Why don't heavy objects fall more SLOWLY?

[TurtleMeister]

Ah, yes, I see now the issue with active gravitational mass. Gravity is caused by the presence of matter, but this matter may have different compositions (different internal arrangements) and the question is whether different arrangements of the same quantity of matter (or at least the same inertial mass) might have different gravitational active capacities...

Thanks for taking the time to interpret my ramblings Saw. I am a little lacking in math and communication skills, so you've made my day in being able to see what I have been puzzled over for so long. One of the questions involved in all of this is whether or not leptons have active gravitational mass. If you find this discussion interesting then you may also find this http://www.gravityresearchfoundation.org/pdf/awarded/2001/unnikrishnan_gillies.pdf" interesting.

I have not thought enough about the experiment with the Moon that DH mentioned and about which you argue that it does not exclude the possibility of the above discrepancy.

I believe that this test is invalid, because it is based on Newton's laws, which as you have already discovered, breaks down when there is an inequality between m_a and m_i.

Just a semantic comment for the time being: strictly speaking, what the WEP and the experiments that back it up require is that a given source accelerates all bodies in free fall at the same rate, that is to say, that the passive mass plays no role. This is not threatened by the above mentioned issue: if the passive mass is irrelevant, it’s so in all respects, quantity and composition.

Yes, this has also been an area of confusion for me. I have stated in my previous posts that a test of the equivalence of m_a is also a test of WEP. But I've never been sure of that. Since, as I've already stated, I have not studied GR, I'm not sure as to how this issue affects or threatens it.

As to the need that the two bodies interacting gravitationally meet at the COM of the system… Well, if active gravitational mass were different from inertial mass, then the bodies would meet at their “gravitational” COM, eventually different from their “inertial” COM.

I tend to disagree here. What is gravitational COM? The barycenter of two bodies is determined by their inertial mass. If that were not the case, then it would lead to the same result as the moon experiment, where the third law of motion is violated and the moon self accelerates. And that would open up a Pandora's box of nonsense, such as perpetual motion and unlimited free energy.

The big problem, as you point out, would be for Newton’s Third Law. Choosing units so that G/r^2 = 1, what the Law requires is that:

m0 (inertial) • m1 (gravitational) = - m1 (inertial) • m0 (gravitational)

So if the ratio inertial (m0/m1) <> gravitational (m0/m1), the equation breaks down…

But, that is my point, this and only this would be what is at stake. “Equivalence” can refer to two different things: (i) when faced with an attractive given mass, all objects are attracted with “equal” acceleration and (ii) the ratio of gravitational accelerations is “equal” to the ratio of inertial accelerations in the context of a collision. Only (ii) would be discussed here.

I agree that (i) is not the issue here. I think the issue is whether or not Newton's laws can be used to describe a non equivalence between m_a and m_p / m_i. Either the non equivalence cannot exist, or we must have another equation, or theory, to describe it.

 

[D H]

Is this the test you are referring to: http://prola.aps.org/abstract/PRL/v57/i1/p21_1?

If so, then I think you are wrong. The highlighted portions pertain to Newton's laws. In effect, what they are saying is that if Fe and Al do not attract with the same force (m_a) then there will be a self acceleration of the moon, causing it's orbit to be different from the one predicted by classical mechanics.

That is not what they are saying.

Suppose, for the sake of argument, that m_a=m_p=m_i for iron, while for aluminum, m_p=m_i but m_a=0. The iron core will exert a gravitational force on the aluminum crust, but the crust will exert zero force on the core. (Note well: Newton's third law fails here.) Because the core and crust are structurally attached to one another, the Moon will self-accelerate along the directed line from the crust's center of mass to that of the core. This self-acceleration would make the orbit of this equivalence principle-violating Moon about the Earth differ from that of a Moon that obeys the equivalence principle. This in turn would manifest as an discrepancy between the measured and predicted values of the distance between the Earth and the Moon.

That is just an extreme example. A much smaller deviation between active and passive gravitational mass would still be observable. An extremely small (but still non-zero) deviation would not be observable because there are some uncertainties in both the measurements and predictions. What this experiment did was to place an upper limit (a very, very small upper limit) on what the deviation could be.

A similar situation occurs in experiments to verify that photons have zero rest mass. Physicists cannot absolutely prove this experimentally. What they can do experimentally is to (a) confirm that the theoretical zero rest mass is consistent with experimental results, and (b) establish an upper limit on the photon's rest mass.

Going back to the case at hand, physicists similarly found the strong equivalence principle to be consistent with experimental results and they were able to place an upper limit on $$|1-m_p/m_a|$$.

 

[Saw]

But TurtleMeister seems to be suggesting that (i) gravitational mass <> inertial mass does not necessarily lead to (ii) the absurdity of a self-accelerating Moon, and that we are only led to think that the inequality (i) brings the nonsense (ii) if… we think in terms of Newton’s Laws?

Is it so? Could you be more specific? If you did, I don’t think you’d be breaking forum rules. We are just trying to better understand Newton’s Laws by considering an unlikely hypothesis that would make them fail and discussing whether any alternative Law would be theoretically compatible with logic and experiment. That’s a very orthodox intellectual tool, which makes study amusing…

 

[TurtleMeister]

But TurtleMeister seems to be suggesting that (i) gravitational mass <> inertial mass does not necessarily lead to (ii) the absurdity of a self-accelerating Moon, and that we are only led to think that the inequality (i) brings the nonsense (ii) if… we think in terms of Newton’s Laws?

Perfectly stated.

Is it so? Could you be more specific? If you did, I don’t think you’d be breaking forum rules. We are just trying to better understand Newton’s Laws by considering an unlikely hypothesis that would make them fail and discussing whether any alternative Law would be theoretically compatible with logic and experiment. That’s a very orthodox intellectual tool, which makes study amusing…

$$F = \frac{G(m^\text{g}_1+m^\text{g}_2)}{r^2(\frac{1}{m^\text{i}_1}+\frac{1}{m^\text{i}_2})}$$

$$A_1 = \frac{F}{m^\text{i}_1}$$

$$A_2 = -\frac{F}{m^\text{i}_2}$$

 

[SystemTheory]

Today I watched the first two lectures in this series on General Relativity:

http://www.youtube.com/view_play_list?p=6C8BDEEBA6BDC78D

which I found referenced on another thread. Near the end of the second lecture there is a discussion of the equivalence principle which I understand initially as follows. There may be more implications in further lectures which I'll view soon.

Using the flat Earth assumption gravity field g is constant. A person accelerating at rate g in a remote region of space would observe events within the accelerated frame of reference as equivalent to a field of gravity. Notice in the flat Earth assumption there is no divergence of the gravitational field at a distance from the earth.

The equivalence principle holds if there is no way to know whether one is accelerating through space or experiencing a gravitational field. But gravitational fields around a sphere have a divergence of the field lines at a large radius, while an accelerating frame of reference has no such divergence in the observer's experience. What can be detected in a divergent gravity field is "tidal forces." So, for example, the way the moon imposes tidal forces on the oceans would not be reproduced by accelerating the Earth at a rate in one direction equal to the moon's gravitational acceleration.

The tidal forces are a barrier to the application of the equivalence principle is how the lecture concludes the analysis. I haven't read the Wiki (it is not necessarily written by the most informed authorities but often includes accurate information) so I can't compare to what is discussed there. Hope this at least brings something informative to the thread.

 

[D H]

But TurtleMeister seems to be suggesting that (i) gravitational mass <> inertial mass does not necessarily lead to (ii) the absurdity of a self-accelerating Moon, and that we are only led to think that the inequality (i) brings the nonsense (ii) if… we think in terms of Newton’s Laws?

It is the difference between passive and active gravitational mass that leads to a self-accelerating Moon, not the difference between gravitational mass and inertial mass. Violations of Newton's third law also arise if gravitational and inertial mass are not equal (or better stated, proportional to one another). These violations would also show up in the lunar laser ranging experiments. What is measured here is the time it takes for a laser pulse to travel to the Moon, hit the targeted retroreflector, and return to the Earth. Multiplying those timing measurements by the speed of light yields the distance between the Earth and the Moon. While those inferred distances do depend on the assumption of the constancy of the speed of light, there is no dependency on the equivalence principle. Because those measurements are so precise and because they are independent of the equivalence principle, they serve as an excellent foil for testing the validity of the equivalence principle.

 

[Saw]

$$F = \frac{G(m^\text{g}_1+m^\text{g}_2)}{r^2(\frac{1}{m^\text{i}_1}+\frac{1}{m^\text{i}_2})}$$

$$A_1 = \frac{F}{m^\text{i}_1}$$

$$A_2 = -\frac{F}{m^\text{i}_2}$$

Ah, you don't play with Newton's 3 Laws of Dynamics, only with the Law of Gravitation. It's hard for me to imagine what this formula might lead to, but I have one early doubt: which reference frame are you considering? Looking at the A formulas, it seems it should be an inertial frame, like the COM frame, just like in Newton’s formula, is it?

On the other hand, regarding:

Suppose, for the sake of argument, that m_a=m_p=m_i for iron, while for aluminum, m_p=m_i but m_a=0. The iron core will exert a gravitational force on the aluminum crust, but the crust will exert zero force on the core. (Note well: Newton's third law fails here.) Because the core and crust are structurally attached to one another, the Moon will self-accelerate along the directed line from the crust's center of mass to that of the core. This self-acceleration would make the orbit of this equivalence principle-violating Moon about the Earth differ from that of a Moon that obeys the equivalence principle. This in turn would manifest as an discrepancy between the measured and predicted values of the distance between the Earth and the Moon.

I have a stupid question: I suppose it’s not possible that a hypothetical difference between the m_a of core and crust originated rotation (or at least served to maintain historical rotation alive)…

 

[Saw]

It is the difference between passive and active gravitational mass that leads to a self-accelerating Moon, not the difference between gravitational mass and inertial mass.

I understand that what would to lead to a self acceleration of the Moon would be a hypothetical difference between the active gravitational masses of different parts of the Moon, assuming the “X” masses of those two portions were equal. What is “X”? You call “X” passive gravitational mass… Well, I called it simply “inertial” mass because of what I said in post 29, which I reproduce here for your convenience:

As to “passive gravitational mass” (PGM)… what is that? The equivalence principle (WEP, let’s leave aside the subtleties of GR) means that the acceleration caused by the Earth (towards an inertial reference) on anybody is equivalent, regardless its mass. That entails that whether there or many or few atoms or particles on that side is irrelevant. No matter how many they are, they do not manage to oppose any concerted action to gravity. Yes, they also cause the acceleration of the other side, but that’s an independent Action, rather than a Reaction. I see here no resistance, no opposition, but mere surrender. So the concept looks void to me: yes, mass, when acted upon by gravity, is passive…, so passive that it does nothing at all, other than fully yielding to an effect that will be stronger or weaker depending on the active mass, but not on the passive one. Can you measure that passivity? The idea looks meaningless: zero is always zero…

A dynamometer or a spring scale measure “weight” (force of gravity on a mass m = GmM/r^2). It’s sometimes said that it can also be used to measure “PGM” (m) by keeping everything else constant: you observe the stretching or compression of the spring (l) with one mass m1; you repeat the experiment with mass m2, also in the Earth (M is constant) and at the same altitude and latitude (r is also constant). If the stretching or compression is 2l, that must be because the PGM of m2 = 2 * m1…

I don’t think this way we have made the concept of PGM more meaningful. It seems that in the end what we are measuring is simply inertial mass. Both m1 and m2 have accelerated due to gravity with identical accelerations. That’s their “reaction” to the Earth’s gravity: nothing. And that’s the end of gravity’s role in the story. Now we have a new episode that is the contest between m and the spring: m pulls the spring and the spring reacts by pulling m. Here we do have resistance or inertia, from both sides. If the spring is more rigid, it will resist more and stretch less. If m is greater, it will resist more the pull of the string and descend further. But this new fight is a pure collision, a match of contact forces, nothing to do with gravity. Hence it seems to me that the instrument measures inertial mass, not PGM, which still remains a void concept…

Probably this dispensing with the concept of passive gravitational mass is wrong, since the more I search, the more I see it’s a well established concept. But really I find it hard to assimilate it, for those reasons.

 

[atyy]

Probably this dispensing with the concept of passive gravitational mass is wrong, since the more I search, the more I see it’s a well established concept. But really I find it hard to assimilate it, for those reasons.

I believe Rindler's discussion was that there was no active/passive distinction in Newtonian gravity, while GR has active/passive but no inertial.

 

[Big O]

This is false. No matter what the mass, all objects accelerate at the rate of 32 ft. per second squared in a vacumn.

Note. If there are only two objects in the entire universe, our sun a me both objects move towards each other. With no other reference there is no reference point. They just get closer together.

 

[thayes93]

This is false. No matter what the mass, all objects accelerate at the rate of 32 ft. per second squared in a vacumn.

SI Units man! They make life so much easier.

 

[TurtleMeister]

It is the difference between passive and active gravitational mass that leads to a self-accelerating Moon, not the difference between gravitational mass and inertial mass.

I am assuming that m_p = m_i. p can be substituted for i in the equation for F. The g represents "active gravitational mass".

Ah, you don't play with Newton's 3 Laws of Dynamics, only with the Law of Gravitation. It's hard for me to imagine what this formula might lead to, but I have one early doubt: which reference frame are you considering? Looking at the A formulas, it seems it should be an inertial frame, like the COM frame, just like in Newton’s formula, is it?

Yes, it is the COM (center of inertial mass). And that is the reason it works¹. Newton's law only preserves the COM when it is used in the manner in which it was meant to be used, where m_a = m_i = m_p.

I believe Rindler's discussion was that there was no active/passive distinction in Newtonian gravity, while GR has active/passive but no inertial.

Yes, I tend to agree with that. Newton's law was not designed to work with masses that have unequal active/passive mass. The equality was assumed.

The mutual gravitational force acting between two bodies should be equal in magnitude and opposite in direction. And indeed, that is what the universal law of gravity shows, except when you try and force it to work with a situation where m_i = m_p <> m_a.

$$F_1 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}$$

$$F_2 = -\frac{Gm^\text{g}_1m^\text{p}_2}{r^2}$$

IF $$m^\text{g}_1 <> m^\text{p}_1$$ THEN ABS(F₁) <> ABS(F₂).

In the example above, the forces are opposite, but not equal in magnitude. Another way to look at it is that the frame of reference is not at the COM. And that is what leads to the violation of the third law.

¹ This equation works in all cases where m_a is less than, greater than, or equal to, m_i and m_p. m_i and m_p are assumed to be equal. The COM is preserved as the frame of reference when m_p = m_i <=> m_a. You may want to go back and review my thought experiments in post #30. This will give a visualization of how the COM is preserved. My equation shows gravity to work the same way, eliminating the nonsensical results when Newton's universal law of gravitation is used in cases where m_a <> m_i = m_p.

 

[Saw]

I believe Rindler's discussion was that there was no active/passive distinction in Newtonian gravity, while GR has active/passive but no inertial.

Thanks for the pointer, Atty.

I googled following your hint and found this quote:

“Because of the symmetry of eqn (1.8) (due to Newton’s third law), no essential difference between active and passive gravitational mass exists in Newton’s theory. In GR, on the other hand, the concept of passive mass does not arise, only that of active mass –the source of the field.”

(Essential relativity: special, general, and cosmological, Wolfgang Rindler, page 16)

So, with regard to GR, he says that passive gravitational mass, as I was arguing, “does not arise”, does not play any role in the film, it’s a superfluous concept.

 

[Saw]

TurtleMeister, I submitted your equation to a little obvious test (namely, whether it reduces to Newton's equation if active gravitational = inertial mass) and of course it does:

$$F = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(\frac{1}{{m_1^{\rm{i}} }} + \frac{1}{{m_2^{\rm{i}} }})}}\frac{G}{{r^2 }} = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(\frac{{m_2^{\rm{i}} }}{{m_1^{\rm{i}} m_2^{\rm{i}} }} + \frac{{m_1^{\rm{i}} }}{{m_1^{\rm{i}} m_2^{\rm{i}} }})}}\frac{G}{{r^2 }} = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(\frac{{m_1^{\rm{i}} + m_2^{\rm{i}} }}{{m_1^{\rm{i}} m_2^{\rm{i}} }})}}\frac{G}{{r^2 }} = \frac{{m_1^{\rm{i}} m_2^{\rm{i}} (m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{G}{{r^2 }}$$

By the way, I've just tried MathType for this purpose and, like your equation, it's a nice toy. I never thought I could be able to do algebra so quickly in a computer...

 

[TurtleMeister]

Thanks for testing the equation Saw. I've been typing my LateX in manually, but I can see how that tool would be very handy if I had a lot of it to do.

Here's a few relavent quotes from a book I've been reading (Concepts Of Mass In Contemporary Physics - by Max Jammer)

Just as a violation of the m_i = m_p, equality would be fatal to Einstein's general relativity, a violation of the m_p = m_a equality would be fatal to Newtonian physics, for it would invalidate Newton's third law of motion.

...It is therefore not surprising that little attention has been paid to an experimental confirmation of the m_p = m_a equality...

In 1992 William B. Bonnor argued that nevertheless the general theory of relativity allows a violation of the equality between m_p and m_a, or rather requires such a violation for massive bodies.

He then goes on to write that other gravitational theories allow for, or predict, a non equivalence of m_p and m_a.

So it appears that the apparent violation of the third law of motion has been a major reason for the lack of laboratory tests for the equivalency of m_p and m_a, even though other theories predict it. And I can agree that it does seem strange to test for something for which a positive result would mean the acceptance of perpetual motion machines and unlimited free energy. So, to me it seems logical to conclude that either an inequality between m_p and m_a is impossible, or something is wrong with the methodology (which includes my argument: that Newton's universal law of gravitation is incompatible in cases where m_a <> m_p).

 

[Saw]

Turtle, I've been looking at your equation under this arrangement:

$$\begin{array}{l} F = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{{Gm_1^{\rm{i}} m_2^{\rm{i}} }}{{r^2 }} \\ A_{m1} = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{{Gm_2^{\rm{i}} }}{{r^2 }} \\ A_{m2} = \frac{{(m_1^{\rm{g}} + m_2^{\rm{g}} )}}{{(m_1^{\rm{i}} + m_2^{\rm{i}} )}}\frac{{Gm_1^{\rm{i}} }}{{r^2 }} \\ \end{array} \]$$

It has an advantage: the part on the right side looks like Newton's Law of Gravitation; you just add a ratio on the left, the ratio between the gravitational and pasive/inertial masses of the system, which would be precisely the correction proposed for the hypothesis under consideration (that one day an experiment proved that gravitational mass may occasionally differ from passive/inertial mass).

The only problem with this is that, like this, it appears as if the cause of the accceleration were passive/inertial mass, which is obviously not the case...

 

[TurtleMeister]

You are correct Saw. The equation is Newton's law of gravitation with a correction for the ratio between active gravitational and passive/inertial mass. In fact, the arrangement that you show for F is my original arrangement of the equation. My rearrangement for the final version was made to eliminate the duplication of the two mⁱ variables.

The only problem with this is that, like this, it appears as if the cause of the accceleration were passive/inertial mass, which is obviously not the case...

There is no provision for m1^g and m2^g in the Newtonian part of the equation. The normal method is to split into two different forces, one for A₁ and another for A₂. But that would lead us right back to the 3rd law violation (when m_a <> m_i). So in effect, what I have done is combine the two equations into one by leaving out m1^g and m2^g for the Newtonian part. This could very well be a flaw for the equation when m_a <> m_i. But I'm not sure. For example, setting a value for m1ⁱ and m2ⁱ and then adjusting the values of m1^g and m2^g for the desired ratio, seems to give the correct results.

 

[Saw]

what I have done is combine the two equations into one by leaving out m1^g and m2^g for the Newtonian part. This could very well be a flaw for the equation when m_a <> m_i. But I'm not sure. For example, setting a value for m1ⁱ and m2ⁱ and then adjusting the values of m1^g and m2^g for the desired ratio, seems to give the correct results.

Why do you need to adjust the values of m1^g and m2^g for the desired ratio, in order to get the correct results? If by "correct results" you mean (i) to comply with your requirement that the two masses must meet at their inertial centre of mass (CM) and (ii) that there's no violation of Newton's 3rd Law (the two forces are equal and of opposite sign), it seems you satisfy those conditions with this equation.

My only problem is to find a physical meaning for the math. The formula is telling us that each mass will get accelerated in proportion to the other's passive/inertial mass (that's why they meet at the inertial CM, as you had ordered) and that, however, if total inertial mass > total gravitational mass, the meeeting will take place later (or earlier in the opposite case). Well, this sounds odd, because what is really causing the acceleration (the gravitational field) does not determine where the objects meet, it simply contributes to determine when, while the resistance of the objects plays the fundamental role.

We noted that small disagreement before. Of course, I do not want to end up with absurd, nonsensical resuls. Of course, the probable end of all this may be that we admit that the assumption that passive = inertial mass is correct. But my point is that I do not see why, to avoid running into nonsensical results, the meeting point must necessarily be the inertial CM.

Take this example. Two identical masses, as measured in a balance scale. In between them, two springs of equal mass and equal length, attached to one another. The CM of the two masses is in the mid-point between them. But the two springs have different elasticity coefficients. Then the two balls are pushed towards each other by identical forces, until they fully compress the springs and meet. Will they meet (i) at the mid-point (CM) or (ii) somewhere closer to the more rigid spring?

It's not a rhetorical question. I really don't know the answer. But my guess is that the ball by the more elastic spring reaches the CM earlier and starts compresing the more rigid one, so the answer would be (ii).

In that case, we could use this quite loose analogy: we consider that each ball plus its respective spring is a "body"; each body has the same mass, in reality, and the same active gravitational mass, by analogy, since we imagine that each of them somehow manages to pull on the other; but the springs have different elasticity coefficients, in reality, and different inertial masses, by analogy. Yes, the idea is convoluted... But of any use?

 

[D H]

These last several posts are stretching the rules of this forum. It is not OK to post personal theories here.

 

[Saw]

These last several posts are stretching the rules of this forum. It is not OK to post personal theories here.

Sorry, DH, but I respectfully disagree. I have little physics, but I am a good lawyer. Nobody is posting personal theories here. We assume and accept mainstream physics and we are just trying to grasp the logic of its principles by noticing how alternative routes lead to a dead-end. That's called reductio ad absurdum and for the benefit of the standard approach. If in the course of this discussion, I made incorrect statements, I would most welcome correction. In particular, I really appreciate your interventions, each of which contains a little jewel of information...

 

[D H]

Those equations are a personal theory.

 

[TurtleMeister]

These last several posts are stretching the rules of this forum. It is not OK to post personal theories here.

Thanks for the warning. I will be more careful with my posts. Also, please feel free to add to the discussion or correct my errors, as you have done in the past. Your input is greatly appreciated.

what is really causing the acceleration (the gravitational field) does not determine where the objects meet, it simply contributes to determine when, while the resistance of the objects plays the fundamental role.

We noted that small disagreement before. Of course, I do not want to end up with absurd, nonsensical resuls. Of course, the probable end of all this may be that we admit that the assumption that passive = inertial mass is correct. But my point is that I do not see why, to avoid running into nonsensical results, the meeting point must necessarily be the inertial CM.

Take this example. Two identical masses, as measured in a balance scale. In between them, two springs of equal mass and equal length, attached to one another. The CM of the two masses is in the mid-point between them. But the two springs have different elasticity coefficients. Then the two balls are pushed towards each other by identical forces, until they fully compress the springs and meet. Will they meet (i) at the mid-point (CM) or (ii) somewhere closer to the more rigid spring?

It's not a rhetorical question. I really don't know the answer. But my guess is that the ball by the more elastic spring reaches the CM earlier and starts compresing the more rigid one, so the answer would be (ii).

If I'm understanding this thought experiment correctly, the answer is (i). You partially answered it yourself in the first highlighted sentence that I quoted from you. The springs only supply the force, they do not determine the meeting point (considering the springs are massless and have no substance). This is true even when using multiple springs with different elasticities. The meeting point is the CM, which is determined by the inertial mass of the two objects. The simple equation for this is:

$$m_1d_1 = m_2d_2$$

where d₁ = the distance from m₁ to the CM and d₂ = the distance from m₂ to the CM.

But my point is that I do not see why, to avoid running into nonsensical results, the meeting point must necessarily be the inertial CM.

Because if they don't meet at the CM then the third law is violated (considering only the two masses are involved). But I guess your question is, how would not meeting at the CM be nonsensical? Well, I'm having trouble thinking of an example for that because you really need to have a good understanding of the third law to get it. And since you're questioning this, I'm assuming you do not have a good understanding of it. But I can tell you that if I had that magic spring from your thought experiment, I could build a device that could propel itself (in the vacuum of space) simply by moving weights around and applying the spring. And that would be nonsensical. You can find lots of stuff on the web about Newton's third law of motion.

In that case, we could use this quite loose analogy: we consider that each ball plus its respective spring is a "body"; each body has the same mass, in reality, and the same active gravitational mass, by analogy, since we imagine that each of them somehow manages to pull on the other; but the springs have different elasticity coefficients, in reality, and different inertial masses, by analogy. Yes, the idea is convoluted... But of any use?

Sorry, I do not understand that.

 

[Big O]

Why all the fuss.

In a vacumn all objects fall at the same rate. If there is an atmosphere it will offer resistance to the object falling just as the planets surface will continously halt the object from falling. If the planet changes in size the object adjust with it.

 

[TurtleMeister]

Why all the fuss.

In a vacumn all objects fall at the same rate. If there is an atmosphere it will offer resistance to the object falling just as the planets surface will continously halt the object from falling. If the planet changes in size the object adjust with it.

That's not what we're talking about. We have gotten a little off topic to the OP, but still related.

In that case, we could use this quite loose analogy: we consider that each ball plus its respective spring is a "body"; each body has the same mass, in reality, and the same active gravitational mass, by analogy, since we imagine that each of them somehow manages to pull on the other; but the springs have different elasticity coefficients, in reality, and different inertial masses, by analogy. Yes, the idea is convoluted... But of any use?

Ok, I think I understand this now. You are saying, by analogy, that the spring can have the same effect as changing the inertial mass of the object it's connected to? No, that is not true. As you have already stated yourself, the springs (source of the force) only has an effect on when they meet, not where. Even though you have two springs with different elasticities connected together, the force applied to each object will be equal in magnitude and opposite in polarity. Where they meet is determined only by their inertial mass.

And by analogy, if we have two objects attracted to each other by gravity, the force on each object will be equal in magnitude and opposite in polarity, even if one has a stronger gravitational field than the other. The source of the force m_a has no effect on where they meet.

And that is the reason the following method of applying Newton's laws fails when m1^g <> m1^p or m2^g <> m2^p

$$F_1 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}$$

$$F_2 =- \frac{Gm^\text{g}_1m^\text{p}_2}{r^2}$$

$$A_1 = \frac{F_1}{m^\text{i}_1}$$

$$A_2 = \frac{F_2}{m^\text{i}_2}$$

Just like in your springs thought experiment, where you have two springs with different elasticities connected together, F_{1 <> F}2, when m1^g <> m1^p or m2^g <> m2^p. This causes a problem if we apply F₁ to the acceleration of m1, and F₂ to the acceleration of m2. If we do that then it would be analogous to your thought experiment where the spring affects the point where they meet. The third law will be violated.

Without inventing new equations for the universal law of gravitation, and for demonstration purposes only, I will attempt to show how this can be done without violating the third law.

$$F_1 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}$$

$$F_2 = \frac{Gm^\text{g}_1m^\text{p}_2}{r^2}$$

$$F = \frac{F_1+F_2}{2}$$

$$A_1 = \frac{F}{m^\text{i}_1}$$

$$A_2 = -\frac{F}{m^\text{i}_2}$$

The force is now equal in magnitude for each object but opposite in polarity. The objects will meet at their COM.

edit: The averaging of the forces has the effect of restoring the frame of reference to the COM.

 

[D H]

$$F_1 = \frac{Gm^\text{g}_1m^\text{p}_2}{r^2}$$

$$F_2 = \frac{Gm^\text{g}_2m^\text{p}_1}{r^2}$$

$$F = \frac{F_1+F_2}{2}$$

$$A_1 = \frac{F}{m^\text{i}_1}$$

$$A_2 = -\frac{F}{m^\text{i}_2}$$

The force is now equal in magnetude for each object but opposite in polarity. The objects will meet at their COM.

What is this F=F₁+F₂ stuff? What justifies that step, and then using that force as the force on each of the two objects? That the result is consistent with Newton's third law is not surprising: You implicitly assumed Newton's third law here.

 

[TurtleMeister]

What is this F=F1+F2 stuff? What justifies that step, and then using that force as the force on each of the two objects?

It's just a demonstration to help us understand the problem. Yes, I am assuming Newton's third law.

 

[D H]

Why didn't you use the obvious $$a_1 = F_2/m_1$$ and $$a_2 = F_1/m_2$$ here? Why that force sum? You did not justify it, and it most certainly does not jibe with reality. If the equivalence principle is true, this summation makes the acceleration be off by a factor of two.

 

[TurtleMeister]

Why didn't you use the obvious $$a_1 = F_2/m_1$$ and $$a_2 = F_1/m_2$$ here? Why that force sum? You did not justify it, and it most certainly does not jibe with reality. If the equivalence principle is true, this summation makes the acceleration be off by a factor of two.

First, I noticed an error in my equations in post #59. I had m1 and m2 reversed for Newton's universal law of gravitation. I have fixed it now, so you may want to change the $$a_1=F_2/m_1$$ and $$a_2=F_1/m_2$$ in your post #62 to $$a_1=F_1/m_1$$ and $$a_2=F_2/m_2$$. I don't think this affects the questions you are asking.

Referring to the first four equations in post #59, and assuming that m1^p = m2^p, do you agree that:

$$A_1 = A_2$$ when $$m^\text{g}_1 = m^\text{p}_1$$ and $$m^\text{g}_2 = m^\text{p}_2$$

$$A_1 <> A_2$$ when $$m^\text{g}_1 <> m^\text{p}_1$$ or $$m^\text{g}_2 <> m^\text{p}_2$$

Referring to the second set of five equations in post #59, and assuming that m1^p = m2^p, do you agree that:

$$A_1 = A_2$$ when $$m^\text{g}_1 = m^\text{p}_1$$ and $$m^\text{g}_2 = m^\text{p}_2$$

$$A_1 = A_2$$ when $$m^\text{g}_1 <> m^\text{p}_1$$ or $$m^\text{g}_2 <> m^\text{p}_2$$

If you do not agree with any of the questions above, then please explain where you think I am wrong. Note: edited to correct errors in above questions.

If you do agree with all of the questions, then you should understand the reason I posted them. The first four equations is the normal way of showing the accelerations of A₁ and A₂ with a variable m^g. It shows that this method causes a violation of the third law of motion (self accelerating moon) when m1^g <> m1^p or m2^g <> m2^p. The second set of five equations show that if the frame of reference is restored to the COM $$F = \frac{F_1+F_2}{2}$$ then the third law of motion is not violated.

If the equivalence principle is true, this summation makes the acceleration be off by a factor of two.

I do not know how you arrived at that conclusion. If the equivalence principle is true then F = F1 = F2.

 

[D H]

What motivates that force averaging? It is far to ad hoc.

 

[TurtleMeister]

What motivates that force averaging? It is far to ad hoc.

It's just a demonstration to show that the cause of the third law violation is the frame of reference not being at the COM. The averaging puts the frame of reference at the COM and eliminates the third law violation.

Also, for you information. I fixed some errors in the questions in post #63. So you may want to review it again.

 

[Saw]

Turtle, I am ready to give up with any arguing based on my thought experiment about the springs. But I would need some further explanation.

When a force is applied to a spring, such force is defined as:

$$F = k\Delta l$$

k being the spring's constant of rigidity and ∆l being its stretching or, in this case, compression

Another way to put it is that the streching / compression is directly proportional to the force and inversely proportional to k

$$\Delta l = \frac{F}{k}$$

Here ∆l seems to play the part of acceleration and it appears that k plays the part of m.

Hence in the experiment where two springs are compressed against each other, I expected that Newton's Third Law would be fulfilled as follows:

$$k'\Delta l' = - k\Delta l$$

What did I miss?

 

[TurtleMeister]

To make sure I'm not misunderstanding you, I've made a little graphic of my interpretation of your thought experiment.

Let's say that A and B have the same inertial mass and are attracted to each other, either gravitationally or magnetically. There are no outside forces involved. Also, we will ignore the mass of the springs. My understanding of your position is that because the spring on the left has a higher k than the spring on the right that the two masses will meet at a point closer to "point1" on the left side of the COM.

Newton's third law of motion states that for every action there is an equal and opposite reaction. So if A moves to the right 1cm, then to satisfy this law, B must move to the left 1cm. Since they both move equally in distance and opposite in direction then they must meet at the COM point. The springs have nothing to do with where they meet. The 1k spring will compress more than the 2k spring and the point where they are attached will move to the right.

Does this help to clear things up, or did I misinterpret your thought experiment?

 

[danielatha4]

I didn't read all the replies to this thread but this picture always helps me understand the difference between acceleration due to gravity and force due to gravity.

now just break F1 up into M1 * A1 and F2 = M2 * A2

 

[Saw]

danielatha4, the drawing is very illustrating, although we were now discussing a further issue:

TurtleMeister wondered what would happen if the composition of the masses had some influence in how objects accelerate due to gravity.

D H pointed out that certain experiments prove that such thing would lead to nonsensical results. If I interpret well, portion A having passive but no active gravitational mass would be attracted to portion B without attracting B in turn; that would mean that A is accelerated towards B and collides with it; thus the body would self-accelerate away from its centre of mass, in the direction of B, breaking Newton’s Laws of Dynamics.

TurtleMeister commented that such nonsensical result could be avoided if the bodies were gravitationally accelerated according to a Law of Gravitation that’d be slightly different from Newton’s: so as to meet at the centre of mass of the system, but at a rate which would depend on the ratio between the total of gravitational masses and the total of inertial masses.

I said: but why is it so important that the bodies meet at the inertial centre of mass? And proposed a thought experiment where it might not be so.

I was wrong due to faulty understanding of Newton’s Laws and TurtleMeister has just taken the trouble to explain why. My thought experiment does not prove what I thought but the opposite: between two balls affected by equal attracting forces there are two massless springs of equal lengths but different spring constants; in spite of that, the balls tend to meet at their centre of mass.

I attach an image with a try at a numerical example and the outcome of the experiment, for correction. (For simplicity, and lack of a better software, I drew the springs as rectangles.)

 

[TurtleMeister]

Looks good to me Saw.