# Interpreting Friedmann’s Equation?

1. Jul 20, 2008

### mysearch

Hi, I am trying to make some assessment and interpretation of Friedmann’s equation, which is often presented in the form:

[1] $$H^2 = \frac{8}{3}\pi G \rho - \frac {kc^2}{a2} + \frac{\Lambda}{3}$$

While I understand that Friedmann’s equation is said to be a solution of Einstein’s field equation within general relativity, it seems that a basic derivation can be made by considering the conservation of energy of the kinetic and potential energy of a unit mass [m] at the surface of an expanding sphere of radius [R] to a central mass [M]. The following equations are only to provide a basic frame of reference:

[2] $$E_T = E_K + E_P = \frac{1}{2}mv^2 + \left(-\frac{GMm}{r}\right)$$

[3] $$E_T = \frac{1}{2}mv^2 - G\left(\frac {4}{3}\pi R^3 \rho\right) \frac {m}{r}$$

[4] $$\frac{E_T}{r^2} = \frac{1}{2}m\left(\frac{v}{r}\right)^2 - \frac {4}{3}\pi G \rho m$$

[5] $$H^2 = \frac {8}{3}\pi G \rho + \frac{2 E_T}{mr^2}$$

My first question is whether the last 2 terms in [1] can be considered comparable to the last term in [5]? However, if we drop $$+ \frac{\Lambda}{3}$$ for the moment, we would have to equate :

[6] $$- \frac {kc^2}{a^2} = + \frac{2 E_T}{mr^2}$$

To be compatible to equation [1], the units must resolve to $$1/sec^2$$. Now on the assumption that (k) has no units and the units of [c] are known, then [a] must have units of distance as per [r]. In fact, if I equate [a=r] and describe this distance as a function of expansion of the universe over time, can we rationalise [k] as follows:

[7] $$k = \frac{2E_T}{mc^2}$$

The units seem to cancel out correctly, but I am left wondering about the value of [k]. It seems that $$mc^2$$ would correspond to the rest energy of the unit mass [m], but what figure can be put on the total energy and would it give k->0?

Taking equations [1] & [5] as a whole, the value of [H] has been established from redshift measurements. If I extend the linearity of [H] to the edge of a visible universe, the value of [r=c/H=13.9 billion light-years]. As such, I appear to be left with 2 unknowns, i.e. density $$[\rho]$$ and total energy [Et]. While the mass-density is often substituted into the Friedmann equation, I would have thought this would be relatively meaningless given the speculation that matter only accounts for about 4% of the required energy density of a universe where r=13.9 billion light-years?

Therefore, would greatly appreciate any technical insights regarding any of the issues raised? Thanks

P.S. In part this is an extension of another thread about 'Distance & Hubble Constant'

Last edited: Jul 20, 2008
2. Jul 21, 2008

### mysearch

In post #1, of this thread, some questions regarding the derivation of the Friedmann equation were raised. Specifically, the interpretation of [k] in terms of classical units. The link below is related, but presented in another thread, which is trying to ascertain the validity of a very, very basic expansion model using the Friedmann equation:

https://www.physicsforums.com/showpost.php?p=1808882&postcount=13

Realise that some of these question may seem a bit elementary, but just trying to build a basic model of understanding, so any technical insights would be appreciated.

3. Jul 21, 2008

### neutralseer

As it turns out, your Newtonian derivation only gives the idea of what the Friedmann equation is all about; it says the expansion of the universe is related by the self gravity of all the mass in the universe. You could go further in connecting (by analogy only) the total energy per unit mass of the test particle with the curvature of the universe. However, you can't simply equate the terms in the equations since they are fundamentally different. Basicly, your derivation is correct for a newtonian universe, but it appears we live in an Einsteinian one, so you shouldn't go to far in relating the energy of the particle to the curvature.

In General Relativity (GR) it isn't just the mass that curves space, it is the total energy of a particle that curves space, $E= \sqrt{m^2p^2+p^2c^2}$. This means you should actually be focusing on the energy density, not the mass density of the universe. Thus, when you worry about the mass energy the universe being only 4%, that's okay, the other 94% of the energy density is in the form of dark matter and dark energy, making the universe have zero curvature (k = 0).

Finally, I'm not sure what you mean by r=13.9 billion light years. That's not the radius of the universe. It is true that the CMB photons we see have been traveling about that long (actually 13.9 billion - 400,000 ~ 13.9 billion), but the farthest thing that we could see in principle is much further (about 46 billion light years).

This is all confusing stuff for me too, let me know if you think I'm wrong or unclear on something.

4. Jul 22, 2008

### mysearch

Response to #3

Neutralseer, thanks for the response:

I entirely agree, but I find it useful to create basic models to see where the discrepancies appear.

Good point. In part, some of my questions on this subject have drifted over into another thread https://www.physicsforums.com/showthread.php?t=245769 or more specifically https://www.physicsforums.com/showpost.php?p=1809935&postcount=19

One of the questions I was trying to resolved was the validity of an expansion model based on Friedmann’s equation when k=0 in a matter-dominated universe:

$$H^2 = \frac{8}{3}\pi G \rho = \frac{8}{3}\pi G \left(\frac{M}{4/3\pi r^3}\right)$$

$$H = \frac{1}{t} = \sqrt{ \frac{2GM}{r^3}}$$

You’re right. Part of this is taken up in the thread mentioned above. One of the members highlighted the following reference, which I have only started to read, but appears to be very useful: http://arxiv.org/PS_cache/astro-ph/p.../0012376v1.pdf

I have only recently taken a serious interest in this subject. However, most of the members of this forum are very helpful and will point you in the direction of useful information.

5. Jul 22, 2008

### neutralseer

Re: Response to #3

Your equations seem to give the correct scaling for the scale factor $a(t)$, where $d(t)=a(t)d(t_0)$, and $a(t_0)=1$. That is,

$$H = \frac{\dot a}{a} = \sqrt{ \frac{2GM}{(ad(t_0))^3}}$$,

implying that

$$\dot a=a^{-1/2}$$.

Solving for $a(t)$ we get,

$$a(t)=Ca^{2/3}$$,

where C is a some constant. Note this yields the correct scaling for a flat matter dominated universe.

An immediate discrepancy that I can see relates to recessional velocities,

$$v_{rec}=\frac{\dot a}{a}d(t)=H(t)d(t)$$.

At a given time, $t$, there is a distance beyond which any receding particle (galaxy) has a recessional velocity exceeding the speed of light, $d(t)=c/H(t)$. In GR this wouldn't be a problem since you and the receding galaxy do not occupy a flat spacetime (not in an inertial reference frame). However, under the implicit assumptions of the Newtonian friedmann equation you are in an inertial reference frame with all the expanding particles, thus the Newtonian friedmann equation is inconsistent with special relativity.

6. Jul 23, 2008

### mysearch

Response to #5

Hi, I would like to bounce a few ideas around, but first wanted to make it clear that I am not trying to propose any wacky new idea in contradiction to either special or general relativity. However, I would suggest that in the context of a cosmological model, Einstein’s field equations of general relativity are not a solution, in isolation, because they require certain assumptions about spacetime to made before any solution can proceed. So, first some general points for consideration, which are partly intended to provoke a degree of debate, but are open for correction:

1) On the large scale, the universe is homogeneous and the mass-density is very low, in the order of 10-20 particles/cu.m. Therefore, as a large-scale generalisation, the Schwarzschild’s metric of general relativity would collapse towards the flat spacetime of special relativity, unless there was some large-scale centre of gravity at work.

2) If the universe was infinite, a theorem called Newton’s Shells implies that no point would feel the net effect of gravity, i.e. there would be no centre of gravity within the universe.

3) Of course, if the universe were finite, the same theorem would suggest that the universe would have a centre of mass. However, many people may counter this assumption based on the nature of the finite geometry being assumed.

4) Accepting the verified corrections of general relativity, under extremes of gravitation force/curvature, Newtonian physics should still remain a useful approximation.

OK, with these points tabled for discussion, I agree with your derivation, but would like to highlight a similar relationship based on Newtonian physics:

$$g = \frac{GM}{r^2} = a = \frac{v}{t} = \frac{r}{t^2}$$

$$r^3 = GMt^2$$

$$r = (GM)^{1/3}t^{2/3}$$

In a wider context, the Friedmann, fluid and acceleration equations of cosmology all suggest that H=2/3t for a matter dominant universe. However, returning to what I believe is the key point of your response:

Certainly, there is a suggestion with a derivation based on the conservation of energy, see post #1 for details, that the equation is viewing the outward velocity of the unit mass (m) from some inertial frame of reference. However, this frame is conceptual, especially when the mass (m) exceeds the speed of light [c], as predicted when r=c/H. It is highlighted that a photon in this conceptual frame of reference would also exceed [c] at a given radius. As such, I not sure that it is inconsistent with special relativity, but again this point is open for correction. So some questions:

1) Many sources make it clear that the Big Bang should not be viewed as an explosion, which in some ways seems to be the assumption of the conservation of energy derivation in post #7. So why does this derivation lead to an accepted form of the Friedmann equation?

2) If the Big Bang is not an explosion, but rather an expansion of each unit volume of space, then it seems to raise some awkward questions about the change in the kinetic and potential energy associated with mass (m), i.e. is this energy change zero? In part, equation 7 in the Newtonian derivation in post #1 was alluding to this issue, i.e.

$$k = \frac{2E_T}{mc^2}$$

3) Typically, mass (m) might be described as co-moving with the expansion of the universe. However, it is my understanding that the size of an atom is not affected by this expansion. If so, does this mean that aspect of an atom’s mass do physically move through expanding space? Are there any implications on particle physics if the structure of an atom is changing with respect to expanding spacetime?

Again, I would like to highlight that the nature of these questions are simply an attempt to better understand some of the underlying principles at work. Thanks

7. Jul 23, 2008

### neutralseer

Oops, my last post contained some errors, with all mysearch's equations, you actually get:

$$\dot a(t)=Ka^{-1/2}$$

$$a(t)=Ct^{2/3}$$

mysearch,
I don't have enough time now to answer all your questions, but let me try to answer a few off the top of my head.

1) In a homogeneous isotropic universe, there is no center of gravity, the correct metric is the FLWR metric.

2) I was thinking about this shell's thing a few days ago. Imagine your the test mass on the edge of the expanding sphere, you only feel the gravitational pull from that sphere. Now add on a layer of mass around the sphere, that extra layer won't affect the test mass according to the shell's theorem (or gauss' law for gravity). Now keep on adding those layers of mass ad infinitum, and you get an infinite universe, yet your newtonian friedmann equation still applies. That's my answer for now, but I'm still a little uncomfortable with it.

One addition to this. If all the galaxies were expanding in a newtonian universe it wouldn't make any sense. You would have to set the initial conditions for each particle velocity to get an effect similar the expansion of space in GR. In GR you only have a handful of initial conditions. Thus a newtonian description of expansion suffers from a huge fine tuning problem.

3) You're right about curvature. If the universe were finite it's 3D spatial structure would be like a sphere. Is there any unique point on a sphere? Nope.

4) GR isn't just important in the strong field limit. In some simulations of the universe people do assume the particles (galaxies) interact gravitationally via the inverse square law. The important corrections they add are: the gravitational interaction can only propagate at the speed of light and the universe is expanding in accordance with the Friedmann equations.

Good questions, I'll see if I can try to answer the rest later.

8. Jul 23, 2008

### mysearch

Quick Response to #7

I will raise comments on your feedback in #7, but will wait until you have had more time to consider the issues. The point I did want to highlight again was that I am not trying to propose a Newtonian-Friedmann solution, simply exploring some of the implications. Therefore, I have appreciated your feedback to-date. Thanks

9. Jul 24, 2008

### neutralseer

Re: Response to #5

I’m not sure what you mean by “this frame is conceptual.” You derived your equations from an inertial frame, and from that inertial frame they violate special relativity.

Now for the second set of numbered questions:

1) The Newtonian friedmann equation you derived does not assume the big bang was an explosion. An explosion causes a blast wave -- a spherical shell (at least initially) of material with a Bulk kinetic energy roughly equal to the energy of the blast. An elementary analysis of this wave in terms of conservation of energy, will yield the taylor-sedov solution. This solution gives different results for how the radius of the blast wave depends on time than the Friedmann equation. Thus the Newtonian friedmann equation does not describe an explosion because it gives a different time dependence of the scale factor, and because it describes a uniformly expanding sphere (as opposed to the explosively expanding spherical shell).

In the end, the reason the sources make clear the big bang is not an explosion is because the correct Friedmann derived from General Relativity does not talk about particles moving through space, like an explosion. It describes how space itself expands.

2) Once again you’re overinterpreting the Newtonian Friedmann equation. A particle has kinetic energy when it moves through space, but in the big bang frame work, particles (or galaxies) are not moving through space. Thus, a straightforward interpretation of at least kinetic energy isn’t as obvious as the usual $KE=1/2mv^2$.

In fact the whole idea of energy and conservation of energy is a tricky concept when looking at the universe as a whole. Here’s what Sean Carroll (Cal Tech cosmologist) has to say about this question:

“Is energy conserved in an expanding universe?
This is a tricky question, depending on what you mean by "energy." Usually we ascribe energy to the different components of the universe (radiation, matter, dark energy), not including gravity itself. In that case the total energy, given by adding up the energy density in each component, is certainly not conserved. The most dramatic example occurs with dark energy -- the energy density (energy per unit volume) remains approximately constant, while the volume increases as the universe expands, so the total energy increases. But even ordinary radiation exhibits similar behavior; the number of photons remains constant, while each individual photon loses energy as it redshifts, so the total energy in radiation decreases. (A decrease in energy is just as much a violation of energy conservation as an increase would be.) In a sense, the energy in "stuff" is being transferred to the energy of the gravitational field, as manifested in the expansion of the universe. But there is no exact definition of "the energy of the gravitational field," so this explanation is imperfect. Nevertheless, although energy is not really conserved in an expanding universe, there is a very strict rule that is obeyed by the total energy, which reduces to perfect conservation when the expansion rate goes to zero; the expansion changes the rules, but that doesn't mean that anything goes.”

3) I have to say this question confuses me too. The answer I always find is (from Sean Carrol again):

“No. Any system that is bound together by internal forces -- whether it is a table, the solar system, or the galaxy -- does not expand along with the universe. (Not just that it only expands slightly; it really doesn't expand at all, or at least not because of the expansion of the universe.) To observe the expansion, we need to study objects that are very distant, not directly bound to us by gravity or anything else.”

Here's what my intuition (incorrectly?) tells me:

Let's say space is initially static and not expanding. Then two stars form a (bound) binary system (let's not talk about atoms since we'd have to worry about quantum mechanics) with a seperation $d$. Now let the space start expanding described by a Hubble constant. It seems to me the binary system of stars would expand a bit, and then reach a new equilibrium distance of $d+\epsilon$. Notice that as long as the Hubble constant is truly constant, then the distance between the stars would not change. However, the Hubble constant isn't really constant, it changes with time (People actually call the time dependent version the Hubble parameter). Thus, according to my intuition, the distance should be slowly changing. Since Sean Carroll is the expert here, I assume I'm wrong.

These are all great questions, many of which I have asked before (and sometimes am still asking).

10. Jul 24, 2008

### neutralseer

One more thing regarding the "expanding binary star system" issue. The general relativistic way of looking at is this is:

When you do GR on the entire universe, pretending that matter is perfectly smoothly distributed you get the expanding friedmann solutions. However, when you do GR on our solar system, you get a completely different metric that doesn't include space expanding. Nonetheless, my question remains, what does the true metric look like? That is, what does the metric, Schwarzschild + FLWR look like? I know the Schwarzschild metric dominates in the case of the binary star system, but does FLWR metric not affect the binary system at all?

11. Jul 24, 2008

### MeJennifer

Then you are misinformed the Schwarzschild metric does not even remotely model the vacuum solution of two point masses.

12. Jul 24, 2008

### neutralseer

MeJennifer,
Thanks for the correction. It sounds like you know what your doing, maybe you could help us out here. Let me try to rephrase my question.

Instead of considering a binary system of stars, let's consider one star (spherical, non-rotating) with a very light point mass orbiting the star. This situation would give us, approximately, a Schwarzschild metric.

Let's call the metric of this bound system g* (which for the above case is approximately a Schwarzschild one) and the metric for the universe g(FLWR). Obviously, on the scale of this bound system, g* dominates over g(FLWR), and g* doesn't include an expanding component, while g(FLWR) does. This seems to be the usual reason people give for why the universe isn't expanding.

But doesn't the true metric (where g(true)~g*) include an expanding component? Is my error that I'm implicitly assuming some sort of superposition of the metrics?

Here's my guess at my own question:

g(FLWR) is only valid for regions of space that have energy densities much less than the average energy density of the universe (which is probably the critical density). The density of the above mentioned bound system far exceeds the average density of the universe. Thus this system drops out of the Hubble flow, meaning g(FLWR) just doesn't apply to it at all, and g(true)~g(schwarzschild) takes hold.

What do you think, does that answer my question?

13. Jul 24, 2008

### MeJennifer

The problem is that due to the fact that GR is a non linear theory you cannot simply "add" two or more solutions. For instance some of the rotating two mass solutions have gravitational waves coming from infinity. However there are some exceptions, for instance one can paste multiple Schwarzschild funnels together.

14. Jul 25, 2008

### mysearch

Response to #7

For cross-reference:
In post #6, 4 general points were raised for consideration along with 3 questions. Neutralseer has responded to the 4 general points in post #7 and the 3 questions in post #9. This post is a response to #7:

I tend to agree with you, but that still leaves the issue of whether the universe is infinite or finite in spatial terms. The Newton shells theorem was simply to illustrate how an infinite universe might not have any centre of gravity. However, would a nearly infinite spatial universe approximate to this theorem? See link to reference to the particle horizon, estimated at around 46 billion lightyears minus inflation:
https://www.physicsforums.com/showpost.php?p=1811258&postcount=22
If you set k=0 in the FLWR metric, you appear to be back to the flat spacetime of SR, is this significant to bullet 3) below?

Not sure I agree with the first sentence. As I understand it, there would be no effective gravitational pull towards the cavity shell for any mass (m) within the cavity, irrespective of whether it was expanding or not. In an infinite universe, my assumption was that you could consider the shell to be infinitely thick from any point.

I assume you are referring to the issue that every unit mass (m) would have to be given a unique amount of kinetic energy so that the expansion of the universe was both homogeneous and isotropic from every point of observation. This does seem to be a bit of a problem, but as stated, I wasn’t seriously proposing this as a solution, only trying to understand the issues associated with it. However, I not sure that I understand how the accepted theory explains the uniform expansion of each unit volume of space given that the rate is assumed linear with distance, but variable with time? ( see later comments about dark energy in following post)

Again, I am assuming you are making reference to the 2-D/3-D balloon analogy of expanding spacetime. To be honest, I am not always sure what people are really trying to imply with this analogy. First, it is only an analogy, not a serious description of 3-D/4-D spacetime. Is there any subliminal implication that you can’t travel to the edge of the universe without coming back to where you started? If k=0 and the FLWR metric collapses towards SR flat spacetime, I not sure how you avoid the centre of gravity issue. Again, this is an inquiry not a statement of fact.

4) GR isn't just important in the strong field limit. In some simulations of the universe people do assume the particles (galaxies) interact gravitationally via the inverse square law. The important corrections they add are: the gravitational interaction can only propagate at the speed of light and the universe is expanding in accordance with the Friedmann equations.

Wasn’t really sure what specific point was being raised here, so my comments may be a bit off base and I need to double-check my maths, but here goes. Today, you can calculate the Hubble radius = c/H, which comes out at about 13.7 billion lightyears. Its significance being that it is the point at which the recession velocity associated with the expansion of space exceeds [c]. If you could mark this point in space and then reverse time, the radius would reduce as we wind back the expansion of the universe. However, as we do, the Friedmann equation suggests that the rate of expansion would increase. If we wind back the age of the universe to 1 billion years, which is still within the matter-dominated era, I believe the value of (H) increases by over a factor of 10. This suggests that our point in space could no longer be called a Hubble radius, because the rate of expansion would be greater than [c] by over a factor of 2. If my assumptions are correct, then the expansion of the universe was outrunning gravity, which is confined to [c], as you point out. Given that Friedmann’s equation is predicated on the gravitational constant (G), does this cause other problem?

15. Jul 25, 2008

### mysearch

Response to #9

For cross-reference:
In post #6, 4 general points were raised for consideration along with 3 questions. Neutralseer has responded to the 4 general points in post #7 and the 3 questions in post #9. This post is a response to #9, whilst an earlier post responded to #7.

I realised I was on shaky ground here, but let me try to clarify what I was think about. It is a postulate of SR that nothing can go faster than light [c]. At the Hubble radius, the expansion of the universe exceeds [c]. Cosmology and SR are resolved by highlighting that at no point in space does any massive particle exceed the speed of light [c]. However, ‘conceptually’ to an inertial frame of reference the massive particle and a photon could both be receding faster than [c].

Thanks for the information about the taylor-sedov solution; I had not heard of this. Does this solution require the concept of space for the blast wave to expand into, which presumably didn’t exist in this unique case? You point out, GR defines the expansion of space, but as raised in my previous post, what I am not clear about is how the accepted theory explains what causes this uniform expansion of each unit volume of space given that the rate is assumed linear with distance, but variable with time? I known dark energy is usually mention about here, but I not sure that we can describe the physics of something we don’t even know really exists, but I am pre-empting your Sean Carroll quote.

More than likely, but I was really only trying understand why the conservation of energy approach to the derivation in post #1 comes up with the same form as GR, especially in light of Sean Carroll comments about the large scale issues associated with energy conservation. I was particular intrigued by what is implied by the following quote:

As cited by Carroll, there is the issue of where energy is lost by photons due to the $$[1/r^4]$$ factor. There is the more obvious implication that kinetic energy can be converted into gravitational potential energy. You also mentioned earlier that:

So I was wondering whether gravitational potential is lost as parts of the universe expands away faster than the speed of light? Coming back to the issue of dark energy, I grant you that it provides an explanation of the uniformity of expansion per unit volume, but the fact that it stays constant or even increases with time in order to drive an increasing rate of expansion presumably requires some unknown ‘quantum’ reservoir of energy, e.g. zero-point energy?

Now the interesting thing about this statement is that it seems to imply that there is a threshold at which expansion overpowers the ‘internal forces’ of a given structure. Apparently, it is not big enough to overpower nuclear force holding together the atomic nuclei, nor the electromagnetic force between nuclei and electrons or even the gravitational force within a galaxy, but is big enough to ‘overpower’ the weaker gravitational attraction between galaxies, which presumably helped form the large-scale structures of the universe?

I appreciate that it is unfair to keep raising questions that may not yet have an answer, but all the same, it sometimes helps to put things in some sort of perspective

16. Jul 25, 2008

### neutralseer

mysearch,
I'll respond to your questions soon. Trying to answer these questions is actually helping me understand cosmology a lot better since this isn't my area of expertise. For an excellent introduction to cosmology I highly recommend reading "Introduction to Cosmology," by Barbara Ryden. For now, let me correct some embarrassing errors I've made:

Post #3 I should have written, $E=\sqrt{p^2c^2+m^2c^4}$.

Post #12 I said (at the end of the third paragraph), "This seems to be the usual reason people give for why the universe isn't expanding." I should have said, "This seems to be the usual reason people give for why the bound systems aren't expanding."

Cheers,
neutralseer