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I noticed in this thread:

Implications of a comoving Friedmann equation

Aside from the error of him having curvature term on the wrong Friedmann equation, I'm still interested in what literature speaks of this Planck length term arising, who derived it? I ask because I fell upon a derivation of similar structure. For instance, it's not difficult to show:

$m\dot{R}^2\ \psi = \frac{8\pi R}{3}\ \hbar c\ \Box^2\ \psi$

From it, we rearrange the mass and make use of

$\frac{R}{m} = \frac{G}{c^2}$

Giving

$\dot{R}^2\ \psi = \frac{8\pi G\hbar}{3c}\ \Box^2\ \psi$

And further divide through by $ c^2$

$(\frac{\dot{R}}{c})^2\ \psi = \frac{8\pi G\hbar}{3c^3}\ \Box^2\ \psi$

The object

$\frac{G\hbar}{c^3} =\ell^2_P$

Is the Planck length squared. By a reparameterization

$\ell^2_P \rightarrow R_0$

Allows us to construct from rearrangement, now in $c=1$ units

$(\frac{\dot{R}}{R_0})^2\ \psi = \frac{8\pi}{3}\ \Box^2\ \psi$

Further, you can express the four operator in curved spacetime. Chosing a simple one dimensional case on a curved background, it is well-known that it is given as

$(\frac{\dot{R}}{R_0})^2\ \psi = \frac{8\pi}{3}\ \Box^2\ \psi$

$=\mathbf{k} \frac{1}{\sqrt{-g}} (\frac{\partial}{\partial x}[\sqrt{-g} g \frac{\partial}{\partial x}])\psi$

The potential $\phi$ differs that it falls out of this simpler theory with dimensions consistent to the derivatives in question. You can keep it in like

$\Box^2 \phi = \frac{1}{\sqrt{-g}} (\frac{\partial}{\partial x}[\sqrt{-g} g \frac{\partial}{\partial x}])\phi$

And I'll point out that

$G\cdot\rho = \Box^2 \phi$

Such that all terms are replaced accordingly

$(\frac{\dot{R}}{R})^2\ \psi = \frac{8\pi G}{3} \rho\ \psi$

$ = \frac{8\pi \phi}{3} \Box^2\ \psi$

The subtle way they change depends on a factor of the speed of light ^2. For instance, this equation

$\frac{1}{c^2}(\frac{\dot{R}}{R_0})^2\ \psi = \frac{8\pi}{3}\ \Box^2\ \psi$

And $\phi$ the potential has units exact to velocity squared. This is why

$(\frac{\dot{R}}{R_0})^2\ \psi = \frac{8\pi \phi} {3}\ \Box^2\ \psi$

The reason why I found it interesting as a result, gives us some ball park figure to calculate the vacuum energy in absence of the cosmological constant.

$(\frac{\dot{R}}{c})^2\ \psi = \frac{8\pi G\hbar}{3c^3}\ \Box^2\ \psi$

Can in a way provide an energy measure of the vacuum. In order to do so, we must define the cosmological vacuum energy carefully. Suppose that the order of some loop diagram cuts off at each Planck "cell" in spacetime. We then have the integration from;

$k~=~0$

to

$k~=~1/\ell_{p},$

for

$\ell_P~=~\sqrt{G\hbar/c^3}$

That would be

$=~1.6\times 10^{-35}m$

which is of course the Planck length. It is said then that the divergence is of order 4 or

$\simeq~10^{140}m^{-4}$

Then the expected cosmological constant would then be

$~\simeq~10^{70}m^{-2}$

making this $122$ orders of magnitude larger than the measured cosmological constant. But there is some subtlety to the physics as well. It may be such that this vacuum energy is indeed there, and that the biggest mistake we made is not recognising that off-shell particles are not described by Hermitian matrices, and by default the vacuum energy cannot be described by an observable. The only special case is when a virtual particle becomes long-lived by virtue of background curvature as shown by Sakharov. There's also a complicated issue that the universe could have began with a single Planck length, or squared length, and the expansion saw the creation of spacetime by virtue of the creation operator: Yes! We use creation and annihilation operators to construct or destroy spacetime just as we use them for the particles to describe the creation or annihilation of particles. The wave number calculating the energy flux of off-shell to on-shell is given by Sakharov as

$\mathcal{L} = \hbar c \int\ k\ dk \cdot R + \hbar c \int \frac{\text{d}k}{k} + C$

Let's write the equation for the wave number directly with equivalence,

$(\frac{\dot{R}}{c})^2\ \psi = \frac{8\pi G\hbar}{3c^3}\ \Box^2\ \psi =\frac{8\pi \ell_P^2}{3}\ \Box^2\ \psi = \frac{8\pi k^{-2}}{3}\ \Box^2\ \psi$

Of course,

$\frac{8\pi G\hbar}{3c^3}$

Is not the CC term, but it's interesting we can theorise the energy this way. So can anyone provide insight, or a link, to the derivation of this length^2 appearing the Freidmann equation

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2 Answers 2

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In units where $c$, $\hbar$, and $G$ are not set equal to $1$, the Friedmann equation (for a spatially flat Universe) is \begin{equation} \frac{3 H^2 c^2}{8\pi G} = \epsilon \end{equation} where $\epsilon$ is the combined energy density of matter, radiation, and a cosmological constant.

Now the Planck length is defined by \begin{equation} \ell_P = \sqrt{\frac{\hbar G}{c^3}} \end{equation} Therefore, we can also rewrite the Friedmann equation in terms of the Planck length \begin{equation} \frac{3 \hbar H^2}{8\pi c \ell_p^2} = \epsilon \end{equation}

There's nothing deep about this, it is just an algebraic substitution. From the point of view of classical, non-quantum physics, this is a strange thing to do, and even obscures the physics, because $\hbar$ only appears here to cancel out the quantum mechanical nature of $\ell_P$.

Now in units where $c=\hbar=1$, the Friedman equation becomes \begin{equation} \frac{3 H^2}{8\pi \ell_p^2} = \epsilon \end{equation} If you plug in the energy density of a scalar field, that should correspond to the form of the Friedmann equation you found.

To emphasize, there's nothing particularly deep about $\ell_P$ appearing in this equation. It's just a valid way of expressing a combination of constants in a particular unit system. The intent of introducing $\ell_P$ in this way is to suggest that the natural scale of a cosmological constant term on the right hand side is related to $\ell_P$. But the fact that the constant appears in the equation is not a proof of this scale, and you need more sophisticated (and not universally accepted) arguments to justify the "natural" scale of the cosmological constant.

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  • $\begingroup$ I see, thank you very much! $\endgroup$ Sep 30, 2022 at 16:13
  • $\begingroup$ The only issue I have is that we don't need a cosmological constant as my work has shown. $\endgroup$ Oct 20, 2022 at 13:05
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I agree the answer by Andrew, and wish to point out that this is an example of something more general.

The equations of General Relativity do not involve Planck's constant. But if you put Planck's constant in somewhere, say in a numerator, then it will also appear somewhere else, in a denominator, such that it cancels out. Then you could gather up one of these $\hbar$ with some other bits to make a Planck length, and claim you have found a Planck length scale in the physics. But since you could do the same to any other equation, this is not really offering any insight.

More generally, you could take any well-defined length you like. Say, the length of the first caterpillar you find in your garden. Then you could make this your unit of length and express the equations of physics using this 'natural unit' of length. You might then get the impression that the universe has the caterpillar-length somehow built into its basic structure. That would, of course, be a false impression.

There are natural length scales built into the universe, but we find them not by any such artificial approach. Rather we notice combinations of constants such as the Planck length or the various Compton lengths and so on, and try to understand what is the physical phenomenon which that length is telling us about, or is characteristic of.

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  • $\begingroup$ Yes good points also. I do understand that the Planck constant doesn't come about in a completely natural way. Indeed, the starter equation was come about by careful observations on the dimensions. I didn't show how I arrived at it, but as I said it isn't too difficult to construct it from some first principles based on dimensional analysis alone. $\endgroup$ Sep 30, 2022 at 19:06

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