Inertia and Relativistic Mass

Discussion in 'Pseudoscience Archive' started by Reiku, Sep 15, 2008.

  1. Reiku Banned Banned

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    Inertia and Relativistic Mass

    Inertia is a property of matter which opposes changes in velocity, and relativistic mass is a change in energy as matter increases with velocity, so there may indeed be a relation.

    I state, that the inertia of a system is the resistance to an increase of energy due to the acceleration of a system. And the resistance to change is related to the system not willing to use up energy unless acted upon by some external force. So inertia is also the resistance to a deceleration due to reserving the energy of its local system.

    The Idea’s

    In this following work, I attempt to put my ideas down to math. But first a quick summery of the predictions that are justly made: The gravitational charge which is an innate property of inertial matter acts on the inner structure of a system when a moving or sitting still. It generates the relativistic mass of a system, and is theorized that a system has a natural force of resistance to acceleration unless provoked by some external force.

    So in effect, it basically means that inertia is the force of resistance to a change in relativistic mass, which must arise from the gravitational potential. It is logically evaluated this way based on two major rules and that is, where, if the gravitational charge of a system must give rise to matter and therefore inertia. It must also refer to a relativistic mass as well. And that the relativistic energy of a system also adds to the mass of the system, as exampled by having a tiny relativistic photon, with a relativistic mass, that can add to the mass of a system, if it is confined within the structure of a box. If it increases the mass, it equivalently adds to inertia as well.

    Inertia – What is it?

    The basic line of inertia is that matter tends not to decelerate or accelerate unless pushed by some force, so there is a natural resistance in matter, in all forms of matter.

    So the acceleration is never accomplished unless an external force is acting on it, because of a relativistic mass increase. The deceleration is never witnessed, because it also takes energy to slow a thing down. The main idea here, is that a system does not want to use up energy to increase of decrease in speed.

    Three main equations will be used in the calculations. These being \(F=Ma\), \(F=Mg\) and \(E=pc\) (two being major players in the theory of inertia), all needed components to suggest the main base of the theory: gravitational charge creates the relativistic mass of a moving object, due to these inertial laws.

    \(F_{g}=-M_{g}\nabla \Phi\)

    And since \(M_{i}=M_{g}\), we can incorporate the usage of inertia as a property of relativistic mass.

    \(\sum \gamma Mc^{2}=I_{t}=M_{g}\)

    Where \(Mc^{2}\) is the energy content and \(I_{t}\) is total inertia.

    The Inertial System of Increasing Energy

    All objects resist changes in their state of motion. All objects have this tendency - they have inertia. But do some objects have more of a tendency to resist changes than others? Absolutely yes! The tendency of an object to resist changes in its state of motion varies with mass. This varying property is evidence that inertia may be dependant on the mass of a system, and ultimately, the relativistic energy of a system.

    So it can be said that inertia can be measurable, so the more mass a system has, the more inertia it has. Then it can be said, that inertia is dependant on the mass content of a body.

    But mass is normally considered an unchanging property of systems, unless the mass is moving. In this case, the mass that moves has more mass content than the stationary system. This increase in relativistic mass directly alters the inertia of a system. Unless acted upon by some foreign force, the inertia of the system will remain the same, and hence the resistance to change.

    My Theory on Inertial Matter and Relativistic Matter

    My theory in a nutshell, is to say there is a force of resistance in a moving system of mass, so that there is a natural resistance to change of speed. Just like inertia, a system that moves tends to continue in the same speed, and never increases or decreases unless affected by some external force. But there is more to this, because we must ask why a system would have a resistance to a change of speed.

    These equations to start off, are very simple, where I link inertia to mass and also with momentum;

    \(I=M+p\)

    \(I=M+M(v+1)\)

    \(I=M(v+1)\)

    And in terms of relativistic mass I can say:

    \(I=\gamma M(v+1)\)

    If \(M=\gamma M\)

    Linking Inertial Mass with Relativistic Mass and Momentum

    In the Weak Equivalence Principle, it states that ‘inertial mass’’ is the same as the ‘’gravitational mass.’’ You can understand this, using Newton’s formula;

    \(F=M_{i}a\)

    Where the inertia mass is given here as \(M_{i}\), and since \(M_{i}=M_{g}\) in the Weak Equivalence Principle then the Newtonian equation \(F=Ma\) can become \(F_{g}=-M_{g}\nabla \Phi\). The gradient is the gravitational potential, and the constant of proportionality in this case is the gravitational mass.

    It is then surmised that the charge of a system, the gravitational charge that is, is proportional to the inertial mass of a system, but I am wanting to incorporate the idea with more detail, putting inertial conditions right down to a relativistic increase of mass. If you increase the relativistic mass of a system (that is within the system in question), you also increase the gravitational mass of that system, as found with the photon in a box analogy. A photon may not have any mass but it has a relativistic mass. If the photon is captured inside a box, the box now has more mass than what it had without the tiny little photon.

    \(M_{i} = p + Mc^{2}\)

    \(M_{i}-p = Mc^{2}\)

    Momentum must be expressed \(p\), because \(E=pc\).

    \(M_{i}-Mv=Mc^{2}\)

    Which we can now express as \(M_{i}=\Delta \gamma Mc^{2}\) . In this equation, we can see that the prediction of having an inertial force of resistance is equivalent to a change of the relativistic mass content.

    Saying inertial matter is equivalent to gravitational mass and the sum of the relativistic energy content \((Mc^{2})\). If you are wondering why \(\gamma Mc\) is being used, because in relativity relativistic mass can be expressed as \(E=\gamma Mc^{2}\)

    Gravitational Charge and Force of Resistance

    Originally when I initiated this theory, I included a symbol called the force of resistance \(F_{r}\), and I originally believed that the equation \(F_{r}=\gamma M+ \gamma M_{added}(v+1)\), was inconsistent as an explanation, because it did not take into account the specific field responsible for any generation of relativistic mass: the gravitational potential. It must increase the structure of the particle lets say, as it accelerates and the gravitational potential becomes excited and provides the transformation and the effects of inertia. We can even end up using energy density equations to measure the inertia and density of the energy within a system.

    We can say that the gradient of the gravitational mass, which can be said to be the gravitational charge, can be linked with energy in the following example:

    \(F_{r}=-\nabla \Phi Mg + \Delta pc\)

    So that

    \(F_{r}-\Delta E=\Phi Mg\)

    The force of resistance can be thought of the total inertial mass.

    Total Values over Time

    In this equation, we can mathematically evaluate the evolution of some supposed time of a system that experiences the inertial effects of matter. Since it was really a measure of a change in energy, the equation which satisfies;

    \(Mg=\Delta \gamma Mc^{2}/t\)

    In supposing some vary over time. And can be simplified to;

    \(\sum M_{i}=\Delta E/t\)

    Solving for Relativistic Mass, Momentum and the Gravitational Potential

    With the equations derived;

    \(\Delta M_{i}=\Delta \Phi Mg\)

    And

    \(\sum Mg = \Delta pc / t \)

    Gravitational Charge

    Where \(M_{i}=M_{g}\) in Newtonian Mechanics, we can involve the gravitational potential in an equation defining both the change in energy and change in momentum.

    The potential energy in question of the gravitational field gradient \(\Phi(x)\), and we can state that:

    \(\Delta PE=\Delta \Phi Mg\)

    And to find the total change in energy;

    \(\Delta PE=\Delta \Phi\)

    When I evaluated this theory just recently, I realized something important. We are universally taught that all forms of matter are but spacetime distortions of fluctuating and accelerating materials. Inertia seems to have an important relationship to the generation of the mass itself, since when we compare lets say a massless particle like a gluon to an electron, one has a small mass, while the other generates no mass at all. The reason why the gluon does not experience inertial effects is because the innate properties do not generate mass from any excited field within itself, but since energy is related to momentum, it can couple to gravity, and matter.

    But a gluon and even a photon do not experience inertial effects. They feel absolutely no resistance and of course, they don’t even interpret a lifetime. They experience no time pass, or any volume of space. Only an external observer issues those properties. But electrons, protons and neutrons all experience inertial effects. They have a mass, and this is what links them.

    Since matter is nothing but distortions of a gravitational vacuum, weak one at that, somehow manifests the fluctuations, and the energy content of the system (as Einstein hypothesized), depended on the inertia of the system (1).

    (1) – I prefer to say that the energy content \((Mc^{2})\) of the system causes a coupling with the gravitational charge of the system, and both implement each others properties, the total measure of inertial mass when \(\Delta M_{i}\).

    Gravitational Charge and Mass

    In using the gravitational potential field \(\Phi(x)\) to describe the excitation of relativistic energy in a moving body, is just as radical as believing that the gravitational charge may somehow generate matter itself. To quickly demonstrate the last set of equations, I wanted to cover the total gravitational charge (in some time), which is related to the total inertia of a system, since the gravitational charge adds yet more energy to a system.

    \(M_{i}_{t}=g_{\mu}_{t}+t\)

    I believe that the gravitational charge is the excitation of matter itself, and evolves from the gravitational potential field. The momentum of the system, will increase in relativistic mass, and mass in local content, so the gravitational charge must also increase with magnitude as momentum is added.

    Observing Evidence

    The electromagnetic inertia in the direct influence of omission by radiation from charged particles in a gravitational field as it accelerates was speculated upon by Feynman in a famous lecture. He made it known that when an electron accelerates through spacetime, it gives off and increasing value of photons, and referred to it as a type of electromagnetic inertia, because it would also require an increasing force to accelerate the particle.

    Why should the electron radiate more photons, as it accelerates, if it also requires more energy to get it to that position to start with? Because an a electron requires more force to accelerate it due to emitting photons, it seemed that the idea I brought forth would suggest it was not only due to an increasing inertia of the system, but also an increase in energy density \(\Delta \epsilon |E_{g}|^{2}\) of the gravitational field. The increase of density was due to the field have a stronger interaction with the moving inertial mass, and therefore more relativistic mass was being added to the system, increasing the inertial effects of the accelerating electron.
     
    Last edited: Sep 15, 2008
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  3. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    This is nonsense, and can be seen as nonsense on purely dimensional grounds.

    In line (1), if \(F_g\) has units of force, then \(M_g\) is dimensionless. Note that \(\Phi\) is the gravitational potential, as you refer to it later. This means that either \(F_g\) doesn't have units of force, or \(M_g\) doesn't have units of mass. At the very least, this is incredibly confusing.

    In line (2), you clearly state that \(M_g\) is a mass by identifying it with \(M_i\). This means that \(M_g\) is a mass, or Equation (2) is wrong.

    In line (3), you state that \(M_g\) is an energy. This is completely inconsistent with line (2), and probably inconsistent with line (1).

    Note that there is no way for lines (1), (2), and (3) to be consistent with each other.

    There are other glaring inconsistencies further down, but how about we start with these three. First, you never define \(F_g\) or \(M_g\).
     
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  5. Reiku Banned Banned

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    Can i explain then, because its clear there was misunderstanding. Give me time to explain. I will be back soon.
     
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  7. Reiku Banned Banned

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    But of course, you have moved this anyway without wanting an explanation. I will still explain later.
     
  8. John99 Banned Banned

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    Looks good to me.
     
  9. Reiku Banned Banned

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    Anyway, i'm back now Ben.

    ''In line (1), if \(F_g\) has units of force, then \(M_g\) is dimensionless. Note that \(\Phi\) is the gravitational potential, as you refer to it later. This means that either \(F_g\) doesn't have units of force, or \(M_g\) doesn't have units of mass. At the very least, this is incredibly confusing.''

    >> It is right from how i calculate it. I don't quite get your arguement, because the only modification on this equation, is \(F_g\) from the normal equation derived \(F=\nabla \Phi Mg\) from general relativity. So, i still am not sure.

    In line (2), you clearly state that \(M_g\) is a mass by identifying it with \(M_i\). This means that \(M_g\) is a mass, or Equation (2) is wrong.

    >> Again, i think i made it clear in the work, that the equation \(M_i=M_g\) is the normal equation derived from applying Newtonian Mechanics. So it must hold true as far as i am concerned. So if it has a mass, then the above equation is still true in my eyes, since both equations so far are used universally.

    In line (3), you state that \(M_g\) is an energy. This is completely inconsistent with line (2), and probably inconsistent with line (1).

    >> Not at all. To start with, i explain further in the work, that when we speak of a Mass in the equations, then we are to take mass for a relativistic mass adding to some system. That would mean i was following the relativistic rules following \(M=\gamma M\) -- I think Tolkan notation. If i take this following equation: \(\gamma M=E/c^{2}\), what am i saying about a hypothetical system with a mass? I can say that the mass has a change of energy, or a property of changing energy. The consequently can be said about any of the equations i show when i talk about Mass having an energy. Just like a relativistic photon can add to the mass of a box, the relativistic mass increase also increases the mass of that system. In any equations deriving the motion of a particle through some field, then i can relate relativistic effects also.

    \(F=\nabla \Phi \gamma M_{g}\)

    Its really just about using your imagination.
     
  10. James R Just this guy, you know? Staff Member

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    Reiku:

    Last warning.
     
  11. Montec Registered Senior Member

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    Hello Reiku

    I think you would get a better insight on "mass" and the many pseudo forces that affect it if you incorporate "time gradients" into your thinking.

    Please Register or Log in to view the hidden image!

     
  12. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    I haven't got a clue what you're going on about, but I did have a look at the equations.
    Don't they only hold up if \(M=0\)?
     
  13. James R Just this guy, you know? Staff Member

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    Worse. Those two equations directly contradict each other. And it's blindingly obvious.

    And the first one is dimensionally incorrect, as well.
     
  14. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    No. The proper definition of force in terms of gravitational potential is

    \(F = -\nabla \Phi\).

    This is the definition of ANY force, given a potential. Simply, the equation has the wrong units. If \(F = Mg\) and \(F= -\nabla \Phi\), how can \(F=\nabla \Phi Mg\) ?

    Do you think that this is correct?

    If you tell me that \(M_g\) has units of mass, then you are admitting that the Equation (1), as I've labeled it, is wrong on dimensional grounds. Period.

    True or false... The following equation appears in your work below:

    \(\sum_i\gamma M_i c^2 = M_g\)?

    How can \(M_i = M_g\) AND \(\gamma M_i c^2 = M_g\)?

    None of this has anything to do with the fact that your equations contradict each other mathematically. It is impossible to talk about physical interpretations of things if the equations aren't mathematically consistent.
     
  15. Read-Only Valued Senior Member

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    And the wannabe world class mathematician strikes out yet again.

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    How many times does that make to date? 30? 40? Or more?
     
  16. Reiku Banned Banned

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    Maybe this time around?

    For starters, you are completely wrong. You have derived my equations from falsehood.

    Sean Carrol defines the equation \(F=\nabla \Phi Mg\), where he says the form \(Mg\) is seen in the context of the gravitational charge.

    This is in an excellent set of lecture notes, so you are calling him wrong, and really isn't a matter of whether i believe is right or not. I take it that it is.

    So don't tell me it is wrong, when a certified scientist uses it.

    Secondly, yes, i did use the equation

    \(\sum \gamma M_{i}_{t}=M_g\)

    Reason? Simple.

    I said that in Newtonian Mechanics, relativity defines \(M_i=M_g\). And my theory says that inertia is the resistance to a change in relativistic mass. If inertia is somehow the resistance to a change in relativistic, then the inertia must also equal \(M_i=\gamma M\), so \(M_g=\gamma M\) in Tolkan notation. Do you get this?

    Why wasn't this understood? I explained that these equations talk for a mass. For this mass, we also talk about the change in relativistic mass.

    So, you have basically said the first equation was flawed, which flawed the rest. The first equation is actually universally-used in relativity and is not a creation of mine. So you are arguing with Sean Carrol, not me.
     
  17. Reiku Banned Banned

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    Read Only,

    Only if Ben was absolutely correct, which he wasn't. He defines equations as nonesense, when they are used by physicists. This doesn't make any accurate conclusions drawn from my work, which i still stand by as being correct.
     
  18. Reiku Banned Banned

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    I made equations where the mass equalled zero, and they must refer to particles without any inertia or gravitational charge.

    So as far as we are concerned in the equations given, if we are using \(\gamma M_ic^{2}\) we are not talking about the energy alone of the system, but we are talking about a mass as well, because \(M_i\).If it has a mass, it has some countable mass.
     
  19. Reiku Banned Banned

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    Taking Bens corner arguement are we? Well, you'd be wrong too. Unless of course, you have evidence Sean Carrols equation was wrong?
     
  20. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    So from Dr. Wolf to Dr. Carroll? A certified scientist---maybe he has business cards that say that.

    Why don't YOU tell ME what the equation means, and why it should be correct. In physics, we strive to understand things for ourselves, not simply accept as doctrine things that we copy from others' work.

    I guess you're referring to Carroll's GR lecture notes? Can you cite an equation and page number so that I can look up this mystical relationship?
     
  21. Reiku Banned Banned

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    ''So from Dr. Wolf to Dr. Carroll? A certified scientist---maybe he has business cards that say that.''

    I'm stoned, so i found that pretty funny.

    ''Why don't YOU tell ME what the equation means, and why it should be correct. In physics, we strive to understand things for ourselves, not simply accept as doctrine things that we copy from others' work.

    I guess you're referring to Carroll's GR lecture notes? Can you cite an equation and page number so that I can look up this mystical relationship?''

    The equation seems to mean that a force in Newtonian Mechanics is derivable under the gravitational field as \(F=\nabla \Phi Mg\), where the charge is related as a constant of proportionality. I personally accept it for my theory, since inertia is related to F=Ma, and that the gravitational charge would be proportional to the relativistic mass of a system.

    Page 104.

    \(F=-Mg \nabla \Phi\)
     
  22. Reiku Banned Banned

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    I'm now working on relationships between gravitational charge and a thing moving through the zero point field. So like a cart that has a sensor attached to a long pole that scrapes over the electric ''bumpy'' wire, the particle as it moves through the vacuum is always in interaction with virtual particles in the potential spacetime.

    Do you dislike that too?
     
  23. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Right.

    First, you did copy the first equation correctly. In Equation (4.4) Carroll defines

    \(a = -\nabla \Phi\)

    which is ok, but a bit different than I have seen it before, but makes sense, in light of the way forces and potentials are defined in electromagnetism. Perhaps this is the proper way to define gravitational potential, so that the analogy between electrical charges and gravitational "charges" is clear.

    Second, your equation \(F = Mg\nabla\Phi\) is still wrong. The correct Equation, from the lecture notes Eq. (4.2) is \(F = - M_g \nabla\Phi\), which is the same as the first equation. For future reference, \(M_g \neq M g\).

    to all who are following, note that \(M_g\) has units of mass, while \(M g\) has units of force. This is important, because Reiku's next equation

    \(\sum_i \gamma M_i c^2 = I_t = M_g\)

    is definitely wrong. No matter WHAT \(I_t\) is, the left hand side has units of energy, and the right hand side has units of mass.
     

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