Would an inverse-cube law obey Newton's shell theorem?

What in

did you fail to understand?

Err, you can't follow simple math (yet you put on airs in "discussing" GR, black holes, etc). Here is again:

If \(F =k / r^2\) the resultant force is identically zero everywhere inside the shell since
\(F(r) = \frac{GMm}{4r^2 R} \int_{R-r}^{R+r} \left( 1 + \frac{r^2 - R^2}{s^2} \right) \, ds=\frac{GMm}{4r^2 R} (2r-2r)=0.\)

 

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None of it. I understand it just fine and I agree with it. The question is, what about it did you fail to understand when you said

??

You were wrong Tach, just accept it. Perhaps you could get Syne to proclaim that you were right but refuse to discuss why.:thumbsup:

 

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I am glad that you agree because , as opposed to natural language, math is very precise. But , contrary to your claim above, obviously you don't understand the math since you posted the nonsense:

 

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It's convenient that you cut off my reference to 1-spheres. Perhaps you'd like to try your calculations again with 1-spheres? It is apparent to me that you try to revise history and, when that cannot be done, you try to obfuscate the conversation so that others reading the thread will be confused.

 

Err, the calculation I showed you applies to "1-spheres" just the same. As a matter of fact, the problem is solved by reducing the 2d statement to a 1d statement

Err, the following is true for 1-spheres (you can stop puffing yourself and call them "circles"):
If \(F =k / r^2\) the resultant force is identically zero everywhere inside since
\(F(r) = \frac{GMm}{4r^2 R} \int_{R-r}^{R+r} \left( 1 + \frac{r^2 - R^2}{s^2} \right) \, ds=\frac{GMm}{4r^2 R} (2r-2r)=0.\)

The fact that you are unable to perform any calculations bites you in the a$$ again. "Armchair philosophers" cannot become physicists, no matter how much they huff and puff themselves up.

 

This is correct. Reread my explanation and it should be obvious I was referring to 0-spheres.

 

lol

 

No. I said: "but but it would be at least 100 billion times larger than that caused by gravity in truth not being exactly inverse square. " Read again and see that the "it" is refering back to the small instrument´s gravity field making a 100 billion times greater dievation from constant gravitational potential inside the hollow sphere than GR effects do.

 

I guess I need to hold your hand on this.

Now...reread my text:

 

And that figure was not just pulled from thin air? I don't recall seeing any parameters involved - mass density or radius of shell, or mass and radial location of 'test instrument'. So exactly what parameters did you have in mind in making that 100 billion times claim?

 

I see. But you're referring to a hollow, Earth sized, Earth-massed planet?

 

You should have quit while you still had a chance:

 

I just guessed based on fact sun´s mass with a light beam grazing past it* made only a very tiny fraction of a degree bend in the photon´s path. Yes I was as indicated by idea of a tiny unicorn running around inside the hollow sphere, thinking of a possible sphere in some lab. Then to a very high order (or pehaps not at all?) the " mass density or radius of shell" make no differences as spherical shell was stated to be uniform.

* Note there is by either GR or Newtonian gravity a very strong** gradient near the sun but essentially none inside the hollow sphere.

** You seem to like my WAGs so here´s another for you: I bet if you could not evaporate, and stood on the sun´s surface that the gravity graient there might pull you apart (neglecting fact just the gravity alone would have made you pancake in shape). That gradient stress is not hard to calculate on a 6 foot tall man but I´m too lazy and like making WAGs.

 

About what I thought you were thinking. Let's give the unicorn a break from running

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. Seems the finding of zero field interior to a spherical shell is a consequence of Birkhoff's theorem - see 2nd passage here: http://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity)#Implications
But that's just an assertion there and I wonder what's behind it, given inverse squared law does not hold in GR.

Caught this addition after I started replying. Won't even dare ask for an expansion of WAG - but I can guess.

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Yes Tach. When I write something, even a mistake, I do not change it once it has been responded to. I call it integrity.

I erred in referring to a 0-sphere as a 1-sphere. I meant a 1-ball because we're actually talking about the interior of the shell, even though the reason for the misstep should obvious

This is all a red-herring anyway. You're trying to distract from the fact that you made a generalization which was flat wrong, and you're hoping that it equates to my simple semantic blunder.

 

Coming from you, this is rich.

Nothing that you post is obvious since you are incapable to express yourself in the language of physics : math.
On the other hand , it is quite obvious to anyone reading your posts that you don't know what you are talking about. With each post you dig yourself deeper.

 

Do you maintain this statement?

 

You have been given the exact mathematical formulation of the answer. Since you are unable to calculate it yourself, you readily agreed with it, so you can stop trolling now.

 

Here's some quoteception to clarify things.

So RJ asked about whether Newton's Shell theorem would hold in hyperspace. Tach said that it does due to spherical symmetry. But Alpha disagreed.
Subsequently Tach clarified that he was referring to zero net force inside the shell. That zero net force would apply in hyperspace.

Zero net force is only half of Newton's Shell theorem. Afterwards Tach elaborated that zero net force only holds for a non inverse-square law if you're in the center of the sphere.


Unfortunately whatever they were discussing on Pg 2 didn't make much sense. Heh

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"Corrects", also known as "backtracks without admitting error". As I said, I'm only harping on this point because he's so hypercritical of others.