METEORITE SHOOTING AS A DIFFUSION PROBLEM

Page 1

Journal for Research | Volume 03| Issue 01 | March 2017 ISSN: 2395-7549

Meteorite Shooting as a Diffusion Problem Dr. Prof. Rashmi R. Keshvani Professor Department of Mathematics Sarvajanik College of Engineering & Technology, Surat, Gujarat, India

Prof. Maulik S. joshi Assist Professor Department of Mathematics Aditya Silver oak Institute of Technology, Ahmedabad, Gujarat, India

Abstract Diffusion problems have been problems of great interest with various initial and boundary conditions. Among those, infinite domain problems have been more interesting. Many of such problems can be solved by various methods but those which can be used for various initial functions with minor changes in the solution obtained are more attractive and efficient. Fourier transforms method and methods obtaining Gauss- Weierstrass kernel play such role among various such methods. To show this feature here in this paper, first the consequences of a local injection of heat to an infinite domain are being discussed. Solutions to such problems at different time are discussed in terms of Gaussian distributions. The theory is then extended to a meteorite shooting problem. Keywords: Fourier Transform, Inverse Fourier Transform, Dirac Delta Function, Gaussian distribution, Mean and variance of a probability distribution, Meteorites, Refraction index _______________________________________________________________________________________________________ I. It is known and can be verified that đ?‘’ −đ?‘–đ?œ”đ?‘Ľ đ?‘’ −đ?‘˜đ?œ” principle of super position, it can be shown that 2 ∞ đ?‘˘(đ?‘Ľ, đ?‘Ą) = âˆŤâˆ’âˆž đ?‘?(đ?œ”) đ?‘’ −đ?‘–đ?œ”đ?‘Ľ đ?‘’ −đ?‘˜đ?œ” đ?‘Ą đ?‘‘đ?œ” is solution of heat equation,

đ?œ•đ?‘˘ đ?œ•đ?‘Ą

=đ?‘˜

đ?œ•2 đ?‘˘ đ?œ•đ?‘Ľ 2

2đ?‘Ą

INTRODUCTION

satisfy heat equation

đ?œ•đ?‘˘ đ?œ•đ?‘Ą

=đ?‘˜

đ?œ•2 đ?‘˘ đ?œ•đ?‘Ľ 2

for all values of đ?œ” So using generalized (1)

where −∞ < đ?‘Ľ < ∞. ∞

The initial condition đ?‘˘(đ?‘Ľ, 0) = đ?‘“(đ?‘Ľ), will be satisfied, if đ?‘˘(đ?‘Ľ, 0) = đ?‘“(đ?‘Ľ) = âˆŤâˆ’âˆž đ?‘?(đ?œ”) đ?‘’ −đ?‘–đ?œ”đ?‘Ľ đ?‘‘đ?œ”. From definitions of Fourier transform and

inverse Fourier transform [1],

∞ âˆŤâˆ’âˆž đ?‘?(đ?œ”) đ?‘’ −đ?‘–đ?œ”đ?‘Ľ đ?‘‘đ?œ”

∞

đ?‘“(đ?‘Ľ) = âˆŤâˆ’âˆž đ?‘?(đ?œ”) đ?‘’ −đ?‘–đ?œ”đ?‘Ľ đ?‘‘đ?œ”

implies that

1 ∞ âˆŤ đ?‘“(đ?‘Ľ)đ?‘’ đ?‘–đ?œ”đ?‘Ľ đ?‘‘đ?‘Ľ 2đ?œ‹ −∞

is inverse Fourier transform of đ?‘“(đ?‘Ľ) and đ?‘?(đ?œ”) = is the Fourier transform of initial temperature function đ?‘“(đ?‘Ľ). 1 ∞ Substituting đ?‘?(đ?œ”) = âˆŤâˆ’âˆž đ?‘“(đ?‘Ľ)đ?‘’ đ?‘–đ?œ”đ?‘Ľ đ?‘‘đ?‘Ľ in (1), and changing dummy variable đ?‘Ľ to đ?‘ĽĚ… in expression for đ?‘?(đ?œ”), now (1) 2đ?œ‹ becomes ∞ 1 ∞ 2 đ?‘˘(đ?‘Ľ, đ?‘Ą) = âˆŤ ( âˆŤ đ?‘“(đ?‘ĽĚ… )đ?‘’ đ?‘–đ?œ”đ?‘ĽĚ… đ?‘‘đ?‘ĽĚ… ) đ?‘’ −đ?‘–đ?œ”đ?‘Ľ đ?‘’ −đ?‘˜đ?œ” đ?‘Ą đ?‘‘đ?œ” 2đ?œ‹ −∞ −∞ 2 ∞ 1 ∞ â&#x;š đ?‘˘(đ?‘Ľ, đ?‘Ą) = âˆŤâˆ’âˆž đ?‘“(đ?‘ĽĚ… ) ( âˆŤâˆ’âˆž đ?‘’ −đ?‘–đ?œ”(đ?‘Ľâˆ’đ?‘ĽĚ… ) đ?‘’ −đ?‘˜đ?œ” đ?‘Ą đ?‘‘đ?œ”) đ?‘‘đ?‘ĽĚ… (2) 2đ?œ‹ ∞

2

2

Also it is known that đ?‘”(đ?‘Ľ) = âˆŤâˆ’âˆž đ?‘’ −đ?‘–đ?œ”đ?‘Ľ đ?‘’ −đ?‘˜đ?œ” đ?‘Ą đ?‘‘đ?œ” is inverse Fourier transform of đ?‘’ −đ?‘˜đ?œ” đ?‘Ą , ∞

2

(a Gaussian Curve). So, đ?‘”(đ?‘Ľ − đ?‘ĽĚ… ) = âˆŤâˆ’âˆž đ?‘’ −đ?‘–đ?œ”(đ?‘Ľâˆ’đ?‘ĽĚ… ) đ?‘’ −đ?‘˜đ?œ” đ?‘Ą đ?‘‘đ?œ” = Substituting this in (2), one gets đ?‘˘(đ?‘Ľ, đ?‘Ą) = It can be shown [1] that lim √ đ?‘Ąâ†’0+

1 4đ?œ‹đ?‘˜đ?‘Ą

√đ?œ‹ √đ?‘˜đ?‘Ą

đ?‘’

Ě… )2 −(đ?‘Ľâˆ’đ?‘Ľ 4đ?‘˜đ?‘Ą

.

∞ −(đ?‘Ľâˆ’đ?‘ĽĚ… )2 1 ∞ 1 √đ?œ‹ −(đ?‘Ľâˆ’đ?‘ĽĚ… )2 âˆŤ đ?‘“(đ?‘ĽĚ… ) ( đ?‘’ 4đ?‘˜đ?‘Ą ) đ?‘‘đ?‘ĽĚ… = âˆŤ đ?‘“(đ?‘ĽĚ… ) √ đ?‘’ 4đ?‘˜đ?‘Ą đ?‘‘đ?‘ĽĚ… 2đ?œ‹ −∞ 4đ?œ‹đ?‘˜đ?‘Ą √đ?‘˜đ?‘Ą −∞

đ?‘’

Ě… )2 −(đ?‘Ľâˆ’đ?‘Ľ 4đ?‘˜đ?‘Ą

= đ?›ż(đ?‘Ľ − đ?‘ĽĚ… ), where đ?›ż(đ?‘Ľ) is the Dirac delta function (Impulse function).[1] The

Dirac delta function, [2] denoted by đ?›ż(đ?‘Ľ), also known as impulse function, is defined as ∞ 0 if đ?‘Ľ ≠0 đ?›ż(đ?‘Ľ) = { ensuring âˆŤâˆ’âˆž đ?›ż(đ?‘Ľ)đ?‘‘đ?‘Ľ = 1 . ∞ if đ?‘Ľ = 0 ∞ Also âˆŤâˆ’âˆž đ?‘“(đ?‘Ľ)đ?›ż(đ?‘Ľ − đ?‘Ž)đ?‘‘đ?‘Ľ = đ?‘“(đ?‘Ž) where đ?‘“ is any continuous function?

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32

Meteorite Shooting as a Diffusion Problem (J4R/ Volume 03 / Issue 01 / 007)

II. SOLUTION OF THE PROBLEM 1) If đ?‘‰(đ?‘Ľ) denote temperature in a bar in which the heat can flow only in the Âą đ?‘Ľ directions, then heat flow will be only đ?œ•đ?‘‰ there, where gradient of temperature is. The amount of heat per second, đ??ź, which can be urged along the bar , is đ?œ•đ?‘Ľ proportional to the temperature gradient and is inversely proportional to the thermal resistance đ?‘&#x; of the material of the bar per unit length. 1 đ?œ•đ?‘‰ That is đ??ź = − đ?‘&#x; đ?œ•đ?‘Ľ

The amount of heat accumulated in unit length per second is the difference between what flows in and what flows out, i.e. The temperature rise is inversely proportional to thermal capacitance đ?‘? per unit length. So đ?œ•đ?‘‰ 1 đ?œ•đ??ź 1 đ?œ•2đ?‘‰ = − = đ?œ•đ?‘Ą đ?‘? đ?œ•đ?‘Ľ đ?‘&#x;đ?‘? đ?œ•đ?‘Ľ 2 So here diffusion problem, is as follows đ?œ•2 đ?‘‰

= đ?‘&#x;đ?‘?

đ?œ•đ?‘Ľ 2

đ?œ•đ?‘‰ đ?œ•đ?‘Ą

−∞<đ?‘Ľ <∞

đ?‘Ą>0,

,

đ?œ•đ??ź đ?œ•đ?‘Ľ

.

(4)

Suppose initial condition is given as đ?‘‰(đ?‘Ľ, 0) = đ??´ đ?›ż(đ?‘Ľ). Here A is some constant. So đ?‘‰(đ?‘Ľ, 0) = đ??´ đ?›ż(đ?‘Ľ) implies that it is case of local injection of heat at a point.[3] đ?œ•2 đ?‘‰

1

1 đ?œ•đ?‘‰

If = đ??ž , we have = , đ?‘Ą > 0 , −∞ < đ?‘Ľ < ∞ with initial condition đ?‘‰(đ?‘Ľ, 0) = đ?‘“(đ?‘Ľ) = đ??´ đ?›ż(đ?‘Ľ) đ?‘&#x;đ?‘? đ?œ•đ?‘Ľ 2 đ?‘˜ đ?œ•đ?‘Ą For infinite domain diffusion problems, as discussed above, solution will be −(đ?‘Ľâˆ’đ?œ€)2

�(�, �) =

∞ đ?‘’ 4đ??žđ?‘Ą âˆŤâˆ’âˆž đ?‘“(đ?œ€) √4đ?œ‹đ??žđ?‘Ą

đ?‘‘đ?œ€

(5)

where đ?‘“(đ?‘Ľ) = đ?‘‰(đ?‘Ľ. 0) is initial function. So, if, for a fixed đ?‘Ą, đ?‘‰(đ?‘Ľ, đ?‘Ą), is denoted by đ?‘‰đ?‘Ą (đ?‘Ľ), then −(đ?‘Ľâˆ’đ?œ€)2

�� (�) =

∞ đ?‘’ 4đ??žđ?‘Ą âˆŤâˆ’âˆž đ?‘“(đ?œ€) √4đ?œ‹đ??žđ?‘Ą

As for any continuous function đ?‘“(đ?‘Ľ),

đ?‘‘đ?œ€ =

−(đ?‘Ľâˆ’đ?œ€)2 4đ??žđ?‘Ą

đ?‘‰đ?‘Ą (đ?‘Ľ) = âˆŤ đ?›ż(đ?œ€) đ?‘’ √4đ?œ‹đ??žđ?‘Ą −∞ Thus for fixed time t, the temperature function đ?‘&#x;đ?‘?

∞

âˆŤâˆ’âˆž đ?›ż(đ?œ€) đ?‘’

√4đ?œ‹đ??žđ?‘Ą ∞ âˆŤâˆ’âˆž đ?‘“(đ?‘Ľ)đ?›ż(đ?‘Ľ)đ?‘‘đ?‘Ľ

∞

đ??´

đ??´

1â „ 2

đ?‘‘đ?œ€ =

−(đ?‘Ľâˆ’đ?œ€)2 4đ??žđ?‘Ą

đ?‘‘đ?œ€

(6)

= đ?‘“(0) , đ??´ √4đ?œ‹đ??žđ?‘Ą

đ?‘’

−đ?‘Ľ2 đ?‘&#x;đ?‘? 4đ?‘Ą

= đ??´(

đ?‘&#x;đ?‘? 4đ?œ‹đ?‘Ą

)

1â „ 2

đ?‘’

−đ?‘Ľ2 đ?‘&#x;đ?‘? 4đ?‘Ą

1 đ?œŽâˆš2đ?œ‹

đ?‘’

−(đ?‘Ľâˆ’đ?œ‡)2 2đ?œŽ2

,

2đ?‘Ą đ?‘&#x;đ?‘?

.

(7)

and variance đ?œŽ 2 =

the curve đ?‘‰đ?‘Ą (đ?‘Ľ) has mean đ?œ‡ = 0

deviation for this curve is đ?œŽ = √

1 đ?‘&#x;đ?‘?

−đ?‘Ľ2 đ?‘&#x;đ?‘?

đ?‘‰đ?‘Ą (đ?‘Ľ) = đ??´ ( ) đ?‘’ 4đ?‘Ą 4đ?œ‹đ?‘Ą This implies đ?‘‰đ?‘Ą (đ?‘Ľ) is a Gaussian distribution. As Gaussian distribution with mean đ?œ‡ and standard deviation đ?œŽ, is defined as đ?‘“(đ?‘Ľ, đ?œ‡, đ?œŽ) =

as đ??ž =

2đ?‘Ą đ?‘&#x;đ?‘?

. That is, the standard

. So curves broaden as √đ?‘Ą .[4]

The same discussion can be done for meteorite shooting also. 2) Meteorites are pieces of other bodies in our solar system that make it to the ground when a meteor or "shooting star" flashes through earth’s atmosphere at speeds of 15 to 70 kilometers per second (roughly 32,000 to 150,000 miles per hour). The majority originate from asteroids shattered by impacts with other asteroids. In a few cases they come from the Moon and, presumably, comets and the planet Mars. Meteorites that are found after a meteoric event has been witnessed are called a "fall," while those found by chance are called a "find." Meteorites are usually named after a town or a large geographic landmark closest to the fall or find, collectively termed localities. The word "meteorite" can refer to an individual specimen, to those collected within a strewn field, or to a specific locality. [5] From mathematical point of view, a meteorite shooting through the earth’s atmosphere leaves a trail of đ?›ź electrons and positive ions per meter, which diffuse away with a diffusion coefficient đ??ž. [3] To find electron density đ?‘ per cubic meter at a distance đ?‘&#x; from a point on the meteor trail at a time đ?‘Ą after the meteor passes the point, one may imagine the trail of ionization created as diffusing cylindrically as time elapses.[3] The general three dimensional diffusion equation is

đ?œ•2 đ?‘ đ?œ•đ?‘Ľ 2

+

đ?œ•2 đ?‘ đ?œ•đ?‘Ś 2

+

đ?œ•2 đ?‘ đ?œ•đ?‘§ 2

=

1 đ?œ•đ?‘ đ??ž đ?œ•đ?‘Ą

(8)

Converting to cylindrical co-ordinate system, equation will be đ?œ•2 đ?‘

1 đ?œ•2 đ?‘

1 đ?œ•đ?‘

đ?œ•2 đ?‘

1 đ?œ•đ?‘

+ + 2 2+ 2= đ?‘&#x; đ?œ•đ?‘&#x; đ?‘&#x; đ?œ•đ?œƒ đ?œ•đ?‘§ đ??ž đ?œ•đ?‘Ą Considering particular value of đ?‘§, and applying circular symmetry, (9) becomes đ?œ•đ?‘&#x; 2

đ?œ•2 đ?‘

as

đ?œ•đ?‘&#x; 2 đ?œ•đ?‘ đ?œ•đ?œƒ

+

1 đ?œ•đ?‘ đ?‘&#x; đ?œ•đ?‘&#x;

=

= 0 and

1 đ?œ•đ?‘

(10)

đ??ž đ?œ•đ?‘Ą đ?œ•đ?‘ đ?œ•đ?‘§

(9)

=0

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33

Meteorite Shooting as a Diffusion Problem (J4R/ Volume 03 / Issue 01 / 007)

As meteor enters earth’s atmosphere suddenly, here again situation is of point injection, so solution must be of the form �

đ?‘ =

−đ?‘&#x;2 4đ??žđ?‘Ą

đ?‘?đ?‘Ą đ?‘’ where đ?‘&#x; is distance from a point on the meteor trail at a particular time đ?‘Ą. đ?‘š and đ?‘? are constants, which can be determined depending upon the differential equation (10) and the situations. −đ?‘&#x;2

−đ?‘&#x;2

đ?œ•đ?‘

−đ?‘&#x;2

−2đ?‘&#x;

−đ?‘&#x;

As đ?‘ = đ?‘? đ?‘Ą đ?‘š đ?‘’ 4đ??žđ?‘Ą , = đ?‘? đ?‘Ą đ?‘š đ?‘’ 4đ??žđ?‘Ą ( ) = đ?‘? đ?‘Ą đ?‘š đ?‘’ 4đ??žđ?‘Ą ( ) , đ?œ•đ?‘&#x; 4đ??žđ?‘Ą 2đ??žđ?‘Ą −đ?‘&#x; 2 đ?œ•2đ?‘ đ?‘&#x;2 1 đ?‘š 4đ??žđ?‘Ą â&#x;š = đ?‘?đ?‘Ą đ?‘’ ( − ) đ?œ•đ?‘&#x; 2 4đ??ž 2 đ?‘Ą 2 2đ??žđ?‘Ą −đ?‘&#x;2

đ?œ•đ?‘

−đ?‘&#x;2

đ?‘&#x;2

−đ?‘&#x;2

đ?‘&#x;2

đ?‘š

and = đ?‘? (đ?‘Ą đ?‘š đ?‘’ 4đ??žđ?‘Ą ( 2) + đ?‘’ 4đ??žđ?‘Ą (đ?‘šđ?‘Ą đ?‘šâˆ’1 )) = đ?‘?đ?‘’ 4đ??žđ?‘Ą đ?‘Ą đ?‘š ( 2 + ) đ?œ•đ?‘Ą 4đ??žđ?‘Ą 4đ??žđ?‘Ą đ?‘Ą Substituting these expressions in (10), one gets −đ?‘&#x; 2 −đ?‘&#x; 2 đ?‘? đ?‘š −đ?‘&#x; 2 đ?‘&#x; 2 đ?‘š đ?‘&#x;2 1 1 −đ?‘&#x; đ?‘Ą đ?‘’ 4đ??žđ?‘Ą ( + ) = đ?‘? đ?‘Ą đ?‘š đ?‘’ 4đ??žđ?‘Ą ( − ) + (đ?‘?đ?‘Ą đ?‘š đ?‘’ 4đ??žđ?‘Ą ( ) 2 2 2 đ??ž 4đ??žđ?‘Ą đ?‘Ą 4đ??ž đ?‘Ą 2đ??žđ?‘Ą đ?‘&#x; 2đ??žđ?‘Ą đ?‘&#x;2 đ?‘š đ?‘&#x;2 1 1 đ?‘š 1 â&#x;š( + ) = ( − )− â&#x;š = − â&#x;š đ?‘š = −1 2 2 4đ??žđ?‘Ą đ?‘Ą 4đ??žđ?‘Ą 2đ?‘Ą 2đ?‘Ą đ?‘Ą đ?‘Ą −đ?‘&#x;2

That is đ?‘ = đ?‘? đ?‘Ą −1 đ?‘’ 4đ??žđ?‘Ą đ?›ź The central electron density will be đ?‘ (0, đ?‘Ą) = đ?‘? đ?‘Ą −1 , obtained on substituting đ?‘&#x; = 0 , must be some constant times of , so đ?œ‹đ??ž đ?›ź đ?›ź 1 say đ?‘ (0, đ?‘Ą) = â„Ž where â„Ž is some constant. No harm, if â„Ž is selected as â„Ž = â „4 â&#x;š đ?‘ (0, đ?‘Ą) = will be central đ?œ‹đ??žđ?‘Ą 4đ?œ‹đ??žđ?‘Ą electron density. 4đ?œ‹đ??žđ?‘Ą

�

For constant đ?‘Ą, say đ?‘Ą = đ?‘Ą0 đ?‘ (đ?‘&#x;, đ?‘Ą0 ) =

−đ?‘&#x;2

�

đ?‘ (đ?‘&#x;, đ?‘Ą) =

That is, the solution of (10) is,

4đ?œ‹đ??žđ?‘Ą0

đ?‘’

đ?‘’ 4đ??žđ?‘Ą

−đ?‘&#x;2 4đ??žđ?‘Ą0

(11)

is function of đ?‘&#x; only and it is Gaussian with variance 2đ??žđ?‘Ą0 , that is with

standard deviation √2đ??žđ?‘Ą0 . So curves broaden as √đ?‘Ą . Larger the value of đ?‘Ą, broader the curve will be. For a fixed đ?‘Ą, peak đ?›ź density, that is central ordinate will be đ?‘ = , for that particular value of đ?‘Ą. That is the peak density, for family of curves for 4đ?œ‹đ??žđ?‘Ą −1 different values of đ?‘Ą, diminishes as đ?‘Ą . Larger the value of đ?‘Ą, smaller the peak density. While observing such showers, refractive index of air also play vital role. The refractive index of air, denoted by đ?‘›, containing đ?‘ electrons per cubic meter, is given by đ?‘› = (1 −

81đ?‘ 1 đ?‘“2

)2 ,

where đ?‘“ denotes frequency of radio waves. [2] One would be interested to find surface of zero refractive index. As đ?‘› = (1 −

81đ?‘ 1 2 đ?‘“2

) â&#x;š đ?‘› =(1− đ?‘“2

đ?‘› will be zero if đ?‘ =

81

−đ?‘&#x;2 81đ?›źđ?‘’ 4đ??žđ?‘Ą

4đ?œ‹đ??žđ?‘Ąđ?‘“ 2

1 2

) ,

−đ?‘&#x;2

, that is if

81đ?›źđ?‘’ 4đ??žđ?‘Ą 4đ?œ‹đ??žđ?‘Ąđ?‘“ 2

= 1.

−đ?‘&#x;2 81đ?›źđ?‘’ 4đ??žđ?‘Ą

2

−đ?‘&#x;2

81đ?›źđ?‘’ 4đ??žđ?‘Ą 1

â&#x;š đ?‘› = 0 đ?‘–đ?‘“ đ?‘“ = â&#x;š đ?‘› = 0 if radio frequency đ?‘“ = ( 4đ?œ‹đ??žđ?‘Ą So to find maximum value of đ?‘&#x; at which đ?‘› = 0, one should differentiate đ?‘› =(1−

−đ?‘&#x;2 81đ?›źđ?‘’ 4đ??žđ?‘Ą

4đ?œ‹đ??žđ?‘Ąđ?‘“ 2

1 2

−đ?‘&#x;2

) =0â&#x;š

Upon differentiating,

)2

4đ?œ‹đ??žđ?‘Ą

81đ?›źđ?‘’ 4đ??žđ?‘Ą 4đ?œ‹đ??žđ?‘Ąđ?‘“ 2

− 1 = 0, with respect to đ?‘Ą, and should equate

−đ?‘&#x;2

−đ?‘&#x;2

81đ?›źđ?‘’ 4đ??žđ?‘Ą

81đ?›źđ?‘’ 4đ??žđ?‘Ą

4đ?œ‹đ??žđ?‘Ąđ?‘“ 2

â&#x;š

=1

−đ?‘&#x;2 81đ?›źđ?‘’ 4đ??žđ?‘Ą

4đ?œ‹đ??žđ?‘Ąđ?‘“ 2

(

with respect to đ?‘Ą, one gets −2đ?‘&#x; 4đ??žđ?‘Ą

Ă—0 +

đ?‘&#x;2 4đ??žđ?‘Ą 2

−

1 đ?‘Ą

)=0

4đ?œ‹đ??žđ?‘Ąđ?‘“ 2

â&#x;š

đ?‘&#x;2 4đ??žđ?‘Ą 2

(

−2đ?‘&#x; đ?œ•đ?‘&#x; 4đ??žđ?‘Ą đ?œ•đ?‘Ą

đ?œ•đ?‘&#x;

to zero.

đ?œ•đ?‘Ą

+

đ?‘&#x;2 4đ??žđ?‘Ą 2

1

1

− )=0 đ?‘Ą

đ?‘&#x; 2 = 4đ??žđ?‘Ą .

− =0 â&#x;š đ?‘Ą

This means maximum value of đ?‘&#x;, at which đ?‘› is zero, is đ?‘&#x; = √4đ??žđ?‘Ą. −đ?‘&#x;2

81đ?›źđ?‘’ 4đ??žđ?‘Ą 1

Upon substituting đ?‘&#x; = √4đ??žđ?‘Ą, in đ?‘“ = ( 81đ?›źđ?‘’ −1 1 ( )2

4đ?œ‹đ??žđ?‘Ą

)2 , one gets 1â „ 2

81đ?›ź Ă— 0.36778 1 29.79018đ?›ź )2 = ( ) 4đ?œ‹đ??žđ?‘Ą 4 Ă— 3.14159 Ă— đ??žđ?‘Ą 12.56636 Ă— đ??žđ?‘Ą đ?›ź 1â „ So maximum radius is obtained if radio frequency is đ?‘“ = 1.53968 Ă— ( ) 2 đ?‘“ =

â&#x;š đ?‘“= (

đ??žđ?‘Ą

Upon substituting

đ?‘&#x; = √4đ??žđ?‘Ą , in đ?‘ =

đ?›ź 4đ?œ‹đ??žđ?‘Ą

−đ?‘&#x;2

đ?‘’ 4đ??žđ?‘Ą , one gets

= 1.53968 Ă— (

đ?‘ =

đ?›ź 4đ?œ‹đ??žđ?‘Ąđ?‘’

đ?›ź 1â „ ) 2 đ??žđ?‘Ą

,

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34

Meteorite Shooting as a Diffusion Problem (J4R/ Volume 03 / Issue 01 / 007)

which implies at time đ?‘Ą = When đ?‘ =

đ?‘“2 81

,

đ?‘Ą =

đ?›ź 4đ?œ‹đ??žđ?‘’đ?‘

81đ?›ź 4đ?œ‹đ??žđ?‘’đ?‘“ 2

, radius đ?‘&#x;, will be maximum as đ?‘&#x; = √4đ??žđ?‘Ą = √

�

đ?œ‹đ?‘’đ?‘

�

81�

and đ?‘&#x; = √ = √ 2 = 3.0798 đ?œ‹đ?‘’đ?‘ đ?œ‹đ?‘’đ?‘“

in general.

1 � ⠄2

đ?‘“

III. CONCLUSION 1) Family of curves for temperature đ?‘‰đ?‘Ą (đ?‘Ľ) and electron density đ?‘ (đ?‘&#x;, đ?‘Ą) for different constant values of đ?‘Ą consist of Gaussian curves, which broaden as √đ?‘Ą . Hence for larger values of đ?‘Ą, curves become flatter and central ordinate become smaller to have constant area under the curve for that specific value of đ?‘Ą. 2) The cylinder of zero refractive index can have maximum radius đ?‘&#x; =

1 3.0798� ⠄2

đ?‘“

.

3) Radius of cylinder of zero refractive index, will shrink to zero, when the central electron density đ?‘ = So after time đ?‘Ą =

6.445đ?›ź đ??žđ?‘“ 2

đ?›ź 4đ?œ‹đ??žđ?‘Ą

falls to

đ?‘ =

đ?‘“2 81

=

đ?›ź 4đ?œ‹đ??žđ?‘Ą

. That is, when đ?‘Ą =

81đ?›ź 4đ?œ‹đ??žđ?‘“ 2

=

6.445đ?›ź đ??žđ?‘“ 2

.

, the column of electrons ceases to act as a sharply bounded reflector. REFERENCES

[1] [2] [3] [4] [5]

Richard Habberman, “Elementary Applied Partial Differential Equations with Fourier series and Boundary value problems�, Prentice Hall Inc. G.F.Roach, “Green’s functions: Introductory Theory with Applications�, Van Nostrand Reinhold Company. Ronald N. Bracewell, “The Fourier Transform and its applications�, International edition 2000, McGraw-Hill Education. Richard A. Johnson, “ Miller & Freund’s Probability and Statistics For Engineers�, sixth edition www.meteorlab.com/METEORLAB2001dev/whatmeteorites.htm

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