### Powerpoint

```How it is possible to infer too much physical meaning
from mathematical tools
By J.P. Baugher
October 2012
Ver 1.0
Let me know if you catch any mistakes!

From Poisson Equation
   f
2
For a function f
given on

2
or
3

On boundary
of 

  
g


“First, since u = constant satisﬁes the
homogeneous problem with f = 0, g = 0, it is
clear that a solution to a Neumann problem can
only be unique up to an additive constant.”

This means take the derivative of g
 
( ) ' 

g
'
2

2
 
0 

g
'
2

2
 1
   g'   f
2
Minus sign switched
normal vectors
pointing out from a
volume
   f
2
Typical field
drawing…

Presence of au seems irrelevant since disappears when
derivative taken, arbitrary value would give same
solution
 
( ) ' 

g
'
2

2
 
0 

g
'
2

2

What does solution “only unique up to an additive
constant” mean? Example given as Euclidean 2D slice
of scalar values in R3 space for center point.

  
g

10 10 10
10 10 10
10 10 10
1
1
2
1
2
3
2
3
3
11 11 12
11 12 13
12 13 13

Second derivative…
 
( ) ' 

g
'
2

2
10 10 10
10 10 10
10 10 10
Regardless of
magnitudes, no
1
1
2
1
2
3
2
3
3
11 11 12
11 12 13
12 13 13
These become
identical second
derivatives

But our understanding is metaphysical. We are
implying more certainty than our tool allows…

  
 g


  
 g


Second derivative…

  
 g

10 10 10
10 10 10
10 10 10
Regardless of
magnitudes, no
1
1
2
1
2
3
2
3
3
11 11 12
11 12 13
12 13 13
These are identical
second derivatives

  
 g

10 10 10
10 10 10
10 10 10
Regardless of
magnitudes, no   0
but we could model
this as….
9
9
8
1
1
2
9
8
7
1
2
3
8
7
7
2
3
3
…however the directional
derivatives are still identical
whether au is taken into
consideration or not.

2

      f
 


Theorem also implies too much certainty…
1


 
2
(
C

)


Integration of a constant over a
differential volume has no effect…
   
2



f
Not a problem since second derivatives
identical…but…
There is another relationship that
must hold to demonstrate that there
are no unique solutions…so not only...
1


 
2
(
C

)


…but also (need to preserve total u in
volume, scalar values of u important)...
1

 
2
(
C

)


This means that although the theorems are formulaically
correct, the most we can agree on is that, by convention, the
“field” is directed towards or away from the source since it
could be considered as either attracting…
…or reducing a repulsion. If au exists, what
does it physically represent?
0
To state definitively that some physical parameter
represented by function f causes, through action at a
distance, an “attraction” represented by the Poisson equation
cannot be supported by the equations themselves. This is a
metaphysical description brought about by ignoring au.
   f
2
Is f over “there”causal
to the appearance (from
nothing) of this vector
“here”? or…
0
is f over “there” causal
to a change to what is
Why is this important?

General Relativity…

For regular General Relativity, the Einstein tensor
Guv is equated to the stress energy tensor of matter
kTuv. The T00 term is modeled to be energy density
from a perfect fluid tensor. In the weak field
approximation to Newtonian gravity, we now must
use the Poisson equation to derive our
understanding of how gravity is “attractive”. So…
1
R  Rg   G   T
2
Why is this important?

T00 term only for velocities much slower than c
we can infer an approximation…
T  T00
…so that we get Poisson’s equation for gravity…
 T00  4 G
2
Why is this important?


Great approximation to “attractive” Newtonian
gravity for most of 20th century as long as one
does not worry about “dark matter”…
Cosmological Constant problem

General Relativity is derived so that in the presence
of matter (energy, mass) there is curvature, but if this
mass or energy vanishes, so too must the
curvature…
R  0
Why is this important?

Although there is much debate about the
cosmological constant, in Unimodular
relativity it is known simply as a constant of
integration…
1
R  Rg   G  g 
2
For the derivation of GR, this  term must be set
kTuv.
Why is this important?

It became apparent through observations
though, as accurate as GR is, for some reason
there appears to be a tiny value for .
And so, in spite of the mathematical rigor used to
derive the field equation using Ruv=0, our
understanding demanded that we must consider
it possible for  to have a value. In the late 1960s,
Zel’dovich postulated that perhaps it represented
an energy density of the vacuum. Although it is
possible to measure the value there are no accepted
methods to understand either its magnitude or even
physical presence.
Why is this important?

Using our previous argument of the Poisson
equation and Unimodular Relativity, we ask:
Using Unimodular theory, let us define a multiple
of the metric as
g  G  L
where Guv is the Einstein tensor and Luv is a
tensor we name the Lorentz tensor.
Why is this important?
We can see that if =0, then the negative of the
Lorentz tensor is equivalent to the Einstein tensor
(reduces to GR)..
L  G
Thus let us enforce Ruv=0, so that the field
equation can be stated as
1
R  Rg   G  g   L
2
Why is this important?
Note that since we are enforcing Ruv=0, no
extraneous multiple of the metric, such as  is
allowed. Therefore no “vacuum solutions” or
cosmological models such as deSitter space exists
within this field equation…
This model presents the same challenge that was
required of Guv, explaining the Newtonian
approximation. Although I have made much
progress, there will probably be great push back
on explanation of the perfect fluid tensor. The
end goal is to compare rigorously…
G  G00
g  L  g00  L00
…approximates to…
…approximates to…

  
 g


  
g

…for example…
10 10 10
10 10 10
10 10 10
  0
…for example…
0
1
1
2
1
1
2
1
2
3
1
2
3
2
3
3
2
3
3
9
9
8
1
1
2
9
8
7
1
2
3
8
7
7
2
3
3
Between these two approximations, we
can develop an understanding of why
Equation from General Relativity: An Introduction for Physicists
, Hobson, Efstathiou, Lasenby.
For a spherical mass M (note that the r vectors are unit vectors)
GM
c r
g     2 r 
r
r
3
2
We have  from modified EFE. Why is the first vector
so many orders of magnitude larger than the second?
Why isn’t the vector from  exactly zero? What does it
represent? What is dark energy?
g   
A
0
is not derived as being completely
independent of radius. If this also applies
to  then vector A may decrease in magnitude at a
greater rate than Newtonian. When a mass
achieves the radius of where vector A is zero, what
occurs at radius greater than this?
Does gravity become repulsive after:
6G  V 1/3
r(
)
2
c
res
```