1) The document discusses solving diffusion problems using Fourier transforms. It provides the solution for the temperature distribution over time resulting from an initial localized injection of heat into an infinite domain. 2) The solution is a Gaussian distribution that broadens over time according to the square root of time. This analysis is extended to model the diffusion of ionization from a meteorite shooting through the earth's atmosphere, treating it as a cylindrical diffusion problem. 3) The solution for electron density as a function of distance from the meteor trail and time since passage has the form of a Gaussian distribution, indicating the ionization diffuses away from the meteor trail over time.

1. Journal for Research | Volume 03| Issue 01 | March 2017
ISSN: 2395-7549
All rights reserved by www.journal4research.org 32
Meteorite Shooting as a Diffusion Problem
Dr. Prof. Rashmi R. Keshvani Prof. Maulik S. joshi
Professor Assist Professor
Department of Mathematics Department of Mathematics
Sarvajanik College of Engineering & Technology, Surat,
Gujarat, India
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. INTRODUCTION
It is known and can be verified that 𝑒−𝑖𝜔𝑥
𝑒−𝑘𝜔2 𝑡
satisfy heat equation
𝜕𝑢
𝜕𝑡
= 𝑘
𝜕2 𝑢
𝜕𝑥2 for all values of 𝜔 So using generalized
principle of super position, it can be shown that
𝑢(𝑥, 𝑡) = ∫ 𝑐(𝜔)
∞
−∞
𝑒−𝑖𝜔𝑥
𝑒−𝑘𝜔2 𝑡
𝑑𝜔 (1)
is solution of heat equation,
𝜕𝑢
𝜕𝑡
= 𝑘
𝜕2 𝑢
𝜕𝑥2 where −∞ < 𝑥 < ∞.
The initial condition 𝑢(𝑥, 0) = 𝑓(𝑥), will be satisfied, if 𝑢(𝑥, 0) = 𝑓(𝑥) = ∫ 𝑐(𝜔)
∞
−∞
𝑒−𝑖𝜔𝑥
𝑑𝜔.
From definitions of Fourier transform and inverse Fourier transform [1], 𝑓(𝑥) = ∫ 𝑐(𝜔)
∞
−∞
𝑒−𝑖𝜔𝑥
𝑑𝜔 implies that
∫ 𝑐(𝜔)
∞
−∞
𝑒−𝑖𝜔𝑥
𝑑𝜔 is inverse Fourier transform of 𝑓(𝑥) and 𝑐(𝜔) =
1
2𝜋
∫ 𝑓(𝑥)𝑒 𝑖𝜔𝑥
𝑑𝑥
∞
−∞
is the Fourier transform of initial
temperature function 𝑓(𝑥).
Substituting 𝑐(𝜔) =
1
2𝜋
∫ 𝑓(𝑥)𝑒 𝑖𝜔𝑥
𝑑𝑥
∞
−∞
in (1), and changing dummy variable 𝑥 to 𝑥̅ in expression for 𝑐(𝜔), now (1)
becomes
𝑢(𝑥, 𝑡) = ∫ (
1
2𝜋
∫ 𝑓(𝑥̅) 𝑒 𝑖𝜔𝑥̅
𝑑𝑥̅
∞
−∞
)
∞
−∞
𝑒−𝑖𝜔𝑥
𝑒−𝑘𝜔2 𝑡
𝑑𝜔
⟹ 𝑢(𝑥, 𝑡) =
1
2𝜋
∫ 𝑓(𝑥̅) ( ∫ 𝑒−𝑖𝜔(𝑥−𝑥̅)∞
−∞
∞
−∞
𝑒−𝑘𝜔2 𝑡
𝑑𝜔) 𝑑𝑥̅ (2)
Also it is known that 𝑔(𝑥) = ∫ 𝑒−𝑖𝜔𝑥
𝑒−𝑘𝜔2 𝑡
𝑑𝜔
∞
−∞
is inverse Fourier transform of 𝑒−𝑘𝜔2 𝑡
,
(a Gaussian Curve). So, 𝑔(𝑥 − 𝑥̅) = ∫ 𝑒−𝑖𝜔(𝑥−𝑥̅)∞
−∞
𝑒−𝑘𝜔2 𝑡
𝑑𝜔 =
√𝜋
√𝑘𝑡
𝑒
−(𝑥−𝑥̅)2
4𝑘𝑡 .
Substituting this in (2), one gets
𝑢(𝑥, 𝑡) =
1
2𝜋
∫ 𝑓(𝑥̅) (
√ 𝜋
√𝑘𝑡
𝑒
−(𝑥−𝑥̅)2
4𝑘𝑡
∞
−∞
) 𝑑𝑥̅ = ∫ 𝑓(𝑥̅) √
1
4𝜋𝑘𝑡
𝑒
−(𝑥−𝑥̅)2
4𝑘𝑡
∞
−∞
𝑑𝑥̅
It can be shown [1] that lim
𝑡→0+
√
1
4𝜋𝑘𝑡
𝑒
−(𝑥−𝑥̅)2
4𝑘𝑡 = 𝛿(𝑥 − 𝑥̅), where 𝛿(𝑥) is the Dirac delta function (Impulse function).[1] The
Dirac delta function, [2] denoted by 𝛿(𝑥), also known as impulse function, is defined as
𝛿(𝑥) = {
0 if 𝑥 ≠ 0
∞ if 𝑥 = 0
ensuring ∫ 𝛿(𝑥)𝑑𝑥 = 1
∞
−∞
.
Also ∫ 𝑓(𝑥)𝛿(𝑥 − 𝑎)𝑑𝑥 = 𝑓(𝑎)
∞
−∞
where 𝑓 is any continuous function?

2. Meteorite Shooting as a Diffusion Problem
(J4R/ Volume 03 / Issue 01 / 007)
All rights reserved by www.journal4research.org 33
II. SOLUTION OF THE PROBLEM
1) If 𝑉(𝑥) denote temperature in a bar in which the heat can flow only in the ± 𝑥 directions, then heat flow will be only
there, where gradient of temperature
𝜕𝑉
𝜕𝑥
is. The amount of heat per second, 𝐼, which can be urged along the bar , is
proportional to the temperature gradient and is inversely proportional to the thermal resistance 𝑟 of the material of the bar
per unit length.
That is 𝐼 = −
1
𝑟
𝜕𝑉
𝜕𝑥
The amount of heat accumulated in unit length per second is the difference between what flows in and what flows out, i.e.
𝜕𝐼
𝜕𝑥
.
The temperature rise is inversely proportional to thermal capacitance 𝑐 per unit length. So
𝜕𝑉
𝜕𝑡
= −
1
𝑐
𝜕𝐼
𝜕𝑥
=
1
𝑟𝑐
𝜕2
𝑉
𝜕𝑥2
So here diffusion problem, is as follows
𝜕2 𝑉
𝜕𝑥2 = 𝑟𝑐
𝜕𝑉
𝜕𝑡
, 𝑡 > 0 , − ∞ < 𝑥 < ∞ (4)
Suppose initial condition is given as 𝑉(𝑥, 0) = 𝐴 𝛿(𝑥). Here A is some constant.
So 𝑉(𝑥, 0) = 𝐴 𝛿(𝑥) implies that it is case of local injection of heat at a point.[3]
If
1
𝑟𝑐
= 𝐾 , we have
𝜕2 𝑉
𝜕𝑥2 =
1
𝑘
𝜕𝑉
𝜕𝑡
, 𝑡 > 0 , −∞ < 𝑥 < ∞ with initial condition 𝑉(𝑥, 0) = 𝑓(𝑥) = 𝐴 𝛿(𝑥)
For infinite domain diffusion problems, as discussed above, solution will be
𝑉(𝑥, 𝑡) = ∫ 𝑓(𝜀)
𝑒
−(𝑥−𝜀)2
4𝐾𝑡
√4𝜋𝐾𝑡
𝑑𝜀
∞
−∞
(5)
where 𝑓(𝑥) = 𝑉(𝑥. 0) is initial function.
So, if, for a fixed 𝑡, 𝑉(𝑥, 𝑡), is denoted by 𝑉𝑡(𝑥), then
𝑉𝑡(𝑥) = ∫ 𝑓(𝜀)
𝑒
−(𝑥−𝜀)2
4𝐾𝑡
√4𝜋𝐾𝑡
𝑑𝜀
∞
−∞
=
𝐴
√4𝜋𝐾𝑡
∫ 𝛿(𝜀) 𝑒
−(𝑥−𝜀)2
4𝐾𝑡 𝑑𝜀
∞
−∞
(6)
As for any continuous function 𝑓(𝑥), ∫ 𝑓(𝑥)𝛿(𝑥)𝑑𝑥 = 𝑓(0)
∞
−∞
,
𝑉𝑡(𝑥) =
𝐴
√4𝜋𝐾𝑡
∫ 𝛿(𝜀) 𝑒
−(𝑥−𝜀)2
4𝐾𝑡 𝑑𝜀
∞
−∞
=
𝐴
√4𝜋𝐾𝑡
𝑒
−𝑥2 𝑟𝑐
4𝑡 = 𝐴 (
𝑟𝑐
4𝜋𝑡
)
1
2⁄
𝑒
−𝑥2 𝑟𝑐
4𝑡 as 𝐾 =
1
𝑟𝑐
.
Thus for fixed time t, the temperature function
𝑉𝑡(𝑥) = 𝐴 (
𝑟𝑐
4𝜋𝑡
)
1
2⁄
𝑒
−𝑥2 𝑟𝑐
4𝑡 (7)
This implies 𝑉𝑡(𝑥) is a Gaussian distribution.
As Gaussian distribution with mean 𝜇 and standard deviation 𝜎, is defined as
𝑓(𝑥, 𝜇, 𝜎) =
1
𝜎√2𝜋
𝑒
−(𝑥−𝜇)2
2𝜎2
, the curve 𝑉𝑡(𝑥) has mean 𝜇 = 0 and variance 𝜎2
=
2𝑡
𝑟𝑐
. That is, the standard
deviation for this curve is 𝜎 = √
2𝑡
𝑟𝑐
. So curves broaden as √ 𝑡 .[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 𝛼 electrons and
positive ions per meter, which diffuse away with a diffusion coefficient 𝐾. [3]
To find electron density 𝑁 per cubic meter at a distance 𝑟 from a point on the meteor trail at a time 𝑡 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
𝜕2 𝑁
𝜕𝑥2 +
𝜕2 𝑁
𝜕𝑦2 +
𝜕2 𝑁
𝜕𝑧2 =
1
𝐾
𝜕𝑁
𝜕𝑡
(8)
Converting to cylindrical co-ordinate system, equation will be
𝜕2 𝑁
𝜕𝑟2 +
1
𝑟
𝜕𝑁
𝜕𝑟
+
1
𝑟2
𝜕2 𝑁
𝜕𝜃2 +
𝜕2 𝑁
𝜕𝑧2 =
1
𝐾
𝜕𝑁
𝜕𝑡
(9)
Considering particular value of 𝑧, and applying circular symmetry, (9) becomes
𝜕2 𝑁
𝜕𝑟2 +
1
𝑟
𝜕𝑁
𝜕𝑟
=
1
𝐾
𝜕𝑁
𝜕𝑡
(10)
as
𝜕𝑁
𝜕𝜃
= 0 and
𝜕𝑁
𝜕𝑧
= 0

3. Meteorite Shooting as a Diffusion Problem
(J4R/ Volume 03 / Issue 01 / 007)
All rights reserved by www.journal4research.org 34
As meteor enters earth’s atmosphere suddenly, here again situation is of point injection, so solution must be of the form 𝑁 =
𝑐 𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡 where 𝑟 is distance from a point on the meteor trail at a particular time 𝑡. 𝑚 and 𝑐 are constants, which can be
determined depending upon the differential equation (10) and the situations.
As 𝑁 = 𝑐 𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡,
𝜕𝑁
𝜕𝑟
= 𝑐 𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡 (
−2𝑟
4𝐾𝑡
) = 𝑐 𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡 (
−𝑟
2𝐾𝑡
) ,
⟹
𝜕2
𝑁
𝜕𝑟2
= 𝑐 𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡 (
𝑟2
4𝐾2 𝑡2
−
1
2𝐾𝑡
)
and
𝜕𝑁
𝜕𝑡
= 𝑐 (𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡 (
𝑟2
4𝐾𝑡2) + 𝑒
−𝑟2
4𝐾𝑡(𝑚𝑡 𝑚−1)) = 𝑐𝑒
−𝑟2
4𝐾𝑡 𝑡 𝑚
(
𝑟2
4𝐾𝑡2 +
𝑚
𝑡
)
Substituting these expressions in (10), one gets
𝑐
𝐾
𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡 (
𝑟2
4𝐾𝑡2
+
𝑚
𝑡
) = 𝑐 𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡 (
𝑟2
4𝐾2 𝑡2
−
1
2𝐾𝑡
) +
1
𝑟
(𝑐𝑡 𝑚
𝑒
−𝑟2
4𝐾𝑡 (
−𝑟
2𝐾𝑡
)
⟹ (
𝑟2
4𝐾𝑡2
+
𝑚
𝑡
) = (
𝑟2
4𝐾𝑡2
−
1
2𝑡
) −
1
2𝑡
⟹
𝑚
𝑡
= −
1
𝑡
⟹ 𝑚 = −1
That is 𝑁 = 𝑐 𝑡−1
𝑒
−𝑟2
4𝐾𝑡
The central electron density will be 𝑁(0, 𝑡) = 𝑐 𝑡−1
, obtained on substituting 𝑟 = 0 , must be some constant times of
𝛼
𝜋𝐾
, so
say 𝑁(0, 𝑡) = ℎ
𝛼
𝜋𝐾𝑡
where ℎ is some constant. No harm, if ℎ is selected as ℎ = 1
4⁄ ⟹ 𝑁(0, 𝑡) =
𝛼
4𝜋𝐾𝑡
will be central
electron density.
That is, the solution of (10) is, 𝑁(𝑟, 𝑡) =
𝛼
4𝜋𝐾𝑡
𝑒
−𝑟2
4𝐾𝑡 (11)
For constant 𝑡, say 𝑡 = 𝑡0 𝑁(𝑟, 𝑡0) =
𝛼
4𝜋𝐾𝑡0
𝑒
−𝑟2
4𝐾𝑡0 is function of 𝑟 only and it is Gaussian with variance 2𝐾𝑡0, that is with
standard deviation √2𝐾𝑡0 . So curves broaden as √ 𝑡 . Larger the value of 𝑡, broader the curve will be. For a fixed 𝑡, peak
density, that is central ordinate will be 𝑁 =
𝛼
4𝜋𝐾𝑡
, for that particular value of 𝑡. That is the peak density, for family of curves for
different values of 𝑡, diminishes as 𝑡−1
. Larger the value of 𝑡, smaller the peak density.
While observing such showers, refractive index of air also play vital role. The refractive index of air, denoted by 𝑛,
containing 𝑁 electrons per cubic meter, is given by 𝑛 = (1 −
81𝑁
𝑓2 )
1
2 ,
where 𝑓 denotes frequency of radio waves. [2]
One would be interested to find surface of zero refractive index.
As 𝑛 = (1 −
81𝑁
𝑓2 )
1
2 ⟹ 𝑛 = ( 1 −
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡𝑓2 )
1
2
,
𝑛 will be zero if 𝑁 =
𝑓2
81
, that is if
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡𝑓2 = 1.
⟹ 𝑛 = 0 𝑖𝑓 𝑓2
=
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡
⟹ 𝑛 = 0 if radio frequency 𝑓 = (
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡
)
1
2
So to find maximum value of 𝑟 at which 𝑛 = 0, one should differentiate
𝑛 = ( 1 −
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡𝑓2 )
1
2
= 0 ⟹
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡𝑓2 − 1 = 0, with respect to 𝑡, and should equate
𝜕𝑟
𝜕𝑡
to zero.
Upon differentiating,
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡𝑓2 = 1 with respect to 𝑡, one gets
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡𝑓2 (
−2𝑟
4𝐾𝑡
𝜕𝑟
𝜕𝑡
+
𝑟2
4𝐾𝑡2 −
1
𝑡
) = 0
⟹
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡𝑓2 (
−2𝑟
4𝐾𝑡
× 0 +
𝑟2
4𝐾𝑡2 −
1
𝑡
) = 0 ⟹
𝑟2
4𝐾𝑡2 −
1
𝑡
= 0 ⟹ 𝑟2
= 4𝐾𝑡 .
This means maximum value of 𝑟, at which 𝑛 is zero, is 𝑟 = √4𝐾𝑡.
Upon substituting 𝑟 = √4𝐾𝑡, in 𝑓 = (
81𝛼𝑒
−𝑟2
4𝐾𝑡
4𝜋𝐾𝑡
)
1
2 , one gets
𝑓 = (
81𝛼𝑒−1
4𝜋𝐾𝑡
)
1
2 ⟹ 𝑓 = (
81𝛼 × 0.36778
4 × 3.14159 × 𝐾𝑡
)
1
2 = (
29.79018𝛼
12.56636 × 𝐾𝑡
)
1
2⁄
= 1.53968 × (
𝛼
𝐾𝑡
)
1
2⁄
So maximum radius is obtained if radio frequency is 𝑓 = 1.53968 × (
𝛼
𝐾𝑡
)
1
2⁄
Upon substituting 𝑟 = √4𝐾𝑡 , in 𝑁 =
𝛼
4𝜋𝐾𝑡
𝑒
−𝑟2
4𝐾𝑡, one gets 𝑁 =
𝛼
4𝜋𝐾𝑡𝑒
,

4. Meteorite Shooting as a Diffusion Problem
(J4R/ Volume 03 / Issue 01 / 007)
All rights reserved by www.journal4research.org 35
which implies at time 𝑡 =
𝛼
4𝜋𝐾𝑒𝑁
, radius 𝑟, will be maximum as 𝑟 = √4𝐾𝑡 = √
𝛼
𝜋𝑒𝑁
in general.
When 𝑁 =
𝑓2
81
, 𝑡 =
81𝛼
4𝜋𝐾𝑒𝑓2 and 𝑟 = √
𝛼
𝜋𝑒𝑁
= √
81𝛼
𝜋𝑒𝑓2 = 3.0798
𝛼
1
2⁄
𝑓
III. CONCLUSION
1) Family of curves for temperature 𝑉𝑡(𝑥) and electron density 𝑁(𝑟, 𝑡) for different constant values of 𝑡 consist of Gaussian
curves, which broaden as √ 𝑡 . Hence for larger values of 𝑡, curves become flatter and central ordinate become smaller to
have constant area under the curve for that specific value of 𝑡.
2) The cylinder of zero refractive index can have maximum radius 𝑟 =
3.0798𝛼
1
2⁄
𝑓
.
3) Radius of cylinder of zero refractive index, will shrink to zero, when the central electron density
𝑁 =
𝛼
4𝜋𝐾𝑡
falls to 𝑁 =
𝑓2
81
=
𝛼
4𝜋𝐾𝑡
. That is, when 𝑡 =
81𝛼
4𝜋𝐾𝑓2 =
6.445𝛼
𝐾𝑓2 .
So after time 𝑡 =
6.445𝛼
𝐾𝑓2 , the column of electrons ceases to act as a sharply bounded reflector.
REFERENCES
[1] Richard Habberman, “Elementary Applied Partial Differential Equations with Fourier series and Boundary value problems”, Prentice Hall Inc.
[2] G.F.Roach, “Green’s functions: Introductory Theory with Applications”, Van Nostrand Reinhold Company.
[3] Ronald N. Bracewell, “The Fourier Transform and its applications”, International edition 2000, McGraw-Hill Education.
[4] Richard A. Johnson, “ Miller & Freund’s Probability and Statistics For Engineers”, sixth edition
[5] www.meteorlab.com/METEORLAB2001dev/whatmeteorites.htm