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

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    $\begingroup$ This would presumably be the volume density of the dark energy multiplied by the volume of the cup. Is that what you are asking? $\endgroup$ – John Rennie Jul 30 '19 at 11:54
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    $\begingroup$ 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? $\endgroup$ – JMac Jul 30 '19 at 12:35
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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

enter image description here

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 $\beta = 0.1$

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

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