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Imagine an arbitrary point in space. It is within the gravitational 'potential' of every mass (although billions of ly away) in the entire universe.

Since every mass adds a tiny fraction, what is the total gravitational 'potential' energy in this point?

Edit:

Let point masses be located distance $r_i$ from the point and have masses $m_i$, then the 'potential' is

$$\Phi = - G \sum_{i} \frac{m_i}{r_i}.$$

I'm looking for this value averaged over all points in space. How does this depend on the shape of our universe or can we measure it?

  • For example gravity on my location is given by $$ 9.81 m/s^2 \text{(earth )} + 6 mm/s^2 \text{(sun)} + 200 pm/s^2 \text{(milky way)} + ? \text{(rest of the universe)}.$$
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The universe cannot be described at cosmological scales using newtonian physics. We have to use general relativity.

How does this depend on the shape of our universe or can we measure it?

Ideas like the "shape of the universe" don't make sense in newtonian gravity, so for this reason as well we need general relativity in order to answer your question.

In general relativity, a gravitational potential exists only for static spacetimes. Realistic cosmological models are not static, and therefore we can't define a potential for them.

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  • $\begingroup$ Thanks, the best answer so far. Though, the pure formula for $\Phi$ seems not to be limited to Newtonian physics. $\endgroup$ – fwgb Oct 26 '19 at 21:13
  • $\begingroup$ Though, the pure formula for Φ seems not to be limited to Newtonian physics. Not sure what you mean by this. $\Phi=- Gm/r$ is newtonian. E.g., it's not the same as the potential that gives the exterior Schwarzschild spacetime. It's a weak-field approximation to that potential. And summing it is also a newtonian approximation -- GR is nonlinear. $\endgroup$ – user4552 Oct 26 '19 at 22:32
  • $\begingroup$ The universe cannot be described at cosmological scales using newtonian physics That is not true. Newtonian cosmology (with $\Lambda$) provides a surprisingly good description for large scale cosmological evolution at least from photon decoupling epoch. It also serves 1) as a useful datapoint for Bayesian evaluation of various DM/DE alternative models, 2) as a conceptually simple starting point for various post-Newtonian cosmological frameworks (such as structure formation simulation codes). $\endgroup$ – A.V.S. Oct 27 '19 at 5:02
  • $\begingroup$ @BenCrowell In GR we also have $m$ (or $E$) and $r$ (with respect to another metric), thus we can calculate $-G m/r$ as well, but it has a different meaning. $\endgroup$ – fwgb Oct 27 '19 at 8:40
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Since every mass adds a tiny fraction, what is the total gravitational potential energy in this point?

There is no unique value of gravitational potential energy, or gravitational potential, at a point in space. It is always measured relative to another point in space.

A mass $m$ in space has gravitational potential energy $U$ with respect to another mass $M$ where the two masses are separated by a distance $r$ of

$$U=\frac{-GMm}{r}$$

Since gravitational potential energy is inversely proportional to $r$ it make sense to choose the zero of gravitational potential energy of $m$ with respect to $M$ for infinitely large $r$.

However, to the extent that there may be a net gravitational force acting on $m$ due to all the masses $M$ acting on $m$, the mass $m$ will experience an acceleration. If we were able to measure the change in velocity of $m$ between two points, we could determine the difference in gravitational potential energy between the two points since the loss in gravitational potential energy between the points will equal the increase in kinetic energy between the points, neglecting any friction in space.

Hope this helps.

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  • $\begingroup$ Okay, but formally the formula of $\Phi$ above does not contain any reference point, although the derivation maybe relies on a reference point at infinity. $\endgroup$ – fwgb Oct 24 '19 at 12:15
  • $\begingroup$ @fwgb I drafted my answer before seeing your edit. Even so, with respect to what point in space is phi measured? As far as I am aware, gravitational potential and potential energy is always relative to some point in space. Please convince me otherwise. $\endgroup$ – Bob D Oct 24 '19 at 12:26
  • $\begingroup$ The formula derives form assuming an infinite reference point. But mathematically, one can calculate $\Phi$ even without that assumption. $\endgroup$ – fwgb Oct 24 '19 at 12:34
  • $\begingroup$ It is always measured relative to another point in space this is not true in situations where there are no inertial coordinate systems. And this is precisely the case in Newtonian cosmological models. So gravitational potential has a whole new level of ambiguity and to fix it one would need more than specifying “a point”. $\endgroup$ – A.V.S. Oct 26 '19 at 18:45
  • $\begingroup$ @A.V.S actually my answer was intended in the context of an inertial coordinate system. $\endgroup$ – Bob D Oct 26 '19 at 18:48
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It seems to me you are asking about gravitational potential, not gravitational potential energy, since gravitational potential energy is property of some object and not space.

First of all, potential must be specified against some reference point. The Universe as we understand it today is homogeneous and isotropic at large scales. Therefore at large scales all points of space are equivalent and no matter which reference point you will choose (if you average the inhomogeneities at small scales) the difference will be zero. So the distant sources do not matter for the gravitational potential, all that matters is only close surrounding.

You can imagine it this way: the potential tells you how much work you need to do to move some object of unit mass from the reference point to the point in question. But since universe is everywhere (on the large scales) the same and every point on the path is as good as previous one, no work is needed.

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  • $\begingroup$ "...So the distant sources do not matter for the gravitational potential..." I don't think so. Surely the potential on large scales is flat, but not zero. $\endgroup$ – fwgb Oct 24 '19 at 10:42
  • $\begingroup$ as i said, potential has no absolute meaning. It has meaning only to specific reference point. In island universe the point is usually taken to be infinity, but in homogeneous universe what point would you choose? and why would another point be different, since the universe is homogeneous? $\endgroup$ – Umaxo Oct 24 '19 at 11:01
  • $\begingroup$ See the edit. Its maybe better to say 'potential of the distribution' what I mean. $\endgroup$ – fwgb Oct 24 '19 at 11:16
  • $\begingroup$ @fwgb where did you get the formula for potential from? because as far as i know, that formula is computed for the island universe and cannot be used for our universe. $\endgroup$ – Umaxo Oct 24 '19 at 11:47
  • $\begingroup$ @fwgb In a homogeneous model of the universe the potential is the same everywhere, and can be taken to be zero. Deviations of value and gradient are due to deviations from homogeneity. I suppose that we know enough about nearby galaxies to do a calculation, but it would still be an approximation. "We" excludes "me", of course. :-) $\endgroup$ – garyp Oct 26 '19 at 14:00

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