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?

  • $\begingroup$ Hi! How did you find your first and second equations? Citing a source and/or explaining your derivation is optimal. $\endgroup$ Jun 21, 2021 at 12:50
  • $\begingroup$ If mass can be negative, so can energy $\endgroup$ Jun 21, 2021 at 14:40
  • $\begingroup$ This question should be superseded by the more overarching question: "Is the law of energy conservation valid for the universe as a whole, when it is considered as an isolated system?" $\endgroup$
    – Alex
    Jul 29, 2021 at 19:23

1 Answer 1


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).

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    $\begingroup$ 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 $\endgroup$ Jun 21, 2021 at 22:30
  • $\begingroup$ @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. $\endgroup$ Jun 22, 2021 at 9:45
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    $\begingroup$ 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. $\endgroup$
    – pela
    Jun 22, 2021 at 14:45
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    $\begingroup$ 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, and is analogous to the momentum$\leftrightarrow$homogeneity and the angular momentum$\leftrightarrow$isotropy connection. $\endgroup$
    – pela
    Jun 22, 2021 at 19:24
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    $\begingroup$ 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. $\endgroup$
    – pela
    Jun 22, 2021 at 21:08

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