## Abstract

In a beam of monochromatic unpolarized light the electric field vector at a point traces out an ellipse whose size, eccentricity, and orientation are slowly varying functions of time. The statistical properties of the parameters of this ellipse are investigated. It is shown that the quantity S which is defined as twice the product of the principle axes of the ellipse divided by the sum of the squares is uniformly distributed between zero and one. It therefore has median value $12$ which corresponds to a ratio of minor to major axis equal to .268. Hence fairly thin ellipses predominate. The square root of the sum of the squares of the semi-major and semi-minor axes, R, is statistically independent of S and has the distribution function (r3/2p4) exp (−r2/2p2) where 2p2 is the average value of R2.

© 1945 Optical Society of America

### References

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1. We assume that the electric field is constant over the part of the cross section of the beam under consideration.
2. H. Hurwitz and M. Kac, Ann. Math. Stat. 15, 173 (1944); S. O. Rice, Bell Sys. Tech. J. 23, 282 (1944), Bell Sys. Tech. J. 24, 46 (1945).Some insight as to why the quantities in Eq. (5) are Gaussianly distributed can be had from the following considerations. LetF(t)=∑1Nf(t-ti)=∑k=0N(ak cos ωkt+bk sin ωkt).From Eq. (4) ak=∑1N(αk2+βk2)12 cos (ωkti-φk),bk=∑1N(αk2+βk2)12 sin (ωkti-φk),         φk=tan-1βkαk.If the ti’s are randomly distributed the quantities ak and bk can be regarded as the two components of a vector which is the sum of N two-dimensional vectors of constant length but random direction. Hence from the familiar results of the problem of “random walk” we can immediately conclude that ak and bk are Gaussianly distributed in the limit that N becomes infinite.
[CrossRef]
3. The assumption that Ex and Ey are statistically independent can be justified in terms of the somewhat more physical assumption that the statistical properties of an intense beam of unpolarized light cannot be changed by passing the light through a wave plate. By the use of a wave plate of suitable thickness, one field component can be retarded with respect to the other by any desired amount. It is evident that no possible type of correlation between Ex and Ey could remain invariant under this general class of transformations.
4. M. Kac and H. Steinhaus, Studia Mathematica 6, 89 (1936).
5. L. Brillouin, Die Quantenstatistik (Julius Springer, Berlin, 1931), p. 111, 177.
6. M. Born, Optik (Julius Springer, Berlin, 1933), p. 23.
7. Private communication.

#### 1944 (1)

H. Hurwitz and M. Kac, Ann. Math. Stat. 15, 173 (1944); S. O. Rice, Bell Sys. Tech. J. 23, 282 (1944), Bell Sys. Tech. J. 24, 46 (1945).Some insight as to why the quantities in Eq. (5) are Gaussianly distributed can be had from the following considerations. LetF(t)=∑1Nf(t-ti)=∑k=0N(ak cos ωkt+bk sin ωkt).From Eq. (4) ak=∑1N(αk2+βk2)12 cos (ωkti-φk),bk=∑1N(αk2+βk2)12 sin (ωkti-φk),         φk=tan-1βkαk.If the ti’s are randomly distributed the quantities ak and bk can be regarded as the two components of a vector which is the sum of N two-dimensional vectors of constant length but random direction. Hence from the familiar results of the problem of “random walk” we can immediately conclude that ak and bk are Gaussianly distributed in the limit that N becomes infinite.
[CrossRef]

#### 1936 (1)

M. Kac and H. Steinhaus, Studia Mathematica 6, 89 (1936).

#### Born, M.

M. Born, Optik (Julius Springer, Berlin, 1933), p. 23.

#### Brillouin, L.

L. Brillouin, Die Quantenstatistik (Julius Springer, Berlin, 1931), p. 111, 177.

#### Hurwitz, H.

H. Hurwitz and M. Kac, Ann. Math. Stat. 15, 173 (1944); S. O. Rice, Bell Sys. Tech. J. 23, 282 (1944), Bell Sys. Tech. J. 24, 46 (1945).Some insight as to why the quantities in Eq. (5) are Gaussianly distributed can be had from the following considerations. LetF(t)=∑1Nf(t-ti)=∑k=0N(ak cos ωkt+bk sin ωkt).From Eq. (4) ak=∑1N(αk2+βk2)12 cos (ωkti-φk),bk=∑1N(αk2+βk2)12 sin (ωkti-φk),         φk=tan-1βkαk.If the ti’s are randomly distributed the quantities ak and bk can be regarded as the two components of a vector which is the sum of N two-dimensional vectors of constant length but random direction. Hence from the familiar results of the problem of “random walk” we can immediately conclude that ak and bk are Gaussianly distributed in the limit that N becomes infinite.
[CrossRef]

#### Kac, M.

H. Hurwitz and M. Kac, Ann. Math. Stat. 15, 173 (1944); S. O. Rice, Bell Sys. Tech. J. 23, 282 (1944), Bell Sys. Tech. J. 24, 46 (1945).Some insight as to why the quantities in Eq. (5) are Gaussianly distributed can be had from the following considerations. LetF(t)=∑1Nf(t-ti)=∑k=0N(ak cos ωkt+bk sin ωkt).From Eq. (4) ak=∑1N(αk2+βk2)12 cos (ωkti-φk),bk=∑1N(αk2+βk2)12 sin (ωkti-φk),         φk=tan-1βkαk.If the ti’s are randomly distributed the quantities ak and bk can be regarded as the two components of a vector which is the sum of N two-dimensional vectors of constant length but random direction. Hence from the familiar results of the problem of “random walk” we can immediately conclude that ak and bk are Gaussianly distributed in the limit that N becomes infinite.
[CrossRef]

M. Kac and H. Steinhaus, Studia Mathematica 6, 89 (1936).

#### Steinhaus, H.

M. Kac and H. Steinhaus, Studia Mathematica 6, 89 (1936).

#### Ann. Math. Stat. (1)

H. Hurwitz and M. Kac, Ann. Math. Stat. 15, 173 (1944); S. O. Rice, Bell Sys. Tech. J. 23, 282 (1944), Bell Sys. Tech. J. 24, 46 (1945).Some insight as to why the quantities in Eq. (5) are Gaussianly distributed can be had from the following considerations. LetF(t)=∑1Nf(t-ti)=∑k=0N(ak cos ωkt+bk sin ωkt).From Eq. (4) ak=∑1N(αk2+βk2)12 cos (ωkti-φk),bk=∑1N(αk2+βk2)12 sin (ωkti-φk),         φk=tan-1βkαk.If the ti’s are randomly distributed the quantities ak and bk can be regarded as the two components of a vector which is the sum of N two-dimensional vectors of constant length but random direction. Hence from the familiar results of the problem of “random walk” we can immediately conclude that ak and bk are Gaussianly distributed in the limit that N becomes infinite.
[CrossRef]

#### Studia Mathematica (1)

M. Kac and H. Steinhaus, Studia Mathematica 6, 89 (1936).

#### Other (5)

L. Brillouin, Die Quantenstatistik (Julius Springer, Berlin, 1931), p. 111, 177.

M. Born, Optik (Julius Springer, Berlin, 1933), p. 23.

Private communication.

The assumption that Ex and Ey are statistically independent can be justified in terms of the somewhat more physical assumption that the statistical properties of an intense beam of unpolarized light cannot be changed by passing the light through a wave plate. By the use of a wave plate of suitable thickness, one field component can be retarded with respect to the other by any desired amount. It is evident that no possible type of correlation between Ex and Ey could remain invariant under this general class of transformations.

We assume that the electric field is constant over the part of the cross section of the beam under consideration.

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### Equations (45)

$E x = X ( t ) cos ω 0 t + U ( t ) sin ω 0 t = A ( t ) cos ( ω 0 t - δ x ( t ) ) , E y = Y ( t ) cos ω 0 t + V ( t ) sin ω 0 t = B ( t ) cos ( ω 0 t - δ y ( t ) ) , A ( t ) = [ X 2 ( t ) + U 2 ( t ) ] 1 2 ; B ( t ) = [ Y 2 ( t ) + V 2 ( t ) ] 1 2 ; δ x ( t ) = tan - 1 [ U ( t ) / X ( t ) ] ; δ y ( t ) = tan - 1 [ V ( t ) / Y ( t ) ] ; ω 0 = 2 π ν 0 .$
$P X ( x ) d x = 1 p P G ( x p ) d x = 1 ( 2 π ) 1 2 exp ( - x 2 2 p 2 ) d x p ,$
$P A ( x ) d x = P B ( x ) d x = x p 2 exp ( - x 2 2 p 2 ) d x .$
$f ( t ) = ∑ k = 0 ∞ ( α k cos ω k t + β k sin ω k t ) , ω k = 2 π k / T .$
$E x = ∑ k = 0 ∞ p k ( X k cos ω k t + U k sin ω k t ) , E y = ∑ k = 0 ∞ p k ( Y k cos ω k t + V k sin ω k t ) .$
$〈 X k 〉 Av = 〈 Y k 〉 Av = 〈 U k 〉 Av = 〈 V k 〉 Av = 0 , 〈 X k 2 〉 Av = 〈 Y k 2 〉 Av = 〈 U k 2 〉 Av = 〈 V k 2 〉 Av = 1.$
$P G ( x ) d x = [ 1 / ( 2 π ) 1 2 ] exp ( - x 2 / 2 ) d x .$
$X ( t ) = ∑ p k [ X k cos ( ω k - ω 0 ) t + U k sin ( ω k - ω 0 ) t ] , U ( t ) = ∑ p k [ - X k sin ( ω k - ω 0 ) t + U k cos ( ω k - ω 0 ) t ] , Y ( t ) = ∑ p k [ Y k cos ( ω k - ω 0 ) t + V k sin ( ω k - ω 0 ) t ] , V ( t ) = ∑ p k [ - Y k sin ( ω k - ω 0 ) t + V k cos ( ω k - ω 0 ) t ] .$
$p 2 = ∑ k = 0 ∞ p k 2 .$
$f ( ξ ) = 〈 e i ξ A 〉 Av = 〈 e i ξ ( A cos θ + B sin θ ) 〉 Av .$
$f ( ξ ) = 〈 e i ξ A cos θ 〉 Av 〈 e i ξ B sin θ 〉 Av = f ( ξ cos θ ) f ( ξ sin θ ) ;$
$f ( ( x 2 + y 2 ) 1 2 ) = f ( x ) f ( y ) ,$
$x = ξ cos θ ; y = ξ sin θ .$
$f ( ξ ) = exp ( λ ξ 2 ) .$
$∣ f ( ξ ) ∣ = ∣ ( e i ξ A 〉 Av ∣ ⩽ 1.$
$P X , U ( x , u ) d x d u = P A ( ( x 2 + u 2 ) 1 2 ) 2 π ( x 2 + u 2 ) 1 2 d x d u = g ( x 2 + u 2 ) d x d u .$
$P X ( x ) = 1 p ( 2 π ) 1 2 exp ( - x 2 2 p 2 ) = ∫ - ∞ + ∞ P X , U ( x , u ) d u = ∫ - ∞ + ∞ g ( x 2 + u 2 ) d u .$
$1 p ( 2 π ) 1 2 exp ( - ( x 2 + z 2 ) 2 p 2 ) = ∫ - ∞ + ∞ g ( x 2 + z 2 + u 2 ) d u ;$
$1 p ( 2 π ) 1 2 exp ( - x 2 2 p 2 ) ∫ - ∞ + ∞ exp ( - z 2 2 p 2 ) d z = ∫ - ∞ + ∞ ∫ - ∞ + ∞ g ( x 2 + z 2 + u 2 ) d u d z ,$
$exp ( - x 2 2 p 2 ) = π ∫ x 2 ∞ g ( y ) d y .$
$1 2 p 2 exp ( - x 2 2 p 2 ) = π g ( x 2 ) ,$
$P X , U ( x , u ) d x d u = 1 2 π p 2 exp ( - ( x 2 + u 2 ) 2 p 2 ) d x d u .$
$W τ ( t ) = ∫ t t + τ ( A 2 ( t ′ ) + B 2 ( t ′ ) ) d t ′ .$
$Δ = 〈 W τ 2 ( t ) 〉 Av - 〈 W τ ( t ) 〉 Av 2 〈 W τ ( t ) 〉 Av 2$
$M 2 ( t ) + N 2 ( t ) = A 2 ( t ) + B 2 ( t ) = R 2 ( t ) , S ( t ) = 2 M ( t ) N ( t ) R 2 ( t ) = 2 A ( t ) B ( t ) R 2 ( t ) ∣ sin δ ( t ) ∣ , δ ( t ) = δ x ( t ) - δ y ( t ) .$
$Φ ( t ) = tan - 1 B ( t ) A ( t ) 0 ⩽ Φ ( t ) ⩽ π 2 .$
$P R , Φ ( r , φ ) d r d φ = P A ( r cos φ ) P B ( r sin φ ) r d r d φ = r 3 2 p 4 sin 2 φ exp ( - r 2 2 p 2 ) d r d φ .$
$S ( t ) = sin 2 Φ ( t ) ∣ sin δ ( t ) ∣ .$
$P ∣ sin δ ∣ ( σ ) d σ = 2 π d sin - 1 σ = 2 π d σ ( 1 - σ 2 ) 1 2 .$
$P ∣ sin δ ∣ ( s sin 2 Φ ) d s sin 2 Φ = 2 π d s sin 2 Φ ( 1 - s 2 sin 2 2 Φ ) 1 2 .$
$P S , R ( s , r ) d s d r = d s d r ∫ 2 π P R , Φ ( r , φ ) d φ sin 2 φ ( 1 - s 2 sin 2 2 φ ) 1 2 .$
$P S , R ( s , r ) d s d r = d s d r r 3 π p 2 exp ( - r 2 2 p 2 ) × ∫ d φ ( 1 - s 2 sin 2 2 φ ) 1 2 .$
$∫ d φ ( 1 - s 2 sin 2 2 φ ) 1 2 = - 1 2 ∫ d y ( 1 - s 2 - y 2 ) 1 2 = π 2 y = cos 2 φ$
$P S , R ( s , r ) d s d r = r 3 2 p 4 exp ( - r 2 2 p 2 ) d s d r .$
$P R ( r ) d r = r 3 2 p 4 exp ( - r 2 / 2 p 2 ) d r .$
$L ( t ) = N ( t ) / M ( t )$
$S = 2 L / ( 1 + L 2 ) ,$
$d S = 2 ( 1 - L 2 ) d L / ( 1 + L 2 ) 2 .$
$P L , R ( l , r ) d l d r = ( 1 - l 2 ) ( 1 + l 2 ) 2 r 3 p 4 × exp ( - r 2 2 p 2 ) d l d r ,$
$P L ( l ) d l = [ 2 ( 1 - l 2 ) / ( 1 + l 2 ) 2 ] d l .$
$L m = .268.$
$P L , M ( l , m ) d l d m = ( 1 - l 2 ) m 3 p 4 exp [ - m 2 ( 1 + l 2 ) 2 p 2 ] d l d m .$
$P M , N ( m , n ) d m d n = ( m 2 - n 2 ) p 4 exp [ - ( m 2 + n 2 ) 2 p 2 ] d m d n .$
$F(t)=∑1Nf(t-ti)=∑k=0N(ak cos ωkt+bk sin ωkt).$
$ak=∑1N(αk2+βk2)12 cos (ωkti-φk),bk =∑1N(αk2+βk2)12 sin (ωkti-φk), φk=tan-1βkαk.$