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midtskogen
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### Science and Astronomy Questions

The time dilation at play for things falling in isn't very confusing, but that it doesn't apply to the evaporation is.  Intuitively, if we wait a long time for all black holes to evaporate, they will in some sense still exist as ghosts observed from the outside - all the things that fell into it frozen in time at their former horizons.  Evaporation takes a very long time, though, and it may be long enough for the things that fell into it and any information about them to become undetectable and so there may be no practical paradox.
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Watsisname
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### Science and Astronomy Questions

The time dilation does apply to the evaporation.  The photons of Hawking radiation are redshifted as they climb out of the gravity well.  The information of the timer we dropped in is radiated away, but not before 02:00 as measured by our clocks far away.  We must hang around until the hole evaporates completely.  Then at that point nothing will remain as a ghost.  All information about everything that ever fell in to the black hole will have been radiated away.  We could intersect these photons and do a count.  We'll find that it is all accounted for, all the mass and all the information is conserved.  Just completely scrambled, utterly useless for determining what the object originally was.  The black hole has maximally increased the entropy.

Hornblower
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### Science and Astronomy Questions

Was my question too hard?
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DoctorOfSpace
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### Science and Astronomy Questions

Watsisname, given that in our interactions with the vacuum of space friction is not really an issue I have been wondering about something.

Imagine I am in deep space, perhaps even intergalactic space or the far future long after galaxies.

I have a standard baseball

a major league baseball weighs between 5 and 5 1⁄4 ounces (142 and 149 g), and is 9 to 9 1⁄4 inches (229–235 mm) in circumference (2 7⁄8–3 in or 73–76 mm in diameter).

I throw this baseball out in the void of space. In classical Newtonian mechanics the baseball should continue moving forever in a friction-less environment. However given that gravitational waves are generated by anything moving in space, how long would it take for that baseball to lose it's momentum to the generation of gravitational waves?
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FastFourierTransform
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### Science and Astronomy Questions

DoctorOfSpace wrote:
Source of the post how long would it take for that baseball to lose it's momentum to the generation of gravitational waves?

I'm probably wrong but wouldn't the gravitational waves form only if there was an acelleration?

Watsisname
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### Science and Astronomy Questions

DoctorOfSpace wrote:
Source of the post However given that gravitational waves are generated by anything moving in space, how long would it take for that baseball to lose it's momentum to the generation of gravitational waves?

It never will.  Gravitational waves require a particular type of motion of the source called a changing quadrupole moment.  This means to change its shape or motion in some way which does not preserve cylindrical symmetry.  So an object moving in a straight line at constant speed will not emit any gravitational waves.  Nor will an object accelerating in a straight line.  But two orbiting objects will, and so their orbits will slowly decay.  A rotating sphere with a bump on its equator will also radiate, and continue to do so until either the bump is smoothed out or the rotation stops.  (This is why we may expect to see some gravitational radiation from solitary neutron stars, if they have some deformities on their surface).

Perhaps an easier way to convince yourself that momentum alone cannot be responsible for emitting gravitational waves -- that your thrown baseball moving at constant velocity will never slow down -- is to apply principles of special relativity.  We can easily shift into a frame of reference in which the baseball is not moving, and this is an equally valid frame of reference.  Then we should not expect the baseball to lose momentum, because it has no momentum.

Whereas for two co-orbiting masses, we cannot find a frame of refernece where they are both at rest.  (Not being able to find a frame where they are both at rest does not prove they must emit gravitational waves, but it helps show why they are not equivalent situations, that we can't apply the same logic as for the baseball to two orbiting masses to conclude that those ones do not radiate.)

Hornblower, sorry, I don't think anyone meant to ignore your question; it just requires a bit of research to find a good and complete answer.  The surface temperature on Venus is about 460C with a pressure of about 90atm, which is definitely fatal to sensors and electronic components.  But most of the structural components of the landers have significantly higher melting points and so they should still be pretty much intact.  Corrosion should also not be a problem because the sulfuric acids are present at higher altitude -- they evaporate before reaching the ground.  So I would imagine if some future astronaut walks around on Venus they could still find the metal husks of those landers sitting on the ground.

DoctorOfSpace
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### Science and Astronomy Questions

Watsisname, so in the flat example it does not.  Would it's spin imparted by the individual throwing it radiate gravitational waves over time?  Especially given that the ball has protruding ridges?
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Watsisname
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### Science and Astronomy Questions

From the spin, yes -- very weak ones.   This won't change the baseball's linear speed, but it will slowly reduce its spin.

How long will it take?  To give a very rough estimate for the baseball's spindown, we can apply equation 24 in this article.  Basically we're treating the baseball as an incredibly puny neutron star and applying the same physics.  The formula is

Keep in mind this formula isn't completely rigorous (for example it assumes the gravitational wave luminosity is constant, whereas in reality it will decrease as it spins down), but it is still quite useful for getting an approximate timescale for the spindown, in a sense similar to a half-life in radioactive decay.  And we only care about rough estimates anyway.

This formula uses geometrized units, so the mass M of the baseball (about .15kg) must be converted to meters by multiplying by G/c^2.  That will be about 10-28 meters.  The radius R (about .037 meters) can be left alone.  The speed v that it is spinning must also be expressed in terms of the speed of light (divide by c).

The parameter is a measure of how much the object's mass is not spherically symmetric ( where M is the total mass and m is the mass of a bump on the surface).  For the baseball, this will be quite tiny since the ridges and stitching is not only a small fraction of the ball's mass, but it is also fairly symmetric.  Probably it is something like .01 or .001.  Let's choose .01 to be generous.  That is, we'll say the baseball's mass is spherically symmetric to 1%.  Finally, is the rotational frequency.  According to a quick google search, typical MLB spin rates are around 2000RPM, or about 30Hz.  We'll assume your throw is similar.

A spin rate of 30Hz means the surface of the ball is rotating about the center at around 7m/s, or 2x10-8 of c.

Let's pause for a moment to contemplate this number and how it appears in the formula.  2x10-8 is small.  It shows up in the denominator (negative exponent), which will make it big.  But it's also cubed, which will make it really big.  Furthermore, the ratio R/M (the radius of the baseball compared to half of its Schwarzschild radius) is also really big.  Strong gravitational waves require very massive and very compact sources (R/M not very much bigger than 1) which are spinning very rapidly (v/c not very much less than 1), and the baseball is neither of these.  So we expect a very long timescale for it to spin down.

Plug in the numbers, and we get about 1051 seconds.  That's 1043 years, or 1033 times longer than the universe has existed.

This is quite a long time.  It goes beyond the end of star formation, beyond the time for all stars to die, and beyond the time for the orbits of all remnants to either be ejected or decayed.  Also by an interesting coincidence, 1043 years is the theorized upper limit for the timescale for all nucleons in the universe to decay!  The baseball itself will literally evaporate into radiation at least as fast as it can spindown due to gravitational wave emission!

HarbingerDawn
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### Science and Astronomy Questions

Watsisname, for this spindown time, is this referring to the time it takes for the baseball's rotation rate to decrease to 0, or to decrease by some set amount?
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Watsisname
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### Science and Astronomy Questions

By some set amount.  The derivation assumes the power emitted in gravitational waves is constant at the initial value, whereas in reality it will decrease as the rotation slows, so the spinrate vs. time follows a decaying exponential.  So treat the spindown time as being analogous to a radioactive decay half-life (it isn't exactly the half-life, but a similar concept.)

(Edited the initial post to emphasize this).

Gnargenox
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### Science and Astronomy Questions

Then we should be able to detect constant gravitational waves from Neutron stars with binaries, instead of only watching for Black Hole mergers or Neutron star mergers. Fast spinning Neutron stars must be slowing down alot faster than that baseball, since they are so dense, and they are actually asymmetrical too. Some emit Radio waves and then switch to emitting X-rays, basically when they start feeding off companion stars. This material creates enormous mountains nearly a few millimeters high, that crush the atoms below, fusing heavier elements and producing the X-rays. Their spindown times vary by as much as 30% more when emitting x-rays. So, I guess if you feed a Neutron star fast enough, you might halt it's spin, down from 1000 of times a second.
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FastFourierTransform
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### Science and Astronomy Questions

Gnargenox wrote:
Source of the post Then we should be able to detect constant gravitational waves from Neutron stars with binaries, instead of only watching for Black Hole mergers or Neutron star mergers.

Supermassive close binary systems indeed produce gravitational waves. Well also normal binary stars etc...
We haven't been able to detect them because of the amplitude of the wave been below the signal to noise ratio of current gravitational waves observatories, but those waves are there. We have detected massive mergers just because the amplitude gets quite huge in the last seconds, relative to the last inspiraling.... years for example.

But I lied we have also detected gravitational waves coming from binary pulsars. Just in an indirect way. Those weren't mergers at all, just slowly (very very slowly) inspiraling pulsars in a binary system. It has been estimated that the merger will occur in 300 million years.

Just to compare the power of each signal:
• The power of the gravitational radiation emmited by the Solar System (the orbital motion of all the planets) is around 5000 Watt of which Earth's motion accounts for 200 Watt.
• The binary pulsar PSR B1913+16 emmits continuously 7.35×1024 Watts in gravitational waves (1.9% of the Solar output in electromagnetic radiation but been injected in space-time).
• The first black hole merger detected by LIGO, GW150914, emmited 3.6×1049 watts during a very brief time-period.  More than the combined power of all light radiated by all the stars in the observable universe (but as I said in this case it was not continuous).
This is not the whole story. The power output alone doesn't account for the "detectability". The amplitude of the waves (associated with this power) has to be related to distance to know if it's detectable. GW150914 happened 1.4 billion light years away so the huge power was dispersed across much more space when it arrived than a possible signal from PSR B1913+16 (that is 21000 light years away). Even with that the difference in power is huge so GW150914 was detected and PSR B1913+16 not (we make an indirect detection).

Just to make some more numbers; GW150914 is 66700 times farther than PSR B1913+16, so (using and inverse square law), the intensity of the signal is 4.4 billion times weaker. But since GW150914 is 5×1024 times more powerfull than PSR B1913+16, the problem is totally overcomed been 1015easier to detect GW150914 from Earth than PSR B1913+16.

Watsisname
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### Science and Astronomy Questions

Gnargenox wrote:
Source of the post Then we should be able to detect constant gravitational waves from Neutron stars with binaries, instead of only watching for Black Hole mergers or Neutron star mergers.

That's right.  This was understood from calculations back in the early days of the theoretical gravitational wave research in the 60s and 70s, and is also a major topic of the article I linked.  There are many possible detectable sources of gravitational waves besides mergers (asymmetric core collapse supernovae may be another), and our ability to detect those waves depends on the details of their emission and of the detector.  For example, the intensity of the waves from solitary spinning neutron stars depends a lot on how much internal variation there is in their mass distribution from spherical symmetry.  Or going from the route of their observed spin-downs, how much of that spin-down is due to gravitational wave emission or other processes like magnetic braking.

As FFT says, the other requirement for being able to detect them is that their strain amplitude at Earth must be large enough to be picked out over noise.  Researchers initially expected 10-21 to be the threshold for reliably detecting events on a reasonable timescale in the first generation of detectors, and this turned out to be pretty much correct.  Another factor is their frequency, since the noise threshold of the detector depends a lot on frequency.  A larger detector (like a space-based interferometer) can probe much lower frequencies, like inspirals long before the merger event.

FastFourierTransform wrote:
Source of the post Just to make some more numbers; GW150914 is 66700 times farther than PSR B1913+16, so (using and inverse square law), the intensity of the signal is 4.4 billion times weaker.

A quick modification to your calculation:  The intensity of the gravitational wave obeys the inverse square law, but what the interferometer actually measures is the amplitude or strain (an important difference from electromagnetic observations).  The intensity is proportional to the square of the amplitude (as for any wave), so the strain and thus detectability of the gravitational wave decreases in proportion to the distance, not as the inverse square!

This is a very good thing for being able to detect them.   It means tripling the detector's sensitivity increases the effective range by a factor of 3, and thus the volume and number of expected events per unit time by a factor of 27!

FastFourierTransform
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### Science and Astronomy Questions

Watsisname wrote:
Source of the post so the strain and thus detectability of the gravitational wave decreases in proportion to the distance, not as the inverse square!

Wow!! that's awesome. Finally nature makes things easier for us hahaha. Thanks Watsisname you are trully a wise man.

I'm now thinking about the idea of detecting exoplanets one day. Maybe the sensitivity of far future gravitational observatories permits this. We would have the same bias for low inclination planets as in the eclipsing method (since gravitational waves radiate non-symetrically in this configuration)? How we would have to increase sensitivity in order to start detecting nearby exoplanets? Do you think that by the time we have gravitational observatories capable of doing this all the exoplanets of the galaxy would have been detected or there is hope for this new kind of observation to contribute in planetary search?

Watsisname
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### Science and Astronomy Questions

In principle the concept works.  The difficulty, as usual, is in practice.  To get some insight, we can do another back of the envelope calculation for how large the amplitude should be from exoplanet sources.  Written in conventional units this time,

where is the reduced mass of the planet-star system and is the total mass.  If the mass of the star is much larger than the mass of the planet, then is approximately the planet's mass and is approximately the star's mass.  is the orbital radius of the planet, and is the distance of the system from Earth.

Consider a 1 Jovian mass planet in a 0.01AU orbit of a 1 solar mass star at 10pc. This yields a strain of about 10-24.  This seems promising at first glance!  It's comparable to the design sensitivity of LIGO and other sensible Earth-based detectors.

The problem?  The frequency!  LIGO's useful sensitivity range at this strain level is from about 1 to 1000 Hz.  But the frequency of gravitational waves from an orbiting exoplanet is much smaller.  Reframing Kepler's Third law in terms of orbital frequency,

the frequency of gravitational waves (which is twice the orbital frequency) from this planet will be about 10-5 Hz.  Planets in larger, slower orbits will have even lower frequencies.

To get a sensitivity to gravitational waves of this low frequency (and thus larger wavelength), we need longer baseline distances in the interferometer.  This points us to a space-based observing platform, like the planned LISA mission.  And as it turns out, a lot of researchers have looked into this in great detail.  The gravitational waves from individual exoplanets, if detectable, could tell us a lot about the masses and orbits and of planets that we would otherwise not be able to see electromagnetically.

We would have the same bias for low inclination planets as in the eclipsing method (since gravitational waves radiate non-symetrically in this configuration)?

Actually we'll get a bias for high inclination (face-on) orbits.  The strongest gravitational wave emission is directed near the axis of rotation, and overall the emission has a pretty complex shape.  Here's a good video of the 3D shape of the waves and their effects from a close binary, which will be somewhat similar for an exoplanet.  Best part to watch is from about 0:30 to 1:00.  Notice the "detector" located directly above the axis of rotation has the greatest stretch and squeeze, but instead of the stretch and squeeze alternating (as it does along the plane of the orbit), here it only rotates.

How we would have to increase sensitivity in order to start detecting nearby exoplanets? Do you think that by the time we have gravitational observatories capable of doing this all the exoplanets of the galaxy would have been detected or there is hope for this new kind of observation to contribute in planetary search?

The best answer I can find for that is from this paper which investigates not just the gravitational waves from planets, but how they add up from all sources in the galaxy (and other galaxies) to form a component of the stochastic background of gravitational wave noise.  In particular, see figure 2:

We see that LISA comes tantalizingly close to being able to detect exoplanet signals at the ~10-4 Hz range.  But on the log-log scale, tantalizingly close means two to three orders of magnitude.    We would need an even more sensitive detector than LISA to pick these up -- possibly doable and a challenge for even further generations of space-based interferometers.

Another interesting concept is to use pulsar timing arrays, included in the figure as the PTA curves in blue.  These would be more sensitive to lower frequencies around 10-8 Hz.  Unfortunately even these are still not good enough in the strain to pick up exoplanet sources.  What they could do is provide upper limits on the stochastic background noise, such as from inflation.

In summary, there's no principle of physics that prevents us from studying exoplanets by gravitational waves, and I was very surprised to find that their amplitudes could be this large.  But it's still a tremendous experimental challenge to design detectors with the sensitivity at the required frequencies to pick them up.  Finally, with so many exoplanets all over the sky, studying any individual planet in this manner would be difficult because their signals blend together to form a background noise, in addition to the collective noise from other cosmological sources.

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