<p>Energy is always conserved. The stretching of the wavelengths of photons is due to the fact that we don't stretch out in space while photon wavelengths do. So it's an an illusionarry loss of energy. This is expressed by a negative energy or a negative pressure (dark energy).</p>

CC BY-SA 4.0
2021-06-21T22:30:14.890Z
Energy is actually not conserved in the Universe. The basis of energy conservation is time-translation invariance which is violated in general relativity (including cosmology). More here : [Energy conservation in universe](https://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/)
Aryan Bansal
CC BY-SA 4.0
2021-06-22T09:45:59.633Z
@AryanBansal It depends on the frame of reference. In our frame it is not conserved. In another it is. Which is the true one? I think its the one where it is.
Deschele Schilder
CC BY-SA 4.0
2021-06-22T09:50:37.260Z
Also, dark energy is nothing but an illusionary energy. The universe isnt really expanding. It are the galaxies moving away from each other in negatively curved spacecake. Space cannot expand. This is falsely assumed in the metric above. It might be a sution to the Einstein equafions but so is negative mass/energy.
Deschele Schilder
CC BY-SA 4.0
2021-06-22T14:45:28.077Z
Energy is not necessarily conserved in systems that are not time-symmetric, such as an expanding Universe. You're right that conservation of energy can depend on the frame of reference, but there is no frame of reference in which the energy of the CMB is conserved globally.
pela
CC BY-SA 4.0
2021-06-22T14:48:16.263Z
@pela What about the co-expanding frame?
Deschele Schilder
CC BY-SA 4.0
2021-06-22T19:24:54.243Z
No, not in that either. In comoving coordinates their number density is conserved, but they're still redshifted, so their comoving energy density decreases as $1/(1+z)$, which is why it decreases as $1/(1+z)^4$ in physical coordinates. The connection between energy conservation and time symmetry is given by [Noether's theorem](https://en.wikipedia.org/wiki/Noether%27s_theorem), and is analogous to the momentum$\leftrightarrow$homogeneity and the angular momentum$\leftrightarrow$isotropy connection.
pela
CC BY-SA 4.0
2021-06-22T19:34:34.573Z
@pela I think the photon wavelength will stay the same if the frame expands along in every direction. The number of photons will stay the same too. Just as the wavelength will stay the same in a falling frame on earth.
Deschele Schilder
CC BY-SA 4.0
2021-06-22T19:47:04.557Z
You're right that the number of photons will stay the same, but the wavelength won't. You can show this by integrating the FLRW metric over the journey of a photon. The theoretical derivation is consistent with observations.
pela
CC BY-SA 4.0
2021-06-22T20:00:16.327Z
@pela. Suppose I fall freely in the gravity field of the earth. Then a photons wavelength will not be blue shifted. Likewise if I fall freely (expand freely) I will see no change in the photons wavelength. The metric will be that of flat spacetime.
Deschele Schilder
CC BY-SA 4.0
2021-06-22T20:06:59.833Z
The difference is that space doesn't expand while you fall freely toward Earth. The metric of spacetime around Earth is the Schwarzschild metric, while the metric of the expanding Universe is the FLRW metric. It's the little $a(t)$ that makes all the difference.
pela
CC BY-SA 4.0
2021-06-22T20:11:47.683Z
@pela It's the question if spacetime really expands. Galaxies are really moving away from each other in negatively curved spacetime. Its not space that is expanding, its negatively curved spacetime that is stationary (or static). Particles move away from each other in such a space. If I fall freely in such a spacetime it would be very strange!
Deschele Schilder
CC BY-SA 4.0
2021-06-22T21:08:15.610Z
I don't think we're talking about the same universe. Why do you think it's negatively curved? A flat space is consistent with observations, and an expanding space is a waaay better fit to observations than a static. Also, how would you keep your universe static? Unless GR is wrong, it would be unstable. And GR is pretty well-tested.
pela
CC BY-SA 4.0
2021-06-22T23:35:25.227Z
@pela I think its negatively curved because the galaxies move away from each other accelerated (since 5 billion years or so). The positive curvature due to mass/ energy is getting less. In the very esrly universe there was no mass/energy yet, only negative curvature which made virtual particles move away from each other. Inflation. When the virtual excitations became real, positive curvature overtook, but they had still the motion due to negative curvature.
Deschele Schilder
CC BY-SA 4.0
2021-06-23T07:16:20.967Z
You can't change your global geometry from positively to negative curved, or vice versa. One is finite, the other is infinite. Moreover, what mechanism would make particles move away from each other in a negatively curved space? I'm sorry, Deschele, but (with all respect) I think you're making your own conclusions, based on misunderstandings of physics and cosmology. At the very least, let's agree that your ideas are not mainstream physics, which is what this site promotes.
pela
CC BY-SA 4.0
2021-06-23T07:27:47.133Z
@pela Cant you imagine particles move away from each other on a negatively curved spacetime? Its exactly the opposite from particles moving towards each other. Locally spacetime can be positively curved while globally irs negatively curved (dips , or bumps, in the overal negatively curved structure). Nothing non main stream. Quite the opposite. Like a cut open torus  that has negative curvature everywhere. With both sides stretching to infinity and a Planck sizes hole (where the big bang takes place).
Deschele Schilder
CC BY-SA 4.0
2021-06-23T08:33:33.823Z
No, a globally negative curvature would not cause particles to move away from each other _through_ space. You need a differential curvature to do that, i.e. a non-homogeneous / non-isotropic space, which is observationally ruled out.
pela
CC BY-SA 4.0
2021-06-23T08:38:01.497Z
@pela Imagine the saddle shape (like present on the inside of a torus).  Geodesics are diverging.
Deschele Schilder
CC BY-SA 4.0
2021-06-21T19:00:04.990Z
2021-06-21T19:00:04.990Z
Deschele Schilder
<p>I have tried to derive an equation for the total energy of the universe.
I have found that,
<span class="math-container">$$E(t)= \delta\dot a(t)^2a(t)\Omega(t)$$</span>
Where <span class="math-container">$\delta$</span> is just a positive constant, a(t) is the scale factor and <span class="math-container">$\Omega$</span> is the density parameter.
From there I have taken the derivative with respect to time of both sides and I get,
<span class="math-container">$$\dot E(t)= \delta\dot a(t)^3(2q(t)-\Omega(t))$$</span>
Where <span class="math-container">$q(t)$</span> is the deceleration parameter.
Since the universe isn't static, <span class="math-container">$\dot a(t)\neq0$</span>. Therefore, the only way for which the 1st law of thermodynamics (energy cannot be created nor destroyed) not to be violated is for <span class="math-container">$2q(t)=\Omega(t)$</span> for all time in R+.</p>
<p>The problem is that today's value for the deceleration parameter is <span class="math-container">$q(t_0)\approx-0.55$</span>, but by definition, the density parameter cannot be negative.</p>
<p>So is it possible that there exist other types of energies (of which we don't know of)  that have different <span class="math-container">$w$</span> state parameters that make the real value of <span class="math-container">$q(t_0)&gt;0$</span> for all time in R+ and thus allowing the conservation of energy? Or is the total energy of our universe just not constant?</p>

CC BY-SA 4.0
2021-06-21T12:50:38.053Z
Hi! How did you find your first and second equations? Citing a source and/or explaining your derivation is optimal.
Daddy Kropotkin
CC BY-SA 4.0
2021-06-21T14:40:10.007Z
If mass can be negative, so can energy
Deschele Schilder
CC BY-SA 4.0
2021-06-21T19:50:06.170Z
See https://astronomy.stackexchange.com/questions/1498/expansion-again-where-does-the-energy-come-from https://astronomy.stackexchange.com/questions/18613/where-does-the-energy-of-light-go-when-it-red-shifts
ProfRob
CC BY-SA 4.0
2021-07-29T19:23:01.013Z
This question should be superseded by the more overarching question: "Is the law of energy conservation valid for the universe as a whole, when it is considered as an isolated system?"
Alex
<cosmology><time><dark-energy><density><thermodynamics>
Energy conservation of the universe
Stan
# Energy conservation of the universe

(Asked by: username_0 on 2021-06-21)

I have tried to derive an equation for the total energy of the universe.
I have found that,
$$E(t)= \delta\dot a(t)^2a(t)\Omega(t)$$
Where $\delta$ is just a positive constant, a(t) is the scale factor and $\Omega$ is the density parameter.
From there I have taken the derivative with respect to time of both sides and I get,
$$\dot E(t)= \delta\dot a(t)^3(2q(t)-\Omega(t))$$
Where $q(t)$ is the deceleration parameter.
Since the universe isn't static, $\dot a(t)\neq0$. Therefore, the only way for which the 1st law of thermodynamics (energy cannot be created nor destroyed) not to be violated is for $2q(t)=\Omega(t)$ for all time in R+.

The problem is that today's value for the deceleration parameter is $q(t\_0)\approx-0.55$, but by definition, the density parameter cannot be negative.

So is it possible that there exist other types of energies (of which we don't know of) that have different $w$ state parameters that make the real value of $q(t\_0)>0$ for all time in R+ and thus allowing the conservation of energy? Or is the total energy of our universe just not constant?

**Tags:** cosmology, time, dark-energy, density, thermodynamics

> **Comments:**
> **username_1**: Hi! How did you find your first and second equations? Citing a source and/or explaining your derivation is optimal.  
> **username_2**: If mass can be negative, so can energy  
> **username_3**: See https://astronomy.stackexchange.com/questions/1498/expansion-again-where-does-the-energy-come-from https://astronomy.stackexchange.com/questions/18613/where-does-the-energy-of-light-go-when-it-red-shifts  
> **username_4**: This question should be superseded by the more overarching question: "Is the law of energy conservation valid for the universe as a whole, when it is considered as an isolated system?"  

---

## 1 answers

### Answer by username_2 on 2021-06-21. Score: -3

Energy is always conserved. The stretching of the wavelengths of photons is due to the fact that we don't stretch out in space while photon wavelengths do. So it's an an illusionarry loss of energy. This is expressed by a negative energy or a negative pressure (dark energy).

> **Comments:**
> **username_5**: Energy is actually not conserved in the Universe. The basis of energy conservation is time-translation invariance which is violated in general relativity (including cosmology). More here : [Energy conservation in universe](https://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/)  
> **username_2**: @username_5 It depends on the frame of reference. In our frame it is not conserved. In another it is. Which is the true one? I think its the one where it is.  
> **username_2**: Also, dark energy is nothing but an illusionary energy. The universe isnt really expanding. It are the galaxies moving away from each other in negatively curved spacecake. Space cannot expand. This is falsely assumed in the metric above. It might be a sution to the Einstein equafions but so is negative mass/energy.  
> **username_6**: Energy is not necessarily conserved in systems that are not time-symmetric, such as an expanding Universe. You're right that conservation of energy can depend on the frame of reference, but there is no frame of reference in which the energy of the CMB is conserved globally.  
> **username_2**: @username_6 What about the co-expanding frame?  
> **username_6**: No, not in that either. In comoving coordinates their number density is conserved, but they're still redshifted, so their comoving energy density decreases as $1/(1+z)$, which is why it decreases as $1/(1+z)^4$ in physical coordinates. The connection between energy conservation and time symmetry is given by [Noether's theorem](https://en.wikipedia.org/wiki/Noether%27s_theorem), and is analogous to the momentum$\leftrightarrow$homogeneity and the angular momentum$\leftrightarrow$isotropy connection.  
> **username_2**: @username_6 I think the photon wavelength will stay the same if the frame expands along in every direction. The number of photons will stay the same too. Just as the wavelength will stay the same in a falling frame on earth.  
> **username_6**: You're right that the number of photons will stay the same, but the wavelength won't. You can show this by integrating the FLRW metric over the journey of a photon. The theoretical derivation is consistent with observations.  
> **username_2**: @username_6. Suppose I fall freely in the gravity field of the earth. Then a photons wavelength will not be blue shifted. Likewise if I fall freely (expand freely) I will see no change in the photons wavelength. The metric will be that of flat spacetime.  
> **username_6**: The difference is that space doesn't expand while you fall freely toward Earth. The metric of spacetime around Earth is the Schwarzschild metric, while the metric of the expanding Universe is the FLRW metric. It's the little $a(t)$ that makes all the difference.  
> **username_2**: @username_6 It's the question if spacetime really expands. Galaxies are really moving away from each other in negatively curved spacetime. Its not space that is expanding, its negatively curved spacetime that is stationary (or static). Particles move away from each other in such a space. If I fall freely in such a spacetime it would be very strange!  
> **username_6**: I don't think we're talking about the same universe. Why do you think it's negatively curved? A flat space is consistent with observations, and an expanding space is a waaay better fit to observations than a static. Also, how would you keep your universe static? Unless GR is wrong, it would be unstable. And GR is pretty well-tested.  
> **username_2**: @username_6 I think its negatively curved because the galaxies move away from each other accelerated (since 5 billion years or so). The positive curvature due to mass/ energy is getting less. In the very esrly universe there was no mass/energy yet, only negative curvature which made virtual particles move away from each other. Inflation. When the virtual excitations became real, positive curvature overtook, but they had still the motion due to negative curvature.  
> **username_6**: You can't change your global geometry from positively to negative curved, or vice versa. One is finite, the other is infinite. Moreover, what mechanism would make particles move away from each other in a negatively curved space? I'm sorry, username_2, but (with all respect) I think you're making your own conclusions, based on misunderstandings of physics and cosmology. At the very least, let's agree that your ideas are not mainstream physics, which is what this site promotes.  
> **username_2**: @username_6 Cant you imagine particles move away from each other on a negatively curved spacetime? Its exactly the opposite from particles moving towards each other. Locally spacetime can be positively curved while globally irs negatively curved (dips , or bumps, in the overal negatively curved structure). Nothing non main stream. Quite the opposite. Like a cut open torus  that has negative curvature everywhere. With both sides stretching to infinity and a Planck sizes hole (where the big bang takes place).  
> **username_6**: No, a globally negative curvature would not cause particles to move away from each other _through_ space. You need a differential curvature to do that, i.e. a non-homogeneous / non-isotropic space, which is observationally ruled out.  
> **username_2**: @username_6 Imagine the saddle shape (like present on the inside of a torus).  Geodesics are diverging.  

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