Gravity is BS
Tom's 4th Ball Surprise
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Hey Tom, what do you think?: I propose that there is a minimum unit of matter which as part of its nature throws out a minimum unit or quantum of gravity, let's call that latter a graviton, distributed spherically and superficially around itself and pulling toward itself all of space and incidentally all mass in the universe but only out to a fixed radius, call it Rg, and outside the sphere of that radius, not at all. This radius Rg is say a third of a galactic radius for an average galaxy, or so. By "superficially", I mean that the pull on a bit of space at some radius is in proportion to the relative amount of the sphere's surface at the given radius occluded by that bit of space. Like a shadow effect, although another bit of space behind the first is no less pulled or squeezed because of the effect on the first.
Then the force of gravity on a particle will be in the direction of the average inverse-distance-squared-weighted center of gravity of the mass within the radius Rg of the particle. A triple integral over all space summing up these quanta of gravitational effects, zero outside Rg, will calculate the local effect of gravity.
Then matter far from the center of a galaxy will not be gravitationally influenced by the black hole at the center of the galaxy nor the other half of the galaxy across the other side nor even some part of the near side of the galaxy nor anywhere outside a sphere around it.
Then arms will form with matter orbiting, or rather slowly spiraling inward around, the center-lines of the arms. Further, bars will form, swallowing up galactic arms as they spiral toward the greater mass in the bar end relative to the dispersed arm.
Then we have an explanation for galactic spiral arms and bars, which otherwise should spin out and de-cohere like cream in coffee after a few rotations, if the effect of gravity were not quantized and local.
Tom said: Keep up the good work veatch.
Spinning coagulating masses, like forming galaxies or planetary systems or funnel spirals like tornadoes or hurricanes or spiral-flow toilets, seem to first acquire angular momentum equally distributed through the mass, or at least not more distributed on the edges than in the middle.
Then, (considering galaxies and planetary systems) under gravity, they flatten into a disc in the plane of the average rotation of the whole mass, because as you may notice the orbits are not typically polar to the disc, so things gradually tend to fall into the disc and get captured there, thus we have disc shaped distributions of planets and galaxies.
That's why they are called planets, because they are little thingies in a disc, like a plane of thingies, little plane-ettes. Makes sense to me. Has nobody else noticed this?
Well, okay, then under local fluctuations the distributed masses coagulate into bodies like planets or galactic arms. If you're near a small peak of local density, you'll tend to fall toward that direction as you all proceed through the sky, so scattered dust becomes bigger and bigger more or less coagulated chunks.
Then if the distribution of angular momentum is constant or, with the same effect, if it is random around some non-zero mean, then angular velocity is greater closer to the center, and even more so if the angular momentum per unit mass is greater in the interior. Therefore inner masses spin much faster than the outer ones, as the inner parts spin past the outer parts.
Thus the planetary year of Mercury or actually of each inner planet is shorter than each outer planet. Inner planets make a full round long before outer planets do. Isn't this a universal fact about planets? Please tell me I'm wrong.
Let me give you a somewhat similar example. Suppose you spin a cup of coffee with your spoon, and drizzle cream quickly in a line diametrically across the center. Very soon the line becomes a spiral, like a spiral galaxy, but after a few rotations the spiral spins out and loses coherency, either looking more like a bunch of circles or circle-arcs, or a merged creamy coffee. In this case the edges slow down by friction against the cup walls rather than by an equal distribution of angular momentum, but the point is quite analogous. Cream soon loses its spirality; why not galaxies?
When a galaxy has arms curving a quarter, half, or rarely a full turn per arm, how old must that galaxy be, in full-galaxy rotations?
Let's say the arms start, as makes sense, as a straight elongation of a newly-formed rotating galactic disc, then the interior should gradually turn a lot more than the tips of the arms. Half a radius inward, given constant angular momentum, should yield half a turn slip per full turn of the disk. The amount of slip and the age of the formation are inversely related. If the spiral formation was created ten turns back, the spirals should have the half-radius part of the arms 5 turns ahead of the tips of the arms, by this reasoning. 5 full spirals for each arm.
If the universe began with random distributions of galaxy- and star-forming masses coagulating by gravity alone keeping everything together and on center, then a half-turn spiral arm galaxy should be half a rotation in its age, since the elongation began.
Therefore all these galaxies are brand new.
But astrophysicists claim these typical spiral galaxies have spun many times. A universe of 12B years, and a galaxy rotation time of 100M years, maybe? I forgot the exact numbers. But many times, lets say at least 10 times.
Well, that could only be true if they are something more like rigid plates than like gravity-controlled, center-orbiting masses.
Because, yes, a spiral painted on a frisbee maintains its spiral shape irrespective of spinning, but independent orbiting masses may only very temporarily form a coherent spiral shape. Therefore, some force imparting rigidity of mutual location, on the scale of galactic radii, other than gravity, must be responsible for spiral arm galaxies. QED.
SOLVE ME THAT ONE, my pretties!
Math friends have me thinking a different way. Newton's law of universal gravitation says:
F = g m1 m2 / r^2(g is a very small constant, m1 and m2 are the two masses pulling on each other, and r is the radius or distance between them. If you substitute out the famous equation F=m1*a you will see that we accelerate toward the remote mass, m2, at a rate, g m2 / r^2. That rate, times our own mass, is the Force of gravity on us.)
Note that this quantity is NEVER zero if m1 and m2 are >0 and no matter how large is r. Because the law of universal gravity is universal.
This is also called the inverse square law because the pull of gravity is proportional to the inverse square of the radius between the objects (planet and sun). If you go out to twice the radius, the pull of gravity drops off to one quarter.
So consider two equal-mass planets on circular stable orbits around a much larger star, one at radius R, the other at radius NR. Are their speeds related to the radii? Let's pick a start time, and observe that after a short time, dT, the first planet has travelled a short distance, dC, around the circumference and now travels in a direction that has been turned by some small number of degrees, dD=360*(dC/(2*pi*R)), and has adjusted its radius (or height) by some small radially-oriented height, dH, deflected toward its star from its original trajectory, due to the gravitational pull, F1, of the star at distance R. Its speed is distance / time = dT / dC.
Now the second planet, at radius NR, has an orbit N times the size of the first, which is identical geometrically except for being scaled by N in its radius, its diameter, its circumference, and any corresponding similar part thereof. Let's see how to get the same corresponding values for force and speed and time etc., when gravity is attenuated by the inverse square law.
Okay: Let's think about the correspondence of similar wedges of the orbital circle. After travelling a distance of N*dC around its N* scaled circle, the second planet has been similarly turned by dD degrees, because the arc has the same shape in two similar wedges. Further, its trajectory has been deflected from a straight path by a fall toward the star of N*dH, which was correspondingly been effected on planet 1 in dT time. How long did it take on planet 2? If it were travelling at the same speed as planet 1, it would take N*dT to go N*dC distance. E.g., twice as long to go twice as far. The force pulling it downward would then have N times longer to operate, but it is 1/N^2 times weaker, and indeed it has to do sqrt(N) more work to achieve the same deflection because the deflection work is also scaled by N. So a planet at a radius of NR should take N*dT to make its similar wedge, and N times its neighbor's year to make its own full circle around the star. Anyway a galaxy ought to be the same, and a half-radius spiral arm segment ought to make a full rotation around its galactic center in 1/2 the time the outer segment does. I have to check my arithmetic. Would you?
But I say, after two rotations of the spiral arm tips, the half-radius segments ought to be multiple full rotations ahead, which to me would be hard to identify as a spiral galaxy any more, it would be all spread out and unorganizedly disc-like.
What's your conclusion?
Dark matter? No, stuff would be rotating around the dark matter just the same as it does around matter we can see, you just wouldn't see the dark parts. It seems more like stiff space. Could space really get stiff, or you might say, might space itself be chasing around the matter that is in it, at some appropriate scale? I don't know, I'm just a plumber.
But gravity sure doesn't seem to be true in spiral galaxies that are older than a rotation or two. Which means it's not actually true, where I come from, because true means true everywhere. As in, true.
Am I wrong? Happy to be wrong. Email me! Tell me why!
So my friends have all been making fun of me over this, but today I checked the Wikipedia:Spiral Galaxy page, which confirms and elaborates as follows:
If I had to guess I'd say gravity stops working quite the same when it's at the scale of galaxies. Stuff seems sticks to itself more than it falls inward to the average local center of gravity when that center is a fat part of a galactic radius away. Thus bars form and remain coherent rather than falling inward to their galactic centers, and thus spiral arms also form and remain coherent and hold together as they spin around the galactic center without being stretched and "wound" as much as they would if the gravity of the center were actually pulling on them as much as it is supposed to.
Perhaps the force of gravity has fallen off so far when the radius part of the gravity equation, 1/r^2, brings the force so close to zero that maybe only a zero or one quantum of gravity can occasionally get that far out, or maybe mostly zero quanta of gravity get that far out, they are so reduced in concentration at galactic radii. Or perhaps the gravity that gets out to such a distance has to be entrained and supported by more gravity (or mass or ...) in the area, providing it a sort of minimum paved surface or a road or a travelling trough or even an amplifying or a multiplier effect of other gravitons in the area or something whereby it can only barely reach out to such enormous distances as across a galaxy to the far bits of a spiral arm. That is, this might be a quantum round-off error in the effect of gravity from such large radii.
Just imagining possibilities here, but the density wave concept offered on Wikipedia also seems like BS to me; suspended particles don't accelerate and decelerate in a density wave unless there is a medium pushing them back and forth or stars somehow push each other away like masses of air in a resonating column, for which there appears to be no physical basis, since what they actually do is pull each other together using gravity. Is the interstellar medium of an ion every 100 cubic meters supposed to make space springy for stars? No. And anyway, that doesn't explain why it's in a spiral formation, or a bar, either. No, spiral arms, and bars, are just stickier than they seemingly ought to be. If there's another parameter for gravity that limits its distance of application, that would seem right to me. Then galactic bars and arms can hold together while the galactic center, or center of gravity doesn't hold their parts individually as tightly. Maybe.
On the other hand, isn't the issue here the rapidity of the tips rather than the slowness of spiral arm midpoints? That is, the tips are able to fall toward the center more rapidly than expected, and thus to maintain their circular orbits at a higher velocity than expected. Then the galaxy is almost a circumferentially-driven vortex, or like a railed track held together more tightly than gravity for the outer bits.
So perhaps gravity is accentuated rather then attenuated at longer distances.
These are mere idlings of a curious character, who is willing to be considered a fool while still figuring stuff out. But if your first thought was to disagree with and contemptuously dismiss the title, Gravity is BS, well, the fool was not me, as it merely restates and clarifies Wikipedia's comment: "higher than expected from Newtonian dynamics". Looks like something needs fixed, and we are all fools until it is.