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Thread: fully covariant Gravitoelectromagnetism

  1. #1 fully covariant Gravitoelectromagnetism 
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    anybody has ever seen the fully covariant form of
    gravitoelectromagnetism (GEM)???
    Apparently, the Maxwell like GEM equations are missing acceleration.

    GEM is taking electromagnetism equations and adapting them for gravity.
    Yes i know they aren't exact.

    Can we use the Liénard–Wiechert potential for that? Right?


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  3. #2  
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    Quote Originally Posted by Quantum immortal View Post
    anybody has ever seen the fully covariant form of gravitoelectromagnetism (GEM)???
    I have seen GEM formalisms of various kinds, but none of them was fully covariant. Would you have a reference ?


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  4. #3  
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    i assumed it existed.
    The usual form, is covariant, as long as there are no accelerations.
    Its almost there...

    So i deduce, that it only missing acceleration to become covariant.
    Are they Maxwell like equations with acceleration in them?
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  5. #4  
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    Quote Originally Posted by Quantum immortal View Post
    i assumed it existed.
    The usual form, is covariant, as long as there are no accelerations.
    All formulations of GEM that I am aware of contain a mass current ( density ) term, which destroys the general covariance. But if you happen to have a link to a fully covariant version, I would be interested in seeing it.
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  6. #5  
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    you mean that the source is the 4 momentum instaid of the stress energy tensor?
    It would just need to add the missing elements to complete the stress energy tensor and the corresponding equations....
    Right?
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  7. #6  
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    Quote Originally Posted by Quantum immortal View Post
    you mean that the source is the 4 momentum instaid of the stress energy tensor?
    No, it is mass current density.

    It would just need to add the missing elements to complete the stress energy tensor and the corresponding equations....
    How exactly would you do that ?
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  8. #7  
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    mass current density is just momentum
    the definition from electricity
    J = ro*v

    in GEM this becomes momentum density

    we are missing normal force and tork, so that the stress energy tensor would be complete...
    they can be deduced from accelerations. If the stress energy tensor is complete, that would make the new set of equations fully covariant...
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  9. #8  
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    Quote Originally Posted by Quantum immortal View Post
    in GEM this becomes momentum density
    The problem is that in GEM this object is a 3-vector, and hence it isn't covariant.

    we are missing normal force and tork, so that the stress energy tensor would be complete...
    they can be deduced from accelerations.
    To construct the full energy-momentum tensor, you need the following :

    = mass-energy density
    = 4-momentum per unit volume
    = k-th component of energy flux
    = j,k component of classical stress tensor

    I do not see the need to try and incorporate accelerations in this.

    If the stress energy tensor is complete, that would make the new set of equations fully covariant...
    Suppose you have constructed an energy-momentum tensor - how would you re-write the GEM equations to turn them into a tensor relation ? That's what my original question was aiming at.
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  10. #9  
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    i though you were an expert in general relativity
    this is just unit conventions .... >

    the 0 component of the 4momentum is energy....
    in GEM this corresponds to the mass density( charge density in EM)
    its the source of the first equation....
    here it is, all 4 components of the 4 momentum

    the components of classical stress tensor are basically forces per area
    pressure is normal force per unit area
    shear stress is force parallel to the unit area (or tork)

    ..... if you have the acceleration, you can deduce the forces....

    The components of the stress energy tensor are energy, momentum and forces.
    The divergence of the stress enegy tensor is simply all of relativistic mechanics
    The continuity equation of matter/energy is just .... all of mechanics

    we already have a set of equations for the 4 momentum, it can't be that hard to derive the equations for the full stress energy tensor
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  11. #10  
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    Quote Originally Posted by Quantum immortal View Post
    this is just unit conventions .... >
    An object not being covariant has nothing to do with unit conventions; I am not sure what you are trying to say here.

    i though you were an expert in general relativity
    If I was an expert in GR I wouldn't be spending time on amateur forums, but instead would work on publishing papers, or teaching. So no, obviously I am not an expert, merely an interested amateur, just like ( almost ) everyone else here.

    we already have a set of equations for the 4 momentum, it can't be that hard to derive the equations for the full stress energy tensor
    So how would you do it ? The stress-energy tensor couples to the Cartan moment of rotation enclosed within an elementary 4-volume - can you suggest a way to formulate this in such a way as to be covariant, while still being different from the already known EFEs ?

    The divergence of the stress enegy tensor is simply all of relativistic mechanics
    The divergence of the stress-energy tensor is zero :



    You can interpret this conservation law mechanically, I prefer to look at it geometrically.
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    "So how would you do it ? The stress-energy tensor couples to the Cartan moment of rotation enclosed within an elementary 4-volume - can you suggest a way to formulate this in such a way as to be covariant, while still being different from the already known EFEs ?"

    didn't get what you said.
    The current GEM equations are covariant, if they are no forces acting on the source.
    If they were fully covariant, they would still be lineair.
    The EFEs are unique? But i don't quite understand what they put in the derivation. It's too abstract....

    with this expression of the stress energy tensor
    Tab=gamma*po*Ua*Ub = Pa*Ub

    we should be sble to express the elements that are needed. Right?
    We alredy have 4momentum and 4velocity

    is there a variant of maxwell equations with accelerated charges?
    the best i found was Liénard–Wiechert potential
    Last edited by Quantum immortal; February 15th, 2014 at 01:01 AM.
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  13. #12  
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    Quote Originally Posted by Quantum immortal View Post
    The current GEM equations are covariant, if they are no forces acting on the source.
    Can you provide a link to a source for this ? The GEM versions I have seen are all non-covariant, the most commonly used being this one due to their formal resemblance to Maxwell's equations : Gravitoelectromagnetism - Wikipedia, the free encyclopedia. Since both mass density and mass current do not transform as 4-vectors, the resulting equations are not Lorentz invariant, hence the theory is not generally covariant, unlike Maxwell's equations.

    is there a variant of maxwell equations with accelerated charges?
    You can express Maxwell's equations in terms of differential forms; to do so you start not with charges, but with a more general potential A. You can then define the Faraday 2-form F as



    Poincare's lemma then immediately yields



    The dual of the Faraday 2-form is called the Maxwell 2-form, and obeys the relation



    Together these two form Maxwell's equations written in terms of differential forms in 4-dimensional space-time. This formulation is fully covariant, and valid also for accelerated charges, in which case you get electromagnetic waves as their vacuum solutions, as expected. Not sure though if this what you were looking for.
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  14. #13  
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    you replace mass density and mass current with the 4 momentum. or if you prefer, you consider gamma*m as the replacement of the gravitational charge. Better still, when GEM is derived from GR, then the source really is the 4-momentum. Do you see now that GEM is covariant when no forces act on the charges?
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  15. #14  
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    Quote Originally Posted by Quantum immortal View Post
    Do you see now that GEM is covariant when no forces act on the charges?
    No, I don't. Please show me mathematically.
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    If you take the manifestly covariant formulation of electromagnetism. Replaced the 4current with the 4momentum and changed the constants.... And take in to account the sources aren't accelerating. The whole thing, is composed only of covariant elements, so the whole is also covariant. Right?
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    Whenever a conversation similar to this takes place in person, it is customary for me to nod my head, and pretend I have the slightest idea what's going on.
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  18. #17  
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    Quote Originally Posted by shlunka View Post
    Whenever a conversation similar to this takes place in person, it is customary for me to nod my head, and pretend I have the slightest idea what's going on.

    you take the equations of electromagnetism and observe that the coulomb law is similar to the newton law. From the derivation of electromagnetism, you know that all the equations are derived by combining relativity and coulomb law. So its not stupid to adapt the EM equations for gravity, by replacing the electric constant with the gravitational equivalent, replacing charge with mass and taking in to account that its always attractive. This model is called gravitoelectromagnetism, or GEM for short. It actually works beater then plain newton law, but still has some flaws.


    A point of important difference, is that charge is invariant. When you change reference frames, the charge doesn't change. Relativistic mass/energy however changes with a change of reference frame. So its not a perfect equivalent, need to think about some details.


    Now, i'm prety sure, that GEM is covariant at the absence of forces, and i would like to see what it looks like with forces included. Covariant means, that the law would stay valid, if you change reference frame. If you prefer, it has build in the fact that there is a maximum speed, the speed of light. Newton law simply say that the force is instantaneous, thats not covariant. Schroedinger equation is not covariant either, it allows for arbitrarily large speeds. Normal mechanics also allows for arbitrary speeds, not covariant either. The honorable gentleman over there doubts that is covariant even if you consider the absence of forces.


    Now that i think about it. The equations have build in the fact that the field go at the speed of light. So i start to think that its purely and simply covariant all the time, and that the honorable gentleman over there is just wrong. And i was splitting hear needlessly. Radiation from acceleration could be deduced by analogy with EM.


    GEM has other flaws beyond covariance (maybe), its linear, while gravity is not. In practice this means, that phenomena of very high energy are just wrong under GEM. Still a very good approximation, and you can apply on gravity what you know about electromagnetism. Bear in mind that in practice, even newton is a good approximation, normally gravity is very week except for a few pathological cases.


    Now, can you nod your head for real?
    I don't understand either what he is saying when he gets in to too abstract mathematics. Whats a 2-form? :P
    But he doesn't really understands me when i get too imprecise....
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  19. #18  
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    Quote Originally Posted by Quantum immortal View Post
    The whole thing, is composed only of covariant elements, so the whole is also covariant. Right?
    Can you be more specific and write down the maths for me; then I will be able to tell you whether the resulting equations are covariant or not.

    But he doesn't really understands me when i get too imprecise....
    That's why it is best to use mathematics.
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  20. #19  
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    you take the manifestly covariant representation of EM. You can readily replace the 4 current by the 4momentum. You also change the constants and take in to account that its always attractive.


    Covariant formulation of classical electromagnetism - Wikipedia, the free encyclopedia


    You have the electromagnetic tensor F
    the 4 current
    EM is equations of these elements


    GEM would use the tensor F and the 4 momentum.
    For accelerations, you can use the relativistic relations for accelerated charges.
    The missing elements of the stress energy tensor is actually accelerated mater/energy.
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  21. #20  
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    Quote Originally Posted by Quantum immortal View Post
    GEM would use the tensor F and the 4 momentum.
    For accelerations, you can use the relativistic relations for accelerated charges.
    The missing elements of the stress energy tensor is actually accelerated mater/energy.
    Can you please write down for me the end result you are proposing ? Please note that



    and



    are valid only in flat space-time ( i.e. locally Lorentz invariant ), but they are not generally covariant. Furthermore, I am not clear what you are going to replace the Faraday tensor with, and how to incorporate the full energy-momentum tensor ( which is the source term of gravity ). I would appreciate if you can quite simply write down the equation you are proposing, so that we can move an and discuss it.
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  22. #21  
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    what you just written. with j as 4momentum








    Yes, they are valid only in flat space time. You want to express GEM in GR perhaps??? It makes no sense.
    GEM is not a perfect bijection with GR, all the non linear stuff are cut off.


    Faraday tensor? You mean electromagnetic stress energy tensor? Why not use the same? Yes, non linear stuff are cut off.


    For the case that they are no forces. The stress energy tensor reduces to just the 4 momentum. The divergence of the stress energy tensor gives the divergence of the 4momentum (equivalent to the divergence of the 4 current) plus dp/dt=0 , that is what we assumed at the beginning.


    For the case of forces( the rest of the stress energy tensor). You use the equations of EM with accelerated charges. EM equations them selves don't explicitly treat accelerated charges, but that doesn't mean they can't. Similarly in special relativity, special relativity can treat acceleration, but it's not explicit in the normal equations. Because the force terms don't appear explicitly, it doesn't mean that you just ignore forces, accelerated charges emit radiation, so here too we have emission of gravitational radiation.


    You just use EM equations as if its gravity. The charge Q is energy, they are always attractive and use the gravitational constant.
    I don't see where you get stuck.
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  23. #22  
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    Quote Originally Posted by Quantum immortal View Post
    what you just written. with j as 4momentum
    4-momentum is not the source term for the Faraday tensor. I really don't know what you are trying to say here.

    Yes, they are valid only in flat space time. You want to express GEM in GR perhaps??? It makes no sense.
    GEM is not a perfect bijection with GR, all the non linear stuff are cut off.
    That's exactly what I have been trying to point out all along - GEM is merely a linear approximation to the full theory. It is also not generally covariant, despite what you appear to think.

    Faraday tensor? You mean electromagnetic stress energy tensor?
    No, I mean Faraday tensor ( also called electromagnetic field tensor ); that is what appears in Maxwell's equations, and these two tensors aren't the same thing. You cannot just swap the Faraday tensor for the energy-momentum tensor, and current density for 4-momentum, and expect everything to still make sense. This is why I asked you to write down the mathematics of what you propose.

    I don't see where you get stuck.
    I get stuck on not being sure just what it is you are trying to do. It would be so much easier if you could just write down the equations of what you propose.
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  24. #23  
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    This is the problem with physics. You cannot learn physics just by reading books about physics. You need very solid mathematical background (usually better than mathematicians) to be able to fully comprehend what is going on. And only after you know meanings of terms like covariance and general covariance MATHEMATICALLY (eg. you see what will happen to equations when you lorentz transform them) you will see where the problem is. It is admirable that amateurs are willing to learn something about this deep field by reading obviously a lot of physics books but sometimes it will result in misunderstandings and misinterpretations. Problem is many of more popular physics books for general audience (Hawkings especially) tend to subtly decept the reader with sensational statements (which are usually trivial mathematical abstractions but reader has no chance of knowing) or flat out lie.
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    the relativistic mass density is just relativistic energy density
    the relativistic mass current density is just the relativistic momentum density
    the relativistic 4current of mass density is just the 4momentum density


    this is just conversion of units. You replace Q with the relativistic mass.


    When i said, that GEM cuts of non lineair stuff, i also implied that it was an approximation. I even say, that in practice, even newton is a good approximation.


    I can make the difference between the stress energy tensor anf the electromagnetic tensor. I didn't knew it was also called the Faraday tensor.
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  26. #25  
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    Quote Originally Posted by Quantum immortal View Post
    When i said, that GEM cuts of non lineair stuff, i also implied that it was an approximation. I even say, that in practice, even newton is a good approximation.
    Ok, agreed.

    the relativistic mass density is just relativistic energy density
    the relativistic mass current density is just the relativistic momentum density
    the relativistic 4current of mass density is just the 4momentum density
    this is just conversion of units. You replace Q with the relativistic mass.
    I'll leave you to it at this stage, since, despite asking you several times, you appear to be unable or unwilling to write down the equations you propose. I cannot help you unless I see your maths, because, to be honest, I have no idea what you are trying to say or do here. Given the title of this thread, I presumed you were presenting a generally covariant version of GEM ( which would have been interesting ), but if you aren't actually showing your work, there isn't much I can meaningfully comment on. I can only point out that none of the versions of GEM I am familiar with is generally covariant, and that there are good reasons to suspect that such a formalism in fact cannot exist, as pointed out here as well :

    Gravitoelectromagnetism - Wikipedia, the free encyclopedia
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    Read the initial message. I was expecting some one to show me the result.


    i thought it over. I think what i was saying is of by a factor of gamma.


    We use rest mass as the replacement for Q. Because rest mass is invariant, its a drop in replacement of the invariant Q, leaving the equation Lorentz covariant ( nor general covariant, don't mix up GR in here, i mean simple lorentz covariance like in EM).


    You could have pointed out, that the equations are covariant with the rest mass, so they can't be covariant with the relativistic mass because of a factor of gamma.
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    For who ever is interested


    By using the rest mass in place of Q, GEM is Lorentz covariant. The rest of the energy is kinetic energy, been part of the current. All the energy is there. Some times, kinetic energy can be part of the rest mass, this is why i say it like this.


    If you think about it, it makes sense. If you just accelerate an object enough, at some point he will get enough energy to collapse in to a black hole, but in its reference frame you are going at grate speed in the opposite direction, possibly becoming a black hole your self. Contradiction.


    Consider an immobile point mass m . It has a field m/R^2 , we consider this to be basically true. When you change reference frame, the field is messed up by special relativity. In the new reference fame, this field is the solution of the GEM equations, with m as the charge. This is just a consequence of special relativity. We only assumed that m/R^2 was true in its own reference frame. I'm trying to say, that the rest mass as the charge is a consequence from minimal assumptions.


    If you consider elementary particles this is straight forward. For composite objects, its a bit more complicated. For planets of the solar system, its the sum of there particles rest masses, minus the binding energy. Thats good enough for the solar system. For super massive stars, neutron stars and white dwarfs, need to add the energy of internal pressure and i think the heat.


    This is kind of interesting, i wasn't expecting that. Its like the reference frame with the lowest energy is "special".


    What does General relativity says? I suspect something similar is happening, from the simple solutions, m appears in the metrics, not gamma*m . Also other similar effects with photons....
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    Quote Originally Posted by Quantum immortal View Post
    This is kind of interesting, i wasn't expecting that. Its like the reference frame with the lowest energy is "special".
    And this is why it isn`t covariant.

    Quote Originally Posted by Quantum immortal View Post
    The rest of the energy is kinetic energy, been part of the current.
    That`s not a vector.

    Quote Originally Posted by Quantum immortal View Post
    If you just accelerate an object enough, at some point he will get enough energy to collapse in to a black hole
    Not true.

    Quote Originally Posted by Quantum immortal View Post
    For planets of the solar system, its the sum of there particles rest masses, minus the binding energy.
    Why?

    Quote Originally Posted by Quantum immortal View Post
    If you think about it, it makes sense.
    No.
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    rest mass is invariant. If you plug that in place of the charge, the equations stay covariant.


    I was using the term kinetic energy loosely.


    On black holes, just finish reading what i wrote....


    The rest mass of a system, is the total energy minus the kinetic energy of the system as a whole. A single photon has no rest mass, two anti parallel photons have a rest mass. For the planets, the sum of the rest masses of the particles is the greatest contribution. But during formation some energy was radiated due to the gravitational potential well. The other contributions are not significant.


    I want to make some comparisons with GR. What happen with single photons in GR? Why fast moving things don't become black holes in GR? Whats the gravitational atraction between two photons in GR?
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    Quote Originally Posted by Quantum immortal View Post
    Why fast moving things don't become black holes in GR?
    Because the source of gravity is not "relativistic mass" ( I deeply detest this term, for this very reason ), but the full energy-momentum tensor, which is a covariant geometric object and hence the same for all observers, regardless of their states of relative motion.
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    How exactly? What is the gravitation of a single electromagnetic wave? Of two anti parallel electromagnetic waves?


    You think that GEM is covariant now with rest mass density and its relativistic current? Its just impossible to do simpler then that.
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    Quote Originally Posted by Quantum immortal View Post
    How exactly? What is the gravitation of a single electromagnetic wave?
    First, find the energy-momentum tensor for your wave. Then solve for the metric tensor. That will give you the geometry of the region of space-time containing your wave.

    You think that GEM is covariant now with rest mass density and its relativistic current?
    No, for reasons already explained multiple times.
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    In my mind without realizing, i was thinking in term of photons, with h*f . It wasn't really doing the trick.


    Now i'm thinking something different. Just replace Q with rest mass density (and the constants). Its invariant, its a one to one replacement for invariant Q. Its just impossible that the equations aren't covariant now, its trivially true. Its basically just changing some names in the EM equations. Its covariant for the same reasons the EM equations are covariant. If its not covariant, i would like to see the explanation of it.


    At first i was thinking in terms of the rest mass of particles only. This seamed too limited so didn't think about it very seriously. However rest mass is well defined for infinitly small cubes of space. rest mass, and rest mass current contain enough information to fully characterize the flow of matter. If you consider an EM-like proper derivation of GEM, then you are forced to use rest mass in the newton law.


    Its not my fault you are so literal B[
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    Quote Originally Posted by Quantum immortal View Post
    Its just impossible that the equations aren't covariant now, its trivially true.
    The GEM equations are









    Neither mass density nor mass flux form 4-vectors; the above equations are invariant under spatial translations and rotations, but not under boosts, and not in curvilinear coordinate systems. Therefore they are not Lorentz invariant, unlike their EM counterparts, let alone covariant. It is irrelevant how many times you try to claim the opposite, the mathematics are clear, and are in accord with all sources I know of, such as : Gravitoelectromagnetism - Wikipedia, the free encyclopedia

    If you think you can present the maths for a covariant version, then now is the time to do so.
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    You are using these equations from the start?
    The 4 in the rotational of B is wrong. It says the speed of light is c/2....
    Just use the EM equations. One misunderstanding solved.


    which mass you are talking about? relativistic mass? or rest mass? I switched to rest mass now.


    rest mass is invariant, even more convenient then covariant. If you change reference frames, it behaves like the electric charge. so the 4 current of rest mass is covariant, exactly like the electric 4 current. Q and the rest mass are both invariant, what exactly prevents the 4 current of the rest mass to be analogus to the 4 current of electricity? Use the definitions of electric 4 current and replace Q with rest mass.
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    Quote Originally Posted by Quantum immortal View Post
    You are using these equations from the start?
    The 4 in the rotational of B is wrong. It says the speed of light is c/2....
    Just use the EM equations. One misunderstanding solved.
    Look at the link. Everything I gave is completely correct.
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    just use the same form as the EM equations.

    ∇·E=-4pi*G*ro
    ∇·B=0
    ∇×E=-∂B/∂t
    ∇×B=-4pi*G*j/c^2 + (∂E/∂t)*(1/c^2)

    invariant mass
    ro = E^2/c^2-px^2-py^2-pz^2
    3 current of invariant mass
    j = ro*v

    It simply replaced Q with invariant mass. The equations are still covariant.

    4 current of invariant mass
    J = ro*(c,v) (EDIT)

    The same form for electric current (EDIT)
    Last edited by Quantum immortal; February 22nd, 2014 at 10:08 PM.
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  39. #38  
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    Quote Originally Posted by Quantum immortal View Post
    The equations are still covariant.
    No they aren't, we've already been through this. E and B in the equations you gave are 3-vectors; to make them covariant you have to generalise these equations to contain only Minkowski 4-vectors or tensors, in which case you can no longer use the cross product, because that isn't defined for 4-vectors. That is why the covariant formulation of EM uses the Faraday and Maxwell tensors, which arise via the exterior derivative ( and the dual thereof ) from an underlying 4-potential. Such a generalisation is not possible in the case of GEM, since neither nor form 4-vectors. This is also clearly stated in the reference I linked to, and I have explained it twice before. If you still don't believe it, despite having been shown appropriate textbook references, then I cannot help you any further. GEM is no more covariant than linearised gravity is, for the same underlying reasons.
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    it seams that i did a mistake earlier. I didn't take relativistic volume properly in to account. More on that later.


    Define me the 4 current density for electric charges.

    Is it this?
    J = ro*(c,v)

    ro is electric charge density for an observer at rest.
    v is 3 velocity
    Last edited by Quantum immortal; February 22nd, 2014 at 10:23 PM.
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    Is this thing electric 4current or not ?
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    a definition for 4current for an observer at rest.
    J=(dQ/dV)*(c,v)

    Q is invariant. If you replace it with an other invariant, the above stays a 4vector
    if we use the rest mass m that is invariant. defined as
    m=Er^2/c^2-px^2-py^2-pz^2

    The above stays a 4vector

    in other words
    J=(dm/dV)*(c,v)
    is a 4 vector also


    What is wrong in what i said here?
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    Quote Originally Posted by Quantum immortal View Post
    r
    if we use the rest mass m that is invariant. defined as
    m=Er^2/c^2-px^2-py^2-pz^2

    The above stays a 4vector

    What is wrong in what i said here?
    Your dimensions are all wrong. You seem to be a well known internet troll.
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    Quote Originally Posted by Quantum immortal View Post
    .

    Is it this?
    J = ro*(c,v)
    Nope. You need to stop pretending that you know what you are talking about.
    PhDemon likes this.
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  45. #44  
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    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    r
    if we use the rest mass m that is invariant. defined as
    m=Er^2/c^2-px^2-py^2-pz^2

    The above stays a 4vector

    What is wrong in what i said here?
    Your dimensions are all wrong. You seem to be a well known internet troll.
    If you really believe that, then ask an admin to ban me or don't make accusation like these.

    I forgot the other stuff on the left.
    m^2c^2=Er^2/c^2-px^2-py^2-pz^2

    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    .

    Is it this?
    J = ro*(c,v)
    Nope. You need to stop pretending that you know what you are talking about.
    Ok. What is it then? Common tell me.
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    Quote Originally Posted by Quantum immortal View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    r
    if we use the rest mass m that is invariant. defined as
    m=Er^2/c^2-px^2-py^2-pz^2

    The above stays a 4vector

    What is wrong in what i said here?
    Your dimensions are all wrong. You seem to be a well known internet troll.
    If you really believe that, then ask an admin to ban me or don't make accusation like these.

    I forgot the other stuff on the left.
    m^2c^2=Er^2/c^2-px^2-py^2-pz^2
    Basic stuff that you missed.

    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    .

    Is it this?
    J = ro*(c,v)
    Nope. You need to stop pretending that you know what you are talking about.
    Ok. What is it then? Common tell me.
    Try reading a book , beats pretending.
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    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    r
    if we use the rest mass m that is invariant. defined as
    m=Er^2/c^2-px^2-py^2-pz^2

    The above stays a 4vector

    What is wrong in what i said here?
    Your dimensions are all wrong. You seem to be a well known internet troll.
    If you really believe that, then ask an admin to ban me or don't make accusation like these.

    I forgot the other stuff on the left.
    m^2c^2=Er^2/c^2-px^2-py^2-pz^2
    Basic stuff that you missed.

    I love tachs blatant hypocrisy. Knowing fine well he can't calculate dimensions on an equation properly, there is a funny sense of irony about his statement.
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    Quote Originally Posted by Chesslonesome View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    r
    if we use the rest mass m that is invariant. defined as
    m=Er^2/c^2-px^2-py^2-pz^2

    The above stays a 4vector

    What is wrong in what i said here?
    Your dimensions are all wrong. You seem to be a well known internet troll.
    If you really believe that, then ask an admin to ban me or don't make accusation like these.

    I forgot the other stuff on the left.
    m^2c^2=Er^2/c^2-px^2-py^2-pz^2
    Basic stuff that you missed.

    I love tachs blatant hypocrisy. Knowing fine well he can't calculate dimensions on an equation properly, there is a funny sense of irony about his statement.
    You have been reduced to outright trolling, Reiku.
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  49. #48  
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    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Chesslonesome View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    r
    if we use the rest mass m that is invariant. defined as
    m=Er^2/c^2-px^2-py^2-pz^2

    The above stays a 4vector

    What is wrong in what i said here?
    Your dimensions are all wrong. You seem to be a well known internet troll.
    If you really believe that, then ask an admin to ban me or don't make accusation like these.

    I forgot the other stuff on the left.
    m^2c^2=Er^2/c^2-px^2-py^2-pz^2
    Basic stuff that you missed.

    I love tachs blatant hypocrisy. Knowing fine well he can't calculate dimensions on an equation properly, there is a funny sense of irony about his statement.
    You have been reduced to outright trolling, Reiku.

    Yeah cause you don't do this sort of stuff constantly, TACH. You shouldn't be talking about anyone, until you no longer live in a glass house.
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  50. #49  
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    Quote Originally Posted by Chesslonesome View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Chesslonesome View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by Quantum immortal View Post
    r
    if we use the rest mass m that is invariant. defined as
    m=Er^2/c^2-px^2-py^2-pz^2

    The above stays a 4vector

    What is wrong in what i said here?
    Your dimensions are all wrong. You seem to be a well known internet troll.
    If you really believe that, then ask an admin to ban me or don't make accusation like these.

    I forgot the other stuff on the left.
    m^2c^2=Er^2/c^2-px^2-py^2-pz^2
    Basic stuff that you missed.

    I love tachs blatant hypocrisy. Knowing fine well he can't calculate dimensions on an equation properly, there is a funny sense of irony about his statement.
    You have been reduced to outright trolling, Reiku.

    Yeah cause you don't do this sort of stuff constantly, TACH. You shouldn't be talking about anyone, until you no longer live in a glass house.
    Still trolling, Reiku? Or are you just trying to bump up the number of your posts?
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  51. #50  
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    ** TEST - DO NOT REPLY **

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