by Crazyguy 280 Replies latest watchtower bible
Think about it this way. Let's say you were a computer scientist working on solving P/NP. Someone who has never heard of that problem decides to wade into a debate on it armed with literally what they just read on wikipedia and is constantly confusing terms.
Would you take that person seriously? Of course not, bohm. That's what you are doing. You were genuinely confused by the relationship between acceleration and gravity. How do you expect to be taken seriously in this discussion?
Actually I would say there was no gravity inside the shell (or at the center of the earth), not just that it "cancelled" out, however clearly the answer depends on what we mean by there "being gravity"
At the exact center of a perfect shell there is gravity, exactly balanced gravitational acceleration in all directions.
If you disagree, please please please explain how mass can avoid warping space because it's inside of a shell. I await your math.
Because of any friction, the almost weightless floater would start falling from lower and lower heights from the center, would be moving slower and slower, come to a standstill or hangstill at the center, ceased to be accelerated, for there is no gravitational pull in either direction at the center of such an ideal sphere.
Friction? Show me, mathematically, what that has to do with it? Friction from what? Air? Now the shell isn't hollow, there is mass inside and you're talking about a completely different thing that you were before. And, gravitational acceleration doesn't cease to exist, it's simply perfectly balanced in all directions.
If you disagree, please show me how 1)air is not mass and 2) how mass can avoid warping space because it's inside of a shell.
Bohm: There is furthermore no "other" radius as you seem to suggest ("that's the radius you were missing, the second body") since the geometry of the other object does not matter insofar as the shell theorem is concerned, as long as it is either inside or outside the hollow sphere (if it is on the boundary the shell theorem still applies to the parts on either side of the boundary seperately).
Viv: Well, you were getting it. Read a paper on it rather than wikipedia.
So you disagree with this point? Can you eleborate then in what sense the radius of the second body enters into the derivation not covered by my description? Specifically, does the shell theorem in your oppinion not cover the case where a non-spherical body is inside a spherical shell?
Viv: Because it's not "no acceleration on an object inside the sphere", it's "total acceleration is equal to zero". Very different things.
Interesting. There is some basic confusion here on the meaning of acceleration. Try to give a counter example. How does the "total" acceleration of a body differ from it's acceleration?
So you disagree with this point? Can you eleborate then in what sense the radius of the second body enters into the derivation not covered by my description? Specifically, does the shell theorem in your oppinion not cover the case where a non-spherical body is inside a spherical shell?
I already did. I'm not re-typing it because you didn't grasp, read my post it or read a paper on it.
Interesting. There is some basic confusion here on the meaning of acceleration. Try to give a counter example. How does the "total" acceleration of a body differ from it's acceleration?
I already gave an example in the form of an alalogy using the elctrostatic version of the shell theorem. I'm not re-typing it because you didn't read or get it, whichever it is.
Bohm: Actually I would say there was no gravity inside the shell (or at the center of the earth), not just that it "cancelled" out, however clearly the answer depends on what we mean by there "being gravity"
Viv: If you disagree, please please please explain how mass can avoid warping space because it's inside of a shell. I await your math.
I notice you re-phrased what I wrote so as to create a strawman. Firstly, as is apparent in the context, I was treating the newtonian/relativistic example seperately. Secondly, you are talking about the effect of the mass *of an object inside the shell* whereas my statement was (and this is apparent if you read it) about the gravitational field *inside a hollow shell*. That you insert an object inside the hollow shell and discuss the gravitational effect *of that object* is the strawman
Bohm: Can you eleborate then in what sense the radius of the second body enters into the derivation not covered by my description? Specifically, does the shell theorem in your oppinion not cover the case where a non-spherical body is inside a spherical shell?
Viv: I already did. I'm not re-typing it because you didn't grasp, read my post it or read a paper on it.
You quite plainly did not. That you insist you did, and are unable to substantiate your claims except insisting it is so is quite revealing.
I notice you re-phrased what I wrote so as to create a strawman. Firstly, as is apparent in the context, I was treating the newtonian/relativistic example seperately. Secondly, you are talking about the effect of the mass *of an object inside the shell* whereas my statement was (and this is apparent if you read it) about the gravitational field *inside a hollow shell*. That you insert an object inside the hollow shell and discuss the gravitational effect *of that object* is the strawman
Please show my strawman and the math you promised to help with. The shell theorem deals with BOTH mass internal and external mass. You specifically earlier claimed I was moving the goalposts by even mentioning GR, now I'm creating a strawman by literally quoting you and pointing out your error?
Please, please, please show me where I, in any way, changed your words to create a strawman. The ENTIRE conversation has been about people saying "there is no gravity inside a hollow shell". The FIRST person to even mention that the whole point of the shell theorem is to be able to do grvitational computations between external bodies was me.
So, your claim is that now, somehow, me talking about the point I brought up to explain shell theorem to you is somehow me creating a strawman?
I would love to see the chain of logic that allows you to think that me bringing up my own point is turning your argument into a strawman.
And you promised math and haven't delivered.
You quite plainly did not. That you insist you did, and are unable to substantiate your claims except insisting it is so is quite revealing.
It reveals that you are out of your depth. However, since I'm such a giving person, I quote myself again for you. Seriously, try reading next time. You'll look less foolish. This is the second time you've claimed I didn't write something I did and you've been proven wrong.
The shell theorem shows that gravitational acceleration between two bodies can be calculated using the center of those bodies because, effectively, the TOTAL gravitational acceleration inside nets out to zero as can be shown using two perfectly spherical but hollow bodies and how, relative to the internal coordinates, gravitational acceleration is calculated relative to ANOTHER body either inside or external to the spherical hollow mass (that's the radius you were missing, the second body). That doesn't mean there IS NO GRAVITY or that it is not stronger inside the body in one corrdinate than in another place. It just means that it NETS to zero.
There is also a shell theorem application for electrostatic systems. A rough analogy would be that, inside the bubble of the universe, the total amount of energy = zero yet we can still utilize energy to get work done.
Also, you insinutated you could do all the math on this and that I couldn't. Please show your math.
Bohm,
I would say that the question of whether there is no gravity or it being cancelled out is at the outer edges of my knowledge. The reason I would tend to go with it being cancelled is because I am looking at it from an engineering problem point of view. Since gravity acts as if from the point centre of the mass then at a delta distance away from that centre there would be some very small gravitational force. This is because as per the shell theorem that mass beyond the radius of your test dummy does not act upon it. In other words if your test dummy was halfway to the centre of the earth all of the earth's mass that sits above that radius has no net effect on it.
On that point I would agree with Viviane that it is merely that your acceleration is zero. It would seem to me to be a fairly meaningless distinction from a practical point of view however since it is a single point in space.
