Why don't heavy objects fall more SLOWLY?

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In summary: The principle of equivalence states that the more massive object will resist acceleration more than the less massive object. However, the more massive object has a greater force acting on it. The two are perfectly balanced so that the effect is that all objects will accelerate at the same rate regardless of their mass.
  • #1
chudd88
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This is a simple question, and I'm sure it has a trivial answer, but this thought occurred to me just now.

How is it that gravity is able to accelerate things at the same rate regardless of its mass, while one definition of mass is the measure of resistance to acceleration? If two objects are 100 miles from Earth, I would expect that the Earth's gravity would be able to accelerate the lighter object more quickly, while the heavier object would resist the acceleration and fall more slowly. Why is this not the case?
 
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  • #2
I would expect that the Earth's gravity would be able to accelerate the lighter object more quickly, while the heavier object would resist the acceleration and fall more slowly. Why is this not the case?
It is true that the more massive object will resist acceleration more than the less massive object. However, the more massive object has a greater force acting on it. The two are perfectly balanced so that the effect is that all objects will accelerate at the same rate regardless of their mass.

I'm sure this question has been asked many times at PF, so you may be able to use the search function to learn more. Or, please feel free to ask more questions here if you like.
 
  • #3
You may notice that the equation for acceleration due to a force and the equation for weight are just two different forms of the same equation...

f=ma
a=f/m

"a" is the acceleration due to gravity.
 
  • #4
chudd88 said:
This is a simple question, and I'm sure it has a trivial answer, but this thought occurred to me just now.

How is it that gravity is able to accelerate things at the same rate regardless of its mass, while one definition of mass is the measure of resistance to acceleration? If two objects are 100 miles from Earth, I would expect that the Earth's gravity would be able to accelerate the lighter object more quickly, while the heavier object would resist the acceleration and fall more slowly. Why is this not the case?

Because the two objects do not experience the same force.

Gravity is a puzzling force; it is proportional in magnitude to the mass of the object ACTED UPON.
(Very few other forces have this strange property)


Thus, the more massive object is acted upon with a greater force than the lighter object, the net result being that both experience the same acceleration.
 
  • #5
arildno said:
Because the two objects do not experience the same force.

Gravity is a puzzling force; it is proportional in magnitude to the mass of the object ACTED UPON.
(Very few other forces have this strange property)Thus, the more massive object is acted upon with a greater force than the lighter object, the net result being that both experience the same acceleration.

But this rule (principle of equivalence) is established judging… from which reference frame?

I see two possibilities:

(a) From the perspective of an inertial frame not participating in the interaction. For example, the centre of mass of the system composed by bodies with masses M and m, which are gravitationally interacting.

(b) From the perspective of a frame participating in the interaction and hence an accelerated one.

If we judge (measure) from (a), the inertial frame, the centre of mass (CM) of the system, it seems to me that the principle of equivalence should not apply: the more massive object M would cover the shorter distance to the CM with lesser acceleration and the less massive should traverse its own longer path with more acceleration, so that both meet precisely at the said CM. This would be in line with what the OP suggested and seems to be the natural intuition following the example of collisions (i.e., when the relevant force is contact force).

Given this, the principle of equivalence would apply only with regard to or as measured from (b), i.e., from the reference frame of one of the attracted objects. In the frame of M, for example, the acceleration of m1 would be the sum of m1's acceleration as measured in (a) plus the acceleration of M itself as measured also in (a), in the opposite direction. If we now choose m2, more massive than m1, the first component would be lesser (m2 resists more) but the second would be greater (M is accelerated more). One thing would offset the other and thus, as judged from M’s reference frame, all objects would fall towards it with the same acceleration, regardless their respective masses. The same would apply to the judgment from m1, m2…

Did I guess it right?

Note: If you read this post before, I've edited it to make it clearer. Thanks for any comment.
 
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  • #6
Saw

Its a 3 body Problem .. no CM. CM situs are not i think what the op is talking about.

Huge mass.. the effector

One small mass [sm] and one large mass[lm]..Both > to effector
Why does lm accelerate same dv.
" Why does a feather fall as fast as a lead balloon"

Louis
________________________________
PS i have no formal 'education', everything i say i taught myself.
So pls be patient. I tend to incorporate real life into the crap jargon so many are fond of spouting. Albert tried this and i does not work. I persevere
 
  • #7
Saw said:
But this rule (principle of equivalence) is established judging… from which reference frame?
Any frame of reference.

If we judge (measure) from (a), the inertial frame, the centre of mass (CM) of the system, it seems to me that the principle of equivalence should not apply: the more massive object M would cover the shorter distance to the CM with lesser acceleration and the less massive should traverse its own longer path with more acceleration, so that both meet precisely at the said CM. This would be in line with what the OP suggested and seems to be the natural intuition following the example of collisions (i.e., when the relevant force is contact force).
It would be silly for the equivalence principle to say something that contraindicates well-explained phenomena. Experiment one: Place an Earth-sized mass and a pea-sized mass at rest with respect to one another but separated by 384,400 km (the Earth-Moon distance). Measure the time to collision, which will be several days. Experiment two: Replace the pea-sized mass with a neutron star. These objects will collide on a much smaller time scale.

Fortunately, the equivalence principle does not say that the time to collision will be the same. It instead says that the initial accelerations of the pea and the neutron star toward the Earth will be exactly the same.
 
  • #8
D H said:
It would be silly for the equivalence principle to say something that contraindicates well-explained phenomena. Experiment one: Place an Earth-sized mass and a pea-sized mass at rest with respect to one another but separated by 384,400 km (the Earth-Moon distance). Measure the time to collision, which will be several days. Experiment two: Replace the pea-sized mass with a neutron star. These objects will collide on a much smaller time scale.

Well, yes, that sounds logical.

D H said:
Fortunately, the equivalence principle does not say that the time to collision will be the same.

Yes… I suppose I had not grasped what it really says. But you’ll maybe concede that there’s something confusing in all this. Take for example the formulation in Wikipedia:

The weak equivalence principle, also known as the universality of free fall:

The trajectory of a falling test particle depends only on its initial position and velocity, and is independent of its composition.

or

All test particles at the alike spacetime point in a given gravitational field will undergo the same acceleration, independent of their properties, including their rest mass.

And also, a little above in the page:

The simplest way to test the weak equivalence principle is to drop two objects of different masses or compositions in a vacuum, and see if they hit the ground at the same time.

Reading this, wouldn’t you be led to think that the pie and the neutron star, if "dropped" from the same place", should hit the Earth ground at the same time? Of course, for clarity, we should be assuming that each object is dropped at a different moment, in a different experiment, because otherwise what we have (maybe that’s what Louis was pointing at) is a 3-body problem. My understanding was thus that the principle requires that in each experiment (done from the same place but in a different moment) the fall time is identical. But then if the principle does not entail so, what does it state?

D H said:
It instead says that the initial accelerations of the pea and the neutron star toward the Earth will be exactly the same.

So… only the initial accelerations would be equivalent? And why and when should they start differing? As the fall progresses, the distance is reduced and the acceleration augments…, but that is independent of the mass of the object acted upon…
 
  • #9
So… only the initial accelerations would be equivalent? And why and when should they start differing?
The acceleration of the Earth due to the neutron star's pull would be WAY stronger, so the Earth would swiftly accelerate towards it, decreasing the distance swiftly.
 
  • #10
the heavier object falling faster than the lighter in the Earth's surface is due to the air resistance acting on the objects not due to the gravity as far as i know.
 
  • #11
mikhailpavel said:
the heavier object falling faster than the lighter in the Earth's surface is due to the air resistance acting on the objects not due to the gravity as far as i know.
Not if the heavy object is a neutron star as in the given example.

Even in vacuum, a (n extremely) heavier object will, RELATIVE to the Earth surface, accelerate faster than a lighter object, due to its stronger attractive pull on the Earth.

Think about it...:smile:
 
  • #12
what about the black holes? am confused?
 
  • #13
arildno said:
The acceleration of the Earth due to the neutron star's pull would be WAY stronger, so the Earth would swiftly accelerate towards it, decreasing the distance swiftly.

Yes, I understand that. Certainly, since there is an interaction (both sides suffer effects = acceleration towards each other), it's not indifferent that the Earth faces a pie or a neutron star: if it is the latter, the Earth itself is accelerated more. My problem is how to reconcile that with the equivalence principle, in the formulation quoted above from Wikipedia. In the classical Galileo's (real or thought) experiment two lead balls of different sizes dropped from the Tower of Pisa reach the ground (neglecting air resistance) at the same time. It is said that an austronaut dropped a feather and a hammer on the Moon (where there's no atmosphere) and the two fell at the same rate. What changes if you substitute the neutron star for one of the leaden balls or the hammer? Unless... the difference is that the masses of the balls, feather and hammer are all ridiculous in comparison with the mass of the Earth and so they are all equivalent to a good approximation, for practical purposes... But I have not read that interpretation anywhere, it looks even less orthodox and it would convert the equivalence principle into a mere pragmatic rule of thumb for use under certain conditions...
 
  • #14
Newton's Law of Universal Gravitation:

[tex]F = G\frac{m_{1}m_{2}}{r^{2}}[/tex]

where each body has its own acceleration a = F/m toward the other body:

[tex]a_{1} = G\frac{m_{2}}{r^{2}}[/tex]

[tex]a_{2} = G\frac{m_{1}}{r^{2}}[/tex]

so then the time of acceleration should be established by taking an integral of a differential equation after applying some convenient frame of reference. It is not clear to me that the times will differ as a function of mass.
 
  • #15
SystemTheory said:
Newton's Law of Universal Gravitation:

[tex]F = G\frac{m_{1}m_{2}}{r^{2}}[/tex]

where each body has its own acceleration a = F/m toward the other body:

[tex]a_{1} = G\frac{m_{2}}{r^{2}}[/tex]

[tex]a_{2} = G\frac{m_{1}}{r^{2}}[/tex]

so then the time of acceleration should be established by taking an integral of a differential equation after applying some convenient frame of reference. It is not clear to me that the times will differ as a function of mass.

Yes, that was the next step for discussion. It's not only the formulation in words of the equivalence principle, it's Newton's Laws what seems to require that the neutron star reaches the Earth ground at the same time as the pie...
 
  • #16
SystemTheory said:
[tex]a_{1} = G\frac{m_{2}}{r^{2}}[/tex]

[tex]a_{2} = G\frac{m_{1}}{r^{2}}[/tex]
Correct. Note that the acceleration of object 1 is independent of the mass of object 1, and the acceleration of object 2 is independent of the mass of object 2.

It is not clear to me that the times will differ as a function of mass.
The acceleration of objects 1 and 2 toward each other is

[tex]a_{\text{rel}} = \frac{G(m_1+m_2)}{r^2}[/tex]

and this obviously depends on the masses of both objects.

If one of those masses is many orders of magnitude larger than the other, the acceleration for all practical purposes depends only on the mass of the larger object. Suppose that in a vacuum you lift a 1 gram pea 4.9 meters high and release it. It will hit the Earth one second later. Now do the same with a 10 metric ton stone. It too will hit the Earth one second later. Because the Earth's acceleration toward the stone is 10 million times that toward the pea, there will be a very slight difference in the timing. How much? About 1.7×10-21]/sup] seconds. This is of course immeasurably small.

The Moon's mass is about 1/81 that of the Earth. This is enough to affect the Moon's orbit about the Earth. Suppose the Moon's mass was a couple of orders of magnitude smaller than it is. This reduced mass Moon would take about 20 minutes longer to orbit the Earth than does our big honkin' Moon.
 
  • #17
D H said:
Fortunately, the equivalence principle does not say that the time to collision will be the same. It instead says that the initial accelerations of the pea and the neutron star toward the Earth will be exactly the same.
I think the way this is worded may have caused some confusion. It is true that the accelerations of the pea and the neutron star will be the same. However, for clarification, it should be pointed out that the accelerations are in relation to the barycenter and not to either one of the objects. Or in other words, the frame of reference is the barycenter.
Saw said:
The simplest way to test the weak equivalence principle is to drop two objects of different masses or compositions in a vacuum, and see if they hit the ground at the same time.
Reading this, wouldn’t you be led to think that the pie and the neutron star, if "dropped" from the same place", should hit the Earth ground at the same time? Of course, for clarity, we should be assuming that each object is dropped at a different moment, in a different experiment, because otherwise what we have (maybe that’s what Louis was pointing at) is a 3-body problem. My understanding was thus that the principle requires that in each experiment (done from the same place but in a different moment) the fall time is identical. But then if the principle does not entail so, what does it state?
The simple test that you refer to assumes that the masses of the objects dropped are so small compared to the mass of the Earth that the Earth's acceleration can be ignored. If the objects dropped are of significant mass in relation to the earth, then of course you cannot ignore the acceleration of the earth. Also, the frame of reference for this test is the earth. So the acceleration will be the relative acceleration of the Earth and the object toward each other.
 
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  • #18
D H said:
Correct. Note that the acceleration of object 1 is independent of the mass of object 1, and the acceleration of object 2 is independent of the mass of object 2.


The acceleration of objects 1 and 2 toward each other is

[tex]a_{\text{rel}} = \frac{G(m_1+m_2)}{r^2}[/tex]

and this obviously depends on the masses of both objects.

If one of those masses is many orders of magnitude larger than the other, the acceleration for all practical purposes depends only on the mass of the larger object. Suppose that in a vacuum you lift a 1 gram pea 4.9 meters high and release it. It will hit the Earth one second later. Now do the same with a 10 metric ton stone. It too will hit the Earth one second later. Because the Earth's acceleration toward the stone is 10 million times that toward the pea, there will be a very slight difference in the timing. How much? About 1.7×10-21]/sup] seconds. This is of course immeasurably small.

The Moon's mass is about 1/81 that of the Earth. This is enough to affect the Moon's orbit about the Earth. Suppose the Moon's mass was a couple of orders of magnitude smaller than it is. This reduced mass Moon would take about 20 minutes longer to orbit the Earth than does our big honkin' Moon.


I didn't see this post before I made my previous post. The first equations by SystemTheory are the ones relevant to the equivalence principle. It assumes the reference frame to be the barycenter. The second equation by DH for relative acceleration is the one used for the simple test quoted by "Saw". This one assumes that one object (such as the earth) is of much greater mass than the other object.
 
  • #19
Then if I understand you all correctly:

(a) If the acceleration is measured from the frame of one of the objects participating in the gravitational interaction (i.e., an accelerated frame, say the Earth in our example)..., in other words, if we refer to the relative acceleration between the two interacting objects, the equivalence principle does not strictly speaking apply: it does only under certain circumstances (if one mass is negligible wrt the other), to a good approximation, for most practical purposes.

Accordingly here the formula is:

[tex]a_{\text{rel}} = \frac{G(M+m)}{r^2}[/tex]

(b) If the acceleration is measured from an inertial frame (like the center of mass of the system formed by the two bodies or barycentre)…, in other words, if we refer to the individual accelerations towards the origin of such frame, then:

those individual accelerations are different for each body:

the more massive body accelerates at the rate of

[tex]a_{M} = G\frac{m}{r^{2}}[/tex]

which is less than the acceleration rate of the less massive body

[tex]a_{m} = G\frac{M}{r^{2}}[/tex]

and the equivalence principle applies in the sense that if you put any other body in the side of, for example, m (the neutron star in the place of the pea), the acceleration of the star towards the barycentre will be identical to that of the pea, since in both cases it's a function of the mass of M (apart from G and r).

Did I get it right now?
 
  • #20
Saw said:
Did I get it right now?
Yes. At least to the best of my understanding. I agree with everything in your post.
 
  • #21
TurtleMeister said:
Yes. At least to the best of my understanding. I agree with everything in your post.

Thanks a lot for your help.

Now that the scope of the equivalence principle (EP) is clear, one could wonder WHY it is so, which is what the OP asked.

Admitting that usually in science it's enough learning experimentally how things work and modelling that behaviour mathematically, it's also true that speculation about root-causes may turn out to be helpful.

My understanding (maybe faulty) is that Einstein's GR explains the EP as a result of the fact that mass curves spacetime. But I gather from answers in the SR and GR forum that this "geometrical" approach, while quite fortunate as a description (in fact, it brings about mathematical changes that match better with experiments), is after all an analogy, a good analogy, but it does not exclude the possibility to search for a more "physical" answer.

In this line, what I have heard and read sometimes is that gravity, whatever its ultimate cause (gravitons emitted by mass or whatever), acts directly on individual atoms. I mean, in a collision the atoms of the shell of body A interact with the atoms of the shell of body B, but then the effect is transmitted by the internal atoms of each body through a pressure wave. Instead, when faced with gravity, bodies would be transparent, accepting its effect in each of their individual atoms, without the need for internal propagation.

Is this the standard explanation or speculation of some authors or, even worse, non-accepted theory?
 
  • #22
General relativity does not prove the equivalence principle any more than does Newton's law of gravity. The equivalence principle is an implicit assumption in Newtonian mechanics and is an explicit assumption (i.e., a postulate) in general relativity.
 
  • #23
Saw said:
In this line, what I have heard and read sometimes is that gravity, whatever its ultimate cause (gravitons emitted by mass or whatever), acts directly on individual atoms. I mean, in a collision the atoms of the shell of body A interact with the atoms of the shell of body B, but then the effect is transmitted by the internal atoms of each body through a pressure wave. Instead, when faced with gravity, bodies would be transparent, accepting its effect in each of their individual atoms, without the need for internal propagation.

Is this the standard explanation or speculation of some authors or, even worse, non-accepted theory?
I am not a physicist, nor am I an expert in this field. But from my recent studies in this area I can tell you that laboratory tests have shown (to an accuracy of 1 part in 1012) that gravity (active gravitational mass) affects all matter (passive gravitational mass), including leptons, equally. So your assessment agrees with current experimental evidence.

The one area where experimental evidence is lacking is the equivalence of active gravitational mass and passive gravitational mass / inertial. And this is something that I have been pondering for some time.
 
  • #24
When you step on the accelerator in your car, the giant redwood trees and the people standing by the side of the road accelerate at the same rate. If you are in a rocket ship out in space undergoing acceleration and you "drop" a marble and a bowling ball in your ship, they will "fall" with the same acceleration. Einstein's equivalence principle is saying that gravity is the same thing as the "force" that pushes you back in your car seat when you step on the gas. This is probably the single most brilliant and yet insane idea I have ever encountered in physics...basically Einstein is saying that things fall at the same rate not because some magical force called gravity reaches out and pulls them down, but because the ground (and everything sitting on it) rushes up to meet the "dropped" objects. Now I can hear your objections...how can the ground be rushing up to meet the ball I drop?...after all, a guy on the other side of the world could drop a ball too... doesn't that mean that the Earth should be in a constant state of expansion?...since the ground everywhere is always rushing upward...? Well good news! Einstein gets rid of this problem by saying that the very structure of space and time is bent and twisted in such a way as to counteract this expansion. Physicists would word this as, "The equivalence principle describes gravitation locally as a special relativistic pseudoforce. Gravity manifests itself globally as curverature of the spacetime manifold." The conceptual beauty of this idea cannot be understated...wouldn't it be nice if all the "forces that act at a distance" could be understood in terms of the "inertial force" and spacetime curverature? There were some attempts at constructing a theory of electromagnetism where it would be viewed this way, but it required an additional fifth dimension and made predictions that seriously conflicted with experiment.
 
  • #25
D H said:
General relativity does not prove the equivalence principle any more than does Newton's law of gravity. The equivalence principle is an implicit assumption in Newtonian mechanics and is an explicit assumption (i.e., a postulate) in general relativity.

Hrm... I'm not quite sure what assumptions are inherent in this, and one of them might be the EP itself, but I think you can actually "derive" the equivalence principle from GR. Once you have the geodesic equations, can't you make some study of geodesic deviation and conclude that for a small enough initial geodesic separation and small enough time scales, the geodesic deviation vanishes? This would show that the space is locally minkowski, which is (I suppose) one formulation of the equivalence principle.

Reading the wiki article for EP, I see that there are many different formulations of it though. Which are we specifically discussing?
 
  • #26
You can "derive" the constancy of the speed of light from the Lorentz transformation. That is exactly what Lorentz and Poincaré did. Einstein postulated the constancy of the speed of light and derived the Lorentz contraction. One of the key reasons why Einstein's formulation won the day is because he started with a set of simple postulates and let the math fall out from them.

The same goes for general relativity. The equivalence principle was one of the motivating factors in its development. The reason you can "derive" the equivalence principle from general relativity is because it is, by design, built into GR.
 
  • #27
Well I know it was one of Einstein's motivations for coming up with the theory, but what I'd like to see spelled out explicitly is how it is used in the derivation. I realize Einstein of course wanted to have a theory that obeyed the EP, but it seems that the EP is something Einstein originally thought of, then constructed GR, and just one small thing in GR is the EP.

Perhaps you could point me to an article/paper/something that shows how one "let's the math fall out of" the EP, because I can't think of anything off the top of my head.

Is my concern clear? I feel like I'm being vague.
 
  • #28
Nabeshin said:
Reading the wiki article for EP, I see that there are many different formulations of it though. Which are we specifically discussing?
The weak equivalence principle (WEP).
Nabeshin said:
I realize Einstein of course wanted to have a theory that obeyed the EP, but it seems that the EP is something Einstein originally thought of, then constructed GR, and just one small thing in GR is the EP.
But not WEP. I think that originated with Galileo. Both Newton mechanics and GR assume it to be true. But I cannot speak with certainty about GR since I have not studied it to any great length.
 
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  • #29
TurtleMeister said:
laboratory tests have shown (to an accuracy of 1 part in 1012) that gravity (active gravitational mass) affects all matter (passive gravitational mass), including leptons, equally

TurtleMeister said:
The one area where experimental evidence is lacking is the equivalence of active gravitational mass and passive gravitational mass / inertial. And this is something that I have been pondering for some time.

I’ve just read the Wikipedia page on Mass, dealing with that subject and, as many times before, the issue still looks confusing to me.

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…

As to "active gravitational mass", what is the question? Whether it is the same thing that reacts with inertia in a collision? Well, to me that amounts to asking: is gravity caused by the presence of matter, which is supposed to be out of question...

Am I missing anything?
 
  • #30
Saw said:
I’ve just read the Wikipedia page on Mass, dealing with that subject and, as many times before, the issue still looks confusing to me.
You're not alone. I've been trying to make sense of this for a long time. The more I learn, the more confused I am. If I'm understanding you correctly, then I agree with you that the distinction between passive gravitational mass (mp) and inertial mass (mi) seems pointless. They seem to be truly one and the same thing. And current experimental evidence supports it, to a high degree of accuracy. However, the same cannot be said for the equivalence, or more precisely, the equivalence of proportionality between active gravitational mass (ma) and passive gravitational mass (mp) / inertial mass (mi). The best accuracy to support this equivalency is the Kreuzer experiment of 1966 (1 part in 5x105).

Saw said:
As to "active gravitational mass", what is the question? Whether it is the same thing that reacts with inertia in a collision? Well, to me that amounts to asking: is gravity caused by the presence of matter, which is supposed to be out of question...
ma is the magnitude of force that is produced by a quantity of matter. The question of the equivalency of ma to mp/mi also involves another postulate of the EP. Phrased in my own words: The generation of, and the response to gravity are the same regardless of the composition of the masses involved. A violation of this principle is sometimes called "composition dependent gravity". And a test of it is also a test of the WEP. So, if I have 1kg (mi/mp) of Cu and 1kg (mi/mp) of Pb, then the gravitational force (ma) produced by the Cu should be the same as the gravitational force (ma) produced by the Pb. Believe it or not, there is very little laboratory test evidence to support this type of equivalency, even though the technology exists which could extend the constraints by many orders of magnitude.

So where is my confusion? It is in the way that the universal law of gravitation is used as a reason to explain the equivalency of ma. Or, in some cases, the way it is used as a way to test the equivalency. It is usually presented in the following way:

[tex]F_1 = \frac{G m^\text{act}_0 m^\text{pass}_1}{r^2}[/tex]

[tex]F_0 = -\frac{G m^\text{act}_1 m^\text{pass}_0}{r^2}[/tex]

[tex]A_1 = \frac{F_1}{m_1} = \frac{G m^\text{act}_0}{r^2}[/tex]

[tex]A_0 = -\frac{F_0}{m_0} = -\frac{G m^\text{act}_1}{r^2}[/tex]

Notice that if m0 is of a different composition than m1, and if m0 has a different active gravitational mass but the same inertial/passive mass as m1, then A0 <> A1. This would lead to a violation of the third law of motion, the reaction would not equal the action. So obviously an inequality of ma and mp / mi is impossible. Or is it?

Consider the following thought experiment:

I am sitting in a chair on a frictionless surface. You are sitting in a chair across from me on the same frictionless surface. I reach out and pull you toward me. The result is that we both meet at our COM, or barycenter. Now we repeat the experiment, except this time you reach out and pull me toward you. The result is the same. We both meet at our COM. We repeat the experiment again, except this time we both pull simultaneously, me pulling with a slightly greater force than you. The result is exactly the same. We will always meet at our COM.

And, consider this thought experiment:

Two magnets, M1 and M2, are sitting on a frictionless surface with their opposite poles facing each other. Both magnets have the same inertial mass mi, but M1 produces a stronger magnetic field than M2. Now, will M2 have a higher acceleration than M1? Where will they meet? Of course they will have the same acceleration and they will meet at their COM (which is determined by their mi).

Now my question is, referring to the previous equations, why shouldn't gravity work the same as the thought experiments? Why should there be a violation of the third law when the two masses do not have equivalent ratios of mi and ma? It is the inertial mass mi and the inertial mass only, that determines the barycenter. The ma has nothing to do with it.

My thoughts on this are that Newton's universal law of gravitation assumes that WEP is true. To try and use this equation in a world where WEP is not true, will not work. So contrary to what I have read, I do not believe that this equation can be used as proof that ma = mp / mi. I have some other ideas about this, but I am reluctant to post them here because of forum rules.
 
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  • #31
TurtleMeister said:
You're not alone. I've been trying to make sense of this for a long time. The more I learn, the more confused I am. If I'm understanding you correctly, then I agree with you that the distinction between passive gravitational mass (mp) and inertial mass (mi) seems pointless.
The distinction becomes clearer if you were to represent one of the object's gravitational field in the form of a potential.

The best accuracy to support this equivalency is the Kreuzer experiment of 1966 (1 part in 5x105).
Direct test, yes. However, lunar laser ranging provides an indirect but much more precise test of the equivalence between passive and active gravitational mass. The centers of mass of the Moon's crust and core are not collocated. The Moon would self-accelerate if passive and active gravitational mass were not equivalent (at least for the distinctly different material that form the Moon's crust and core). That self-acceleration was not observed, at least to within the incredible accuracy of the range measurements.
Consider the following thought experiment: ... We will always meet at our COM.
This is not true if passive and active mass are not equivalent. The non-equivalence would make gravity will disobey Newton's third law, or alternatively, the conservation laws.
 
  • #32
D H said:
Turtle said:
You're not alone. I've been trying to make sense of this for a long time. The more I learn, the more confused I am. If I'm understanding you correctly, then I agree with you that the distinction between passive gravitational mass (mp) and inertial mass (mi) seems pointless.
The distinction becomes clearer if you were to represent one of the object's gravitational field in the form of a potential.
A gravitational field is the product of active gravitational mass (ma). We were talking about mp and mi. Are you saying that mp has gravitational potential?
D H said:
Turtle said:
The best accuracy to support this equivalency is the Kreuzer experiment of 1966 (1 part in 5x105).
Direct test, yes. However, lunar laser ranging provides an indirect but much more precise test of the equivalence between passive and active gravitational mass. The centers of mass of the Moon's crust and core are not collocated. The Moon would self-accelerate if passive and active gravitational mass were not equivalent (at least for the distinctly different material that form the Moon's crust and core). That self-acceleration was not observed, at least to within the incredible accuracy of the range measurements.
Yes, I am aware of this test. It is based on Newtons laws, which assume WEP to be true. Why should we test WEP based on an equation that already assumes it to be true?
D H said:
Turtle said:
Consider the following thought experiment: ... We will always meet at our COM.
This is not true if passive and active mass are not equivalent. The non-equivalence would make gravity will disobey Newton's third law, or alternatively, the conservation laws.
Again, this can also be used to show that Newton's laws only work when WEP is true. One can easily look at the equations and see that they were not designed to work with mass that has an unequal ratio of ma to mp / mi.

Newton did not write the laws and then have the universe obey. He built his laws based on observation and experimentation. And modern experimentation has already proven them wrong (relativity).
 
  • #33
TurtleMeister said:
A gravitational field is the product of active gravitational mass (ma). We were talking about mp and mi. Are you saying that mp has gravitational potential?
Oops. My response was off-topic to that particular issue. Describing an object's gravitational potential applies to active gravitational mass.

Yes, I am aware of this test. It is based on Newtons laws, which assume WEP to be true.
That test did not assume Newton's laws. It was (is) a test of the strong equivalence principle: general relativity. The measurements are measurements of the distance between the Earth and Moon; about the only thing assumed in those measurements is the constancy of the speed of light and the correctness of the clock. There is nothing in those measurements that assumes any form for gravity. Gravitation, and particularly the equivalence principle, predicts values for those measurements. In particular, a particular wobble would have been observed if the equivalence principle was not true. No such wobble was observed. What this means is that the the equivalence principle is consistent with the observed data. That is the best science can do. While scientific theories can be disproved (all it takes is one stupid experiment), they cannot proven true.

If you're interested, you want to read up on falsification and statistical hypothesis testing.
 
  • #34
D H said:
That test did not assume Newton's laws. It was (is) a test of the strong equivalence principle: general relativity. The measurements are measurements of the distance between the Earth and Moon; about the only thing assumed in those measurements is the constancy of the speed of light and the correctness of the clock. There is nothing in those measurements that assumes any form for gravity. Gravitation, and particularly the equivalence principle, predicts values for those measurements. In particular, a particular wobble would have been observed if the equivalence principle was not true. No such wobble was observed. What this means is that the the equivalence principle is consistent with the observed data.

Is this the test you are referring to: http://prola.aps.org/abstract/PRL/v57/i1/p21_1?
A limit is established for the violation of the equality of passive and active gravitational mass. Our test is based on an asymmetry in the composition of the moon. We hypothesize that the 2-km offset between the moon’s center of figure and center of mass indicates an asymmetry in the distribution of Fe and Al. Unless the Fe on one side and the Al on the other attract each other with the same force, the moon will not be in the orbit predicted by classical mechanics. Using the results from laser ranging and a model for the moon’s interior we find that the ratios of active to passive mass for Fe and Al are the same to a precision of 4×10-12.
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 (ma) then there will be a self acceleration of the moon, causing it's orbit to be different from the one predicted by classical mechanics.

This what I am talking about in my post #30. Please look at the equations and the three sentences that precede and follow it.
 
  • #35
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...

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.

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.

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.

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.
 
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