The Death of the Centre of Mass Theorem.

[Rogue Physicist]

Many people believe that the Centre of Mass Theorem is a powerful and useful tool in Newtonian Mechanics. In fact it is a farce.

(1) It is trivially true at distances in which the massive object is virtually a point-mass, such as between distant stars.

(2) It is completely incoherent and self-contradictory at distances in which the size of the object(s) approaches 1/20 the distance between them.

Try it yourself. I'll post the simple proof that it is nonsense after a few people try to guess what is wrong.

 

[quantumdude]

Try it yourself. I'll post the simple proof that it is nonsense after a few people try to guess what is wrong.

Just get on with it already. This isn't sciforums, and we don't play those kinds of games here.

 

[panthera]

prove it then...i use this many times and its one of the easiest method to solve complex problems

 

[mezarashi]

Many people believe that the Centre of Mass Theorem is a powerful and useful tool in Newtonian Mechanics. In fact it is a farce.

(1) It is trivially true at distances in which the massive object is virtually a point-mass, such as between distant stars.

(2) It is completely incoherent and self-contradictory at distances in which the size of the object(s) approaches 1/20 the distance between them.

Try it yourself. I'll post the simple proof that it is nonsense after a few people try to guess what is wrong.

First off, I would like to point out that there is nothing wrong with approximations. Newtonian Mechanics is an "estimate" now with the Relativistic framework.

Secondly, I don't see how it would compromise the theorem which states that the center mass of a system cannot change in the absence of an external force.

 

[Rogue Physicist]

Consider a sphere. The center of mass is located at the Geometrical center. It is a fixed point. Let us place a test-particle a few diameters away from a solid sphere on the right along the x-axis. (make the radius of the sphere 1 unit, and put the sphere at the origin). According to the CMT, the mass acts as if it were concentrated at the centre and the distance will be measured centre to centre for purposes of using Newton's Gmm/d^2 formula.

(1) divide the sphere logically into two halves vertically. Each half will have its own centre of mass, located 3/8 radians down the axis of symmetry. The actual position isn't important, but because it is an average mean of the atoms in the solid, it is fixed relative to the geometric skin of the half-sphere. So the near half will be 3/8 rads toward the test-particle along the x-axis.

(2) Naturally, the CM for the other half will be the same distance away from the geometric centre of the sphere in the opposite direction away from the test-particle. The total force will be the vector addition of the two halves. But by inspection this is impossible. The increase in force for one half the mass now located closer cannot balance the decrease for the other half, because while the distances are equal, the forces have changed by an unequal amount. Gravity is an inverse exponential force.

(3) The actual force calculated by summing the halves separately will be larger than the one calculated treating the sphere as a whole. Conversely, the force calculated by dividing the sphere horizontally will be weaker than the 'whole' force, since vertical components from each half will cancel.

(4) The Centre of Mass Theorem contradicts itself, and also the Sphere Theorem as well, which is a special case of the CM.

 

[Rogue Physicist]

Secondly, I don't see how it would compromise the theorem which states that the center mass of a system cannot change in the absence of an external force.

This may be one side-effect resulting from the definition chosen for the Centre of Mass (i.e., the arithmetic mean average of position of particles weighted by the mass of each particle) but it is not really a formal part of the Theorem.

The main statement of the Theorem is that objects act under gravitational and intertial forces as if their mass was a point-mass located at the Centre of Mass so defined. Since the procedure for calculating the CM is not ambiguous but always produces a unique and fixed answer for any given configuration of particles (whose mass cannot change) in 3-space, Naturally a corollary of the definition or rather a direct effect of it is that the CM cannot change unless the mass distribution itself changes. The corollary to this is NOT true however. It is perfectly possible to rearrange the distribution of mass symmetrically *without* altering the position of the CM.

But this triviality is not what is being challenged here. The point is that the CM is not just 'an approximation'. To be reliable it has to have clear constraints on when it is inaccurate and an accompanying formula for the error. This of course is not in your physics textbook. When the proper constraints are applied to the CM Theorem (i.e., it's only really valid between masses whose distance dwarfs their size) it becomes a mere truism in any case, and is not needed.

For relatively close distances, such as at the surface of the earth, the Theorem and method for calculating the force using the CM is grossly inaccurate, and results in an actual miscalculation of the 'constant' portion of the Gravitational Constant (G). The current G in Newton's Gmm/d^2 formula for the force is functional as a coordinator of units, but is in fact not the real Gravitational Constant. That must be teased out of the apparent constant after the corrections have been made for the inaccuracy caused by using the CM to calculate the force.

 

[Meir Achuz]

Newton proved that large spheres with uniform mass distribuitions behave like point masses in gravity. Since you don't believe Newton's geometric proofs (They are tricky, but Newton was smarter than us.), you cause yourself, and maybe others, a lot of wasted time.

 

[panthera]

(2) Naturally, the CM for the other half will be the same distance away from the geometric centre of the sphere in the opposite direction away from the test-particle. The total force will be the vector addition of the two halves. But by inspection this is impossible. The increase in force for one half the mass now located closer cannot balance the decrease for the other half, because while the distances are equal, the forces have changed by an unequal amount. Gravity is an inverse exponential force.

this isn't clear! distances of two CMs from test particle are not equal...u first draw the diagram...if distances are equal then forces MUST also be equal considering it as a homogenous solid sphere.u are contradicting yourself...forces here
differ because of diff in dist only...its the only changeable quantity and rest are const...gravity is inverse square law NOT inverse exponential one...hey if u don't mind are u kidding...i think u r doing this purposefully as whole world knows that its a square law! there is no q of increase or decrease...the resultant force on the test particle will remain same even if the sphere is split into two halves with two diff CMs...

(3) The actual force calculated by summing the halves separately will be larger than the one calculated treating the sphere as a whole. Conversely, the force calculated by dividing the sphere horizontally will be weaker than the 'whole' force, since vertical components from each half will cancel.

the test particle is not within the sphere but outside...from where u are getting vertical and horizontal components of a three dimensional object? there are r,phi,theta as coordinates and not X and Y that u can treat it as a two dimensional object!

(4) The Centre of Mass Theorem contradicts itself, and also the Sphere Theorem as well, which is a special case of the CM.

I am now bound to think that either u are kidding OR u need to clear the concepts...

Regards

 

[Doc Al]

The main statement of the Theorem is that objects act under gravitational and intertial forces as if their mass was a point-mass located at the Centre of Mass so defined.

Nope. Your "center of mass theorem" only holds (in general) for objects in a uniform gravitational field. The net gravitational force on an arbitrarily-shaped object in a uniform gravitational field acts at the object's center of mass.

Newton showed that objects with spherical mass distributions attract each other as if their mass were concentrated at the center of the sphere. This doesn't apply to hemispheres.

 

[Rogue Physicist]

Your responses are kind of funny:

Newton proved that large spheres with uniform mass distribuitions behave like point masses in gravity.

No, this was just one of Newton's claims, in proposing his theory of gravity. Try to follow the discussion.

this isn't clear! distances of two CMs from test particle are not equal...u first draw the diagram...if distances are equal then forces MUST also be equal considering it as a homogenous solid sphere.u are contradicting yourself...forces here

Read again carefully. We've divided the sphere into a near half and far half, without moving either half. Each has its own Centre of Mass, which by symmetry will be along the axis passing through both. (in this case the x-axis itself.) Since the halves are identical, each Centre of Mass will be an EQUAL distance from the original Centre of Mass of the whole sphere. The distance to the test-particle is independant, and irrelevant for this part of the argument.

Now consider the distance from each CM to the test-particle. In this case, the change in distance away from the original CM of the whole sphere for one half of the mass is equal and opposite to the change in distance for the other half of the mass. But the change in forces upon the two CMs is not equal, since (if you like it said this way) it varies as 1/d^2, the inverse square of the distance.

Your hairsplitting over whether the law of gravity is 'exponential' in the general sense versus calling it 'inverse square' is moot. Since the denominator has an exponent, it is an exponential law, in contrast to a linear law with an exponent of one. That was my point in referring to it as exponential in the general sense. Only a linear law would vary in force equally with the variation in distance, and only a linear law of gravity could be consistent with the Centre of Mass theorem.

 

[Rogue Physicist]

Nope. Your "center of mass theorem" only holds (in general) for objects in a uniform gravitational field. The net gravitational force on an arbitrarily-shaped object in a uniform gravitational field acts at the object's center of mass.

What you are talking about is the Centre of Gravity concept. That is entirely different from the Centre of Mass theorem. The Sphere Theorem is different again. Three names, three theorems.

Newton showed that objects with spherical mass distributions attract each other as if their mass were concentrated at the center of the sphere. This doesn't apply to hemispheres.

The Centre of Mass Theorem (which is what we are discussing) applies to objects of any shape. What you are talking about here is the Sphere Theorem. Two names: Two different theorems.

 

[quantumdude]

The Centre of Mass Theorem (which is what we are discussing) applies to objects of any shape. What you are talking about here is the Sphere Theorem. Two names: Two different theorems.

I have never heard of anything called "The Center of Mass Theorem". Why don't you type it out mathematically? The only real clue I have as to what you mean is here...

According to the CMT, the mass acts as if it were concentrated at the centre and the distance will be measured centre to centre for purposes of using Newton's Gmm/d^2 formula.

That is not a consequence of any theorem related to the center of mass of a system of particles. At least not a theorem of which I am aware. I know that a the center of mass of any system of particles of total mass M follows the same trajectory that a single particle of mass M would follow. And I know that a uniform spherical mass is gravitationally equivalent to a point mass M, located at the center, for all points outside the mass.

The former theorem applies to all mass distributions, but says nothing of gravitation. The latter theorem speaks of gravity, but only applies to uniform spherical distributions. But you seem to be citing a theorem that is supposed to be about gravitation, and apply to nonspherical objects. As far as I know, that is not a theorem at all, but a false statement.

 

[Q_Goest]

Gravitational attraction between two point masses is commonly given as:

1) F=G*m1*m2/r^2

If we looked at any two symetrical points within a mass (m1), you might think their combined gravitational atraction to some other mass nearby (m2) would be as if the two symetrical points in mass m1 were both at the center of mass m1. Assume for a moment the distance from the center of mass m1 for these two points is x, then the force on mass m2 is:

2) F=G*.5m1*m2/(r+x)^2 + G*.5m1*m2/(r-x)^2

In other words, imagine a dumbell with two spherical masses connected by a massless bar who's length is 2x. (Note also the axis of the dumbell is aligned with the axis between m1 and m2.) Imagine the total mass of this dumbell is m1, so each of the two spherical masses is .5*m1 The center of mass of this object is at the geometrical center, half way between the two masses. One would then have to say that if the mass of this dumbell was m1, then the force it produces on another mass, m2, should be equal to a single mass of m1 at the center of mass of the dumbell. You could equally suggest each mass is just a differential mass, dm. In this case, each differential mass should counteract the equal amount of mass that it is symetrically opposite to in m1. So:

Let:
a = r-x
b = r+x
C = G*.5m1*m2

Then putting these into equation 2 above gives:

F = C/b^2 + C/a^2

F = (a^2*C) / (a^2*b^2) + (b^2*C) / (a^2*b^2)

F = (C*a^2 + C*b^2) / (a^2*b^2)

multiply top and bottom by 2 and replacing C:

F = 2*(G*.5*m1*m2)*(a^2+b^2)/(2*a^2*b^2)

F = G*m1*m2* ((a^2+b^2)/(2*a^2*b^2))

So 1/r^2 should equal ((a^2+b^2)/(2*a^2*b^2)) if the two point masses in the dumbell have the same gravitational attraction as a single mass at the center of the dumbell. But these two are not mathematically the same, so the two point masses do NOT have the same gravitational attraction to m2 as a single mass at the geometrical center. Note that the geometrical center is generally called the "center of mass". I think this is what RP is saying is the center of mass theorem as opposed to the center of gravity, but not sure.

It would seem there's a discrepency here that I'm sure someone can explain, but I believe you'll need to do a lot more math than I did. Obviously, the gravitational field around a dumbell is different than the gravitational field around a spherical object, but this doesn't answer the question RP raises.

RP, I think you need to do the math on this including the calculus to put your thoughts into better perspective. I believe you're right about the gravitational field inside a sphere, but I also have to believe this has all been done before and I'd bet someone here understands this enough to explain better than I.

 

[Doc Al]

In other words, imagine a dumbell with two spherical masses connected by a massless bar who's length is 2x. (Note also the axis of the dumbell is aligned with the axis between m1 and m2.) Imagine the total mass of this dumbell is m1, so each of the two spherical masses is .5*m1 The center of mass of this object is at the geometrical center, half way between the two masses. One would then have to say that if the mass of this dumbell was m1, then the force it produces on another mass, m2, should be equal to a single mass of m1 at the center of mass of the dumbell.

I believe this is what RP is stating as his "center of mass theorem". It's just not true.

So 1/r^2 should equal ((a^2+b^2)/(2*a^2*b^2)) if the two point masses in the dumbell have the same gravitational attraction as a single mass at the center of the dumbell. But these two are not mathematically the same, so the two point masses do NOT have the same gravitational attraction to m2 as a single mass at the geometrical center. Note that the geometrical center is generally called the "center of mass". I think this is what RP is saying is the center of mass theorem as opposed to the center of gravity, but not sure.

RP is confusing "center of gravity" with "center of mass". They are only coincident in a uniform gravitational field.

 

[Q_Goest]

Problem is, the "dumbell" I'm referring to could equally be two halves of a sphere. There's something wrong with equation 1. Is equation 1 only an aproximation? I seem to remember seeing someone mention it is, and that the exact equation is an infinite series. If so, that would help clear up the confusion.

 

[Doc Al]

Problem is, the "dumbell" I'm referring to could equally be two halves of a sphere. There's something wrong with equation 1. Is equation 1 only an aproximation? I seem to remember seeing someone mention it is, and that the exact equation is an infinite series. If so, that would help clear up the confusion.

I'm not getting your point. Equation 1 is just Newton's law of gravity for point masses. No problem there. To find the net force on any extended body, like the dumbell or sphere, you must add up the forces on each element. In general it is not true that the net force equals what you'd get by treating all the mass as concentrated at the center of mass. (It happens to be true for spherical mass distributions.)

Maybe you can restate what you think is the confusion.

PS: You cannot model a sphere as a two-piece dumbell and expect to get the correct answer! If you replace the halves of a sphere by two point masses--it's no longer a sphere.

 

[Q_Goest]

We treat any 3 dimensional geometry as a point mass when calculating the force of gravity at a large distance, so this should apply to a hemispherical shape as well as a spherical shape.

If we consider a spherical shape is the same as two hemispherical shapes, then at any distance from the sphere, regardless of whether we consider the sphere a single element or two hemispherical elements, the gravitational force calculated should be the same.

I created a spreadsheet that calculated the force and difference between these two given the math I did above. Regardless of how far one goes out, even if we consider the force of gravity of Mars on Earth for example, there is a very slight difference in the two calculations. The difference will be larger, the closer you get, but it always remains and is calculable.

I might suggest two possible reasons.
1) The equation given (equation 1) is not exact. I seem to remember someone mentioning this here, but haven't see it anywhere else.
2) One might be able to mathematically show that even at a significant distance such as from Earth to Mars, the gravitational field of a hemisphere is not equal to that of a point mass, so the slight deviation of both hemispheres from the assumption of being point masses rectifies the error that's calculated.

EDIT:

PS: You cannot model a sphere as a two-piece dumbell and expect to get the correct answer! If you replace the halves of a sphere by two point masses--it's no longer a sphere.

Yes, I think that's it exactly. That's what I mean by #2 above and I think that's where RP is thinking there's a paradox. The gravitational field of a hemisphere is not perfectly symetrical, so although equation 1 will be a close aproximation at a large distance, it won't be exact. There's some deviation in the n'th decimal place depending on how far you go out, and when calculating gravitational force at a distant point far from these two hemispheres, you still need to take into account that the hemisphere's gravitational field is not symetrical.

 

[Doc Al]

We treat any 3 dimensional geometry as a point mass when calculating the force of gravity at a large distance, so this should apply to a hemispherical shape as well as a spherical shape.

If the object is sufficiently far enough from the gravitating body to be considered a point mass, then its geometry is irrelevant. (Note that if you far enough away, the gravitational field can be treated as uniform.) In estimating the gravitational force that the moon exerts on an object I hold in my hand, the shape doesn't matter.

If, on the other hand, you are talking about the object's gravitational field, then, once again, if you are far enough away then the shape doesn't matter. For smaller distances, it does. The field of a two-piece dumbell is a dipole field; as distance goes to infinity, it becomes the field of a point mass. The sphere is a special case: Its field is the same as that of a point mass at any distance (beyond its surface). At close distances, a dumbell is a poor model for a sphere.

If we consider a spherical shape is the same as two hemispherical shapes, then at any distance from the sphere, regardless of whether we consider the sphere a single element or two hemispherical elements, the gravitational force calculated should be the same.

Right. Two hemispheres equals a sphere. But two hemispheres does not equal a dumbell.

I will look over your math when I get a chance. If your expression for the field from a dipole does not approach that of a point mass as distance goes to infinity, you did something wrong. (Either way, a dipole is not a sphere!)

 

[Q_Goest]

If your expression for the field from a dipole does not approach that of a point mass as distance goes to infinity, you did something wrong.

Yes, as the distance goes to infinity, the equation for a point mass equals the equation I came up with for a dipole. Similarly, as distance goes to infinity for a hemisphere, the gravitational field becomes uniform.

If you model two hemispheres as a dipole, (which seems on the surface to be a reasonable assumption) you find it is only an estimate of the actual field created by a hemisphere. The slight deviations then, if taken into account, should cancel out the error of the initial assumption.

I'm afraid Newton's law of gravity will have to remain standing...

 

[Doc Al]

Gravitational attraction between two point masses is commonly given as:

1) F=G*m1*m2/r^2

If we looked at any two symetrical points within a mass (m1), you might think their combined gravitational atraction to some other mass nearby (m2) would be as if the two symetrical points in mass m1 were both at the center of mass m1. Assume for a moment the distance from the center of mass m1 for these two points is x, then the force on mass m2 is:

2) F=G*.5m1*m2/(r+x)^2 + G*.5m1*m2/(r-x)^2

OK, let's expand this to look at the on-axis field at some distance from this dumbell:
F = \frac{0.5 G m_1 m_2}{r^2} [\frac{1}{(1 +x/r)^2} + \frac{1}{(1 -x/r)^2}]
Taking the first few terms of a binomial expansion (for r >> x):
F = \frac{0.5 G m_1 m_2}{r^2} [\frac{1}{(1 + 2x/r + (x/r)^2)} + \frac{1}{(1 - 2x/r + (x/r)^2)}]
Which becomes:
F = \frac{G m_1 m_2}{r^2} (1 + (x/r)^2)
Which approaches your equation 1 as r goes to infinity.

PS: Looks like you already figured this out while I was typing my post.

 

[Rogue Physicist]

For you younger folks, all three distinguishable theorems/concepts have a long and interesting history inside and outside physics textbooks.

(1) The Sphere Theorem. A special case of the Centre of Mass Theorem. It is currently still (by some physicists such as astronomers) believed to be an exact and accurate result based upon the integration of a hollow sphere.

Briefly, the theorem states that the gravity field inside a hollow sphere is zero or neutral everywhere, since all gravitational forces from the sphere are said to balance. It was originally supported by a faulty argument from Newton, but is currently defended by a reasonably straightforward integration. (It does *not* claim the sphere is capable of shielding a test-mass inside from other forces.)

(2) The Centre of Mass Theorem. Another practical method of simplifying the calculation of gravity between two massive objects. The centre of mass itself is carefully defined as the arithmetic mean average of all the point-mass positions weighted by their individual masses in three dimensions. This is just the definition and calculation of the location itself.

The theorem states that any 'system' of point-masses (massive object) can be treated as if it were a point-mass of equivalent mass located at this point. It was known to be an 'approximation' even when Newton originally proposed it. But it was assumed to be a good approximation, with only a minor error factor at closer ranges.

(3) The Centre of Mass Concept Itself. This has been a useful idea with many practical applications not just in gravity problems, but also in the application of forces, inertia, and also circular motion.

Two simple cases for rigid bodies: (a) To move an object efficiently without rotating it (wasting energy), you apply force in a straight line through the Centre of Mass. (b) If you rotate an object in the absence of other forces, it will naturally rotate on an axis passing through the Centre of Mass.

In both of these concepts, application to rigid bodies is preferred since plastic or flexible bodies complicate both the exchange of energy and can cause changes in the position of the Centre of Mass itself.

(4) The Centre of Gravity Concept. This is a statement about how bodies behave in a gravitational field (assumed to be homogenous in the local area of interest, such as near the Earth's surface). It uses the Centre of Mass concept also (the definition of this never changes).

Briefly, the Centre of Mass of bodies rotating and/or flying through the air or through space will indeed trace a parabola or move in a straight line, implying that the object's motion as a system is strictly obeying Newton's laws of force and gravity. This is the only theory which mentions or requires a uniform gravitational field over the area or volume of interest.

The reason many of your new (especially basic) textbooks don't clearly distinguish or define these separate theorems and ideas is that for a while now physicists have known that there are more problems than previously thought about all of them, and so textbooks have been quietly discussing them in more cautious terms and saying less about them. You can trace the interesting history of both the theorems and the textbook writers by comparing physics texts from the sixties, eighties, and now.

 

[Rogue Physicist]

For the 'dumbbell' case, the equations are quite simple and accurate (assuming point-masses for the dumbell ends, and a connecting rod of negligible mass):

For a vertical dumbell centred on the origin and a test particle on the x-axis:

F = Gm1m2 * cos^3(theta)
...d^2

Where theta is the angle between either dumbell end and the x-axis, and
d is the distance between the test-particle and the Centre of Mass of the dumbell (0,0).

The force decreases as the cube of the cos of the angle as the length of the dumbbell increases.
------------------------------------
The horizontal dumbbell centred on the x-axis is equivalent to treating the two halves of the sphere separately by cutting it vertically.

The three methods of calculating the force give three different answers, and the error is dependant upon the ratio of the radius of the barbell/sphere to the distance between the barbell/sphere and the test-particle. It is independant of size.

This shows clearly that the Centre of Mass Theorem is self-contradictory at close distances relative to radius.

 

[Rogue Physicist]

PS: You cannot model a sphere as a two-piece dumbell and expect to get the correct answer! If you replace the halves of a sphere by two point masses--it's no longer a sphere.

Yes, I think that's it exactly. That's what I mean by #2 above and I think that's where RP is thinking there's a paradox. The gravitational field of a hemisphere is not perfectly symetrical, so although equation 1 will be a close aproximation at a large distance, it won't be exact.

The point is, in this thread we are talking about the failure of the Centre of Mass theorem, not the failure of the Sphere Theorem, which fails for different reasons entirely.
I haven't 'imagined a paradox'. The Centre of Mass Theorem is inaccurate and self-contradictory if applied in different ways to the same problem at close ranges.
The Sphere Theorem is just as inaccurate because it can't properly handle discrete distributions of mass. (see my other thread.)

 

[Doc Al]

The imaginary "center of mass" theorem

(2) The Centre of Mass Theorem. Another practical method of simplifying the calculation of gravity between two massive objects. The centre of mass itself is carefully defined as the arithmetic mean average of all the point-mass positions weighted by their individual masses in three dimensions. This is just the definition and calculation of the location itself.

The theorem states that any 'system' of point-masses (massive object) can be treated as if it were a point-mass of equivalent mass located at this point. It was known to be an 'approximation' even when Newton originally proposed it. But it was assumed to be a good approximation, with only a minor error factor at closer ranges.

This "theorem" is not true in general, only in a uniform gravitational field. (Or for objects far enough apart.) It certainly doesn't hold at close distances! (Please tell us who claims otherwise?)

Since you claim that this is a "theorem", why don't you derive it for us and point out the erroneous assumptions that must underlie it?

 

[Rogue Physicist]

There's no point in getting into semantics. It's more than just a 'concept' and less than a 'theory'. It is best described by the usual grey-zone in the middle handle, a 'theorem'. A fairly short statement of some definitions, an axiom or two, some hidden assumptions, and an optionally syllogistic assertion. It has been described well enough here and elsewhere, as well as in physics textbooks for the last 50 years.

Again, I'll reiterate, but it is clear from the previous post. The Centre of Mass theorem has nothing to do with uniform gravitational fields. You are confusing it with the Centre of Gravity concept.

I don't have a copy of "Latex" or "Mathematica" handy, and I can't draw you any nice diagrams, because no one will tell me how to post a picture here.
Perhaps you can help. That way I could post some equations and pictures properly.

P.S. There is no point in belabouring the idea of a 'uniform field'. Since all primal fields are radially symmetric in three dimensions and defined by F = 1/r^2, and the resultant field between a sphere or other object is simply the superposition (vector addition) of the field from each object. This always results in a certain type of field even for the simplest case of two particles. There is no such thing as a 'uniform field' except as regards the component of the resultant field due to the other (far away) objects: the actual field is never close to flat in which the particle sits, from the point of view of any nearby test-particle.

According to Newton's view (modernized using vector calculus) the only 'field' any particle can experience is the resultant field *minus* the contribution to the field from itself. Thus each particle in the real objective field experiences its own unique field based upon the resultant minus its own component.

 

[delton]

I agree with Rouge Physicist, and have come across this problem before.
An interesting example to consider is an L shaped bar. Under the center of mass "theorum", an object would be attracted to a point outside the bar, but of course, this is not what really happens. To actually calculate the motion of something attracted to the bar, one must use calculus to sum the forces of the bar, correct??

Anyways, that's my humble say in all this.

 

[rachmaninoff]

An interesting example to consider is an L shaped bar. Under the center of mass "theorum", an object would be attracted to a point outside the bar, but of course, this is not what really happens.

It's not? How do you reason?

 

[Hurkyl]

I remember, back when I was in college, I made the same observation you did, Rogue Physicist. However, rather than jump to the conclusion that Newtonian gravity is self-contradictory, I decided that I should understand what's going on.

As is often true, you can't really understand stuff until you do the math. In all its splendor, the Newtonian law of gravitation gives the infinitessimal force on an infinitessimal volume of matter as the integral of an inverse square law over the mass distribution across the entire universe.


So why the heck do we use F = Gmm/R^2 if that's not what Newton's law of gravitation says? Because, when we can find an upper bound on the error introduced by using this approximation, so we can use it in situations where the error is tolerable. The domain of applicability is precisely when the sizes of the objects is negligable as compared to the distance between them. (Thus we say it's valid for "point masses") It should not be surprising that this formula loses its validity when the sizes are not negligable compared to the distance between the objects, and happily you observe exactly that!


What about the "Central Mass Theorem" you describe? Again, we can appeal to the math to find out exactly what it says:

Given a spherically symmetric body, we can evaluate the relevant integrals to prove:

Outside of the body, the gravitational field it generates is identical to that of a point mass located at its center with the same mass.


Notice, in particular, that it does not say that the gravitational force felt by a spherical mass is identical to the force felt by a point mass located at its center with the same mass.

 

[Rogue Physicist]

Once again I am mildly surprised by the responses:

(1) Hurkyl, you obviously believe that F=Gmm/d^2 is only an approximation of an integral formulation of gravitational potential fields. I am sorry to say, (if mass is treated as a continuum as it is in the integration), it is not an approximation but an exact result identical to the integral calculation.

You were right to be suspicious of Newton's argument, and the weakly supported claims regarding the integration and its physical application. You should have gone with your first instinct.

(2) Once again, someone has confused another theorem or concept with the one we are discussing. In your case, you have confused the Centre of Mass Theorem with the Sphere Theorem. (re-read post #21) This is probably not your fault, as almost every two years, the theorems have changed form and presentation in standard textbooks, until no two graduation classes have a clear understanding or even the same understanding of the history and value of these theorems.

Your only hope to clear your head of all this muddle is to sit down with a half-dozen texts from different half-decades and watch how the theorems have evolved.

 

[Hurkyl]

Or, I could work out the math myself. *shrug*

I am saying that, according to Newton's law of Gravitation, the force F experienced by a body with mass density f caused by gravitational attraction to a body with mass density g is given by:

<br /> \vec{F} = \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br />

It is fairly straightforward to deduce from this that if the bodies are localized in spatial volumes that are small compared to their separation, then we have:

<br /> \vec{F} \approx \frac{G m_1 m_2}{|\vec{r}|^3} \vec{r}<br />

I understand this fact because I've actually done the calculus needed to prove it at one time in my life. It's a basic application of the mean value theorem for integrals -- the spatial locality says that the term:

<br /> \frac{1}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x})<br />

is approximately constant where f and g are nonzero. More precisely, you can find upper and lower bounds for each of its components and apply the mean value theorem.

You were right to be suspicious of Newton's argument

Really? Quite frankly, I have no idea what Newton actually argued, and it's irrelevant to whether his end results are valid.

In your case, you have confused the Centre of Mass Theorem with the Sphere Theorem. (re-read post #21)

I may have confused what you meant by the terms, but I tried to avoid any confusion about what I meant by actually stating a theorem.

 

[Rogue Physicist]

Testing upload feature...
Okay. I have to admit the innuendo here is low key enough not to make a big fuss about it. We haven't strayed into insults or name-calling, for which I am grateful. Having prefaced my next remark with that, it still is a bit annoying to have people constantly imply that either I don't know how to integrate a simple summation equation, or that if only I would do this, miraculously all the theoretical problems with Newtonian gravitational theory would vanish, like an error term in an approximation.

For the purposes of establishing this small but annoying point once and for all, I have taken the time to convert several pages of one of my books on the flaws of Newtonian gravitational theory into an html file (the word file was too large). The graphics for the equations are a bit primitive (I did this 5 years ago) but I hope you will satisfy yourselves that I can both do the integration and I know what it means when I look at the result. Again I repeat that the integration results mathematically were never in question for the Sphere theorem. If someone can help me convert this into a .pdf document, I can attach it to this post.

 

[Hurkyl]

Arg, I edited out one of the main questions I had wanted to ask in my last post.


When I had said that Newton's law of gravitation was:

<br /> \vec{F} = \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br />

but doing it verbally rather than formulaically, you responded with:

(1) Hurkyl, you obviously believe that F=Gmm/d^2 is only an approximation of an integral formulation of gravitational potential fields. I am sorry to say, (if mass is treated as a continuum as it is in the integration), it is not an approximation but an exact result identical to the integral calculation.

I had wanted to ask if you were asserting that the above integral was not Newton's law of gravitation.

But now that I've reread the quoted section, it sounds like you are asserting the following equality:

<br /> \frac{G m_1 m_2}{|\vec{r}|^3} \vec{r} = \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br />

Where \vec{r} is the displacement vector between the center of mass of the two bodies in question, and m_1, m_2 are their masses. Is my interpretation of that quote correct?

 

[Doc Al]

There's no point in getting into semantics. It's more than just a 'concept' and less than a 'theory'. It is best described by the usual grey-zone in the middle handle, a 'theorem'. A fairly short statement of some definitions, an axiom or two, some hidden assumptions, and an optionally syllogistic assertion. It has been described well enough here and elsewhere, as well as in physics textbooks for the last 50 years.

And yet you seem unable to give a simple, clear statement of this "theorem".

The most I can figure from your posts is that you are saying that Newton's law of gravity, written as:
F = \frac{Gm_1m_2}{r^2}
is only approximately true applied to macroscopic objects. Well... of course! That law applies to point masses only. For extended bodies one must integrate.

If you are saying somehow that that law is meant to apply exactly to macroscopic objects where r is the distance between the center of masses... well, no, that's not generally true.

Again, I'll reiterate, but it is clear from the previous post. The Centre of Mass theorem has nothing to do with uniform gravitational fields. You are confusing it with the Centre of Gravity concept.

You are the one talking about "center of mass" and gravity together. Please explain the connection.

 

[Dr.Brain]

(It does *not* claim the sphere is capable of shielding a test-mass inside from other forces.)

It does claimly directly but its understandable.This phenomena is known as "Internal Shielding" , it says that a chrge enclosed inside a conductor is shielded from all other outside electrical influences.

(2) The Centre of Mass Theorem. Another practical method of simplifying the calculation of gravity between two massive objects. The centre of mass itself is carefully defined as the arithmetic mean average of all the point-mass positions weighted by their individual masses in three dimensions. This is just the definition and calculation of the location itself.

The theorem states that any 'system' of point-masses (massive object) can be treated as if it were a point-mass of equivalent mass located at this point. It was known to be an 'approximation' even when Newton originally proposed it. But it was assumed to be a good approximation, with only a minor error factor at closer ranges.

As per my information , this error is larger in non-symmetrical objects and negligible in round objects .These approximations don't violate anything , they were made to make problem-solving and motion-analysing a bit simple.

(3) The Centre of Mass Concept Itself. This has been a useful idea with many practical applications not just in gravity problems, but also in the application of forces, inertia, and also circular motion.

Two simple cases for rigid bodies: (a) To move an object efficiently without rotating it (wasting energy), you apply force in a straight line through the Centre of Mass. (b) If you rotate an object in the absence of other forces, it will naturally rotate on an axis passing through the Centre of Mass.

In both of these concepts, application to rigid bodies is preferred since plastic or flexible bodies complicate both the exchange of energy and can cause changes in the position of the Centre of Mass itself.

The assumption that if force is applied at a point randomly then the object will rotate around the centre of mass is not correct. Tests have shown that there are intermediate-points around which the whole rod/rigid body rotates in between the rotations.


QUOTE]Briefly, the Centre of Mass of bodies rotating and/or flying through the air or through space will indeed trace a parabola or move in a straight line, implying that the object's motion as a system is strictly obeying Newton's laws of force and gravity. This is the only theory which mentions or requires a uniform gravitational field over the area or volume of interest.

[/QUOTE]

I think you can refer to the example of the parabolic-projectile motion which follows the rule that c.m keep son moving the parabolic path even if there is a clast within the ball which is going along the parabolic path.Yes, its true that projectiles being fired near the Earth , are analysed assuming the force on it does not change for that period of time, which is an approximation and results in minor errors.

I should say you pretty much summed up the setbacks in centre of mass concept but as per me , I feel it was more for a problem solving and motion-analysing theory and it didnt ensure that it will give correct results.This does not mean end of the theory .It is a theory of approximations accurate enough to give us near about good results.

BJ.

 

[arun_mid]

Hi, I'm new to this forum, and just out of high school, and I didn't even know that the Center of Mass Theorem was an approximation (is this because gravitation is not included as one of the internal forces acting in a system during its derivation? this could have been countered by Newton's third law, but that was proved wrong-I don't know whether later derivations included the gravitational force or not), but coming back to RP's reasoning with the sphere, if his reasoning were correct, I could apply the same reasoning to a ring of uniform mass M and radius R; then the field at a point on its axis at a distance 'd' from its center would experience a field E=GM/(d^2) (because its CM is at its center) while the actual answer is E=GM/(d^2+R^2). I think this point has been implied by many time and again, in even plainer words. Can RP explain this before going on to the complex aspects of it?

 

[Stingray]

Rogue, please look up what a multipole expansion is. I agree that the force between two objects in Newtonian gravity (or electrostatics) is not given by the inverse square law between their center's of mass. This is hardly a great conspiracy, though. It is just the monopole approximation to the infinite series that actually describes the forces between two extended bodies. It is an approximation valid only when the distance between the center's of mass is much greater than the radii of either object. Here's a quick lesson:

Hurkyl gave the exact expression for the force, but I'm going to put it into a more convenient form. The potential due to body A is equal to (using slightly different notation)
<br /> V_{A}(x) = G \int \frac{\rho_{A}(x&#039;)}{|x - x&#039;|} d^{3} x&#039;<br />

Then the force on body B is
<br /> F_{AB} = - \int d^{3} x \rho_{B}(x) \nabla V_{A}(x)<br />

Looking at the potential in some more detail, we can use the identity
<br /> \frac{1}{|x-x&#039;|} = 4 \pi \sum_{l,m} \frac{1}{2 l +1} \left( \frac{ |x|_{&lt;}^{l} }{ |x|_{&gt;}^{l+1} } \right) Y_{lm}^{*} ( \theta, \phi ) Y_{lm} (\theta&#039;,\phi&#039;),<br />
where the Y_{lm}'s are spherical harmonics, and |x|_{&lt;} is the smaller of |x| and |x&#039;|.

For simplicity, center the coordinate system on body A's center of mass. Also assume that every point in body B is at a greater radius than any point in body A. Then
<br /> V_{A}(x) = 4\pi G \sum_{l,m} \frac{ M_{lm} }{2 l +1} \frac{ Y^{*}_{lm} (\theta, \phi) }{ |x|^{l+1} }<br />
where M_{lm} are the multipole moments of body A:
<br /> M_{lm} := \int d^{3} x&#039; \rho_{A}(x&#039;) |x&#039;|^{l} Y_{lm} (\theta&#039;, \phi&#039;)<br />
In spherical symmetry, the only moment that survives is the monopole, and we find that the potential is given by the standard point mass formula. In general, though, this does not occur. I'll leave it as an exercise to figure out what the full force equation is, why the point mass (monopole) approximation is usually very good.

By the way, a theorem is not something of "intermediate" certainty. It is something which is proven with complete rigor from a set of axioms (whether the axioms are useful ones or not is irrelevant). If something is a theorem, it has a precise proof.

 

[arun_mid]

whoops, sorry, the field due to the ring at the point on the axis is
E=GMd/(R^2+d^2)^3/2; it is however not E=GM/d^2, but in the sphere problem(suppose it has mass M), RP has split the sphere into two hemispheres and taken the fields due to the hemispheres at an axial point to be equal to that due to mass M/2 present at the CM of the hemispheres. This is an incorrect approach. Nowhere does the CM theorem justify this. It cannot be applied like this.

 

[Rogue Physicist]

First I'd like to thank everyone for their excellent replies. This has been an interesting thread for me, and hopefully to others too. Now I would like to clarify my position as is my obligation in response to your efforts and questions:

And yet you seem unable to give a simple, clear statement of this "theorem".

I probably deserve this, on the following analogy: Just as one cannot claim not to have stepped on someone's foot simply because you didn't know you did (or might have), but you have to take into consideration the experience of the person who feels the pain of a squashed toe, I have not been as clear as I could have. We can judge this partly by the results.

But now that I've reread the quoted section, it sounds like you are asserting the following equality:
<br /> \frac{G m_1 m_2}{|\vec{r}|^3} \vec{r} = \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br />

I am going to try to reply to this more clearly:
What I was doing was applying the premise of the approximation known as the Centre of Mass "method" (we will avoid calling it a 'theorem' so as not to disturb mathematicians who like precision of descriptive terms). I applied it in a way so as to show that it is not just a simple approximation with a predictable 'error' term, but one that produces arbitrary and self-contradictory results when applied in different ways. The purpose was educational, not an attempt to use the Centre of Mass 'method' to realistically calculate the forces. If I had the choice with the solid sphere example, I'd prefer the Sphere Theorem because it will naturally be more accurate and consistent at least with a large mass that itself approximates a continuum distribution of mass. The Sphere Theorem fails for different reasons when applied to discrete distributions of mass in cases like the ones discussed in my other thread.

Dr. Brain was correct that in the Electrostatic case, it is possible for charges to 'shield' one another or counter effects. This is not the case for gravitational forces (at least according to Newtonian Gravitational Theory as normally proposed).

The assumption that if force is applied at a point randomly then the object will rotate around the centre of mass is not correct. Tests have shown that there are intermediate-points around which the whole rod/rigid body rotates in between the rotations.

This is quite interesting: Can you articulate this in more detail? I am not sure exactly what effect is being discussed here.

I could apply the same reasoning to a ring of uniform mass M and radius R; ...Can RP explain this before going on to the complex aspects of it?

In fact, with a ring the force experienced by a particle moving along the axis of symmetry (make that the x-axis) is actually interesting: there is an angle to the ring edge at which the force is maximum, then it fades to zero as the particle moves even closer to the Geometric Centre, then it gradually increases again (still attractive) as the particle continues its path, (the force pulling in the opposite direction) and finally fading out like a 1/d^2 law. I will post the equation and the graph shortly for you.
Again the Centre of Mass 'approximation' will be in error. This time it will be a quantifiable error, due to symmetry, even if the mass is arbitrarily divided up and summed.

Stingray's response is excellent, however I have one fault with it, and that is, it is impenetrable to people like arun_mid who have no advanced engineering methods at their fingertips. Perhaps you could break it down into more steps with some real discussion for ordinary humans: otherwise it will appear to be more like 'showing off' than illuminating a physics problem. You've got DEL operators, spherical harmonics, unidentified angles (or more exotic objects) and no diagrams. It seems a bit unfair to just say 'go find a book on dipole analysis techniques'.

 

[Hurkyl]

The error term is predictable. It's exactly:

<br /> \int \int \frac{G f(\vec{x}) g(\vec{y})}{|\vec{y} - \vec{x}|^3} (\vec{y} - \vec{x}) d\vec{x}^3 d\vec{y}^3<br /> -<br /> \frac{G m_1 m_2}{|\vec{r}|^3} \vec{r}<br />




I'm not sure what you think is so revolutionary about discovering that two different approximations of the same thing aren't equal -- one should only expect them to be approximately equal. (The size of the error there is at most the sum of the sizes of the errors of the two individual approximations)


BTW, I've caught some mistakes in post #5 that I can't believe I missed:

Each half will have its own centre of mass, located 3/8 radians down the axis of symmetry.

"radian" is an angle measure, yet you're using it to talk about a length.

Gravity is an inverse exponential force.

No, gravity is an inverse square force. Inverse exponential looks like: f(x) = K/e^(ax)

 

[Stingray]

What I was doing was applying the premise of the approximation known as the Centre of Mass "method" (we will avoid calling it a 'theorem' so as not to disturb mathematicians who like precision of descriptive terms). I applied it in a way so as to show that it is not just a simple approximation with a predictable 'error' term, but one that produces arbitrary and self-contradictory results when applied in different ways. The purpose was educational, not an attempt to use the Centre of Mass 'method' to realistically calculate the forces.

If your intention in this thread was purely educational, then I think you need to work on how you word things a bit. It sounded more like you were trying to claim that Newtonian gravity contradicted itself. Anyway, I agree that your main point is often a source of confusion for beginners, and it is important to correct them.

This might just be another instance of imprecise wording, but I disagree with your claim that the "center of mass method" does not have a predictable error term. This is not correct. You can estimate errors from the expansion I gave before. You'll also find that those errors reach 100% (or more) if you try to look at center of mass separations comparable to the radii in the problem. These are exactly the cases where you're pointing out that the method fails.

Stingray's response is excellent, however I have one fault with it, and that is, it is impenetrable to people like arun_mid who have no advanced engineering methods at their fingertips. Perhaps you could break it down into more steps with some real discussion for ordinary humans: otherwise it will appear to be more like 'showing off' than illuminating a physics problem. You've got DEL operators, spherical harmonics, unidentified angles (or more exotic objects) and no diagrams. It seems a bit unfair to just say 'go find a book on dipole analysis techniques'.

I apologize for being terse, but I cannot possibly explain all of that in an internet posting. I think some people followed it. The rest can (and should) go learn about these things at some point. For more reference, these types of multipole moments are commonly explained in E&M books. Any applied mathematics book discussing the Poisson equation in detail will do the same.

The basic conclusion is explainable, though. The gravitational potential generated by anybody can be expanded in an infinite series. The first term is the same as the point-mass potential. Each higher term falls off faster and faster with radius (they're also not spherically symmetric). The point mass potential is therefore a good approximation only when very far away from the source.

More specifically, let the source's approximate radius be equal to r (it's not necessarily a sphere, so 'radius' is being used roughly), and say that we want the potential a distance x away from the source's center of mass. Then the nth term in the potential is going to be at most a factor of (r/x)^n times the point-mass potential. The "center of mass method" is therefore only reasonable when r/x is extremely small.

All I've done is given a way of writing down the full integral expression of the force (as given by Hurkyl) in a more convenient way for talking about this problem.

 

[arun_mid]

ok, this is what I meant:
for your argument about the sphere, you divided the sphere(kept at the origin), suppose of mass m and radius R, into two equal halves, and considered the CMs of the two hemispheres. You assumed that the sphere could be considered as the system of these two point masses at equal distances 3R/8 from the center of the sphere, at opposite ends. Now you kept a point test mass m at a distance 'd' from the center of the sphere. Suppose the frame is such that the sphere was cut into two halves by the plane x=0, and that this test mass is on the line y=0,z=0. Then you calculated the grav. force as follows:
F = (GMm/(d-3R/8)^2)+GMm/(d+3R/8)^2)
If this is what you meant, then I could use the same approach to calculate the field due to a ring of uniform mass M and radius R at a point on its axis at distance 'd' from its center. The field thus calculated would be
E = GM/d^2
as the CM of the ring is at its center. This formula is not correct except at very large distances. So this approach is wrong. If you mean that this contradicts CM theorem, I disagree as CM theorem was never designed for gravitational force. it was assumed that each differential mass element of the body did not interact; that mass was just that-mass, with no gravitational field --a valid assumption, given that the gravitational force is negligibly small between such particles, and this is true, even the gravitational force between two atoms can't bind them in a molecule.

 

[Rogue Physicist]

First to respond to Hurkl:

No integral or algebraic expression properly explains the basic fact that in my examples, it is critically important to know *how* one has bisected the sphere, that is, what the orientation of the cut was, in order to predict both the direction of the error as a geometrical vector term in 3-space, and its magnitude. Your equation is fine, but completely un-descriptive for humans trying to understand the effect of orientation.

Something cannot be classified as a 'mistake' if it is done with purpose, successfully meeting the desired goal.

I chose radians deliberately to both normalize the equations and descriptions, and legitimately talk of radians according to their original definition and essential meaning in Euclidean 2 or 3-space: The distance along the perimeter of a unit circle in the axis units of distance, the distance of a radius unit. Radians are abstracted into angles as a consequence, but were fundamentally defined as distances and remain true in Euclidean space (Newtonian Space). To me this reminds the reader of our 'roots', and it remains perfectly legitimate to speak of and use radians as distances in Euclidean space, handily tying the units of measure to the geometrical scale of relevance. Plainly the effects discussed directly depend upon the 'radius' of the sphere and its ratio to the test-mass distance, which should naturally be in this unit.



'exponential' is a natural adjective formed from 'exponent', and quite a general term, used both historically, and still widely, to mean any and all expressions and polynomials that take a non-trivial exponent. Of course it has also taken on a specific reference to that special number e, always itself expressed with an exponent term. In order to be more precise, not less, I always use the capital 'Exponential', often as a noun or substantive phrase when speaking of e. This is less ambiguous and retains the ordinary adjective 'exponential' for its natural meaning according to English usage as 'of or having the qualities of an exponent'. Again, hardly a 'mistake' but rather a useful educational procedure for human beings whose first language is English.

 

[Rogue Physicist]

To respond to Arun Mid:

...I disagree as CM theorem was never designed for gravitational force. it was assumed that each differential mass element of the body did not interact; that mass was just that-mass, with no gravitational field --a valid assumption, given that the gravitational force is negligibly small between such particles, and this is true, even the gravitational force between two atoms can't bind them in a molecule.

In the first part here you appear to think that the CM theorem was never associated with the calculation of gravitational forces. Of course this is teleologically unsound, as even when used to handle problems involving translational energy of projectiles and balancing forces, the forces are of course gravitational. Perhaps you have only come across the concept in a discussion of inertia or angular momentum. I can assure you that this is not the whole story, nor does it reflect the history of ideas in Newtonian Mechanics and gravity.

Next you refer to what appears to be a discussion of independant (free-floating) systems of particles, again a common Newtonian application of both the CM concept and the Newton's 3rd Law of action/reaction, in which typically only the kinetic energy of collisions is usually discussed, and the (external) gravitational field is assumed uniform. In your (1st year?) textbook these subjects are usually limited to a half-page or so, avoiding all the problems we discuss in this thread.

Another way of looking at things might be to take you as having the General Relativity idea of a *real* gravitational field that can carry mass and energy in the back of your mind as non-applicable to Newtonian gravity, and you would be right there also (two different theories).

But I think the basic problem for you is that your working textbooks are not adequately covering the concepts in the depth necessary to really understand them, and this is not your fault, but actually one of my complaints about University Introductory Physics texts.

 

[James Jackson]

What are you talking about? The radian is definied as the ANGLE subtended at the centre of a circle by an arc of equal length to the radius. Are you getting confused as it is a dimensionless quantity?

 

[Rogue Physicist]

...an arc of equal length to the radius. Are you getting confused as it is a dimensionless quantity?

I don't know why you are having trouble with the fact that the radian is defined as a length in physical space describing a distance along an arc or circle's perimeter. You yourself were forced to use the word length to describe it.

Yes it is dimensionless, and that is why it is the ideal measure to describe a dimensionless effect, the failure of the CM 'method', which fails in circumstances having to do with the ratio or proportion of the distance to the radius. Describing the CM 'method' in terms of units of the radius is the best and only way to generalize and concisely convey the effect discussed.

 

[James Jackson]

No, the radian is the ratio of two lengths and henceforth itself is not a length, it is a dimensionless quantity. Dimensions used to 'create' a property and performing dimensional analysis on that quantity are separate entities.

 

[Hurkyl]

Describing the CM 'method' in terms of units of the radius is the best and only way to generalize and concisely convey the effect discussed.

And you could say that -- you could say something like "3/8ths of the way along the radius", or "at a distance of 3r/8 from the center (where r is the radius)". I've even seen "radius" used directly as a unit of measure, as in "3/8 radii from the center". But using radians as a unit of measure is patently incorrect.

Of course it has also taken on a specific reference to that special number e

No, it has not. It is merely customary to write an exponential function in terms of e, but any base can be used. Specifically, for any nonzero a and b, any function of the form f(t) = a^rt can be written in the form f(t) = b^st.

(edit: I think, now, you were referring to the fact exp(x) is called the exponential function, as opposed to talking about exponential functions)

used both historically, and still widely

The graphite in your pencil is still widely referred to as "lead", but it's still wrong to say that you're writing with the element Pb. Whether or not your claim that the use of "exponential" is widely used to describe polynomial growth is correct, that usage is still incorrect.

 

[arun_mid]

definitely you can apply the 'fact' that the body acts as if its mass were concentrated at it's CM for a uniform gravitational field, so I did apply CM to gravity there without much problem, although with large objects you would consider CG instead because the grav field is not uniform anymore.
But this loose definition of CM does NOT hold everywhere. However the fact that mass times the acceleration of the CM equals the net external force is UNIVERSAL. So if you had a free uniform hemispherical mass you could calculate the net grav force on it by a test mass on its axis you can say that the mass times the acceleration of CM equals the net grav force. but you CANNOT say that the entire mass is concentrated at that point, so apply the grav. force law for particles at the CM.

 

[arun_mid]

This definition of CM theorem is true universally:
mass*acceleration of CM = Net external force on system

Try to prove THAT wrong. If you are simply saying that the previous loose definition was wrong, then you are right, and if this is the only definition in textbooks, that is confusing--yet I'm sure that's not the case, I didn't study it that way.

 

[Rogue Physicist]

I guess all good threads must eventually deteriorate into semantic quibbling, so this should be no exception. I'd like to thank everyone for participating, and say how much I appreciate the theoretical contributions to the discussion.

Whether or not your claim that the use of "exponential" is widely used to describe polynomial growth is correct, that usage is still incorrect.

I think we can all chuckle a little bit and lighten up after this line of reasoning.

Since word definitions are based upon usage (at least in real languages like English), if there is any argument to be had for one definition over another it would be based on the convention of statistical usage. The purpose of language is communication, and its success is determined by convention. Six million Welshmen can be stupid, but they can't be 'incorrect' about their own usage of a word.

Correct me if I'm incorrect, but I believe I am probably just stupid. And I fear I'll go on referring to 'pencil lead' to the end of my days, and have to suffer being understood perfectly by my peers, and by 'graphite' men too.

I guess we're both right: its a candymint and a breathmint. Longer *and* milder. Tastes good, really works. You say tomatoe I say tomatoe with a different accent.

Henceforth radian is not a length!
It shall be defined as the portion of my breast exposed during the intermission of a football game.