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Gnargenox
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### Science and Astronomy Questions

If I have 2 bowling balls and they are attached by non-stretchable string.

o ~~~~~~ o

What happens when spacetime stretches or expands everywhere including of course in-between the balls?

a) The string snaps.

b) The string becomes loose.

c) Nothing happens to the string.

d) The balls shrink in size.

e) The balls become larger.

f) Nothing happens to the balls.

g) The distance between balls increases.

h) The distance between balls decreases.

i) Nothing happens to the distance between.

The string is just a measuring device -and very easily snapped. It is NOT a bound system. More than one answer could be possible.
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Watsisname
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### Science and Astronomy Questions

Interesting thought experiment.

Assuming the string is made of discrete particles, I think that it will snap between every particle, leaving you with a wisp of atomic vapor.  My reasoning is that we are asserting it is non-stretchable, meaning there is zero width to the stable equilibrium position between the particles that make up the string.  And you also assert that it can be snapped, meaning the strength of the force that maintains this equilibrium position is also not infinite.  Therefore even the smallest perturbation, like the expansion of the space, will immediately break the connections.  Not only is it very easily snapped, it is infinitely easily snapped.  So I think the most correct answers are (a) (very many times), and (g).  Possibly also (e), depending on the expansion rate over the bowling balls compared to the forces that are holding them together.

Possibly there are other ways to argue the answer(s) and it'd be interesting to hear more responses, but that's my interpretation of it.

Gnargenox
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### Science and Astronomy Questions

Yes, those things keep me up at night after reading too much Wikipedia.

Is not Gravity the force that keeps the bowling balls from becoming larger? Gravity has also been keeping all other matter from being ripped apart, ever since the expansion rate started accelerating 7 billion years after the universe came into being. But, does that mean the force of Gravity also changed since things before this rate change where still the same size as after it accelerated?

Also around this same time the universe switched from being dominated by photons and neutrinos to being dominated by both normal matter and Dark Matter. The universe was fairly equal in density in all directions, but can it not stay stable without expansion?

As the universe expanded, materials remained the same size while the volume of space increased, at some point (around 800,000 more years) matter density (mass) became so low, a more subtle contribution to the Universe's energy density began to show up; Dark Energy. For some reason when Dark Energy reached 33% of the total energy density in the Universe, the expansion rate began accelerating.

Since that time, some 6 billion years ago, the matter density has continued to drop, while dark energy has remained a constant. As time goes on in the future, the matter density will continue to drop, while the dark energy density will remain constant, meaning dark energy becomes more and more dominant. Does perhaps Dark Energy also keep things from "snapping" or turning into atomic vapor?

Another neat thing about Expansion is that, on average, 20,000 stars transition every second from being reachable by us at the speed of light to being unreachable. The light they emitted a second ago will someday reach us, but the light they emit this very second never will.
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Watsisname
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### Science and Astronomy Questions

Cool post, and a lot to cover with it so apologies in advance for the length!

Gnargenox wrote:
Source of the post Is not Gravity the force that keeps the bowling balls from becoming larger? Gravity has also been keeping all other matter from being ripped apart, ever since the expansion rate started accelerating 7 billion years after the universe came into being.

The bowling balls are held together by the inter-atomic forces, which is far stronger of a binding than their gravitation.

Gravitation becomes the dominant binding for much larger masses like planets, stars, galaxies, and galactic clusters.  The space within galactic clusters is also not expanding anyway, because the gravitation causes it to change from the expanding FLRW metric to the non-expanding Schwarzschild metric.  In a sense, the space within these regions has "broken off" from the expansion.

But, does that mean the force of Gravity also changed since things before this rate change where still the same size as after it accelerated?

Not quite.  The gravitational force stays the same.  G is a constant, and the laws of gravitation (whether Newtonian or general relativistic) also stay the same.

Something else does happen with this shift of the universe from being radiation dominated to matter dominated to dark energy dominated: the rate of change of the expansion changes.  That is, the acceleration changes.  For a radiation dominated universe, the scale factor is proportional to the square root of time.  For matter dominated it's proportional to time to the 2/3rd power, and for dark energy dominated it's exponential.  The reason for this has to do with how the density of matter, radiation, and dark energy change as the universe expands.

Gnargenox wrote:
Source of the post Also around this same time the universe switched from being dominated by photons and neutrinos to being dominated by both normal matter and Dark Matter. The universe was fairly equal in density in all directions, but can it not stay stable without expansion?

The universe being equal in density in all directions is always true, as far as data can show.  This is the cosmological principle: on large enough scales the universe looks the same no matter where you are (homogeneity) or which direction you look (isotropy), and it was as true very close to the Big Bang as it is true today.

But I'm not quite sure what you mean about it being stable.  Are you talking of being able to be static -- remaining neither expanding or contracting?  This is possible if the dark energy density exactly balances the tendency for matter and radiation to pull the universe back together, but it is an unstable equilibrium.  Einstein himself tried to model a static universe by introducing the cosmological constant, but alas a universe balanced this way would be like balancing a pencil on its point.  Even the smallest fluctuation would immediately result in the universe collapsing or expanding.

Gnargenox wrote:
Source of the post For some reason when Dark Energy reached 33% of the total energy density in the Universe, the expansion rate began accelerating.

That's right, and we can show that this follows from the Friedmann equations, which describe the universe's expansion rate as governed by general relativity and its contents (matter, radiation, and dark energy).  Let's do a bit of math.  We can write the first equation as

The term on the left side describes the expansion rate of the universe.  Specifically, a is the scale factor ("size" of the universe), and a with a dot over it is the time derivative (its expansion velocity).  So is the expansion velocity per distance, which is Hubble's constant.

The right side of the equation describes the universe's contents.  Each of the Ω's represents a type of mass or energy density.  The one with subscript m represents matter density, r is for radiation, and Λ is for dark energy.  These are all expressed in terms of the critical density (the specific density which would make the spatial curvature of the universe flat.)  In fact the spatial curvature is very close to flat, so total density is close to the critical density and the sum of these Ω's is about 1.  The actual values today are about 10-4 for radiation (pretty negligible), 0.3 for matter, and 0.7 for dark energy.

Let's use this equation to figure out something about the universe's acceleration (in effect, deriving the acceleration equation).  To do this, start by moving the denominator a2 over to the right hand side:

Now take the time derivative of both sides [remember to use the chain rule: e.g. derivative of a2 is 2a(da/dt)].

Finally, solve for the acceleration (the a with two dots over it, meaning 2nd time derivative of the scale factor)

The universe switching from decelerating to accelerating corresponds with .  So we can set this equation to zero to see what conditions on the dark energy density will satisfy that.  Let a=1 across the board, and also let Ωbe zero, since by the time dark energy can start to dominate, the radiation density will already be negligible.  We then get ΩΛ = 1/2Ωm.  But the total density is ΩΛ + Ωm, which is 3ΩΛ.  So the dark energy density at the onset of acceleration is 1/3 of the total density.  Neat!

Gnargenox wrote:
Source of the post Since that time, some 6 billion years ago, the matter density has continued to drop, while dark energy has remained a constant. As time goes on in the future, the matter density will continue to drop, while the dark energy density will remain constant, meaning dark energy becomes more and more dominant. Does perhaps Dark Energy also keep things from "snapping" or turning into atomic vapor?

Dark energy is the agent that would snap them, if it is strong enough.

For an analogy (which is not quite correct, but a decent starting point), you can think of dark energy as acting like a small repulsive force between particles.  In a solid, the atoms or molecules or whatever are bound together like masses attached by springs.  What then happens if you introduce a small repulsive force between two masses connected by springs?  It simply increases the equilibrium position.  If the force becomes too strong, it may tear them apart.

In reality dark energy's effect is, like the basic expansion, only really meaningful on large scales -- larger than galactic clusters.  However, in the hypothetical "Big Rip" scenario, where dark energy is assumed to have an equation of state that leads to it getting stronger over time, it could very well overwhelm all structures, tearing things apart from top down.  Galaxies in clusters would be separated first, then the stars from each other, then the planets from their stars, and finally down to the atoms.

I think most cosmologists view this scenario as unlikely on physical grounds, though the best we can do with data is constrain how far into the future it would have to happen, if it happens at all.  For more detail and data about it I can refer to the second half of my post here.

Gnargenox wrote:
Source of the post Another neat thing about Expansion is that, on average, 20,000 stars transition every second from being reachable by us at the speed of light to being unreachable. The light they emitted a second ago will someday reach us, but the light they emit this very second never will.

I've never checked the actual number, but that is a neat thing to think about.   Or maybe depressing, I'm not sure...  Another interesting (and often confusing) thing is that we can continue to see those objects, even though their recession speeds are faster than light.  It just means we don't receive signals they are releasing "now", and we can't ever reach or send signals to them.

We look outward and see all these distant galaxies, yet as time goes on they are pulled out of our reach, though not from view.  What a cruel thing.  The size of our playground gets bigger, but the amount of stuff we are allowed to play with grows smaller.

An'shur
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### Science and Astronomy Questions

So I chose to plot a curve of a variable star for homework at university before realizing that it may be more complicated...

Star's radius and temperature changes in time. I know that the surface temperature is highest about the time of the smallest radius. I am talking about Chi Cygni here. It's luminosity is said to change from 6000 to 9000 times Solar luminosity. I get something close to that in Matlab, even the curve looks believable, but the calculated absolute magnitude does not plummet down by 8 to 11 magnitudes. Instead, it lowers only by about 0.25 mag. I believe that the mistake I may be committing is using black body radiation equation for this.

The equation takes whole EM spectrum into account, as we can see in the definite integral bounds. I am not sure but I think that when we say that the star's magnitude varies by 8-11, we mean visible light. If I change the bounds to the visible light boundaries, I should be able to get closer to the magnitude difference of real Chi Cygni, or so I assume. Given that the star is relatively cold, most of it's output is in infrared. Are there other variables that could influence the magnitude difference so drastically? So what is the Lλ function? I do not know the full formula.

Watsisname
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### Science and Astronomy Questions

Yeah, this calculation is quite a bit more complicated.  When we talk of a star's magnitude we do usually mean in the visible range, or as measured with a V filter, which roughly corresponds to how our eyes see.  The figure you link is labelled mV which is the magnitude in the V band.  So you're correct that your calculation needs to account for this range of wavelengths.  Changing your limits to the visible range (400-700nm) would a decent first correction, though to be more precise you would also need to consider how sensitivity depends on wavelength.  I'll give the details on that in a moment.

First let's go over the computation of the luminosity L of the star starting from the blackbody distribution.  Each bit of surface area (call it dA) of the star emits radiation outward to what it sees as a hemisphere, with solid angle element .  So to find the total luminosity we have to do this angular integral as well.  Then the energy per unit time emitted by the entire surface, between wavelengths and is

where is the Planck function (blackbody distribution).

The integral over the surface gives , but the angular integral also gives a value of pi.  It's very easy to forget that extra value of pi or where it comes from.  So altogether,

This is the "monochromatic luminosity".  If you integrate it over all wavelengths and relate to the Stefan Boltzmann Law, you'll get

Finally, if we want apparent magnitude (seen from Earth) we use inverse square law to relate the luminosity to the flux observed:

How then to get the apparent visual magnitude?  Here is where we account for the range of wavelengths and sensitivity over them.  Multiply by a "sensitivity function", and then integrate over all wavelengths.  Taking -2.5log10 of the result will convert to magnitude, and we also add a constant at the end of the calculation, with the constant being defined so that the magnitude of Vega in any filter is zero.

Again as first approximation you could just integrate from 400nm to 700nm, and let the sensitivity function be "1" over that range.  That's quick and easy enough, but I'm not sure exactly how good of an approximation this will be.  At least it should be a lot closer.

To be more precise, you can try approximating the sensitivity function.  The sensitivity functions for the standard filters (Johnson curves) can be found here, and I'll copy the values for the V filter below.  The first column is wavelength in Angstroms, and the third column is the normalized sensitivity.

   4450    0.000    0.000   4500    0.010    0.002   4550    0.020    0.005   4600    0.020    0.005   4650    0.040    0.009   4700    0.050    0.012   4750    0.070    0.016   4800    0.100    0.023   4850    0.170    0.039   4900    0.430    0.100   4950    0.920    0.213   5000    1.600    0.371   5050    2.360    0.548   5100    3.040    0.705   5150    3.580    0.831   5200    3.950    0.916   5250    4.190    0.972   5300    4.300    0.998   5350    4.310    1.000   5400    4.240    0.984   5450    4.110    0.954   5500    3.950    0.916   5550    3.760    0.872   5600    3.560    0.826   5650    3.340    0.775   5700    3.110    0.722   5750    2.880    0.668   5800    2.640    0.613   5850    2.410    0.559   5900    2.170    0.503   5950    1.940    0.450   6000    1.720    0.399   6050    1.490    0.346   6100    1.280    0.297   6150    1.080    0.251   6200    0.900    0.209   6250    0.730    0.169   6300    0.580    0.135   6350    0.450    0.104   6400    0.350    0.081   6450    0.270    0.063   6500    0.210    0.049   6550    0.180    0.042   6600    0.160    0.037   6650    0.150    0.035   6700    0.130    0.030   6750    0.120    0.028   6800    0.100    0.023   6850    0.080    0.019   6900    0.070    0.016   6950    0.060    0.014   7000    0.050    0.012   7050    0.040    0.009   7100    0.030    0.007

For ease of integration, you could try fitting a simple curve to them.  Or even a Gaussian with the central maximum and FWHM (full width at half the maximum), though this wont be as precise.  You can find those values here.  Alternatively you could try converting the integral to a summation.  Whatever you think is best.  I'm not sure how precise you need to be for this, or how much work you want to put into it.

I hope that helps.   Good luck!

Watsisname
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### Science and Astronomy Questions

To summarize all of that a little more briefly in formulas:

The Luminosity of a star is

where is the Planck function:

The flux we observe is the luminosity divided by the surface area of a sphere whose radius is the distance to the object.  And the apparent magnitude is -2.5log10 of the flux, with an offset such that Vega's magnitude is zero.  So the apparent magnitude in the V band is:

Where CV is the offset so that Vega is zero, and is the sensitivity function that corrects for the sensitivity over your range of wavelengths.  So in practice, you do this computation for Vega, and subtract that value from the same computation for the star of interest.

An'shur
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### Science and Astronomy Questions

Watsisname, thank you for swift reply, I will look into it this week.

Hornblower
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### Science and Astronomy Questions

With all the geological activity happening recently, specifically Puʻu ʻŌʻō, I've started to learn about volcanoes. However, I have a couple questions that I couldn't find a good clear answer to online. Hopefully, there's some geologist here who can answer them for me.
1. What's at the end of a lava tube/lava cave? Is it just lava or is there some sort of wall or rubble?
2. Is there any way to map magma vents and lava tubes while magma is still in them? For example, the magma chamber of Yellowstone is mapped out below from seismic data. Could a similar technique be used to map magma features closer to the surface like dikes, vents, and large active lava tubes just a few meters below the surface?
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midtskogen
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### Science and Astronomy Questions

I'm not a geologist, but here are my thoughts anyway.

Hornblower wrote:
Source of the post 1. What's at the end of a lava tube/lava cave? Is it just lava or is there some sort of wall or rubble?

You mean after the eruption has ended?  If you follow a lava tube the end will either be where the lava solidified to fill the entire cave, or the end will be rubble at a point where the cave collapsed.

Hornblower wrote:
Source of the post Is there any way to map magma vents and lava tubes while magma is still in them? For example, the magma chamber of Yellowstone is mapped out below from seismic data. Could a similar technique be used to map magma features closer to the surface like dikes, vents, and large active lava tubes just a few meters below the surface?

It's possible to use georadars (ground penetrating radars), if you need meter resolution.  But they don't go deep.  To see things further down, you need to rely on seismic surveys (artificial sources or natural).
NIL DIFFICILE VOLENTI

pzampella
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### Science and Astronomy Questions

Yesterday I had a thought experiment, and I would like to share it and ask about it.

Let's assume that we are in a place next to a random planet, and in that point the gravitational acceleration is 1 m/s2 (just to make calculations easier).

In that place, Dr. Stephen Strange created two portals: the first leads to the second portal, just 1 meter behind the first one and facing in the same direction, where the planet is. That way, any object that goes through that portal would be "moved" one meter behind.

Now, the experiment would be to put a cannonball between the portals and wait. What should happen?

Easy: the cannonball will accelerate due to gravity and go through the first portal, just to find itself one meter back. That process will repeat, increasing the cannonball speed linearly.

An easy math calculation will tell us that after 9.5129376 years, that cannonball should "reach" lightspeed... but, what now?

I know that reaching lightspeed is supposed to be impossible, but:

1.- What would happen to the cannonball after that?
2.- How would an external observer see when it happens?
3.- If instead of a cannonball, there was an observer instead of it: What would that observer see?

Gnargenox
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### Science and Astronomy Questions

Could it be possible that they are bowling balls rather than cannonballs perhaps?
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pzampella
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### Science and Astronomy Questions

Gnargenox wrote:
Could it be possible that they are bowling balls rather than cannonballs perhaps?

Hahahahaha yes it is. Let's just assume that the portals are wide enough (and that they are all perfectly align, of course).

Gnargenox
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### Science and Astronomy Questions

Haha great now I can think through this properly. My first thought is frame-dragging might destroy the portals but perhaps they are indestructible somehow
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Mr. Abner
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### Science and Astronomy Questions

Does not length and distance also change when nearing light-speed? And also mass. So I would say that the acceleration is actually a little slower with each pass through the portal, and hence the bowling/cannonball never actually reaches the speed of light.

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