- #1

ObsessiveMathsFreak

- 406

- 8

Ok the above statement is not STRICTLY true. It should read...

"The gravity of a light emmitting star will appear to decrease as 1/r, at large distances"

I think this is what MOND theory holds but has never actually said why. This is a proof I discovered to possibly explain this phenomenon.

God I hope these scripts work.

----------------------------------------------------------------------

OK my bold assumption is as follows. Light, or photons, exerts a gravitational force on the universe around it. A photon with energy E has associated with it a gravitational acceleration

a

where m = E/c

I'll sight gravitational lensing and redshift as examples of this phenomenon. In both cases the momentum of the light photons has been changed. Direction has changed in the former, and magnitude has changed in the latter.

In both phenomena the momentum of the photons has been changed and so by Newtons second law the mass that shifted them, usually a star, galaxy, or black hole, will experience an equal and opposite change in momentum.

This is my justification, flimsy as it may be for setting light to have gravity.

Imagine a star,alone in space, no planets or space dust about it. Now imagine a sphere of radius

Star..............Mass

000

000-------------------------------------------------------------0

000

<-------------------------------a------------------------------->

We know that the Force of Gravity on the mass

F

Imagine the sphere, of radius

Our first responce, nothing! Empty space. But of course if we look more closely we realize that there is indeed something in this void. The light from the star!

If light has gravity as we have supposed then what is the gravity on the object on the edge of the sphere, due to the "mass" of the light within this sphere.

Let us call this supposed force F

N.B. From here on in the maths gets really hairy so if you hate calculus the just skip to the end.

It is known that

M = ∫∫∫ ρ dV , where ρ is the density at a point.

Thus M

but what is ρ

We can regard ρ

So ρ

Where I is the Intensity of the light at that point. Convince yourself of this, or convince me otherwise! :E

Note : I tried to find a better symbol for proportionality but all I came up with was α

So take it that

y α x means

y = kx , for some constant k.

We know that I α 1/r

where r is the distance from the light source.

thus ρ

ρ

k is an arbitrary constant

thus

M

M

it is more convinient for us to use spherical co-ordinates in this integral so we use the fact that

dxdydz = r

r - radius from origin

Θ - co-latitude

φ - latitude

----------------------------------------------------------------------------

Note to the Side:

I'm unable to find a good script for the limits of integration so I'm using this convention. The integral from x=a and x=b is

∫

----------------------------------------------------------------------------

and so

M

M

Subing in the usual limits for φ and Θ and r ranges from 0 to

M

M

M

M

M

M

M

M

M

So the mass of the sphere of light is 4ak[pi]

where a is the radius, and k is the coefficient of E.M. intensity

So the gravity due to this collection of shells is the gravity due to a point mass of the same mass as the sphere, at its centre. This is due to Newton's shell theorem.

Thus

F

F

F

And since 4,k, and m are constant

FL α 1/a

but Fs α 1/a2

Thus F

For our own sun this occurs at about 7.537 * 10

Or maybe you should take this whole post with a pinch of salt, a very large one at that! :E

"The gravity of a light emmitting star will appear to decrease as 1/r, at large distances"

I think this is what MOND theory holds but has never actually said why. This is a proof I discovered to possibly explain this phenomenon.

God I hope these scripts work.

----------------------------------------------------------------------

OK my bold assumption is as follows. Light, or photons, exerts a gravitational force on the universe around it. A photon with energy E has associated with it a gravitational acceleration

a

_{p}= Gm/d^{2}where m = E/c

^{2}because E=mc^{2}I'll sight gravitational lensing and redshift as examples of this phenomenon. In both cases the momentum of the light photons has been changed. Direction has changed in the former, and magnitude has changed in the latter.

In both phenomena the momentum of the photons has been changed and so by Newtons second law the mass that shifted them, usually a star, galaxy, or black hole, will experience an equal and opposite change in momentum.

This is my justification, flimsy as it may be for setting light to have gravity.

Imagine a star,alone in space, no planets or space dust about it. Now imagine a sphere of radius

**a**surrounding the star and a mass just hanging on the edge of this sphere, as badly shown.Star..............Mass

000

000-------------------------------------------------------------0

000

<-------------------------------a------------------------------->

We know that the Force of Gravity on the mass

**m**due to the star of mass**M**is_{s}F

_{s}= GM_{s}m/a^{2}Imagine the sphere, of radius

**a**. What is contained within this sphere?(Apart from the star at its center of course.Our first responce, nothing! Empty space. But of course if we look more closely we realize that there is indeed something in this void. The light from the star!

If light has gravity as we have supposed then what is the gravity on the object on the edge of the sphere, due to the "mass" of the light within this sphere.

Let us call this supposed force F

_{L}N.B. From here on in the maths gets really hairy so if you hate calculus the just skip to the end.

It is known that

M = ∫∫∫ ρ dV , where ρ is the density at a point.

Thus M

_{L}= ∫∫∫ ρ_{L}dVbut what is ρ

_{L}, the density of the light at a point.We can regard ρ

_{L}as either the mass or energy density at a point. But the energy density is the same as the Intensity of the light at that point.So ρ

_{L}= IWhere I is the Intensity of the light at that point. Convince yourself of this, or convince me otherwise! :E

Note : I tried to find a better symbol for proportionality but all I came up with was α

So take it that

y α x means

y = kx , for some constant k.

We know that I α 1/r

^{2}where r is the distance from the light source.

thus ρ

_{L}α 1/r^{2}ρ

_{L}= k/r^{2}k is an arbitrary constant

thus

M

_{L}= ∫∫∫ k/r^{2}dVM

_{L}= ∫∫∫ k/r^{2}dxdydzit is more convinient for us to use spherical co-ordinates in this integral so we use the fact that

dxdydz = r

^{2}SinΘ drdΘdφr - radius from origin

Θ - co-latitude

φ - latitude

----------------------------------------------------------------------------

Note to the Side:

I'm unable to find a good script for the limits of integration so I'm using this convention. The integral from x=a and x=b is

∫

_{x=a..b}dx----------------------------------------------------------------------------

and so

M

_{L}= ∫∫∫ kr^{-2}r^{2}SinΘ drdΘdφM

_{L}= ∫∫∫ k*SinΘ drdΘdφSubing in the usual limits for φ and Θ and r ranges from 0 to

**a**M

_{L}= ∫_{φ=0..2[pi]}∫_{Θ=0..[pi]}∫_{r=0..a}k*SinΘ drdΘdφM

_{L}= ∫_{φ=0..2[pi]}∫_{Θ=0..[pi]}r*k*SinΘ |_{r=0..a}dΘdφM

_{L}= ∫_{φ=0..2[pi]}∫_{Θ=0..[pi]}ak*SinΘ dΘdφM

_{L}= ∫_{φ=0..2[pi]}-ak*CosΘ|_{Θ=0..[pi]}dφM

_{L}= ∫_{φ=0..2[pi]}-ak*Cos([pi]) + ak*Cos(0) dφM

_{L}= ∫_{φ=0..2[pi]}2ak dφM

_{L}= 2akφ |_{φ=0..2[pi]}M

_{L}= 4ak[pi] - 2ak(0)M

_{L}= 4ak[pi]So the mass of the sphere of light is 4ak[pi]

where a is the radius, and k is the coefficient of E.M. intensity

So the gravity due to this collection of shells is the gravity due to a point mass of the same mass as the sphere, at its centre. This is due to Newton's shell theorem.

Thus

F

_{L}= GM_{L}m/a^{2}F

_{L}= 4ak[pi]m/a^{2}F

_{L}= 4k[pi]m/aAnd since 4,k, and m are constant

FL α 1/a

but Fs α 1/a2

Thus F

_{L}although weaker initally will eventually become greater than F_{m}For our own sun this occurs at about 7.537 * 10

^{14}m from the sun, but my calculations are VERY rough so take that with a pinch of salt.Or maybe you should take this whole post with a pinch of salt, a very large one at that! :E

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