Thank you for responding noAxioms

Nevertheless, the mirror thing can be falsified. You just have two mirrors in a vacuum in the same place moving relative to each other. Shine a light pulse at it and detect the reflected light from each. If they arrive at the same time (but different wavelength/frequency), then light speed is not a function of the motion of the mirrors. If light from the approaching mirror gets there first, then we need to rewrite the last 130 years of physics. — noAxioms

Did they ever do this experiment? What is it called? It seems that a very simple version would be a cylindrical mirror rotating , where two laser light beams reflecting the rotating mirror, one on each side (mirror moving away on one side and forward the other side).
My concern would be whether we would have the technology accurate enough to be able to observe
whether the two light beams would have the same speed or not.

Grampa Dee

Thank you for responding noAxioms

I sent a post concerning this in the “The Newtonian gravitational equation seems a bit odd to me" thread.
— Gampa Dee
Which I did not immediately see because you didn't reference me (reply to something of mine say) anywhere in it. — noAxioms

Oops; sorry...my mistake;

Therefore, it would have predicted the nul result because of this....the light was going to be c relative to the whole experiment
— Gampa Dee
OK, so the M&M setup isn't the optimal experiment to falsify this particle theory. — noAxioms

Well, if they would have observed some fringe interference, the ether would have been validated, and the particle theory disregarded.
But, since no interferences were observed, this was to be explained; and so we had Fitzgerald who claimed the ether contracted the measuring rods, or there was the speed of light as being invariant, but the particle theory of light would have also explained the null result.

If the combined velocity of reflected light, in the reference frame of the laboratory, is (c + v), then the ballistic theory, in question, is a new-source theory, in which starlight loses its initial velocities. By contrast, if the combined velocity of reflected light, in the reference frame of the laboratory, is (c + 2v) instead, then the ballistic theory, in question, is an elastic-impact theory, in which starlight does not lose its initial velocities. — Foraj

OK, I got that. I know the difference between the two now. They're both wrong, but they didn't know it at the time. Not sure if the spectra of binaries can falsify both since apparently the new-source theory produces spectra very similar to relativity theory (reflected light speed is neither c+v nor c+2v, but just c. — noAxioms

Could you show me the experiments which proves this (reflected light has a speed of c)? I'd be interested..

Check the copyright. Is it legal to paste the whole thing here? You already pasted an email address, which is against the rules for some forums.
What question do you need answered? — noAxioms

There is no copyright on the copy I have...but I understand that copy and paste might involve some errors...so I won’t copy and paste the whole document.... I’ll continue to read the document and get back at you if I have any questions....thanks..

Grampa Dee

Thank you for replying,noAxioms

So,the addition of vectors in this case is 0...but relative to what?
As I said, it doesn't yield anything meaningful. I don't like the example text. It obfuscates more than it clarifies anything.
It also mixes metric (kg mass) with American units (mi/hr). That's unforgivable in a physics text.
Their momentum vectors seem to have a ratio of about 4, but the text says the ratio is closer to 7. That's poor graphics. — noAxioms

That’s exactly what I feel; there’s nothing meaningful...I don’t have any problems with the addition of the two vectors (tip to tail) being 0...but I don’t see anything relevant to the problem itself. I do understand , for example, that a force vector acting on a mass will cause an acceleration in the direction of the force, and adding an opposite force having the same magnitude will give an acceleration of 0.However, this is about two opposite forces acting on a single body.

Again,we have a momentum,of -111 kgm/s....what does that even mean?
Just what it says. For instance, if, in space (no friction with road), the car were to hit the truck and stick to it in a tangled wreck, the new 5200 kg mass would be moving left at about 21 m/sec to the left, the total momentum / total mass. Momentum is conserved in a closed system. I put them in space to keep it closed since the road would very much be exerting forces if it was there. — noAxioms

Very good; thank you....it made sense but only when you brought both masses together with the new velocity, the total having preserved the conservation of momentum....

Car 700kg * 29.05m/s = 20,340kgm/s
Truck 4500kg * -29.05m/s = -130,725 kgm/s
-110,385kgm/s being the new momentum after the crash....I think I understand.

I saw this as being a potential scenario instead, where the car and the truck were about to collide...but had not yet collided.

Grampa Dee

ok, that didn't work....for now, I will just copy some extract until we can figure out what can be done.


The general science journal
The Double-Star Experiment:
A Comprehensive Review of de Sitter's 1913 Demonstration

A. A. Faraj
a_a_farajathotmaildotcom

Abstract:
In this review, the key points of the debate between W. de Sitter and M. la Rosa are discussed; and the
main conclusions of de Sitter's 1913 demonstration are re-examined and compared to the theoretical
predictions computed on the ballistic assumption, with regard to the velocity of starlight and the orbital
velocity of binary stars, in accordance with the kinematic rules of the elastic-impact emission theory.

Introduction:

The Double-Star Experiment was the primary subject of a long series of exchange between W. de Sitter
and M. la Rosa, which started in 1913 and ended, based on the published literature, more than a decade later in 1924.

In spite of the poorly done translations that tend to make the arguments of both sides, and those of la
Rosa in particular, appear weaker than they actually are, and contrary to the widely held view that de
Sitter won it, the de Sitter-la-Rosa debate ended up in a tie, in which the supposed winner was forced to move from his initial simple position that the prediction of the Ritz theory violates the laws of Kepler
[Ref. 1.a], and to take the complicated and less clear position that the Ritz theory, according to its
Doppler equations, predicts fundamental changes in the spectra of binary stars [Ref. 1.e].

Throughout the whole debate, W. de Sitter and, to some extent, M. la Rosa as well, had taken it for
granted that starlight retains, based upon the formal Ritz theory, its original velocity resultant for the
entire duration of its journey from binary stars to distant observers.

However, recent and more thorough investigations of this subject indicate quite clearly that the formal
Ritz theory, based on the published works of W. Ritz himself, is, in fact, a new-source theory, according
to which starlight loses its initial velocities upon refraction and reflection by intervening materials.
The distinguishing criterion, here, between a ballistic theory, in which starlight retains its original
velocities, and a ballistic theory, in which it does not, is very simple:

Is light, from a stationary source, reflected, according to the ballistic theory in question, from a directly
approaching mirror, at the combined velocity of (c + v), or at the combined velocity of (c + 2v)?
If the combined velocity of reflected light, in the reference frame of the laboratory, is (c + v), then the
ballistic theory, in question, is a new-source theory, in which starlight loses its initial velocities.
By contrast, if the combined velocity of reflected light, in the reference frame of the laboratory, is (c +
2v) instead, then the ballistic theory, in question, is an elastic-impact theory, in which starlight does
not lose its initial velocities.

Since the combined velocity of reflected light from an approaching mirror, according to W. Ritz, is (c
+ v) relative to the reference frame of the laboratory, the predictions of his formal theory are very
similar to the predictions of the classical wave theory and those of the Lorentz theory and Einstein's
theory, with regard to light from binary stars. And so, the entire lengthy debate between de Sitter and
la Rosa is irrelevant as far as the theory of W. Ritz is concerned.
Nonetheless, the de Sitter-la Rosa debate is, in itself, interesting and still quite relevant within the
framework of the elastic-impact emission theory.


I'm going to stop here, and I could either continue to give you bits and pieces until we can figure out how I can send the whole thing.

Grampa Dee

Thank you for replying,noAxioms

I’ve read some things concerning vector additions that I just don’t get, which maybe you could help me out with.
— Gampa Dee
Maybe. Don't know the problem.
You mean the equation a = GM/r² ? I suppose that would need a unit vector to make it into a vector acceleration and not just a scalar. Nothing on the right side as I wrong it is a vector. — noAxioms

.... I sent a post concerning this in the “The Newtonian gravitational equation seems a bit odd to me" thread.

It seems that this would imply the light as having a speed of .5c relative to the mirror
Well, no. In the scenario I outlined, when moving up it has a speed of .134c relative to the mirror, and in the reverse direction the relative speed would be 1.866. That still presumes light is independent of emitter speed. — noAxioms

Ok, I see.

With emission theory, you'd have to specify the speed of emission, not obvious with a light clock which just reflects the pulse back and forth and has no obvious emission event. I am also unsure what emission theory says about how the speed gets altered when hitting a moving mirror. — noAxioms

I believe that one view is the reflected light (back towards the source from the mirror) having a speed of c relative to the mirror ( new source theory), and therefore a speed of (c+v) relative to the source while going back. However, The other option, I believe, hypnotizes that the light would have a reflected speed of (c+v) relative to the mirror, since it was also the speed of light going towards the mirror. This would make it a speed of (c+2v) relative to the source while going back towards the same.I’m not sure which would be correct through observation.
Since my thought experiment was the mirror remaining stationary relative to the source, being the same in the M&M experiment, both option would agree, I think.

If you’re speaking of an observer moving at .866c, relative to the frame of the clock
I wasn't. I was speaking of the clock moving at .866c relative to the ether. Neither the observer nor the frame plays any role in the predictions. That's the general model that the M&M experiment was trying to measure. — noAxioms

Yes; I understand better now.
I know that the M&M experiment was to try to measure the movement of the earth through the ether, but the particle theory does not have an ether or some other light medium model. Therefore, it would have predicted the nul result because of this....the light was going to be c relative to the whole experiment

I would be interested in learning more about the scientific jargon...I will try to read up on this more.
If you accelerate at 10 m/sec² for 100 million seconds, you achieve a rapidity (or proper velocity) of a billion m/sec. You just add 10 a hundred million times.
But to compute velocity relative to the frame in which you were initially stationary, you add 10 using relativistic addition, all those times. The former adds up to about 3.3c, meaning at that rapidity you move 3.3 light years for every year of your travel. But the velocity is .997c relative to Earth. That sort of illustrates the difference. So if your ship is fast enough, you can cross the 100,000 LY galaxy before you die because there's no upper limit to rapidity. — noAxioms

Ok; thank you; I think I see.now.

here's the link that I told you about concerning the "double star experiement"....I hope it works.
Doesn't work. It's just a pdf file name without a website in front of it. I tried searching the web for any site containing that file name and got nothing.
I am interested. Tried googling it, but the name is too generic to get to what you're talking about.
Sure, 2 orbiting stars will alternate approaching and receding, but that just results in redshift and blueshift. I don't know how they'd decide that the images being looked at departed at the same time, so to speak. — noAxioms

I'm going to try the post the whole thing on a separate post and see what happens.

I looked up on how to solve vector addition problems
Already,I don't seem to understand this addition of vectors

http://www.lon-capa.org/~mmp/kap6/cd149.htm

Adding the velocity vectors yields vt + v c = 0 mi/h.

For me, velocity is having a speed,in a certain direction. relative to something else,
So,the addition of vectors in this case is 0...but relative to what?

Since the velocities add up to 0, will the same be true for the momentum vectors? Of course not!
The momentum of the car is c = mc c= 20,000 kg m/s
The momentum of the truck is t = mt t= -131,000 kg m/s (negative because v is to the left)
The total momentum is therefore = c + t = -111,000 kg m/s

Again,we have a momentum,of -111 kgm/s....what does that even mean?

Thank you for responding noAxioms

From what I understand, in the M&M experiment, the velocity of the light would be c through all paths within a particle theory of light.
— Gampa Dee
Special relativity theory (early 20th century) posited the frame independent fixed speed (not velocity, which is frame dependent) of light. The M&M experiment (late 19th century) neither presumed nor demonstrated the fixed frame independent speed of light.
The speed of light is not dependent on whether one uses a particle or wave model for it. — noAxioms

Yes, I do have a hard time with the “velocity” concept, but the reason I use the term is due to the fact that I need it in context with the light going in a specific direction.....I will try to go back to the old thread “The Newtonian gravitational equation seems a bit odd to me” sometimes because I’ve read some things concerning vector additions that I just don’t get, which maybe you could help me out with.

I don’t understand why you say one path would be longer than the other?
Picture a light clock moving at 0.866c with mirrors separated by a distance of 1. Presume no length contraction. Move the clock with the mirrors to the sides. Light travels a distance of 1 to the left and 1.732 up to get to the other side, a total distance of 2. Another 2 to get back. So it runs at half speed since it has a distance of 4 to go instead of 2 when the clock is stationary. — noAxioms

It seems that this would imply the light as having a speed of .5c relative to the mirror, whereas, if I’m not mistaken, the emission theory would claim the light as having a speed of c. If you’re speaking of an observer moving at .866c, relative to the frame of the clock, then, accordingly, the observer should measure the light as having a speed of 1.886c relative to himself and still c relative to the mirror. However, there might be indeed some “apparent” speeds due to the Doppler Effect, but those aren’t real.

that is, the receding galaxies with a velocity greater than c would not be interpreted as having those velocities.
Technically, they're rapidities, not velocities. The former adds the normal way (a+b) as opposed to velocity with adds the relativistic way, in natural units: (a+b)/(1+ab) — noAxioms

... I would be interested in learning more about the scientific jargon...I will try to read up on this more.

The better empirical evidence would be something like seeing incoming ejecta before seeing the explosion that caused it, or seeing a star explode well before the neutrinos hit instead of the observed neutrinos coming just before the light. — noAxioms

I think I see what you mean.... here's the link that I told you about concerning the "double star experiement"....I hope it works... just started to read it; still got lots to go through, and you might be interested in this.

[url]http://research_papers_astrophysics_science_journal_5693.pdf [/url]

Thank you for responding noAxioms, I appreciate it.

I agree with what you wrote, except, the Newtonian model would have predicted a null result as Newton believed that light was made up of particles.
— Gampa Dee
Yes and no. Particles would also have taken longer to go the greater distance with the grain than the shorter distance against it.
But the interferometer used by M&M leverages the wave nature of light, something known back in Newton's time since particles don't explain rainbows. I don't know when interferometers were invented. — noAxioms

From what I understand, in the M&M experiment, the velocity of the light would be c through all paths within a particle theory of light.

The Project Gutenberg EBook of The Theory of the Relativity of Motion, by
Richard Chace Tolman

Suggested Alternative to the Postulate of the Independence of the Velocity of Light and the Velocity of the Source.

20. Because of the extraordinary conclusions derived by combining
the principle of the relativity of motion with the postulate that the
velocity of light is independent of the velocity of its source, a number
of attempts have been made to develop so-called emission theories of
relativity based on the principle of the relativity of motion and the
further postulate that the velocity of light and the velocity of its source
are additive...
As a particular example of the simplicity of emission theories we may show, for instance, how easily they would account for the negative result of the Michelson-Morley experiment.

Einstein claimed the light’s velocity is invariant without any specific reason why

Empirical evidence? Einstein didn't originate the claim. He just ran with it without dragging in the baggage that everybody else tried to keep. — noAxioms

First, I hope that I’m not sounding as if I think little of the genius of any/all physicists who were at the same time developing QM.
I was just wondering why Einstein, who did mention the particle characteristic of light for QM, did not think that this could also be the solution for the M&M experiment

it seems to me that Fitzgerald allowed a mechanism for the length contraction to exist, being the ether, whereas Einstein did not have any mechanism

The heck he didn't. It was explained via Minkowskian geometry. The contraction (and the underlying 4D geometry) derives directly from the frame-invariant speed of light, even if there was a preferred frame. The geometry and contraction were both a byproduct of the work of Minkowski and Lorentz, so that too wasn't something Einstein originated. Lorentz was first, but clung to the 3D ether model like Fitzgerald. That model added complications preventing the special version of the theory coming out before Einstein's, and preventing a general version from coming out until nearly a century after Einstein's. — noAxioms

I understand, but it was in consequence of the postulate in the first place; in other words, if the 2nd postulate is true, then certainly, there will be some modifications involved when dealing with space and time, since velocity is distance over time.

As I said, there are some empirical tests one can perform to see who is right, but not that one where results can be physically published in a journal. — noAxioms

I have started reading a journal named “double star experiment”, which I will try to leave as a link.,if I
cannot, I will write what I have read later on.

for what I understand...and for me, it seems that the postulate of the invariant speed of light would fall into the "extraordinary claims require extraordinary evidence” category.
OK. For me it falls under Occam's razor: The simpler model is the more likely one, proposing the fewest additions and complications. — noAxioms

I personally don’t see the 2nd postulate as being simple in essence.However, If it is indeed correct, well then yes, we don’t have any choice but to use it

My “personal” opinion would be that the particle theory, which would have predicted a null result

It predicted no such thing since the particle would have longer to go this way than that way. The contraction (which both theories describe, but Newton does not) explains the null result of M&M. — noAxioms

I don’t understand why you say one path would be longer than the other?

But, what if the light speed was c relative to the source (sort of particle theory)

Then an easy experiment would show it. As I said, this is easily falsified.
The binary star thing doesn't work since there is no way to know when the light you're looking at was emitted. Both stars continuously emit light. You need two relatively moving sources that simultaneously, in each other's presence, emit a short pulse. The further away the emitter the better since it would give one pulse more time to outrun the other, showing up as two pulses at different times at the detector. No such thing is seen. — noAxioms

I need to ponder more on this one... I’m not sure what you’re saying.

Distant galaxies receding over c would not be visible due to the light approaching too slowly to stay inside our event horizon, but there they are, with Webb telescope finding ever more distant ones. The visible universe would be far smaller if light speed was dependent on emitter motion. — noAxioms

Yes, but on the other hand, if the light speed is dependent on the emitter, there would simply be a different interpretation... that is, the receding galaxies with a velocity greater than c would not be interpreted as having those velocities....for example, could some of the red shifts be interpreted with distance, as well as velocities?

Grampa Dee

Thank you for replying PhilosophyRunner

The length contractions in Fitzgerald's theory is the same as that in Einstein's special relativity - they are both Lotentz transformations. The difference is that in Fitzgerald's theory the frame of reference of the ether was a privileged frame of reference in which light traveled, while Einstein showed that the ether was not needed in the theory and that the frame of reference can be any inertial frame of reference. — PhilosophyRunner

However, by taking out the ether, it seems that Einstein was sort of claiming at the same time that light might not be a wave after all....I think he did so when he explained the photoelectric effect.

To put is more simply:

According to Fitzgerald: Light traveled in the ether frame of reference. This frame of reference was moving relative to an observer on Earth. Lorentz transformations can be used to explain length contractions due to the observer and ether being in different frames of references.

According to Einstein: The ether is not needed in the theory. Length contractions due to Lorentz transformations are observed when the observer is in a different inertial frame of reference to the experiment, but this is not relevant to the M&M experiment as both the observer and experiment are in the same frame of reference. — PhilosophyRunner

I agree with most of what you wrote, except, If the ether was not needed, then, why was the length contraction and time dilation needed? what did it explain if it wasn't needed to explain the null result?

Thank you for your time
Grampa Dee

Thank you for responding, noAxiom

The reason why I brought up this problem was due to it resembling the M&M experiment.
— Gampa Dee
It very much does. Just like with a light clock, without length contraction, the M&M experiment would show it taking more time for light to make the circuit with and against the motion, and less time when it moves perpendicular to the motion. The difference should have been noticed and the Newtonian models were falsified when it wasn't. — noAxioms

I agree with what you wrote, except, the Newtonian model would have predicted a null result as Newton believed that light was made up of particles.

Einstein claimed the light’s velocity is invariant without any specific reason why

It's a postulate, not something that can be known. Special relativity used a fairly strong version of the postulate, that light actually goes the same speed regardless of inertial frame choice. Some later papers took much of that metaphysical assertion away and used a weaker statement, that the laws of physics (including any measurement of light speed) are the same relative to any inertial frame.
The latter wording allows for the existence of a preferred frame despite no local way to detect it. — noAxioms

it seems to me that Fitzgerald allowed a mechanism for the length contraction to exist, being the ether, whereas Einstein did not have any mechanism...for what I understand...and for me, it seems that the postulate of the invariant speed of light would fall into the "extraordinary claims require extraordinary evidence” category... I am not saying that Fitzgerald was correct and Einstein was not, only that Fitzgerald had a cause for the length contraction.
My “personal” opinion would be that the particle theory, which would have predicted a null result, may have been disregarded prematurely. .

So in the case of the M&M experiment, Einstein would claim that there was no length contractions nor time dilations involved because there was no different inertial frames to measure....

I assure you that the M&M experiment was performed in many different inertial frames. The statement above is false and Einstein would certainly not have said anything to that effect. — noAxioms

I agree that my statement is false in the sense that Einstein did not write about the M&M experiment. But what I am saying is that Einstein “would have said” the M&M experiment did not contain any time dilation or length contraction because the light source and observer were on the same inertial frame, whereas Fitzgerald claimed the experiment did indeed experience a length contraction in the direction of motion.
As for the M&M experiment performed on different frames, I think that the sun (being on a different inertial frame) was used as a source of light; however some have identified the light passing through the mirrors involved a “new source” for the light which brought the experiment back on the same frame as the observer.

But, what if the light speed was c relative to the source (sort of particle theory)

Immediately falsifiable by having two light sources moving at different speeds emit a flash when they pass each other. A distant observer would see one flash from the approaching source sooner than the one from the receding source, thus falsifying Einstein's postulate. Such a result is not observed. Light speed is empirically demonstrated to be independent of the speed of the light source. — noAxioms

Yes, but then again, it seems that there is still a possibility for the new source theory. Within a double star system, the light from a receding star and the one advancing might interact within a gas medium before leaving the system, which would cause the effect to disappear.

The observer outside of the frame
An observer cannot be outside any frame. He's in all of them, just not stationary in them all. — noAxioms

Yes; true...I am having problems in explaining scientific theories properly... :)

Thank you for your time
Grampa Dee

Thank you for responding, Alan:

On the other hand, the monoclastic neutrinal differentiation of the autosomatically-determined spin value inherent in all such equitational bivalent transmogrifications cannot be entirely ignored, wouldn't you agree? — alan1000

I’m sorry, I don’t quite understand what you’ve written down. I haven’t read about spin values within any M&M experiment explanations.
Are you speaking about the g factor? Could you expand?

Particularly if we keep in mind the various M&M chromatic values which may be obtained at little expense in our own time. — alan1000

However, the null result of the M&M experiment can be explain by the particle theory of light, Why was this possibility not considered...or was it? I haven't read anything concerning this

Thank you for your time.
Grampa Dee

noAxioms — noAxioms

Thank you noAxioms, and good post;
I think that I'm not sufficiently prepared when discussing vectors, and so I'll
read more on the issue..

Thank you for your time.

Grampa Dee

Thank you for responding noAxioms

.Galileo’s law ( all bodies fall at the same rate).
— Gampa Dee
That's not the law, and you wording is trivially falsified. I can drop a rock off a building and simultaneously throw another one downward. The thrown one will fall at a greater rate and arrive first. — noAxioms

Maybe it’s not a law, I don’t know. but I did hear of Galileo’s law of falling bodies... if Newton has three laws of motion, I personally don’t see why Galileo’s “description” of freefalling bodies can’t be viewed in terms of a “law”. To throw a rock downwards will make the rock fall at a greater rate, as you mentioned... but we were talking about a freefall caused strictly by gravity.

However, this part does seem to show that there are some “unknown” elements as well.

Well yea. The alternative is some klnd of solipsism where things are gravitationally attracted only to known object, which makes the knower very special. — noAxioms

What I was saying, noAxiom was this:

. But it accelerates at Σ GM/r² ( talking about the moon)

If there is no mistake in this observation ,then this will certainly prove that Galileo’s law is indeed valid.....

If they can indeed account for every other mass in the sigma factor, then it should prove ( I think) that
Galileo was right ... and it should be considered a law, if it isn’t already.

if a car comes towards you as you are driving your car, the measured velocity relative to both of you is v..
.,
That's impossible. If a car is moving relative to me at v, then I am moving relative to it at it at -v by definition. — noAxioms

Sorry; I meant the relative velocity measured by you, the driver, was v

Total v? You want to add velocity of me relative to the road to the velocity of them relative to me? — noAxioms

...the car coming towards me I measure as being v; I also measure my velocity relative to the road as being -.5v...(sorry, I had this wrong...-.5v since I claimed the car coming towards me as having a velocity of +v already) then I calculate (I do not measure) the car as moving at velocity of .5v towards me(now that I know that I also am moving at -. 5v towards him/her)....I don’t know what is wrong with this example apart from my - sign error ...sorry

if a car accelerates towards you as you are accelerating towards it, the total acceleration will “not” be a + a ??
Correct. Nothing is accelerating at a+a (which is zero) nor a-a which is twice something. Doing would violate Newton's laws, F=ma in particular.
Each car is accelerating at 1a, presumably in opposite directions. — noAxioms

So how should we calculate the acceleration between two cars if...

Car 1 .... a1 = 2 m / s^2 .relative to the road
Car 2.... a2 = - 2 m /s^2 relative to the road

At time 1sec the first car has a velocity of 2 m / s , the second car will have a velocity of -2m/s
What is the relative “speed” between the two cars? I see 4m/s as measured by car1
at time 2 sec.the first car has a velocity of 4 m/s the second car will have a velocity of - 4 m/s
at time 3 sec the first car has a velocity of 6m/s the second car will have a velocity of -6 m/s
It seems the acceleration relative to both cars should then be, in my opinion 4 m/s^2 ...as measured by car 1

Having said this, I will also look up for the proper manner to do this right.

Take once again a pair of equal mass planets X&Y orbiting each other with X at coordinate -1, 0 and Y at 1,0..The both trace a unit circle about the origin at 0,0. X has a coordinate acceleration of 1,0 and Y has a coordinate acceleration of -1,0. Nothing is accelerating at 2 anything. The distance between them isn't changing over time, so adding the acceleration magnitudes does not in any way express the rate of change of their separation. — noAxioms

..No,.not in this case, because the acceleration is due completely to the change in direction...
But we were talking about the case of a freefalling body.
However, I don’t understand why one has the negative acceleration of the other..wouldn’t they crash in this case?
Also, I know that I’m mixed up when we’re dealing with vectors...
But, what we are discussing could be discussed in terms of speeds, if you will...we are trying to discuss whether a more massive body falls faster than a less massive one..


Grampa Dee.

Thank you for responding Erich

We may be mixing up speed & velocity. Speed is a scalar but velocity is a vector. — EricH

I may have; although I did mention the direction as being "towards each other"..

In your example, both cars are traveling at a speed of 0.5v relative to the road (again no direction). In order to calculate the relative velocity we need to do vector calculations (which can get complicated). In the most general situation the two objects may be moving in arbitrary directions and may or may not ever collide. In your simplified one dimensional example, the velocity of one car is .5v and the velocity of the other car is -5v. But these are vectors, so to calculate the relative velocity between the two you cannot simply add the .5v & -5v and get 0.[/quote]

I agree, this is indeed what I was saying; I did claim the total velocity as being v.
However,I was actually talking of accelerations, and the velocity example was only to make a point. PhilosophyRunner wrote saying that I will need to read up on inertial and non-inertial frames because this is the part that confuses me ...and this is what I'll be doing.
— EricH

You have to factor in the direction and do vector math. The end result will be a relative speed of v - and the vector math will show a collision between the two cars at some point in time depending on the values of v and the starting distance between the two cars. — EricH

Yes; I will need to look up vector math as well, because I just realized that addition of "tip to tail"
vectors will also give me 0 in this case.

Thank you for your time, Erich
Grampa Dee

Thank you for replaying PhilosophyRunner

↪Gampa Dee I suggest you read up on inertial and non-inertial frames of reference. I think your misunderstanding stems from not knowing the difference between them and treating both as if there is no difference. — PhilosophyRunner

Yes, I will do that;
and thank you again for your time.

Grampa Dee

Thank you for responding noAxioms

Ok; I’ll try to remember to use “a” instead :) ...but I did see some charts which identify planets using g in the same way they use distance in terms of earth distance units (AU).
— Gampa Dee
Those charts would be correct. Gravity (a) on Saturn is ~1.08g and it orbits at about 9.5 AU. Both g and AU are constants. Saturn does not define a different AU any more than it defines a different g.

https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html — noAxioms

While I do understand that “g” as such is a constant, g is used as a form of measurement for other planets as well. Should have I rather written how many “g” s a planet’s surface gravity possess?

Well, for me personally, I ask myself whether there is a meaning behind r^2 besides simply being a distance squared. It would seem logical, for me, to view gravity as being, for example, some sort of energy emanating from the massive body (probably at light`s speed), which would continuously be reducing it`s energy density as it travelled away from the mass source,
That's one way of looking at it. A light shining at intensity proportional to the mass would decrease in brightness at the square of the distance from the light. But gravity doesn't travel, so you can take the analogy only so far. — noAxioms

For me personally, my upper statement would not constitute gravity as such...if we continue...

“to view gravity as being, for example, some sort of energy emanating from the massive body (probably at light`s speed), which would continuously be reducing it`s energy density as it travelled away from the mass source, this would, in my mind, constitute a form of diverging medium which could be responsible for the acceleration. This would mean that r^2 would have something to do with the surface area....”

I would rather identify the sort of diverging medium as having the quality of gravity.

For example, the moon's orbital distance increases by several cm a year as Earth transfers its angular momentum to it. It also transfers about 3% of its angular energy to it, and radiates the remainder away as heat. So energy is lost in the process of the moon's orbital changes. — noAxioms

Interesting... I heard the moon was moving away from the earth, but didn’t know the cause.

If the falling body was taken from the ground, that reduces M, and it will thus accelerate less than a similar object falling from space — noAxioms

I agree, and the way I view this is simply that the mass from outer space will increase the total gravity of the system,

For a small rock, the difference is immeasurable since the mass of Earth changes more per second than any nitpicking about where you got the rock. — noAxioms

The idea of nitpicking, noAxioms, is to try to explain a system’s physical laws...
For example, let us assume that I am correct (just for argument’s sake)...
I understand very well, that measurements have been made many times supporting Galileo’s law ( all bodies fall at the same rate).
However, if this is due to the difference of acceleration between two unequal masses in freefall as being too small to measure, then, the whole law would be incorrect even if our measurements couldn’t prove it.
We do know that stars orbital velocities at the edge of galaxies don’t agree with our laws, unless we add more “dark mass”....we have never detected any dark mass yet...so while the theory of dark matter is still valid, it remains that a possible different gravitational law ought to possible as well.

The moon is like that. It is a big rock that was essentially taken from the ground, not having arrived from space. When they brought back the moon rocks, they were quite surprised to find it was a chunk of Earth and not something else like every other moon out there. But it accelerates at Σ GM/r² where Σ is the summation of the accelerations due to every bit of mass out there, mostly the due to the sun and also Earth (what's left of it after the moon was scooped out), but everything else pulls at it as well. — noAxioms

But it accelerates at Σ GM/r²

If there is no mistake in this observation , then this will certainly prove that I am in error and that Galileo’s law is indeed valid.....

where Σ is the summation of the accelerations due to every bit of mass out there, mostly the due to the sun and also Earth (what's left of it after the moon was scooped out), but everything else pulls at it as well

However, this part does seem to show that there are some “unknown” elements as well.

But the equation would still remain a = GM/r^2 + Gm / r^2

This cannot be right. For two equal stationary masses, it says they will not accelerate towards each other since the sum is zero. This is not the case. — noAxioms

Yes...true; but again, we are dealing with semantics...I did not go to university to learn science, and my mathematical background is indeed limited. I guess we need to add vectors...
I will explain it the way I understand it... if a car comes towards you as you are driving your car, the measured velocity relative to both of you is v..., if you knew your velocity relative to the road as being .5v, then you would say that the other car is coming towards you at –.5v....the total v will not be 0...

Even if you fixed that, the equation there does not express an acceleration, so the 'a=' part in front is blatantly wrong. Nothing accelerates at that rate. — noAxioms

Ok,... maybe this is where I’m off the rail, although I don’t yet understand why...
Are you saying that instead of velocity, in the upper example I wrote, if we used acceleration instead , it wouldn’t work?

... if a car accelerates towards you as you are accelerating towards it, the total acceleration will “not” be a + a ?? I will think about that...

Thank you for your time

Grampa Dee

Thank you for responding noAxioms

g is a constant. Saturn doesn't have a different g, it has a different acceleration a. — noAxioms

Ok; I’ll try to remember to use “a” instead :) ...but I did see some charts which identify planets using g in the same way they use distance in terms of earth distance units (AU).

https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html

The acceleration is dependent on mass and radius and has little direct connection with density, especially since density of any planet/star varies considerably at different depths. You complicating things needlessly by trying to work with area and/or density. — noAxioms

Well, for me personally, I ask myself whether there is a meaning behind r^2 besides simply being a distance squared. It would seem logical, for me, to view gravity as being, for example, some sort of energy emanating from the massive body (probably at light`s speed), which would continuously be reducing it`s energy density as it travelled away from the mass source, This would, in my mind, constitute a form of diverging medium which could be responsible for the acceleration. This would mean that r^2 would have something to do with the surface area....

I do have a hard time with this...is it possible to share this proof?

Nothing in science is actually a proof, but the reasoning goes like this: You have two identical balls of mass M each. They presumably fall at the same rate. Now you connect them by a thin thread or spot of glue, creating one object of mass 2M. Either the connection makes some sort of magical difference, or the 2M mass should fall at the same rate as before. Hence the rate is independent of mass. — noAxioms

The 2 M already side by side(even without the string) while accelerating towards the earth, I believe also accelerates the earth towards themselves equally, so no difference will be seen.... I have also mentioned this in one of the posts to unenlightened.

I would in fact surely agree to viewing the system as a whole(earth and falling bodies)as being invariant in terms of energy “if” the fallen body was taken from the earth (did not come from outer space), for in this case, the falling body was formally on the ground being part of the cause for the g acceleration, and since while the falling body will attract the earth as well, then,for the whole system, the overall acceleration will not change.[/b]


But the equation would still remain a = GM/r^2 + Gm / r^2 .... since what was taken from the ground no longer is part of the earth’s gravity but is now attracting the earth instead.

One can reason that the connection cannot apply a downward force to both objects (violating Newton's laws), but Newton's laws were not available at that time yet. F=ma was unknown, and the theory of impetus and such was still a thing. — noAxioms

Well, I think that Galileo was simply claiming that if a heavier object accelerates faster, then, the two masses tied together ought to accelerate faster than the two masses when separate.
Since, like you said, the idea of every mass attracting every other mass was not yet developed,
it wasn't a bad argument.

Grampa Dee..

thank you for replying noAxioms

I think we can deal with the Newtonian equation using only one dimention.
— Gampa Dee
Any talk of orbital mechanics has velocity that is not parallel with the acceleration, and thus involves at least two dimensions. Orbits can be described in a plane. Only things falling straight up and down can be described in one dimension. — noAxioms

Yes; of course...my mistake in writing 1 dimensional problem as being Newtonian...I was thinking of freefall.

I don’t believe that Newton had a three dimensional frame of reference in mind

All frames of reference have 3 dimensions of space, and require 3 velocity components to describe. — noAxioms

I understand; but Newton developed his equation from Kepler, and Kepler was viewing the problem in a two dimensional fashion.... What time did it take for a planet to complete an orbit, which was viewed as simply an orbital velocity, being 2 dimensional in essence, as the time was proportional to the area . Yet, through mathematical manipulation, his constant R^3 / T^2 .
we can get:

a*r^2 = v^2* r
a =v^2 / r....
and
v^2 = a*r
v = sqrt a*r

We can get both the acceleration towards the sun as well as the orbital velocity.This, one would think,
would need a 3 dimensional coordinate system.

However , to claim that g is not acceleration I still don’t get

g is acceleration, but is simply a constant scalar. So the gravitational pull on the surface of Saturn is 1.08g. It would be wrong to say Saturn has a slightly larger g. — noAxioms

The surface g acceleration is dependent to the mass density...and Saturn, being a gaseous planet,would be much less dense than the earth . Would you have the equation for this?
The reason why I had written M / Area was to have the denominator as being r^2 instead of r^3
as in the case of mass density.

A frame of reference is a starting point, not something thrown in. Without it, any velocity is completely undefined. — noAxioms

I was thinking something like Kepler using only distances and time period to make up his equations, and so not really using a frame of reference although it was implied I am sure.

Orbital accelerations are always changing, even in the circular case. Remember that it is a vector. — noAxioms

I thought that is was constant because the direction is towards the center, but I think I see what you mean. So the orbit magnitude in a circular orbit will remain constant ,while when the orbit is elliptical, the magnitude of acceleration will be changing and a change in acceleration is a 3rd derivative, which makes the calculation more tedious , I would think.

There's nothing special about M since it works with any M. Galileo actually published a tidy proof that the acceleration is independent of the mass of the thing accelerating. — noAxioms

I do have a hard time with this...is it possible to share this proof?

Grampa Dee

Thank you for responding PhilosophyRunner

To use your car example, I am standing on the pavement not accelerating. I see one car (A) moving accelerating at 3m/s2 towards my left and another car (B) accelerating at 5m/s2 towards my right. That is simply the acceleration of the cars. It is meaningless (from a classical mechanics sense) to then add these up and say car A is accelerating at 8m/s2 from the point of view of the car B. Neither car is accelerating at 8m/s2. Car A is accelerating at 3m/s2 and car B at 5m/s2. — --`r

I would agree that car A is accelerating at 3m/^s^2 and car B is accelerating at 5m/sec^2 relative to the road, however, when we view this as being relative to themselves, then I can only detect 1 acceleration, not two.
Here, I am strictly speaking changes in velocities, having nothing to do with the g force that "might" be involved, as in freefall, there is no force that is being felt by the accelerated body.

thank you again for your time
Grampa Dee

Thank you for replying,noAxioms

I suppose you can use [the two added equations] to compute the rate of change in distance between the two objects, only in a 1-dimensional case, but that rate isn't acceleration.
— noAxioms

I agree with you; I certainly am not trying to give a 3 dimensional situation where one might have some partial derivatives and some 3rd derivatives as these can become too complicated.
I think we can deal with the Newtonian equation using only one dimention...I don’t believe that Newton had a three dimensional frame of reference in mind when he was introducing his equation.
However , to claim that g is not acceleration I still don’t get; Im not saying that you are wrong, I’m saying I don’t understand.

It's not that it's simpler. Adding the equations only produces a useful result if there is no motion except along one axis. So for instance, under Newtonian mechanics, the ISS is continuously accelerating (coordinate acceleration) towards Earth at about 8.7 m/s² and yet its distance from Earth is roughly fixed, and its speed relative to Earth is also roughly fixed. This is because it is not a 1d case. The ISS has motion in a direction other than just the axis between it and Earth.
The ISS example also serves as a wonderful example of the difference between the physics definition of acceleration (rate of change in velocity=8.7 m/s² down) vs street definition (zero change in speed). — noAxioms

I fully agree that anything which has an orbit will have a combination of gravitational acceleration and sidereal velocity at the same time...this, I’m certain can become very complicated if one throws in a frame of reference from which everything is calculated. If the orbit isn’t circular (which most aren’t) there will be a change in acceleration (Jerk) being a 3rd derivative.....but our example is 1 dimensional, the distance need not to be great.
It’s very simple,noAnxiom; what I’m asking is this: the acceleration of a body in freefall is GM/r...
What is it about this mass (M), that is so special that the smaller mass (m) cannot have any influence whatsoever on the acceleration?

Grampa Dee

Thank you PhilosophyRunner

For example see: https://orbital-mechanics.space/the-n-body-problem/two-body-inertial-motion.html
— PhilosophyRunner — Gampa Dee

I went through the example, and I will try to explain how I have understood to the best of my ability.what was written.....I'm not claiming to be correct, only that this is how I have interpreted the text.
Also beware of my use of symbols...they're awful as I don't have the use of the correct symbols.
For example R(double dot) simply means acceleration, however, use the text beside you and you will be able to follow......Here goes nothing :)

Two-Body Equations of Motion in an Inertial Frame
The position vectors R1 and R2 of two point masses in this reference frame are:
R1 = X1Ihat + Y1Jhat + Z1Khat
R2 = X2Ihat + Y2Jhat + Z2khat

Here, we are given a 3 dimensional reference frame X,Y,Z where two bodies are positioned at (x1,y1 z1) and (x2,y2,z2)...I don’t understand how they could be identified as vectors though, since position is not a vector.

Let r be the vector pointing from m1 to m2, which we also phrase as m2 relative to m1. Then:
r = r2 – r1
r = (x2 – x1)Ihat + (y2 – y1) Jhat + (Z2 – Z1)Khat

Since the mass is said to be pointing towards the other mass, then ok, r is then a vector.

We also define a unit vector pointing from m1 toward m2:
Uhat*r = r / r

This, being a unit vector, will not do anything to the equation as such except maybe to identify r^2(denominator), which is not a vector,....as being identified as a vector....in my opinion.

where r (denominator) is the magnitude of r ( nominator), or the distance between the two masses.

Forces in the Two-Body System#
The two masses are acted upon only by their mutual gravitational pull. F12 is the force exerted on m1 by m2 and F21 is the force exerted on m2 by m1. By Newton’s third law:
F12 = -F21

We have two forces coming from the two bodies...1 being + and the other -.


Newton’s second law says that the force is equal to the mass times the acceleration:
(17)#
F12 = m1 R(double dot)1
F21 = m2 R(double dot)2
where R¨ is the absolute acceleration of the subscripted mass. Absolute means that the acceleration is taken relative to an inertial reference frame. This is important because Newton’s second law only applies for absolute accelerations.

This is simply, in my opinion,

F1 = m1*a1
F2 = m2*a2
F1 =- F2

Since the only force in this system is the gravitational attraction, the force is also equal to Newton’s law of gravitation, (1). The force of m2 on m1, F12, points in the positive direction of u^r. Because of Newton’s third law, as represented by (16), the force of m1 on m2, F21, points in the negative direction of u^r. This is shown in (18):

F12 = Gm1m2 / r^2 (uhatr)
F21 = - Gm1m2 / r^2 (uhatr)

Here we have two of the Newtonian gravitational equations... one of them is negative identifying the reverse direction of the second force.




Finding the Equations of Motion *********

Here’s where our focus of discussion lies.



Combining (17) and (18), we find:
m1R1(double dot) = Gm1m2 / r^2 (uhatr)
m2R2(double dot) = - Gm1m2 / r^2 (uhatr)

This would mean, in my opinion, m1*a1 = Gm1m2 / r^2 ... u(hat)r is simply a unit vector...I think we can ignore it.The same goes for m2R2(double dot)

Finally, we divide through by the mass on the left side of each equation and replace u^r with its definition, (15) to arrive at the two-body inertial equations of motion:
R1(double dot) = Gm2 r / r^3
R2(double dot) = -Gm1r /r^3

Notice here that there are two accelerations, which are not going to have the same magnitude.
a1 = Gm2 / r^2 ( one of the r in r^3 cancels out with the r in the numerator)
a2 = - Gm1 / r^2

So, now we have 2 accelerations; one is negative, the other positive. What do we do at this point?
If we have two cars having velocities of v1 and –v2(since it is going towards v1), relative to the road, what will the observers within the car measure the relative velocity between themselves?
It’s not going to be v1 – v2, but v1 + v2....I see the same problem for the two accelerations.

I hope this post wasn't too long.

Grampa Dee

Thank you for replying, noAxioms

The symbol for that is 'a', not 'g'. 'a' is a vector variable acceleration. g is a scalar constant acceleration. Neither are a force. Force is measured in Newtons and uses the symbol F. Try to use standard symbols when discussing such things, as personal preferences only lead to confusion. — noAxioms

I will try to do that noAxiom...I’m not well versed in the scientific terminology .

:(
— noAxioms

The acceleration of the moon when it is at radius r is exactly that in Newtonian physics. That does not mean it will hit the ground in 3 1/3 seconds like the 10 kg ball. As I said, a small fraction of a second is more likely. — noAxioms

I agree with you, but don’t understand how Newton’s equation can show this particular case.

I'm assuming that the dense super-mass is rigid, so yes, the acceleration applies to every point in the moon-ball. I am not assuming Earth is sufficiently rigid to not deform under the ungodly tidal stress the ball would apply to it. The sidewalk slab 60 meters below will be yanked up without waiting for Earth to catch up with it. — noAxioms

Sure....but we’re only using thought experiments ...gedank...or whatever

:)
— noAxioms

Acceleration is absolute, not relative, so adding them seems to result in a fairly meaningless value — noAxioms

I agree that there is only one “observed” acceleration.

I suppose you can use it to compute the rate of change in distance between the two objects, only in a 1-dimensional case, but that rate isn't acceleration. For instance, if an apple detaches from a tree and the distance between me and it decreases at 9.8 m/sec², that in no way suggests that I am accelerating at that rate, and in fact I'm accelerating (coordinate acceleration, not proper acceleration) away from it a little bit. — noAxioms

I do agree that the 1 dimensional is much simpler. As for the acceleration of the apple at that rate relative to you, if we identify acceleration as simply a change in the rate of velocity, then, I would say that it’s only a relative situation to claim yourself as the one who is accelerating... and you feel the force to say so :)

If we switch to proper acceleration, both the apple and I have a proper acceleration of about 1g until the apple detaches, at which point I have 1g proper acceleration upward and the apple has none. So it is me that hits the apple, not the other way around. But that isn't really the Newtonian way of looking at it.

I agree, it’s Einstein who recognized that a freefall object didn’t experience any force.

The Newtonian gravitational equation, identifying two masses, has only one acceleration , the one caused by the earth.
But there is an equation for each object, each dependent on only that mass, and not on the mass of the thing accelerating towards it. — noAxioms

Ok; I’m not sure what you’re saying but this might be where I’m not thinking it through.

ok; I think we might be going somewhere. When we measure the acceleration of the rock towards the earth, aren't we not measuring, at the same time, the acceleration of the earth towards the rock? How could you know the difference?
— Gampa Dee
As P-R points out, the coordinate acceleration described by Newton's equations is relative to any inertial coordinate system, and the equations don't work when used with a non-inertial coordinate system such as an accelerating or rotating one (Earth is both).
Mind you, plenty of books treat things like the Earth-moon system in isolation, which is to reference an accelerating coordinate system. It works to an extent, but adds confusion. For instance, few are aware that the moon always accelerates towards the sun since the sun exerts more force on it that does Earth. That means that when the moon is between the two (solar eclipse), the moon is accelerating directly away from Earth.

From the accelerating frame of Earth, the moon seems to maintain a fairly constant distance that isn't much a function of which direction the sun is. Yes the orbit of the moon (like any orbit) is eccentric to a degree, but that again isn't due to the sun. — noAxioms

I’ll ponder on this, noAxiom...thanks again for your time
Grampa Dee

Thank you PhilosophyRunner

For example see: https://orbital-mechanics.space/the-n-body-problem/two-body-inertial-motion.html — PhilosophyRunner

Yes, I will....
Thank you again.

Grampa Dee

Thank you for replaying, PhilosophyRunner

Secondly why would we include the acceleration of the earth caused by the second mass? That is not what we were calculating - we were calculating the acceleration of the rock. If you wanted to calculate the acceleration of the Earth caused by the rock, that is a separate calculation
an hour ago — PhilosophyRunner

ok; I think we might be going somewhere. When we measure the acceleration of the rock towards the earth, aren't we not measuring, at the same time, the acceleration of the earth towards the rock? How could you know the difference?

If you want to model a system where both bodies are accelerating you will have to use the two-body equations of motion, — PhilosophyRunner

ok; so, what would be this equation?

Grampa Dee

I have no idea what you might consider the 'gravitational energy' of Earth. Mass divided by 4 π r² gives you, well, I don't know what. It seems vaguely related to area of a circle. — noAxioms

4 π r², being the surface area of a sphere, would be dependent to the surface Acceleration

M / 4 π r² = A * constant....It was meant to what you have written concerning a high density mass.
While, this is not directly the density of mass, it seems to identify a mass having a certain spherical area as having a certain gravitational surface acceleration... if the radius is small the acceleration will be greater...

Of course, the equation could be erroneous, but I can't see why....that is why I asked you what you thought about the equation

Grampa Dee.

Thank you for replying,PhilosophyRunner

So the acceleration of the rock in earth's gravitational field is not dependent on the mass of the rock. It is dependent on the distance of the rock from the center of mass of the earth, and on the mass of the earth.

You seem to be asking, what if M were different? — PhilosophyRunner

No, I am asking how is it that we do not include the acceleration of the earth caused by the second mass?

Grampa Dee

Thank you for responding noAxiom

First of all, gravity isn't energy. I have no idea what you might consider the 'gravitational energy' of Earth. Mass divided by 4 π r² gives you, well, I don't know what. It seems vaguely related to area of a circle. You equate this to 'k g', but no idea what 'k' is (kilo?). 'g' is the constant 9.8 m/sec², so you're seemingly equating this function of mass and radius to the constant 9800 m/sec² — noAxioms

I view gravity as a potential for movement observed as the weight of a body, if the mass is stationary, located at a certain height above the ground, or the increase of k.e. when it accelerates towards the ground.

Here, if r is small then g can become extremely large.

You seem to be equating g with A. 'g' is a constant magnitude of acceleration (a scalar), so it cannot be smaller or larger. A is a variable, and a vector, not a scalar. A = GM/r², so yes, 'A' becomes quite large if r is small enough. Saying 'g' can be quite large is like saying a meter can be larger if my table is wide enough. — noAxioms

For me, g is the potential gravitational acceleration given to a “test mass” caused by some other mass (usually a large one), not necessarily the earth; it could be the moon, mars or Jupiter. I was not using the letter “a” because we cannot speak of a force as being the cause. However, if you want me to write down “A”, instead of “g”, then, no problem, noAxiom,, I will use the letter “A”.

So , you wrote: . “A = GM/r², so yes, 'A' becomes quite large if r is small enough.”

This is all that I was saying.

Now, concerning the example of the high density ball pulling the earth towards itself ; How would you write the equation, if A = GM/r² is the acceleration caused by the earth?

Any other mass g acceleration would need to be added at every point in space...

'mass g' doesn't parse. I don't know what you mean by this. Are you now adding random objects here and there? Then you need to separately compute the acceleration of each and add those accelerations. Newton showed (via shell theorem) that any spherical distribution of mass of radius r can be treated as a point mass by objects outside of r. — noAxioms

...I am sorry for having used the letter “g”, noAxiom...just replace it by “A” instead, from now on. Now you mentioned that when adding masses we also need to “add” accelerations... this is all that I am saying in this post.
The Newtonian gravitational equation, identifying two masses, has only one acceleration , the one caused by the earth.

Sorry, but I kind of came in late to this discussion and I don't know what you're trying to do beyond what is described by these very simple equations. For small objects, they all fall at the same rate, and the only reason the feather falls slower is due to air friction. They brought a feather to the moon to show it falling at the same rate as a rock. PR stunt..,. tax dollars at work. — noAxioms

I am privileged to have any response at all, noAxiom....I have merely a layman viewpoint on this issue, and am simply trying to explain the problems that I have...
Thank you for your time
Grampa Dee.

In reality, it would take not 3.3 seconds to hit the ground, but probably just a fraction of a second in the dense moon case. You're not going to compute that by just adding the accelerations, which treat both objects as point masses or rigid spheres, which they are not. — noAxioms

I understand the case for the dense moon would be extreme....would you have a problem with the equation I have written to Pantagruel?

M / 4pi * r ^2 = k g

Here, if r is small then g can become extremely large.

However, I would also claim that this is only half of the equation.

Any other mass g acceleration would need to be added at every point in space...

...just a thought.


Grampa Dee

Thank you for responding noAxiom

But now if you increase the density of that ball to say the mass of the moon, it will hit the ground in less time, not because its initial coordinate acceleration is any greater, but because it has enough mass to significantly cause the ground to come up and meet it — noAxioms

I fully agree with your description noanxiom, but to claim that the increased acceleration due to the earth moving upwards is separated from the downward moving mass, I don’t quite get. If someone drives towards you at 20 km/hr and you towards him/her at 20 km/hr, the velocity is indeed 40 km/hr relative to the two cars... we add the two velocities

Initial acceleration of an object due to gravity of a primary is mass independent. I mean, F=ma, which if substituted directly into F=GMm/r² gets you A=GM/r², something independent of m altogether. — noAxioms

Here is the reason why I have a problem with the equation. As it is written it follows Galileo’s axiom, for it doesn’t matter what you put as the small mass, the acceleration will continue as being (GM/r^2).
But then, what do we do for the mass equal to the mass of the moon? The upper equation will calculate the acceleration still as being (GM / r^2)...
It is here that I personally believe that another acceleration needs to be added onto the first.
being A is also = Gm/r^2... if we add both of them, then we get A = (GM / R^2) + (Gm / r^2)
However , this equation does not agree with Galileo since a change in mass for the small m will indeed change the total acceleration of the system.

Thank you for responding Pantagruel

Ok. And they have the same acceleration because they have different inertias. How does this not answer that? — Pantagruel

I don't understand how they can have the same acceleration, Pantagruel.
That's the problem that I have.
Grampa Dee

I'm sorry, Pantagruel, I might very well be misunderstanding what you're saying.

But also yes, the overall force realized does vary if the mass of the smaller object varies — Pantagruel

I do agree that the overall force will vary with different masses.
My post is about reconciling gravity with Galileo’s concept of different masses having exactly the same acceleration in freefall;... this is the problem I’m having.right now.

Grampa Dee

Thank you for replaying Pantagruel

ndeed it is, but the same condition applies with respect to the relationship between force and inertia. The force is a composite product of the two masses, as is the acceleration. — Pantagruel

Maybe you could help me in this area. The way I see it would be that Mass (the earth) gravitational energy will be causing an acceleration g at a certain distance ....the whole sphere at that distance will have the same g...
M / 4pi * r ^2 = k g

...in fact, this is all we need in my opinion to calculate the g acceleration.
However, since the same is said for the second mass, the two accelerations will need to be added, and if the second mass is great or small will make a difference in the overall g, it seems.

Grampa Dee

Thank you for responding Pantagruel:

Different masses accelerate at the same rate towards a reference mass because they also have different inertias, which balances the different forces generated..... — Pantagruel

I personally see this as being only part of the equation since the falling bodies should attract the earth as well, and the difference of the earth’s acceleration towards the falling body will be different with different masses .

Grampa Dee

Thank you for your response, unenlightened

The reason for this is that the mass of a falling object is negligible in relation to the mass of the Earth. — unenlightened

Yes, I would agree with that as well; my concern is the Galileo assumption that the mass of a falling body does not affect the acceleration (g) at all. This opposed the Greek philosopher Aristotle, who was said to claim that heavier objects were falling quicker .
I would in fact surely agree to viewing the system as a whole(earth and falling bodies)as being invariant in terms of energy “if” the fallen body was taken from the earth (did not come from outer space), for in this case, the falling body was formally on the ground being part of the cause for the g acceleration, and since while the falling body will attract the earth as well, then,for the whole system, the overall acceleration will not change, in my mind.

the mass of the Earth cannot probably still be measured to the ton, and if it could, some of the mass would been the air above and exert a negative force. — unenlightened

I agree that the air will indeed attract the earth as all bodies will cause gravity......
Since the air has weight, it still will have an effect in being part of the earth’s gravitational acceleration. However you have pointed out something important, as gravity is caused not only by the mass only, but by it’s density as well; for example, if the earth would be squashed to the size of the moon, it would have a much greater surface gravity....

Grampa Dee

That wouldn't work since this has the dimensions of squared meters per seconds and we want something that has the dimensions of seconds.

To arrive at the simplification, note that gamma can be rewritten as 1c2c2−v2c2√=1(c2−v2)c2√1c2c2−v2c2=1(c2−v2)c2 and therefore γ2γ2 is c2c2−v2c2c2−v2

Our expression 2Lcγ(c2−v2)2Lcγ(c2−v2) can therefore be rewritten as 2L⋅γ2c⋅γ2L⋅γ2c⋅γ, which is 2Lγc2Lγc. As gamma is dimensionless, this expression indeed has the dimensions of seconds. — Pierre-Normand

Thank you Pierre - Normand; it's now clear as a bell :)
I was impressed with you're mathematical juggling
and I understood everything you did.

Hello Pierre Normand; how are you? Thank you again for this interaction.


Pierre Normand wrote ....within our first exchange....I think :)

The time required for the light ray to reach the receding mirror is therefore t1 = d/v1, where d is the distance between the source and the mirror (d = L/gamma) and v1 is the relative velocity between the source and the mirror (v1 = c - v). Similarly, the time required for the light ray to return to the source is therefore t2 = d/v2 where v2 = c + v.

The total time elapsed is therefore t1 + t2 = L/(gamma(c - v)) + L/(gamma(c+v)).

I agree

Pierre Normand wrote:
To simplify, you can multiply the numerator and denominator of the first term by c + v and the numerator and denominator of the second term by c - v.

As you’re multiplying by 1, I fully agree.


Pierre Normand wrote:
You get
t = L(c + v)/(gamma(c^2-v^2))+L(c - v)/(gamma(c^2-v^2)) = 2Lc/(gamma(c^2-v^2))

ok...good

Pierre Normand wrote:
Since gamma = 1/sqrt(1-v^2/c^2), 2Lc/(gamma(c^2-v^2)) simplifies to 2L*gamma/c which is the same time registered by the vertical light clock.

Shouldn’t it be 2Lc*gamma instead?

Grampa Dee (Andre)

Hello Pierre-Normand, nice to hear form you again

Your thinking was correct, but since both the forward and the return path are travelled by the light pulse at the same speed c (in any referential frame), and since the return path is shorter, the time to travel it is also shorter. This is why you must setup two distinct equations:

d1 = L/gamma + vt_1 = ct_1
d2 = L/gamma - vt_2 = ct_2 — Pierre-Normand

I don't think I have any problem with upper statement.

Think of two people walking an unleashed dog at a steady pace v thereby keeping a constant distance L between them. Picture the dog running back and forth between them at constant speed v_2 > v. The forward path for the dog is longer than L since the person ahead keeps moving the 'goalpost' further away until the dog reaches them. Conversely, the person behind moves the 'goalpost' closer during the time when the dog returns. — Pierre-Normand

I was able to visualize the situation using (c+v) (c-v) instead of the increase/decrease of length to the light path thanks to your example of the two people and the dog.However, I must have something missing here as well because I still arrive at the same conclusion. Maybe you will be able to spot my error.

If we have a moving platform with a velocity of v relative to us (moving from left to right),and let’s say that a camera is recording the event and what we see is the platform on a monitor as though it isn’t moving.The source is at the left and the mirror is located to the right of the platform.

Now we will observe a ray of light travelling from the source to the mirror as having a velocity of (c-v) towards the mirror. Afterwards, the light ray will travel in the opposite direction, back towards the source with a velocity of (c+v).
Here is where I might be wrong (if I’m not wrong already :) ), the average velocity the light ray will have is going to be c. However, we are also observing the platform as having been contracted by gamma^-1. This, in my mind, identifies a time contraction of gamma^1 instead of a time dilation of gamma, it seems.

I hope you don’t think that I’m merely trolling here. I sincerely have problems understanding s.r.
The example that you gave (people and dog) was very good and easy to understand, however, my problem stems directly in the invariant speed of light.

Thanks again for your time and patience..
Andre

ok: Would vt_2 be equal to -vt?
— Gampa Dee

No, since vt is the distance travelled by the mirror during the forward path while vt_2 is the distance travelled by the source during the return path, and the return path takes a shorter time. t_2 < t. — Pierre-Normand

Ok...Maybe this is where I’m off the rail.I was visualizing the light going to the mirror as traveling a distance vt further away from the length of the apparatus (distance between source and mirror)and a distance vt less than the apparatus when coming back towards the source being the reason why it took a shorter time.Maybe, somehow I will focus on this part.

From the first equation, t = L/(gamma(c - v))
From the second equation, t_2 = L/(gamma(c + v))

Therefore, the total time

T = t+t_2 = L/(gamma(c - v)) + L/(gamma(c + v))

To simplify this, we can multiply the first term by (c+v)/(c+v) and the second term by (c-v)/(c-v), effectively multiplying each term by 1.

T = L(c + v)/(gamma(c - v)(c + v)) + L(c - v)/(gamma(c - v)(c + v))
= L((c + v) + (c - v))/gamma(c2 - v2) = 2Lc/gamma(c2 - v2)

Next, we can express gamma explicitly as 1/sqrt(1 - v^2/c^2), and divide both the numerator and denominator by c^2 to get

T = (2L/c)sqrt(1 - v^2/c^2)/(1 - v^2/c^2) = 2Lgamma/c

As long as all velocity terms are expressed relative to the same inertial frame of reference, the relative velocities between two material objects (or between a material object and a photon) can be expressed as simple sums or differences (e.g., c + v or c - v). This doesn't involve any Newtonian assumptions. When the light pulse travels at c and the clock at v, their relative velocity is c + v (or c - v). This is because it's the rate at which their separation changes, as measured in the stationary reference frame. — Pierre-Normand

What I will do for now, Pierre-Normand, is try to understand on my own your derivation using (c+v)(c-v)method since it does indeed seem to make sense. Right now, the way I’m visualizing this, I am calculating the distances travelled by the moving apparatus vt and then claiming that the light is travelling with the velocity of "c" in between those distances (L+vt). The other thing I will focus on is my visualizing the return path as being vt shorter than the apparatus. If you think of some way to clarify any of this, I will
certainly look at it; if not, it’s ok; for this is simply a pastime of mine, I don’t feel the “need” to understand any of this.

ps: Your name tells me that you might be a french speaking person? I am as well...my name is Andre

Yes, from the Montreal area, in Quebec, Canada. Nice meeting you, André. — Pierre-Normand

Hello Pierre-Normand, I’m your next door neighbor from northern Ontario.
Take care....we’ll probably chat again sometime.
Grampa Dee
Andre

In the case of the horizontal clock you indeed have d1 = L/gamma + vt for the first path and d2 = L/gamma - vt_2 for the return path. I use "t_2" since the time for traveling the return path is different (shorter). Since in both cases the distance is travelled at c, we have the two equations:

d1 = L/gamma + vt = ct
d2 = L/gamma + vt_2 = ct_2 — Pierre-Normand

ok: Would vt_2 be equal to -vt?

T = L/((gamma(c - v)) + L/(gamma(c + v)) with is the same result I had obtained more directly by considering the relative velocities of the light pulse and clock. — Pierre-Normand

This is where it gets foggy for me; you seem to divide by a relative velocity (c+v) instead of adding another distance vt....I’m not saying that you’re wrong, only that I can’t visualize it . If v causes a foreshortening of L, then why not say T = L / (c – v) + L / ( c+v) ? but here, we would be in Newtonian mechanics wouldn’t we? :)

Would it be possible to give the calculation without having (c+v) or (c –v) in the equation?
I understand a bit why you’re doing this but I’m not comfortable with it. For example: if we take c = 1
and v = .5 , then L /(c+v) = L / 1.5 and L/(c-v) = L/.5...Now let L = 1
We would then have L / 1.5 + L/.5 = .66666 + 2 = 2.66666
But, for ( L +vt) / c and (L – vt) / c, we would have [(1+.5t) + (1 -.5t) / 1] = 2
If we start with what you once wrote:

In the case of the horizontal clock you indeed have d1 = L/gamma + vt for the first path and d2 = L/gamma - vt_2 for the return path. I use "t_2" since the time for traveling the return path is different (shorter). Since in both cases the distance is travelled at c, we have the two equations:

d1 = L/gamma + vt = ct
d2 = L/gamma + vt_2 = ct_2 — Pierre-Normand

Without using (c+v)(c-v), could you derive T = 2L*gamma/c in another way ?

Thank you again for your time and patience.

Grampa Dee.

ps: Your name tells me that you might be a french speaking person? I am as well...my name is Andre



:

Thank you for your time and patience, Pierre Normand

.

Yes, you're right that the Lorentz contraction of the path must also be taken into account. The Lorentz factor, gamma, is always greater than 1, so the contracted length is L/gamma. — Pierre-Normand

Yes; thank you...I always think of gamma as being the SQRT 1 – v^2 / c^2, but it’s actually the inverse.

The time required for the light ray to reach the receding mirror is therefore t1 = d/v1 — Pierre-Normand

I have a problem understanding this though; I rather see the time light displaces any distances as being d / c; I would instead claim that v would be responsible for the length contraction instead, being d or L/ gamma + or - some other distance being vt .Also, isn’t the relative velocity between the source and the mirror 0? Therefore, I, for now, understand the 1st path as being
t1 = d / c or (L/gamma +vt) / c

Similarly, the time required for the light ray to return to the source is therefore t2 = d/v2 where v2 = c + v. — Pierre-Normand

Again, I would assume t2 = d / c, d being equal to L/gamma – vt.in this case.

The total time elapsed is therefore t1 + t2 = L/(gamma(c - v)) + L/(gamma(c+v)). To simplify, you can multiply the numerator and denominator of the first term by c + v and the numerator and denominator of the second term by c - v. — Pierre-Normand

I’ve got
(L / gamma + vt) / c + (L/gamma – vt) / c = 2L/gamma c

You get

t = L(c + v)/(gamma(c^2-v^2))+L(c - v)/(gamma(c^2-v^2)) = 2Lc/(gamma(c^2-v^2))

Since gamma = 1/sqrt(1-v^2/c^2), 2Lc/(gamma(c^2-v^2)) simplifies to 2L*gamma/c which is the same time registered by the vertical light clock. — Pierre-Normand

2L*gamma /c was what I had in my first response to you :); however you correctly pointed to me that gamma was a factor which increased the original value, where I needed one which would reduce the outcome ( length contraction).
Now, what I have is instead 2L/gamma c.
I will try to clarify a bit my point. Since the velocity remains c, I think that all we need is the distance for the light to travel in order to calculate the time.
In the first case, (the vertical light clock) the distances the outside observer calculates are the two hypothenuses SQRT L^2 + (vt)^2, the total being 2 * SQRT L^2 + (vt)^2
In the second case,for the horizontal clock, I have L/gamma. + vt one way, and the return path being L/gamma – vt.
If L/gamma +vt = L* [SQRT 1 –v^2/c^2] +vt ..and.the second path is even shorter.
Here’s my problem...as v tends to c, in the vertical clock the path tends towards L^2 +ct^2 ...a very long path,if not infinite, whereas for the horizontal path , it seems the path tends towards 0.
I seem to have two extreme opposite situations.

(Note that c/(gamma(c^2-v^2) = c*sqrt(1 - v^2/c^2)/c^2-v^2 = sqrt(c^2 - v^2)/(c^2-v^2) = 1/sqrt(c^2-v^2) = 1/(c*sqrt(1-v^2/c^2) = gamma/c.)

Intuitively, as v tends towards c and taking into account the Lorentz contraction of the clock, the duration of the return travel tends towards zero. Meanwhile, the duration of the forward travel tends towards infinity despite the shortened distance, as c - v approaches zero faster than sqrt(1-v^2/c^2) does. — Pierre-Normand

Interesting...it’s what I see in the two clocks...(vertical and horizontal). What I would disagree, for now, is that the duration of the clock for the forward travel (on the horizontal clock) would still tend to 0 due to the length contraction of the path between the source and the mirror

I thank you again for your time.
Grampa Dee.