OK: 1] They're not standing waves, they're propagating. (You know that's a still image, right?) 2] They're already concentric. (Did you mean something else?) Oh god no. Not the vibrating stars again...
Standing waves form around the physical . There are two fundamental properties of the Universe , the physical , and waves .
dave, yeah, each one of the 2 sets iof near circular waves is centered on their respective middle point, but I assume we are talking about merging entities, and that picture would come to the conclusion with only one center. I did not see the caption, but no doubt an interference pattern would emerge either way. It is thought that the multiple sources of stellar pulses create such patterns , harmonics, Eigenschwingungen, Chladni-plate-like. thank you.
yeah, that red line if seen as moving in the direction of the g force action, toward the center, looks like the final phase of a mean tsunami, or tidal bore. Please Register or Log in to view the hidden image!
As I walked up the road to the fishmonger this morning, I found myself wondering why the acceleration due to gravity increases linearly from the centre of a solid sphere to its surface, as shown in the diagram in Origin's post 20, reproduced below for convenience: Please Register or Log in to view the hidden image! The effective mass creating the acceleration, at a given radial distance, is only that portion of the sphere within the radial distance. (By the shell theorem, all the mass at greater distance has no effect.) Assuming uniform density, the mass is proportional to the volume, which is 4πr³. So the portion of the mass creating the acceleration goes up with r³. However, the strength of the acceleration due to gravity is inversely proportional to the square of the distance from the centre of mass, i.e. it is proportional to 1/r². So the net effect is to make gravitational acceleration directly proportional to r. I realised this just as I arrived at the fishmonger's door, and bought some smoked cod roe to celebrate. Please Register or Log in to view the hidden image! Maybe other readers already knew this, but I confess I had not thought about it until now.
ec, Is that insight not reflected by the linear slope of the blue line from o to 1R at the highest point, the surface? Think now how that results in the red line of post# 579 and the explanation at #578 On the inside a direct , halved acceleration is now acting over a surface of a sphere of half the size, with only a quarter of the surface area. There is where you get the curved slope of the red line, the more rapid fall of the "total" accumulated gravity field. Searching for the best word to express that g, sigma? like in the diagram of post# 625? ?
Yes of course. I am explaining WHY it is linear. It was not obvious to me why this should be so, until I went through the thought process I have outlined. You multiple a cube relationship by an inverse square relationship and so the result is direct proportionality.
ec, In a similar process it dawned on me, that the total gravity (sum of field strength) stays the same no matter how great the radius above the surface and the local strength has fallen., and in the interior. at 1/2 the radius, the total combined field strength over the whole inner surface is 1/8 of the outside value. Total Gravity field strength is red lined (to use a racy term) at surface value. but scouring wiki images, and astro sites I have yet to see a precursor to the graphics or the equation, or references to the term. (and still looking for the proper one. g sigma?) Of course the idea was obvious, staring me in the face.
dave. Not to take anything away from ec's "eureka"moment., and his delicious reward (I miss my old word delicacies), but Bill T in post # 3 already dealt with the whys and hows of the linear decline. I posted the red line g total graph to show that the difference between inside and outside gravity field strength is even more pronounced than obvous at first look.
dave, earliest post# 597, here another version, work in progress: Please Register or Log in to view the hidden image!
River, That red line might resemble a tidal bore in a river in cross section. but the similarity ends there. Instead, in a gravity field, When a body expands, -it's surface moving away from the center-, the red line slope of the inner curved portion would lengthen, while the heights of this red line outside will remain constant. (total, global gravity field strength never changes)*. In a contraction, the total, global gravity field strength would still be constant, shown in in red line heights, but the slope steepen, to be vertical for a black hole. so get this: A black hole has not more global gravity field strength than the bigger entity it formed from!. The local gravity field strength, the blue line , changes dramatically in a contraction as seen in post #260 page #13 : Please Register or Log in to view the hidden image! Talking about river waves here, The Severn has such a high tidal waves because the total energy coming in from the Atlantic is focussed into a smaller surface.;- Similarly, the global gravity field strength is constant, but when being exerted over a smaller area, peaks locally. as shown in the above graph. *At 1 AU, the surface gravity of the sun will be ~the same as it acts now on the Earth.
sorry, there is even the graph showing on your "quote" link. is #635 blank too? P.S. My question was: --has anybody seen a graph with a red horizontal line giving the global, total gravity field strength among any of the "IMAGES" sections, or equations of wiki or anywhere? because it could be that nobody has bothered to look at that aspect of gravity.
They are all broken image icons. Everyone but you sees them as broken. You see them because you have authenticated to your own web host. If you log out of Shaw Cable, you too will see broken links.
I am 90 this year and did not understand this computer issue. . here is the latest version of the graph with this seemingly different aspect of global , sum of the gravity field "g":. did you see post 637?