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By Hontas Farmer | February 23rd 2009 10:44 AM | 5 comments | Print | E-mail | Track Comments
A few months ago I did a bit of thinking about gravity from a different angle.  I asked the question... How would the equations of classical General Relativity look if Planck Units were used.  Let me make it clear I am not not talking about the Planck Scale.. just the use of it's units at the classical scale.  Of course anywhere near Planck Scale Quantum effects dominate.  I have already written a book about that.  "Quantum Space-Time Dynamics"
by Hontas Farmer .         This came to me after writing the book and finalizing it.  Consider the following. 

The Planck Units are defined using three universal physical constants.....the reduced Planck constant ħ, the speed of light c, and the gravitational constant G.  If ħ=c=1  then they become  Planck Length = √G,  Planck time= √G, and Planck mass = 1/√G. 

For this blog posting just consider Newton's law of gravity.  Which appears as part of general relativity in the weak field limit, and Schwarzchild geometry, such as in the solar system. (Basically any place but near a massive black hole.)

The potential of Newtonian gravity .  V=-GM/r  is supposed to have units of energy.  But if the Planck units are used. 
V= - G(1/√G)/√G  = -√G/√G=1  !  Or the dimensions disappear!   This is not supposed to happen unless all three base units are set equal to one all quantities should retain their units in the new system of measurement.   So I propose the following modification to classical gravity when planck units are used.

V=-GM/(√G r).

When this is done and planck units are plugged in V=-(G/√G)/G= -√G/G=-1/√G  The correct unit of measure for potential energy.  Now I know it is not a trivial matter to modify such a well tested law of nature.  However the concern I raise is such a simple and fundamental point of science it cannot be totally ignored. 

Comments

One area of dimensionalisation where a classical length, area or volume can be evaluated in Planck units is in the formula (Mc^2)/h. On its own it represents the number of Planck areas [4pi(Planck radius)^2,] on the surface of a sphere where surface gravity is equal to c and the mass is given by M. Beware, we cannot, necessarily, determine escape speed from this formula. To continue, the Planck frequency of such a sphere or its radius in Planck radii is the square root of the given formula above. From this we can determine that the Planck volume is 8pi(r)^3, where r=Planck radius, around, 4.05084x10^-35m. From these two values, c & h, we can build a fascinating assembly of mass models involving Planck units and gravity but tying them down to our own SI system is elusive.

Hfarmer's picture
Yes.  That is why when we see a formula in a text book written with the standard SI units we have to be careful to check that it still works in whatever system of units we actually work in.  In real life SI units are not always convenient. 

For example, last year I took a class on computational physics.  I choose as my project modeling the relativistic perihelion shift in Mercury's orbit.  I found that the simulation would only work and not crash my computer if I switched from SI units... to a system of units where GM=4Pi.   :-?  I know but it worked and I got a shifting perihelion with the right magnitude and everything.

You also cannot miss a point in my blog posting.  In it I choose to set c and Hbar equal to one for convinience.  In that system of units Newton's clasic formula needs a rescaling.  That is the point of this post.  Nothing more. 

"The potential of Newtonian gravity . V=-GM/r is supposed to have units of energy."

That is incorrect. It's energy/mass.

Hfarmer's picture
You would be right but you have ignored the context in which I used the term potential.  In this context the term potential refer's to potential energy.  It is a common practice to not write the value of the so called "test mass" into the formula.  The test mass is the mass that is mostly just feeling the potential of some much larger mass.  (consider a 1 kg lead weight in orbit of the sun...get it).  With the test mass inserted explicitly V=-GMm/r  where "m" is the test mass I speak of.

If you were really smart you would have read one line farther  to where I really did make a mistake
"But if the Planck units are used. V= - G(1/√G)/√G  = -√G/√G=1  !  Or the dimensions disappear! "  Then you would have pointed out that if I insert a appropriate test mass the potential engery then has it's correct units of mass-energy which in Planck units are the same quantity.

The second commentor got the point of the post which is that one has to look carefully at the units used in books (and blog postings) consider their context, and be sure they are on the same page as the person doing the writing.

Wow, this is absolutely brilliant. I nominate Dr. Hontas for the next Nobel in physics!

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