# Galileo was technically WRONG

Discussion in 'Physics & Math' started by RJBeery, Oct 1, 2014.

1. ### RajeshTrivediValued Senior Member

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I won't be so outright dismissive about what Tasja posted about professors...But the fact is she took the opinion of two different professors, suggesting something fishy.

Moreover the response of professors depends on the status of querist also (as they see). If a child or a layman or a non technical staff, asks some complex question, then professors would be very affectionate and his response may not be very technical or very accurate. Imagine a professor telling some one about trillionth of second when that someone is not able to figure out that Carl Lewis lost the 100 m dub by a fraction of second. Professor would be very guarded if the question comes from a scholar or a peer.

Last edited: Oct 7, 2014

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5. ### Fednis48Registered Senior Member

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Ok, I'm glad you responded; I know you're an intellectually honest fellow, so I assume you'll be able to give me a no-nonsense answer here. And even if you think I'm a crank, please don't waste time saying as much or posing Socratic questions. Just point out the error in the following reasoning:
1. Your source says any accelerometer in an inertial frame will read zero.
2. Your source says any two inertial frames are in a constant state of rectilinear motion with respect to one another.
3. Since the Earth is a single object, all parts of it are in a constant state of rectilinear motion with respect to all other parts of it (at least in Newtonian mechanics, which is the context we're dealing with right now).
4. From 3, any point on the surface of the Earth is in a constant state of rectilinear motion with respect to the center of the Earth.
5. From 2 and 4, a frame centered on the center of the Earth is inertial iff a frame centered on any point on the surface of the Earth is inertial.
6. You claim the a frame centered on the center of the Earth is inertial.
7. By 5 and 6, you claim that a frame centered on any point on the surface of the Earth is inertial.
8. By 7 and 1, you claim any accelerometer at any point on the surface of the Earth will read zero. This is false.
As far as I can tell, this proves that either your claim is wrong or you're misinterpreting your sources. Where is my mistake?
I know, right? Even when I was telling off Aqueous, I didn't expect him to go all Joseph McCarthy on me... Still, I'd like to see your analysis, if and when you finish it up.
This is an excellent question - one that gets right to the heart of what I think you've been confused about. Physics doesn't give us any rules about when we have to treat two things as part of the same object. If we really wanted to, we could treat every atom in the Earth as a separate object, and we should still get the right results (although the calculations would be impossibly hard). Instead, the reverse is true: physics tells us when we can treat two things as part of the same object. Specifically, we can treat two things as one if they act on and are acted on by the rest of the system in all the same ways. In our problem, we're assuming the whole rest of the Earth is locked together into a single, rigid body that accelerates as one; therefore, we can treat it as a single object. Same with the golf ball. But since we're exploring the interaction between the Earth and the golf ball, we cannot treat them as the same object. The Earth exerts a net gravitational pull on the golf ball but not on itself, and vice-versa, so we have to treat the two separately to get the right results. Of course, as you've hinted at before, it's not actually correct to treat the Earth as a single, rigid body, which leads to other corrections much larger than the one being discussed here. That's why RJ and I keep throwing around the word "pedantic" so much; we don't want anyone to get the mistaken impression that we're talking about anything more than an interesting technicality.

7. ### exchemistValued Senior Member

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Aq Id, my dear fellow, you seem to be becoming quite paranoid and seeing creationists under the bed!

If you think Fednis is a crank, or an agent of "SPAZ" (love the concept, by the way!), please do me a big favour and read the exchange he or she and I have had on the virtual particles thread I started yesterday.

Personally, I think this is evidence that he or she is quite clearly a practising physical scientist, who can communicate clearly and authoritatively, at any rate in their own field.

8. ### RajeshTrivediValued Senior Member

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Please refer the color comments of yours above..So let me extrapolate and ask you a different question...Only technical aspect....

We drop two balls (B1 and B2) of masses m1 and m2 (M>>>>m1>m2) simultaneously from a height h = 10 meter. As per your assumption above, can we deduce following, with your permission ??

Acceleration of Earth in upward direction = g (m1+m2)/M.
Acceleration of Ball m1 towards Earth = g - gm2/M
Acceleration of Ball m2 towards Earth = g - gm1/M

[Please see the cross reference of m1 and m2 and this components comes because m1 and Earth is locked for m2 motion and....], if so, then your Physics is very selective...

I am quite convinced that you do not have the right argument to quash my assertion (not assumption) that the acceleration of ball in OP shall be (g-gm/M)..

9. ### RajeshTrivediValued Senior Member

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This thread as OPed by RJBeery and well fought by fednis48 has done wonders in science, it proved following..

1. Galileo was wrong by 10^-20 ms.
2. Newton was wrong by 10^20 factor on reaction force.
3. Einstein space time curvature is observed and proved...

1 & 2 already explained in detail in earlier posts, now read on for 3....

I will further extrapolate the above reasoning of fednis48. Drop two balls simultaneously one from location A and another from a location 90 Degree from that (one from pole and another from Equator). So the path of both these balls shall be a curvature not a radial line. I do not think I have to elaborate, its kind of self explanatory if we stick to locking principle of fednis48. This can be cited as wonderful proof of relativity and spacetime curvature. May be few more balls can be strategically added to get a perfect curved motion of BALLS !! [pun intended]..

Last edited: Oct 7, 2014
10. ### selfjeszRegistered Member

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wow.. thats hardly suprising

11. ### rpennerFully WiredStaff Member

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Assuming each body is spherically symmetric, and h is measured surface-to-surface, $m_1$ is the dropped object, and $m_2$ is the much larger planet mass. Neglecting air.

$t(m_2, r_1, r_2, h) = \sqrt{ \frac{(r_1 + r_2 + h)^3}{2 G m_2}} \cos^{\tiny -1} \frac{r_1 + r_2}{r_1 + r_2 + h} + \sqrt{ \frac{(r_1 + r_2 + h)(r_1 +r_2) h}{2 G m_2}}$
$\frac{\partial t}{\partial m_1} = 0$

Example: $m_1 = 1 \textrm{kg}, \; m_2 = 6 \times 10^{24} \textrm{kg}, \; r_1 = 0, \; r_2 = 6 \times 10^6 \textrm{m}, h = 5 \textrm{m}, G = 6.673 \times 10^{-11} \textrm{N} \cdot \textrm{m}^2 \cdot \textrm{kg}^{-2}$ Then $t = 1.14462 \textrm{s}$

This type of calculation is more suited for orbital dynamics than actual analysis of terrestrial experiment.

$t(m_1, m_2, r_1, r_2, h) = \sqrt{ \frac{(r_1 + r_2 + h)^3}{2 G ( m_1 + m_2)}} \cos^{\tiny -1} \frac{r_1 + r_2}{r_1 + r_2 + h} + \sqrt{ \frac{(r_1 + r_2 + h)(r_1 +r_2) h}{2 G ( m_1 + m_2)}}$
$\frac{\partial t}{\partial m_1} = - \frac{t}{2 ( m_1 + m_2 ) } \neq 0$

If $m_1 = 50 \textrm{kg}$ and $m_2 = 6 \times 10^{24} \textrm{kg}$ then $\Delta t_{m_1 \to k m_2} \approx \frac{\partial t}{\partial m_1} \times (k m_1 - m_1 ) = - \frac{(k-1) m_1}{2 (m_1 + m_2)} t$ and for k = 2, $\frac{\Delta t_{m_1 \to 2 m_2}}{t} \approx - \frac{m_1}{2 (m_1 + m_2)} \approx -4 \times 10^{-24}$.

This is far beyond human precision experimental results, especially in the time of Galileo. Thus this thread is dominated by pedantry and talking past each other.

http://en.wikipedia.org/wiki/Galileo's_Leaning_Tower_of_Pisa_experiment

Last edited: Oct 7, 2014
12. ### Fednis48Registered Senior Member

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Thanks for the backup!

Also, I really like the idea of "creationists under the bed." Sounds likes a Goosebumps story, or maybe a Doctor Who episode.
No, we can't deduce this at all! What I said was that we're interested in how the ball interacts with the Earth, so we have to treat the ball and the Earth as separate objects. On the other hand, we don't really care about the dynamics of the system beyond the net acceleration of the ball towards the Earth, so we treat them as a pair of single, rigid bodies and only concern ourselves with the displacement between their centers. Taking a more nuanced treatment of Earth can lead to other corrections - most notably air resistance - many of which are larger than the effect being discussed here. But the effect you describe - where the ball effectively accelerates upward under its own gravity just because it is "part of" the Earth - will never appear, no matter how precisely you treat the system. Why? Because there is not a single law in all of physics that cares whether two objects are "really" part of the same thing. It's just not a meaningful physical statement. If you want to derive an effect like the one you describe by looking at air resistance or tidal forces or something else we've been ignoring, go for it! More power to you. But it is never correct to add in extra force terms just because of how you feel the objects in a problem should be grouped together.
Yes, both balls will curve, albeit with the same caveats of extreme technicality as the difference in fall time.

Also, the 10^20 error in the OP is a typo, probably from dropping a piece of scientific notation somewhere. Don't read too much into it - the important point is just that the effect is nonzero, which is true with or without the typo.

13. ### OnlyMeValued Senior Member

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So, Galileo was not wrong and the underlying point, setting aside the Galileo issue, that RJ was pointing out is accurate, with the caveat that any difference in the fall times for the bowling and golf ball are insignificant and thus would be measured as consistent with Galileo's conclusions.., even where a theoretical deviation from those measurements, might be present.

You are correct, in that this discussion has largely been arguments talking past eachother. I don't think that the two body issue was ever the real argument, even theoretically. It from my point of view is only how the insignificant application of the theoretical aspects were applied to a conclussion that Galileo was technically wrong.

And I played as big a role in that talking past RJ's argument, as anyone.

Again, I am often impressed by the clarity of many of your posts. I seldom spend the time and reflection when posting that it seems you invest, or that I would require to even approach a similar clarity of intent and purpose.

14. ### TrippyALEA IACTA ESTStaff Member

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This is the exact same point I made (as well as the number I came to). The distance is something like 1/2000th the radius of a proton, or something comparably ridiculously small. Is such a small measurement even meaningful?

Such a pity that RJBeery has left the conversation already, and such a pity that I hadn't checked out his source (Irony: I lacked the time) before he did so. This is the exact point that I was trying to make to him, that the discussion of the time was around falling rates rather than falling times.

Last edited: Oct 7, 2014
15. ### Aqueous Idflat Earth skepticValued Senior Member

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Thanks for that feedback, exchem. Lately I've been willing to forego overreacting to posters I don't know. If Fednis really has training, then it means I should be able to arrive at some common ground, even if only by definition of terms. My rationale went like this: first, I've seen RJ's posts quite a bit, and it never dawned on me that he was a crank until now. Then when I saw RJ summoning Fednis to the rescue, my SPAZometer started pegging. That's one of the benefits of having "sleeper" sock puppets that can be summoned to shore up the crank argument. Having seen how hard it has been for the mods to root out and put an end to the SPAZzes, and how manic they are, it seems a worthy cause for me to risk being abrasive to people I would ordinarily have only wanted to reach out to.

Again, thanks. I value your opinions a lot, first of all because you knew who Josquin des Prez was--which may seem a little silly to you, but these are the bits of common ground that serve to "identify friend or foe". Plus, it speaks to another world we don't get to talk about much here, one that's just freaking awesome.

Your mistake is trying too hard to nail me with "any" accelerometer, since I did not clarify that we calibrate our accelerometers against g. But first this:

Based on the positive character reference from exchem I'm going to zeroize my SPAZometer reading of you. Let me restate a few things that set it off and we can put this behind us. First, I saw what I thought was an attempt to arrive at relative acceleration, by unnecessary and superficial application of the Hamiltonian. One of the markers of SPAZdom is to wave jargon around like this for no purpose but to try to hack the speech of science itself in order to get traction in a thread, so they can set their bait traps out. My point was that you only needed to state the allegations directly, something like this:

$F\quad =\quad ma\quad =\quad \frac { G{ m }_{ \oplus }{ m }_{ 0 } }{ { r }^{ 2 } } \\ \\ { \therefore \quad a }_{ rel }\quad \equiv \quad { a }_{ 0 }\quad -\quad (-{ a }_{ \oplus })\quad =\quad \frac { G({ m }_{ \oplus }+{ m }_{ 0 }) }{ { r }^{ 2 } } \\ \\ r\quad =\quad \frac { 1 }{ 2 } \quad { a }_{ rel }\quad { t }_{ 0 }^{ 2 }\\ \\ \therefore \quad { t }_{ 0 }\quad =\quad \sqrt { \frac { 2r }{ { a }_{ rel } } } =\quad \sqrt { \frac { 2h{ r }^{ 3 } }{ G({ m }_{ \oplus }+{ m }_{ 0 }) } }$

Now we just need to decide whether this is valid. I say no, that the notion of relative acceleration is an artifact of a non-inertial reference frame. To arrive at the truth of the matter, we should be able to dispense with the confusion over what constitutes an inertial or non inertial frame, and we should be able to test the truth of which opinion is correct by using the correct coordinate transformation that ports the vectors between Galileo's frame and the thing I'm labelling non-inertial, which I call RJ-space. I will try to avoid the unnecessary calculations, although I might come up with a less tedious way to do it. I just prefer the transform approach, since it seems to be the main deficit of cranks and SPAZzes, and a little education might go a long way to disarming them the minute they pop up.

So for now I'll simply rely on the more scholarly method, which rejects vector math done outside of inertial reference frames as something "warped" by the nature of the coordinate transformations involved. That way, at least the educated people will follow my logic. And it gives RJ an incentive to feast on the mathematical analysis he's laid on his plate, should he be a glutton for punishment.

First, let's restate the operative assumptions. RJ is assuming the Earth is a perfect sphere of uniform density, free from any rotational or orbital acceleration.

Now let's pick a reference frame. The acceleration at the center of the Earth is ~0.05%g with an additional ~0.3%g at the equator. RJ is neglecting these. We begin by placing a Cartesian coordinate system at the center of an Earth-sized sphere. We pick the point where the z-axis intersects the sphere and we designate this the position of impact.

A frame at the center of the sphere is inertial because, in RJ's model, it is not accelerating. An accelerometer placed there reads zero. An accelerometer placed on the surface reads -1g in the z-direction. We calibrate the accelerometer to remove the -1g bias, and thereafter it serves as a benchmark, against which we can validate all other reference frames. We can now place the accelerometer anywhere on the surface of our spherical Earth and it will continue to read zero. Further, we can place the accelerometer in constant rectilinear motion and after a brief spike, the accelerometer will read zero. Thus we find Galileo is in our inertial reference frame, at a distance h from our origin directly below him at the surface of the sphere, which is the vertical height from which he drops the object. We run the experiment and conclude that the time it takes from release until impact is

${ t }_{ 0 }\quad =\quad \sqrt { \frac { 2h }{ g } }$

But RJ is not in an inertial reference frame. He has chosen as his origin the center of mass of the Earth-object system. When we place our accelerometer at the origin of RJ-space, it at first reads zero, but then, as the experiment proceeds, his origin begins to accelerate. The accelerometer reads nonzero. Alternatively, we can say RJ-space is non-inertial since it is accelerating against a known inertial reference, the center of our idealized spherical Earth.

Conclusion: the relative acceleration method of calculating the time of impact is invalid since it relies on a non-inertial reference frame. Galileo was right and RJ was wrong.

Last edited: Oct 7, 2014

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Thank you!

17. ### OnlyMeValued Senior Member

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Aqueous, I have no issue with anything upto the paragraph above or the conclusion that followed. I don't quite see the two body center of mass as anymore non-inertial than the earth's mass would be, if it were the FoR. It seems that as long as the hypothetical begins with the to be dropped object in a fixed postion relative to the earth, those two centers of mass would both begin as inertial... Then as the object is dropped the earth's center of mass undergoes an acceleration. In RJ-space both the object and the earth, once the object is dropped, accelerate toward the center of his inertial frame of reference.(?) As long as this is a two body problem with no orbits, axial rotation or other external forces, the two body center of mass should remain inertial, even when it is indistinguishable from the earth's center of mass.

As rpenner has pointed out twice, at least indirectly, this detail would not change any conclusions unless the dropped object was of significant mass. Most of my (coherent) objections (if ther were any), were intended to focus on the point that in terms of Galileo's experiment, both centers of mass have to be regarded as inertial, since they would be indistinguishable.

Edit: re: the conclusion.., I don't think the FoR issue changes the conclusion that RJ's two body problem does not falsify Galileo's conclusion..

Last edited: Oct 7, 2014
18. ### rpennerFully WiredStaff Member

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What external force operates on the Earth-object system? None. Then by Newton's first law, the center of mass moves inertially and thus is valid as a standard of rest for an inertial reference frame.

19. ### Aqueous Idflat Earth skepticValued Senior Member

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No, thank you! (The old school oriental dudes trying 1-downsmanship with all the bowing heh heh).

I think I see your point: I think you are thinking that since the center of mass is the place where forces cancel, it should be a good choice of inertial reference frame. So far so good. But when the the ball is released, that origin begins to leave (I would say "diverge" if we were talking about SR) and to return to the center of Earth at at a rate that we know is not uniform, since the thing causing it to shift is the falling ball, and that happens at the nonlinear rate g. Consequently we have to declare their frame to be non-inertial, since it is accelerating against Galileo's origin, which, if you accept it as inertial, becomes the benchmark.

Notice how my analysis, unless I screwed up somewhere, is immune to the absurdity of scale that comes from the idea of moving the Earth by the gravity of a feather. I'm allowing for very large and very small numbers to be mixed. I'm only examining the question of whether RJ's origin went into acceleration, and I deduce that it does, since it moves at a rate which is a function of the gravitational acceleration of the featherweight against my benchmark, which is the spot on the ground where it lands, where I've placed Galileo's origin, and which I've concluded is an inertial reference frame. Once I detect that a second frame is accelerating against an inertial benchmark, then I have no choice but to declare it as an invalid, non-inertial reference frame. That's my argument, anyway.

20. ### Neddy BateValued Senior Member

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If you want a truly inertial frame, why not choose one which is very far away from all gravitating bodies. Choose it such that both masses $m_1$ and $m_2$ are stationary in the instant before the test mass is dropped. Do you see that both objects experience an attractive force toward one another?

$F=\dfrac{G m_1 m_2}{r^2}$

Do you see that both objects will accelerate toward one another, as measured by the inertial frame I chose to start in?

$a_1=\dfrac{F}{m_1}$

$a_2=\dfrac{F}{m_2}$

And finally do you see that an observer on earth, taking the earth as his stationary reference frame, must therefore measure the acceleration of the falling object to be the sum of both accelerations?

$a = a_1 + a_2$

Neglecting air resistance of course. Also, I'm treating these as scalars for simplicity.

21. ### OnlyMeValued Senior Member

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Take another look at the issue, when the object is dropped both the earth and the object fall toward their common center of mass. That center of mass does not change. The accelerations experienced by the earth and the object change depending on their relative mass.

RJ's thought experiment is technically correct, his use of it as a means to claim that Galileo's conclusion is technically wrong, is not. For small objects the two FoR would be indistinguishable throughtout. It does not matter whether you are examining the issue from a Newtionian or GR perspective.

22. ### Neddy BateValued Senior Member

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Yes, thank you.

Correct, but RJBeery was not limiting his scope to small objects.

23. ### OnlyMeValued Senior Member

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For dropped objects of large mass that would be fine. The problem with a small mass being dropped is that the inertial center of mass of the two body system is indistinguishable from the earth's center of mass and any acceleration the earth might undergo is so insignificant, it might even be lost in the zitterbewegung motion of the fundamental particles, at the center of mass, the two frames of reference share.