The Balloon Analogy .... the Good, the Bad, and the Ugly - Comments

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  • #36
Hornbein
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I agree, it's a good analogy. Why don't you write an insights article on it? I'm sure Greg would be happy to have that.

Hmmm, how would I go about doing that? The key thing that people miss is that the early Universe had no empty space whatsoever and stayed that way for quite a long time. I think it would be entertaining to trace an imaginary eye witness of the Universe looked as it grew. At first it is unimaginable, then like starting at the center of the Sun and traveling outward. Then some empty space appears. The color goes down the spectrum to dark red then the whole thing turns black though still very hot. It appears that it will stay black forever, then the previously negligible force of gravity very slowly saves the day. Giant suns form and quickly explode to make iron and such. Neutron stars collide to produce the heavy elements. Rocky planets form.
 
  • #37
phinds
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Hmmm, how would I go about doing that? The key thing that people miss is that the early Universe had no empty space whatsoever and stayed that way for quite a long time. I think it would be entertaining to trace an imaginary eye witness of the Universe looked as it grew. At first it is unimaginable, then like starting at the center of the Sun and traveling outward. Then some empty space appears. The color goes down the spectrum to dark red then the whole thing turns black though still very hot. It appears that it will stay black forever, then the previously negligible force of gravity very slowly saves the day. Giant suns form and quickly explode to make iron and such. Neutron stars collide to produce the heavy elements. Rocky planets form.
I think we must be talking about different "baking bread" analogies. The one I'm talking about has noting to do, really, with all that you just said. Rather, it is an alternate to the balloon analogy and talks about the universe NOW, not starting just after the singularity.
 
  • #38
Hornbein
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I think we must be talking about different "baking bread" analogies. The one I'm talking about has noting to do, really, with all that you just said. Rather, it is an alternate to the balloon analogy and talks about the universe NOW, not starting just after the singularity.

Right. It took a while for the "raisins" to form.

I was thinking I couldn't fill up a whole article with the raisin bread thing. But I guess I can if I try.
 
  • #39
phinds
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Right. It took a while for the "raisins" to form.
Yes, it did, but that has nothing to do with the raisin bread analogy, which is just a way of talking about what the "raisins" are doing now and in the future.
 
  • #40
Imager
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@phinds

Thumbs up! As beginner, I find the analogy very helpful and I look forward to your next Insights article!
 
  • #41
Edriven
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Could you please explain this free fall to me. I thiught sun was in orbit at 828,000 mph and completes one orbit in 230 million years.
 
  • #42
Edriven
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I agree just wish we had a better model
 
  • #43
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Thx for helping me with this topic. I thought there is no way that it could be applied evenly because of solar busts and storms happen unevenly Most of its power or wind goes in space but when a another star is "near" it's solar winds push on our entire universe.
 
  • #44
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Could you please explain this free fall to me.

Free fall means zero proper acceleration--it means the object feels no force. It is weightless.

I thiught sun was in orbit at 828,000 mph and completes one orbit in 230 million years.

It's in orbit about the center of the galaxy, yes. But that's perfectly consistent with it being in free fall, just as the International Space Station, in orbit about the Earth, is in free fall--it feels no force, and is weightless.
 
  • #45
Edriven
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Free fall means zero proper acceleration--it means the object feels no force. It is weightless.



It's in orbit about the center of the galaxy, yes. But that's perfectly consistent with it being in free fall, just as the International Space Station, in orbit about the Earth, is in free fall--it feels no force, and is weightless.
Thanks for the explanation. But the difference in the space station and the sun is that the space station is not accelerating away from the earth. Due to red shift we know that we are accelerating away from other stars. Right?
 
  • #46
phinds
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Thanks for the explanation. But the difference in the space station and the sun is that the space station is not accelerating away from the earth. Due to red shift we know that we are accelerating away from other stars. Right?
You seem to keep changing the subject. What is it that you want to know, exactly? Please be as precise as you can with your question.
 
  • #47
Edriven
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You seem to keep changing the subject. What is it that you want to know, exactly? Please be as precise as you can with your question.
I'm not sure how I changed the subject. I was responding to a post discribing acceleration of the universe. My point is, that red shift tells us that the universe is accelerating. Therefore the sun can not be in a balanced free fall because it is accelerating. Acceration means that forces are not equal and solar thrust is pushing stars apart. It's my idea that solar thrust is the energy described in dark matter. What better place to get energy from than the stars? What would change the direction of solar thrust? Gravity and collision of other matter.
 
  • #48
phinds
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I'm not sure how I changed the subject. I was responding to a post discribing acceleration of the universe. My point is, that red shift tells us that the universe is accelerating. Therefore the sun can not be in a balanced free fall because it is accelerating. Acceration means that forces are not equal and solar thrust is pushing stars apart. It's my idea that solar thrust is the energy described in dark matter. What better place to get energy from than the stars? What would change the direction of solar thrust? Gravity and collision of other matter.
Ah, good. Now I understand your question.

No, red shift does NOT tell us that the sun is accelerating. You are making a very common mistake of confusing recession velocity with proper velocity. Google "Metric Expansion" for more discussion.
 
  • #50
phinds
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http://news.nationalgeographic.com/...iverse-expansion-what-is-dark-energy-science/
I'm sorry I might have confused you with my red shift comparison. Nonetheless the universe is expanding. Which means unequal force and acceleration
No, it does not. Yes, the universe is expanding, and at an accelerating rate, but what "unequal force" are you talking about? The sun in not accelerating in the sense that you think it is, nor are the distant galaxies that are receding from us at an accelerating rate.
 
  • #51
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the difference in the space station and the sun is that the space station is not accelerating away from the earth. Due to red shift we know that we are accelerating away from other stars. Right?

Not in the sense I have been using the term "acceleration". You will notice that I have tried to say "proper acceleration", to make it clear that I am talking about acceleration that is actually felt--or, in the case of the space station and the sun and distant galaxies, not felt. All of those objects are in free fall, feeling zero acceleration.

The kind of "acceleration" you are referring to when you say that the sun is accelerating away from other stars is more precisely called "coordinate acceleration". (Actually, the red shifts you are referring to are from other galaxies, not other stars; we can't see individual stars at the distances at which the cosmological redshift becomes measurable. So it's more correct to say that our galaxy as a whole sees other galaxies' light as redshifted.) The key point about coordinate acceleration is that you can change it by changing coordinates--i.e., by a mathematical abstraction that doesn't change anything physical. So if you are trying to understand the actual physics going on, coordinate acceleration is the wrong thing to focus on.

red shift tells us that the universe is accelerating.

No, they don't. Red shifts, in and of themselves, only tell us that the universe is expanding. They do not tell us that the expansion is accelerating. For that, we need not only the observation that there are red shifts, but detailed correlations between red shifts and other observations, like the brightness and angular size of distant galaxies.

Also, once again, when we say the universe's expansion is "accelerating", we mean this in the sense of coordinate acceleration, not proper acceleration. All of the galaxies have zero proper acceleration. See below.

Therefore the sun can not be in a balanced free fall because it is accelerating.

You are confusing coordinate acceleration and proper acceleration here. The sun has zero proper acceleration, and that means there is no net force acting on the sun. The sun does have nonzero coordinate acceleration in certain coordinates, but coordinate acceleration does not tell you whether there is an unbalanced net force. Only proper acceleration does. As above, all of the galaxies have zero proper acceleration, so there is no unbalanced net force acting on any of them.
 
  • #52
1oldman2
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As a "layman"(barely) I find the balloon analogy and related conversations very useful and interesting in regards to my understanding of the principles discussed.I must add at this point that I'm no scholar, we won't be collaborating on any papers and the Nobel prize is safe from me. in fact if intelligence is relative then compared to yours mine would be measured on the Planck scale.My point is when someone takes the time to develop good analogies such as the balloon one here then a much broader segment of society is able to learn these concepts and hopefully get interested and excited about the sciences in general. Good job ! looking forward to more writing in this manner.
 
  • #53
RUTA
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So is there a valid feature of the Balloon analogy that actually contradicts with Doppler recession?

Narlikar once told me you can derive the cosmological redshift using the Doppler shift between local frames, integrated from emission event to reception event. I've always meant to check that, but never got around to it ...
 
  • #54
marcus
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Narlikar once told me you can derive the cosmological redshift using the Doppler shift between local frames, integrated from emission event to reception event. I've always meant to check that, but never got around to it ...
http://arxiv.org/abs/0808.1081
The kinematic origin of the cosmological redshift
Emory F. Bunn, David W. Hogg
(Submitted on 7 Aug 2008 (v1), last revised 14 Apr 2009 (this version, v2))
A common belief about big-bang cosmology is that the cosmological redshift cannot be properly viewed as a Doppler shift (that is, as evidence for a recession velocity), but must be viewed in terms of the stretching of space. We argue that, contrary to this view, the most natural interpretation of the redshift is as a Doppler shift, or rather as the accumulation of many infinitesimal Doppler shifts. The stretching-of-space interpretation obscures a central idea of relativity, namely that it is always valid to choose a coordinate system that is locally Minkowskian. We show that an observed frequency shift in any spacetime can be interpreted either as a kinematic (Doppler) shift or a gravitational shift by imagining a suitable family of observers along the photon's path. In the context of the expanding universe the kinematic interpretation corresponds to a family of comoving observers and hence is more natural.
 
  • #55
RUTA
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That looks familiar, marcus. I think we've had this exchange before :-)
 
  • #56
RUTA
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I finally took the time to derive the cosmological redshift [itex]z+1 = \frac{a_o}{a_e}[/itex] using the accumulated nonrelativistic redshifts along the path of the light, as in Eqs (4) & (5) of http://arxiv.org/pdf/0808.1081v2.pdf. It's straightforward and nicely shows how the local flat frames can be pieced together along the light path in the global curved spacetime. I also liked the authors' derivation of their Eq (6) [itex] \sqrt{\frac{c + v_{rel}}{c - v_{rel}}} = \frac{a_o}{a_e}[/itex] using the SR velocity addition equation in local flat frames along the light path, i.e., parallel transport. Very cool. I will add the contents of this paper to my GR course.

That being said, I don't agree with their view that [itex]v_{rel}[/itex] constitutes a "natural" notion of velocity for cosmology. "... the velocity of the galaxy at the time of light emission relative to the observer at the present time" is about as unnatural as I can imagine haha. The natural notion is clearly v = Hr, i.e., proper time rate of change of proper distance, which holds not only locally, they use it to get both Eqs (5) and (6), but globally in the "expanding bread" or "stretching rubber sheet" analogies. That view is very Newtonian, so it's very intuitive and as Rindler once told me, "It's not just an analogy, it's exact!" But, it is certainly the case that there is no unique way to define velocity between objects that don't share a local flat frame in curved spacetime, so to each his own :-)

Thanks again for sharing that article, marcus. Hopefully, now that I've gone through the derivation myself, I won't keep requiring you to post it :-)
 
  • #57
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Someone who wishes to remain anonymous sent me the following link: http://arxiv.org/abs/1111.6704. In this paper, Dag Østvang (O) shows that Bunn & Hogg's (B&H's) claim that a pair of co-moving observers (COs) who have small enough Hubble velocities (v = Hr) relative to each can be considered to occupy the same flat spacetime frame is false. O shows specifically that in the flat or closed RW models, it is not possible to have H nonzero if the curvature is zero. Thus, O concludes that the redshift caused by v = Hr is 100% attributable to spacetime curvature, so it is best understood as a "gravitational redshift" not a "kinematic redshift" as B&H claim. Further, one cannot use the SR Doppler formula between observers with very small v = Hr, since the SR Doppler formula can only be applied between observers who share the same flat frame and any nonzero v = Hr entails nonzero spacetime curvature. O does agree that the SR Doppler formula can be used on the parallel transported 4-velocity of the emitter at emission, which is the Nalikar-Synge computation. He also eschews the notion of "expanding space," so I don't think O would disagree with B&H's conclusion that the redshift is a "Doppler shift" not an "expanding space shift." But, O's analysis refutes B&H's argument using SR with small v = Hr between COs.

None of this causes me to change what I have my students compute as far as kinematics in RW cosmology (although I will definitely add this material to my GR course). I have them compute recession velocity of and proper distance to the source at time of reception and time of emission as a function of z in the Einstein-deSitter model. [itex]r = 3ct_o \left(1 - \frac{1}{\sqrt{1 + z}}\right)[/itex], [itex]v = 2c \left(1 - \frac{1}{\sqrt{1 + z}}\right)[/itex], [itex]v_e = v \sqrt{1+z} [/itex], [itex]r_e = \frac{r}{1+z}[/itex], and [itex]t_e = \frac{t_o}{\left(1+z\right)^{1.5}}[/itex]. Then we add [itex]\Lambda[/itex] and obtain [itex]\dot{a} = H_o \sqrt{\frac{\Omega _m}{a} + \Omega _\Lambda a^2}[/itex] with [itex]\Omega _m + \Omega _\Lambda = 1[/itex] which I have them use to verify [itex]\frac{t_e}{t_o} = 0.0366[/itex] for z = 9.6, [itex]\Omega _m = 0.3[/itex] and [itex]\Omega _\Lambda = 0.7[/itex] found in http://arxiv.org/abs/1204.2305. Then I have them obtain [itex]\ddot{a} = 0[/itex] at z = 0.671 [itex]\left(\frac{t_e}{t_o} = 0.544 \right)[/itex] with [itex]\Omega _m = 0.3[/itex] for comparison with http://www.ptep-online.com/index_files/2012/PP-29-02.PDF [Broken].
 
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  • #58
Jorrie
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But, O's analysis refutes B&H's argument using SR with small v = Hr between COs.
I've always been skeptical of the B&H argument, since it would imply that a string of COs could not all be at the same cosmological time. It may not be a sufficient argument, but to me it made SR redshift as an interpretation for cosmological redshift uncomfortable, to say the least.
 
  • #59
timmdeeg
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I've always been skeptical of the B&H argument, since it would imply that a string of COs could not all be at the same cosmological time. It may not be a sufficient argument, but to me it made SR redshift as an interpretation for cosmological redshift uncomfortable, to say the least.
Peacock and Chodorowski argue that interpreting the cosmological redshift one should take gravity into account. And I think this makes sense because in curved space-time the redshift isn't purely dopplerian per se. But at the end, as we talk about interpretations the answer to this question is not wright or wrong but might be rather a matter of personal taste.
http://arxiv.org/abs/0809.4573
http://arxiv.org/abs/0911.3536
 

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