# Is Dark Matter the Unruh effect applied to the acceleration of Gravity?

1. Nov 10, 2005

### Mike2

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":

$$${\rm{T = }}\frac{{\hbar a}}{{2{\rm{\pi }}ck}}\,$$$

where,

$$$\hbar \, = \,\frac{h}{{2{\rm{\pi }}}}$$$,

$$$\begin{array}{l} h = 6.626 \cdot 10^{ - 34} \,(J \cdot \sec ) \\ c = 2.998 \cdot 10^8 \,(m/s) \\ k = 1.381 \cdot 10^{ - 23} \,(J/^ \circ {\rm{K)}} \\ \end{array}$$$

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:

$$${\rm{p}}_{\rm{T}} {\rm{(\nu ) = }}\frac{{{\rm{8\pi }}h}}{{c^3 }}\,\frac{{{\rm{\nu }}^{\rm{3}} }}{{e^{h\nu {\rm{/}}k{\rm{T}}} \, - \,1}}$$$

This can be done by noting that:

$$$\int_0^\infty {\frac{{x^3 }}{{e^x - 1}}\,\,dx} \,\, = \,\,\frac{{\,\,\,{\rm{\pi }}^4 }}{{15}}$$$

found at: http://en.wikipedia.org/wiki/Table_of_integrals

Then,

$$$E_{\rm{T}} \, = \,\int_0^\infty {\frac{{8{\rm{\pi }}h}}{{c^3 }}\frac{{{\rm{\nu }}^3 }}{{e^{h{\rm{\nu /}}k{\rm{T}}} - 1}}\,\,d{\rm{\nu }}} \,\, = \,\,\frac{{8{\rm{\pi }}h}}{{c^3 }}(\frac{{k{\rm{T}}}}{h})^3 (\frac{{k{\rm{T}}}}{h})\int_0^\infty {(\frac{{h{\rm{\nu }}}}{{k{\rm{T}}}})^3 \frac{1}{{e^{h{\rm{\nu /}}k{\rm{T}}} - 1}}} \,(\frac{{h\,d{\rm{\nu }}}}{{k{\rm{T}}}}{\rm{) = }}\frac{{8{\rm{\pi }}h}}{{c^3 }}(\frac{{k{\rm{T}}}}{h})^4 \,\frac{{\,{\rm{\pi }}^4 }}{{15}}$$$

Or,

$$$E_{\rm{T}} = \,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 {\rm{T}}^{\rm{4}} }}{{{\rm{15}}c^3 h^3 }}$$$

When divided by $$$c^2$$$, this gives a mass density as a function of temperature.

But to find the acceleration due to a mass density, we can start with

$$$\Phi _{\rm{S}} \,\, = \,\,\,\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\,{\rm{ = }}\,\,\int_{\rm{V}} {\vec \nabla \cdot {\rm{\vec g }}d{\rm{V}}} \, = \,\int_{\rm{V}} {{\rm{\rho (r) }}d{\rm{V}}}$$$

where $$$\Phi _{\rm{S}}$$$ 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 $$${\rm{\rho }}$$$ mass density.

From observation, we know that $$${\rm{g(r) = }} - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}$$$

And for spherically symmetric g, we have:

$$$\Phi _{\rm{S}} \,\, = \,\,\,\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\, = \,\,\oint_{\rm{S}} {{\rm{g }}dA} \,\, = \,\,\oint_{\rm{S}} { - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}} \,dA\,\, = \,\, - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }}\oint_{\rm{S}} {dA}$$$

Or,

$$$\Phi _{\rm{S}} \, = \,\, - \frac{{{\rm{GM}}}}{{{\rm{r}}^{\rm{2}} }} \cdot 4{\rm{\pi r}}^{\rm{2}} {\rm{ = }} - 4{\rm{\pi GM}}$$$

And since the gravitational flux lines continue out to infinity and never cross each other, $$$\Phi _{\rm{S}}$$$ will not change for any surface surrounding M which produces g.

So $$$\Phi _{\rm{S}} \, = \, - 4{\rm{\pi GM}}$$$ for any surface S surrounding mass M.

In other words, we may take as a general rule:

$$$\oint_{\rm{S}} {{\rm{\vec g}}\, \cdot \,d{\rm{\vec S}}} \,\, = \,\, - 4{\rm{\pi GM}}$$$

Let us apply this to an initial mass density $$${\rm{\rho }}_{\rm{0}} {\rm{(r)}}$$$ that is spherically symmetrical and constant out to a radius, R. Or:

$$${\rm{\rho }}_{\rm{0}} {\rm{(r) = }}\frac{{{\rm{M}}_{\rm{0}} }}{{\rm{V}}}\,\, = \,\,\frac{{{\rm{M}}_{\rm{0}} }}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,\, = \,\,{\rm{constant}}$$$ for r < R, and

$$${\rm{\rho }}_{\rm{0}} {\rm{(r) = 0}}$$$ for r > R.

Then the initial mass density $$$ms_0$$$, for r < R is:

$$$ms_0 {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{\rho }}_{\rm{0}} {\rm{(r)}}} \,d{\rm{V = }}\int_{{\rm{V in S}}} {(\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }})\,} d{\rm{V = }}\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,\int_{{\rm{V in S}}} \, d{\rm{V = }}\frac{{\rm{M}}}{{(\frac{4}{3}){\rm{\pi R}}^{\rm{3}} }}\,(\frac{4}{3}){\rm{\pi r}}^{\rm{3}}$$$

Or, for r < R
$$$ms_0 = \,\frac{{{\rm{Mr}}^{\rm{3}} }}{{{\rm{R}}^{\rm{3}} }}$$$

And for r > R, the initial mass density is zero.

Or more generally, for spherical symmetry:

$$${\rm{g}}_{\rm{n}} {\rm{(r)}}4{\rm{\pi r}}^{\rm{2}} \, = \,\, - 4{\rm{\pi G}}ms_n {\rm{(r)}}$$$

Or,

$$${\rm{g}}_{\rm{n}} {\rm{(r) = }} - \frac{{{\rm{G}}ms_n {\rm{(r)}}}}{{{\rm{r}}^{\rm{2}} }}$$$

When this gravitational acceleration is applied to the Unruh temperature, we get:

$$${\rm{T}}_n \, = \,\frac{{\hbar {\rm{g}}_{\rm{n}} {\rm{(r)}}}}{{2{\rm{\pi }}ck}}\,\, = \,\,\frac{{hG}}{{4{\rm{\pi }}^{\rm{2}} ck}}\frac{{ms_n ({\rm{r}})}}{{{\rm{r}}^2 }}$$$

and the energy density is:

$$$E_{\rm{T}} = \,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 }}{{{\rm{15}}c^3 h^3 }}(\frac{{hG}}{{4{\rm{\pi }}^{\rm{2}} ck}}\frac{{ms_n ({\rm{r}})}}{{{\rm{r}}^2 }})^4 \,\, = \,\,\frac{{8{\rm{\pi }}^{\rm{5}} k^4 }}{{{\rm{15}}c^3 h^3 }}\,\frac{{h^4 G^4 }}{{{\rm{256\pi }}^{\rm{8}} c^4 k^4 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}\,\, = \,\,\frac{{G^4 h\,}}{{480{\rm{\pi }}^{\rm{3}} c^7 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}$$$

Then dividing by $$$c^2$$$, we get a mass density:

$$${\rm{\rho }}_{new} ({\rm{r}}) = \frac{{G^4 h\,}}{{480{\rm{\pi }}^{\rm{3}} c^9 }}\,\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}\,\, = \,N\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}$$$

This new mass density (n+1) can be added to the old mass density (n) to give us an iterative process:

$$${\rm{\rho }}_{{\rm{n + 1}}} ({\rm{r}}) = {\rm{\rho }}_{\rm{n}} ({\rm{r}}) + N\frac{{ms_n ^4 ({\rm{r}})}}{{{\rm{r}}^8 }}$$$

where,

$$$ms_n {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{\rho }}_{\rm{n}} {\rm{(r)}}} \,d{\rm{V}}$$$

which for spherical symmetry we get:

$$$ms_n {\rm{(r) = }}\int_{{\rm{V in S}}} {{\rm{4\pi r}}^2 {\rm{\rho }}_{\rm{n}} {\rm{(r)}}} \,d{\rm{V}}$$$

2. Nov 11, 2005

### hellfire

It seams that current observations invalidate any form of radiation as a good canditate for dark matter. Furthermore, to talk about photons as a form of “dark matter” seams a paradoxon to me and I don’t think that you can get ten times more energy density than baryonic matter with unobserved photons. But besides of this there are some issues I did already comment in an older thread.

First, when you consider gravitational systems you have to switch over to general relativity, instead of special relativity. In principle, you have a similar effect there: a change of the frame of reference (e.g. from free fall to stationary) may imply the definition of a new vacuum, having linear combinations of the old basis functions in the Hilbert space. Due to the different definitions of vacuum and number operators, particle production may occur. Such scenarios are considered for example in cosmology, where one has to deal with time dependent Hamiltonians. This makes it impossible to define a time independent vacuum and therefore the vacuum at one cosmological time is a state with particles at another cosmological time. However, the particles that appear due to different vacua are not always in thermal equilibrium. As far as I know, thermal equilibrium arises only if there exist horizons (Hawking radiation or Unruh radiation in special relativity) and this is not the case for the scenario you are considering here.

Second, I think this radiation would not be localized in any region of space, but would permeate space homogeneously.

3. Nov 11, 2005

### Mike2

In any event, it seems like an easy enough calculation to do. Though it may be difficult because the next iterations will accumulate more and more terms since you are then raising a polynomial to the power of 4. But since all the terms are positive, perhaps a lower bound can be determined by just taking the first term after the raising the polynomial to the forth power. If this goes to infinty, then the theory is wrong.

Also, it would seem that my scenario can be iterated only for mass densities to a desired radius - anything outside the radius can be ignored in the iterative process. This allows one to iterate first out to the radius of the initial mass density. And then use that result to iterate from R out to any desired radius.

As I understand it, the Unruh effect is exactly what causes the Hawking radiation whenever a horizon exists. But locally no observer knows where the horizon is. Local observes only feel a tidal force. So the potential to create Hawking radiation (the Unruh effect) must exist everywhere and is simply more noticeable near horizons. Also, there is a horizon, somewhere, the cosmological event horizon. Perhaps it plays a role here.

The Unruh effect is another zero point energy effect. The ZPE, if it exists, does not effect photon travel now. So neither should the Unruh effect.

I suspect this Unruh effect is just another way of describing the zero point energy of QFT in curved space. It may even be the cause of particle creation in the Big Bang. With Inflation comes hyper-acceleration which reduces the cosmologial event horizon to such a small distance that permanent particles are pulled out of the Unruh effect.

4. Nov 11, 2005

### Phobos

Staff Emeritus