# The Naked Scientists Forum

### Author Topic: Newton made an error and Einstein copied it  (Read 24734 times)

#### gem

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##### Newton made an error and Einstein copied it
« Reply #25 on: 26/04/2010 23:22:34 »

OK Geezer in diagram 1 we have a overhead view of two one newton forces acting on a one KG weight on a frictionless surface at 90 degrees to each other giving a net resulting acceleration of 1.4 metres per second squared so 70 percent of the gross input potential.

And in diagram 2 we have the same 1 newton forces acting on the 1 KG weight at a angle of 45 degrees  to each other giving a net resulting acceleration of 1.8 metres per second squared so 90 percent of the gross input potential.

And in diagram 3 we have a example of the angle of interaction getting more acute and any mass that is not in line with the centre line C/L Note, As the distance between the mass of the two bodies changes only the particles that are in a direct line with the centre to centre actually travel the distance prescribed by inverse square law relative to each others centres.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #26 on: 27/04/2010 01:03:02 »
Ah! Nice pix. Thanks.

From your previous description I thought only one of the forces acted along a frictionless surface. I see what you were getting at now.

I'll get back to you.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #27 on: 27/04/2010 04:47:48 »
And in diagram 3 we have a example of the angle of interaction getting more acute and any mass that is not in line with the centre line C/L Note, As the distance between the mass of the two bodies changes only the particles that are in a direct line with the centre to centre actually travel the distance prescribed by inverse square law relative to each others centres.

This is all gravitational force?  The forces between points of mass all still follow a r-2 law, which I think we're agreeing on.  If all your masses are tiny, then the r you use is the distance between the point masses.  If the masses are large, then you have to sum over all the tiny elements making each of them up, making sure to add the vectors appropriately as you indicated.

Do you disagree with any of this?

I think what people (and Newton) were saying is that the addition of all these vectors greatly simplifies if one mass is really tiny and the other is a uniform sphere, since all this vector addition reduces to treating all the large-sphere mass as if its concentrated at the center of the sphere.  The vector addition isn't wrong, but you can do the vector addition in a much simpler way.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #28 on: 27/04/2010 05:13:51 »
I think what Gem is getting at is that the angles between all the individual gravitational forces that sum to the total force vector vary with the distance between the two objects. At a great distance, the angles would hardly matter at all. As the distance diminishes, the angles have a greater effect.

I'm pretty sure Newton's model takes that into account, but I'm not that familiar with how he actually did it.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #29 on: 27/04/2010 05:23:09 »
Newton's model tells you to sum* gravitational forces as vectors.  It's only in very particular cases that the symmetry of things lets you eliminate the vector sums.  (You remove them from the sums by making symmetry arguments to account for the vector nature of the forces rather than actually having to sum over the vector components.)

Although I haven't read Newton's original work, so this is all based on how they teach Newtonian gravity these days.  I think Newton did something similar.

*In most problems you're dealing with continuous objects, so instead of summing, you integrate.
« Last Edit: 27/04/2010 05:25:14 by JP »

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #30 on: 27/04/2010 05:43:05 »
*In most problems you're dealing with continuous objects, so instead of summing, you integrate.

Well, yes. But isn't integration really just a way of adding all the little bits together without having to do the actual adding?

#### JP

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##### Newton made an error and Einstein copied it
« Reply #31 on: 27/04/2010 05:45:31 »
*In most problems you're dealing with continuous objects, so instead of summing, you integrate.

Well, yes. But isn't integration really just a way of adding all the little bits together without having to do the actual adding?

Sssshhhh!  If you give away the secret that we're actually just adding, calculus won't seem so fancy anymore!

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #32 on: 27/04/2010 06:13:16 »
Wait a minute! The mists are finally beginning to clear.

If you want to calculate the gravitational force at the surface of the Earth, you have to integrate all the "individual" forces acting on the body using the inverse square law, taking into account the force vectors. If you do that, you'll get a number that is the same as the observed gravitational force. If you assume the forces all act towards the center of the Earth, you'll calculate a value that is substantially greater than the observed value.

Newton never said you can neglect the angles did he? Only when the two objects are sufficiently distant can you use an approximation and assume that all the forces act between the centers of mass of the objects.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #33 on: 27/04/2010 09:10:23 »
That's what I was getting at.  Summing (or integrating) all the individual masses using the r-2 law along with the proper vectors will always get you the right answer.  Newton didn't say you could neglect the angles/vectors.  You can simplify the vector sums by utilizing symmetry and/or vector calculus tricks.

To determine forces between two extended objects, you have to sum over the forces between all pairs of points on both objects, making it a very complicated equation.  It's much easier if we treat one object as a point (which is a good approximation if it's much smaller than the other object).  Then we only need to vector sum over the forces generated by all the points of the large object on the small object.

There are a variety of cases in which this simplifies.  If the objects are very far apart, then all the vectors point more or less in the same direction and you can just assume they all point towards the center of the large object without much error.  If the large object is a uniform sphere, then you know from symmetry that the total force on the small object has to point directly to its center, and you can take care of the vector sum using some mathematical tricks.  The result in that case is that mathematically you can treat all the mass of the large object as being concentrated at its center and use a 1/r2 law from there.  The vector addition still holds, but because of the symmetry you can do the addition ahead of time and get a simple looking answer.  In the uniform sphere case, your objects don't have to be far apart to make the simplification.  It's the symmetry of the problem that allows you to do it.

You should get similar simplifications in other symmetric cases, but since planets/stars tend to form into roughly spherical objects, that's the case that often gets cited.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #34 on: 27/04/2010 09:24:27 »
I can even tell you basically how to do it.  Think about a ring of mass and an object lying along the ring's axis, like I show in the figure.  You know that two points on opposite sides of the ring (shown in red) each pull towards each other with an inverse square law.  When you add the vectors, because of symmetry, the resulting vector points directly towards the center of the ring.  Now you can add up all points around the ring.  Since each point has a corresponding point on the opposite side, the entire force is directed exactly towards the ring's center.  The magnitude of it is pretty easy to calculate as well.

Next, you can consider a disk of mass by shading in the ring with mass.  Since a disk can be sliced into a bunch of rings, and each ring contributes a force pointing along its axis, the entire disk contributes to a force pointing along its axis.

Finally, you consider the whole earth to be made up of a bunch of disks of different sizes, stacked along the axis.

Anyway, if you didn't follow the whole argument, I hope at least the bit about the ring made sense to you.  You can do the entire ring calculation and get a force without having to actually do vector addition of each point on the ring.  You need to know how two points on opposite sides of the ring sum to give you a force along the ring's axis, and then you simply have to sum around the ring, knowing that your final force has to point along the axis, from symmetry.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #35 on: 27/04/2010 21:39:00 »
If you want to calculate the gravitational force at the surface of the Earth, you have to integrate all the "individual" forces acting on the body using the inverse square law, taking into account the force vectors. If you do that, you'll get a number that is the same as the observed gravitational force. If you assume the forces all act towards the center of the Earth, you'll calculate a value that is substantially greater than the observed value.

I take it back! According to http://en.wikipedia.org/wiki/Earth%27s_gravity

"Note that this formula only works because of the mathematical fact that the gravity of a uniform spherical body, as measured on or above its surface, is the same as if all its mass were concentrated at a point at its centre."

I suspect that it's because of this

If the object is sitting at the "North Pole", it will be subjected to equal attractive forces from either side of the "equator". The B force is attenuated by distance whereas the A force is attenuated by angle. I think that the two forces acting along the vertical axis will be equal.

I have left the proof as an exercise for the reader (because the algebra defeated me!)

#### gem

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##### Newton made an error and Einstein copied it
« Reply #36 on: 27/04/2010 21:58:16 »

This is all gravitational force?  The forces between points of mass all still follow a r-2 law, which I think we're agreeing on.

Yes every particle follows inverse square law

Newton never said you can neglect the angles did he? Only when the two objects are sufficiently distant can you use an approximation and assume that all the forces act between the centers of mass of the objects.
There is no allowance for the change in angle of interaction from earths surface to a hundred earths radius's away in the way the strength of earths gravity field is calculated at present.

For the inverse square law to work at different distances away, the force arrows in black have to stay at the same ratio to the vector sum 'arrow in red' for every pair of particles at what ever distance away.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #37 on: 28/04/2010 01:00:54 »
Well, I figured out why I could not produce a general proof that the two forces in my diagram are equal. It's because they are not! It's more complicated than said.

However, I still think the statement in Wiki is correct, but I'd like to see the proof.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #38 on: 28/04/2010 01:13:36 »
http://en.wikipedia.org/wiki/Shell_theorem

Derivation is there.  They eliminate explicit vector sums by using the symmetry of the sphere to do them.  Rather than each force being
,
they use
,
where the sine accounts for the vector nature of the force and uses the fact that that is the only part of the force (pointing towards the earth's center) that isn't canceled by another force.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #39 on: 28/04/2010 20:17:55 »
Brilliant! That's a really elegant proof. (Why didn't I think of that?  )

#### gem

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##### Newton made an error and Einstein copied it
« Reply #40 on: 28/04/2010 21:37:54 »
I can even tell you basically how to do it.  Think about a ring of mass and an object lying along the ring's axis, like I show in the figure.  You know that two points on opposite sides of the ring (shown in red) each pull towards each other with an inverse square law.  When you add the vectors, because of symmetry, the resulting vector points directly towards the center of the ring.  Now you can add up all points around the ring.  Since each point has a corresponding point on the opposite side, the entire force is directed exactly towards the ring's center.  The magnitude of it is pretty easy to calculate as well.

'pretty Easy'

For the inverse square law to work at different distances away, the force arrows in black have to stay at the same ratio to the vector sum 'arrow in red' for every pair of particles at what ever distance away.

The above statement is correct otherwise you would have a variation in the value of where the sine accounts for the vector nature of the force and uses the fact that that is the only part of the force (pointing towards the earth's center) that isn't canceled by another force.

But i will make it even easier could anyone point out any pair of point masses on any of the rings that make up a sphere where  the force arrows in black stay at the same ratio to the vector sum 'arrow in red' at what ever distance away.
[other than on the rings axis]

I think that you will find the vector sum arrow in red will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all your calculations.

Making a mockery of applying inverse square law to the value of attraction at the surface of the sphere.

And also can anyone highlight any pairs of point masses on any of the rings that make up a sphere that travel the same distance from the object mass as the centre of the ring. [other than on the rings axis]

Also Making a mockery of the symmetry of the sphere argument as well as applying inverse square law to the value of attraction at the surface of the sphere.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #41 on: 28/04/2010 22:17:01 »

But i will make it even easier could anyone point out any pair of point masses on any of the rings that make up a sphere where  the force arrows in black stay at the same ratio to the vector sum 'arrow in red' at what ever distance away.
[other than on the rings axis]

Of course they don't. The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance. The "ratio" that you speak of will only be the same for points that have the same angle.

But how does that disprove the shell theorem?

#### gem

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##### Newton made an error and Einstein copied it
« Reply #42 on: 29/04/2010 20:40:56 »

The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.

Absolutely hold that thought

But how does that disprove the shell theorem?

Because of this statement which is physically impossible
For the inverse square law to work at different distances away, the force arrows in black have to stay at the same ratio to the vector sum 'arrow in red' for every pair of particles at what ever distance away.

This is because when newton applied inverse square law to the strength of earths gravitational attraction at earths surface he fixed that value as the one to be used at all distances from the surface.

And if you take a close look at all the pairs of point masses even the ones on the far side of the sphere [which we call earth] they are all contributing to the value of the gravitational force that is measured on earths surface at a angle to the axis,
so the attraction is along the axis of the centre of the rings.

Not from the centre of the sphere, IE each point mass attracts from where it is in reality.

Which means the value of gravitational attraction at earths surface is a vector sum only, [ie a net force]  of all the pairs of point masses that make up the rings and disks all the way through the earth.

So if we then bring your statement back in to consideration.

The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.

Remembering that what we are measuring on earths surface is only vector sum so only a proportion of the gross force.

So the error is Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass,when he set the value at earths surface to be the one to be used in all calculations at all distances for the inverse square law.

#### PhysBang

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##### Newton made an error and Einstein copied it
« Reply #43 on: 29/04/2010 21:45:16 »
This is because when newton applied inverse square law to the strength of earths gravitational attraction at earths surface he fixed that value as the one to be used at all distances from the surface.
Well, no he didn't. One has only to look at the famous "moon test" of Book III of the Principia to see that the force of gravity on the moon from the Earth is less than that at the surface.
Quote
And if you take a close look at all the pairs of point masses even the ones on the far side of the sphere [which we call earth] they are all contributing to the value of the gravitational force that is measured on earths surface at a angle to the axis,
so the attraction is along the axis of the centre of the rings.
As basic geometry tells us it should be.
Quote
Not from the centre of the sphere, IE each point mass attracts from where it is in reality.
Except that the centre of the sphere is on the axis of the centre of the rings and, geometrically, it doesn't matter whether or not all the mass is along the axis or concentrated at the centre.
Quote
Remembering that what we are measuring on earths surface is only vector sum so only a proportion of the gross force.

So the error is Newton did not allow for the variation in the amount of cancellation due to the vector nature of gravitational attraction when at different distances in space from a mass,when he set the value at earths surface to be the one to be used in all calculations at all distances for the inverse square law.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #44 on: 29/04/2010 22:42:38 »
Gem,

Because of symmetry, the gravitational force acting on a body is going to act along a line from that body towards the center of the Earth. I think (at least I hope) we all agree about that.

So, the only question is, how far along that line does the mass of the Earth appear to act when all the forces are resolved?

Shell theory proves it's half way across the diameter of the Earth along that line.

If our point mass is at the "north pole" that means the sum of all the forces along the axis produced by the upper hemisphere is equal to the sum of all the forces produced by the lower hemisphere.

It's not so difficult to understand why that might be. The forces that the matter in the Earth exert near the body are large because the distance is small, however, as many of them act at a large angle relative to the axis, the resultant force that they produce along the axis is greatly reduced.

On the other hand, the forces coming from the lower hemisphere are reduced because the distance is greater, but the angles are also much smaller, so a much greater component of the force acts along the axis.

The theorem proves that without a doubt. If you want to convince anyone that Newton was wrong, you'll have to explain what is wrong with the math.

#### gem

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##### Newton made an error and Einstein copied it
« Reply #45 on: 30/04/2010 08:27:08 »

Except that the centre of the sphere is on the axis of the centre of the rings and, geometrically, it doesn't matter whether or not all the mass is along the axis or concentrated at the centre.

Shell theory proves it's half way across the diameter of the Earth along that line.

If you pretend that all of earths mass is at the centre in a point mass you eliminate this effect from being taken in to account
The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.

for every point mass contained within the earth as you leave earths surface.

If our point mass is at the "north pole" that means the sum of all the forces along the axis produced by the upper hemisphere is equal to the sum of all the forces produced by the lower hemisphere.

The theorem proves that without a doubt. If you want to convince anyone that Newton was wrong, you'll have to explain what is wrong with the math.

As you point out it is only the vector sums that equal to the centre ,and as you put space between a point mass and the earth this effect

The force in the direction of the axis is proportional to the cosine of the angle, and the angle varies with distance.

Actually continues to happen.
[because all of earths mass is not concentrated at the centre]

so to show what is wrong with the maths.
I think that you will find the vector sum arrow in red in JP's diagram , will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations done for any pair of point Masses contained within the sphere acting on the said  point mass object. IE a greater percentage of the potential gross force

And this will cause inverse square law violation.

#### gem

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##### Newton made an error and Einstein copied it
« Reply #46 on: 30/04/2010 12:43:04 »
This is because when newton applied inverse square law to the strength of earths gravitational attraction at earths surface he fixed that value as the one to be used at all distances from the surface.
Well, no he didn't. One has only to look at the famous "moon test" of Book III of the Principia to see that the force of gravity on the moon from the Earth is less than that at the surface.

Its my understanding Newton had worked backwards from a known and relatively accurate value for the rate of fall of objects on Earth and applied inverse square law from earths centre using that value. are you saying he did not?
« Last Edit: 30/04/2010 12:48:19 by gem »

#### PhysBang

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##### Newton made an error and Einstein copied it
« Reply #47 on: 30/04/2010 17:19:00 »
Newton took the measured rate of fall of the moon and applied an inverse square law to it, determining what it's rate of fall would be at the surface of the Earth. Thus he is comparing the strength of terrestrial gravity at the surface of the Earth to the strength of an inverse square force holding the moon in orbit. He is explicitly assuming that the force holding the moon in orbit obeys the inverse square law.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #48 on: 30/04/2010 18:03:35 »
Gem,

I'm still thoroughly confused as to what you're saying Newton claimed.  So let's assume the earth is a uniform sphere.

1) Are you saying that Newton claimed that the earth's gravity measured above the earth's surface is proportional to the inverse square of the distance between the center of the earth and the object?

or

2) Are you saying that what Newton did was to compute the force of gravity of something at the earth's surface and then say that the force of gravity on an object raised off that surface is proportional to that force multiplied by the inverse square distance between the earth's surface and the object?

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #49 on: 30/04/2010 21:18:19 »
so to show what is wrong with the maths.
I think that you will find the vector sum arrow in red in JP's diagram , will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations done for any pair of point Masses contained within the sphere acting on the said  point mass object. IE a greater percentage of the potential gross force

And this will cause inverse square law violation.

I think what you are saying here is that as the object gets further from the earth, the angles will reduce, so the component of the force acting between the centers (the attractive force times the cosine of the angle) will increase with distance.

Assuming I got that right, then yes, I agree it will. However, the gravitational attraction will also  diminish while the distance increases as the inverse of the square of the distance, so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.

The theorem takes that into account, and, it turns out that, regardless of the distance between the Earth and the object, when the forces are all integrated, they produce the same force that would be produced if all the mass of the Earth was concentrated in one point at its center.

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##### Newton made an error and Einstein copied it
« Reply #49 on: 30/04/2010 21:18:19 »