Gravity as a consequence of conservation of energy

[Simplico]

Theory of zero

[StringJunky]

2 minutes ago, Simplico said:

Theory of zero

You need to post it here or summarize it.

[swansont]

!

Moderator Note

What StringJunky said, with moderator confirmation. Rule 2.7 mandates that you post it here.

 

[Markus Hanke]

In some sense, gravity is indeed the result of a conservation law of sorts, but it has nothing to do with energy. The law in question is the topological principle that the boundary of a boundary is zero. Misner/Thorne/Wheeler have described this very nicely in “Gravitation”.

So far as energy is concerned though, it should be remembered that the energy associated with a region of curved spacetime is a difficult to define concept. It is also not localisable.

[geordief]

21 minutes ago, Markus Hanke said:

In some sense, gravity is indeed the result of a conservation law of sorts, but it has nothing to do with energy. The law in question is the topological principle that the boundary of a boundary is zero. Misner/Thorne/Wheeler have described this very nicely in “Gravitation”.

So far as energy is concerned though, it should be remembered that the energy associated with a region of curved spacetime is a difficult to define concept. It is also not localisable.

You earlier today(?) said that  things (inc gravity?) were fundamentally quantum rather than classical.

Does this hierarchy also apply to the "boundary of a boundary "? ie it applies if it does in a classical context and so cannot be considered  fundamental?

 

Ps I don't want to give the impression that I was able to understand the "boundary of a boundary" idea which you tried to help me with in an earlier thread elsewhere a good while previously(although I feel I may have appreciated its importance)

[Markus Hanke]

2 hours ago, geordief said:

Does this hierarchy also apply to the "boundary of a boundary "? ie it applies if it does in a classical context and so cannot be considered  fundamental?

Yes, I did indeed say that the universe is fundamentally quantum. Nonetheless, in this thread I am only talking about classical gravity, being the kind we know about and can mathematically describe. Whether aforementioned topological principle plays a role in quantum gravity is not a question I can answer, since we don’t have a model for quantum gravity just yet. It remains to be seen.

[Simplico]

Theory of zero (part 1) - Inertial motion.

Abstract

  The purpose of this essay is to show that gravity is a consequence of the law of conservation of energy, and not an agency  in it's own right.

Introduction

This essay postulates that the law of conservation of energy is the only law of physics that exists at the macro level. All others  - including gravity - are the local manifestations of this one universal law. This essay therefore defines  common units in terms of energy; distance between two bodies is defined as the work required to bring them to a common centre  of mass (barycentre); mass is the amount of work a body can do per unit of distance; velocity is the difference in kinetic energy between two bodies.

A gun fired in space

If a gun is fired in space, the bullet will move in one direction (relative to their barycentre), with the gun moving in the opposite  direction (recoil). They move with velocities that are proportional to their mass, so that the work done (mass x distance)  in each direction is equal. This leaves the nett work done as zero at the barycentre. Using the definition of mass as resistance  to change in velocity means the work done by the individual components of this system (gun and bullet) may increase indefinitely,  even though the nett work done remains zero. If one imagines reversing the procedure, then the gun and bullet would be moving  towards each other and would stop moving relative to each other when they had collided and formed a common centre of mass.  Thus, all motion between two bodies maybe considered to occur relative to their barycentre. If we say that mass is the  amount of work a body can do per unit of distance then mass reduces as distance increases (in order to keep the work done in a given  time the same). This means that both the energy levels of all components of the system and the nett work are unchanged with time.

 

.

Inertial motion

Inertial motion is motion which does not cause the energy levels of any body in the system to change.  Because it does not cause change it cannot be detected. The energy of a body (with respect to another body)   in inertial motion can be given by the formula: e=m*d*G where e is the kinetic energy of the body, m is it's mass, d is it's distance from it's barycentre with another body, and G is the inertial constant (how much the mass of a body changes in a given period of time).

Path of equilibrium

The path of equilibrium is the path along which the mass point of a body in inertial motion travels in order to conserve the energy of the system.    It is given by the formula:

 where x and y represent the centre of mass of a body in a co-ordinate system whose origin is the barycentre of said  body and the body it is moving relative to. e is it's kinetic energy (again, relative to the body it is moving towards or away from)  and m is it's mass (again relative). The only constant is G, as a body in inertial motion must do equal work  in equal time, or violate the law of conservation of energy.

Neutral and real observers

 The neutral observer is an observer who is at the barycentre of two bodies moving relative to each other. It is s/he who sees the bodies move in opposite directions with velocities relative to their mass. But the neutral observer is an imaginary observer only; any real observer must have mass of their own,  and therefore can never be at the barycentre of themselves and any body they are observing.  Additionally, any real observer will be unaware of their own motion relative to another body and will always see the  other body as the one in motion. In the ollowing example, there are two spacecraft, A and B, who are 15km apart,  and are going to dock with each other. Spacecraft A has a mass of 5000kg, and craft B a mass of 10000kg.  The neutral observer will therefore be 10km from craft A and 5km from craft B. The neutral observer sees both craft moving towards him. An observer on craft A will see craft B as being 15km from him,  and will not see his own craft as moving, while the same will be true for an observer on craft B.

I now introduce a third craft spacecraft, C into the proceedings. This craft is moving relative to both craft A and B.   So, observer A will see craft B and C move, observer B will see A and C move, while observer C sees A and B move.  Because there is no interaction between the crafts, all 3 observers will see the energy of the system  conserved - but none will agree as to what this energy is. This is because they cannot measure their own energy levels. Only the imaginary neutral observer sees the energy of the system as zero;  real observers must see a non-zero value for the energy levels of the system - but this is different for all of them.

Acceleration

Acceleration is the transfer of energy from one body to another. This can be detected by all bodies involved in an interaction because it changes the energy levels of the bodies involved. All observers will agree on the change in the energy levels of the bodies involved. This transfer of energy takes place in waves.

Gravity

Two bodies in motion relative to each other will follow the path of equilibrium around their common barycentre. Depending on their energy levels, this path may take one inside the physical radius of one or the other body,   in which case they will collide. This will cause them convert or lose some of their energy (accelerate) until they come to rest relative to each other (have a common centre of mass). This means however that the equation e=m*d*G is no longer true as the point of zero energy is now common to both. This imbalance causes all mass in the body to attempt to resume inertial motion. However, this is prevented by the other mass in the body, which is what we feel as gravity. When we lose contact with the ground,  inertial motion resumes so gravity 'disappears'. Gravity is caused by preventing the inertial motion -  which is needed to conserve the energy of the system - when the cause disappears, so does gravity.

Prediction

If e=d*m*G it then the distance (d) any body is from it's barycentre with another body is: d=e/mG. Velocity is distance over time, so that means that velocity (v) is e/mGt, where t is the time elapsed.  As the energy of a body in inertial motion doesn't change and neither does G  but mass (m) does, an observer will see the velocity  of a body in inertial motion increase as it moves away from him (assuming he is measuring the distance between himself and the body he is observing as a straight line in a co-ordinate system).  If an observer sees a body of mass mb, moving away from at a velocity v, this will be made up of the sum of the body's movement  away from the barycentre, and the observers own motion in the opposite direction. In a given time (t) he will see the velocity of the  body he is observing increase by tm/tm1 where tm is the total mass of the observer and the body he is observing, and tm1 is the same  after one second has elapsed. In essence, the path the body is following straightens out the further away it is from the barycentre.  So the observer will see the body increase in velocity, but the work done by the body in a given time remains constant.

Conclusions

• Inertial motion is motion which does not change the energy of any body in the system. Inertial motion is orbital in nature.
• All observers see the energy of an inertial system conserved, but no two will agree as to what that energy is.
• The mass of a body is the work it can do per unit of distance from the barycentre.
• Distance is the amount of work that needs to be done to bring two bodies to a common centre of mass.
• Gravity is what happens when inertial motion is prevented.
• It is not possible to distinguish between inertial mass and gravitational mass because there is no gravitational mass - only inertial mass.
• Acceleration is the transfer of energy from one body to another.
• Time is the conversion of energy from one form to another.

[studiot]

1 hour ago, Simplico said:

Acceleration is the transfer of energy from one body to another

What energy is transferred to the accelerating body executing constant circular motion?

[Mordred]

along with what orbital component is involved in a strictly linear momentum system such as a spring ?

[swansont]

 

Quote

If a gun is fired in space, the bullet will move in one direction (relative to their barycentre), with the gun moving in the opposite  direction (recoil). They move with velocities that are proportional to their mass, so that the work done (mass x distance)  in each direction is equal.

mass x distance is not work. The product would be the same because it's related to momentum, which would have the same magnitude.

work is related to energy, and if the masses are unequal, the kinetic energy will be as well.

[Simplico]

•  
• Senior Members
• 1332
• 8481 posts
• Location: Somerset, England

• Report post
•  

Posted Saturday at 09:15 AM

On ‎16‎/‎06‎/‎2018 at 7:47 AM, Simplico said:

Acceleration is the transfer of energy from one body to another

What energy is transferred to the accelerating body executing constant circular motion?

A body only moves in constant circular motion if a force is applied to it. As force results in the transfer of energy, the source of energy is from the force being applied.

[Strange]

47 minutes ago, Simplico said:

•  
• Senior Members
• 1332
• 8481 posts
• Location: Somerset, England

• Report post
•  

Posted Saturday at 09:15 AM

What energy is transferred to the accelerating body executing constant circular motion?

A body only moves in constant circular motion if a force is applied to it. As force results in the transfer of energy, the source of energy is from the force being applied.

Please use the "Quote" button. It will make your posts more readable.

[swansont]

2 hours ago, Simplico said:

• What energy is transferred to the accelerating body executing constant circular motion?

A body only moves in constant circular motion if a force is applied to it. As force results in the transfer of energy, the source of energy is from the force being applied.

The radius is constant, so the potential energy doesn't change. The speed is constant, so the kinetic energy doesn't change.

Where does this energy go?

[Simplico]

I think you are using the definition of acceleration as being any change in speed or direction, so circular motion is considered acceleration. I am not using that definition. I am saying that if there is no transfer of energy there is no acceleration. Consider the case of a rocket ship far out in space, who wants to circle a particular point in space at a constant velocity. In order to do this, the pilot must continuously apply the thrusters. If s/he does not, the rocket will no longer travel in a circle. The rocket is not being accelerated because it is moving in a circle, it is moving in a circle because it is being accelerated. The transfer of energy here is not the potential energy of the rocket as measured by it's radius, or its kinetic energy, as both remain constant (ignoring the loss of mass from the fuel burned). The transfer of energy is from the rocket fuel to empty space. 

[Mordred]

So in order for your model of gravity being a consequence of the laws of inertia, to work you must reinvent those laws to begin with ?

 

Got it. You have your work really cut out for you proving those laws are invalid. Better get cranking out the mathematics including inner and outer products under differential geometry. (Not to mention every physics model out there, as they all apply kinematics lol.

So here is the definition of the very word inertia.

"Inertia is the resistance of any physical object to any change in its state of motion. This includes changes to the object's speed, direction, or state of rest. Inertia is also defined as the tendency of objects to keep moving in a straight line at a constant velocity."

https://en.wikipedia.org/wiki/Inertia

You obviously have a completely different definition involved in your article via requiring orbital motion around some central potential. So you then change the meaning of acceleration. Yet your paper only has two formulas that neither one applies force to calculate work done on a central potential system.

however I am curious what equation you used to plot your image for path of equilibrium as a plot of the equation you have on wolframalpha is significantly different. (I You can plug values into your variables if you like... won't get the same curve as you have in your article.

Edited June 20, 2018 by Mordred

[Mordred]

However lets look at the first equation shall we [latex] E=D*m*G,[/latex] well the consequence of this equation setting G at some constant is that as you increase the distance the energy increases. So where does your conservation apply ?

Edited June 20, 2018 by Mordred

[Markus Hanke]

There is one fundamental issue in this thread that hasn’t even been acknowledged - “gravity” cannot be described as a scalar field; it can’t even consistently be modelled as a vector field. It’s a rank-2 tensor - and it has to be, in order to capture all relevant degrees of freedom. What’s more, real-world gravity is self-coupling, and hence non-linear, unlike any of the concepts bandied about in this discussion. It is therefore futile to play around with notions of acceleration and energy, since at the very best this would produce something akin to linear Newtonian gravity, which is not a complete description of what we see in the real world.

So what is the point in all of this?

[Mordred]

5 minutes ago, Markus Hanke said:

There is one fundamental issue in this thread that hasn’t even been acknowledged - “gravity” cannot be described as a scalar field; it can’t even consistently be modelled as a vector field. It’s a rank-2 tensor - and it has to be, in order to capture all relevant degrees of freedom. What’s more, real-world gravity is self-coupling, and hence non-linear, unlike any of the concepts bandied about in this discussion. It is therefore futile to play around with notions of acceleration and energy, since at the very best this would produce something akin to linear Newtonian gravity, which is not a complete description of what we see in the real world.

lol hadn't got that far yet

[swansont]

12 hours ago, Simplico said:

I think you are using the definition of acceleration as being any change in speed or direction, so circular motion is considered acceleration. I am not using that definition.

That's the definition of acceleration. If you aren't using it, you need to pick another word that describes your situation.

12 hours ago, Simplico said:

I am saying that if there is no transfer of energy there is no acceleration. Consider the case of a rocket ship far out in space, who wants to circle a particular point in space at a constant velocity. In order to do this, the pilot must continuously apply the thrusters. If s/he does not, the rocket will no longer travel in a circle.

However, there is more than one way to do this. If it so happens that point in space is a massive object, one can let gravity apply the force. No work is being done, and the motion is circular.

12 hours ago, Simplico said:

The rocket is not being accelerated because it is moving in a circle, it is moving in a circle because it is being accelerated. The transfer of energy here is not the potential energy of the rocket as measured by it's radius, or its kinetic energy, as both remain constant (ignoring the loss of mass from the fuel burned). The transfer of energy is from the rocket fuel to empty space. 

That's not consistent with the application of conservation of energy. If you want to do physics, you have to follow its rules.

[studiot]

This thread is IMHO a prime example of one that would benefit from this book.

I posted a description here.

 

https://www.scienceforums.net/topic/115096-non-mathematical-reasoning-and-physics/?tab=comments#comment-1057686

 

 

[swansont]

6 hours ago, Markus Hanke said:

There is one fundamental issue in this thread that hasn’t even been acknowledged - “gravity” cannot be described as a scalar field; it can’t even consistently be modelled as a vector field. It’s a rank-2 tensor - and it has to be, in order to capture all relevant degrees of freedom. What’s more, real-world gravity is self-coupling, and hence non-linear, unlike any of the concepts bandied about in this discussion. It is therefore futile to play around with notions of acceleration and energy, since at the very best this would produce something akin to linear Newtonian gravity, which is not a complete description of what we see in the real world.

I think we're still stuck on terminology and getting the physics 101 stuff sorted out. And later on there will be special relativity and QM to acknowledge ("All observers will agree on the change in the energy levels of the bodies involved")

We're a long way from GR objections.

6 hours ago, Markus Hanke said:

So what is the point in all of this?

Indeed.

[Markus Hanke]

40 minutes ago, swansont said:

I think we're still stuck on terminology and getting the physics 101 stuff sorted out. And later on there will be special relativity and QM to acknowledge ("All observers will agree on the change in the energy levels of the bodies involved")

We're a long way from GR objections.

That means the OP is approaching this subject from the wrong angle, because GR is the proper description of gravity; Newtonian physics are only the weak field limit. And issues about who observes what don’t arise if the proper formalism is used right from the start.

Let’s think about this for a minute. First of all, “energy” - when taken in isolation - is an observer-dependent concept, so observers will in general not agree on its conservation, or lack thereof. It’s therefore unsuitable as the source of gravity. So before we can do anything else, we need to ask ourselves just what it is that is actually conserved. To answer this, we consider a small patch of spacetime, and apply Noether’s theorem - out of all the possible symmetries, we find that it is time translation invariance that gives us a conserved quantity related to energy; that is the stress-energy-momentum tensor. It is locally (!) conserved in the sense that

[math]\displaystyle{\triangledown_{\mu}T^{\mu \nu}=0}[/math]

Two things need to be noted here: first of all, the fundamental concept here is not energy-momentum, but spacetime. It’s from spacetime and its local symmetries that, via Noether’s theorem, the notion of energy-momentum arises. Therefore, saying that gravity is the result of the conservation of energy-momentum is missing the point - gravity is the result of the geometry of spacetime. Energy-momentum just forms the source term in the dynamical laws; that’s not the same thing. Secondly, energy-momentum is a purely local concept, and as a tensor it is a purely local mathematical object. Its conservation cannot be easily extended across regions of curved spacetime, because if you integrate the above expression across some volume of curved spacetime, you are left with non-covariant curvature terms that do not identically vanish. So physically, energy-momentum is conserved everywhere locally, but in general no conservation law exists for global regions. Since gravity is clearly a global concept, this will again not get us very far.

This being said, the above expression is one of the constraints that must hold when we construct the field equations from the source term. If the energy-momentum tensor is conserved in the above manner, then so must be the entire other side of the field equation. When we bring together the entire list of mathematical and physical requirements needed to make everything self-consistent, we find that the only object that fulfills them all is a tensor constructed from the Ricci tensor, less its trace; this is called the Einstein tensor. Up to a proportionality constant (and a term of the form const*metric, which we ignore here), this fixes the field equations:

[math]\displaystyle{G^{\mu \nu}=\kappa T^{\mu \nu}}[/math]

And we find that the Einstein tensor is conserved just in the same way as the energy-momentum tensor:

[math]\displaystyle{\triangledown_{\mu}G^{\mu \nu}=0}[/math]

To summarise: the conservation of energy-momentum plays a role in determining the form of the field equations, but it isn’t in itself where gravity comes from. This should be rather obvious, because in vacuum the above equations (after trace-reversal) reduce to

[math]\displaystyle{R^{\mu \nu}=0}[/math]

There is no energy-momentum, but we still get non-trivial solutions (i.e. gravity) to these equations.

Edited June 20, 2018 by Markus Hanke

[Simplico]

If the rules worked, we would have a complete understanding of the universe.

[swansont]

17 minutes ago, Simplico said:

If the rules worked, we would have a complete understanding of the universe.

There's some truth there, but the issue is that the rules we know work pretty well, over a wide range of application. So in that sense, they work. The rules break down where quantum mechanics comes into play. Having an incomplete understanding is not the same as having no understanding.

And the points that have been brought up reflect that you are trying to "fix" an area that works really well, with no hope of addressing where the actual issues are.

[Mordred]

14 hours ago, Markus Hanke said:

That means the OP is approaching this subject from the wrong angle, because GR is the proper description of gravity; Newtonian physics are only the weak field limit. And issues about who observes what don’t arise if the proper formalism is used right from the start.

Let’s think about this for a minute. First of all, “energy” - when taken in isolation - is an observer-dependent concept, so observers will in general not agree on its conservation, or lack thereof. It’s therefore unsuitable as the source of gravity. So before we can do anything else, we need to ask ourselves just what it is that is actually conserved. To answer this, we consider a small patch of spacetime, and apply Noether’s theorem - out of all the possible symmetries, we find that it is time translation invariance that gives us a conserved quantity related to energy; that is the stress-energy-momentum tensor. It is locally (!) conserved in the sense that

▽μTμν=0

Two things need to be noted here: first of all, the fundamental concept here is not energy-momentum, but spacetime. It’s from spacetime and its local symmetries that, via Noether’s theorem, the notion of energy-momentum arises. Therefore, saying that gravity is the result of the conservation of energy-momentum is missing the point - gravity is the result of the geometry of spacetime. Energy-momentum just forms the source term in the dynamical laws; that’s not the same thing. Secondly, energy-momentum is a purely local concept, and as a tensor it is a purely local mathematical object. Its conservation cannot be easily extended across regions of curved spacetime, because if you integrate the above expression across some volume of curved spacetime, you are left with non-covariant curvature terms that do not identically vanish. So physically, energy-momentum is conserved everywhere locally, but in general no conservation law exists for global regions. Since gravity is clearly a global concept, this will again not get us very far.

This being said, the above expression is one of the constraints that must hold when we construct the field equations from the source term. If the energy-momentum tensor is conserved in the above manner, then so must be the entire other side of the field equation. When we bring together the entire list of mathematical and physical requirements needed to make everything self-consistent, we find that the only object that fulfills them all is a tensor constructed from the Ricci tensor, less its trace; this is called the Einstein tensor. Up to a proportionality constant (and a term of the form const*metric, which we ignore here), this fixes the field equations:

Gμν=κTμν

And we find that the Einstein tensor is conserved just in the same way as the energy-momentum tensor:

▽μGμν=0

To summarise: the conservation of energy-momentum plays a role in determining the form of the field equations, but it isn’t in itself where gravity comes from. This should be rather obvious, because in vacuum the above equations (after trace-reversal) reduce to

Rμν=0

There is no energy-momentum, but we still get non-trivial solutions (i.e. gravity) to these equations.

excellent post Marcus

8 hours ago, Simplico said:

If the rules worked, we would have a complete understanding of the universe.

Those rules work far better than what you have presented to us. You should study those rules to see where your errors are. Perhaps then you might have a chance presenting something that makes sense under physics instead of the wishy washy haphazard presentation you provided where you have to change every rule under physics.

If that is your only response to all the issues we have raised then this thread is going nowhere nor accomplishing anything.

In the interest of adding something useful to this thread and to go along with Marcus excellent post who has already covered the stress tensor and curvature aspects. There is under GR and SR a particular relation. The inertial mass and gravitational mass are equivalent. So lets examine this.

[latex] m_i=m_g[/latex]

Ok so lets use Newtons laws of force for a central potential system for gravity.

[latex]F=\frac{GMm}{r^2}[/latex] then set up Newtons second law. [latex]F=\frac{GMm}{r^2}=ma[/latex] so the mass of the object factors out leaving the gravitatonal field potential.

[latex]a=\frac{GM}{r^2}[/latex].

What an interesting conversation piece this makes if one things about it. If the mass or composition of the object does not matter to freefall motion, then there is no way to shield gravity. Unlike the electromagnetic force which definitely relies upon the charge of the particle. This is what GR realizes, not only is there no way to shield gravity but it also couples to everything...unlike other fields which only interact with specific particles.

Now due to this freefall consequence we can set up a series of geodesics that describe the motion of all particles simply by looking at its momentum compared to c (which is a constant). when the particles velocity equals c the geodesic becomes null (timelike).

This means that the gravitational field is a consequence of spacetime curvature not a field into an of itself. One where the absence of gravity is the absence of curvature...

Food for thought considering I used a central potential force equation under Newton...to reflect where GR starts to step in. ie adding the time dimension to understand how curvature works. It s a heuristic view I so far as we have not examined parallel transport of two force vectors and the loss of parallel transport defined by tidal forces for a central potential system. (simple to describe take a circle and from the centre look at the degrees, the distance between these two vectors decrease as one approaches CoM.

However just to be more complete the geodesic (striaghtline for flat no curvature) is

[latex]\frac{d^2x^\mu}{d\tau}^2=0[/latex] were [latex]x^\mu[/latex] has coordinates [latex]( x^0,x^1,X^2,x^3)=(ct,x,y,z)[/latex] the ct gives time unit dimenionality of length where all coordinates have the same unit of length. (See Natural unit).

the curved spacetime geodesic is

[latex]\frac{d^2x^\alpha}{d\tau^2}+\Gamma^{\alpha}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0[/latex]

[latex]G_{\mu\nu}=kT^{\mu\nu}[/latex] can be thought of as gravity=curvature= energy via e=mc^2 where [latex]G_{\mu\nu}[/latex] is the Einstein tensor, k is some proportionality constant [latex]T_{\mu\nu}[/latex] energy momentum tensor.

 

Edited June 21, 2018 by Mordred