# How much dark energy will fit in an average cup of coffee? [closed]

I am looking for the answer in Joules for obvious reasons.

• This would presumably be the volume density of the dark energy multiplied by the volume of the cup. Is that what you are asking? – John Rennie Jul 30 '19 at 11:54
• It's not clear to me what "fit in" means in this context. Do you just mean how much dark energy would we expect in that volume? – JMac Jul 30 '19 at 12:35

Dark energy as expressed by the cosmological constant is, as the name implies, a constant of space. Therefore, in a cup of coffee, we get, for some static observer $$t$$, and a spacelike hypersurface $$\Sigma$$ (I'm assuming that in our universe, there exists a neighbourhood that can be foliated in spacelike hypersurfaces large enough to accommodate a coffee cup) on which we do the actual volume integration,

$$\begin{eqnarray} E &=& \int_☕ T_{\mu\nu} t^\mu t^\nu d\mu[g_\Sigma]\\ &=& \int_☕ - \frac{c^4}{8\pi G} \Lambda g(t,t) d\mu[g_\Sigma] \end{eqnarray}$$

If we consider the cosmological constant as part of the stress-energy tensor, $$T'_{\mu\nu} = T_{\mu\nu} - \frac{c^4}{8\pi G} \Lambda g_{\mu\nu}$$. A cup of coffee is fairly small We can without much loss of experimental precision consider some Riemann normal coordinates around the center of the cup, so that $$g \approx \eta$$ (and, on $$\Sigma$$, that it is just the Euclidian metric) in the neighbourhood of the cup (Any extra term will be $$\approx \mathcal{O}(l^3)$$ here, with $$l$$ the characteristic dimension of the cup). Therefore, picking the canonical static observer $$t^\mu = (1,0,0,0)$$, this gives us

$$E = \frac{c^4}{8\pi G} \Lambda \int_☕ d^3x = \frac{c^4}{8\pi G} V \Lambda$$

In other words, we just have the volume by the cosmological constant. Given the current Lambda-CDM model of our universe, $$\Lambda$$ is estimated at

$$\Lambda = 1.1056 \times 10^{-52}\ \text{m}^{-2}$$

Unfortunately, the cosmological constant doesn't seem to have the uncertainty written down. This is due to the fact that in actual cosmology work, people generally use the dark energy density instead, $$\Omega_\Lambda$$, which we have as (cf particle data group)

$$\Omega_\Lambda = 0.692 \pm 0.012$$

The general formula relating the density parameter to its density, in the $$\Lambda$$CDM model, is

$$\Omega_\Lambda = \frac{8\pi G \rho_\Lambda(t = t_0)}{3 H_0^2}$$

So

$$\rho_\Lambda(t = t_0) = \frac{3 \Omega_\Lambda H_0^2}{8\pi G}$$

Where we have

$$\begin{eqnarray} \pi &=& 3.141592653 \pm 0.0000000005\\ G &=& (6.674 08 \pm 0.00031) \times 10^{-11} \text{m}^3 \cdot \text{kg}^{−1}\cdot \text{s}^{−2}\\ H_0 &=& (0.2197 \pm 0.027) \times 10^{-17} s^{-1} \end{eqnarray}$$

Using rough uncertainty propagation, this gives us

$$\begin{equation} (\Delta \rho_\Lambda)^2 = \rho_\Lambda^2 \left[(\frac{\Delta \pi}{\pi})^2 + 4 (\frac{\Delta H_0}{H_0})^2 + (\frac{\Delta \Omega_\Lambda}{\Omega_\Lambda})^2\right] \end{equation}$$

so that

$$\begin{equation} \rho_\Lambda = (0.59739 \pm 0.0734) \times 10^{-26} \text{m}^{-3} \cdot \text{kg} \end{equation}$$

For some reason this formula doesn't actually give us the energy density as it's only equivalent to our formula up to a factor of $$c^2$$, so we get

$$\begin{eqnarray} \frac{c^4}{8\pi G} \Lambda &=& c^2 \rho_\Lambda &=& (5.36907 \pm 0.65968) \times 10^{-10}\ \text{J}\cdot\text{m}^{-3} \end{eqnarray}$$

That's roughly the same value we'd get from our value of $$\Lambda$$, but with uncertainty.

A medium coffee cup, as shown here, is about (assuming an error of every dimension of about $$\approx 0.5 mm$$), $$(0.34 \pm 0.0015)\ \text{L}$$, or $$(0.34 \pm 0.0015)\times 10^{-3}\ \text{m}^3$$, so this gives us

$$E_{\Lambda ☕} = (1.810220805 \pm 0.22255)\times 10^{-13}\ \text{J}$$

We've dragged around a lot of digits for the calculations, now let's cut them off to significant figures : the smallest number of significant figures in our values is the dark energy density, at 3 significant figures. Therefore, we can cut off everything at that point.

$$E_{\Lambda ☕} = (1.81 \pm 0.22)\times 10^{-13}\ \text{J}$$

As an exercise left to the reader, compute the energy as measured by an observer running to a coffee cup with speed $$\beta = 0.1$$

• Great answer! Every student should know this facts. – Stephan Januar Aug 12 '19 at 20:56