Pi and universe structure, gravity

[PatrickGarrow17]

Was pondering gravity as it relates to orbit...

 

Since, most orbits are circular in nature (or better, elliptical), wouldn't it follow that somehow the ratio pi is inherent in the force of gravity.

 

I've seen pi present in the equations for general relativity and the cosmological constant, and was hoping to get some insight on the significance of the number as it relates to the structure of the universe.

 

• The cosmological constant:

• Einstein's field equation of general relativity:

 

A circular rotational motion and spherical structure seems to be a pattern from the atom to the star to the galaxy...

 

Can anyone here tie this all a little tighter for me?

Edited August 20, 2012 by PatrickGarrow17

[mathematic]

I can't comment on the expressions you showed. However when π appears, it usually means something has been averaged over a circle or a sphere.

[O'Nero Samuel]

[quote name=PatrickGarrow17'

timestamp='1345415048' post='697938]

Was pondering gravity as it relates to orbit...

 

Since, most orbits are circular in nature (or better, elliptical),

wouldn't it follow that somehow the ratio pi is inherent in the force of

gravity.

 

I've seen pi present in the equations for general relativity and the

cosmological constant, and was hoping to get some insight on the

significance of the number as it relates to the structure of the

universe.

 

• The
  cosmological
  constant:

• Einstein's
  field equation of
  general
  relativity:

 

A

circular rotational motion and spherical structure seems to be a

pattern from the atom to the star to the galaxy...

 

Can anyone here tie this all a little tighter for me?

To add to your speculations, if you square the value of pi you'd get approximately g(9.876...). Why? Coincidence? I doubt.

[PatrickGarrow17]

To add to your speculations, if you square the value of pi you'd get approximately g(9.876...). Why? Coincidence? I doubt.

 

Holy Christmas Batman. Interesting, although doesn't that value apply only on Earth? Looking for answers on a larger scale.

 

Did a bit of reading today, a book called New Theories of Everything by John Barrow. Last chapter is called "Is 'pi' really in the sky?"

 

Ended up being a disappointment as it relates to pi specifically, but was a nice argument for mathematics in the quest for universal formula.

[mathematic]

Pi squared = g is a coincidence. Pi is a pure number, g depends on the units used for distance, etc. For example in "English" units, g ~ 32.

[elfmotat]

The reason the Einstein Field Equations (and therefore the Cosmological Constant equation you give) have a factor of pi is merely a matter of (historical) convention. When Newton first wrote down his equation for the force of gravity, [math]F=Gm_1m_2/r^2[/math], he included a proportionality constant "G." Instead of breaking this constant up into multiple other constants, he used the logical (at the time anyway) choice of only one constant.

 

Contrast this with Coulomb's Law in electrostatics: [math]F=q_1q_2/4\pi \epsilon_0 r^2[/math]. Instead of using one constant of proportionality, it uses three: 4, pi, and the electrostatic constant (although some introductory textbooks absorb them into a single constant, [math]k=1/4\pi \epsilon_0[/math]).

 

The reason for using three constants instead one is because it simplifies the following Maxwell equation: [math]\nabla \cdot \bold{E}=\rho_q /\epsilon_0[/math]. This is equivalent to Coulomb's Law in simplifying cases, but it's also much more general. The factor of 4*pi appears in Coulombs law when the above Maxwell equation is applied to a spherically symmetric surface (because the surface area of a sphere is 4*pi*radius²).

 

Now look at the analogous equation for gravity: [math]\nabla \cdot \bold{g}=4\pi G\rho_m[/math]. This equation is, again, more general than Newton's force law. The choice of a single constant simplified his force law, but that means an additional factor of 4*pi must be added to the above equation so that when it is applied to a spherically symmetric surface, the 4*pi cancels.

 

When the Einstein Field Equations were first written down (taking the cosmological constant to be zero for simplicity), they looked as follows: [math]R_{\mu \nu }-g_{\mu \nu }R/2=\kappa T_{\mu \nu }[/math]. [math]\kappa[/math] is again some constant of proportionality. Since General Relativity is, in a sense, a generalization of Newtonian gravity, this constant must be chosen so that the EFE's reduce to the gravity equation I wrote above in the low-energy limit. What you find is that [math]\kappa=8\pi G/c^4[/math].

 

So if Newton had originally chosen to break up his proportionality constant into something like [math]F=Gm_1m_2/\pi r^2[/math], then the EFE's would look like [math]R_{\mu \nu }-g_{\mu \nu }R/2=8G T_{\mu \nu }/c^4[/math]. So, in summary, the pi appears because of an arbitrary choice for the value of G. If G were chosen differently, the factor of pi wouldn't be there.

[PatrickGarrow17]

If I'm following correctly, your saying that changing G would take the character pi out of the equation, but pi would still be present as a factor within the alternate G?

 

Thanks for the reply.

[O'Nero Samuel]

The reason the Einstein Field Equations (and therefore the Cosmological Constant equation you give) have a factor of pi is merely a matter of (historical) convention. When Newton first wrote down his equation for the force of gravity, [math]F=Gm_1m_2/r^2[/math], he included a proportionality constant "G." Instead of breaking this constant up into multiple other constants, he used the logical (at the time anyway) choice of only one constant.

 

Contrast this with Coulomb's Law in electrostatics: [math]F=q_1q_2/4\pi \epsilon_0 r^2[/math]. Instead of using one constant of proportionality, it uses three: 4, pi, and the electrostatic constant (although some introductory textbooks absorb them into a single constant, [math]k=1/4\pi \epsilon_0[/math]).

 

The reason for using three constants instead one is because it simplifies the following Maxwell equation: [math]\nabla \cdot \bold{E}=\rho_q /\epsilon_0[/math]. This is equivalent to Coulomb's Law in simplifying cases, but it's also much more general. The factor of 4*pi appears in Coulombs law when the above Maxwell equation is applied to a spherically symmetric surface (because the surface area of a sphere is 4*pi*radius²).

 

Now look at the analogous equation for gravity: [math]\nabla \cdot \bold{g}=4\pi G\rho_m[/math]. This equation is, again, more general than Newton's force law. The choice of a single constant simplified his force law, but that means an additional factor of 4*pi must be added to the above equation so that when it is applied to a spherically symmetric surface, the 4*pi cancels.

 

When the Einstein Field Equations were first written down (taking the cosmological constant to be zero for simplicity), they looked as follows: [math]R_{\mu \nu }-g_{\mu \nu }R/2=\kappa T_{\mu \nu }[/math]. [math]\kappa[/math] is again some constant of proportionality. Since General Relativity is, in a sense, a generalization of Newtonian gravity, this constant must be chosen so that the EFE's reduce to the gravity equation I wrote above in the low-energy limit. What you find is that [math]\kappa=8\pi G/c^4[/math].

 

So if Newton had originally chosen to break up his proportionality constant into something like [math]F=Gm_1m_2/\pi r^2[/math], then the EFE's would look like [math]R_{\mu \nu }-g_{\mu \nu }R/2=8G T_{\mu \nu }/c^4[/math]. So, in summary, the pi appears because of an arbitrary choice for the value of G. If G were chosen differently, the factor of pi wouldn't be there.

You mean G is an arbitrary constant? Mean as a matter of choice, it would have been able to do away with pi? And at the same time, 4 pi Square makes up part of the three constant of the Coulomb force? That is really something to look into. From general outlook, pi comes to play whenever we have to resolve orbits. Gravitation is resolved in orbits, how does pi play an arbitrary role in the gravitational constant? And, what could it be replaced with if we were to use "three constants"?