Galactic Rotation, Outboard Mass Included?

[v4theory]

I thought I understood how to calculate mass based on Newton vis a vis circular velocity and radius. If I know the circular velocity and the radius of an orbiting object I can compute the mass inside the orbit.

In effect the circular velocity at radius causes an outward vector that must be resisted by a specific mass inside the radius in order for an object to orbit at that radius.

Most of what I've been studying about galactic rotation agrees with what I think I understand. However many sites claim that symetrical disk outboard mass (outside the radius) must be included...that outboard mass causes a second (gravitational) vector outward. (symetrical sphere outside the radius can be ignored).

The problem I'm having is that if we assume that I calculate for the unknown mass inboard by the circular velocity and radius and there is no outside of radius mass then end of story case closed I have correctly solved for the unknown mass inside the radius. But if there is an addiitional unknown outside mass causing an additional unknown gravitational vector outward besides the circular velocity vector then my solution for the inside mass would be in error. There would have to be more mass inboard to compensate for the additional outward vector. There could be a hundred or a thousand times the mass inside and outside the radius I calculated under the assumtion that there is no ouside mass and still have the same circular velocity and radius as in the original case. As far as I can understand including outside mass would render Newton entirely irrelavant to an accurate calculation of either the inboard or outboard mass.

The sites I mentioned have been...passing the buck in this issue. At present the buck has stopped here at physics forums. The last passer of the buck said that I should post in the classical physics forum.

Must outside radius mass be included or not? How is the issue I mentioned above resolved? Is it a Newtonian solution?

 

[Bill_K]

Calculating the Newtonian potential is made easy in the spherically symmetric case by several important facts: a) the potential outside a spherical shell is equal to the potential of an equivalent point mass at the origin and b) the potential inside a spherical shell is constant. Thus if you are located at radius R you can ignore the exterior matter at r > R and easily calculate the potential due to the matter at r < R using Kepler's laws.

Unfortunately none of these assumptions works for an axially symmetric disc. The potential outside a ring of matter is not equivalent to a point mass at the origin. Kepler's laws do not hold. And yes you must include in your calculation the potential due to the exterior matter at r > R.

 

[v4theory]

And?

 

[Bill_K]

And therefore the potential for a given mass distribution has to be calculated numerically. You can then easily derive the velocity curve from the potential. However you want to do the inverse - derive the mass distribution from the velocity curve - and the only way to do that will be iteration: take a mass distribution and keep adjusting it until the velocity comes out right.

 

[v4theory]

How is the fact I illustrated above handled? That if outboard mass is included then Newton only sets a lower limit for mass. Any outboard mass can be counter balanced by inboard mass to result in the same velocity.

I agree your solution is the correct solution for calculating mass through the rotation curve but it works through Newton (Kepler) when outboard mass is ignored.

Point of detail. My OP was about calculating mass from rotation in a disk not calculating mass distribution through rotation curve. Though, as I said, a correct formula for calculating mass from rotation can be used to calculate mass distribution from rotation curve. Using Newton (calculating for unknown mass inside the radius of rotation with outboard mass ignored) I can compute a mass distribution from the rotation curve but it can't be the correct mass if unknown outboard mass must be included.

I should point out that calculating a mass distribution curve based on a rotation curve (outboard mass ignored) does (with certain nuclear fusion based caveats) fit the observed distribution of visible luminant matter.

 

[Bill_K]

Using Newton (calculating for unknown mass inside the radius of rotation with outboard mass ignored) I can compute a mass distribution from the rotation curve but it can't be the correct mass if unknown outboard mass must be included.

It isn't correct whether you include the outside mass or not. Since you don't have spherical symmetry, the orbital velocity and the 'mass inside' are not even related.

 

[v4theory]

I have satified myself that the OP question is answered as the gravitational field is a scaler field but can also be treated as a vector field.

With a sphericly symetrical mass the vector lines outside the mass all point to the center point of the mass. For an object anywhere in 3d space outside the mass the acceleration vector is to the center point of the mass. In effect the pull of the "north pole", the "south pole", the equator and all points in between have a net vector on an object outside the mass toward the center point of the mass. For objects anywhere inside a hollow sphere the net vectors on the inside result in a zero vector on an object inside so its mass can be ignored.

For a Gaussian cylinder (a cylinder of infinite length) all gravitational vectors point to a center line rather than a center point in the cylinder. An object outside the Gaussian cylinder will have a vector towards the center line of the cylinder at 90 degrees to the line. So as in the sphericly symetrical case the pull of the "north pole", the "south pole", the equator and all points in between have a net vector on an object outside toward the center line of the cylindrical mass. Objects anywhere inside a hollow Gaussian cylinder will like the symetrical sphere have a net zero gravitational vector.

For the disk all vectors anywhere in space outside the disk do not point to the center point of the disk. Objects above the plane of the disk will have a vector towards the center point and vectors toward all points of the plane some of which do not net to the center point.

However like the hollow sphere and the Gausian cylinder for all points in the space of the plane of the disk all the vectors do point to the center point of the disk. An object outside the disk in the plane will have no polar vectors and all remaining net vectors will be to the center point of the disk.

So for an object anywhere inside a hollow ring on the plane of the ring the vectors of the ring result in a net zero vector on the object inside in the plane of the ring. The outboard mass can be ignored.

By the way for reasons I posted to the mods this will be my last post on this forum.

 

[Bill_K]

However like the hollow sphere and the Gausian cylinder for all points in the space of the plane of the disk all the vectors do point to the center point of the disk. An object outside the disk in the plane will have no polar vectors and all remaining net vectors will be to the center point of the disk. So for an object anywhere inside a hollow ring on the plane of the ring the vectors of the ring result in a net zero vector on the object inside in the plane of the ring. The outboard mass can be ignored.

Yes the vectors point to the center of the disk, but no they're not zero.

 

[v4theory]

My last post was a little sloppy and incomplete. This one is still incomplete. It's not in greek. Though I'm sure it will be greek to some people. I think it's in plain enough language to be understadable to anyone of moderate intelligence and recognizable to to anyone skilled in the subject.

The gravitational field is a scaler field but can also be treated as a vector field.

Sphericly symetrical mass

Geometric exposition;

There is no preferred (same orientation for all points in space) axis or equator of a sphere. Only the center point has the same orientation for all points in space. Any point anywhere in 3d space inside or outside a sphere can define a point on a non preferred equatorial plane of a sphere bisecting the sphere equaly across the equatorial plane through the center point of the sphere into "north" and "south" halfs.

Vector exposition;

The "net" vectorization function convetionalizes treatment of multiple vectors through the additive properties of vectors.

Vectors point from points in space to the sources of gravitational force. Net vectors are the additive result of two or more vectors.

For all points anywhere in 3d space outside the spherical mass the vectors toward the "north half" plus the vectors toward the "south half", plus the vectors toward the equator plus the vectors toward all points in between, "east" and "west", have a single net vector toward the center point of the mass.

For all points anywhere inside a hollow sphere the (non preferred) "north" vectors plus the (non preferred) south vectors net zero polar vector. All vertical "north south" vectors can be ignored leaving only the horizontal vectors which net to equatorial vectors. The equatorial plane vectors can be treated as toward a two dimensional ring around the equatorial plane. The equatorial vectors from the center point to all sides of the equatorial "ring" equal in number and strength. All vectors from off center points pointing toward the "far side" of the equatorial "ring" increase in number and decrease in strength and the vectors toward the "near side" Of the equatorial "ring" increase in strength and decrease in number by coresponding amounts resulting in net zero horizontal (equatorial) vector. All possible vectors net zero on every point inside a hollow sphere so the mass of a hollow sphere can be ignored for all points inside.

Gaussian cylinder (a cylinder of infinite length)

Geometric exposition;

There is no preferred equator of a Gaussian cylinder. The axis of a cylinder is preferred (oriented the same for all points in space). Any point anywhere in 3d space inside or outside a Gausian cylinder can define a point on a non preferred equatorial plane through the center line of the cylinder at 90 degrees to the center line bisecting the cylinder equaly across the equatorial plane through the center line of the cylinder into "north" and "south" halfs.

Vector exposition;
For all points anywhere in 3d space outside the Gaussian cylinder the vectors toward the "north half" plus the vectors toward the "south half", plus the vectors toward the equator plus the vectors toward all points in between, "east" and "west" have a single net vector 90 degrees to the center line of the mass toward the center point of the "equatorial" disk of the mass.

For all points anywhere inside a hollow Gaussian cylinder the "north" vectors plus the "south" vectors net zero polar vector. All vertical "north south" vectors can be ignored leaving only the horizontal vectors which net to equatorial vectors. The equatorial plane vectors can be treated as toward a two dimensional ring around the equatorial plane. The equatorial vectors net zero on the center point as all horizontal vectors equal in number and strength from the center point. All vectors from off center points pointing toward the "far side" of the equatorial ring increase in number and decrease in strength and the vectors toward the "near side" increase in strength and decrease in number by coresponding amounts resulting in net zero horizontal (equatorial) vector. All possible vectors net zero on every point inside a hollow Gaussian cylinder so the mass of a hollow Gaussian cylinder can be ignored for all points inside.

Disk

Geometric exposition;

In a disk there is only one preferred equator, only one preferred axis and only one preferred center point.

The vector exposition only treats for points in the preferred equatorial plane oriented to the preferred axis from the preferred center point.

Vector exposition
For the disk all vectors anywhere in space outside the disk do not point to the center point of the disk. Points above or below the plane of the disk will have vectors towards the center point and vectors toward all points of the plane which do not net to the center point.

However like the hollow sphere and the Gausian cylinder all vectors from all points in the plane of the equatorial disk outside the disk net to the center point of the disk. A point outside the disk in the plane has no north or south polar vectors. The polar vectors net zero.

For all points anywhere inside a hollow ring on the plane of the ring all points anywhere inside, like a hollow Gaussian cylinder and like the symetrical sphere, have north vectors plus the south vectors equal net zero polar vector (as there is no north or south mass there are no north or south vectors so the polar vectors net zero). The equatorial plane vectors can be treated as toward a two dimensional ring around the equatorial plane. Like the equatorial disk of the sphere and the Gaussian cylinder the vectors from the center point to all sides of the equatorial ring are egual in strength and equal in number so net zero vector from the center point. All vectors from off center points to the sides reduce in strength and increase in number to the far side of the ring and vectors to the near sides increase in strength and decrease in number by coresponding amounts to the far side vectors resulting in net zero horizontal (equatorial) vector. All possible vectors net zero from every point inside a hollow ring in the plane of the ring so the mass of a hollow ring can be ignored for all points inside the ring in the plane of the ring.

Thus all mass outside the radius of any point in a disk can be ignored.

 

[Doc Al]

Thus all mass outside the radius of any point in a disk can be ignored.

If you're claiming that a disk-shaped mass distribution has the property that mass outside a given radius has no gravitational effect on points within, that's not true. A flat disk and a spherically symmetric mass distribution have different properties.

 

[v4theory]

If you're claiming that a disk-shaped mass distribution has the property that mass outside a given radius has no gravitational effect on points within, that's not true. A flat disk and a spherically symmetric mass distribution have different properties.

That's what he said. Means a much now as it did then.

 

[Doc Al]

That's what he said. Means a much now as it did then.

How about this: Learn how Newton derived the result for a spherical distribution of mass. Look up the Shell Theorems. (Or use calculus.) Then see if you can do the same for a disk. You'll find that you cannot.

 

[v4theory]

How about this: Learn how Newton derived the result for a spherical distribution of mass. Look up the Shell Theorems. (Or use calculus.) Then see if you can do the same for a disk. You'll find that you cannot.

Well this is so basic, high school stuff really, you probably just don't remember it.

You did read my posts didn't you? I wrote a lengthy exposition of the issue above. I think it's correct even though it isn't in greek. It uses concepts and terms that have been mainstream for the field for hundreds of years since Newton. My treatment of the sphere and Gaussian cylinder is virtually verbatim from Newton and Gauss. They didn't treat explicitly for a disk as the disk is a priori fundamental to both. Since you claim the mantle of expert and mentor, if you think I made a mistake, perhaps you could write a more detailed critique pointing them out rather than one composed of such brief unsupported assertions.

And no, I don't think "i am an expert" is good support for assertions. Take another look at my above post. That is what I call good support. At least it's detailed and Illustrates all the relavant aspects.

So far you are just reiterating what I believe I have proven is simply a common misconception.

How about I make it easy for you and take the points one by one?

The gravitational field is a scaler field but can also be treated as a vector field.

True or false?

 

[Doc Al]

Well this is so basic, high school stuff really, you probably just don't remember it.

Nice try.

You did read my posts didn't you? I wrote a lengthy exposition of the issue above. I think it's correct even though it isn't in greek. It uses concepts and terms that have been mainstream for the field for hundreds of years since Newton.

While you did use some recognizable terms, the post itself seemed like word salad.

My treatment of the sphere and Gaussian cylinder is virtually verbatim from Newton and Gauss. They didn't treat explicitly for a disk as the disk is a priori fundamental to both. Since you claim the mantle of expert and mentor, if you think I made a mistake, perhaps you could write a more detailed critique pointing them out rather than one composed of such brief unsupported assertions.

Since you claim to understand Gauss's law, please describe in plain English how you'd apply it to a sphere (no problem), an infinite cylinder (no problem), and then a disk (big problem).

Please, no baloney about vector and scalar fields.

 

[Doc Al]

Disk

Geometric exposition;

In a disk there is only one preferred equator, only one preferred axis and only one preferred center point.

The vector exposition only treats for points in the preferred equatorial plane oriented to the preferred axis from the preferred center point.

Vector exposition
For the disk all vectors anywhere in space outside the disk do not point to the center point of the disk. Points above or below the plane of the disk will have vectors towards the center point and vectors toward all points of the plane which do not net to the center point.

However like the hollow sphere and the Gausian cylinder all vectors from all points in the plane of the equatorial disk outside the disk net to the center point of the disk. A point outside the disk in the plane has no north or south polar vectors. The polar vectors net zero.

Translation: The field at any point within the plane of the disk cannot have a component perpendicular to the disk. Good!

For all points anywhere inside a hollow ring on the plane of the ring all points anywhere inside, like a hollow Gaussian cylinder and like the symetrical sphere, have north vectors plus the south vectors equal net zero polar vector (as there is no north or south mass there are no north or south vectors so the polar vectors net zero). The equatorial plane vectors can be treated as toward a two dimensional ring around the equatorial plane. Like the equatorial disk of the sphere and the Gaussian cylinder the vectors from the center point to all sides of the equatorial ring are egual in strength and equal in number so net zero vector from the center point.

Translation: The field at the center of a ring must be zero. Good!

All vectors from off center points to the sides reduce in strength and increase in number to the far side of the ring and vectors to the near sides increase in strength and decrease in number by coresponding amounts to the far side vectors resulting in net zero horizontal (equatorial) vector. All possible vectors net zero from every point inside a hollow ring in the plane of the ring so the mass of a hollow ring can be ignored for all points inside the ring in the plane of the ring.

Here's where your argument fails. 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. (See the discussion "Field Inside a Spherical Shell" towards the bottom of this page: http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm" )

This argument does not work for a circular ring. For any point within the ring you can divide the ring into segments of equal angle. Unfortunately, the mass subtended is proportional to the distance (not the distance squared) while the field is still inversely proportional to the distance squared. Thus the far end exerts less force than the near end, proving that there is a net force radially outward at any off-center point within the ring.

Thus all mass outside the radius of any point in a disk can be ignored.

As shown above, no.

 

[v4theory]

Now we're getting somewhere!

Your mistake was to apply a conic area section to a treatment that had already treated for the vertical component of the vectors to the "north south" polar mass. North component plus south component nets zero vertical vectors. This leaves only the horizontal components of the vectors to the vertical mass which net to equatorial vectors.A ring, a circular line, not an area. This is how it is treated in the sphere, the cylinder and the disk. The treatment of the equatorial vectors is the same for the sphere, cylinder and disk because they are the same. A ring of vectors netting toward equatorial mass. The horizontal component of the vectors in the sphere and cylinder cases are netting toward a virtual ring mass. In the disk ring case it an actual ring mass. A net vector to a virtual mass shouldn't be too hard to understand. We do it when we take two masses in front and treat the vectors to them as vectors to virtual points in the centers and then add those vectors to net a vector to a virtual mass point between the two masses where there is no actual mass. The horizontal components of the vertical mass can be treated as if they are equatorial vectors because that's the direction the horizonatl component's vectors net.

The quoted part of my ealier post that you took dispute to was taken out of a context that had already treated for the vertical component.

 

[Doc Al]

Now we're getting somewhere!

Your mistake was to apply a conic area section to a treatment that had already treated for the vertical component of the vectors to the "north south" polar mass. North component plus south component nets zero vertical vectors. This leaves only the horizontal components of the vectors to the vertical mass which net to equatorial vectors.A ring, a circular line, not an area. This is how it is treated in the sphere, the cylinder and the disk. The treatment of the equatorial vectors is the same for the sphere, cylinder and disk because they are the same. A ring of vectors netting toward equatorial mass. The horizontal component of the vectors in the sphere and cylinder cases are netting toward a virtual ring mass. In the disk ring case it an actual ring mass. A net vector to a virtual mass shouldn't be too hard to understand. We do it when we take two masses in front and treat the vectors to them as vectors to virtual points in the centers and then add those vectors to net a vector to a virtual mass point between the two masses where there is no actual mass. The horizontal components of the vertical mass can be treated as if they are equatorial vectors because that's the direction the horizonatl component's vectors net.

The quoted part of my ealier post that you took dispute to was taken out of a context that had already treated for the vertical component.

Sorry, but I can't understand what you're saying here. As I showed above, the argument that allows you to conclude that the gravitational field within a spherical shell is everywhere zero does not work for a ring of mass.

Just stating that the vectors must add to zero is not enough--you must actually provide an argument.

If you know calculus, actually do the calculation. (If you attempt to apply Gauss's law, you'll find that the ring lacks sufficient symmetry to get anywhere. Recall that the Gaussian surface is a closed surface.)

 

[v4theory]

You're conflating scaler field terms in a vector field treatment. The surface area of the conic section scales up by the square of distance while the strength per unit area scales down by the square of the distance. See? Scaler field.

Vectors don't scale. Their functions are strength, direction and number. Even if two vectors are on the same line they are not the same vectors. You can't treat vectors as "the strength of the vector increases or decreases with distance". You can only say a vector from a point closer to a mass has a greater net magnitude than a net vector from a point farther from a mass. Each point in space has its own unique collection of vectors and has to be treated that way in a vector field treatment. The closer the point is to a mass (not treated as a point source) the greater the number of vectors to the mass. You can explicitly add and subtract vectors and components of vectors for net vectors.

You never did answer the question.

Gravity is a scaler field and can also be treated as a vector field.

True or false.

 

[Doc Al]

You're conflating scaler field terms in a vector field treatment.

Not me. I've only spoken about the gravitational field, which is a vector. (The gravitational potential is a scalar, but I haven't mentioned that.)

The surface area of the conic section scales up by the square of distance

Right, for a spherical shell.

while the strength per unit area scales down by the square of the distance.

The strength per unit area of what? What are you talking about?

See? Scaler field.

Huh?

Vectors don't scale. Their functions are strength, direction and number. Even if two vectors are on the same line they are not the same vectors. You can't treat vectors as "the strength of the vector increases or decreases with distance". You can only say a vector from a point closer to a mass has a greater magnitude than a vector from a point farther from a mass. Each point in space has its own unique collection of vectors and has to be treated that way in a vector field treatment.

Huh? What are you talking about? You can certainly say that the gravitational field (a vector) of a point mass decreases with distance. What's the problem with that?

You never did answer the question.

Gravity is a scaler field and can also be treated as a vector field.

True or false.

The gravitational potential is a scalar field; the gravitational field is a vector field. They are related, but not the same.

 

[v4theory]

The surface area of the conic section scales up by the square of distance while the strength per unit area scales down by the square of the distance.

The strength per unit area of what? What are you talking about?

The strength per unit area of the shell.

In the conic section the area of the shell in the section. Double the distance from the surface of the shell square the area in the conic section (4 times more area in the conic section), inverse square the strength per unit area of the conic section (quarter the strength per unit area). 4 times more units at a quarter per unit equals same total strength per unit distance. Balanced by a coresponding reduction of area and increase in strength per unit area in a coresponding conic section of the other side of the sphere. Net zero

In the ring, in vector terms. The vectors (potentials for gravitational acceleration) from a point equal in number for all directions from the point regardless of the distance from any other point.
In the ring it is an angle section of a line rather than a conic section of an area. You can't treat it as an area because it doesn't occupy an area. In the section of the ring covered by the angle section if you you double the distance you double the length of the section of the ring in the angle section (2 times more length in the angle section), Half the number of vectors (potentials for gravitational acceleration) pointing at that section per unit length of that section. (half the number of vectors per unit length) Two times the length of the section at half the number of vectors pointing to each unit of length equals same total number of vectors (potentials for gravitational acceleration) pointing at the longer section for a longer distance to it. Balanced by a corresponding decrease in length of the ring and increase in number of vectors per unit length in a coresponding angle of ring section on the other side of the ring. Net zero.

The sphere considered in vector terms. As distance from the surface of the shell increases the area in the conic section increases by the square of the distance and the number of vectors (potentials for gravitational acceleration) to the area per unit area reduces by the square of the distance. Four times the area one quarter the number of vectors per unit area. Four times the area at one quarter the number of vectors per unit area equals same number of vectors (potentials for gravitational acceleration) for total area. Therefore the gravitational potential remains the same.

I suspect you are considering the vectorization in terms of what happens outside a spherical mass.

Say a point has 4 vectors (4 potentials for gravitational acceleration) pointing at the whole area of massive sphere at one unit of distance from the sphere. If you double the distance to the mass the area the mass presents to the point is reduced by the square of the distance. It occupies a quarter of the conic section from the farther point than it did at the closer distance. It therefore now has only one vector pointing at it. The number of vectors pointing at it has scaled down by the square of the distance. The potential for gravitational acceleration of the vectors has not scaled down. The number of vectors (potentials for gravitational acceleration) pointing at the mass has scaled down.

It would be incorrect to say the potential for gravitational acceleration for the original 4 vectors has scaled down. You could say the net potential for gravitational acceleration for each of the original 4 vectors has scaled down by a quarter as the full magnitude of the one vector pointing at the mass is then being divided by the three others vectors not pointing at the mass that have no potential for gravitational acceleration.

You can certainly say that the gravitational field (a vector) of a point mass decreases with distance. What's the problem with that?

That's incorrect.

The gravitational field in vector terms is a field of points with vectors (potentials for gravitational acceleration) from the points pointing in all directions. It's conventional to treat for the vector that has positive gravitational acceleration potential on it toward mass (not neccesarily a point mass) because gravitation as a treatment treats for the effect of gravity on the space around mass. In vector terms gravity produces a potintial for acceleration on the vector of a point that is pointing to a mass. The other vectors from the point are still there. They just don't have a positive gravitational potential for acceleration on them. The gravitational field is not "(a vector)" it is a vector field.

The gravitational potential is a scalar field; the gravitational field is a vector field. They are related, but not the same.

No the gravitational potential not a scaler field. The gravitational potential treats for specific points in a gravitational field. The gravitational field can be treated as a scaler field or a vector field.

By the way, it seems the further we get into this the shorter (and less...accurate?) your responses are getting...again. I understand that brevity and some inaccuracy may be a common habit teachers can develop with their students. However it is also a common dodge by the uninformed on internet bulletin boards.

 

[Doc Al]

The strength per unit area of the shell.

In the conic section the area of the shell in the section. Double the distance from the surface of the shell square the area in the conic section (4 times more area in the conic section), inverse square the strength per unit area of the conic section (quarter the strength per unit area). 4 times more units at a quarter per unit equals same total strength per unit distance. Balanced by a compensating reduction of area and increase in strength of the other side of the sphere. Net zero

I suspect that this is your version of the argument I gave in post #15 (first given by Newton). Note that it only works for spherical shells.

In the ring, in vector terms. The vectors (potentials for gravitational acceleration) from a point equal in number for all directions from the point regardless of the distance from any other point.
In the angle section the length of the ring in the section. Double the distance double the length of the ring in the angle section (2 times more length in the angle section), Half the number of vectors (potentials for gravitational acceleration) pointing at that section per unit length of that section. (half the number of vectors per unit length) Two times the length of the section at half the number of vectors pointing to each unit of length equals same total number of vectors (potentials for gravitational acceleration) pointing at the longer section for a longer distance to it. Therefore the gravitational potential remains the same.

Please rewrite this using standard terminology. You say vectors, then immediately mention potential--which is a scalar. We are talking vectors here. Also, I don't know what you mean when you refer to the 'number of vectors'.

The gravitational field produced by a small segment of mass at some distance is a vector. It has magnitude and direction; the magnitude varies with the mass of the segment and the distance from the segment. Unlike with the spherical shell, the field from the 'cone' (really just an angle, since we are talking about a ring not a shell) of mass does not balance on each side, since the mass increases linearly with distance from a point (for a given angle) while the field per unit mass decreases with the square of the distance.

Go back to post #15 and reread my arguments carefully.

 

[Doc Al]

That's incorrect.

The gravitational field in vector terms is a field of points with vectors pointing in all directions. It's conventional to point a vector from some of the points toward mass (not neccesarily a point mass) because gravitation as a treatment treats for the effect of gravity on the space around mass. In vector terms gravity produces a potintial for acceleration on the vector of a point that is pointing to a mass. The gravitational field is not "(a vector)" it is a vector field.

Give me a break.

At any point in the space surrounding the mass you can represent the strength and direction of the gravitational field by a vector. (After all, a vector field just means that you've assigned a vector to every point over some space.)

 

[Doc Al]

No the gravitational potential not a scaler field. The gravitational potential treats for specific points in a gravitational field.

Apparently you have no clue as to what gravitational potential is. Either that, or you don't know what a scalar field is. (Or both.)

The gravitational field can be treated as a scaler field or a vector field.

You can certainly represent gravity by the gravitational potential (a scalar) or gravitational field strength (a vector). When talking about the gravitational field, most people mean the vector field representing the force per unit test mass at each point.

Pick one and get on with it. Don't mix them up.

By the way, seems the further we get into this the shorter (and less...accurate?) your responses are getting...again.

And you continue to serve up the same word salad.

(How many times will you edit your post?)

 

[v4theory]

How many times will you edit your post?

You're right I should take more time to compose before posting. So should you. Three posts since my last post?

I'm going to give you more time too to think over what I wrote before I post again.

 

[Doc Al]

Three posts since my last post?

You keep adding to it; my last posts were responses to those additions.

I'm going to give you more time too to think over what I wrote before I post again.

I'm done.

 

[v4theory]

Here's where I'm at.

I can't understand how I can treat the sphere and the cylinder, both inside and out with a vector treatment and get the exact same result as your scaler treatment. But when I apply the exact same procedure on the disk you disagree with the result.

I think I'm done too.

 

[v4theory]

On second thought I think I can show how your conic section treatment of the ring falls apart as I can produce two different results with it. One that fits your result and one that fits mine using your exact same method.

Would you agree that your treatment of the ring with conic sections that can produce two different results is the wrong treatment?

Wait for a day or two to compose.

 

[Doc Al]

I can't understand how I can treat the sphere and the cylinder, both inside and out with a vector treatment and get the exact same result as your scaler treatment. But when I apply the exact same procedure on the disk you disagree with the result.

Your 'vector treatment' is nothing more than vague handwaving.

Please look up the word 'scalar'. To use a 'scalar treatment' for this problem, you'd calculate the potential function and then take the gradient to find the field. Why don't you try it? (This has nothing to do with my simple, yet precise argument.)

On second thought I think I can show how your conic section treatment of the ring falls apart as I can produce two different results with it. One that fits your result and one that fits mine using your exact same method.

Would you agree that your treatment of the ring with conic sections that can produce two different results is the wrong treatment?

No, it just means that you do not understand the simple argument given.

 

[sankalpmittal]

Your 'vector treatment' is nothing more than vague handwaving.

Please look up the word 'scalar'. To use a 'scalar treatment' for this problem, you'd calculate the potential function and then take the gradient to find the field. Why don't you try it? (This has nothing to do with my simple, yet precise argument.)


No, it just means that you do not understand the simple argument given.

The universal law of gravitation works for every aspect . However in the case of ring the centre is in the middle where there is no mass . But still the vectors and scalars both will work .

But in this case its sheer calculation . Here you just can't take vectors of velocity .

 

[v4theory]

Your 'vector treatment' is nothing more than vague handwaving.

Please look up the word 'scalar'. To use a 'scalar treatment' for this problem, you'd calculate the potential function and then take the gradient to find the field. Why don't you try it? (This has nothing to do with my simple, yet precise argument.)


No, it just means that you do not understand the simple argument given.

That looks like an air tight case of a closed...loop.

Why should you want me to use a "scaler treatment" for this problem? You and I both know that is a complicated, lengthy, time consuming method. You wouldn't want me to use a treatment that I "don't understand" to produce a complicated "word salad" that produces a result that disagrees with yours would you?

Though I know it won't convince you as you know, even without seeing it, it must be wrong. I think, first, for now, I'll stick to composing my disasembly of your (handwaving?) conic section treatment. Maybe it'll convince sombody else.

Should you diegn to produce a "scaler treatment" for this problem I'll have a look and if it's right maybe we'll all learn something. If it's wrong I think you know I'm prepared to disasemble it. Though I'm sure you already know that if I did it would just be because I "didn't understand" it and produced another "handwaving" "word salad" that disagrees with your result.

 

[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&lt;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&lt;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 boundry 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 boundry of the ring and empty space.
The solution for treating the boundry 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 boundry 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 boundry 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 boundry 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?