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The lost energy of Cosmic Background Radiation |
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| Jan6-13, 10:51 AM | #1 |
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The lost energy of Cosmic Background Radiation
Could anyone clarify where is gone the energy lost by CBR as the universe expands?
Thank you in advance and happy new year |
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| Jan6-13, 04:57 PM | #2 |
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General relativity has no global energy conservation, energy can get lost.
Those photons lead to a pressure, which influences expansion of the universe - you could view this as "the energy went into expansion of space". |
| Jan6-13, 06:08 PM | #3 |
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I've never seen it put quite like that MFB. Can you elaborate? Are you referring to dark energy or something else?
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| Jan6-13, 06:26 PM | #4 |
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The lost energy of Cosmic Background RadiationThat is, the current expansion has what you could think of as a base value which has nothing to do with dark energy and then acceleration due to dark energy. It is the presence of dark energy that was such a surprise when first discovered because the belief was that only the base expansion existed and it would have been slowing down. |
| Jan6-13, 06:28 PM | #5 |
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Yes but how do photons lead to expansion pressure?
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| Jan6-13, 10:00 PM | #6 |
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| Jan6-13, 10:43 PM | #7 |
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| Jan7-13, 07:47 AM | #8 |
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The FLRW metric depends - apart from initial conditions - on a cosmological constant, energy density and pressure (see section "solution") only. Both energy density and pressure slow expansion - but in the current universe, pressure is negligible.
Energy density and pressure are related via thermodynamics - as radiation energy drops with the 4th power of space expansion, we "lose" energy if space expands. |
| Jan7-13, 08:49 AM | #9 |
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| Jan7-13, 03:47 PM | #10 |
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| Jan7-13, 03:55 PM | #11 |
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[tex]\dot{H} + H^2 = {\ddot{a} \over a}= -{4 \pi G \over 3}\left(\rho + 3p\right)[/tex] For normal matter, the pressure [itex]p = 0[/itex], and the energy density [itex]\rho[/itex] is positive. So the right hand side is always negative: normal matter always acts to slow the expansion by a factor of [itex]-4\pi \rho G / 3[/itex]. By contrast, with photons, [itex]p = \rho/3[/itex], so that we pick up an extra factor reducing the expansion, leading to a photon gas reducing the expansion by [itex]-8\pi \rho G/3[/itex], or twice as rapid a deceleration compared with normal matter for the same energy density. |
| Jan7-13, 03:57 PM | #12 |
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What exactly is meant by "photon pressure"? Is this the same thing as normal Radiation pressure?
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| Jan7-13, 04:19 PM | #13 |
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As far as I can tell yes every search I do using Photon pressure returns radiation pressure
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| Jan7-13, 06:08 PM | #14 |
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| Jan7-13, 06:25 PM | #15 |
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We have a FAQ about this: http://www.physicsforums.com/showthread.php?t=506985
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| Jan7-13, 06:27 PM | #16 |
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One way to understand why it has this effect is to consider a somewhat different scenario:
Imagine that we have an enclosed box, and within that box is a gas of photons (you can simply imagine the box as having some temperature, which causes it to be filled with radiation). That gas of photons exerts radiation pressure on each wall of the box equal to [itex]\rho/3[/itex]. Now, what happens if we cause this box to expand in size? Well, if the box expands by a factor of [itex]a[/itex], then the photon pressure on each side of the box exerts work on the box. Because the work is in the direction of the motion of the walls of the box, this amounts to a transfer of energy from the photon gas to the walls of the box. In fact, if you calculate the energy transfer, you exactly get the loss of energy of the photon gas that we see as a redshift. As to why this pressure leads to a faster slowdown of the expansion, well, that's a bit harder to explain. But suffice it to say that pressure is sort of a kind of energy density, and gravity responds just as well to this sort of energy density as it responds to mass energy. |
| Jan7-13, 11:33 PM | #17 |
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