some exotic thought experiments

[sandor]

Well, my earlier post, the one I care about at:

http://www.overunity.com/index.php/topic,5121.0.html

is being ignored so I'll post something that perhaps will get more people's interest and perhaps the other will get some interest by being along for the ride.

Ok, I have thought about physics for a very long time and there are two thought experiments which still hold some interest and you might find interesting as well.

#1: Suppose you had a very, very, very tough string. so tough in fact that its tensile strength is far greater than its mass density times the speed of light squared. So for instance, if it weighed one gram per meter, .001 kg, it would have a tensile strength of over 9x10^13 newtons, aka kilogram meters/second^2.

Now you proceed to lower this string into a black hole. Let's say there's some kind of toroid built around the black hole which you are standing on as you lower this string. Obviously this string will be under tension. As you let out more line, soon it will be under a great deal of tension. At some point it will be under more tension than the mass density of the string times c^2 because there is no limit to how high the gravity gets as you lower it down into the infinite pit in the center of the torus. Note that contrary to some people's understandings, a black hole bends space in such a way that even if the diameter of the torus is 1 mile (say the Schwarzschild radius is a tenth of a mile) and you have a million miles of string, you could lower down all million miles of it and the other end will not be touching the event horizon. You can never reach the event horizon in this way. Of course, if I let go of the ball of string altogether it's another story. Ok, so eventually the tension in this string passes its mass density times c^2. However, in letting out one meter of string, lowering it down, in its being lowered, it can be made to do work. In fact, the amount of work I can make it do for me, where I am, is the tension it's under times the distance I let out. Force times distance. However, the tension t is more than the mass density times c^2, so in letting out one meter of string, the amount of work it does for me is t while the mass of string I let out was m, which is less than t/c^2. In other words, for every kilogram of string I let out once the tension reaches this level, it does more work for me than c^2, more than the energy I'd get by converting that kilogram of string directly into energy. So it would seem that I can let out one meter of string, and use the work it does, the energy it releases, to create another meter of string out of that energy, and have some left over! Of course it requires a string which is stronger than any real substance that exists could be but that is not the issue at the moment.

So, the question is, is this free energy? The answer is no, because something truly bizarre and spectacular happens. Once you hit that point where the tension passes the mass density times c^2, you are actually lowering the mass of the black hole by giving it more string! Because once you've let out the string to that distance, in your perspective, any measurements you try to perform on the other end of the string determine that its mass is negative. In other words, being held that close to the black hole, its mass has actually gone from positive to negative. Of course if a spider were to climb down the string and take note of the mass at the end of the string, the spider would decide the string's mass was most definitely positive. Of course, you, standing far out, would now see the SPIDER as having a negative mass as it is past that critical distance where it switches from a positive to a negative mass. Think about this scenario and you will see it actually makes perfect sense. The only difference before and after letting out one more meter of string is that the other end now has an extra meter at the end of it. And in order for you to have gotten t joules of energy, where t is the tension (actually the integral of the variable tension as it will only increase as you let out another meter) out of your mass m 1 meter length of string for m*c^2<t is if the difference of t-m*c^2>0 came from somewhere, and where is it? There is now a mass of m*c^2-t in that last meter at the end of the string, and that is a negative mass. You've created a negative mass and positive energy. Or at least that's what I thought.

Until I realized the REAL explanation! Did I have you going there? It certainly had me going for many months when I was thinking of this. The real explanation is almost as weird. The problem with the old explanation is that if in your perspective, the last meter of the other end of the string dangling over the black hole actually had negative mass, then being under positive gravity, its being there should actually yield NEGATIVE weight and actually lift the string above that point up, making the tension in the string less. That is, as you let out another meter of string, even though you're lowering it further into the black hole, the tension it's under actually goes down.

So what is the real explanation you ask? The real explanation is that the mass goes down and approaches 0 as it gets closer to the black hole. True, the force of gravity from the black hole goes up as you get closer and closer but don't confuse mass with weight. As the string gets put under more and more gravitational acceleration, you might think that it gets heavier, but in fact it gets lighter, because though its weight to mass ratio is going up, its actual mass is going DOWN. Remember it's getting lowered into an infinite abyss. If you reach the event horizon of a black hole, you've lost ALL your energy. And that literally means that you have also lost all your mass. In high school physics class they told you that a kilogram on the moon is the same mass as a kilogram on Earth but it weighs 6 times more on Earth, but in fact the kilogram weight on Earth, brought to the moon, wouldn't have the same mass, it would actually have a little bit MORE mass! It would weigh a lot less, but have a little more mass. The amount it loses again in coming back to Earth exactly equals the amount of gravitational potential energy it loses going down Earth's gravitational well, divided by c^2, of course. So back to the black hole. As you lower this string down, as the other end approaches what might be this hypothetical distance where the tension on your end reaches the mass density times c^2, were it to reach such a distance, what you observe to be the mass of the string at that distance would be ZERO. However, as I said, if the string reaches this 0 mass point, the same point where the tension on your end hits m*c^2 and goes beyond it, letting out more line would decrease the tension because on the other end it would be negative mass. So if it were to hit a tension of m*c^2, letting out more would lower it again. What's really going on is that there's no such distance. You lower it and lower it and the tension on your end only gets closer and closer and closer to m*c^2, no matter how far you lower it, because the actual mass of the string on the other end goes down and down and down even if the force on it goes up and up and up. The funny thing is, that nowhere in the string does the tension ever exceed its mass density times the speed of light squared. So if you COULD make a material which had this quite finite tensile strength, you could lower a spool of it down into a black hole and there's no limit to how far you could lower it because it would never break as there's no way to get the tension to go over the mass density times c^2. Even though the gravitational acceleration goes to infinity, and it doesn't matter how big or small the black hole is and how FAST the gravitational acceleration goes to infinity, a tensile strength of m*c^2 is always enough to take the best the black hole can dish out. The same is also true with an incredibly long string stretching across the universe. Since the discovery that the expansion of the universe is accelerating, it is known that the cosmological constant is nonzero. That is, if you had, say, two ball bearings one light-year apart (so that their gravity for one another is negligible) with no initial speed (say they were joined by a string and you simply cut the string), they will accelerate apart as the universe literally creates empty space between the two objects and one meter inflates to 2, 4, 8, etc. Well, if you had a very long string stretching across the universe, this tendency to drift apart as space expands will put the string under tension. Think of it as reverse gravity. One end of the string is pulled one way, and wants to accelerate away from the other end, but they're held together. The tension is of course the greatest in the center. But what this means is that no matter how long you make the string, the tension in the center will never exceed the string's mass density times c^2. Unfortunately I am afraid even carbon nanotubes are nowhere near this tensile strength to mass density ratio. It's a real shame, too. If only the first explanation panned out, while with the black hole there's the mass of the black hole to take away from, but with the universe, there's nothing whose mass could decrease. No black hole, just the edge of the universe. If the negative mass explanation held up, while doing this with a black hole may rob the black hole of its mass, doing it across the universe with the cosmological constant would have nothing to rob of its mass and that really would be true perpetual motion.

#2 Ok, number one was practically a book so I'll try to make this one a LITTLE shorter. Constructive interference. As you are possibly aware, interference can be constructive or destructive. If two waves add together, the result is twice the amplitude, if they are 180 degrees out of phase though, they cancel out. It is also important to note that as the energy of a wave is proportional to the SQUARE of its intensity, when the two waves add together to produce a wave of twice the amplitude, that wave actually has 4 times the power. So constructive: 4 times the power. Destructive: 0 times the power. The question is, is there some scenario where you can have constructive interference that ISN'T accompanied by destructive interference? If they always go hand in hand in equal proportions, then half the time you get 4 and the other half you get 0. 2 on average. You get 1 wave plus 1 wave in and you get 2 out on average. But if you could somehow set it up so that all the interference is constructive, it would be 1+1=4. So, can you have pure constructive interference that isn't accompanied by destructive? Well, I think you know the answer to this from the phrasing of the question. At one point I calculated by hand what happens when you hit a dielectric from both sides with an electromagnetic wave. I found that regardless of the angle of incidence, dielectric constant and permeability of the dielectric film, the thickness of the film, the dielectric constant and permeability of the medium it's embedded in (we're assuming not electrically conductive, so no electrical currents), and finally without regard to the polarization, that for a single incoming wave, whatever is reflected back up from a wave coming from above is always, without regard to any of these parameters, 90 degrees out of phase with respect to whatever is refracted on through the bottom, relative to the middle of the film. That is, if the middle of the film is the plane z=0, if you take the formula for the reflected wave valid for z>d/2 (if the thickness of the film is d) and the formula for the refracted wave valid for z<-d/2, and plug in z=0 into both of these formulas, you would find that they would be two sinusoids that are functions of x, y and t that are exactly 90 degrees out of phase. That is, one will be A*cos(ux+vy+wt+phi) and the other will be B*sin(ux++vy+wt+phi). And it turns out this is exactly the condition required to guarantee that if you hit it from both sides by an EM wave with the same angle of incidence, whatever bounces up, which will be a composite of the reflected part of the top wave and the refracted part of the bottom wave, and whatever bounces down, which will be a composite of the refracted part of the top wave and the reflected part of the bottom, will always add up exactly to the input power. That is, if one experiences pure constructive interference, the other experiences pure destructive and if one is halfway between, so is the other, There's no way to get this scenario to yield more constructive than destructive interference (or more destructive than constructive if you wanted to destroy energy - which could have its uses - you could create energy at one speed, reflect it off a mirror, and then destroy it while it's traveling in another direction and therefore change the momentum of a closed system). Now, the question is, is there another setup, another scenario, maybe some special substance which does NOT follow the behavior expected of Maxwell's equations solved for a specific dielectric constant and permeability, so that somehow, you CAN get more constructive interference than destructive? For instance, some sort of special splitter made of some unknown substance (call it upsiedasium) which is not just a perfect splitter (hit it with an electromagnetic wave at a 45 degree angle and half the power goes down and half goes up, so you have field strength E in the incoming wave and E divided by the square root of 2 in each resultant wave) but DOESN'T always make the reflected wave 90 degrees out of phase with the refracted wave but perhaps doesn't change the phase at all? Then you could do it. I can tell you the results of simulating it with metals. That didn't affect the phase more than one part in many million - but what you got out wasn't the original field strength divided by the square root of 2, but rather just 2. The field strength of the relfected ADDED to the field strength of the refracted wave equaled the field strength of the incoming wave, rather than the sum of their squares equaling the square of the incoming wave field strength. So it's screwed either way. But is there a substance which splits it well without messing up the phase? That would be something, wouldn't it?

[triffid]

Your ideas are interesting but I don't agree about your idea that the mass of one kilogram is different when its located on the moon or the earth.I was always taught to think of mass as the number of atoms in an object(which could be anything you want it to be).Weight of that same object could change depending on which gravitiy field its in.This past week I saw a timing device that depends upon temperature and consists of an aluminum rod,thumbtacks,wax,and a candle.The flame is at one end of the rod,the thumbtacks are stuck on the rod using wax,and as the heat travels down the rod( or rather from one end of the rod to the other) the thumbtacks fall off the rod when the wax melts.Thereby creating one of the most unusal clocks I've ever seen(a clock since it measures time).Triffid

[hansvonlieven]

Old hat triffid.

They used candles that had a metal nail stuck in it every inch or so. As the candle burned down the hails broke free and dropped onto a bell at the base thus ringing the time intervals. Really goes back to antiquity.

Hans von Lieven

[sandor]

Well, triffid, since you so kindly replied to my post about a year after I made it, I thought I would return the favor.

As a matter of fact, a "one kilogram" object on the Earth WOULD have greater mass if brought to the moon. And here is WHY. In order to lift it out of the Earth's gravitational field, you must put energy into lifting it. The moon's escape velocity is much less than the Earth's so the energy it releases in falling into the moon's gravitational well is much less than that required to lift it out of the Earth's. You have therefore done a large net amount of work in moving that object from the Earth to the moon. And that object will therefore actually be more massive on the moon. Not MUCH more massive.

The escape velocity of the Earth is 11186 meters/second. This is 0.0000373 times the speed of light.
The escape velocity of the moon is 2380 meters/second. This is 0.00000794 times the speed of light.
relativistic kinetic energy is m*v^2/(1-v^2/c^2+(1-v^2/c^2)^.5), but I don't need to use that for non-relativistic speeds, I'll just use .5*m*v^2 and divide it by c^2 to get the mass change. And I get 665 parts per trillion when I do this. So by my estimates, if an object that was lifted from the Earth to the moon decreased the Earth's mass by its absence by 1 kilogram, it would raise the mass of the moon by 1.000000000665 kilograms. Of course, at the surface of a neutron star, it would be a big difference. If you dropped 1 kilogram of mass onto the surface of a neutron star and waited for it to radiate the heat from the impact into space, the mass of the neutron star may only be 700 grams more than it was before. The object actually lost 300 grams of its mass by going to the neutron star's surface, and this mass was liberated in the form of kinetic energy as it fell, and a massive amount of heat on impact. Indeed if Newton's apple had been dropped onto a neutron star, it would strike with approximately the energy of an Ivy Mike type hydrogen bomb. Of course, nothing in the vicinity would be able to appreciate the power of the impact, I don't even think that it would bother anything hardy enough to live at the surface of a neutron star if there was such a thing, since it lives under 100 billion times Earth's gravity anyway.

[sandor]

But the really funny thing about all this is, that if you DID move the 1 kilogram object from the Earth to the moon, even if you had a scale that was theoretically sensitive enough to measure a change in mass of 665 parts per trillion, you would be unable to do so, because the reference masses of the scale would also increase in mass by 665 parts per trillion. And indeed, if you took a trip to a neutron star, you may measure your weight and that of any object you bring with you to be 100 billion times what it is on Earth, but even though you and everything else actually only has 70% as much mass, you would be very hard pressed (that's just a stupid pun - get it? You're at the surface of a neutron star, and you're hard-pressed) to measure even a 30% change in mass, because your entire concept of mass is of comparing it to other masses. Even if you took one kilogram of matter and one kilogram of antimatter and smooshed them together at the surface of the neutron star, you would see 2*c^2 joules of electromagnetic energy come out. But anyone far from the neutron star would see the electromagnetic radiation redshifted by the gravity, and would say that the total amount of energy was in fact only 1.4*c^2 because the 2 masses were only 700 grams each to begin with. In other words, wherever you go, 1 kilogram will always SEEM to be 1 kilogram, even though it actually won't be.