Quantum Physics Quantum Physics Help Forum 
Nov 13th 2009, 06:43 PM

#1  Junior Member
Join Date: Nov 2009
Posts: 4
 Is my Math ok?
Just need to know if the math is ok. It's good i think to check work like this with alot of people, because someone might see something another won't.
F_g=▼φM_g (1)
(F_gvt)²=▼²φ²(ћ(c/G))_g v²t² (2)
M²=ћ(c/G) (3)
ћc=GM² (4)
β=v/c (5)
pc=E(v/c) (6)
i²=ξ=+1 (7)
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξε_0(M²ψM²ψ]+½[ξε_g(M²ψ*M²ψ*]) (8)  Charged equation in curved spacetime
A=e^i ∫d^4 x(½[ξε_0(M²ψM²ψ]+½[ξε_g(M²ψ*M²ψ*]) (9)
(vortex) k=Dφ*,Dφ* (9)
and ∫k=D_{φ*,φ*}(x,1... x,2)... D_φ*(x_n) which is also equal to =
∞
∏ D_{φ*,φ*}(x_i) (6)
n=i
so the fourvortex is given as:
k²=Dφ*Dφ*DφDφ (7)
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i∫d^4 x(½[ξε_0(∂t(ψ)+▼²ψ(∂t(ψ)+▼²ψ)]+½[ξε_g(∂t(ψ)+▼²ψ*(∂t(ψ)+▼²ψ)]) (8)  plane wave solutions
ξ1,ξ1_{ε_g,ε_0} (8)
Substution
μ1,μ2_{g,e,0} (9)
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i∫d^4 x(½[μ_0(∂t(ψ)+▼²ψ)(∂t(ψ)+▼²ψ)]+½[μ_g(∂t(ψ)+▼²ψ*(∂t(ψ)+▼²ψ)]) (10)  electromagnetic and graviatational charge
Epsilon ε in the plane wave solution is itself a tensor so has a factor of √g' which is respectively negative. In mathematical terms, its the determinent of the metric tensor.
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξ√g'_0(M²ψM²ψ)]+½[ξ√g'_g(M²ψ*M²ψ*)]) (11)
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξ√g'_0((∂²/∂t²Δ + M²)ψ(∂²/∂t²Δ + M²)ψ)]+½[ξ√g'_g((∂²/∂t²Δ + M²)ψ*(∂²/∂t²Δ + M²)ψ*]) (12)
Where (∂²/∂t²Δ + M²)ψ = 0 (13)
→ = g_μν∂^ν∂^μ (14)
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 δ(xx')(½[ξ√g'_0((∂²/∂δt²Δ + M²)ψ(∂²/∂δt²Δ + M²)ψ)]+½[ξ√g'_g((∂²/∂δt²Δ + M²)ψ*(∂²/∂δt²Δ + M²)ψ*]) (14)
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξε_0(M²ψM²ψ]+½[ξε_g(M²ψ*M²ψ*]) (15)
is balanced, because it takes into respect the electromagnetic permittivity added with that of the gravitational permittivity with a Langrangian term for M². More interestingly enough, M²ψ is similar to the KleinGorden relationship. Here are some interesting reationships:
M²ψ=∂t(ψ)+ ▼²ψ (16)
which results in plane wave solutions. By substitution, you can reconfigurate eq.(1) into:
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξε_0(∂t(ψ)+ ▼²ψ)(∂t(ψ)+ ▼²ψ)]+½[ξε_g(∂t(ψ)+ ▼²ψ*)(∂t(ψ)+ ▼²ψ)]) (17)
Which is very attractive as a wave equation.
We could manipulate the equation even more to have nuetral components after taking ino account, from a KleinGorden relationship, where for manipulative convenience we can rewrite the plane wave solutions in quantized form as:
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξε_0((∂²M²)ψ*(∂²M²)]+½[ξε_g((∂²M²)ψ*(∂²M²)ψ*]) [18]
This is I suppose the mechanism which would cancel them out, or at least, this is my interpretation of the equation.
Nuetral Charge in Linear Metric Space
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i∫d^4 x(½[μ_0(∂t(ψ)+▼²ψ)(∂t(ψ)+▼²ψ]+½[μ_g(∂t(ψ)+▼²ψ*(∂t(ψ)+▼²ψ]) [19]
Assuming not only a small box space d^3, we are using the notion of d^4 which takes into respect an equally tiny amount of time. This will be given the form as:
(∫F_g vt)²_<A_k²>=∫▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 δ(xx')(½[ξ√g'_0((∂²/∂δt²Δ + M²)ψ(∂²/∂δt²Δ + M²)ψ)]+½[ξ√g'_g((∂²/∂δt²Δ + M²)ψ*(∂²/∂δt²Δ + M²)ψ*]) (20)
Any actions that can come from this are restricted by the presence of d^4 δ(xx'). Finding this in the equation however has itself a bit of interesting qualities. The term d^4(x) in the equation above Some has the products of a 0th, 1st, 2nd and 3rd component, or dx^0dx^1dx^2dx^3 which inexorably makes it a zeroeth rank tensor. However, integral to it is in fact displacement from one position x to another x'. This immediately makes this part a vector quantity, and thus, the entire configuaration d^4 δ(xx') as rank 1 tensor.
There is also a rank 3 tensor in the equation
eq.1
.........∞
E²/c² ∏_(3)p²(x_i)=U(xn)
........n=i
Where U is a scalar quantity, definiting presicely the rest mass on the left hand side under Lorentz Covariantrelations. Since the expression U(x) describes the scalar mass quantity at some point, then the contraction of the tensor is calculated as:
eq.2
.........i
U(x)=∑ T
......i..i
Using Einstein notation. The right hand side of equation 1 can now be equated to exactly the value of equation 2, with the interesting geometrical formation of the left hand side of equation 1.
The rank 3 tensor can be found when scaling out a matrix for the threemomentum as a noncovariant vector:
(p,x)
(p,y)=η_νμ U^μ
(p,z)
The final equation involving the substitutions of equations 1. and eq 2 we can derive a much more simpler form of:
= ∫▼²φ²(ћ(c/G))_g β²t²(e^i U(x_{1,2})A
Where A is simply
A=(½[ξ√g'_0((∂²/∂δt²Δ + M²)ψ(∂²/∂δt²Δ + M²)ψ)]+½[ξ√g'_g((∂²/∂δt²Δ + M²)ψ*(∂²/∂δt²Δ + M²)ψ*])
Thanks in advance

 
Nov 13th 2009, 10:18 PM

#2  Physics Team
Join Date: Feb 2009
Posts: 1,425

You had better send a P.M to pmb for this. Such stuff is right up his sleeve!

 
Nov 14th 2009, 12:10 AM

#3  Junior Member
Join Date: Nov 2009
Posts: 4

Originally Posted by physicsquest You had better send a P.M to pmb for this. Such stuff is right up his sleeve! 
It' my first time here. I am not sure who he is.

 
Nov 14th 2009, 01:49 AM

#4  Physics Team
Join Date: Feb 2009
Posts: 1,425

He is one of the helpers and is very proficient in gen rel and the like.
From what i know, he is best suited to help you out on this

 
Nov 18th 2009, 12:39 PM

#5  Physics Team
Join Date: Apr 2009 Location: Boston's North Shore
Posts: 1,573

Originally Posted by physicsquest He is one of the helpers and is very proficient in gen rel and the like.
From what i know, he is best suited to help you out on this 
I'll take a look at it. Seems like some heavy stuff and as such I may not be a lot of help here. I.e. I never heard of a 4vortex before etc.
Perhaps if you can walk me through what you're doing I can take a look. I.e. walk me through the formulas about and let me know what they mean. I.e. I don't know what the following mean/are
F_g=▼φM_g (1)
(F_gvt)²=▼²φ²(ћ(c/G))_g v²t² (2)
M²=ћ(c/G) (3)
ћc=GM² (4)
Whre did these come from? What do they mean? I.e. F_g seems to be a force of some kind. But in relativity the 4force can't always be written as you have it. Let's start there. I.e. we'll discuss these 4 equations and hopefully that will get me in the swing of the derivation.

 
Nov 21st 2009, 11:13 PM

#6  Junior Member
Join Date: Nov 2009
Posts: 4

Originally Posted by Pmb I'll take a look at it. Seems like some heavy stuff and as such I may not be a lot of help here. I.e. I never heard of a 4vortex before etc.
Perhaps if you can walk me through what you're doing I can take a look. I.e. walk me through the formulas about and let me know what they mean. I.e. I don't know what the following mean/are
F_g=▼φM_g (1)
(F_gvt)²=▼²φ²(ћ(c/G))_g v²t² (2)
M²=ћ(c/G) (3)
ћc=GM² (4)
Whre did these come from? What do they mean? I.e. F_g seems to be a force of some kind. But in relativity the 4force can't always be written as you have it. Let's start there. I.e. we'll discuss these 4 equations and hopefully that will get me in the swing of the derivation. 
F_g  force due to gravity
Since F_g is explicitely a formulation of General Relativity, (notes by sean carrol) it can be seen as the force required to move a gravitational charged mass, or can be seen as the gravitational charge on the particle itself.
ћc=GM²  this is a plankian equation. He used these forms when making his Planck Constants. The basic thing needed to be known is simple that they can easily be made to equate the units of energy, since a little rearrangement:
M²=ћ(c/G)
c√ћ(c/G) = Mc²
In the generalized particle moving in a gravitational field, i have simple multiplied (vt)(vt) on both sides,
(F_gvt)²=▼²φ²(ћ(c/G))_g v²t² (2)
Where Fvt has units of energy, so (F_gvt)² is the energy due to gravity squared. Is that ok for now?

 
Nov 23rd 2009, 05:16 PM

#7  Physics Team
Join Date: Apr 2009 Location: Boston's North Shore
Posts: 1,573

Originally Posted by Singularitarian F_g  force due to gravity 
Originally Posted by Singularitarian Since F_g is explicitely a formulation of General Relativity, (notes by sean carrol) it can be seen as the force required to move a gravitational charged mass, or can be seen as the gravitational charge on the particle itself. Where in Carroll’s notes did you see this expression? What is Phi? You never defined it.  Where did you get that expression from? The gravitational force on a particle is given by (See Basic Relativity, Richard A. Mould, page 262, Eq. (8.65) $\displaystyle G_{\mu} = m\Gamma^{\alpha}_{\mu\beta}v_{\alpha}v^{\beta}$ If you don’t have Mould then I can put up my own version of the derivation. I’m moving from Geocicities to my own site so I can easily put that derivation up.
Originally Posted by Singularitarian ћc=GM²  this is a plankian equation.  “a” Planckian equation? Where did it come from? Am I to assume you postulated this? Did someone else postulate this? Was it derived somewhere? When you mention Planck you sound like you’re getting into quantum mechanics. Force is a concept from Classical Mechanics, not quantum mechanics. The reason being is that for force to have meaning a particle must move on a classical trajectory and that doesn’t happen in quantum mechanics. Note that I don’t know quantum general relativity so this might be a very short derivation.
Originally Posted by Singularitarian He used these forms when making his Planck Constants.  That sentence is not meaningful to me. I take it that English is not your primary language? Who is “He”? Surely not Planck!
Originally Posted by Singularitarian The basic thing needed to be known is simple that they can easily be made to equate the units of energy, since a little rearrangement: M²=ћ(c/G) c√ћ(c/G) = Mc²  This too makes no sense. Why are you making these substitutions? Where did you get the idea that these things can “easily be made to equate the units of energy”???
Originally Posted by Singularitarian Is that ok for now?  Sure. But you’re not making any sense so far. You haven’t even told us whether this is classical or quantum physics yet. And if so you haven’t justified making those equalities just because the units are the same. E.g. I could multiple one side by 10^(1200000) and the units would still be right. 
 
Nov 23rd 2009, 05:40 PM

#8  Physics Team
Join Date: Apr 2009 Location: Boston's North Shore
Posts: 1,573

Singularitarian  I'm going to bow out of this conversation. To me it appears that you're attempting to dabble in quantum gravity. Since
I have yet to study quantum gravity yet I choose not to attempt to decipher someone's attempt at some sort of equation. If you want help all you have to do is to go to the physics newsgroup
sci.physics.research
and post the link to your derivation there. There is someone who posts there who is an expert, i.e. John Baez, who I may want to read your post. Warning  When you go to the internet and ask someone to review your work then it will be very hard to get help if it's a long derivation with no documentation.

 
Nov 24th 2009, 11:06 PM

#9  Junior Member
Join Date: Nov 2009
Posts: 4

Originally Posted by Pmb Singularitarian  I'm going to bow out of this conversation. To me it appears that you're attempting to dabble in quantum gravity. Since
I have yet to study quantum gravity yet I choose not to attempt to decipher someone's attempt at some sort of equation. If you want help all you have to do is to go to the physics newsgroup
sci.physics.research
and post the link to your derivation there. There is someone who posts there who is an expert, i.e. John Baez, who I may want to read your post. Warning  When you go to the internet and ask someone to review your work then it will be very hard to get help if it's a long derivation with no documentation. 
Fair do's  thank you again.

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