Galactic Rotation, Outboard Mass Included?

[Doc Al]

Do the following exercise.

Given: A ring of linear mass density λ and radius R.

Pick a point within the plane enclosed by the ring that is a distance r from the center. Call that point P. Consider the point on the ring closest to P (call it P') and the corresponding point on the opposite side of the ring that is furthest from P (call it P''). Just to be clear, the distance from P to P' is R-r and the distance from P to P'' is R+r.

Now imagine a pair of lines crossing at P subtending small vertical angles dθ. Let the bisector of those angles intersect P' and P''.

Please calculate the gravitational force on a unit test mass placed at point P from each of the two opposite segments of the ring that are defined by those crossing lines separated by dθ.

This is the two-dimensional version (required for a ring) of the three-dimensional construction used to discuss the spherical shell. Of course, this just uses two opposite points on the ring. But it makes the point.

 

[vanhees71]

For the shell or mass distribution of spherical symmetry, it's much easier to solve the Poisson equation,

\Delta V=4 \pi \Gamma \rho.

in spherical coordinates. Since by assumption \rho depends only on the distance from the center, also V is a function only of the distance, and thus the equation reduces to a simple differential equation,

\frac{1}{r} \frac{\mathrm{d}}{\mathrm{d} r}[r V(r)]=4 \pi \Gamma \rho(r).

If you have \rho(r)=0 for r<R, then one has

\frac{1}{r} \frac{\mathrm{d}^2}{\mathrm{d} r^2}[r V(r)]=0.

Integration gives

\frac{\mathrm{d}}{\mathrm{d} r}[r V(r)]=C_1, \; \Rightarrow \; V(r)=C_1+\frac{C_2}{r}.

Since for r=0 the potential should be nonsingular, we have necessarily C_2=0 and thus

V(r)=\text{const for} \quad r<R.

Thus, the force in the interior of the hole vanishes.

 

[Doc Al]

Sure, that's equivalent to applying Gauss's law. That's the easy way to prove that the field within a spherically symmetric shell is zero, but the point here is to demonstrate why it doesn't work for a ring of mass.

 

[vanhees71]

It's not spherically symmetric!

 

[Doc Al]

It's not spherically symmetric!

Please read this (rather silly) thread from the beginning. (If you have time to waste, like I apparently do.)

 

[v4theory]

Sure, that's equivalent to applying Gauss's law. That's the easy way to prove that the field within a spherically symmetric shell is zero, but the point here is to demonstrate why it doesn't work for a ring of mass.

Or, dismissing a preconcieved notion, demonstrating that it does.

 

[Doc Al]

Or, dismissing a preconcieved notion, demonstrating that it does.

You've already demonstrated that you don't understand Gauss's law and the need for symmetry. Let's move on.

Please do the exercise described in post #31.

 

[v4theory]

For any point within a uniform spherical shell you can divide the shell into cones of equal angle and show that the contributions from opposite sides must cancel. This is because the area of the mass subtended is proportional to the distance squared while the field is inversely proportional to distance squared.

I agree. You just applied it wrong.

Consider a ring that is cylindrical in shape. It can be any thickness greater than zero. Let's call it one inch thick (or in cylindrical terms one inch long). It could as well be one foot, a thousand light years or infinite in length like the Gaussian cylinder. It can be any arbitrary thickness to get any arbitrary degree of precision close to a ring of zero thickness (2 dimensional). It can be 1/99999...infinity. Is it above zero or zero? A purely philisophical question at best.


"divide the ring into cones of equal angle"
Case one;
In the case of a one inch thick cylinder ring we treat for a point one inch from one side of a 10 foot diameter ring. The set of "cones of equal angles" cover (or "subtend") one square inch of ring on the far side of the cylinder and a smaller amount on the near side. We then take a point 2 inches from the near side. The "cone of equal angle" of the near side now covers the original area squared by the distance. Two inches distance means four times the original area. The area covered by the far side of the opposite cone has shrunk and the gravitational force has increased by a comensuarate amount. While the area of the near side covered by the cone has squared the gravitational effect has reduced by the inverse square. It has reduced by a factor of four. It is a quarter of the strength. The area under the cone has increased by 4 the strength has decreased by a quarter per unit of the original area (the area covered by the cone at one inch). 4 X .25 = 1. The gravitational force covered by the cone is unchanged at a distance twice as far from the near side. The far side cone has undergone a simmilar but reverse condition. The forces from the mass covered by both cones remain equally balanced. You can treat all the points from one side of the ring to the other and the cones will always have an area of ring mass covered by them.

case two;

If you start from a point one inch from one side of the cylinder and apply a cone one inch across to the near side of the ring one inch thick then the whole cone will cover an area of surface of the ring on the near side. But the whole cone of equal angle to the other side will not be covering area of surface. Part of the area will be empty. The cone from a point twice as far from the near side is no longer covering an area that is the square of the distance. It's covering an area directly proportinal to the distance. Two inches away two inches of ring. As you treat for points farther from the near side the whole cone to the near side will no longer cover area of mass. Only part of it does. That's the wrong way to do it.

Case two is what you did. That's a mistake. As long as both "cones of equal angle" to both sides are of angles small enough to cover a uniform area of mass, no matter how thin, greater than zero, the disk can be treated with conic sections same as the sphere and Gausian cylinder and shown not to apply a net force in any direction on any point in the plane of the disk inside the ring. The sphere and Gaussian cylinder conic sections are treated A priori as having area of mass enclosed by the cone.The cones can be any angle as they are a priori enclosing a uniform area of mass. Why should it be different for the "disk" ring?

Outboard mass can be ignored.

 

[Doc Al]

Consider a ring that is cylindrical in shape. It can be any thickness greater than zero. Let's call it one inch thick (or in cylindrical terms one inch long). It could as well be one foot, a thousand light years or infinite in length like the Gaussian cylinder. It can be any arbitrary thickness to get any arbitrary degree of precision close to a ring of zero thickness (2 dimensional). It can be 1/99999...infinity. Is it above zero or zero? A purely philisophical question at best.

I strongly recommend that you use a cylinder of zero thickness--the thin ring of mass that I suggested in post #31. Much easier to do the complete calculation.

"divide the ring into cones of equal angle"
Case one;
In the case of a one inch thick cylinder ring we treat for a point one inch from one side of a 10 foot diameter ring. The set of "cones of equal angles" cover (or "subtend") one square inch of ring on the far side of the cylinder and a smaller amount on the near side. We then take a point 2 inches from the near side. The "cone of equal angle" of the near side now covers the original area squared by the distance. Two inches distance means four times the original area. The area covered by the far side of the opposite cone has shrunk and the gravitational force has increased by a comensuarate amount. While the area of the near side covered by the cone has squared the gravitational effect has reduced by the inverse square. It has reduced by a factor of four. It is a quarter of the strength. The area under the cone has increased by 4 the strength has decreased by a quarter per unit of the original area (the area covered by the cone at one inch). 4 X .25 = 1. The gravitational force covered by the cone is unchanged at a distance twice as far from the near side. The far side cone has undergone a simmilar but reverse condition. The forces from the mass covered by both cones remain equally balanced. You can treat all the points from one side of the ring to the other and the cones will always have an area of ring mass covered by them.

The big mistake you are making is trying to apply equal solid angle cones when the ring is only of small thickness. The solid angle on the far side will be hugely greater than the thickness of the cylinder, so your argument that the subtended mass scales by the square of the distance falls apart. Instead of solid cones, you must use flat angles.

Fix this error and we can continue. Again, I strongly recommend that you do the exercise that I laid out in post #31.

 

[Doc Al]

Let me express it a bit differently. In order to apply the symmetry used for a spherical shell, you need to consider opposite sides in every direction. But you can't do that with your 1" cylinder. Say the point in question is one inch away from one side of the cylinder (as per your description). Draw a line from the top piece of the cylinder, through that point, to the other side. When it gets to the other side (10 feet away) that line is about 5 feet out of the plane of the 1" thick cylinder. No mass there!

 

[v4theory]

I strongly suggest you read my previous post more carefully. I see you missed the part about uniform mass (or non mass) in both sides of the cones. As long as you adhere to that rule you can point it in any direction that the cones have a uniform area of mass in them from any point in the disk and find zero net force.

 

[Doc Al]

I strongly suggest you read my previous post more carefully. I see you missed the part about uniform mass (or non mass) in both sides of the cones. As long as you adhere to that rule you can point it in any direction that the cones have a uniform area of mass in them from any point in the disk and find zero net force.

You must include all the mass! You can't just pick a few teeny tiny points that happen to have zero net force. Please reread my previous post carefully.

You must show that any and every element of mass through any point has a corresponding element on the other side that provides an equal and opposite force. I gave an obvious example in my last post where this is not true.

It is true for a spherical shell. Not so for a ring.

(Students of freshman electrostatics learn this early on, since the electrostatic field has a 1/r² dependence just like gravity.)

 

[v4theory]

Can you define exactly everything you mean by "any and every element of mass"? Specificly focus on "element of mass".

In case #1 I treated for every bit of mass in the ring relative to horizontal force on points in the plane. You must meant something else besides showing that every bit of mass has a coresponding bit on the other side.

 

[sankalpmittal]

You must include all the mass! You can't just pick a few teeny tiny points that happen to have zero net force. Please reread my previous post carefully.

You must show that any and every element of mass through any point has a corresponding element on the other side that provides an equal and opposite force. I gave an obvious example in my last post where this is not true.

It is true for a spherical shell. Not so for a ring.

(Students of freshman electrostatics learn this early on, since the electrostatic field has a 1/r² dependence just like gravity.)

F=GMm/r²

This universal law of gravitation formula will not work in this case , mind you .
The F in above formula is not the same as the gravitational potential .
Firstly gravitational potential is a scalar quantity not a vector because it need not be worried on direction. It is just the tendency of an object to attract another object (always positive) .

Here all you'll have to do is the vector calculation of ring in case its rotating . IN case of centripetal or fictional force and centrifugal force apply this formula : mv²/r .

Gravitational field is a vector field because in case of rotation vectors of velocity change their direction every unit time .


Doc Al says :
"You must show that any and every element of mass through any point has a corresponding element on the other side that provides an equal and opposite force. I gave an obvious example in my last post where this is not true.

It is true for a spherical shell. Not so for a ring. "


Yeah . Newtons laws are not applicable everywhere . If Newtons third law works everywhere then no work can be done by anybody .

"Newton's third law is only applicable when the force applied by one body would be equal to the resistive force of another body . "

 

[Doc Al]

F=GMm/r²

This universal law of gravitation formula will not work in this case , mind you .

Sure it will, applied correctly. (Using infinitesimal mass elements.)

The F in above formula is not the same as the gravitational potential .

Who are you addressing these comments to?

Firstly gravitational potential is a scalar quantity not a vector because it need not be worried on direction. It is just the tendency of an object to attract another object (always positive) .

Here all you'll have to do is the vector calculation of ring in case its rotating . IN case of centripetal or fictional force and centrifugal force apply this formula : mv²/r .

Gravitational field is a vector field because in case of rotation vectors of velocity change their direction every unit time .

Gravitational field is a vector field, but not because velocity vectors change.

Doc Al says :
"You must show that any and every element of mass through any point has a corresponding element on the other side that provides an equal and opposite force. I gave an obvious example in my last post where this is not true.

It is true for a spherical shell. Not so for a ring. "


Yeah . Newtons laws are not applicable everywhere . If Newtons third law works everywhere then no work can be done by anybody .

Nonsense. You can add Newton's 3rd law to the list of things you don't understand.

 

[Doc Al]

Can you define exactly everything you mean by "any and every element of mass"? Specificly focus on "element of mass".

In case #1 I treated for every bit of mass in the ring relative to horizontal force on points in the plane. You must meant something else besides showing that every bit of mass has a coresponding bit on the other side.

Again, the problem is that you are attempting to apply three-dimensional thinking to a flat ring. Of course you'll end up with nonsense.

Please do the math! Try this. Using your example of a 10 foot diameter ring that is 1" thick and a point 1" from the ring. What solid angle cone will just cover the entire thickness of the cylinder at the near end? Now find the cross-sectional area of that same cone on the far side. Sure, the cross-section of that cone expands with the square of the distance. But the cylinder does not! At the far end, that cone subtends way more area than the cylinder does. So you cannot claim that the mass goes up with the square of the distance.

(You can claim that for the spherical shell, at least for small angles, which is all we care about. The shell goes all the way around in all directions; your 1" thick ring does not.)

 

[v4theory]

Upon what basis do you insist that your way of taking a one inch cone to the near side is correct and my way of taking a one inch cone to the far side is incorrect?

In the sphere the cones always cover a uniform area of mass. That alone is sufficient show that the force is zero along that line and any rotation of that line. In my way the cones cover a uniform area of mass all the way across the plane and rotated at any angle in the plane of the disk at any point in the disk. The disk is not a sphere. There is a boundary area to be treated. Its not a difficult treatment. That's my basis for insisting my way is correct.

That is a sufficient condition to arrive at the conclusion that the horizontal force on the points in the plane are ballanced.

I can treat for your third dispute that the cones in the sphere can be at any angle and solve to conclude zero force at any angle but you have to understand that this solution is correct for the horizontal force first.

The solution obviously also covers all angles above the plane of the ring except for angles pointing at the boundary of the ring and empty space.
The solution for treating the boundary between the area of mass of the ring and empty is much more complex than the treatment for just uniform mass. It involves infinitesimals. It is perfectly standard. But there's little point treating it for you if you can't refuse to understand that by covering a one inch area to the far side solves for zero force in the horizontal direction. Even if you did understand the treatment of the horizontal force, considering how inconsistant and thin your treatments have been in this discussion, you can't possibly have the further prerequisites for understanding the boundary treatment.

I think you should really study up on this instead of just denying what is an obvious and easily visualizably correct solution.

The only reason I'm discussing this with you so extensively is because you are a mod/mentor? and if it can't be resolved I can't continue here. This is basic physics and you are treating it with denials that lack substance.
P.S.
You didn't answer the question. about what you mean by element of mass.

 

[Doc Al]

In the sphere the cones always cover a uniform area of mass.

No they don't. The area covered is proportional to the square of the distance from the cone's origin.

That alone is sufficient show that the force is zero along that line and any rotation of that line.

For a spherical shell, that's true. (After correcting the error you made regarding 'equal areas'.)

In my way the cones cover a uniform area of mass all the way across the plane and rotated at any angle in the plane of the disk at any point in the disk.

No they don't. In the case of a ring, the mass subtended is proportional to the distance from the cone's origin. (Not the square of the distance.)

The disk is not a sphere. There is a boundary area to be treated. Its not a difficult treatment. That's my basis for insisting my way is correct.

But you failed to treat the boundary! You just assumed that it works just like the sphere.

I think you should really study up on this instead of just denying what is an obvious and easily visualizably correct solution.

Why don't you actually do the calculation? Start by drawing a diagram, showing your cones.

The only reason I'm discussing this with you so extensively is because you are a mod/mentor? and if it can't be resolved I can't continue here. This is basic physics and you are treating it with denials that lack substance.

This is basic physics, which I can only assume you've never formally studied.

 

[Doc Al]

Sorry, but I can't help myself...

The solution obviously also covers all angles above the plane of the ring except for angles pointing at the boundary of the ring and empty space.

Uh... realize that if there's no corresponding mass there is an unbalanced force which has a horizontal component. Once again, you've proven yourself wrong. You can't just 'ignore' the boundary or handwave it away as being 'too complicated'. (Note: Solving for the field at an off-center point in the ring is complicated. Of course, if it really were zero as you claim, then it should be easy to show. But it's not. It is easy to show that your reasoning fails, which I've done several times over now.)

Again, I can only suggest that you solve the problem I outlined in post #31. Actually solve it mathematically! The thickness is zero, so there's no boundary conditions to worry about. If I understand you correctly, you claim that the force from each side must cancel for equal 'cones' of angles. So... prove it! It's a trivial exercise so get to it.

 

[v4theory]

Once again of many misquotes and misrepresentations I didn't say it was "too complicated".

Instead of skimming why don't you read the thing?

 

[Doc Al]

How about this: Do the exercise in post #31. It should take you about 1 minute.

If you think that your result (that the field everywhere within a thin ring is zero) is "basic physics", then you should have no trouble whatsoever in finding a standard reference stating that. After all, rings of charge are discussed all the time in electrostatics. And such a simple result should be well known, just like the result for a spherical shell is well known.

I'll wait.

 

[sankalpmittal]

Nonsense. You can add Newton's 3rd law to the list of things you don't understand.

If you can apply then Newton 3rd law everywhere then Newton could have not derived his 3rd law . To derive it he took two balls of same mass and applied equal force on them and so they covered same displacement (after collision).

(mass x)Ball 1 ---------------------><-------------------Ball 2 ( mass x)
Same force of 1 kg ms^-2 on two balls .

<-----------------------Ball1 Ball2------------------------------------>
Same distance y covered after the collision of two balls in opposite direction .

"Newton's third law is only applicable when the force applied by one body would be equal to the resistive force of another body . "

No they don't. The area covered is proportional to the square of the distance from the cone's origin. For a spherical shell, that's true. (After correcting the error you made regarding 'equal areas'.)No they don't. In the case of a ring, the mass subtended is proportional to the distance from the cone's origin. (Not the square of the distance.) But you failed to treat the boundary! You just assumed that it works just like the sphere.Why don't you actually do the calculation? Start by drawing a diagram, showing your cones.

Yes , This turns out to be surprisingly simple! We imagine the shell to be very thin, with a mass density kg per square meter of surface. Begin by drawing a two-way cone radiating out from the point P, so that it includes two small areas of the shell on opposite sides: these two areas will exert gravitational attraction on a mass at P in opposite directions. It turns out that they exactly cancel.



This is because the ratio of the areas A1 and A2 at distances r1 and r2 are given by : since the cones have the same angle, if one cone has twice the height of the other, its base will have twice the diameter, and therefore four times the area. Since the masses of the bits of the shell are proportional to the areas, the ratio of the masses of the cone bases is also . But the gravitational attraction at P from these masses goes as , and that r2 term cancels the one in the areas, so the two opposite areas have equal and opposite gravitational forces at P.



In fact, the gravitational pull from every small part of the shell is balanced by a part on the opposite side—you just have to construct a lot of cones going through P to see this. (There is one slightly tricky point—the line from P to the sphere’s surface will in general cut the surface at an angle. However, it will cut the opposite bit of sphere at the same angle, because any line passing through a sphere hits the two surfaces at the same angle, so the effects balance, and the base areas of the two opposite small cones are still in the ratio of the squares of the distances r1, r2.) Source : http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm

 

[Doc Al]

If you can apply then Newton 3rd law everywhere then Newton could have not derived his 3rd law .

Please explain this sentence. Give me an example of two balls colliding where Newton's 3rd law is not satisfied.

"Newton's third law is only applicable when the force applied by one body would be equal to the resistive force of another body . "

You may as well try to explain this sentence also.

After reading the following from your earlier post, it's hard for me to take you seriously:

If Newtons third law works everywhere then no work can be done by anybody .

And what does Newton's 3rd law have to do with the topic of this thread anyway?

At least the stuff you wrote about the shell theorem was correct. Of course I discussed that several pages ago--I even gave the same reference. Note added: Of course your 'cutting and pasting' from that website misses the point entirely, which is that the same reasoning does not apply to the thin ring.

 

[sankalpmittal]

Please explain this sentence. Give me an example of two balls colliding where Newton's 3rd law is not satisfied.

You may as well try to explain this sentence also.

After reading the following from your earlier post, it's hard for me to take you seriously:And what does Newton's 3rd law have to do with the topic of this thread anyway?

At least the stuff you wrote about the shell theorem was correct. Of course I discussed that several pages ago--I even gave the same reference.

Dude , How can you push a chair ? Answer me .

Because action>reaction .

Another example :
If someone slaps you hard then who will get more hurt ??
Of course you not the person because when person slaps his hands possesses acceleration or dv/dt . Thus the force of slap become more than the resistive force of your cheeks .

Rather difficult for you to convince the person . :)

 

[Doc Al]

Dude , How can you push a chair ? Answer me .

Because action>reaction .

Nope. "Action" always equals "reaction". You push a chair, the chair pushes back on you.

To move a chair, the net force must be non-zero. That's Newton's 2nd law. Nothing to do with the 3rd law.

Another example :
If someone slaps you hard then who will get more hurt ??
Of course you not the person because when person slaps his hands possesses acceleration or dv/dt . Thus the force of slap become more than the resistive force of your cheeks .

More nonsense. The force that your hand exerts on the cheek is equal and opposite to the force that the cheek exerts on your hand.

Same thing if you bang your thumb with a hammer--the force that hammer and thumb exert on each other is equal and opposite. The effect of the same force is different for each object--the hammer is undamaged, while your thumb is crushed. Nonetheless, Newton's 3rd law still applies, like always.

 

[sankalpmittal]

Nope. "Action" always equals "reaction". You push a chair, the chair pushes back on you.

To move a chair, the net force must be non-zero. That's Newton's 2nd law. Nothing to do with the 3rd law.


More nonsense. The force that your hand exerts on the cheek is equal and opposite to the force that the cheek exerts on your hand.

Same thing if you bang your thumb with a hammer--the force that hammer and thumb exert on each other is equal and opposite. The effect of the same force is different for each object--the hammer is undamaged, while your thumb is crushed. Nonetheless, Newton's 3rd law still applies, like always.

Thanks for your reply but please do illustrate an example mathematically .

 

[Doc Al]

Thanks for your reply but please do illustrate an example mathematically .

Illustrate what mathematically? That action equals reaction? F_a/b = -F_b/a?

 

[sankalpmittal]

Illustrate what mathematically? That action equals reaction? F_a/b = -F_b/a?

That i know
F₁=-F₂

Well ok ok ok !

No need !