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How much gravitational effect do we experience (e.g. maybe -.00001 G or smaller) from the edge of the visible universe? By edge of the visible universe, I am talking about the region of the cosmic microwave background and beyond. In the past, the universe was very hot and dense, and we are seeing the light from this time period as the Cosmic Microwave Background. It took slightly less than the age of the universe for that light to reach us, and it has become extremely redshifted with the expansion of space. Since gravity apparently travels at the speed of light, we should be experiencing a gravitational effect from this shell around us as well. As I understand, the universe was extremely dense back then, and it is around us in every direction, which would intensify the gravity greatly. However, it is extremely distant, which would weaken the gravity greatly. I don't know if redshift would affect gravity as it has the light from the cosmic microwave background, but if so, this would weaken the effect greatly as well. With all this in mind, I am curious, what would be the effect of gravity from the edge of the universe on us?

Note: I would expect this to be a very small outward effect in all directions, something like -.00000001 g or something like that, but not 0.

Jonathan
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    I think answering this is going to be difficult because you seem to be conflating both the gravitational wave background and the ambient gravitational field into one idea. Those two things have to be understood to be separate fields before explaining an answer. – zephyr Apr 23 '17 at 22:44
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    Just out of curiosity, do you mean, all together, which might be close to zero due to uniformity, or if, theoretically you could remove 1/2 of the observable universe and just feel the tug from the other half. – userLTK Apr 23 '17 at 23:52
  • @userLTK I would indeed like to know what force would come from just half the universe, as you describe. That would certainly be helpful. I was originally asking how much of a stretch we would experience from the nearly uniform distribution of matter. Perhaps I should ask as a separate question. – Jonathan Apr 24 '17 at 20:36

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In the past, the universe was very hot and dense, and we are seeing the light from this time period as the Cosmic Microwave Background.

This statement is (partially) correct.
However....

Since gravity apparently travels at the speed of light, we should be experiencing a gravitational effect from this shell around us as well. As I understand, the universe was extremely dense back then, and it is around us in every direction, which would intensify the gravity greatly.

This statement perhaps indicates that there are some misconceptions about the present state of the universe and also about what Cosmic Microwave Background Radiation really is (emphases added to identify the problem areas). So, in order to give a reasonable answer to the question asked, it would require addressing these misconceptions too.


The Big Bang model suggests that everything we see in our Observable Universe was once all concentrated in an infinitesimally small, hot and dense region of space. However, the Big Bang happened everywhere, which was followed by a super-luminal expansion of the space by a process called Inflation. The universe remained too hot still for any photons to roam around freely. It wasn't until the next 380,000 years that it continued to expand and eventually got cooled enough for the first hydrogen atoms to form. This is when the light first became visible. The remnant of this light is what we now now call Cosmic Microwave Background Radiation.

Again, it's important to note that the space was expanding in all possible directions continuously when the first light was emitted everywhere. In other words, the "dense" region from which the spacetime had begun, had now become comparatively much less dense overall. All through these years, this expansion hasn't stopped and every region where the light was first emitted, has now receded to great distances (46 billion light years in any direction). All the regions have cooled down further still to evolve into stars and planets, galaxies, clusters and any other celestial objects imaginable.


So, to answer the questions asked (for simplicity, we'll stick to CMB as the "edge")....

  • There is no space outside our Observable Universe which is surrounding us as a concrete shell. To be precise, it is merely a limit to how far back into the past and how deep we can peek into the space with our current technologies. Also, this has got nothing to do with the present state of the universe there. The universe is homogeneous and isotropic — meaning, you could go to any other place in the universe and it would still appear the same as it does from here — stars and galaxies everywhere. Your vantage point would determine which galaxy appears as a distant blob to you and which region of space would merely appear as a CMB radiation.

  • The regions which we now see as CMB, 46 billion light years away, are indeed one of the farthest reaches of space from where anything, including gravity, would have had any effects on us. By now, everything that we see as CMB would have coalesced to form celestial objects. But, there's a catch.... Remember, if the Sun was to disappear from the Solar System now, it would still take approximately 8 minutes for the Earth to be set free from its orbit. Meaning, the state of the Sun 8 minutes ago determines what gravity we feel because of it now.

  • So, to compute the gravitational tug from CMB (46 billion light years away), we need to look at the state of the matter from 13.8 billion years ago minus 380,000 years (when the CMB was emitted). We can roughly assume that, back then, it was just Hydrogen atoms all over. The overall mass of all those atoms combined would have been equivalent to the mass of the Observable Universe, i.e. $10^{53}kg$.

  • But again, the CMB was emitted everywhere and the isotropic nature of CMB tells us that departures from uniformity are only up to seven parts in a million! Meaning, when the CMB was emitted 13.8 billion years minus 380,000 years ago, the mass of the Observable Universe was distributed almost perfectly uniformly in every direction possible. From the distance of 46 billion light years then, this translates into almost equal gravitational pull from everywhere that light (and gravity) has reached us since then. Hence, the gravitational effects of the matter that is (i.e. was) spread uniformly in a radius of 46 billion light years from us, would be equivalent to what one would feel if they were to somehow reach the center of the earth — i.e. experiencing a nearly perfect equilibrium, with a net force of zero.


PS: Should there be any discrepancies above, I'd wholeheartedly welcome the suggestions on correcting or improving the answer.

Dhruv Saxena
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  • Very nice detailed answer! I believe that as you mention, the gravity would be in equilibrium because it would be in every direction. However, I believe there would be a "stretching effect", possibly the explanation for dark energy! How strong would that stretch be? In the case of being in a chamber at the center of the earth, if there were dust in that chamber, I believe it would slowly gravitate to the edges of that chamber, slowly accelerating outward as the galaxies do in our universe. – Jonathan Apr 24 '17 at 01:29
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    Thank you! I just don't think that Dark Energy can be thought of as something that is a consequence of the presence of matter (or of something outside the Observable Universe), exerting a pull on the objects visible to us. Rather, it is often described as a property of the space itself, effects of which start becoming prominent at cosmological scales, i.e. between galaxy clusters. The following Q&As might be of some interest: 1, 2 and 3 – Dhruv Saxena Apr 24 '17 at 02:02
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    @Jonathan In Newtonian physics, the gravitational field inside a hollow uniform sphere is zero. That's not just at the centre of the sphere, it's zero everywhere inside the sphere. See the Shell theorem. In general relativity, the conclusion is the same: the spacetime inside a hollow uniform sphere is flat, according to Birkhoff's theorem. – PM 2Ring Apr 24 '17 at 07:43
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    @Jonathan The effect of gravity is exactly the opposite to that of dark energy. It acts to slow the expansion and ultimately if there were enough matter (there isn't) would cause space to contract again. You cannot (correctly) consider the evolution of the Universe in a Newtonian framework, you require GR and the Robertson-Walker metric. – ProfRob Apr 24 '17 at 08:21
  • @Rob Jeffries Considering the things near us, you are right, gravity does slow the expansion of the universe. However, what would the gravity from the extremely dense portion of the universe from the cosmic microwave background, and beyond do? This would effectively be an extremely dense but distant spherical shell that surrounds everything. To me, it would seem that should gravitationally stretch out the universe. – Jonathan Apr 24 '17 at 20:01
  • @Jonathan It is indisputably the case that gravitating matter slows the expansion everywhere in the universe and everywhen. You are trying to imagine a Newtonian shell-like structure and it just doesn't work like that. – ProfRob Apr 24 '17 at 22:47
  • @PM 2Ring Has the shell theorem been experimentally verified? I would expect that something near the inner edge of a sphere would gravitate to the closest edge, particularly if it were extremely close. I do realize the "shell of the universe" would be equally distant from the objects within from their frame of reference. – Jonathan Apr 24 '17 at 23:59
  • @Jonathan Apologies if I sound a bit repetitive. Frankly speaking, it's really a misconception to think that there's some sort of dense shell outside our horizon. The homogeneity and isotropy of the universe ensures that everything is nearly the same density everywhere. If the universe was as dense as iron at some point in its early moments, then it was so everywhere. And since it appears to be $10^{-30}$ times less dense than water now, it should be the same everywhere else too — from its center to its edge (if center / edge could even ever exist!). Because the distances are so vast... – Dhruv Saxena Apr 25 '17 at 02:45
  • ...and because the light can only travel a finite distance in a second, looking at larger scales becomes equivalent to looking back into the time. Effectively then, CMB is merely a leftover from (and a proof of) that dense past, from the regions which were pretty close to us then, but have drifted far too away now due to expansion. The physical shape of the universe is as yet unconfirmed. A "sphere" is quite simply the limits to what we can observe (and hence influence) from where we are. The "edge" therefore is the boundary of distance from where anything can ever influence us in return. – Dhruv Saxena Apr 25 '17 at 02:46
  • @Jonathan I doubt that a direct empirical test of the shell theorem has been performed. I guess it'd be fun to put a large uniform spherical shell in orbit and do gravity tests inside it, but it would be a rather expensive undertaking with current technology. ;) OTOH, there's plenty of indirect evidence. The most important consequence of the shell theorem is that it predicts that the gravitational field above a uniform sphere is independent of the radius, i.e. the sphere acts gravitationally as if all its mass were concentrated at its centre, and we have lots of evidence that that's true. – PM 2Ring Apr 25 '17 at 09:00
  • @Jonathan Also, if you measure gravity at the bottom of a mine shaft, the shell theorem says that you can ignore the shell of stuff above you. Now, the Earth is certainly not a uniform sphere, it's a spinning inhomogeneous oblate spheroid that's much denser near the core than elsewhere, but once you take that stuff into account gravimeter measurements at the bottom of mine shafts are consistent with the shell theorem. OTOH, maybe the shell theorem is wrong, and our theories about the Earth's internal structure are bogus. :) – PM 2Ring Apr 25 '17 at 09:06
  • @Jonathan You said: "something near the inner edge of a sphere would gravitate to the closest edge". No, that doesn't happen with a perfectly uniform spherical shell. Of course, in the real world stuff is made of atoms, so if you get close enough to the inner surface of the shell it starts looking a bit lumpy, but at that distance you're close enough for the electromagnetic forces between the atoms in the shell and the atoms in your test particle to be more important than gravity. See https://physics.stackexchange.com/questions/158757/is-the-shell-theorem-only-an-approximation – PM 2Ring Apr 25 '17 at 09:14
  • @PM 2Ring Very interesting! I at first had difficulty with the shell theorem, but now I am convinced! – Jonathan Apr 25 '17 at 12:06