Archive for the ‘Space’ Category

Does the Sun have a Positive Charge?

September 23, 2018

While I was reading about Venus’ atmosphere, and how the tremendous heat had split apart the hydrogen from water molecules and sent them off into space at escape velocity, I started thinking about the Sun’s plasma.

Hot gases are interesting, in relationship to how fast each molecule is moving. Small, light molecules move very quickly, while heavier molecules slog along at a slower pace in the same gas. So, in the formula  KEAVG=½ mv2=3/2*k*T, where k is Boltzmann’s constant and KEAVG is the average molecular kinetic energy, you can see right away that for a given temperature, the smaller the mass of the particle, the higher the velocity. So, if a particle in a gas is 100 times larger than another particle, you’d expect the smaller particle to be moving √100 = 10 times as fast.

So, back to the Sun. The Sun has a plasma atmosphere, that is, it’s mostly dissociated electrons and protons and other stripped atomic nuclei. Electrons are about 1/2000 the mass of a proton, so we expect that their average velocity in the atmosphere of the Sun is going to be √2000 faster that the average proton, or roughly 45 times as fast.

What this suggests to me is that, in the solar wind, most of the particles that actually reach escape velocity from the Sun (all suns) are going to be the electrons by a large margin. This also tells me that most of the particles that reach escape velocity from our galaxy are also going to be electrons.

So, some general figures to keep in mind; the escape velocity from our general region of the galaxy is about 537 km/s. Our Sun happens to be moving about 220 km/s around the perimeter, so particles would only have to be leaving our Sun’s heliosphere at about 317 km/s to escape the galaxy. The escape velocity from the Sun’s surface is around 618 km/s, and the solar wind (protons and electrons) supposedly passes by the Earth at about 400 km/s, though as we may discover, this isn’t exactly true. The escape velocity from the Solar System, if you start from Earth orbit, is only 42 km/s, much lower than the 400 km/s stream of particles flying by our planet.

I’m speculating that the solar wind consists of very fast electrons and much slower protons; if they were moving at the same speed, they would recombine into hydrogen. And since the electrons are moving much, much faster than the protons, I’m also speculating that a lot more electrons escape from the Sun than protons.

Over a long period of time, the Sun should become positively charged as more electrons than protons escape into interstellar space and intergalactic space.

One might look at the velocities, and think, well, hey, if the protons are moving at 400 km/s, they’re all going to escape the Sun’s gravity, too, and the balance of charge will be maintained! But they aren’t all traveling at this high speed; there’s something called the Boltzmann velocity distribution curve for particles in a gas, and some fraction of those protons aren’t going to make the necessary 42 km/s as they pass by Earth; they’re going to fall back into the sun. The electrons, as we noted, are probably moving 42 times as fast, on average, as the protons, so a much smaller number of them are going to be trapped by the Sun’s gravity. Likewise, a lot more electrons will escape our galaxy than protons.

Wow, the number 42 sure does pop up a lot. I wonder if that means anything?

Anyway, we speculate that a lot more electrons will escape from the Sun than protons. This would have some interesting side-effects.

The Sun, being positively charged, is going to be pushing and accelerating protons in the solar wind. The surplus interstellar electrons will be pulling on these same protons. Likewise, the motion of the electrons in the solar wind will be retarded due to the positive charge of the Sun. Eventually, I would expect some sort of balance, while still maintaining a net positive charge on the Sun. Somewhere out there in the heliosphere, the proton and electron velocities would finally match up, allowing them to merge into hydrogen.

I read recently that there is a yet-unexplained acceleration of the solar wind away from the Sun. Perhaps this charge imbalance is related to that.

So, we have a cloud of surplus electrons in between the galaxies. We have another cloud, probably denser, of interstellar electrons within our galaxy, between the suns. And we have positively charges suns stuck in this rotating cloud like a plum pudding. Over billions of years, I’d expect the intergalactic cloud to get denser, pushing the galaxies apart as one high-velocity electron wind smashed into others, and the field pushed the galaxies apart. Could this be interpreted as the acceleration of the separation of galaxies in the universe that we see and attribute to dark energy? I have no idea. Could the plum-pudding of positive charges (stars) imbedded in a rotating negatively-charged galactic cloud appear to be more massive than it really is, as it rotates within the universe’s own electron field? I also have no idea. I’m not an astrophysicist, or a plasma physicist, or any of the other useful fields that could actually answer these questions.

ADDENDA: I tried to look up velocity distributions for electrons and protons in a solar wind. The new Parker Probe, just launched, will probably be measuring this, and the GOES satellite and ACE measured this. Looking at the ACE SWEPAM experiment live data, it looks like electrons and protons have, on average, roughly the same electron-volt values, which, as I suggested earlier, would mean that electrons are moving a lot faster than the protons, which would give us results as described. But I don’t know how good the data is, or if I interpreted it correctly.

If you like this speculation, be sure to check out my SF short stories listed at my website.

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Zero Gradient Gravity Fields, Dark Matter, and the Formation of Stars

May 17, 2018

We’ve mentioned in the past (to ourselves) that the formula for the Schwarzschild radius for a black hole, c2=2GM/rs tells us that no matter how thinly distributed a mass is, (such as 1 atom per cubic centimeter), if you have a large enough sphere of it, it will have a Schwarzschild radius when viewed from outside that volume. You can see this just by shuffling the equation around a little, so that c2/2G, which is a constant, equals M/r, the mass over the radius. For any given density, the mass, M increases with the cube of the radius, so for any given density, you can always find a radius that contains enough mass to equal the value c2/2G. Cute, huh?

I struggled for awhile wondering if an infinite 3D field of particles (which would appear to be flat gravitationally, that is, not have a gradient), would allow for overlapping apparent black hole horizons; everywhere you looked, there would be large, overlapping, spherical volumes that had enough mass to become black holes. Could this be our apparent cosmological horizon? But today (5/12/18) it occurred to me that the key feature of a black hole is that it has a gravitational gradient. You have to work to get out of the gravity well, or the idea of an event horizon is meaningless. But an infinite field of equally distributed mass has no gradient. It appears flat. Ergo, no event horizon, no matter the density.

Cruising along in deep space, there is, in essence, the same amount of mass pulling on you from all sides, that tenuous 1 atom per cm3. It could just as well be 10 atoms, or 100, or a million, with no noticeable effect. Once we attained a velocity, we would maintain that velocity – an object in motion remaining in motion. The interstellar gas would eventually slow you down, but it would take a very ong time.

Working with the 1 atom/cm3 extending to infinity, let’s say we superimpose another huge sphere of 1 atom/cm3 gas on top of that, so huge that it provides you with an event horizon (if I’ve done my math right, it would amount to roughly 1.5×105 light years in radius, or a ball 0.3 million light years in diameter). Now there is a mass and a very small gravity gradient. Is the event horizon based on the 2 atoms/cm3, or the 1 atom/cm3 density? We’ve already seen that the original 1 atom/cm3 field provides no gradient, so it would make sense that the only effect to the observer is to see the event horizon created by the new 1 atom/cm3 superimposed on the existing field; the other previously existing field is completely flat and cancels out.

However, the new field created by the new mass is going to affect both the old mass (1 atom of hydrogen per cm3 everywhere) and the new mass (1 atom per cm3 in the giant sphere). The object will form with twice the mass (in this case) predicted by the theory. When it’s first put in front of us, we will measure a mass represented by the 1 atom/cm3 in that volume. As it collapses and takes the background mass with it, it will finally produce a mass that accounts for the 2 atoms/cm3 that we actually started with. While it’s doing this, it will also be backfilling the area that it vacated with more interstellar gas, as that gas is also being pulled in by the gravity of the developing black hole, so the overall density of the universe will appear mostly unchanged, even around the black hole.

Practically speaking, this would be more likely to happen in a nebula, where the density is much higher.

One of the most interesting things about this process is that if there is an undetectable mass-type in the universe (like dark matter) that only interacts with regular matter through gravitation, and it’s distributed equally everywhere, then objects that form (planets, Suns, black holes) will also pull in this other mystery mass. As described above, the tenuous gas (1 particle per cc) that we currently measure may actually mass 2 or 10 or 100 particles per cc. We wouldn’t know since the field is flat. Since this new mass doesn’t react with normal matter, it will clump in the center of the object (although it may have its own chemistry and volume that prevent excessive density). Small objects existing on a larger mass (like humans), would have the dark matter pulled out of them, and when we performed tests like The Cavendish experiment to measure the gravitational constant, it would give us a good value for G for normal matter, and would give us erroneous results for the masses of the planets and the Sun. We would think the core is made of denser matter than it really is, both in the Sun and Earth. We know the mass of the Earth, but a substantial chunk could be dark matter and we’d never know it. Perhaps the iron core is made of silicon at half the molecular weight (which is interesting, because magma is 50% silicon dioxide, and only 9% ferrous oxide).

However, most objects that have formed in the last few million years are going to have some dark matter as part of their core. They have gravity, and any dark matter out there will be attracted to it just like regular matter, until an object forms which is part dark and light matter. This includes asteroids. Eventually, we’re going to move an asteroid, and when we do, the acceleration is going to leave the dark core behind. We may not notice it unless we’re looking for it, or if it’s a substantial enough part of the mass that we detect a mass-change in the object as it’s propelled. We would end up with two objects; the obvious light-matter asteroid, and the invisible dark-matter asteroid that could only be detected with a gravitational gradiometer. It would change the way we thought about the universe.

Black Hole Evaporation versus CMBR

May 5, 2018

Black holes evaporate. At least, that’s what most physicists tell us.

What I stumbled into recently was a conjecture that they can’t evaporate beyond a certain point if the input is greater or equal to their output. This was mentioned on Quora by some physicist as a response to a related question. I thought it was interesting enough to mention it here.

Really large black holes are colder than the cosmic microwave background radiation (CMBR), which is about 2.73 degrees. The radiation going into a black hole is actually greater than the radiation leaving the black hole. The only way a black hole could radiate is if it’s very small and already radiating hotter than the CMBR (plus whatever particles fall into it, adding to its mass).

The limit where the size/mass of the black hole is equal to CMBR input is about 1% Earth mass, about 4×10^22 kilograms, based on Susskind’s formula and Hawking’s formula. This would create a black hole smaller than a millimeter. But black holes can’t even form without at least three or more Solar masses to begin with.

So, any black hole larger than a millimeter is going to keep growing. Presumably, primordial black holes smaller than 10^11 kg, created during the Big Bang, would have evaporated by now. This leaves a range of possible primordial black holes from 10^11 to 10^22 kg as possible existing evaporators, since they would be hotter than the background radiation.

However….

Primordial black holes would form because of high density and radiation. It would be crazy to think that their mass wouldn’t quickly grow far beyond Earth mass when surrounded by a buffet of dense gas and radiation. Just the nature of the formative process suggests that they will never radiate faster than mass/energy is added to them from their environment, and will always grow larger in size.

I really WANT them to exist, however. My next SF story kind of depends on it.

The Spacetime Diet: Redistributing the Fat of the Universe

October 29, 2017

If you’ve had your nose in spacetime physics at all, you’re familiar with the idea that when you move really fast, other objects look thinner. Or, relatively, you look thinner from the viewpoint of someone else’s reference frame.

This is called the Lorentz-FitzGerald contraction. When you observe a moving object, it appears shorter, or thinner, along the axis of its motion relative to you. Likewise, you appear thinner relative to the moving object (who does not feel thin at all). There’s a formula for this, but it’s irrelevant for the discussion below.

You also appear to gain mass (using a similar formula). This is also irrelevant for the following discussion, but I thought I’d toss it out there.

So here’s the rub. You will often hear someone say or write, “When an object nears the speed of light, the universe flattens to a thin sheet from the viewpoint of an observer on that object.”

Just to clarify, this is bullshit. If you are cruising along at near-light speed, then all matter, relative to your frame of reference, is moving in the opposite direction at near-lightspeed. That’s okay so far. Except, the universe is expanding. And the farther out you go, the faster it’s expanding, such that there are regions of space expanding away faster than the speed of light (the expansion of “space” is apparently able to ignore the whole “speed of light” limit thing; go figure).

So when you attain a certain velocity, you become stationary relative to another part of the universe that is moving away from Earth at the same speed. There is no shortening of length or thickness for that object, that part of the universe.

Take the Andromeda Galaxy for example, moving toward us at 110 kilometers per second. When we measure the galaxy in the direction of its travel, along its axis of motion, it’s foreshortened in that direction. Now, fire up your rockets so you’re traveling at 100 kilometers per second in the same direction, and Shazam! The entire galaxy poofs back out to its real shape in its own frame of reference that happens to coincide with your own. Relative to you, the Andromeda Galaxy is no longer moving.

So, back to the expanding universe; as your spaceship speeds up more and more, there’s always a part of the universe that’s moving at the same speed at which you are traveling (a comoving reference frame). It won’t look compressed or thinner or foreshortened at all. In fact, if we take the viewpoint that all parts of the universe are essentially equal, (that is, there is no “center” of the universe) then the universe doesn’t compress into a pancake at all as you near the speed of light; it’s just that the non-foreshortened part of it, the part that matches your current velocity, is farther and farther away from you. But the overall volume will appear unchanged.

 

A Variant Geometry for Spacetime

September 18, 2017

There are a lot of odd characteristics of existing spacetime physics, creating a lot of questions that are difficult or impossible to answer, such as, “Why is the speed of light approximately 300,000 km/s?” or “Why can’t you go faster than the speed of light?” or “Is there such a thing as a tachyon?” Or, “If you can’t go faster than the speed of light, how is it possible to age only 4 months due to time dilation while traveling 4 light years?” I hope to offer an alternate geometry to provide some reasonable answers to these questions.

Let’s start with some fundamental concepts about photons. It’s generally believed that photons are their own antiparticles, and also that the speed of light is the ultimate speed past which nothing can travel. Also, in a photon’s frame of reference, the distance from source to destination appears to be zero, and it takes zero time to travel that distance. This led me to speculate that light might, in fact, travel at an infinite speed, and that somewhere out there, there is a geometry in 4D space-time where that makes sense, where, when we try to measure it, we see light ambling along at a tedious 300,000 km/s. It would also explain why you can’t travel faster than the speed of light; the speed is, in fact, infinite. It’s really hard to go faster than that. The difficulty lies in finding that geometry. My second supposition regarding photons is that they are always emitted perpendicular to the path of travel (through time) of the originating particle. In a standard space-time diagram, assuming a velocity of c, this leads to the light cone diagram. In the new geometry, assuming an infinite photon speed, the picture is a little different, but still leads to the well-known equations we are used to.

Figure1

Looking at Figure 1, the vertical line A represents the source of photons a and b, which travel instantaneously to the observer on line B. Line B observes the two events a and b separated by time ct2, and from B’s perspective, the object has moved away a distance x, which equals a-b. In A’s proper time, the time between the two events is merely c time t, and the distance is zero, so the interval is s=ct1. From B’s proper time, the duration is ct2 and the distance A has traveled away from him is x, so the measured interval between the two events is s=√(ct22-x2), which should be familiar to everyone.

Figure2

In Figure 2, we see what happens as B gets closer to the X axis. But it still produces the common formula for the interval. What the diagram does not explain is why the speed of light appears to be roughly 300,000 km/s.

However, what Figure 1 can do is allow us to derive the standard time dilation formula:

∆t1=∆t2/√(1-v2/c2)

How do we get there?

Note the velocity of B away from A is

v=x/t2

 So x2=(vt2)2

From before, we had s=ct1 in A’s reference frame and s=√(ct22-x2) from B’s perspective. Set the two equations equal, and we get

(ct1)2=(ct2)2-x2

Substituting for x2 we get

(ct1)2=(ct2)2-(vt2)2

 Divide it all by c2 and pull the t2 out of the two terms on the right gives us

(t1)2=(t2)2•(1-v2/c2)

Or, ∆t1=∆t2•√(1-v2/c2), which should be familiar to a lot of you out there. It’s the standard formula for time dilation due to relative velocity.

So Figure 1 works out that if the line is at a 45 degree angle, then v=c/√2, which shouldn’t be a surprise. And as v gets closer and closer to c, then the graph gets flatter and flatter. But this graph is based on the idea that c is infinite. Why is it that when we measure it, it’s always 300,000 km/s? Clearly, if you set up an experiment that bounces light back and forth between mountain tops, and c=∞, then your test should show that light moves instantaneously. Not 300,000 km/s. And yet, we always measure c at the same boring 300,000 km/s (yeah, I know this isn’t exact value-it’s 299792.458 km/s. Get over it-this is easier to type).

So what is it that makes an infinitely fast photon measure as a finite number in all frames of reference? Is it the expansion of spacetime? Is it the curvature of spacetime due to gravity affecting the value of c that we measure? Is it that this theory is just wrong? I don’t know yet – this would obviously have to be resolved before this model made any sense, and may or may not appear in a later entry. Any speculations about this are welcome.

 

Background Gravity – Time Dilation in a Flat Field

August 11, 2017

I got into an argument with a physics buddy not that long ago (a year, maybe), about gravity. We have an intermittent arrangement where we go drink beer and talk about physics at two or three pubs in San Luis Obispo. Usually, the physics becomes a little less coherent as the evening wears on.

One of the discussions centered on whether there is a “background field” of gravity or not, or whether it’s even sensible to discuss such a thing since, in an infinite field of equally distributed mass (or gas, or 1 atom per cubic light year, whatever), all the forces around you seem to cancel out. The mass of the universe to your left is equal in size to the mass of the universe on your right; you feel a net acceleration of zero. I argued that even though the field was “flat”, there was still a field there. He argued that a field implied a gradient; there is always a force.

We did not come to a satisfactory conclusion. It might merely have been the fact that we were defining the same terms in different ways in our heads. I’m not sure. I thought my argument was rock-solid.

So, here is my side of it.

Some of you are probably familiar with Newton’s Shell Theorem. It’s in his Principia Mathematica, and if I remember right, he solved it without using calculus. Basically, what it says is this; if you are inside a spherically symmetric shell of mass, then you feel no gravity pulling you any direction. It’s a bit non-intuitive. Let’s say the Earth is hollow, and the entire planet’s mass has been compressed into a thin spherical shell a few centimeters (or meters, it doesn’t matter) thick. If you are floating around in your Nike Space Suit inside this shell, you will not be pulled toward the center, or the inner surface of the shell, or anywhere else inside the shell. Wherever you are put, you will remain.

Personally, I think this is one of the coolest theorems ever.

It’s also true that if you are outside the surface of a spherically symmetric planet, then it doesn’t matter how dense it is, at a given radius you will feel the pull of a certain amount of gravity. If you are in orbit above the Earth, and the Earth suddenly becomes a black hole of the exact same mass, you will remain in orbit, totally unaffected by that change. That’s pretty cool, too. Given that the gravitational force is based on F=GMm/r2, this should be kind of obvious. Neither your mass (m) or the mass of the Earth (M) has changed, your orbital radius is the same, and G is the gravitational constant. Ergo, the density of the object you are orbiting at a radius “r” from the center of the object is irrelevant.

So, that was me drifting from the actual subject. Shell Theorem—let’s get back to that.

As you might know, the clock of anyone in a gravity field runs slower than that of a clock outside of that gravity field. This is called gravitational time dilation (and is equal to ∆t’ = ∆t √(1-2MG/rc2) for a non-rotating sphere). A person on Earth actually ages slower than a person in deep space, according to relativity. This was verified with clocks flying around the Earth in the Hafele-Keating experiment. Before you ask, yes, they took into account Earth’s rotational speed, the speed of the airplane (in both directions relative to Earth’s rotation) with regard to Special Relativity’s time dilation due to velocity. It was a nice experiment.

Let’s say we’re using a hollow Earth from Newton’s shell theorem. As you get closer to the Earth, you are in a deeper gravity well, and the outside observer sees your clock slow down. There’s a small hole in the planet, and you pass into the planet, where everything is pulling you in opposite directions equally, so you seem to feel no force. And yet, your time dilation effect does not suffer a discontinuity, jumping suddenly to that of the outside observer. You are in a denser gravity field, but a flat gravity field. [to the physics majors out there, for god’s sake, if my terminology sucks, please correct me]. Your time dilation will be just the same as if you were standing on the surface of the planet.

So now you have a flat gravitational field (no “force” pulling you in one direction, that is, all forces pulling you equally in all directions). And yet, even in this apparent lack of gravity, where you can’t actually tell that you’re in a gravity well, your time runs slower than the time of someone far from the planet.

I extended this argument to the rest of the universe. If mass was distributed equally around you, even though you felt no force one way or the other, there would still be a background gravitational field. Gravitational time dilation implies a gradient; for time dilation to be relevant, you need someone in a weaker gravitational field measuring your time. However, both the measured and the measurer can be in locally flat gravitational fields.

Does a flat gravitational field curve spacetime by itself? Or is it only the gradient between two different gravitational fields that curves spacetime? My general opinion is that you don’t need the gradient for the curvature of spacetime. If you have an infinite universe with equally distributed mass, then from some arbitrary center, it will appear to curve spacetime until it closes the dimensions of universe into closed loops, like the inside surface of an event horizon (though other arbitrary centers will have different, yet overlapping event horizons – a subject I will touch upon another day). Likewise, if you have a large, thin flat sheet of soapy water in the air, fluctuations are going to cause it to form bubbles, closing up the edges. In spacetime, there may be a similar tension (gravity?) that closes the edges together into a 4D hypersphere.

How would you test the curvature of space inside a shell? I’m not entirely sure. I think the universe we have is a good test case, however.

Particle Pair Production in Deep Space

August 6, 2017

Many of you know that a matter-antimatter reaction results in a pair of gamma rays. Fewer of you will know that you can take a couple of gamma rays, run them into each other, and get a pair of matter-antimatter particles. This has been done experimentally, and there’s a bit of data about it under “Two Photon Physics” in Wikipedia. Generally, if a subatomic reaction can occur, then it’s reversible. Maybe not statistically probable, but still reversible. This is a concept I used in a story I recently sold to Analog SF. In an area of space with high-density, high energy gamma rays, you’ll get a lot of positrons and anti-protons produced (more positrons, since they are 1/2000th the mass, of course), but there will also be some small production of antihydrogen if the antimatter doesn’t recombine right away with normal matter. And the antihydrogen may be neutral enough to survive and drift in deep space for a while, maybe long enough to be used as a resource.

Some reactions result in the release of more than two photons. A particle and antiparticle meet, three photons are emitted. The photons are lower energy, but the reverse reaction, 3 photons meeting, is a much, much lower probability than 2 photons (gamma rays) meeting. Still, on rare occasions, it might happen.

In fact, it’s my belief that if you have enough photons, even low-energy photons, passing through the same bit of space at the same time, you can also have pair-production, spitting out particles and antiparticles. One calculation for photons from the cosmic microwave background radiation (CMB) estimates 400 photons per cubic centimeter, average, plus whatever higher-energy visible light and gamma rays pass through from billions of stars. And there are a lot of cubic centimeters in a light-year (about 4.9 x 1050). Even if the probability of pair production is very, very low, I still imagine that it would happen on occasion.

As a side-note, the probability of a positron and electron meeting in deep space is very high, since they attract one another, while the probability of two gamma rays meeting at just the right time in just the right way is fairly low. The reaction looks symmetric, but the probability of it happening in a certain direction is much higher one way than the other. Ditto for any two-particle reaction that creates three particles. This contributes to the increased entropy of the universe and the “arrow of time”; there’s a preferred direction for these subatomic reactions to occur.

Looking All the Way Back in Time

July 31, 2017

If you look back in time, (up in the night sky, at the light emitted from galaxies billions of years ago), you are actually looking at an earlier version of the universe when it was smaller.

images-3

Due to the nature of how light moves through space, when you look back 14 billion years to the farthest reaches of the universe, you are actually looking at a very small volume. But the image, warped as it is, is spread out and fills the farthest regions of the sky, like a view through a concave lens. If you were able to look all the way back to the tiny point of the Big Bang, the image would be smeared and spread out across the 14 billion light-year shell, any detail of the event washed out by turning the fine detail of a tiny event into a picture the size of the universe.

So, when you look up at the night sky, everywhere you look in the blackness of deep space, 14 billion light years away, is actually the same small point.

Does this make sense? I’m trying to think of a good analogy for this, but it just isn’t coming to me. Maybe like starting with a tiny drawing on the surface of a tiny sheet of rubber, then stretching it out so that the sheet of rubber stretches all around you in a sphere, like the inside of a balloon, and then trying to figure out what the original picture looked like.

This, of course, begs the question of what physicists are calculating when they measure the accelerating expansion of the universe. If the universe was physically smaller 14 billion years ago, and now the remaining image of it is spread out over a sphere with a radius of 14 billion light years, that’s going to come off as an acceleration; the farther you look, the smaller the original volume and the more the image is spread out over the apparent warped view of the current volume. And, of course, 14 billion years ago, the universe actually was expanding a lot faster than it is now. It’s a double-whammy of accelerations. Most physicists are a hell of a lot smarter than me, so I’m guessing that both these accelerations have been calculated into the “accelerating expansion of the universe” equation. I can only speculate that there is a third element. I wish I could find out without wading through a lot of really obscure math.

Photons Dancing on the Head of a Match

May 1, 2014

Who cares about angels on the end of a pin? Let’s get real; how many photons are there on the end of a match?

There’s a lot of data out there to help us calculate this; one article says that the human eye can detect a candle in the dark at 30 miles. Isn’t that something? The same article says it takes at least 54 photons just for the human eye to register an event. So that match (or candle) has to get 54 photons into that fraction of your eye that actually receives and focuses the photons. But the whole eye doesn’t actually receive the photons; it’s just the black opening in the middle. The largest it ever gets is about 7 millimeters diameter, which is 38.5 square millimeters in area, or .385 square centimeters.

How many events can the human eye see in one second? If we’re looking at the match from 30 miles away, and it looks continuous, then we are receiving over 50 frames a second (though the human eye has been recorded as being able to discern and identify an image in 1/255th of a second, we’ll be conservative). If the image were less than 50 times per second, we would detect a choppiness in the image; still, overlap in the match’s photon emissions could turn a choppy image into a smooth one. But lets assume we get a continuous 54 photons, all the time, at least 50 times a second; anything less would look like a flicker off.

Now we have everything we need to calculate how many photons are coming off the head of a match!

Just put an eyeball…or just the iris, the bit that receives the light, in every spot in a 30 mile radius, add up the total number of irises, multiply by the 50 times-per-second, times the 54 photons per eyeball, and we should have the number we need.

The sphere of irises is 4*pi*r-squared. Or 4*3.14*30*30 = 11,310 square miles, or 29,292 square kilometers. Or 29,292,000,000 square meters. Or 292,920,000,000,000 square centimeters. Since each dissected eyeball (just the iris, you see) only takes up .385 square centimeters, that’s about 760,000,000,000,000 irises stuffed carefully into the perimeter of the sphere to capture all the photons.  Just as a point of interest, that’s about 50,000 times the number of human eyeballs on Earth. Guess we’ll be dissecting all the other animals, too. May as well start with fisheyes; they’re sort of gross to begin with.

Each of those eyes is gathering 54 photons at least 50 times a second, so we get to multiply the 760 million million by another 250-ish, giving us a grand total of about 190,000 million million photons off the head of a match every second. Or, just because I like a lot of zeros, 190,000,000,000,000,000 photons. Every second. From a freakin’ MATCH HEAD. We are awash in a photon bath.

Now, leave the darkness of night, and realize that when you walk outside, you are no longer looking at one tiny spot radiating onto 190 quadrillion eyes, now you have a hemisphere of 30 cubic miles of daylight radiating onto your eyeball. Well, okay, you can’t look at the hemisphere all at the same time. Your turn to do the math! How many photons are hitting your eye every second? Hint; it’s an absolute crapload of photons.

Billionaires of Mars

August 10, 2012

If you could write a check for a spaceship to Mars, would you?

That’s exactly the situation we have right now. There are over a hundred people on Earth right now with personal wealth weighing in at over $10 billion dollars. Gates has over $60 billion bucks available. Any one of these people with any interest at all in putting a colony on Mars, basically, owning Mars, could do so within a decade.

Robert Zubrin, an advocate of a unique mission profile, stated in an article, “…while Mars Direct might cost $30 to $50 billion if implemented by NASA, if done by a private outfit spending its own money, the out-of-pocket cost would probably be in the $5 billion range.” Wow. Five billion. And his mission profile advocates bringing people back, unlike the Mars One group from the Netherlands who wants to do a one-way mission to colonize the planet; of course, now you’re paying to take enough food and infrastructure for people to stay there.

The key point of this is that, given the will of one person (one very rich person), we could be standing on Mars in 10 years time. We could be living on Mars. The Mars rover, Curiosity, just did a radiation measurement that indicated levels are “not a showstopper”. About the same as low-Earth-orbit. Woo-hoo!

And most important of all, if this rich person has a brain, they can make money on the effort. Richard Branson ($4 billion) or Elon Musk ($2 billion) seem to already be heading this direction, building the infrastructure to get to space on their own terms and making money at it as they go (via Burt Rutan’s Scaled Composites, Virgin Galactic and SpaceX). The Mars One people talk about turning the mission into a media extravaganza, a reality show to beat all reality shows, an advertising blitz to beat all others. What would Pepsi pay to have their logo on the first manned lander? What would the first returned samples from Mars be worth to collectors? Can you imagine what actual fossils would be worth if they find them?

And you don’t even have to spend a cent developing the rockets to get you into Earth Orbit. Elon Musk has already done the design work; you can launch 10 Falcon Heavies for a billion dollars, delivering a half-million kg of fuel and hardware to low-Earth orbit.

Let’s look at the potential returns on this for a billionaire entrepreneur;

1. It’s been suggested that a $5-10 billion NASA X-Prize be offered for a private manned mission too Mars. Fine, but likely with lots of strings attached. Still, there it is; your entire mission paid for if you’re successful.

2. Advertising. This starts the moment you actually commit to the project. Just the televised weeding-out process for wannabee astronauts could bring in millions of dollars as a reality show. Competition between countries for seats on the rockets, Olympic fervor and excitement. That could last years. Licensing for games, the official mission logo plastered on every product on Earth, books, autographs from the team members, photo ops, speaking gigs (hey, you own those astronauts, part of the contract), product placements, donations for a variety of perks, or just donations…the list just goes on and on. If I were in advertising, I could make back the $5 billion in the 5 years before the first rocket left the launch pad. How many million-dollar stickers could you get on the side of your lander?

3. Building the rocket to get there, assembled in Earth orbit offers more advertising, more excitement, pay-per-view.

4. The trip there, of course, would be televised. Interviews would be sold. Unlike NASA’s model, nothing is free. Perhaps a bit more reality show programming and product placement advertising. Nothing like a Mars bar when you need a break from your EVA, is there?

5. And when you land, what is that worth? Renting out the copyrighted footage from the first manned landing? Aforementioned ad-space on your hardware? Commercial breaks? The knowledge that you’ll be getting royalties off this footage for the lifetime of the copyright? And the check you got from the country (or company, or person) that paid you off so their guy would be the first on Mars?

6. Experiment space could be sold for your arrival on Mars. Personal items or human ashes carried there and buried there. Designer bacteria could be taken along and tested in the Martian environment. Designer plant species, lichen and such, patented and ready for the colonists to spread around. If the colony was set up inside one of the many known lava tubes on the planet (such as those near Pavonis Mons), with solar collectors and solar pipes channeling the light inside, the colony would be surrounded by rock and safe from radiation. Lava tubes could provide huge living quarters with very little preparation.

7. And once you’re settled? Rent out your colonists. Tell us where you want to go, what samples you want collected, where you want that pickaxe swung. We’ll hop on our Martian bicycles and go check it out. Now that you’re there, anything you sell is just a bonus for you. For that matter, rent out little rovers with cameras. Sure, it’ll take 4 to 20 minutes for Earth-commands to go back and forth, but you know folks would rent time on them. Why stop there; super-light-weight flyers can be programmed to see all sorts of interesting things.

8. Property. Oh yeah, barring international agreements which won’t really apply to you since they can’t reach you, you own Mars. You want to buy 20 acres on Mars? We’ll sell it to you with a deed. It might be underwater when Mars starts to warm up, but that’s the risk you take. Lava tubes, now those are premium property! Create your own Government for your colony and claim it all. Work out deals with Earth governments so they don’t try to steal it all back. Better yet, sell them large swaths of Mars.

9. Sell support services for other colonists and countries. Once you prove it can be done, others will follow, and you can sell some of your infrastructure services (like a communications satellite, if you left one in orbit, you can rent bandwidth. Or a berth in your colony, if the new arrival wants to rent or buy a place to stay). Once they find out you’re trying to claim Mars for yourself, well, there won’t be any lack of newbies clamoring for a piece of the action, and they’ll all be paying you rent for your existing infrastructure.

10. Propellant; assuming you’ve tapped off of some of Zubrin’s brilliant ideas for making propellant from the Martian atmosphere, you can sell that to potential customers. Hey, we have water and propellant for sale! Come as you are. We have the supplies to send you back. For a price.

11. Patents on new minerals, compounds, and materials, and if you’re very, very lucky, microfossils. Unique gemstones on Mars? Who knows. Getting them back to Earth is a problem, but once your infrastructure is in place, heck, that’s a mission you could pay for with pocket change and make your money back ten-fold. Take stamps to Mars and ship them back.

12. And it gets stranger…once your colony is pressurized to 0.5 Earth-atmosphere in 1/3rd G, it’s time to pull out your sports-wings and fly around inside your 200-meter wide lava tube (yeah…they’re huge). You get to start your very own Martian sport. Which team will you bet on? Which Earth-network is going to pay to broadcast it? And Superbowl advertising for $4 million dollars per ad? You ain’t seen nothin’ yet.

Okay, sure I forgot some things. Let me know. Point is, we can go to Mars now; all we need is a billionaire with a dream and a marketing team that makes sure he remains a billionaire. After all, you’ll want to have money left over for that next colony.

UPDATE: Elon Musk (the Paypal billionaire) spoke at the Mars Society Convention in Pasadena last week, and expressed his interest in colonizing Mars, and making seats available for $500,000 a head for would-be colonists. Go, Elon! I might be too old to make the trip by the time this happens (being a spry 58 now), but at least I can make the trip vicariously! If you want to have your spirits bolstered by what he says, you can watch it here. Skip the Zubrin introduction – it’s lengthy.