- #1
Mike2
- 1,313
- 0
I would like to explore the possibility that the gravitational effects of Dark Matter might possibly be accounted for by the Unruh effect applied to the acceleration of gravity. The Unruh effect predicts a temperature associated with any acceleration. And an energy density can be found for this black body radiation temperature using Planck's energy density spectum. So by the equivalence principle, the acceleration of gravity should also have an Unruh temperature whose energy density can be converted to an additional mass density. But once additional mass density has been found, the process must be repeated to account for the new mass density produced by the Unruh effect. It is expected that the continual iteration of this process will converge to some value close to that expected of Dark Matter.
This effort will try to get an order of magnitude by assuming a Newtonian gravitational field with spherical symmetry. The process will produce a series which it is hoped will converge to a recognizable value.
The Unruh effect predicts a temperature associated with any accelerating object. The formula for the temperature T of anything with acceleration "a" is":
[tex]\[
{\rm{T = }}\frac{{\hbar a}}{{2{\rm{\pi }}ck}}\,
\][/tex]
where,
[tex]\[
\hbar \, = \,\frac{h}{{2{\rm{\pi }}}}
\][/tex],
[tex]\[
\begin{array}{l}
h = 6.626 \cdot 10^{ - 34} \,(J \cdot \sec ) \\
c = 2.998 \cdot 10^8 \,(m/s) \\
k = 1.381 \cdot 10^{ - 23} \,(J/^ \circ {\rm{K)}} \\
\end{array}
\][/tex]
Notice that the expression above contains no parameters associated with matter, no variable for mass or charge or spin, etc. This formula does not seem to apply to only accelerating particles but seems to apply to acceleration in general.
The equivalence principle states that there can be no distinction between acceleration and gravitation. So I suppose that the Unruh effect should be applicable to the acceleration due to gravity as well.
So in order to find the equivalent extra mass density because of the temperature increase due to the acceleration of gravity, it is first required to find the energy density due black body radiation temperature. This can be done by integrating over all frequencies Planck's black body energy density spectrum:
[tex]\[
{\rm{p}}_{\rm{T}} {\rm{(\nu ) = }}\frac{{{\rm{8\pi }}h}}{{c^3 }}\,\frac{{{\rm{\nu }}^{\rm{3}} }}{{e^{h\nu {\rm{/}}k{\rm{T}}} \, - \,1}}
\][/tex]
This can be done by noting that:
[tex]\[
\int_0^\infty {\frac{{x^3 }}{{e^x - 1}}\,\,dx} \,\, = \,\,\frac{{\,\,\,{\rm{\pi }}^4 }}{{15}}
\][/tex]
found at: http://en.wikipedia.org/wiki/Table_of_integrals
Then,
[tex]\[
E_{\rm{T}} \, = \,\int_0^\infty {\frac{{8{\rm{\pi }}h}}{{c^3 }}\frac{{{\rm{\nu }}^3 }}{{e^{h{\rm{\nu /}}k{\rm{T}}} - 1}}\,\,d{\rm{\nu }}} \,\, = \,\,\frac{{8{\rm{\pi }}h}}{{c^3 }}(\frac{{k{\rm{T}}}}{h})^3 (\frac{{k{\rm{T}}}}{h})\int_0^\infty {(\frac{{h{\rm{\nu }}}}{{k{\rm{T}}}})^3 \frac{1}{{e^{h{\rm{\nu /}}k{\rm{T}}} - 1}}} \,(\frac{{h\,d{\rm{\nu }}}}{{k{\rm{T}}}}{\rm{) = }}\frac{{8{\rm{\pi }}h}}{{c^3 }}(\frac{{k{\rm{T}}}}{h})^4 \,\frac{{\,{\rm{\pi }}^4 }}{{15}}
\][/tex]
Or,
[tex]\[
E_{\rm{T}} = \,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 {\rm{T}}^{\rm{4}} }}{{{\rm{15}}c^3 h^3 }}
\][/tex]
When divided by [tex]\[
c^2
\][/tex], this gives a mass density as a function of temperature.
But to find the acceleration due to a mass density, we can start with
[tex]\[
\Phi _{\rm{S}} \,\, = \,\,\,\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\,{\rm{ = }}\,\,\int_{\rm{V}} {\vec \nabla \cdot {\rm{\vec g }}d{\rm{V}}} \, = \,\int_{\rm{V}} {{\rm{\rho (r) }}d{\rm{V}}}
\][/tex]
where [tex]\[
\Phi _{\rm{S}}
\][/tex] is the number of gravitational flux lines that pass through the surface S, and where g is the gravitational field in units of meters/second^2, and where [tex]\[
{\rm{\rho }}
\][/tex] mass density.
From observation, we know that [tex]\[
{\rm{g(r) = }} - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}
\][/tex]
And for spherically symmetric g, we have:
[tex]\[
\Phi _{\rm{S}} \,\, = \,\,\,\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\, = \,\,\oint_{\rm{S}} {{\rm{g }}dA} \,\, = \,\,\oint_{\rm{S}} { - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}} \,dA\,\, = \,\, - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}\oint_{\rm{S}} {dA}
\][/tex]
Or,
[tex]\[
\Phi _{\rm{S}} \, = \,\, - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }} \cdot 4{\rm{\pi r}}^{\rm{2}} {\rm{ = }} - 4{\rm{\pi GM}}
\][/tex]
And since the gravitational flux lines continue out to infinity and never cross each other, [tex]\[
\Phi _{\rm{S}}
\][/tex] will not change for any surface surrounding M which produces g.
So [tex]\[
\Phi _{\rm{S}} \, = \, - 4{\rm{\pi GM}}
\][/tex] for any surface S surrounding mass M.
In other words, we may take as a general rule:
[tex]\[
\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\, = \,\, - 4{\rm{\pi GM}}
\][/tex]
Let us apply this to an initial mass density [tex]\[
{\rm{\rho }}_{\rm{0}} {\rm{(r)}}
\][/tex] that is spherically symmetrical and constant out to a radius, R. Or:
[tex]\[
{\rm{\rho }}_{\rm{0}} {\rm{(r) = }}\frac{{{\rm{M}}_{\rm{0}} }}{{\rm{V}}}\,\, = \,\,\frac{{{\rm{M}}_{\rm{0}} }}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,\, = \,\,{\rm{constant}}
\][/tex] for r < R, and
[tex]\[
{\rm{\rho }}_{\rm{0}} {\rm{(r) = 0}}
\][/tex] for r > R.
Then the initial mass density [tex]\[
ms_0
\][/tex], for r < R is:
[tex]\[
ms_0 {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{\rho }}_{\rm{0}} {\rm{(r)}}} \,d{\rm{V = }}\int_{{\rm{V in S}}} {(\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }})\,} d{\rm{V = }}\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,\int_{{\rm{V in S}}} \, d{\rm{V = }}\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,(\frac{4}{3}){\rm{\pi r}}^{\rm{3}}
\][/tex]
Or, for r < R
[tex]\[
ms_0 = \,\frac{{{\rm{Mr}}^{\rm{3}} }}{{{\rm{R}}^{\rm{3}} }}
\]
[/tex]
And for r > R, the initial mass density is zero.
Or more generally, for spherical symmetry:
[tex]\[
{\rm{g}}_{\rm{n}} {\rm{(r)}}4{\rm{\pi r}}^{\rm{2}} \, = \,\, - 4{\rm{\pi G}}ms_n {\rm{(r)}}
\][/tex]
Or,
[tex]\[
{\rm{g}}_{\rm{n}} {\rm{(r) = }} - \frac{{{\rm{G}}ms_n {\rm{(r)}}}}{{{\rm{r}}^{\rm{2}} }}
\][/tex]
When this gravitational acceleration is applied to the Unruh temperature, we get:
[tex]\[
{\rm{T}}_n \, = \,\frac{{\hbar {\rm{g}}_{\rm{n}} {\rm{(r)}}}}{{2{\rm{\pi }}ck}}\,\, = \,\,\frac{{hG}}{{4{\rm{\pi }}^{\rm{2}} ck}}\frac{{ms_n ({\rm{r}})}}{{{\rm{r}}^2 }}
\][/tex]
and the energy density is:
[tex]\[
E_{\rm{T}} = \,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 }}{{{\rm{15}}c^3 h^3 }}(\frac{{hG}}{{4{\rm{\pi }}^{\rm{2}} ck}}\frac{{ms_n ({\rm{r}})}}{{{\rm{r}}^2 }})^4 \,\, = \,\,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 }}{{{\rm{15}}c^3 h^3 }}\,\frac{{h^4 G^4 }}{{{\rm{256\pi }}^{\rm{8}} c^4 k^4 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}\,\, = \,\,\frac{{G^4 h\,}}{{480{\rm{\pi }}^{\rm{3}} c^7 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}
\][/tex]
Then dividing by [tex]\[
c^2
\][/tex], we get a mass density:
[tex]\[
{\rm{\rho }}_{new} ({\rm{r}}) = \frac{{G^4 h\,}}{{480{\rm{\pi }}^{\rm{3}} c^9 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}\,\, = \,N\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}
\][/tex]
This new mass density (n+1) can be added to the old mass density (n) to give us an iterative process:
[tex]\[
{\rm{\rho }}_{{\rm{n + 1}}} ({\rm{r}}) = {\rm{\rho }}_{\rm{n}} ({\rm{r}}) + N\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}
\][/tex]
where,
[tex]\[
ms_n {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{\rho }}_{\rm{n}} {\rm{(r)}}} \,d{\rm{V}}
\][/tex]
which for spherical symmetry we get:
[tex]\[
ms_n {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{4\pi r}}^2 {\rm{\rho }}_{\rm{n}} {\rm{(r)}}} \,d{\rm{V}}
\][/tex]
This effort will try to get an order of magnitude by assuming a Newtonian gravitational field with spherical symmetry. The process will produce a series which it is hoped will converge to a recognizable value.
The Unruh effect predicts a temperature associated with any accelerating object. The formula for the temperature T of anything with acceleration "a" is":
[tex]\[
{\rm{T = }}\frac{{\hbar a}}{{2{\rm{\pi }}ck}}\,
\][/tex]
where,
[tex]\[
\hbar \, = \,\frac{h}{{2{\rm{\pi }}}}
\][/tex],
[tex]\[
\begin{array}{l}
h = 6.626 \cdot 10^{ - 34} \,(J \cdot \sec ) \\
c = 2.998 \cdot 10^8 \,(m/s) \\
k = 1.381 \cdot 10^{ - 23} \,(J/^ \circ {\rm{K)}} \\
\end{array}
\][/tex]
Notice that the expression above contains no parameters associated with matter, no variable for mass or charge or spin, etc. This formula does not seem to apply to only accelerating particles but seems to apply to acceleration in general.
The equivalence principle states that there can be no distinction between acceleration and gravitation. So I suppose that the Unruh effect should be applicable to the acceleration due to gravity as well.
So in order to find the equivalent extra mass density because of the temperature increase due to the acceleration of gravity, it is first required to find the energy density due black body radiation temperature. This can be done by integrating over all frequencies Planck's black body energy density spectrum:
[tex]\[
{\rm{p}}_{\rm{T}} {\rm{(\nu ) = }}\frac{{{\rm{8\pi }}h}}{{c^3 }}\,\frac{{{\rm{\nu }}^{\rm{3}} }}{{e^{h\nu {\rm{/}}k{\rm{T}}} \, - \,1}}
\][/tex]
This can be done by noting that:
[tex]\[
\int_0^\infty {\frac{{x^3 }}{{e^x - 1}}\,\,dx} \,\, = \,\,\frac{{\,\,\,{\rm{\pi }}^4 }}{{15}}
\][/tex]
found at: http://en.wikipedia.org/wiki/Table_of_integrals
Then,
[tex]\[
E_{\rm{T}} \, = \,\int_0^\infty {\frac{{8{\rm{\pi }}h}}{{c^3 }}\frac{{{\rm{\nu }}^3 }}{{e^{h{\rm{\nu /}}k{\rm{T}}} - 1}}\,\,d{\rm{\nu }}} \,\, = \,\,\frac{{8{\rm{\pi }}h}}{{c^3 }}(\frac{{k{\rm{T}}}}{h})^3 (\frac{{k{\rm{T}}}}{h})\int_0^\infty {(\frac{{h{\rm{\nu }}}}{{k{\rm{T}}}})^3 \frac{1}{{e^{h{\rm{\nu /}}k{\rm{T}}} - 1}}} \,(\frac{{h\,d{\rm{\nu }}}}{{k{\rm{T}}}}{\rm{) = }}\frac{{8{\rm{\pi }}h}}{{c^3 }}(\frac{{k{\rm{T}}}}{h})^4 \,\frac{{\,{\rm{\pi }}^4 }}{{15}}
\][/tex]
Or,
[tex]\[
E_{\rm{T}} = \,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 {\rm{T}}^{\rm{4}} }}{{{\rm{15}}c^3 h^3 }}
\][/tex]
When divided by [tex]\[
c^2
\][/tex], this gives a mass density as a function of temperature.
But to find the acceleration due to a mass density, we can start with
[tex]\[
\Phi _{\rm{S}} \,\, = \,\,\,\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\,{\rm{ = }}\,\,\int_{\rm{V}} {\vec \nabla \cdot {\rm{\vec g }}d{\rm{V}}} \, = \,\int_{\rm{V}} {{\rm{\rho (r) }}d{\rm{V}}}
\][/tex]
where [tex]\[
\Phi _{\rm{S}}
\][/tex] is the number of gravitational flux lines that pass through the surface S, and where g is the gravitational field in units of meters/second^2, and where [tex]\[
{\rm{\rho }}
\][/tex] mass density.
From observation, we know that [tex]\[
{\rm{g(r) = }} - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}
\][/tex]
And for spherically symmetric g, we have:
[tex]\[
\Phi _{\rm{S}} \,\, = \,\,\,\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\, = \,\,\oint_{\rm{S}} {{\rm{g }}dA} \,\, = \,\,\oint_{\rm{S}} { - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}} \,dA\,\, = \,\, - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}\oint_{\rm{S}} {dA}
\][/tex]
Or,
[tex]\[
\Phi _{\rm{S}} \, = \,\, - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }} \cdot 4{\rm{\pi r}}^{\rm{2}} {\rm{ = }} - 4{\rm{\pi GM}}
\][/tex]
And since the gravitational flux lines continue out to infinity and never cross each other, [tex]\[
\Phi _{\rm{S}}
\][/tex] will not change for any surface surrounding M which produces g.
So [tex]\[
\Phi _{\rm{S}} \, = \, - 4{\rm{\pi GM}}
\][/tex] for any surface S surrounding mass M.
In other words, we may take as a general rule:
[tex]\[
\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\, = \,\, - 4{\rm{\pi GM}}
\][/tex]
Let us apply this to an initial mass density [tex]\[
{\rm{\rho }}_{\rm{0}} {\rm{(r)}}
\][/tex] that is spherically symmetrical and constant out to a radius, R. Or:
[tex]\[
{\rm{\rho }}_{\rm{0}} {\rm{(r) = }}\frac{{{\rm{M}}_{\rm{0}} }}{{\rm{V}}}\,\, = \,\,\frac{{{\rm{M}}_{\rm{0}} }}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,\, = \,\,{\rm{constant}}
\][/tex] for r < R, and
[tex]\[
{\rm{\rho }}_{\rm{0}} {\rm{(r) = 0}}
\][/tex] for r > R.
Then the initial mass density [tex]\[
ms_0
\][/tex], for r < R is:
[tex]\[
ms_0 {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{\rho }}_{\rm{0}} {\rm{(r)}}} \,d{\rm{V = }}\int_{{\rm{V in S}}} {(\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }})\,} d{\rm{V = }}\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,\int_{{\rm{V in S}}} \, d{\rm{V = }}\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,(\frac{4}{3}){\rm{\pi r}}^{\rm{3}}
\][/tex]
Or, for r < R
[tex]\[
ms_0 = \,\frac{{{\rm{Mr}}^{\rm{3}} }}{{{\rm{R}}^{\rm{3}} }}
\]
[/tex]
And for r > R, the initial mass density is zero.
Or more generally, for spherical symmetry:
[tex]\[
{\rm{g}}_{\rm{n}} {\rm{(r)}}4{\rm{\pi r}}^{\rm{2}} \, = \,\, - 4{\rm{\pi G}}ms_n {\rm{(r)}}
\][/tex]
Or,
[tex]\[
{\rm{g}}_{\rm{n}} {\rm{(r) = }} - \frac{{{\rm{G}}ms_n {\rm{(r)}}}}{{{\rm{r}}^{\rm{2}} }}
\][/tex]
When this gravitational acceleration is applied to the Unruh temperature, we get:
[tex]\[
{\rm{T}}_n \, = \,\frac{{\hbar {\rm{g}}_{\rm{n}} {\rm{(r)}}}}{{2{\rm{\pi }}ck}}\,\, = \,\,\frac{{hG}}{{4{\rm{\pi }}^{\rm{2}} ck}}\frac{{ms_n ({\rm{r}})}}{{{\rm{r}}^2 }}
\][/tex]
and the energy density is:
[tex]\[
E_{\rm{T}} = \,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 }}{{{\rm{15}}c^3 h^3 }}(\frac{{hG}}{{4{\rm{\pi }}^{\rm{2}} ck}}\frac{{ms_n ({\rm{r}})}}{{{\rm{r}}^2 }})^4 \,\, = \,\,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 }}{{{\rm{15}}c^3 h^3 }}\,\frac{{h^4 G^4 }}{{{\rm{256\pi }}^{\rm{8}} c^4 k^4 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}\,\, = \,\,\frac{{G^4 h\,}}{{480{\rm{\pi }}^{\rm{3}} c^7 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}
\][/tex]
Then dividing by [tex]\[
c^2
\][/tex], we get a mass density:
[tex]\[
{\rm{\rho }}_{new} ({\rm{r}}) = \frac{{G^4 h\,}}{{480{\rm{\pi }}^{\rm{3}} c^9 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}\,\, = \,N\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}
\][/tex]
This new mass density (n+1) can be added to the old mass density (n) to give us an iterative process:
[tex]\[
{\rm{\rho }}_{{\rm{n + 1}}} ({\rm{r}}) = {\rm{\rho }}_{\rm{n}} ({\rm{r}}) + N\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}
\][/tex]
where,
[tex]\[
ms_n {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{\rho }}_{\rm{n}} {\rm{(r)}}} \,d{\rm{V}}
\][/tex]
which for spherical symmetry we get:
[tex]\[
ms_n {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{4\pi r}}^2 {\rm{\rho }}_{\rm{n}} {\rm{(r)}}} \,d{\rm{V}}
\][/tex]