Galileo was technically WRONG

[Fednis48]

When debating, I always try to reduce a disagreement to its essence. For Aqueous Id, I think this is it:

Only once you've chosen a coordinate system. I gathered from Fednis48 that you are setting the origin of your system at the c.g. That will invalidate any results you get, unless you take your results and convert them to Galileo's coordinates. Until then, there is no basis for comparing results, since relative quantities can not be taken from one frame of reference to another without a coordinate transformation.

You clearly have a solid grasp of Newtonian mechanics, so I'm sure you understand the difference between frame-dependent and frame-independent quantities. That makes your argument all the more puzzling. Relative position and its derivatives [relative velocity, relative acceleration, etc.] are frame-independent. The time between two events [for instance, dropping a ball and that ball hitting the Earth] is frame-independent. So when I show a fall-time difference in the center-of-mass frame, I necessarily show a difference in all frames. Galileo's mistake/entirely-reasonable-approximation was to work in an inertial, Earth-centric frame. This gives a different result because the Earth is not quite inertial in this problem, so such a frame is technically unphysical.

 

[Fednis48]

It will be interesting to note, how many people on this thread agree to the fact that earth and ball (as in OP) will have different acceleration in free body analysis without violating Newton's any law, as shown in my calculations ??

Those who disagree, along with Fednish48 and RJbeery, should google search the "Pseudo Force" and try to understand the same beyond wiki....

I am quite surprised at the complete lack of understanding of basic Newtonian mechanics as shown by many members, and these very members appear to be quite at ease with Hamiltonian ?? what a farce....

This is science forum guys and you are letting a Tom come, and declare Galileo wrong with some silly and outright incorrect math/Physics. This is not a kind of implicit admission that yeah, may be, poor Galli was technically wrong..My foot...This is a kind of admission that we are ignorant, we could not collectively prove this Tom wrong. As some one said in this thread.. I can only bring water near to the horse mouth, but I cannot make him drink....

The astute observer will notice that your post contains no actual arguments of any kind. Please provide some. And while you're at it, stop talking to me like I'm some naive amateur.

 

[Aqueous Id]

When debating, I always try to reduce a disagreement to its essence. For Aqueous Id, I think this is it:

Only once you've chosen a coordinate system. I gathered from Fednis48 that you are setting the origin of your system at the c.g. That will invalidate any results you get, unless you take your results and convert them to Galileo's coordinates. Until then, there is no basis for comparing results, since relative quantities can not be taken from one frame of reference to another without a coordinate transformation.

Ok so if that lost you we are in a heap of big trouble.

You clearly have a solid grasp of Newtonian mechanics, so I'm sure you understand the difference between frame-dependent and frame-independent quantities.

I thought I did until you said:

That makes your argument all the more puzzling. Relative position and its derivatives [relative velocity, relative acceleration, etc.] are frame-independent.

Well, maybe we need to figure out what you mean by prefixing "relative" to concepts all known to be relative without the prefix.

But let me try this: a tennis player serves the ball from a court on a cruise ship as the ship passes an observer on the docks. The observer sees the same velocity of the ball that the referee at the net sees?

Also: why are you inventing your own language and procedures? What's wrong with the ones already commonly used? Are you winging it or is there some unexpressed purpose to your presentation style?

The time between two events [for instance, dropping a ball and that ball hitting the Earth] is frame-independent.

This is a Galilean relativity problem. No one is talking about SR.

So when I show a fall-time difference in the center-of-mass frame, I necessarily show a difference in all frames.

You just got through telling me your origin is in motion throughout the experiment, relative to Galileo.

Galileo's mistake/entirely-reasonable-approximation was to work in an inertial, Earth-centric frame.

Galileo [so goes the legend] dropped the rocks [or whatever] from the Leaning Tower of Pisa. It was by definition an Earth centered frame.

This gives a different result because the Earth is not quite inertial in this problem,

Cite? We are talking ab0ut the reference frame, not the Earth itself.

so such a frame is technically unphysical.

What does that even mean?
Cite?

 

[OnlyMe]

Galileo did not have a valid reference frame because it was not inertial during the experiment. You're starting to sound like a bit of a blowhard.

RJ, Galileo's FoR is as valid as any FoR. That is one of the lessons that SR teaches. And his conclusions were and are valid.

And as long as you include in your hypothetical objects that originate on the earth, their mass has no valid place in the math other than in determining their inertial resistsance to accelerations.... Unless you find a way to subtract the mass of the dropped object from the earth's total mass, which would give you differnt masses for the earth in each hypothetical where the object originated on the earth.

When you raise a bowling ball 10 meters above the surface of the earth, it does not change the the total mass or the gravitational potential at that point above the earth. You have added no mass to the earth's total gravitational mass. As presented in the OP the gravitational field is solely the result of the earth's graviationnal mass. The field itself is a one body problem.

There is some truth to part of what you have been tring to present but it requires that the mass to be dropped, or originate somewhere other than as a part of the earth's orginal mass.

One further point that would have to be addressed, is how the inertial relationship that exists between any object on earth or held a fixed close distance above the surface of the earth before it is dropped, affects the results of the drop. Anything held in place even ten meters above the earth, has to be considered inertially locked with both the earth's axial rotation and its orbital velocity.

All of the above, other than my position re Galileo, is nit picking. As I said briefly above there is merit to some of your arguement, but not all of your conclusions..., and some of the detail you include, like black holes just further muddies the discussion.

 

[tashja]

I sent a link of the thread to Prof. Lawrence Krauss and this is his reply:

Depends on the question you want to answer.. Clearly Galileo was technically right in that all objects experience the same acceleration toward the earth since the force is proportional to mass and acceleration is proportion to force/mass.. So the acceleration of all objects is the same, as observed in any inertial non-accelerating frame.. But clearly indeed the time it takes to hit the ground is different if one considers the motion of the earth too... But that involves a more subtle question since if one is considering the ground as the reference point then that is a non-inertial frame since the earth is accelerating too..

best
LMK

Lawrence M. Krauss
Foundation Professor
Director, The ASU Origins Project
Co-Director, Cosmology Initiative
Assoc. Dir, Beyond Center
ARIZONA STATE UNIVERSITY

 

[RJBeery]

Note, no one knows if Galileo was even involved in this, it's only a legend.

The reference frame the experiment uses is Earth-based. So by definition it's inertial:

In Newtonian physics and special relativity, an inertial frame of reference [or Galilean reference frame] is a frame of reference in which Newton's first law of motion applies: an object moves at a constant velocity unless acted on by an external force.

http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Inertial_frame_of_reference.html​

I'm just posting facts relevant to the discussion. Feel free to trump them with better facts.

You're making the same mistake that Galileo [supposedly] did; the one that I identified in the OP over 100 posts ago. You can make the Earth a reference frame by definition but that is a convenience, not a technically and infinitely accurate presumption.

The reason most people get this wrong is because it is presumed that the Earth does not also move towards the falling object. It should be obvious that this is the case when the "falling object" is extremely massive (i.e. a black hole) - QED

You tell me to pick a frame of reference. I did. Then you tell me to do it again. I DID. Then you say that we could choose the Earth if we wanted, but many people have already explained many times that additional adjustments to your calculations would be required if you wanted to do this. The simple Newtonian analysis acts as if the Earth did not move as I attempted to illustrate around 60 posts ago with the following picture:

 

[RJBeery]

BRILLIANT, Tashja! Maybe we can eliminate the nay-sayers by inundating them with truth until they just get tired and go away.

Galileo's contribution was debunking the idea that an object's acceleration toward the Earth was a function of its mass, which is without doubt. Inertial mass = gravitational mass.

I frankly don't know if Galileo ever actually said "FALL TIMES TO EARTH ARE INDEPENDENT OF MASS", which strictly differs, but this statement is simply false.

Note that Krauss makes my EXACT POINT.

I sent a link of the thread to Prof. Lawrence Krauss and this is his reply:

Depends on the question you want to answer.. Clearly Galileo was technically right in that all objects experience the same acceleration toward the earth since the force is proportional to mass and acceleration is proportion to force/mass.. So the acceleration of all objects is the same, as observed in any inertial non-accelerating frame.. But clearly indeed the time it takes to hit the ground is different if one considers the motion of the earth too... But that involves a more subtle question since if one is considering the ground as the reference point then that is a non-inertial frame since the earth is accelerating too..

best
LMK

Lawrence M. Krauss
Foundation Professor
Director, The ASU Origins Project
Co-Director, Cosmology Initiative
Assoc. Dir, Beyond Center
ARIZONA STATE UNIVERSITY
======================================
As far as I'm concerned, EVERY POSTER following this message still denying the point I'm trying to make in this thread is either blind, willfully ignorant, or simply playing stupid for attention.

 

[Motor Daddy]

RJ, If the length of the chain is 4.9 meters, and you've been on earth doing this experiment over and over, getting the same result, that it takes 1 second according to your watch for the brick wall to hit the earth, then what seems to be the problem? You've done this over and over and the time is 1 second for a brick wall to hit the earth when dropped from 4.9 meters. That is an acceleration rate of 9.8 m/s^2. The brick wall hit the earth at a velocity of 9.8 m/s 1 second after the release from 4.9 meters above.

If the distance started at 10 meters, then how much time does it take for the wall to hit the earth, using the same watch standing on the same earth as the other tests were performed? How high above the earth is the wall at 1.0345 seconds after being dropped from 10 meters above using an acceleration rate of 9.8 m/s^2? What is the wall's velocity at t=1.0423 seconds after being released?

 

[RJBeery]

As far as I'm concerned, EVERY POSTER following this message still denying the point I'm trying to make in this thread is either blind, willfully ignorant, or simply playing stupid for attention.

Which is it, Motor Daddy? Perhaps a bit if all three for you eh?

Anyway the brick wall cannot move. It's magical, and the chain is massless and virtually infinite in length. This imaginary, absurd scenario is what would be REQUIRED for fall times to Earth to be independent of mass. Now please drop the subject, it has been established beyond all reasonable doubt.

 

[Motor Daddy]

Which is it, Motor Daddy? Perhaps a bit if all three for you eh?

Anyway the brick wall cannot move. It's magical, and the chain is massless and virtually infinite in length. This imaginary, absurd scenario is what would be REQUIRED for fall times to Earth to be independent of mass. Now please drop the subject, it has been established beyond all reasonable doubt.

So no answers, just more BS? How about you answer my questions? I've given you the tools to do so, now get hot!

 

[Fednis48]

There is some truth to part of what you have been tring to present but it requires that the mass to be dropped, or originate somewhere other than as a part of the earth's orginal mass.

I think I can safely speak for RJ when I say we both agree with this. Since you dropped your assertion that the moon would fall at the same rate as a baseball, I've had no disagreement with you. But yours was not the only competing argument in this thread; Aqueous Id is still arguing that fall time depends on your choice of coordinates, and RajeshTrivedi is still arguing that the ball shares the Earth's upward acceleration. http://xkcd.com/386/

Well, maybe we need to figure out what you mean by prefixing "relative" to concepts all known to be relative without the prefix.

But let me try this: a tennis player serves the ball from a court on a cruise ship as the ship passes an observer on the docks. The observer sees the same velocity of the ball that the referee at the net sees?

Of course that's false. The positions and velocities of single objects depend on one's choice of coordinates, so different observers can see the ball moving at different speeds. But relative position, a.k.a. the distance between two objects, mathematically $$x_{rel}= X-x$$, is independent of coordinate choice in Newtonian physics. In your example, for instance, all observers should agree on the velocity of the ball relative to the referee, even as they disagree on the absolute velocity of the ball. Because time and \(x_{rel}\) are both independent of coordinate choice, it is also true that all time-derivatives of \(x_{rel}\) are independent of coordinate choice. My formula gives relative acceleration as a function only of relative position, so it holds true in all coordinate systems.

Also: why are you inventing your own language and procedures? What's wrong with the ones already commonly used? Are you winging it or is there some unexpressed purpose to your presentation style?

You lost me here. What part of my language/procedure is invented? Hamiltonian dynamics are a standard method in classical mechanics; they're probably the single most useful thing I learned from my Classical Mechanics course as an undergrad. Relative coordinates are a long-standing trick, especially for two-body problems, and I use them at work routinely. The rest of what I did was literally just applying Newton's law of gravity and the kinetic energy formula to a falling ball.

This is a Galilean relativity problem. No one is talking about SR.

Nor am I. Maybe "reference frame" was a poor choice of words; I meant Newtonian reference frame, i.e. choice of coordinates.

Galileo [so goes the legend] dropped the rocks [or whatever] from the Leaning Tower of Pisa. It was by definition an Earth centered frame.

Cite? We are talking ab0ut the reference frame, not the Earth itself.

Yes, it was by definition an Earth-centered frame. But in this experiment, the Earth accelerates upward, so it is not inertial. Therefore, any frame centered on the Earth throughout the experiment must also not be inertial. You can't make an accelerating frame inertial by definition, because Newton's laws are modified in a non-inertial frame. When he treated the Earth-centered frame as inertial, Galileo was making a very good but strictly imprecise approximation. My solution, which actually is in an inertial frame, does not suffer the same issue.

RJ, If the length of the chain is 4.9 meters, and you've been on earth doing this experiment over and over, getting the same result, that it takes 1 second according to your watch for the brick wall to hit the earth, then what seems to be the problem? You've done this over and over and the time is 1 second for a brick wall to hit the earth when dropped from 4.9 meters. That is an acceleration rate of 9.8 m/s^2. The brick wall hit the earth at a velocity of 9.8 m/s 1 second after the release from 4.9 meters above.

If the distance started at 10 meters, then how much time does it take for the wall to hit the earth, using the same watch standing on the same earth as the other tests were performed? How high above the earth is the wall at 1.0345 seconds after being dropped from 10 meters above using an acceleration rate of 9.8 m/s^2? What is the wall's velocity at t=1.0423 seconds after being released?

In RJ's illustration, the wall isn't falling to the Earth. It's supposed to act as an anchor point to hold the Earth in place. This is to highlight the contrast with the actual situation, in which the Earth can freely accelerate toward the ball as the ball falls.

 

[Motor Daddy]

In RJ's illustration, the wall isn't falling to the Earth. It's supposed to act as an anchor point to hold the Earth in place. This is to highlight the contrast with the actual situation, in which the Earth can freely accelerate toward the ball as the ball falls.

If gravity is present according to the way you think it works, the wall and the earth will come together. The only questions are how far apart they are at any point in time, and how far they are apart (or making contact) at another point in time. The interval of time between the start point in time and the end point in time is the elapsed time from release to impact. The distance between the two at the start point in time of release is the distance traveled (according to an earth observer using it as an Einstein embankment) until impacting the earth. If the earthling decides to get sneaky and try a trick and use the absolute frame (space coordinates of axis (x,y,z)) then they are only fooling themselves. The TIME doesn't change, but if the earthling decides to change to distance traveled in the absolute frame, then they've done nothing but change what they said from the beginning, ie change from distance in the earth frame to the distance in the absolute frame.

 

[OnlyMe]

I think I can safely speak for RJ when I say we both agree with this. Since you dropped your assertion that the moon would fall at the same rate as a baseball, I've had no disagreement with you. But yours was not the only competing argument in this thread; Aqueous Id is still arguing that fall time depends on your choice of coordinates, and RajeshTrivedi is still arguing that the ball shares the Earth's upward acceleration. http://xkcd.com/386/

Of course that's false. The positions and velocities of single objects depend on one's choice of coordinates, so different observers can see the ball moving at different speeds. But relative position, a.k.a. the distance between two objects, mathematically $$x_{rel}= X-x$$, is independent of coordinate choice in Newtonian physics. In your example, for instance, all observers should agree on the velocity of the ball relative to the referee, even as they disagree on the absolute velocity of the ball. Because time and \(x_{rel}\) are both independent of coordinate choice, it is also true that all time-derivatives of \(x_{rel}\) are independent of coordinate choice. My formula gives relative acceleration as a function only of relative position, so it holds true in all coordinate systems.

You lost me here. What part of my language/procedure is invented? Hamiltonian dynamics are a standard method in classical mechanics; they're probably the single most useful thing I learned from my Classical Mechanics course as an undergrad. Relative coordinates are a long-standing trick, especially for two-body problems, and I use them at work routinely. The rest of what I did was literally just applying Newton's law of gravity and the kinetic energy formula to a falling ball.

Nor am I. Maybe "reference frame" was a poor choice of words; I meant Newtonian reference frame, i.e. choice of coordinates.

Yes, it was by definition an Earth-centered frame. But in this experiment, the Earth accelerates upward, so it is not inertial. Therefore, any frame centered on the Earth throughout the experiment must also not be inertial. You can't make an accelerating frame inertial by definition, because Newton's laws are modified in a non-inertial frame. When he treated the Earth-centered frame as inertial, Galileo was making a very good but strictly imprecise approximation. My solution, which actually is in an inertial frame, does not suffer the same issue.

In RJ's illustration, the wall isn't falling to the Earth. It's supposed to act as an anchor point to hold the Earth in place. This is to highlight the contrast with the actual situation, in which the Earth can freely accelerate toward the ball as the ball falls.

You included three quotes attributed to me that were not from any of my posts. Probably an error in copying the initial quote marker. Maybe you could edit the posts to indicate the quotes true origin/link.

 

[RajeshTrivedi]

But clearly indeed the time it takes to hit the ground is different if one considers the motion of the earth too...... Prof LMK.

It is a difficult situation, but I must respectfully and humbly disagree with professor on this conclusion.

For practical Earth Ball situation [as per OP], I have reason to stick to what I proposed in my earlier post, that is acceleration on the ball will be g-gm/M and acceleration on the Earth will be gm/M, so effective time taken shall be same on this count irrespective of mass of the ball.

The qualitative reasons are...few as came to my mind...

1. Why we think Earth is only up-to surface, it encompasses the ball and beyond.

This gives couple of counters

a. Suppose we create a magnetic floating plate (small rectangular floating at some random 5 cm from the surface), and we drop rubber balls of 0.5 Kg and 6 Kg from a height of 10 meters from the upper surface of this magnetic plate. Whatever distance Earth would travel, will be below this plate only, so the time taken by balls of different mass shall be same in hitting the plate or not ?

b. Extrapolation of (a) above is very simple, the time taken in transit between any two points A and B (before it hits the surface) will be same or not ?

c. The separation between a nearby flying bird from surface will increase or decrease or remain same due to this fall of ball ??

d. It is simply difficult to conceive isolation of any part of earth (the ball included), and give an acceleration of gm/M to the balance excluding this part (ball).

Many more conceivable points can be raised if you think ball and Earth as two different objects. Possibly the presence of air separation between surface and the ball gives a false sense that Earth and ball are two different objects, but from the higher motion perspective (spin and revolution etc), they are not.

I feel professor should not have been quoted in self by the poster...

 

[Trippy]

As far as I'm concerned, EVERY POSTER following this message still denying the point I'm trying to make in this thread is either blind, willfully ignorant, or simply playing stupid for attention.

That's almost amusing.

 

[exchemist]

I'm not "some Tom", and I only consider you ignorant if you blind yourself to facts and reason. This issue has clearly become a litmus test for exposing Physics intuition and the ability to digest new ideas. I'm quite comfortable having rpenner, Fednis, and, most importantly, the truth on my side.

RJB, you can now add me as well. It seems to me that in Newtonian mechanics the outside observer occupies an absolute frame of reference and the "a" of F=ma is defined in terms of this. F acts equally on both bodies independently and if it is such that the larger moves to a non-negligible extent, this will shorten the time from release to contact. But it has certainly made me scratch my head for a few days - good one!

 

[OnlyMe]

Aqueous Id is still arguing that fall time depends on your choice of coordinates,

I agree with Aqueous here, there is no problem with choosing the earth as the frame of reference. This is one of the important things about understanding SR. It does not matter whether you chose the earth, the falling object or as RJ has indicated the center of mass, as your FoR, the result will only differ as to how the observed acceleration is distributed... And that distribution is FoR dependent.

Both objects are in a state of free fall so acceleration can only be determined based on frame of reference.., an accelerometer would not function, as a local indicator of acceleration. In all three FoR the time that the two objects take to fall together, will be the same, only how the associated acceleration is distributed would change. But that may not be the case for all possible FoR, though it would be true for those specifically being discussed.

Though there is a theoretical truth in RJ's hypothetical {which under some circumstances could become a practical or measureable truth}, his choice of including a bowling ball and golf ball, in an argument, perceived as attempting to discrediting Galileo's, choice of FoR and conclusions, was a bad choice, because Galileo's conclusions were based on real world experience and any acceleration the earth would experience when a bowling ball is dropped toward it can only be theoretically defined. The earth's acceleration and distance traveled, during the drop, in that case.., would be insignificant to an extreme.

I could be wrong, but I believe that this was one of the points that rpenner included, in his comments that for me clarified some of my misunderstanding and/or assumptions.

 

[OnlyMe]

RJB, you can now add me as well. It seems to me that in Newtonian mechanics the outside observer occupies an absolute frame of reference and the "a" of F=ma is defined in terms of this. F acts equally on both bodies independently and if it is such that the larger moves to a non-negligible extent, this will shorten the time from release to contact. But it has certainly made me scratch my head for a few days - good one!

Remember, this hypothetical is framed in the context of gravity, which when you start comparring time as measured by multiple frames of reference, starts to get very sticky... You know time dilation and how it affects clocks and time measurements...

When you are talking about a small mass, compared to the total mass of the earth, and an initial separation distance of 10 meters{?}, even Farsight's optical clocks would likely disagree within their indepent margins of error... Even if their individual locations in the gravity well did not make comparrisons completely nonsensical. Even measurements of time are FoR dependent.

The point I am trying to make here, is that for say the bowling ball test, for a clock located at the center of mass of the two body problem and a clock located at the center of mass of the earth, or for practical purpose the surface of the earth and an elevation equal to the distance between those two centers of mass, even Farsight's optical clocks would agree on the elapsed time. Which supports the validity of Galileo's choice of FoR and conclusions.., even though he did not have access to modern clocks, or any understanding of GR.

 

[Neddy Bate]

RJBeery gave an example of a black hole falling to earth, and that makes it pretty clear that he is including non-terrestrial objects in his thought experiment. A black hole certainly would cause the earth to accelerate toward it as it fell toward earth, and the accelerations of both objects should be taken into consideration. (Unless the earth is chained to a magic immovable wall.)

 

[OnlyMe]

RJBeery gave an example of a black hole falling to earth, and that makes it pretty clear that he is including non-terrestrial objects in his thought experiment. A black hole certainly would cause the earth to accelerate toward it as it fell toward earth, and the accelerations of both objects should be taken into consideration. (Unless the earth is chained to a magic immovable wall.)

And I found RJ bringing a black hole into the discussion a bit funny given,

And no, I don't personally believe that black holes exist.

Understanding his position I specifically exchanged a neutron star, for the black hole in an earlier post.., before rpenner's post brought me to my senses.

But yes, a larger mass does change the dynamics as can be seen in how the earth/moon and planets/sun orbital systems affect the systems barycenter or centers of mass.