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Monday, February 20, 2012

voids

inside a void, does time pass faster or slower? by how much?

bonus points: what's $\cos(i)$ with $i = \sqrt{-1}$?

First empirically. We know planetary fields (overdensities) produce gravitational redshift. Underdensities will produce gravitational blueshift. That is an increase in observed frequency, therefore a shrinking of time intervals compared to where the light was sent out. Therefore the clock of the guy inside the underdensity TICKS FASTER.

Then mathematically. I use first post Newtonian approximation of the GR metric, because \Phi/c^2 << 1.What happens to the time part of the metric can be described by

$dt -> dt \sqrt{1+\frac{\Phi}{2c^2}}$

or in the approximation mentioned:

$dt -> dt(1+Phi/c^2)$

Ascribing a positive potential to the inside of the void (ignoring if you want the uniform component due to Birkhoffs theorem), I get an increased dt, or FASTER TICKING CLOCK.

I think we can only give an upper limit. The observer should experience all underdensity enclosed in a radius between himself and the centre of the void. **You might want to check these numbers because they suprise me!**

A)EXTREME CASE: Supervoid with a radius in the order of 100Mpc. This would be larger than the local void!(http://en.wikipedia.org/wiki/Local_Void). It is, though, believed that an even larger one exists in the Northern Galactic hemisphere!

The void in its extreme case has an underdensity equal to the critical one. The critical density equals about 3 protons/ cm^3. I converted that to

yielding - for the void - an enclosed "mass" of $M_{equiv} \approx 10^{38}Kg$. That gave:

$\frac{Phi}{c^2} \approx 0.03$

Which would make a clock wrong by roughly 2seconds every minute!

B)SMALL VOID By looking at a redshift survey of the southern sky (SSRS2, 1994) I tried to identify a small void. I picked one with a radius of

$R \approx 10Mpc\,.$

ARBITRARILY assuming it were critically empty, I can scale the mass with the third and the potential with the first power of $R$, so the effect should be 100 times smaller:

$\frac{\Phi}{c^2} \ approx 3\times 10^-4$

and still this clock would be off several tens of seconds per day!

mmh, I think there might be a square missing somewhere... making things very simple I'd put $\Phi=GM/R$ with $M$ the missing mass of the void and $R$ its size. with the definition $\rho_{crit}=3H^2/8\pi G$ one gets for the missing mass $M=4\pi\rho_{crit}\Omega_m R^3/3$, and therefore $\Phi/c^2 = \Omega_m/2\times(R/\chi_H)^2$ with the Hubble distance $\chi_H=c/H$ ($H$ is the Hubble constant and $c$ the speed of light). Inserting $R=100$ Mpc then gives $\Phi/c^2\simeq 1.4\times10^{-4}$ for $\Omega_m=0.25$. the little void of $10$ Mpc would give $\Phi/c^2\simeq 1.4\times10^{-6}$.

in relation to last week's question I wanted to add that of course $\Phi/c^2$ is equal to $r_S/R/2$ with the Schwarzschild radius $r_S$. equating it with the Hubble radius (for a system with a density close to the critical density as true for the Universe) recovers the above result.

Here's my take:

ReplyDeleteFirst empirically. We know planetary fields (overdensities) produce gravitational redshift. Underdensities will produce gravitational blueshift. That is an increase in observed frequency, therefore a shrinking of time intervals compared to where the light was sent out. Therefore the clock of the guy inside the underdensity TICKS FASTER.

Then mathematically. I use first post Newtonian approximation of the GR metric, because \Phi/c^2 << 1.What happens to the time part of the metric can be described by

$dt -> dt \sqrt{1+\frac{\Phi}{2c^2}}$

or in the approximation mentioned:

$dt -> dt(1+Phi/c^2)$

Ascribing a positive potential to the inside of the void (ignoring if you want the uniform component due to Birkhoffs theorem), I get an increased dt, or FASTER TICKING CLOCK.

$\cos(i) = 0.5(exp(-1) + exp(1)) \approx 1.543$

that's quite right... but by *how much* ticks the clock faster?

ReplyDeleteBased on the above I'd say by the factor of $1+\frac{\Phi}{c^2}$.

Delete:) but how much is $1+\Phi/c^2$ for a large void in actual numbers?

DeleteI think we can only give an upper limit. The observer should experience all underdensity enclosed in a radius between himself and the centre of the void. **You might want to check these numbers because they suprise me!**

DeleteA)EXTREME CASE: Supervoid with a radius in the order of 100Mpc. This would be larger than the local void!(http://en.wikipedia.org/wiki/Local_Void). It is, though, believed that an even larger one exists in the Northern Galactic hemisphere!

The void in its extreme case has an underdensity equal to the critical one. The critical density equals about 3 protons/ cm^3. I converted that to

$\rho_{crit} =\frac{1.5\times 10^{11}M_{solar}}{Mpc^3}\,,$

yielding - for the void - an enclosed "mass" of $M_{equiv} \approx 10^{38}Kg$. That gave:

$\frac{Phi}{c^2} \approx 0.03$

Which would make a clock wrong by roughly 2seconds every minute!

B)SMALL VOID

By looking at a redshift survey of the southern sky (SSRS2, 1994) I tried to identify a small void. I picked one with a radius of

$R \approx 10Mpc\,.$

ARBITRARILY assuming it were critically empty, I can scale the mass with the third and the potential with the first power of $R$, so the effect should be 100 times smaller:

$\frac{\Phi}{c^2} \ approx 3\times 10^-4$

and still this clock would be off several tens of seconds per day!

And... I shouldn't have said the clock is "wrong", because it is a fundamental effect!

Deletemmh, I think there might be a square missing somewhere... making things very simple I'd put $\Phi=GM/R$ with $M$ the missing mass of the void and $R$ its size. with the definition $\rho_{crit}=3H^2/8\pi G$ one gets for the missing mass $M=4\pi\rho_{crit}\Omega_m R^3/3$, and therefore $\Phi/c^2 = \Omega_m/2\times(R/\chi_H)^2$ with the Hubble distance $\chi_H=c/H$ ($H$ is the Hubble constant and $c$ the speed of light). Inserting $R=100$ Mpc then gives $\Phi/c^2\simeq 1.4\times10^{-4}$ for $\Omega_m=0.25$. the little void of $10$ Mpc would give $\Phi/c^2\simeq 1.4\times10^{-6}$.

ReplyDeletein relation to last week's question I wanted to add that of course $\Phi/c^2$ is equal to $r_S/R/2$ with the Schwarzschild radius $r_S$. equating it with the Hubble radius (for a system with a density close to the critical density as true for the Universe) recovers the above result.

Delete