Mining the Universe's expansion?

In summary, Astrophysicist Edward Harrison defends that energy could be extracted from attaching an imaginary cosmologically long string to a receding object from us in an expanding universe. The amount of energy extracted is potentially limited in decelerating universes, but unlimited in accelerating universes due to the difficulty of defining conservation laws on a cosmological scale. However, a physicist points out that even in an accelerating universe, it is not possible to extract unlimited energy from a tethered object due to the limitations of perpetual motion machines. There are also questions raised about the validity of the equations used in the article and the practicality of mining energy from objects over billions of light years away.
  • #1
Suekdccia
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TL;DR Summary
Mining the Universe's expansion?
There is an article written by astrophysicist Edward Harrison [1] which defends that energy could be extracted from attaching an imaginary cosmologically long string to a receding object from us in an expanding universe. He says that the energy extracted is potentially limited (in decelerating universes) or unlimited (in accelerating universes) due to the difficulty of defining conservation laws in cosmological scales.

However, I was discussing this with a physicist and he came out with a good point against the article: Even if our universe is accelerating, and even if there is perpetual tension in the string attached to the receding object, you cannot extract unlimited energy as you can't run a perpetual motion machine on perpetual tension. Even if Λ=0, there's a perpetual negative tension (pressure) in the tether due to gravitational attraction. You can extract work from it only until the tether's length reaches zero. If Λ>0, you can extract work until the tether's length reaches the Hubble length, then you hit the event horizon and the tether breaks. Reeling the mass back in before you hit the horizon costs as much energy as you extracted from letting it spool out. Is this argument right?

Moreover I have another question: in this article [2] it is indicated that if the attached object receding from us is stopped, at the right distance, it may be directed towards our galaxy, go through right it, and join the Hubble flow in the opposite side of us. Then, couldn't we mine some energy from the tension of the object being dragged by the Hubble flow, until putting it at rest (without reeling the object back to us), and then, after being directed towards our galaxy by gravity, going through it and being dragged by the Hubble flow in the opposite side, repeat the process again and again indefinetely...?

[1]: https://adsabs.harvard.edu/full/1995ApJ...446...63H

[2]: https://arxiv.org/abs/astro-ph/0104349
 
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  • #2
Suekdccia said:
Summary: Mining the Universe's expansion?

Moreover I have another question: in this article [2] it is indicated that if the attached object receding from us is stopped, at the right distance, it may be directed towards our galaxy, go through right it, and join the Hubble flow in the opposite side of us. Then, couldn't we mine some energy from the tension of the object being dragged by the Hubble flow,
We are to late for that. For such a scheme you need a decelerating expansion and that ended some 5 to 6 billion years ago. AFAIK, it will never happen again. Any distant object "stopped" today will eventually join the Hubble flow on the same side.

Do you realize the practical "impossibility" of mining expansion energy over billions of light year distances?
 
  • #3
Suekdccia said:
this article [2]
In looking through this article, I see at least three items that I find questionable.

First, the metric in equation (1) is only valid for spatially flat universes; but of the specific cases analyzed in the article, only two, the (1, 0) matter dominated critical density universe, and the (0.3, 0.7) Lambda CDM universe that represents our best current model, are spatially flat. (The condition for spatial flatness, in their notation, is that the numbers in the parentheses must add up to 1.) For the other cases, their equations do not appear to me to be correct.

Second, all time derivatives are taken, according to the paper, with respect to "proper time", but what they actually mean is "FRW coordinate time", which is only the same as proper time for comoving observers. The tethered galaxy is not comoving, so its proper time is not the same; but this does not appear to me to be taken into account anywhere in the article.

Third, a proper analysis of the trajectory of a tethered galaxy when it is released would use the geodesic equation, but I don't see that being done in the article.
 
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  • #4
Jorrie said:
Any distant object "stopped" today will eventually join the Hubble flow on the same side.
But even in an accelerated universe, wouldn't that happen with a tethered galaxy that is close enough to be affected by our gravitational pull if it's "put at rest" by our rope?
Jorrie said:
Do you realize the practical "impossibility" of mining expansion energy over billions of light year distances?
Of course, just being curious with mental experiments
 
  • #5
PeterDonis said:
In looking through this article, I see at least three items that I find questionable.

First, the metric in equation (1) is only valid for spatially flat universes; but of the specific cases analyzed in the article, only two, the (1, 0) matter dominated critical density universe, and the (0.3, 0.7) Lambda CDM universe that represents our best current model, are spatially flat. (The condition for spatial flatness, in their notation, is that the numbers in the parentheses must add up to 1.) For the other cases, their equations do not appear to me to be correct.

Second, all time derivatives are taken, according to the paper, with respect to "proper time", but what they actually mean is "FRW coordinate time", which is only the same as proper time for comoving observers. The tethered galaxy is not comoving, so its proper time is not the same; but this does not appear to me to be taken into account anywhere in the article.

Third, a proper analysis of the trajectory of a tethered galaxy when it is released would use the geodesic equation, but I don't see that being done in the article.
How would this affect the article? Is it completely wrong?
 
  • #6
Also @PeterDonis @Jorrie do you think that the argument that the physicist gave me is correct?:
Suekdccia said:
Summary: Mining the Universe's expansion?

Even if our universe is accelerating, and even if there is perpetual tension in the string attached to the receding object, you cannot extract unlimited energy as you can't run a perpetual motion machine on perpetual tension. Even if Λ=0, there's a perpetual negative tension (pressure) in the tether due to gravitational attraction. You can extract work from it only until the tether's length reaches zero. If Λ>0, you can extract work until the tether's length reaches the Hubble length, then you hit the event horizon and the tether breaks. Reeling the mass back in before you hit the horizon costs as much energy as you extracted from letting it spool out
 
  • #7
Suekdccia said:
But even in an accelerated universe, wouldn't that happen with a tethered galaxy that is close enough to be affected by our gravitational pull if it's "put at rest" by our rope?
I think you misunderstand the "tethered galaxy" scenario. It is worked out for a perfectly homogeneous universe, where all masses are comoving and where there is no residual gravity to move them away from that state. The effect of the 'tether' is to give your object a peculiar radial velocity relative to its local Hubble flow, with a magnitude equal but opposite, from "our perspective" (meaning the perspective of wherever the other end of the tether is "anchored", let's call it the origin).

The purpose of the tether is to forcefully keep the object at a constant proper distance from the origin. When released from the tether the object's deviation from the Hubble flow will decrease for any expanding universe (according to Davis et al.) How it happens depends on the cosmic deceleration parameter q(t).

If q(t) = 0, the object will sit at a constant proper distance from the origin and eventually the Hubble flow at that distance will decay to zero as well. If accelerating (q(t) negative) the object will move away from the origin and join the Hubble flow. If decelerating (q(t) positive) the object will move through the origin and join the Hubble flow on the other side of the origin.

The latter case is not caused by the gravity around the origin, but by the negative peculiar velocity that it acquired by the tether, plus the fact that a decelerating expansion does not counteract that peculiar velocity. Once it has moved to the other side of the origin, the expansion does counteract the peculiar velocity and it eventually joins (an also decreasing) Hubble flow. Remember that a (flat) decelerating expansion eventually results in a zero Hubble constant, unless the universe is closed and H goes negative and a collapse starts.

Of course, any continuous energy mining from the peculiar velocity of the case under discussion will simply result in it not joining the Hubble flow on the other side and probably end up somewhere near the origin.
Whether it is possible to in principle mine energy from an everlasting accelerating expansion, I do not know, but I think it is unlikely.
 
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  • #8
Everyone seems to be rolling with it. So, assuming you can think about this using Newtonian mechanics...
Suekdccia said:
do you think that the argument that the physicist gave me is correct?:
If we treat the universe's acceleration putting tension in the tether as a sort of force doing work, then it'll be conservative if it is independent of time. That answer is then correct insofar as we treat the acceleration of the universe (in the sense of the value of the deceleration parameter) as constant.

That is, however, not exactly true, as the universe is not 100% dark energy dominated, and can never be exactly so - because matter density can always dilute some more. So, on the outbound leg you get work done by the 'acceleration force' of some strength, while on the inbound leg the force is (ever more negligibly, but still) higher while the distance over which work has to be done is the same. I.e., even with 100% efficient extraction and reeling-in process you end up losing more than you gain.
That answer is then overly optimistic on a technicality.

Suekdccia said:
But even in an accelerated universe, wouldn't that happen with a tethered galaxy that is close enough to be affected by our gravitational pull if it's "put at rest" by our rope?
I think you could do this, if you set things up very carefully. But you wouldn't gain anything more than when tethering the object far away from localised sources of gravity and just letting it drift away.

This would work similar to the (late-time integrated) Sachs-Wolfe effect - in a universe with lambda gravity wells evolve to be more shallow (to a finite degree). It's due to the same evolution of the q parameter that makes the 'force' non-conservative, as discussed above.

Normally, in an expanding universe without lambda, if you put an object close enough to a gravity well that it starts falling towards it, then that object and the source of gravity already (by definition) form a bound system that is unaffected by expansion.
You then have a regular case of potential energy being exchanged for kinetic energy, with total mechanical energy conserved.
I.e. the object swings past the centre and ends up just as far on the other side as in the beginning, with zero velocity.
If you ever extract any energy from this, you do so in the same way as you would with an asteroid falling to Earth, or a hydroelectric dam. It's a one-off and the details of the expansion of the universe are irrelevant.

If there's lambda, however, the deceleration parameter evolves (as above, due to matter density going down), and the nett force (gravitational + from acceleration of expansion) varies between the inbound leg and the outbound leg of an infalling object.
So, if set up right, on the outskirts of a gravity well, it's possible to start in a bound state, and emerge unbound on the other side of the - now decayed - gravity well. At which point you could start extracting energy from the acceleration.

But, again, that would be no different than just letting the acceleration of the expansion do the work on a tethered object, only with an added period of traversing the gravity well that does nothing of value.

Unless, of course, I misunderstand the mechanics of the Sachs-Wolfe effect. It works along those lines for light signals, though.
 
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  • #9
Jorrie said:
Whether it is possible to in principle mine energy from an everlasting accelerating expansion, I do not know, but I think it is unlikely.
Why do you think it's unlikely?
 
  • #10
Suekdccia said:
How would this affect the article? Is it completely wrong?
I don't know that it's completely wrong, but I would want to do my own analysis before answering the questions you pose in this thread, instead of just relying on what the article says. That will take a little time.
 
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  • #11
Bandersnatch said:
But, again, that would be no different than just letting the acceleration of the expansion do the work on a tethered object, only with an added period of traversing the gravity well that does nothing of value.
But I think it would be different in the sense that we wouldn't have the horizon problem.

The physicist that I was discussing with said that as long as you would let the rope unwind by the drag of the object going with the Hubble flow energy could be extracted. However, once the object would arrive to the cosmological horizon the rope would break and no more energy could be extracted. It would be similar to having an object attached to a string and letting that object fall into a black hole, once it arrives to the event horizon, the string breaks and no more energy would be harnessed.

So I was thinking that maybe (even if our universe is accelerating), if we attach the rope to an object receding from us but not in a very long distance, we would gain some energy until putting it at rest. Then we detach the rope and let the object be attracted by gravity to us. We let it go right through our galaxy and join the Hubble flow on the other side of the sky. Then we re-attach the string at the same distance as we did before (but in the other side of the observable universe) and extract some energy until putting it at rest and let the process that I've described happen again. We do that again and again, gaining (in principle, and ignoring the endless technical difficulties) unlimited energy, as the object would never cross the cosmological horizon but at the same time it would be always dragged by the Hubble flow
 
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  • #12
Bandersnatch said:
assuming you can think about this using Newtonian mechanics...
You can't, really, particularly not in the case of nonzero lambda. But you can use somewhat similar intuitions to at least get a heuristic grasp on the problem.

The most general approach to a problem of this type is, as I mentioned before, to look at the geodesic equation. I have not done that in detail yet, but here are some quick heuristics.

First, consider the simpler case where we drop an object in the gravity well of a planet. The simplest way conceptually to extract energy from this process is to let the object free fall and then stop it at the bottom and convert its kinetic energy into work (or into some kind of stored energy that we will later convert to work, like a battery). However, we can also extract energy by slowly lowering the object attached to a rope that is also attached to a mechanism at the top that extracts and stores energy as the object lowers. Either way, ideally, we get the same amount of total energy stored for the same distance of fall.

Now, to generalize the above to the cosmological scenarios we are considering, we have to know which direction is "down", i.e., in which direction an object will free-fall when released. If the heuristic that this direction is determined by the sign of the deceleration parameter ##q## is correct, then for the case of zero lambda and matter dominated, "down" will be towards us, whereas for the case of nonzero lambda, "down" will be away from us.

In other words, for the zero lambda matter dominated case, we are at the "bottom", so to speak, and the easiest way to extract energy from the tethered object is to release it, let it fall towards us, and then stop it when it reaches us and convert its kinetic energy relative to us into stored energy that we can use to do work.

Whereas, for the nonzero lambda case, we are at the "top", so the easiest way to extract energy from the tethered object is to let the tether slowly extend outward, corresponding to "slowly lowering" the object, and store energy as it lowers.

Note that the key thing common to both of these approaches is that we can only extract energy by not letting the object follow its "natural" geodesic path; we either have to stop it at the bottom and extract all of its kinetic energy, or we have to slowly lower it from the top instead of just letting it free fall (because if it free falls we have no way of stopping it since it's falling away from us).
 
  • #13
PeterDonis said:
You can't, really, particularly not in the case of nonzero lambda. But you can use somewhat similar intuitions to at least get a heuristic grasp on the problem.
You say that, but the conclusions you draw are the same, and drawn along the same line of reasoning. So, you can?
 
  • #14
Bandersnatch said:
the conclusions you draw are the same
The point is that there is no way to get to those conclusions in Newtonian gravity. I know cosmologists like to use Newtonian reasoning for illustration, but it's not really valid; the scenarios we are discussing do not fit within the domain in which Newtonian gravity is an acceptable approximation to GR.
 
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  • #15
Suekdccia said:
Why do you think it's unlikely?
As you can deduce from the discussions above, none of us has done the detail calculations, so we can only go on relativistic intuition...
 
  • #16
Suekdccia said:
So I was thinking that maybe (even if our universe is accelerating), if we attach the rope to an object receding from us but not in a very long distance, we would gain some energy until putting it at rest. Then we detach the rope and let the object be attracted by gravity to us. We let it go right through our galaxy and join the Hubble flow on the other side of the sky. Then we re-attach the string at the same distance as we did before (but in the other side of the observable universe) and extract some energy until putting it at rest and let the process that I've described happen again. We do that again and again, gaining (in principle, and ignoring the endless technical difficulties) unlimited energy, as the object would never cross the cosmological horizon but at the same time it would be always dragged by the Hubble flow
@PeterDonis @Jorrie @Bandersnatch do you think this may be correct or it would fail?
 
  • #17
I'm not sure how long this will be allowed to continue by the moderators, but as always, 'the devil is in the details'. To form any reliable conclusions, one needs to specify a lot of details about the scenario accurately and then do a lot of analysis.

For example, simply saying "extract some energy until putting it at rest" is hopelessly inadequate. How much effort do you expect members to put into what they may consider as a "perpetual energy fallacy"? The onus is on you to do the calculations and prove the premise, not on members to disprove a questionable idea.
 
  • #18
PeterDonis said:
the metric in equation (1) is only valid for spatially flat universes
In reviewing the math for this, I realized that I was wrong here. The article restricts to radial motion only, which means that the differences between the spatially flat, open, and closed cases do not come into play in the coordinates the paper is using (the ##\chi## radial "angular" coordinate instead of the areal radius ##r##).
 
  • #19
Suekdccia said:
do you think this may be correct or it would fail?
Look at the two methods I described in post #12. Does your proposal match one of them? (As @Jorrie points out, you need to consider carefully how you are going to "bring to rest" the object in order to extract energy.)
 
  • #20
Ok, here is an initial look at the geodesic equation. We are looking at radial motion only, so the relevant portion of the metric is

$$
ds^2 = - dt^2 + a^2 (t) d \chi^2
$$

The relevant Christoffel symbols are

$$
\Gamma^t_{\chi \chi} = a \frac{da}{dt}
$$

$$
\Gamma^\chi_{t \chi} = \frac{1}{a} \frac{da}{dt}
$$

And the relevant geodesic equations are

$$
\frac{d U^t}{d \tau} + \Gamma^t_{\chi \chi} U^\chi U^\chi = 0
$$

$$
\frac{d U^\chi}{d \tau} + \Gamma^\chi_{t \chi} U^t U^\chi = 0
$$

where ##(U^t, U^\chi)## is the object's 4-velocity. Note that the derivatives are with respect to the proper time ##\tau## of the object, which is not the same as the FRW coordinate time ##t## (as I pointed out in an earlier post). To write the equations in terms of derivatives with respect to FRW coordinate time ##t##, we note that ##d t / d \tau = U^t## and use the chain rule, which, if we also fill in the Christoffel symbols from above and reorganize the equations to put only the time derivatives on the LHS, gives

$$
\frac{d U^t}{d t} = - a \frac{da}{dt} \frac{(U^\chi)^2}{U^t}
$$

$$
\frac{d U^\chi}{d t} = - \frac{1}{a} \frac{da}{dt} U^\chi
$$

We next use the fact that ##U^\chi = d \chi / d \tau = U^t d \chi / dt## to rewrite the above equations as:

$$
\frac{1}{U^t} \frac{d U^t}{d t} = - a \frac{da}{dt} \left( \frac{d \chi}{dt} \right)^2
$$

$$
\frac{d^2 \chi}{dt^2} = \left( - \frac{1}{U^t} \frac{d U^t}{d t} - \frac{1}{a} \frac{da}{dt} \right) \frac{d \chi}{dt}
$$

We can now substitute the first equation into the second to obtain

$$
\frac{d^2 \chi}{dt^2} = - \frac{1}{a} \frac{da}{dt} \frac{d \chi}{dt} \left( 1 - a^2 \left( \frac{d \chi}{dt} \right)^2 \right)
$$

We have a limitation that the tethered object must move slower than light relative to comoving observers in its vicinity at the instant it is released (note that this does not have to be the case after it is released), which means we must have ##| a d \chi / dt | < 1## at the instant of release. We also know that ##d \chi / dt < 0## at the instant of release, since the tethered object is not moving away from us with the Hubble flow. We also know that ##a## and ##da / dt## are both always positive, since the universe is expanding.

All of the above means that, at the instant of release, we will have ##d^2 \chi / dt > 0## (the minus sign from ##d \chi / dt## cancels the minus sign out in front on the RHS). This means that ##d \chi / dt## will get less negative. This will keep the signs of all the factors in the above equation the same as long as ##d \chi / dt## remains negative. If it is correct that the object will only asymptotically approach the Hubble flow, then ##d \chi / dt## should remain negative forever.

In my next post I'll examine what the above implies for some specific cases, which means adopting specific functional forms for ##a(t)## and filling those into the above equation.
 
  • #21
Ok, now for some specific cases. The ones I will consider here are:

(1) Matter dominated critical density (the (1, 0) case in the article). For this case, ##a(t) = t^{2/3}##.

(2) de Sitter (which would be (0, 1) in the article's notation--I use this instead of Lambda CDM (0.3, 0.7) because I can write down an exact functional form for ##a(t)##; in any case Lambda CDM should asymptotically have the same behavior). For this case, ##a(t) = e^t##.

(3) Empty universe (the (0, 0) case in the article). For this case, ##a(t) = t##.

The equation we ended up with in the last post takes the following forms for the three cases (where we have defined ##v_\chi = d \chi / dt## in order to make it clear that all of these differential equations are first order):

Case 1:

$$
\frac{d v_\chi}{dt} = - \frac{2}{3 t} v_\chi \left( 1 - t^{4/3} v_\chi^2 \right)
$$

Case 2:

$$
\frac{d v_\chi}{dt} = - v_\chi \left( 1 - e^{2t} v_\chi^2 \right)
$$

Case 3:

$$
\frac{d v_\chi}{dt} = - \frac{1}{t} v_\chi \left( 1 - t^2 v_\chi^2 \right)
$$

Let's focus on Case 3 first. If the claim in the article is correct, that proper distance is constant for this case (where, remember, "proper distance" means distance as measured in a slice of constant comoving time, i.e., along the appropriate hyperbola in Minkowski spacetime), then we should have ##a \chi = \text{const}##, and hence

$$
\frac{da}{dt} \chi + a v_\chi = 0
$$

which rearranges to

$$
v_\chi = - \frac{1}{a} \frac{da}{dt} \chi
$$

Using ##a(t) = t## for this case, we get

$$
v_\chi = - \frac{1}{t} \chi
$$

which obviously integrates to

$$
\chi = \chi_0 - \frac{1}{t}
$$

This gives ##v_\chi = 1 / t^2## and ##d v_\chi / dt = - 2 / t^3##; substituting these into the Case 3 equation above gives

$$
\frac{2}{t^3} = \frac{1}{t^3} \left( 1 - \frac{1}{t^2} \right) = \frac{1}{t^3} - \frac{1}{t^5}
$$

Since this equation is obviously false, we find that it is impossible for proper distance to be constant for this case, contrary to the claim given in the article (but as expected given my earlier comments on this case).

At this point I'm going to cheat and use an online ODE solver to see if closed form solutions for the above equations are obtainable. I'll use the solver here:

https://www.symbolab.com/solver/ordinary-differential-equation-calculator/

The solutions it finds are (##C_1##, ##C_2##, and ##C_3## are integration constants):

Case 1:

$$
v_\chi = - \frac{1}{t^{2/3}} \sqrt{\frac{3}{C_1 - 4 \ln t}}
$$

Case 2:

$$
v_\chi = - e^{-t} \sqrt{\frac{1}{C_2 - 2 t}}
$$

Case 3:

$$
v_\chi = - \frac{1}{t} \sqrt{\frac{1}{C_3 - 2 \ln t}}
$$

The factors out in front of these look fine, but the terms in the denominator under the square root are weird; taken at face value they would indicate that the solution would become invalid at the value of ##t## that would make the quantity under the square root negative. So it looks like I need to review this further, but I'm posting it so others can take a look and maybe spot some place where I've missed something.
 
  • #22
In my opinion, there is a fundamental error in the reasoning. On one side is the expanding universe. On the other hand, there is a string, which is, as it were, separate from the Universe.
When you try to formulate for yourself what an object is - a string or what you attached it to - you begin to have difficulties.
Let me put it simply - your string will "lengthen" synchronously with the Universe.
There is an interesting theory of long strings. Not to be confused with string theory, where strings are multi-dimensional little worms :)
 
  • #23
Conn_coord said:
In my opinion, there is a fundamental error in the reasoning. On one side is the expanding universe. On the other hand, there is a string, which is, as it were, separate from the Universe.
No, this is not what is discussed here. Read the posts and the referenced papers carefully.
 
  • #24
Conn_coord said:
When you try to formulate for yourself what an object is - a string or what you attached it to - you begin to have difficulties.
No, you don't. There are well known ways to do this in relativity.

Conn_coord said:
Let me put it simply - your string will "lengthen" synchronously with the Universe.
No, it won't. It is perfectly possible to have a string that does not expand with the universe. The string will have tension in it, but that doesn't prevent it from having a constant length.
 
  • #25
So a LY is 10^12 km and the mass of the Earth is on the order of 10^24 kg. Say any conceivable material has a mass of 1 kg/km, which would be super light, then at any cosmological distances the gravitation of the tether would counteract the expansion?
 
  • #26
BWV said:
Say any conceivable material has a mass of 1 kg/km, which would be super light, then at any cosmological distances the gravitation of the tether would counteract the expansion?
No. The mass of the Earth would be negligible on cosmological scales.
 
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FAQ: Mining the Universe's expansion?

1. What is the Universe's expansion?

The Universe's expansion refers to the continuous increase in the distance between galaxies, clusters of galaxies, and other cosmic structures. This expansion is believed to have started with the Big Bang and is currently accelerating due to dark energy.

2. How do scientists measure the Universe's expansion?

Scientists use a variety of methods to measure the Universe's expansion, including the redshift of galaxies, the cosmic microwave background radiation, and the brightness of supernovae. These measurements can help determine the rate of expansion and how it has changed over time.

3. What is dark energy and how does it affect the Universe's expansion?

Dark energy is a mysterious force that is believed to make up about 70% of the total energy in the Universe. It is responsible for the acceleration of the Universe's expansion, pushing galaxies and other cosmic structures further apart from each other.

4. Can the Universe's expansion be reversed?

At this time, there is no evidence to suggest that the Universe's expansion can be reversed. In fact, the expansion is believed to be accelerating and will continue to do so in the future. However, there are theories that suggest the expansion may eventually slow down or even reverse in a process called the "Big Crunch."

5. How does the Universe's expansion affect our understanding of the origins of the Universe?

The Universe's expansion is a crucial piece of evidence that supports the Big Bang theory, which is the prevailing scientific explanation for the origins of the Universe. By studying the expansion, scientists can gain insights into the early stages of the Universe and how it has evolved over time.

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