbit characterized by the followingrbital parameters, from which the position and velocityn the ECI frame can be derived when the time is speci-ed:

e = 6.3780000 � 106 m, earth radius, �47�

h = 4.0000000 � 105 m, nominal altitude, �48�

a = Re + h = 6.778000 � 106 m, orbit semimajor axis,

�49�

e = 0.00200 eccentricity, �50�

= 1.600 rad, altitude of perigee, �51�

= − 1.309 rad, angle of ascending line of nodes, �52�

i = 60.00 ° , orbit inclination, �53�

tp = 3400.00 s, time of perigee passage. �54�

. First Lidar Measurementiven the time of perigee passage above, and assuming

he Greenwich meridian of earth rotates in the ECIeference frame with angular velocity e=7.2900010−5 rad/s, the azimuthal angle of the ground point as a

unction of time t (in the ECI reference frame) is

��t� = − 105 � �/180 + et. �55�

he factors � /180 in the first term of the above equationust convert the angle to radians. In a similar manner,ny point with given latitude and longitude on earth’surface can be found in the ECI reference frame.

Also, at any given time, with the Keplerian orbital pa-ameters listed above, the ECI position of the lidar trans-itter can be found by solving the Keplerian equations ofotion [5]. First one solves for the eccentric anomaly E in

he equation

E − e sin E =�GM

a3 �t − tp�, �56�

here GM=3.986004415�1014 m3/s2 is the product ofhe Newtonian gravitational constant and the earth’sass. Then the true anomaly f is found from

cos f =cos E − e

1 − e cos E. �57�

he radius of the spacecraft is obtained from

r = a�1 − e cos E�. �58�

he ECI Cartesian coordinates of the spacecraft are thenound from elementary trigonometry:

x = r�cos cos�f + � − cos I sin sin�f + ��,

y = r�sin cos�f + � + cos I cos sin�f + ��,

z = r sin I sin�f + �. �59�

Assume that a pulse from the lidar apparatus starts tohe ground point at t= tt=13750.000 s exactly. This time ispecified in the ECI frame. The position and velocity ofhe lidar at this moment is obtained by solving the Keple-ian equations of motion, and is found to be

rL = 3490685.517,− 4268379.510,3926595.155� m,

�60�

VL = 1415.72415,5709.87278,4928.19131� m/s. �61�

o find the time ts when the pulse arrives at the groundoint, one uses constancy of the speed of light:

cts = ctt +1

c�rs�ts� − rt�tt��, �62�

here rs�ts� and rt�tt� are the positions of scatterer andransmitter at the scattering and transmission events, re-pectively. An estimate of the time of arrival ts is thensed to recompute the ground point position in the ECIrame; then the distance between the ground point andhe transmission event is recalculated, giving a better es-imate of the arrival time. This process converges afterhree or four steps and gives for the time of arrival

ts = 13750.0023732965 s. �63�

he ground point position and velocity at this instant are

rE = 3296533.996,− 3606135.417,4099699.375� m,

�64�

VE = 262.88727,240.31733,0.00000� m/s. �65�

The z axis is chosen parallel to earth’s rotation axis.)hese data allow us to compute the vector dt and to trans-

orm it into the LRF:

dt� = − 194154.8809,662230.5410,173092.5238� m,

�66�

fiapr

Wt(

a

T

Tbpcscs

Tilaat

afi(tfr

Tcms

BLFpwtnmttrsrF

q

a−gt

l

FEttt


dt� = 711481.8543 m. �67�

Using the relativistic composition law for velocities, werst transform the given ERF wind velocities to the ECI,nd then to the ERF. This is so we can calculate the ex-ected fractional frequency shift. The velocity of the windelative to the lidar apparatus will be

w� = − 1095.2220,− 5447.7191,− 4897.9803� m/s.

�68�

ith these simulated values we can compute the frac-ional frequency shift that should be observed from Eq.4):

�f �

f �= 3.9783903 � 10−5, �69�

nd the quantity

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

= c�− 1.9891555694 � 10−5�. �70�

The vector A defined above is

A = − 0.272888029028,0.93077643041,0.24328452306�.

�71�

hen the vector B may be computed:

B = − 0.272888029028,0.93077643041,0.24328452306�.

�72�

hese two vectors turn out to be essentially equal; this isecause the speed of the ground point is very small com-ared with the speed of light. In a first approximation onean in fact neglect the difference between B and A. Re-olving the vector B into eastward, northward, and verti-al components, we quote only the horizontal componentsince the wind is assumed to be horizontal:

B = − 0.50449226E,0.79099629N�. �73�

hese results can now be used to test Eq. (41). Using thenput values of the wind at the ground point gives for theeft-hand side of Eq. (41) 11.952212693 m/s, while evalu-ting the right-hand side gives 11.952212693 m/s. Thegreement is extremely good, which gives one confidencehat the simulation properly represents the theory.

It is of interest to examine the consequences of makingpproximations leading to Eq. (42). The LOS vector is de-ned in the ECI reference frame. The left-hand side of Eq.42) has the nonrelativistic value 11.953153 m/s, whilehe right-hand side evaluates to 11.950209 m/s. The dif-erence is 2.4 parts in 104; there has been a loss of accu-acy from the subtraction of two large terms on the right:

−c

2

�f �

f �+ �VL − VE� · nt = �− 5.963456998 � 103

+ 5.975359093 � 103� m/s.

�74�

hus the lidar Doppler shift measurement is large be-ause the spacecraft velocity is large. Most of the shiftust be subtracted out to account for velocities of the

pacecraft and ground point in order to measure the wind.

. Second Lidar Measurementet a second measurement be made exactly 150 s later.or this measurement the lidar is 54.31° from the groundoint zenith, whereas for the first measurement the lidaras 59.28° from the zenith. (These conditions were arbi-

rarily chosen and do not correspond to a constant scan-ing angle.) The physical configuration of the two lidareasurements is drawn to scale in Fig. 8; there it is seen

hat the first laser shot occurs as the lidar is approachinghe scatterer, and the second shot occurs as the lidar iseceding. During the time interval between the shots, thecatterer moves due to the combination of wind and earthotation. Relativistic effects are too small to show up inig. 8.For the second laser shot, the measured fractional fre-

uency shift is

�f �

f �= − 3.70437752 � 10−5, �75�

nd the result of evaluating the left side of Eq. (41) is60.22015497 m/s, while evaluating the right-hand sideives −60.22015497 m/s. Again, the agreement is ex-remely good.

These two simulated measurements give the wind ve-ocity relative to the ground point. For the first measure-

ig. 8. Vectors for the simulation in Section 5, viewed from theCI reference frame. Latitude–longitude grid lines are drawn at

he time of the initial measurement 0.5° apart. Displacement ofhe ground point between the measurements is due to earth ro-ation and is along a parallel of latitude.

mwg

B

ace

Am

St

wctm

6Wtctsttlww0adtitd

tthdbadt

a

wstt

imbssfsrT

at

T

A

T

a

Trttopt=


ent, the vector B has been given above. Assuming theind has eastward and northward components �we� ,wn��ives us one equation:

− 0.5044922625we� + 0.7909962943wn�

= 11.952212693 m/s. �76�

For the second measurement, evaluation of the vectorgives

B = − 0.4981724053,− 0.33958394090,− 0.797813889�,

�77�

nd resolving this into eastward, northward, and verticalomponents accounting for the additional rotation ofarth,

B = − 0.3933068009E,− 0.8628684348N�. �78�

ssuming that the wind remains constant in between theeasurements gives

− 0.39330680087we� − 0.86286843481wn�

= − 60.22015497 m/s. �79�

olving the two simultaneous equations (76) and (79) forhe wind velocity gives

�we�,wn�� = �50.000000,46.999999� m/s, �80�

hich agrees very well with the assumed wind velocityomponents �we ,wn�= �50,47� m/s. If the wind has a ver-ical component, a third measurement is needed to deter-ine all three wind components.

. ACCELERATED SPACECRAFTe reanalyze the measurement from the point of view of

he ECI frame, but we include acceleration of the space-raft and the effect of earth’s gravitational potential onhe transmitted and detected signal frequencies. Thepacecraft’s acceleration g in earth’s gravity field changeshe spacecraft position and velocity during the light travelime, so the detected frequency will be affected. For aight travel time of 8 ms, the spacecraft velocity changeill be about 0.07 ms−1. If not accounted for the measuredind velocity could be in error by about half this, or.035 ms−1. To treat this effect it is most convenient tonalyze the measurement in the ECI frame, allowing forifferent spacecraft velocities at the transmission and de-ection events. The calculation will be done without mak-ng any approximations in the relativistic transforma-ions; then contributions of various orders will beiscussed.Let tt be the time of transmission of a wavefront from

he lidar apparatus, and suppose it arrives at the scat-erer at time ts. At the instant of transmission the lidaras velocity VL�tt� and position rL�tt�. The pulse is imme-iately scattered back into a detector, which is assumed toe coincident with the transmitter (monostatic lidar). Therrival time of the scattered wavefront at the detector isenoted by td. During the propagation delay time td− tthe detector velocity will change to

VL�td� = VL�tt� + g�td − tt�, �81�

nd the detector position will change to

rL�td� = rL�tt� + VL�tt��td − tt� +1

2g�td − tt�2, �82�

here g is the acceleration due to earth’s gravity at thepacecraft. We let nt and nd be unit vectors pointing alonghe path of the transmitted and scattered rays, respec-ively.

The interaction time of the radiation with the scatterers assumed to be so short that the positions of the trans-

ission event, scattering event, and detection event cane described as happening at rt�tt�, rs�ts�, and rd�td�, re-pectively. We neglect atmospheric refraction or disper-ion effects and confine this discussion to relativistic ef-ects. For the transmitted radiation, we consider twouccessive wavefronts that start at tt and tt+dtt and ar-ive at the scatterer at times ts and ts+dts, respectively.hen from the constancy of the speed of light we have

ts − tt =1

c�rs�ts� − rt�tt��, �83�

nd differentiating to obtain expressions for the differen-ials dts, dtt, we have

dts − dtt =1

c

�rs�ts� − rt�tt�� · �drs�ts� − drt�tt��

�rs�ts� − rt�tt��. �84�

he unit vector along the LOS of the transmitted ray is

nt =rs�ts� − rt�tt�

�rs�ts� − rt�tt��. �85�

lso, the position increments are related to the velocities

drs =drs

dtsdts = wdts, drt = VL�tt�dtt. �86�

hus

dts − dtt = nt · �wdts − VL�tt�dtt�/c, �87�

nd this may be rearranged to give

dts�1 −nt · w

c � = dtt�1 −nt · VL�tt�

c � . �88�

he coordinate time interval dtt can be related to the cor-esponding proper time d�t� on a standard clock at theransmitter, and hence to the proper frequency of theransmitted signal, by using the ordinary scalar invariantf the gravitational field. We shall suppose that theroper time interval d�t� corresponds to the emission ofwo successive wavefronts by the transmitter so that d�t�1/ f 0�. The proper time is given to sufficient accuracy by

�cd�t��2 = �1 +

2��rt�

c2 ��cdtt�2 − �dx2 + dy2 + dz2�, �89�

we

Fi

a

trl

at

T

st

Ttt

a

S

at

Tdttm

cwmcttcr

atteTsctq

Ilwmf

tdat


=�1 +2��rt�

c2 −VL�tt�2

c2 ��cdtt�2, �90�

here ��rt� is the Newtonian gravitational potential ofarth at the transmission event. Thus

d�t� =�1 +2��rt�

c2 −VL�tt�2

c2 dtt. �91�

or the emission of two successive wavefronts, this times the reciprocal of the frequency, and so

dtt =1

f 0��1 +2��rt�

c2 −VL�tt�2

c2

, �92�

nd therefore using Eq. (88)

dts�1 −nt · w

c � =

�1 −nt · VL�tt�

c �f 0��1 +

2��rt�

c2 −VL�tt�2

c2

. �93�

We perform a similar analysis for the successively scat-ered wavefronts coming back to the detector and use theesult to eliminate dts. From constancy of the speed ofight,

td − ts =1

c�rd�td� − rs�ts��, �94�

nd differentiating to obtain expressions for the differen-ials dtd, dts, we have

dtd − dts =1

c

�rd�td� − rs�ts�� · �drd�td� − drs�ts��

�rd�td� − rs�ts��. �95�

he unit vector along the LOS of the scattered ray is

nd =rd�td� − rs�ts�

�rd�td� − rs�ts��, �96�

o inserting velocities of wind and lidar detector, substi-uting in the definition of nd and rearranging, we obtain

dts�1 −nd · w

c � = dtd�1 −nd · VL�td�

c � . �97�

he coordinate time interval dtd at the detector is relatedo the proper time increment at the detector, and hence tohe detected frequency f �, by

d�d� =1

f �=�1 +

2��rd�

c2 −VL�td�2

c2 dtd, �98�

nd therefore

dtd =1

f ��1 +2��rd�

c2 −VL�td�2

c2

. �99�

ubstituting this into Eq. (97),

dts�1 −nd · w

c � =

�1 −nd · VL�td�

c �f ��1 +

2��rd�

c2 −VL�td�2

c2

. �100�

Dividing Eq. (100) into Eq. (93), canceling dts, and re-rranging, we obtain the following expression for the ra-io of detected to transmitted frequency in the LRF:

f �

f 0�=

�1 +��rt�

c2 −VL�tt�2

c2

�1 +��rd�

c2 −VL�td�2

c2

�1 − nt · w/c�

�1 − nd · w/c�

�1 − nd · VL�td�/c�

�1 − nt · VL�tt�/c�.

�101�

his relationship contains all the information available toetermine the wind velocity. It allows for the possibilityhat VL�td� and rL�td� differ from VL�tt� and rL�tt�, respec-ively. The desired wind velocity w� will be introduced byaking a Lorentz transformation.Let us first consider the ratio of square roots that oc-

urs on the right side of Eq. (101). If the spacecraft orbitere perfectly circular, then to a high degree of approxi-ation the spacecraft speed and altitude would not

hange and the square root factors would cancel. Even ifhe orbit is slightly eccentric, the eccentricity e is expectedo be rather small, say e�0.01. Accounting for thehanges due to earth’s gravity as in Eqs. (81) and (82), theatio is approximately

�1 +��rt�

c2 −VL�tt�2

c2

�1 +��rd�

c2 −VL�td�2

c2

� 1 + 2�td − tt�

c2 VL · g, �102�

nd the last dot product will bring in a factor e. Thus thiserm is small not only because the propagation delay in-erval is small but also because there is another factor ofand there are already factors c−2 in the denominator.he net effect is that under almost all circumstances thequare root ratio differs from unity by less than 10−15 andan be neglected. We shall therefore not consider it fur-her in this paper. The expression for the measured fre-uency reduces to

f �

f 0�=

�1 − nt · w/c�

�1 − nd · w/c�

�1 − nd · VL�td�/c�

�1 − nt · VL�tt�/c�. �103�

f Eq. (103) is transformed to the LRF, assuming the ve-ocity VL is constant, the result can be shown to agreeith Eq. (4). Thus Eq. (103) provides an alternative for-ulation, in terms of quantities specified in the ECI

rame, from which the wind velocity may be derived.With Eq. (103) we can estimate the principal contribu-

ion of spacecraft acceleration to the measurement. Intro-uce the fractional frequency shift by means of Eq. (5)nd expand the right-hand side of Eq. (103), keeping onlyerms of first order in �w /c� and �V /c�. Then
L

1

TAE

Ft�E

bamo

(pcnicnat

E

Tbbi

a

Wt(r

W

Soi

I(a

aaair

7REiie+p

Iaptctt

a

I


+�f �

f 0�= 1 − nt · w/c + nd · w/c − nd · VL�td�/c + nt · VL�tt�/c.

�104�

o a very good approximation, nd�−nt, and w=VE+w�.lso, the acceleration causes a change in velocity given byq. (81). With a little rearrangement we obtain

w� · nt = −��

2�f � + �VL − VE� · nt +

1

2nt · g�td − tt�.

�105�

or reasonable distances of �1260 km between transmit-er and target, the propagation delay is about �td− tt�8.4�10−3 s, and g�9 m s−2. So the acceleration term inq. (105) could contribute to the wind speed as much as

1

29 � 8.4 � 10−3 m s−1 = 0.04 m s−1, �106�

ut this could be further reduced if the angle between ntnd g is large. We have already discussed the approxi-ate cancellations of the two main terms on the right side

f Eq. (105).The next step in the calculation is to transform Eq.

103) so that the wind velocity w� in the ERF appears ex-licitly, without making further approximations. The cal-ulation is tedious, and to complete it with a minimumumber of steps we shall use the Lorentz transformations

n vector form. At any stage in the calculation one mayhoose whether to transform the LOS unit vectors nt andd into some other reference frame. We shall use only par-llel axis transformations from the ECI frame to one orhe other of the LRF and ERF.

With this in mind, the starting point is the expression,q. (103). To save writing, we introduce the abbreviation

F =f �

f 0�

�1 − nt · VL�tt�/c�

�1 − nd · VL�td�/c�. �107�

his quantity includes the measured frequency of theackscattered radiation and Doppler shift factors that cane calculated when the velocities and the LOSs are knownn the ECI. Then Eq. (103) becomes

F =�1 − nt · w/c�

�1 − nd · w/c�, �108�

nd solving for the terms involving w, we have

w · �nt − Fnd� = c�1 − F�. �109�

e next introduce the desired wind velocity w� by substi-uting the velocity transformation, Eq. (19), into Eq.109). Then several dot products involving w� appear. Theesult can be written

c�1 − F��E�1 +VE · w�

c2 � = w� · �nt − Fnd� + ��E − 1�

�w� · VEVE · �nt − Fnd�/VE2

+ � V · �n − Fn �. �110�
E E t d
e collect all terms involving w� on the left:

w� · ��nt − Fnd� − �Ec�1 − F�VE

c2 + ��E − 1�VE

VE2 VE · �nt

− Fnd�� = �Ec�1 − F� − �EVE · �nt − Fnd�. �111�

pacecraft Acceleration. Let us again examine the effectf spacecraft acceleration by assuming the wind velocitys very small. The right side of Eq. (111) reduces to

�c�1 −f �

f 0��1 − nt · VL�tt�/c + nd · VL�td�/c�� − VE · �nt − nd�

� + nt · VL�tt� − nd · Vd�td� − VE · nt + VE · nd + . . . �112�

nserting the spacecraft acceleration, from Eqs. (81) and82), there will be a contribution to the wind speed ofbout

nd · g�td − tt�/c � .08 m s−1, �113�

s discussed in connection with Eq. (105). A factor of 1/2rises from the left side of the equation, so the resultgrees with Eq. (105). In the rest of this paper we shallgnore spacecraft acceleration and concentrate on otherelativistic corrections.

. ALTERNATIVE FORMS FORELATIVISTIC CORRECTIONSquation (111) is an alternative form of the result includ-

ng relativistic corrections, with constant spacecraft veloc-ty. This equation can be reduced to Eq. (41) by dividingach term in Eq. (111) by 2�E�L

2�1−VL ·nt /c��1�f � / �2f 0��. For example, the following identities can beroved:

2�L2�1 −

VL · nt

c ��1 +1

2

�f �

f 0��A = nt − Fnd, �114�

2�L2�1 −

VL · nt

c ��1 +1

2

�f �

f 0�� −

c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+ VL · nt�= c�1 − F�. �115�

n Eq. (111), the quantity into which w� is dotted, as wells the quantity on the right side of Eq. (111), are ex-ressed in terms of the measured frequency ratio f � / f 0� inhe LRF and velocities specified in the ECI frame. Variousases of interest may be discussed; for example, relativis-ic corrections may be obtained by expanding Eq. (111) tohe desired order in �w /c�, �VE /c�, or �VL /c�.

Henceforth we shall neglect spacecraft acceleration andssume that

VL�td� = VL�tt�. �116�

n this case

aaqrl

T

I�

w

pE

Ea

EEp

8RIdLdELoV

Ssottmt

w

ifutLh

a

Swm

B

WT

a


dd + dt = VL�td − tt�, �117�

nd using c�td− tt�=dd+dt, one can exactly solve for ddnd dd in terms of dt. The solutions are useful, so weuote them here. First, calculating the square of dd givesise to a quadratic equation for dd. By an argument simi-ar to that leading to Eq. (27), the physical solution is

dd = dt

1 − 2VL · nt/c + VL2/c2

1 − VL2/c2

. �118�

hen putting this back in to Eq. (117) gives

dd = − dt + 2VLdt

c

1 − VL · nt/c

1 − VL2/c2

. �119�

f all first-order corrections are neglected in Eq. (111), F1+�f � / f 0�, and the equation can be reduced to

w� · nt = −c

2

�f �

f 0�+ VL · nt − VE · nt, �120�

hich is equivalent to Eq. (42).To obtain first-order corrections, we first need to ex-

and nd in terms of other quantities. This comes fromqs. (118) and (119) and is

nd =dd

dd� − nt�1 + 2VL · nt/c� + 2VL/c. �121�

xpanding Eq. (111) to leading order in �w� /c�, �VE /c�,nd �VL /c�, the following result is obtained:

�w� + VE − VL� · �nt�1 + VL · nt/c� − VL/c� = −

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

.

�122�

quation (122) agrees with Eq. (45) to this order. Sinceq. (111) has the same physical content as Eq. (41), ex-ansions to higher order will also agree.

. CORRECTIONS NEGLECTINGELATIVISTIC ROTATION OF AXES

n this section we investigate relativistic corrections un-er the assumption that a direct transformation betweenRF and ERF can be made, which means neglecting theifference in orientation between the two sets of axes.quation (18) allows us to introduce the velocity VL� of theRF relative to the point on the ground. Let the velocityf the lidar apparatus as measured in the ECI frame be

. Substituting this for v in Eq. (18) gives
L
VL�

c=

�1 − VE2 /c2

VL

c+ �1 − �1 − VE

2 /c2�VEVE · VL

VE2 c

−VE

c

1 −VE · VL

c2

.

�123�

ince the positions of spacecraft and earth point are pre-umed known, all quantities appearing on the right sidef this equation are known. We may thus proceed toransform the observable quantity, Eq. (7), to the ERF. Inhe following equations, we make direct Lorentz transfor-ations (recall quantities with primes are measured in

he ERF);

r� = r� + ��L� − 1�VL�VL� · r�

VL�2 + �L�

VL�

cct� �124�

ct� = �L��ct� +VL� · r�

c � , �125�

here �L� is a function of VL� ,

�L� =1

�1 − VL�2/c2

. �126�

Transformation to the ERF. We derive an equation sat-sfied by the wind velocity in the ERF. To do this we trans-orm the observable directly from the LRF into the ERF,sing the ECI frame only as an intermediary for calcula-ion of the relative velocity of the two frames. Using theorentz transformations, for the transmission event weave

rt� = rt� + ��L� − 1�VL�VL� · rt�

VL�2 + �L�

VL�

cctt�, �127�

nd for the arrival event of the pulse at the ground point,

rs� = rs� + ��L� − 1�VL�VL� · rs�

VL�2 + �L�

VL�

ccts�. �128�

ubtracting the first of these equations from the second,e have an expression for the vector dt� from the trans-ission event to the arrival event in the ERF:

dt� = rs� − rt� = dt� + ��L� − 1�VL�VL� · dt�

VL�2 + �L�

VL�

c�cts� − ctt��.

�129�

ut cts�−ctt�=dt� in the LRF, so we find for the path vectors

dt� = dt� + ��L� − 1�VL�VL� · dt�

VL�2 + �L�

VL�

cdt�. �130�

e can find the magnitude of dt� by dotting it into itself.he result is

dt� = �L�dt��1 +VL� · dt�

cdt�� , �131�

nd the inverses are

ss

wrm

d

Tateeit

Te

w


dt� = dt� + ��L� − 1�VL�VL� · dt�

VL�2 − �L�

VL�

cdt�, �132�

dt� = �L�dt��1 −VL� · dt�

cdt�� . �133�

We leave the fractional frequency difference as mea-ured in the LRF unchanged but transform the expres-ion on the left side of Eq. (7). Then we need the quantity

w� · dt�

cdt�, �134�

t t t

skglm

Ta

idptai(nrps

here we have double primes on dt� and w� as in Eq. (7) toemind us that these quantities refer to measurementsade in the LRF.The velocity w� of the wind in the ERF is obtained by

irect Lorentz transformation:

w� =w� + ��L� − 1�VL�VL� · w�/VL�

2 − �L�VL�

�L��1 −VL� · w�

c2 � . �135�

We may now form the observable, Eq. (134):

w� · dt�

dt�=

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

=

�w� + ��L� − 1�VL�VL� · w�

VL�2 − �L�VL�� · �dt� + ��L� − 1�

VL�VL� · dt�

VL�2 − �L�

VL�

cdt��

�L�2dt��1 −

VL� · w�

c2 ��1 −VL� · dt�

cdt�� . �136�

he object of the following derivation is to simplify thebove expression as much as possible, isolating all quan-ities involving w� in the numerator of one side of thequation. Note that everything is presumed known in thisquation except w�. Putting some of the factors that aren the denominator of Eq. (136) on the left, working out allhe terms in the dot product and simplifying, we obtain

�L�2

−c

2

�f �

f 0�

1 +1

2

�f �

f

�1 −VL� · w�

c2 �

=w� · dt�

cdt��1 −VL� · dt�

cdt�� − �L�

2w� · VL�

c+ �L�

2

− VL� · dt�

dt�+

VL�2

c

�1 −VL� · dt�

cdt�� .

�137�

he terms involving w� can be put on one side of thequation:

w� · nt�/�L�2

�1 −VL� · nt�

c � −w� · VL�

c2 c −

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

=

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+

VL� · nt� −VL�

2

c

�1 −VL� · nt�

c � , �138�

here n�=d� /d�. The quantities appearing in this expres-

ion should be known, either from measurement or fromnowledge of the spacecraft orbit and ground positioniven the time of the measurement, except the wind ve-ocity w� in the ERF. Let us introduce a vector Dt� by

eans of

Dt� =nt�

�L�2�1 −

VL� · nt�

c � −VL�

c2 c −

−c

2

�f �

f 0�

1 +1

2

�f �

f 0� . �139�

hen the result for the wind velocity in the ERF involvesdot product:

w� · Dt� =

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+

VL� · nt� −VL�

2

c

�1 −VL� · nt�

c � . �140�

The result gives the component of w�, the wind velocity,n the ERF, along Dt�. Dt� is not quite a unit vector. It alsoepends on the measured frequency shift, so it is a com-licated but presumably known object. Relativistic correc-ions of all orders are contained in this result. For ex-mple, there are several relativistic corrections containedn the expression for the relative velocity VL� , given in Eq.123). However, because the direct transformation has ig-ored the nonparallelism of the ERF and LRF axes, theesults cannot be expected to be accurate to O�c−2�. Ex-anding Eq. (140) to first order in �VL� /c�, the result can behown to agree with Eq. (45).

9ETLpttptEttft

Ibs−t

Al�

TTtm

nEL

OLta

Ean([std(ifiawpeaon

1Idfv(EtE

FT


. NONPARALLELISM OF AXES INRF AND LRFhe theory presented in the previous sections treated theRF, the ECI frame, and the ERF as though they werearallel. This is not strictly true, since it is well knownhat when two noncollinear boosts are involved the netransformation involves a boost and a rotation. In theresent case, the velocities VE and VL are generally not inhe same direction. Therefore the axes of the LRF and theRF are generally not parallel. Equation (123) gives VL� ,

he velocity of the LRF relative to the ERF. The velocity ofhe ERF as measured in the LRF would be of the sameorm as Eq. (123) with VE and VL interchanged. We callhis velocity VE� , and find

VE�

c=

�1 − VL2/c2

VE

c+ �1 − �1 − VL

2/c2�VLVL · VE

VL2c

−VL

c

1 −VL · VE

c2

.

�141�

t is easy to see that VL� and VE� have the same magnitude,ut they do not point in opposite directions in the vectorense. If we let be the angle required to rotate VL� intoVE� , and let N̂ be a unit vector along the rotation axis,hen

N̂ sin� � =VL� � �− VE� �

VL�2 . �142�

fter some algebra, and reducing the expression to itsowest-order contribution by expanding in powers ofVL /c� and �VE /c�, we find

N̂ sin� � =VL � �VE�

2c2 . �143�

his rotation angle is typically only a few parts in 1011.his correction is of order �VL /c� or �VE /c� smaller than

he first-order relativistic corrections discussed earlier. Itay be negligible in the current application.Using the transformations developed in this paper, and

eglecting the difference in direction of VE� and VL� , fromq. (140) one may transform directly from the ERF to theRF and expand to second order and obtain

w� · nt� +VL�

c �1 −VL� · nt�

c � −c

2

�f �

f 0�

1 +1

2

�f �

f 0�

−VL�

c �1 −VL� · nt�

c �=

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

�1 +VL�

2

2c2 −VL� · nt�

c �

+ �1 +VL�

2

2c2�VL� · nt� −VL�

2

c. �144�

n the other hand, one can transform directly from theRF to the ERF using Eq. (7), assuming the velocity of

he LRF relative to the ERF is VE� , expand to second ordernd obtain

w� · nt� −VE�

c �1 +VE� · nt�

c � −c

2

�f �

f 0�

1 +1

2

�f �

f 0�

+VE�

c �1 +VE� · nt�

c �=

−c

2

�f �

f 0�

1 +1

2

�f �

f 0�

�1 +VE�

2

2c2 +VE� · nt�

c �

− �1 +VE�

2

2c2�VE� · nt� −VE�

2

c. �145�

quations (144) and (145) appear at first to be identical,ssuming VL� =−VE� ; however, such an identity ignoresonparallelism of axes of the ERF and LRF. Neither Eq.144) nor Eq. (145) agrees with results derived earlierEqs. (41), (111), and (140)] that were derived without as-uming the axes of the ERF and LRF were parallel; thushey are incorrect and should not be used beyond first or-er in �v /c�, where v is any velocity in the problem. Eqs.144) and (145) provide alternative formulations, depend-ng on different velocity variables, that are valid only torst order in �v /c�. The difference of orientation of thexes in the LRF and in the ERF is illustrated in Fig. 9, inhich the vectors VL� and VE� are drawn in the plane of theaper. These two vectors would be antiparallel if the ref-rence frame axes were parallel. The angle between VL�nd VE� is expressed approximately in Eq. (143) in termsf a vector rotation angle VL�VE / �2c2�, which would beormal to the plane of the paper.

0. ON-BOARD GPS RECEIVERf a GPS receiver is placed on the spacecraft, then the li-ar position, velocity, and time may be available directlyrom the receiver. Ordinarily the GPS receiver will pro-ide position and time in earth-centered earth-fixedECEF) coordinates. (The origins of the ECEF frame andRF are at earth’s center and at the ground point, respec-

ively.) The ECEF frame is rotating with respect to theCI frame; the rotation introduces some new relativistic

ig. 9. The relative velocities VL� and VE� are not antiparallel.he diagram is drawn in the plane formed by VL and VE.

empsngttr

tiatwsnF

Tragta

Ho�t

i

Tb

wee

a

T

T

Trsm

a

TnEp

mftml

1Iatht4ttwntdrdttwtwiDgtp

AIt


ffects. We show here how to transform measurementsade by the GPS receiver to the ECI frame so that the

revious analysis then applies. Fortunately the rotationalpeed is so slow that only effects of first order in �v /c�eed to be considered. The ECEF frame must be distin-uished from the ERF that we have introduced because ofhe time delay between transmission and detection. Inhe noninertial ECEF frame light does not travel in all di-ections with a unique speed c.

In the present section we shall place primes on quanti-ies measured in the rotating frame, keeping in mind thatt is not an inertial frame. Let us imagine freezing thexes of the ECEF frame at the instant a signal pulse isransmitted, and identifying the directions of these axesith those of an ECI frame. GPS time is effectively the

ame as the coordinate time t in the ECI frame, so we doot have to introduce any transformations for the time.or the transmission event, then

rt = rt�. �146�

he scattering event will occur at time ts and at positions. The axes of the ECEF will have rotated through anngle E�ts− tt�, where E is the WGS-84 value of the an-ular rotation rate of earth. Thus the positions of the scat-ering event in the “frozen” ECI frame, and in the ECEF,re related by a rotation:

xs

ys

zs =

1 − E�ts − tt� 0

�E�ts − tt� 1 0

0 0 1

xs�

ys�

zs� . �147�

ere we shall work out the implications of this rotationnly to first order in the small parameter Ed /c�210−7, where L�7000 km is the largest reasonable dis-

ance in the problem.It is then straightforward to compute the ECI distance

n terms of the ECEF distance:

�rt − rs�2 = �rt� − rs��2 + 2�ts − tt�� · �rt� � rs��. �148�

he cross product in the correction term in Eq. (148) cane interpreted as an area:

A� =1

2rt� � rs�, �149�

here A� is the vector area swept out by a vector from thearth’s rotation axis to the signal pulse as the pulse trav-ls along the propagation path. Thus

�rt − rs�2 = �rt� − rs��2 + 4�ts − tt�� · A�, �150�

nd taking a square root,

�rt − rs� = �rt� − rs�� +2�ts − tt�� · A�

�rt� − rs��. �151�

he time interval between transmission and scattering is

ts − tt =1

c�rt − rs� �

1

c�rt� − rs�� +

2�E · A�

c2 . �152�

hen the expression for the distance can be simplified:

�rt − rs� = �rt� − rs�� +2�E · A�

c. �153�

he correction term in Eq. (152) is called the Sagnac cor-ection. In the present application it is only a few nano-econds. The correction term in Eq. (153) is only a feweters.The velocity of the lidar in the ECI and ECEF frames

re related approximately by

VL = VL� + � � rt�. �154�

he relativistic composition of velocities is fortunately noteeded here because the speed of a fixed point in theCEF system is so small compared with c, provided theoint is not too far from the earth’s rotation axis.Equations (152)–(154) can be used to transform infor-ation obtained from an onboard GPS receiver to the ECI

rame, in which the previous analysis is valid. Attemptingo express the equations entirely in terms of quantitieseasured by the GPS receiver leads unproductively to a

arge amount of algebra.

1. SUMMARYn this paper derivations of the Doppler shift measured by

lidar apparatus, when the incident beam is backscat-ered from an aerosol particle carried along in the wind,ave been presented. One of these derivations was fromhe point of view of the rest frame of the lidar [see Section, Eq. (41)], and the other was from the point of view ofhe ECI frame and permitted the calculation of gravita-ional effects [see Section 6, Eq. (101)]. The results agreehen acceleration and gravitational frequency shifts areeglected, and show how relativistic corrections enter intohe fundamental equation for the wind velocity in the li-ar rest frame. Expressions for the wind velocity in theest frame of the ground point, the earth-fixed point un-erneath the wind, were derived. Several ways of defininghe LOS vector were discussed, and relationships betweenhem were derived. Simulations of the lidar measurementere developed and examples were given, demonstrating

he consistency of the approach. Relativistic correctionsere derived; for most purposes only the leading relativ-

stic corrections need be considered in converting lidaroppler measurements to wind velocities, but the theoryiven here [as in Eq. (101)] should be good to all orders inhe small parameters �v /c�, where v is any velocity in theroblem.

CKNOWLEDGMENTam grateful to Michael Kavaya for suggesting this inves-igation and for many helpful discussions.

R


EFERENCES1. C. Møller, The Theory of Relativity, 2nd ed. (Clarendon

Press, 1972) pp. 39–42.2. H. P. Robertson and T. W. Noonan, Relativity and

Cosmology (Saunders, 1968) pp. 45 ff.3. R. D’Inverno, Introducing Einstein’s Relativity (Clarendon

Press, 1992) p. 40.

4. V. S. R. Gudimetla and M. Kavaya, Special RelativityCorrections for Space-Based Lidar, Appl. Opt. 38,6374–6382 (1999).

5. B. W. Parkinson and J. J. Spilker, Jr., Global PositioningSystem: Theory and Applications (American Institute ofAeronautics and Astronautics, 1996) Vol. I, pp. 161–165.

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