J
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H
hecd2
Guest
Let’s continue with the analysis. In my last post I proved that the only acceleration observed from any inertial frame of a body revolving in a circle is the acceleration labelled a[cp] by Jano in his diagrams. When you observe circular motion from a frame moving in the same plane as the circle, the body’s trajectory is a cycloid. I proved that the acceleration of the body following the path of a cycloid is the same in all inertial frames (including the one at rest with respect to the centre of the circle) and it is always dirceted to the centre of the circle. There is no acceleration a[ct]. Wherever Jano shows an acceleration a[ct], for example in his posts #97, #98 and #116, it is erroneous, Such accelerations are not observed from inertial frames. This includes the first diagram in #97 where he shows a[cp] and a[ct] as being equal and opposite. There is no acceleration a[ct] in that and the other diagrams.
Now let’s look at B2 in the same way that we looked at B1:
Compare the acceleration of B2 with that B1:
All the characteristics of B2 are the same as B1 - the acceleration of B2 is a[cp] only, the same in all inertial frames, constant magnitude in time and directed to the centre of the circle (ie along the line of the rod on which B2 is mounted). The only difference between the accelerations of B1 and B2 is that there are in antiphase.
So, in all inertial frames, the accelerations of B2 and B1 are equal and opposite at all times. The forces acting on B1 and B2 which cause these accelerations are given by F=ma, in other words by multiplying the accelerations above by the mass of B1 and B2. Each of B1 and B2 give rise to a reaction force on the axle that is equal and opposite to the force causing their accelerations above. These reaction forces on the axle point towards B1 and B2 respectively in all inertial frames at all times. At the axle these reaction forces are equal and opposite at all times in all inertial frames, and therefore the resultant reaction force on the axle at all times is ZERO.
I have one more post which will discuss the work done and so the changing kinetic energy of B1 and B2 in inertial frames moving with respect to the circle.
Now let’s look at B2 in the same way that we looked at B1:
Compare the acceleration of B2 with that B1:
All the characteristics of B2 are the same as B1 - the acceleration of B2 is a[cp] only, the same in all inertial frames, constant magnitude in time and directed to the centre of the circle (ie along the line of the rod on which B2 is mounted). The only difference between the accelerations of B1 and B2 is that there are in antiphase.
So, in all inertial frames, the accelerations of B2 and B1 are equal and opposite at all times. The forces acting on B1 and B2 which cause these accelerations are given by F=ma, in other words by multiplying the accelerations above by the mass of B1 and B2. Each of B1 and B2 give rise to a reaction force on the axle that is equal and opposite to the force causing their accelerations above. These reaction forces on the axle point towards B1 and B2 respectively in all inertial frames at all times. At the axle these reaction forces are equal and opposite at all times in all inertial frames, and therefore the resultant reaction force on the axle at all times is ZERO.
I have one more post which will discuss the work done and so the changing kinetic energy of B1 and B2 in inertial frames moving with respect to the circle.
H
hecd2
Guest
Let’s now consider the changing kinetic energy of B1 and B2 in time when observed from an inertial frame moving with respect to the circle. Jano believes that a centrifugal acceleration is required to explain this time-varying kinetic energy, but we have seen that no such centrifugal acceleration is present. I’ll prove that the time varying kinetic energy of B1 and B2 is explained by the force provided by the connecting rod on B1 and B2 doing work on B1 and B2.
The work done on a body is defined by the scalar product of the force vector acting on the body and the displacement of that body. The instantaneous work done is therefore the scalar product of the force vector acting on the body and the velocity vector of the body:
What are the characteristics of the instantaneous work (dimensions of Joules/second by the way) done on B1 and B2. First of all, you’ll notice that the analysis using the centripetal acceleration and force only (which are the only accelerations and forces acting on B1 and B2) gives the finite and time-varying work which results in time-varying kinetic energy. Next we can see that if V=0 (if we observe in the rest frame of the circular motion), then the work is zero - ie the kinetic energy of B1 and B2 individually in this frame is constant. Next, the instantaneous work on B1 and B2 is equal and opposite (because their masses are equal) so the summed work on B1 and B2 is zero in all inertial frames. In other words the sum of the kinetic energy of B1 and B2 is constant in all inertial frames.
Jano will ask whether B2 is doing work on B1? If they are connected by a string, yes, B2 is doing positive work on B1 and B1 is doing negative work on B2. (The arrangement he describes is not like that because the axle in his arrangemnet represents an infinite effective mass - infinite because it moves at a constant speed regardless of what forces act on it. The axle is a perfect energy source or sink and so the system is not kinematically complete - the mass of B1 could be a million times the mass of B2 and yet, in his arrangement, the total system would behave exactly the same way - not so if B1 and B2 are simply joined by a string. So in his arrangement, the energy instantaneously lost is sunk and the energy instantaneously gained is sourced in the infinite mass of the axle). It would be instructive for Jano to repeat this analysis with two unequal masses connected by a massless infinitely stiff string.
The conclusion however is incontrovertible. This classical system is internally consistent and predicts exactly what is observed. The accelerations a[ct] as depicled by Jano are not present in inertial frames, and all phenomena including the time-varying kinetic energy of individual masses correctly falls out of the analysis in which the only accelerations and forces acting on B1 and B2 are centripetal.
The work done on a body is defined by the scalar product of the force vector acting on the body and the displacement of that body. The instantaneous work done is therefore the scalar product of the force vector acting on the body and the velocity vector of the body:
What are the characteristics of the instantaneous work (dimensions of Joules/second by the way) done on B1 and B2. First of all, you’ll notice that the analysis using the centripetal acceleration and force only (which are the only accelerations and forces acting on B1 and B2) gives the finite and time-varying work which results in time-varying kinetic energy. Next we can see that if V=0 (if we observe in the rest frame of the circular motion), then the work is zero - ie the kinetic energy of B1 and B2 individually in this frame is constant. Next, the instantaneous work on B1 and B2 is equal and opposite (because their masses are equal) so the summed work on B1 and B2 is zero in all inertial frames. In other words the sum of the kinetic energy of B1 and B2 is constant in all inertial frames.
Jano will ask whether B2 is doing work on B1? If they are connected by a string, yes, B2 is doing positive work on B1 and B1 is doing negative work on B2. (The arrangement he describes is not like that because the axle in his arrangemnet represents an infinite effective mass - infinite because it moves at a constant speed regardless of what forces act on it. The axle is a perfect energy source or sink and so the system is not kinematically complete - the mass of B1 could be a million times the mass of B2 and yet, in his arrangement, the total system would behave exactly the same way - not so if B1 and B2 are simply joined by a string. So in his arrangement, the energy instantaneously lost is sunk and the energy instantaneously gained is sourced in the infinite mass of the axle). It would be instructive for Jano to repeat this analysis with two unequal masses connected by a massless infinitely stiff string.
The conclusion however is incontrovertible. This classical system is internally consistent and predicts exactly what is observed. The accelerations a[ct] as depicled by Jano are not present in inertial frames, and all phenomena including the time-varying kinetic energy of individual masses correctly falls out of the analysis in which the only accelerations and forces acting on B1 and B2 are centripetal.
J
Jaaanosik
Guest
So, “If they are connected by a string, yes, B2 is doing positive work on B1…”
What force is doing work?
The mechanical work can be done only by force. The force can be applied only by contact or field.
Is B2 m*a[cp] pushing the string and doing the work?
… or do you expect B1 with less energy to pull B2 that has more energy?
H
hecd2
Guest
We have seen that the instantaneous work done on B1 (indeed on any mass rotating in a circle) is:
This is identical to the varying component of kinetic energy in Jano’s post #98. This proves that calculating the work using only the acceleration a[cp] gives a consistent result. There is no a[ct]
This is identical to the varying component of kinetic energy in Jano’s post #98. This proves that calculating the work using only the acceleration a[cp] gives a consistent result. There is no a[ct]
H
hecd2
Guest
Now, you’re just trolling.
I’ve spent far more time on this than it’s worth. But I have done you a favour - I’ve proved that the standard understanding of Newtonian mechanics is self-consistent and there are no anomalies or difficulties as you believe. Moreover, I have analysed the two cases you present in detail, so that if you work them through yourself you could actually learn some physics. ( I have little confidence that you’ll do so, given your incorrigible and unjustifiable arrogance.)
And the scenario of two rotating masses connected by a string? I’ll leave that for you as an exercise, with the hint that there is a *proof *in the last few posts that the only forces and the only accelerations in the system are directed along the string as observed in all inertial frames.
Now I’m outta here.
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Jaaanosik
Guest
Hecd2,
I really appreciate your responses. I assure you I am not trolling.
This is a serious question. How B2 m*a[cp] is going to do work against the string? To cut a[ct] out of the inertial frame is wrong in some instances.
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I’ll try to explain it in the following example. Let’s imagine a spaceship in zero gravity with a big cabin like a gym. A floating astronaut in the middle of the cabin.
An engine starts to apply force F1 to follow a circular trajectory, it generates a[cp]. The astronaut sees a wall of the cabin coming towards him and he does not feel any acceleration till he touches the wall.
The engine has to increase output to bigger force F2 in order to stay on the same trajectory or if the force stays at F1 then a[cp] will be lower.
How come a[cp] would go down? Because of the astronaut’s inertial force. That force is real and it’s changing a[cp]. It becomes real when astronaut comes to contact with the ship.
A typical action reaction. It’s interesting that m*g is considered in the inertial systems as a force but the centrifugal force is dismissed.
If a spaceship would try to do an approach to another ship on a half of a cycloid then varying a[ct] and a[t] break the relativity. Your latest integral would not be through the whole cycloid.
Think again about satellite flyby anomalies, varying a[ct] and a[t] on a part of a cycloid.
These are examples when not to consider a[ct] is wrong.
Let me show you some quotes from professor Turok: macleans.ca/politics/ottawa/perimeter-institute-and-the-crisis-in-modern-physics/
“Theoretical physics is at a crossroads right now,”
“In a sense we’ve entered a very deep crisis.”
“But given that everything turned out to be very simple, yet extremely puzzling — puzzling in its simplicity — …”
What do you think, why he is talking in the past tense?
“But given that everything** turned out** to be very simple, yet extremely puzzling…”
I have another point to make but I need to prepare an image for it…
J
Jaaanosik
Guest
Hecd2,
from this book
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When we combine work-energy principle with what you were saying that work is not equal in two inertial frames then the conclusion is that there are two different energy amounts in two different frames.
Does it mean we are going to get energy out of nothing to keep things in balance for the sake of relativity?
Do you see the problem?
Do you see how a spaceship/satellite can run out of fuel if the calculation is not done right?
from this book
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When we combine work-energy principle with what you were saying that work is not equal in two inertial frames then the conclusion is that there are two different energy amounts in two different frames.
Does it mean we are going to get energy out of nothing to keep things in balance for the sake of relativity?
Do you see the problem?
Do you see how a spaceship/satellite can run out of fuel if the calculation is not done right?
J
Jaaanosik
Guest
Let me sum up with a story of a lost astronaut.
There are two space stations S1 and S2 flying around a star at a constant velocity. The distance is a few kilometers between the stations.
An astronaut is flying from S1 to S2 but because some rocks got between the stations, he has to do a half of a circle around them.
He calculates how much fuel is needed to do the trip, he has just enough of it, so he flies.
Oops, he does not make it to the S2 station, he is lost in space.
Had he realized that he is flying in a half a cycloid he would know that more work has to be done on a half of a cycloid and based on the work-energy principle he would know he needs more energy - fuel.
The process of the fuel energy conversion to work (forces) can not be happening at two different rates. The process does not care about inertial frames.
How do we know that the cycloid (outer reference frame) is real?
Going back to my first two post clears the issue. The critical speed and resonance is real, predicted by the observer in the higher frame.
The higher reference frame is the preferred one!
There are two space stations S1 and S2 flying around a star at a constant velocity. The distance is a few kilometers between the stations.
An astronaut is flying from S1 to S2 but because some rocks got between the stations, he has to do a half of a circle around them.
He calculates how much fuel is needed to do the trip, he has just enough of it, so he flies.
Oops, he does not make it to the S2 station, he is lost in space.
Had he realized that he is flying in a half a cycloid he would know that more work has to be done on a half of a cycloid and based on the work-energy principle he would know he needs more energy - fuel.
The process of the fuel energy conversion to work (forces) can not be happening at two different rates. The process does not care about inertial frames.
How do we know that the cycloid (outer reference frame) is real?
Going back to my first two post clears the issue. The critical speed and resonance is real, predicted by the observer in the higher frame.
The higher reference frame is the preferred one!
J
Jaaanosik
Guest
Ignoring inertial forces in the inertial frames of reference is the biggest mistake in physics.
The steel ball mass at the tip of the string is an accelerated frame but it’s connected to the inertial frame via the string and it’s exchanging the kinetic energy with the moving inertial reference frame.
This exchange can happen only with one absolute rate therefore the relativity is not true.
Where does it leave gravity?
Well, this linked pdf of the book is the best possible explanation.
theelectromagneticnatureofthings.com/img/emnature.pdf
The energy/mass density with the fact that the repulsion is slightly smaller than the attraction will end up showing an effect of ‘gravity’.
The steel ball mass at the tip of the string is an accelerated frame but it’s connected to the inertial frame via the string and it’s exchanging the kinetic energy with the moving inertial reference frame.
This exchange can happen only with one absolute rate therefore the relativity is not true.
Where does it leave gravity?
Well, this linked pdf of the book is the best possible explanation.
theelectromagneticnatureofthings.com/img/emnature.pdf
The energy/mass density with the fact that the repulsion is slightly smaller than the attraction will end up showing an effect of ‘gravity’.
J
Jaaanosik
Guest
Hecd2,
This is not correct.
One ball on the rod can be considered as a pendulum in the ground reference frame.
The second ball is another pendulum.
The a[t] accelerations show that this pendulums go against each other.
The angle between the rods will not stay 180 degrees.
This is an experiment to detect a constant linear motion without any signal from outside.
And we should remember how m*g is treated in the pendulum analysis. That’s the force giving the kinetic energy to the pendulum bob.
J
Jaaanosik
Guest
Here is a preview of our latest paper.
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J
Jaaanosik
Guest
H
hecd2
Guest
Gosh - this is something of a zombie thread and one which is rather out of place in the Philosophy forum of Catholic Answers. It would be better on a physics/science forum of some sort in which case you will get excatly the same response as I am about to give you.
What you have linked to is not a paper but a post on a self-promoting crackpot web-site. If it was a paper it would be published in a physics journal having gone through peer review. However that’s never going to happen because any competent referee would tell you what I am going to tell you: it is unmitigated garbage.
The fundamental errors begin as early as equation 4. You cannot calculate acceleration by differentiating the magnitude of the velocity (also known as speed). Why? Because acceleration is *defined *as change in velocity and velocity is a vector. I showed you the correct way to derive the acceleration in both inertial frames of reference in post #139 in this thread and explained that the acceleration is the same in *all *inertial frames. No, I did more than explain it - I proved it because the velocity of the inertial frame with respect to any other inertial frame disappears in the second derivative of position. If you differentiate speed (which is a scalar) you get a scalar quantity which is the change in speed as a function of time - which is *not *acceleration. That is the reason that the paper gets the wrong expression for acceleration in equation 4 and later on in equations 18 through 20.
The same fundamental error is repeated throughout the post you linked. I stress, for anyone on this thread who cares, that I prove in post #139 that the acceleration is the same in all inertial frames, a result that is well known and simple to prove. This linked post is nonsense, and can be seen to be nonsense by any competent physicist with no more than a cursory examination.
J
Jaaanosik
Guest
Normal and Tangential coordinates are equivalent to x, y coordinates.
Vector has magnitude - change in magnitude is tangential acceleration.
Vector has direction - change in direction is normal acceleration.
There are also polar coordinates… any of them can be used.
No physics books says that only x, y coordinates are supposed to be used.
Please, show me if it’s otherwise.
NT calculation matches the x, y calculation for the circle wave. It does not for the cycloid wave.
Well, thank you for your opinion anyway.
A
adamhovey1988
Guest
I don’t know the context of the question so I don’t know how to answer that
H
hecd2
Guest
No, just no - particularly the bolded quotation above.
It is true that you can use any number of co-ordinate systems including Cartesian and a wide range of orthogonal co-ordinates (spherical, cylindrical, ellipsoidal, prolate spherical etc etc). The physics is identical under co-ordinate transformations. You can also determine the instantaneous tangential speed and instantaneous tangential and radial accelerations expressed in any of these co-ordinate systems. However, tangential and radial directions do NOT constitute a co-ordinate system fixed to an inertial frame. They can be expressed in a cartesian co-ordinate system fixed to an accelerating (and thus non-inertial) frame. They are not equivalent to “x, y” (Cartesian) co-ordinates.
As I have pointed out multiple times in this thread, it is easy to prove that, for any arbitrary motion, the acceleration is identical in ALL inertial frames. This is true, independent of orthogonal co-ordinate system. If someone gets a result different from this, you can be sure that they have made either a conceptual or a computational mistake. On the other hand, no-one claims that the tangential and radial components of acceleration are separately identical in different inertial frames, and in fact it is easy to show that they cannot be in any case where the tangential speed in any one frame is time-varying.
This is all conventional Newtonian mechanics - so it’s difficult to see what original result is being claimed in this rather dismal post that you linked to.
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Jaaanosik
Guest
Normal and Tangential coordinates can be used in two different inertial reference frame systems.
I can show you thousands of plots and results when they produce the same acceleration even though the calculations are done in different IRFS.
It has to be like that for the relativity to work.
Look here:
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Your post:
You are absolutely wrong here. Do you admit it?
H
hecd2
Guest
You have been impervious to education from the very first time you posted in this thread.
First of all, we are not talking about relativity - we are talking about Newtonian mechanics (there is a thing called Galilean relativity in Newtonian mechanics, but that’s not what most people mean when they talk about relativity). If we were talking about relativity we would have to use the Lorentz transform rather than the Galilean transform between inertial frames.
Secondly, no-one in their right mind claims that you can’t use path variables to solve problems in physics, but that is NOT the same as saying that they are equivalent to x,y or any other orthogonal co-ordinate system fixed to an inertial frame. It is obvious why that is the case: the path variable co-ordinate system is fixed to the particle and the direction of the tangential axis rotates as the direction of the particle changes. That is so by definition. Therefore for any non-trivial case where the velocity of the particle is time-varying in an inertial frame, the tangential/radial co-ordinate system is not inertial because it is linearly accelerating and/or rotating in an inertial frame. Can you transform from these non-inertial co-ordinates to inertial co-ordinates? Of course you can, but that means taking account of the time-varying direction and speed of the tangential co-ordinate in an inertial frame, which means taking account of the direction and speed of the particle in an inertial frame - so you might as well just work in an inertial frame for a problem like this to begin with.
Which brings me to my last point: the article mixes and confuses vector and scalar quantities. I said you cannot calculate acceleration which is a vector by differentiating the magnitude of the velocity (the speed) and indeed you cannot if you throw away the direction. The post doesn’t properly distinguish between scalar and vector quantities which is fatal as they mean different things. In the section from the textbook that you have linked below vector quantities are bold and scalar quantities are represented by normal text. And you see that quantity e which appears in 2/8. That is a unit vector in the direction of the particle’s motion. e is time-varying and you can’t just throw it away.
I have explained probably 15 times in this thead that the vector acceleration of a particle undergoing any arbitrary motion is the same in all inertial frames - and I have proved why that has to be the case. If you agree with that then I don’t know what all the fuss is about. If you are getting a different result it’s because you have made one or more mistakes, based on the fact that tangential and radial co-ordinates are not fixed to an inertial frame, and that you are conflating vector and scalar quantities.
J
Jaaanosik
Guest
The equation 2/8 has two parts - normal component and tangential component.
It’s in front of your eyes. The second part - tangential component is derivative of what?
What do you see? What does not bolded “v” and the dot above it represent?
Why is it so hard for you to admit that you were wrong?
That linked post does properly account for both: Normal and Tangential components in the ground and in the moving inertial reference frames.
The Galilean/Newtonian relativity is a subset of the Special Relativity.
If the Galilean relativity is screwed up so is the whole relativity.
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