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Unifying Gravity and EM

  1. Sep 1, 2005 #1

    I will try to meet the terms of the 8 guidelines.

    1. The behavior of light is explained with a rank 1 field theory, the
    Maxwell equations. Gravity is explained with a rank 2 field theory,
    general relativity. The two can be combined in one Lagrange density,
    but they are not in any sense unified.

    For my unified field proposal, gravity and EM arise from the same
    4-potential and form a rank 1 field. Here is the Lagrange density
    for my gravity and EM (GEM) unified field proposal:


    [itex]J_{q}^{\mu}[/itex] is the electric charge 4-current density
    [itex]J_{m}^{\mu}[/itex] is the mass charge 4-current density, the standard mass 4-density times [itex]\sqrt{G}[/itex]
    [itex]A_{\mu}[/itex] is a 4-potential for both gravity and EM
    [itex]\nabla_{\mu}[/itex] is a covariant derivative
    [itex]\nabla_{\mu}A^{\nu}[/itex] is the reducible unified field strength tensor
    which is the sum of a symmetric irreducible tensor [itex](\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu})[/itex] for gravity
    and an antisymmetric irreducible tensor [itex](\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})[/itex] for EM which uses an exterior derivative

    The core variance is how one gets a dynamic metric which solves the
    field equations for gravity. With general relativity, one starts with
    the Hilbert action, varies the metric field, and generates the second
    rank field equations. Here, I work with a symmetry of the Lagrange
    density, working directly from the standard definition of a covariant


    Any value contained in the unified field strength tensor could be due
    any combination of the change in the potential or due to a change in
    the metric. One is free to alter the change in the metric so long as
    the change in potential compensates, and likewise the reverse. I
    believe this is called a diffeomorphism symmetry (but my training is
    spotty). Any symmetry in the Lagrange density must be related to a
    conserved charge. For this symmetry, mass is the conserved charge.

    The field equations are generated in the standard way, by varying the
    action with respect to the potential. One ends up with a 4D wave

    J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}

    For the physical situation where the mass density equation is
    effectively zero, one gets the Maxwell equations in the Lorentz gauge.
    If the equations describe a static, neutral system, then the first
    field equation, [itex]\rho_{m}=\nabla^{2}\phi[/itex], is Newton's
    field equation for gravity. If the neutral system is dynamic, then
    the equation transforms like a 4-vector under a Lorentz boost.
    Because this equation is consistent with special relativity, that
    removes a major motivation for general relativity (consistency with

    If the system is neutral, static, and one chooses a gauge such that the
    potential is constant, then the first field equation is the divergence
    of the Christoffel symbol:


    This contains second order derivatives of the metric, a requirement
    for constraining a dynamic metric. The exponential metric solves the
    field equation:

    exp(-2\frac{GM}{c^{2}R}) & 0 & 0 & 0\\
    0 & -exp(2\frac{GM}{c^{2}R}) & 0 & 0\\
    0 & 0 & -exp(2\frac{GM}{c^{2}R}) & 0\\
    0 & 0 & 0 & -exp(2\frac{GM}{c^{2}R})\end{array}\right).

    The easiest way to realize this is that for the definition of a
    Christoffel of the second kind for a static, diagonal metric will only
    involve [itex]g_{00}[/itex] and [itex]g^{0}{}_{0}^{u}[/itex].
    The exponentials will cancel each other, leaving only the divergence
    of the derivative of the exponent, or

    \rho_{m}=\nabla^{2}(GM/c^2 R)

    The 1/R solution should be familiar. This metric gives a point
    singular solution to the field equations.

    One could have chosen a gauge where the metric was flat. With that
    gauge choice, the potential (GM/c^2 R, 0, 0,0) solves the first field
    equation with a point singularity, a good check for logical

    2. The exponential metric solution to the GEM field equations for a
    static, neutral system is consistent with first-order parameterized
    post-Newtonian predictions of weak field theories. The relevant terms
    of the Taylor series expansion are:


    These are identical to those for the Schwarzschild metric of general
    relativity. Therefore all the weak field tests of the metric, and all
    tests of the equivalence principle will be passed. To second-order
    PPN accuracy the metrics are different:



    This will translate into 0.7 microarcseconds more bending of light
    around the Sun according to a paper by Epstein and Shapiro,
    Phys. Rev. D, 22:2947, 1980. We currently can measure bending to 100
    microarcseconds. Clifford Will responding to a question I posed said
    there are _no_ plans in development to get to the 1 microarcsecond
    level of accuracy. Darn!

    The antisymmetric field strength tensor will be represented by the
    spin 1 photon, where like charges repel. These are the transverse
    modes of emission. The symmetric field strength tensor will be
    represented by the spin 2 graviton, where like charges attract. These
    will be the scalar and longitudinal modes of emission. Should we ever
    measure a gravity wave, and then determine its polarization, general
    relativity and the GEM proposal differ on the polarization. If the
    polarization is transverse, GEM is wrong. If the polarization is not
    transverse, general relativity is wrong (Will also made this point in
    his living review article).

    3. Once the Lagrange density is stated, everything else flows from
    that. I have discussed this work as it developed and took misteps on
    sci.physics.research and my own web site, but that should not be
    needed here.

    4. To back up the derivations, I have cranked through all this and a
    bit more in a Mathematica notebook. It is available here:


    [Despite the URL, no quaternions are used in this body of work,
    although they continue to be the wizard behind the curtain.]

    5. This theory is consistent with strong field tests of gravity, such
    as energy loss by binary pulsars. For an isolated mass, the lowest
    mode of emission is a quadrapole moment. This proposal does not have
    extra fields that can store energy or momentum, which is what is
    needed to form a dipole if there is only one sign to the mass charge,
    which the proposal claims.

    6. I know of no physical experiments that contradict this work. There
    are _thought_ experiments that claim that gravity must be non-linear
    (there was a primer on GR by Price I recall as an example). These
    thought experiments appear to always use electrically neutral sources.
    For a unified field theory, one must consider what happens if charge
    is included. What Price did was imagine a pair of boxes with 6
    particles in each. Then the energy of one of the particles in one box
    gets completely converted to kinetic energy of the other 5. Price
    argues that the box with 6 particles should not be able to tell the
    difference between the two boxes, the one with 6 still particles and
    the one with 5 buzzing about. If this is the case, then the field
    equations for gravity must be nonlinear. I argue that if the 6
    particles were charged, there would be no way to destroy an electric
    charge, so the experiment cannot be done in theory. No conclusions
    can be drawn. EM puts new constraints on gravity thought experiments.

    7. It has been my observation that no one is impressed by the
    Mathematica notebook, even people at Wolfram Research. The notebook
    is my best unbiased source that no obvious mathematical errors have
    been made. Earlier versions of this body of work did have errors that
    Mathematica pointed out.

    8. I understand how general relativity works well enough to appreciate
    that a linear, rank 1 field theory is in fundamental conflict with GR.
    That is an observation, nothing more or less. GR works to first order
    PPN accuracy. It is an open question if it will work to second order.
    My money is riding on the exponential metric, because exponentials
    appear to be Nature's favorite function (simple harmonics around the
    identity for small exponents).

    Sorry to be this l o n g, but the guidelines appeared to require it.

    doug sweetser
    Last edited by a moderator: Sep 23, 2005
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  3. Sep 17, 2005 #2


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    The most obvious, although probably naive, "objection" I would have is that we have one and the same field for EM and gravity and that the source is J_tot = J_q-J_m so that two different configurations of J_q and J_m with same J_q-J_m give identical solutions (and hence J_q-J_m should stay identical).
    But we know that true EM or gravitational interactions will make J_q and J_m evolve differently, so that J_tot = J_q-J_m will evolve differently according to the exact composition in J_q and J_m.

    Or maybe I misunderstood how your system works, and there is somehow a different input for J_q and J_m.

    Some simple illustrations:
    if I set up an electrostatic field between two plates, then this should influence neutral matter too, no ? And the other way around, doesn't this mean that electic charges of opposite sign would accelerate essentially in opposite ways in a purely gravitational field ? In order to have each time the same solution for J_q-J_m ?
  4. Sep 17, 2005 #3
    It is natural to think that gravity and EM are too darn close to each other in this theory :-) The total current couples to the 4-potential. If you were to alter the distribution of charges, but not to total charge, a change in the potential would be required.

    Here is one way to see this. Imagine a point source that is a single particle. It will be described by a charge/R potential. If the particle is neutral, then it would be be described by [itex]\sqrt{G}M/R[/itex]. Repeat the exercise, but with a negatively charged particle, that would have a potential [itex](q+\sqrt{G}M)/R[/itex]. Two things to note: all charged particles have mass, so the total effective charge for a negatively charged particle will increase (it is more attractive to a positive charge due to its mass) and the total effective charge for a positively charged particle will decrease (it is less repulsive to another positive electric charge because of gravity). Only because the proposal is linear, one can linearly superimpose solutions. So once you can do one change, you can do an arbitrary number of them, although the actual math gets more complicated. Second, the gravity and EM charges for a proton are more than thirteen orders of magnitude different, but we know electric charge only to ten significant digits, so no electrical engineering needs to change.

    Let me attempt to address a deeper issue, the nature of the relationship between gravity and light. If you focus on the current densities, one observation is that the sign is different between the two is consistent with the well-known fact that like charges attract for gravity and repel for EM. The field strength tensor holds another key. The terms symmetric and antisymmetric are too technical for me to feel comfortable with (yes, they are accurate, but cold). The one for gravity, [itex](\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu})[/itex], I call "The average amount of change in the 4-potential". For EM, [itex](\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})[/itex] I title it as "The deviation from the average amount of change in the 4-potential." I can see how both are describing the same potential in different ways. It is also easy to see why the deviation would have the possibility to be both positive and negative charges. One could imagine the average value being anything greater than zero, suggestive of one sign for the mass charge. Hope this helps.
  5. Sep 19, 2005 #4


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    The Milliken oil drop experiments showed a balance between the gravitational and electric forces. In other words, he used drops where the number of positive and negative charged particles was exactly balanced, and then added a single negative or positive charge to them. The drops were placed in a combination electric and gravitational field and their velocities were observed.

    Your equations would have the net electric charge for a body with balanced positive and negative charges be non zero. The effect would be that certain sizes of oil drops would end up with an electric charge at least equal to e/2. I suspect that this is contrary to the observations of the Milliken experiment.

    Do explain further and keep the fresh ideas coming.

  6. Sep 20, 2005 #5
    Hello Carl:

    Let's agree to a basic point: if my theory disagrees with Millikan's oil drop experiment, it is wrong. The Millikan experiment established the charge on a single electron.

    At first glance I am skeptical Millikan's experiment can in any way effect this proposal, since it is a classical experiment (low speeds) about EM whose forces which vastly exceed gravity. Recall that for our best tests of general relativity, such as light bending around the Sun, the GEM metric gets the identical 1.75 arcseconds of bending around the Sun predicted by general relativity. I have yet to read anyone applying GR to the oil drop experiment, but GR had better be consistent with Millikan :-) If GR at first order PPN accuracy is correct, then GEM is correct. At second order PPN accuracy, the predictions are 10.8 versus 11.5 microarcseconds for GR and GEM. That is the kind of level of accuracy one needs to get at to see any difference. But enough dodging of a direct question.

    This is governed by a balancing of electrical and gravitational forces:
    where [itex]q[/itex] is the electric charge, [itex]V[/itex] is the voltage, [itex]d[/itex] is the distance between the plates applying the voltage, [itex]m[/itex] is the mass of the drop, and [itex]g[/itex] is the Newtonian gravitational acceleration. This looks like classical Newtonian gravity, not any of that fancy PPN stuff. GR and GEM get to Newton in two different ways. The first is as a limit of the metric, and with that approach, there is no difference between GR and GEM. With GEM, if one chooses a metric such that the curvature of the metric makes no contribution, and the apparatus can be described accurately by a static field, one gets exactly Newton's field equations. I am not aware of how one can spot Newton's field equations in Einstein's field equations, but there must be a road. Because GEM has the same field equation as Newton, there is no way it could disagree with the results based on Newtonian gravity analysis.

    I don't see why you think this is so, but will explore the issue anyway. After the oil drops are balanced, the electric field gets shut off, and the oil drop begin to fall. They reach a terminal velocity quickly. Millikan measures the radius and the terminal velocity. The terminal velocity is governed by this equation:
    [tex]v=2 R^{2}\rho g/9 \eta[/tex]
    where small [itex]v[/itex] is the terminal velocity, [itex]R[/itex] is the radius of the drop, [itex]\rho[/itex] is the density of the oil drop, and [itex]\eta[/itex] is the viscosity of the air. So again classical Newtonian gravity is used, and again, I see no way for my proposal can differ. I have yet to point out that the EM part which regenerates the Maxwell equations must also be the same.

    This question brought up a thought experiment that has made my day :-) According to my 1992 Particle Properties Data Booklet, we knew the charge of a single electron is [itex]1.602 177 33\times10^{-19}C[/itex]. Let's say someone somehow calculated the electric charge of a _massless_ particle to 20 significant digits (it would have to be a calculation, there are not massless charged particles), and they found it to be [itex]1.602 177 331 234 567 890\times10^{-19}C[/itex]. Let's further imagine that the Millikan oil drop experiment could measure charge to this level of accuracy (not realistic). Add an electron to a drop. This adds one electric charge, and one mass charge, so that would be [itex]-q_{e}-\sqrt{G}m_{e}[/itex] or [itex]-1.602 177 331 234 567 964\times10^{-19}C[/itex]. Repeat for proton, and its charge would be [itex]+q_{e}-\sqrt{G}m_{p}[/itex] or [itex]+1.602 177 331 234 431 287\times10^{-19}C[/itex]. Not the same!

    This experiment would prove that the mass of an electron is quantized, it comes in discrete clumps of [itex]7.43\times10^{-36}C[/itex] whereas protons, no matter their source, always have a mass charge of [itex]1.36\times10^{-32}C[/itex]. This is different from EM where the quantum electric charge is the same for all charged particles. I have no idea why electric charge is more general that mass charge.

  7. Sep 20, 2005 #6


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    Dear Doug,

    Let me try and explain this better.

    My thinking is that your theory basically amounts to supposing that the gravitational potential is what is left over after the difference in (combined electric/gravitational) potentials between positive and negative charges are added up and cancel.

    Since neutral matter falls, it must be that + and - charges do not exactly cancel. Suppose an object is composed of N atoms and that each atom has as many + as - charges. Then there must be a net electric charge left over. This idea will clearly give the correct gravitational potential, but an object composed of N neutral atoms will not have be exactly neutral electrically. But in a large object like an oil drop we can not count the number of + and - charges literally, so we can neutralize it by adding + or - charges appropriately.

    However, since charge comes in units of e, we cannot precisely cancel the net charge exactly. Consequently, our oil drop will end up with an effective electric charge accurate only to the nearest multiple of the charge of the electron. That means that for some sizes of oil drops, the smallest charge we can get on the oil drop (by adding + or - charges) will be (worst case) e/2. For some other masses of oil drops, we could cancel the charge completely.

  8. Sep 20, 2005 #7
    Hello Carl:

    Thanks for the further clarification. I'm pretty sure I haven't said something like this:
    The 4-potential cannot be observed directly. What is measurable are changes in the potential, the field strength tensor [itex]\bigtriangledown_{\mu}A^{\nu}[/itex]. Fundamental forces are found in irreducible tensors. The GEM tensor is reducible, into the standard one for EM, [itex](\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})[/itex], and the symmetric one I claim is for gravity, [itex](\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu})[/itex]. I believe this is a standard approach, that the potential cannot be measured directly. I try to avoid thinking about things that cannot be measured. Instead it is the average amount of change in the potential that I argue is about gravity, and the deviation from the average amount of change that is about EM by standard EM theory.

    This sounds to me like you are counting the potential, and not either the average value of change in the potential or the deviation from the average amount of change in the potential. If you are dealing directly with the potential, I will argue that is a misunderstanding of this field theory.

  9. Sep 20, 2005 #8


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    I see now that my comment about your theory being that the gravitational potential (and therefore the gravitational force) is simply what is left over after cancelling the electric potentials between + and - is not right and that my objection, which amounted to saying that your theory doesn't quite allow + and - charges to be the same, is incorrect.

    Understanding new theories is very difficult and requires great patience from the person doing the explaining (ask me about it). Let me try and explain your theory back to you and you can correct my misunderstandings. This part got through my skull better: "Instead it is the average amount of change in the potential that I argue is about gravity, and the deviation from the average amount of change that is about EM by standard EM theory."

    Let me try and put it another way. First, I would like to restrict myself to the situation where the underlying metric has no curvature, which I believe you say is acceptable. By the way, I am not a gravity expert (my education is in QM and elementary particles), but my favorite gravity theory is also one that can be expressed by a gauge principle on flat space. Here's a review article:

    Foundations of Physics, 35: 1-67 (2005)
    "Gauge Theory Gravity with Geometric Calculus"
    http://modelingnts.la.asu.edu/pdf/GTG.w.GC.FP.pdf [Broken]

    Above was from this page:
    http://modelingnts.la.asu.edu/html/GCgravity.html [Broken]

    So my limited understanding now is that basically you're using a sort of freedom in the E&M potential (available when you expand its usual potential from antisymmetric to assymmetric) to supply a gravitational potential and you're using the opposite sort of freedom in the gravitational potential (available when you expand its symmetric potential to an assymmetric one) to supply the E&M potential.

    The silly rules on this forum allow only a limited number of posts so I will ask some more stupid questions privately.


    By the way, I looked at the three links in your original post. The first one didn't work for me, but the second was helpful. Joining Newtonian gravitation and E&M is what I will look at, as it is simpler than Misner Thorne and Wheeler.
    Last edited by a moderator: May 2, 2017
  10. Sep 21, 2005 #9
    Hello Carl:

    Wow, real communication, that's tough :-) For parity, let me admit 3 errors I made when working with the Lagrange density for this proposal. First I had the sign of the mass current density be the same as the electric current density. That asserts like mass charges repel. Oops. The second error was trying to talk about [itex]trace(\nabla^{\mu}A^{\nu})[/itex] as a scalar field. This must be a mixed index tensor to take the trace. Duh. And the third error was to write the field strength tensor as [itex]\partial_{\mu}A^{\nu}[/itex] which does not transform as a tensor. Each of these errors were pointed out with a harsh tone by professionals, but they have been repaired.

    To be honest, although I understood the technical difference between symmetric and antisymmetric tensors, I am far more comfortable with these phrases.

    I have to be VERY careful about this. In the newsgroup sci.physics.research, my proposal has been dismissed in one line by Prof. John Baez, a well-respected authority in loop quantum gravity in particular and general relativity in, well, general, because he claims my theory fixes the background metric to be flat Minkowski. What you are doing is a gauge choice. Let me be more precise. Here is the gauge you chose for the GEM Lagrangian, and how it may be changed such that the field equations are unaltered:


    If an expert on GR were to read this, he or she would say, "Oh, that is just Riemann normal coordinates, where the connection happens to be zero at only one point in the spacetime manifold." That is the definition of Riemann normal coordinates, but that is not what is going on here. This is a symmetry of the GEM Lagrange density, not a coordinate choice issue.

    Let me make the opposite gauge choice, where I say all of gravity and EM is the connection, nothing is due to the potential. Here then is the gauge transformation:


    The gauge symmetry here is tough to communicate. It is not between the potential and the metric, it is the change in the potential and the change in metric (via a torsion-free connection which is metric compatible for the GR experts).

    Let me try to explain what is going on compared to general relativity. The connection is almost always introduced the same way. An author introduces the idea of a contravariant vector, say [itex]A^{\nu}[/itex]. He then says that the differential operator also transforms like a vector, [itex]\partial_{\mu}[/itex]. Put the two together, [itex]\partial_{\mu}A^{\nu}[/itex], and the result does NOT transform like a tensor. One needs the connection. The covariant derivative transforms like a tensor, [itex]\nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}+\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}[/itex].

    Now they say, let's divorce the connection from the potential and study it (actually, it is an implicit divorce, not one discussed). The connection does not transform like a tensor. Bummer (OK, they are a bit more formal when the math gets this heavy). One book I recall said you could calculate the divergence of a connection, but there is no reason to because it will not result in a tensor. What is the simplest thing made up of a connection that transforms like a tensor? That is the Riemann curvature tensor, [itex]R^{a}{}_{bcd}[/itex]. From there it is a few steps to Einstein's field equations for gravity only.

    Riemann's curvature tensor is often presented as the simplest thing involving the connection that transforms like a tensor. They forget to add the qualifier: Riemann's curvature tensor is the simplest thing involving the connection that transforms like a tensor so long as you completely and utterly ignore the potential. A covariant derivative has the connection, and it transforms like a tensor. That leads to a big picture idea. Maxwell's theory is all about the potential. One has to actually supply the metric as part of the background structure. GR is all about the changes in the metric, the potential being tossed out when forming the Riemann curvature tensor. The reducible, asymmetric field strength tensor [tex]\partial_{\mu}A^{\nu}+\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}[/tex] has both changes in the potential and changes in the metric to do both gravity and EM, either as a potential or metric theory or some combination of both.


    <Administration>The rule that put limits on the number of posts which will effect me first are probably intended to prevent repetition, folks effectively quoting themselves. If we demonstrate a conversation that moves to different topics (there are plenty here), then we may be able in this situation to get the folks here to change that parameter. I prefer to keep the conversation public as much as possible.

    The first link contained a spurious "\", part of cutting and pasting under x windows to the forum form.

    I hope to keep this thread focused on rank 1 unified field theory. From my scan of Hestene's paper, he hopes to bring new insights to rank 2 field theories of gravity, a good goal, but off-topic.
  11. Sep 21, 2005 #10

    I thought I would point out something cool about the Millikan experiment to 20 digits. If a GR expert were to read the proposal and be in a dismissive mood, he would point out that the theory was linear. Based on GR theory, any linear theory is only an approximation of the real thing. Doing the Millikan experiment for 1, 3, 10, 30, 100 charges would give direct evidence about whether gravity is linear. My proposal says the multiples would be exactly 1, 3, 10,... and general relativity would claim that since gravity fields gravitate, there would be greater than 3, greater than 10,... Another way to test the proposal in theory, if not in practice.

    Nature above all is consistent. If I do the Millikan experiment looking at the electric charge, we know that will be linear. To be consistent, mass charge would behave exactly the same way in my opinion.

  12. Sep 22, 2005 #11


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    Hi sweetser! I like your presentation. My confusion is centered on the way you mix rank 1 and 2 tensors. I have a hard enough time separating the two as is. I'm not saying it's wrong, but, I need my hand held to walk through that mine field.
  13. Sep 22, 2005 #12
    Hello Chronos:

    No problem. It is a simple game of count the Greek letters that do not cancel each other because they are repeats. That is it, honest!

    [itex]\nabla_{\mu}[/itex] is rank 1, 1 Greek letter
    [itex]\nabla_{\mu}A^{\nu}[/itex] is rank 2, 2 different Greek letters (mu and nu)
    [itex]\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu}[/itex] is rank 0, the two sets of same letters cancelling each other out.

    Then there are the names that mean the same thing: rank 0 = scalar, rank 1 = vector, rank 2 = matrix, rank 3 = be scared because it requires three fingers.

    Tensors have a bad reputation as being difficult to understand. I'm sure you have a bunch of other questions, just like I am sure I have the answers for you on my web site, www.TheStandUpPhysicist.com (seriously!). There are two ways to extract information you seek. Click on slide/mp3 link, and then Lecture 1. Slides 8-16 discuss tensors, with words, and better yet, simple pictures. I found that trying to explain physics with only words left the visual part of my brain in a coma, so I decided to have a policy that every slide had to have a simple graphic that somehow explained a little bit of what was going on.

    If you are a person like most people, you will prefer the video. Click on the shows link, show 7. Your computer will have to learn how to play an mp4 file which is 200 MB. By the end, you might have more of a sense of the difference between a regular derivative and a covariant derivative. Now, you will not be a pro, but the fear of this stuff may decrease.

  14. Sep 22, 2005 #13


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    Yes, this is what I don't understand. The lagrangian (if I understand well) determines just as well the action of the potential on the charge (total charge rho - M), as it determines the action of the charge on the potential, right ?
    I mean, for the same total current density, we should find the same solution to the potential problem, or is this not the case ? Is there some extra dynamics then that is NOT described by the lagrangian ? If not, I don't see how IDENTICAL J_total can give rise to DIFFERENT potentials ? And if IDENTICAL J_total give rise to identical potentials, then J_total will be a certain solution to the entire system. Now, if that's the case, this would mean that for two different combinations of J_e and J_M into the SAME J_total, this J_total should be "conserved" independent of how it is split into J_e and J_M, but we know that this is not true.

    Let me be more specific. Again, this might just be a part of your model I simply didn't understand, so please try to find the error in my reasoning.

    Let us have a spacelike sheet with a certain "initial potential" A(sheet) and a certain "initial J_total" J_total(sheet). If I understand lagrangian dynamics sufficiently, this implies that the knowledge of J_total(sheet) and A(sheet) COMPLETELY FIX J_total for all events as well as A for all events. So given these initial conditions on the spacelike (t0-like) sheet, J_total and A are now fixed for all (x,y,z,t).
    But consider now two different physical situations:

    situation 1) J_total (sheet) = J_q1(sheet) - J_M1(sheet)

    This means that somehow at any later time, J_total(t) = J_q1(t) - J_M1(t)

    situation 2) J_total (sheet) = J_q2(sheet) - J_M2(sheet)

    This means that somehow at any later time, J_total(t) = J_q2(t) - J_M2(t)

    So for two DIFFERENT charge and mass configurations 1 and 2, we have that if they sum at a certain moment to J_total, then they will always sum to J_total.

    Consider now two extreme cases:
    1) J_total = J_q1 (so J_M1 = 0) No mass on the sheet

    2) J_total = - J_M1 (so J_q1 = 0) No charge on the sheet

    3) or even other combinations, such as -J_q1 = J_M1 on the sheet.
    They obey exactly the same evolution ?
    I don't think that J_q1 - J_M1 is a conserved quantity when you change the amounts of q and M, no ?

    Probably I simply misunderstood a part of your proposal.
    Last edited: Sep 22, 2005
  15. Sep 23, 2005 #14
    Sorry, but I don't understand what you wrote above. Let me give you my understanding of a Lagrange density. The key for me was learning the units: it is mass per unit volume. So a Lagrange density is an expression which is suppose to contain any and all ways that energy can interact inside a box. The Lagrange density has several uses. First, it is unsurpassed in pointing out things that are conserved. This is done by integrating the Lagrange density over volume and time. Integrating over the volume give back all the energy, but why also do time? One uses the calculus of variations, an important branch of integral calculus. Instead of trying to get a number, one gets a function. Strange, but true. The goal is to find a parameter to vary in Lagrange density that no matter what arbitrary time it is integrated over, the integral remains the same. That parameter that doesn't change a darn thing is a symmetry of the Lagrange density, and is always associated with a conserved quantity. For the EM and GEM Lagrangians, one can vary a time t without changing the integral, and that leads to energy conservation. For the EM and GEM Lagrange densities, there is no distance R or angle [itex]\theta[/itex], so linear and angular momentum are conserved quantities. For EM, there is a rank one gauge symmetry, [itex]A^{\mu}\rightarrow A'^{\mu}=A^{\mu}+\nabla\phi[/itex]. This symmetry is why electric charge is conserved. GEM has a rank two symmetry [itex]\partial_{\mu}A^{\nu}\rightarrow(\partial_{\mu}A^{ \nu})'=\partial_{\mu}A^{\nu}+\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}[/itex]. This probably has to do with both charge and mass conservation, but I am not professional enough to work through those points. If anyone wants a further intro to Lagrange densities, that would be lecture 2 at www.TheStandUpPhysicist.com, slides 11-17, and the corresponding video.

    One can take different derivatives of the Lagrange density to generate the field equations, or the stress-energy tensor, or forces. To get to the field equations, one takes the derivative with respect to the 4-potential and the derivative of the 4-potential. This is called the Euler-Lagrange equation, and can take about an hour and fifteen minutes using 6 blackboards if no shortcuts are taken.

    So now I will try to address the question at hand. The Lagrange density has two sorts of ways for energy interactions. One is known at the charge coupling term, it is [itex]\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}[/itex] The difference in sign for these two is absolutely essential, so that like charges attract for mass and repel for EM. This term however is not very interesting. Applying the Euler-Lagrange equations, one ends up with just the J's.

    The interesting part of the Lagrange density is the field strength tensor contraction, [itex]-\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu}[/itex]. One often hears: "consider the field equations for a vacuum." In that situation, J is zero, and the field equations are all about this field strength tensor contraction, nothing else. The div, grad, and curl stuff is all going to be generated from here. This is the heart of the proposal, but from my reading, vanesch focused exclusively on the charge coupling term.

    So what is going on with the charge coupling? As usual, not much. All it really says is that the normal electric current, whatever it is, is effectively a little be less due to interia. Realize there is nothing radical in that statement, that is standard stuff. Like charges will repel from each other a little less due to their interia. If you calculate how much less, it will make effectively no difference.

    All charged particles have mass. The amount of mass they have is usually small enough that it can be utterly ignored.

    This can happen.

    This proposal is linear for EM (standard) and linear for gravity (in conflict with GR). We know you can add in arbitrary combinations of positive and negative charges, and there is zero problems handling that situations with Maxwell's field equations written in the Lorentz gauge, [tex]J_{q}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}[/tex]
    The field equations for GEM are nearly the same, but there is an additional current density. All that current does is decrease the EM current density by an amount that is difficult to measure unless the system is electrically neutral. The field strength tensor is much more complicated for GEM, since it has 16 terms instead of 6.
    I hope this clarified a point or two, but this proposal necessarily is not easy because it is the Maxwell equations generalized in a way to embrace gravity as a metric theory via a rank 2 Lagrange symmetry.
  16. Sep 25, 2005 #15
    A known problem with GR's Lagrange density

    <The Setup>
    I have only talked with one professor one time three years ago in his office about this project to unify gravity and EM (I am quite independent). I had to pull some strings to talk to a string theorist (an infa-red astronomer who taught a Perl class knew an emeritus MIT professor who knew my grandfather who was the highest ranking American diplomat at the League of Nations and who founded the hight school the emeritus professor and string theorist attended). I emailed the string theorist about my hopes for a rank 1 unification model, and here was his reply:
    The tone was dismissive enough that I had to talk myself into going. The string theorist was pleasant on a personal level, chatting about the International School, wondering how a biologist could ever discuss a Lagrangian. When the topic turned to physics, he was the most arrogant man I have had to deal with.
    </The Setup>

    <The Challenge>
    General Relativity is a beautiful theory. I say that based on my experience studying Sean Carroll's lecture notes. I don't understand every nut and bolt, but the structure is elegant. Is it flawless? The challenge here is to find a specific, entirely mathematical flaw in GR at the Lagrangian level.
    </The Challenge>

    <The Math of Lagrangians>
    I suspect that most people reading this forum do not work with Lagrange densities often, so I'll review a bit. A Lagrange density (mass/volume) encompasses every energy interaction that can happen inside a box. There are two things one can do with a Lagrange density:
    Task 1: Vary a field. This will generate the field equations, or calculate the energy density, or the momentum density.
    Task 2: Look for things that do not vary after integrating the 4-volume and varying something. This will lead to conserved quantities.
    The Maxwell equations, and my unified field equations, are generated by varying the potential field, task 1 work.
    Why is energy conserved? That is a task 2 result, there being no time t in the EM or GEM Lagrangians. Why is momentum conserved? Again that is a task 2 result. What about mass? We know from special relativity that energy and momentum have the following relationship:
    Mass is not only a conserved quantity, but mass is invariant under a Lorentz transformation. In the standard model, a kind of generalization of the Maxwell equations, the mass of every particle is zero. The Higgs mechanism is a way to add mass without breaking the symmetry of the standard model. For GEM, mass is a scalar field, the trace of the symmetric field strength tensor. One can see mass charge is conserved by a task 2 process, because it has to do with the second rank tensor symmetry.
    </The Math of Lagrangians>

    <The Problem with the GR Lagrange Density>
    In the same month that Einstein published his field equations for general relativity, Hilbert did the physics the right way, starting from a Lagrange density, a deceptively simple looking one at that:
    [tex]\mathcal{L}=\frac{c^{4}}{16\pi G} R[/tex]
    where [itex]R[/itex] is the Ricci scalar, formed from two contractions of the Riemann curvature tensor.

    Simple, isn't it? One might imagine that energy conservation will be easy to find. Unfortunately, there is a serious technical roadblock. The action [itex]S[/itex] is the spacetime integral of the Lagrange density, or
    [tex]S=\frac{c^{4}}{16\pi G}\int R\sqrt{-g}dx^{4}[/tex]
    Now vary the metric field [itex]g_{\mu\nu}[/itex]. Oops, that effects both the Ricci scalar, and that factor in the square root which is needed in curved spacetime. The problem gets even more tricky. Recall the Riemann normal coordinates, the coordinates where at only one place in spacetime, the connection is zero. One is free to choose whatever coordinates one wants. For any point in spacetime, one could choose the Riemann normal coordinates, and the energy density there would be zero. That is the problem.

    In defense of GR, it must be noted that the theory conserves both energy and momentum. This is a more subtle problem: that energy is not defined locally. Some people are so familiar with this old problem that they do not consider it an issue: old wounds become proud scars. To me, it is a clear blemish.
    </The Problem with the GR Lagrange Density>

    <The Source>
    Why would GR have the localization of energy problem, but the GEM model not? It is because GR uses the Riemann curvature tensor, and GEM does not. Recall that for GEM, the changes in potential and metric lay in the same bed together, as they should in my opinion. In GR, the connection (a fancy name for a change in the metric for a torsion-free, metric compatible connection) gets a divorce from the potential to become a pure geometric object. The pure object does not transform like a tensor. So one looks at the difference between two paths that are very close to one another to define the Riemann curvature tensor. Fundamentally, the Riemann curvature tensor cannot be about one place in spacetime. The Riemann curvature tensor is about a comparison between two places in spacetime. It is no wonder that a problem occurs for one place in spacetime.
    </The Source>

    <The Solution>
    I don't think there is a way to solve the localization of energy problem if one depends anywhere on the Riemann curvature tensor, which the Ricci scalar is as a pair of contractions of the Riemann tensor. This is a problem with how to handle gravity at the level of the Lagrange density. The GEM model has a chance because it uses the connection, but not the Riemann curvature tensor.
    </The Solution>

    <The End/>
  17. Sep 27, 2005 #16


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    I have a question for my understanding. If one takes J = 0 in the GEM Lagrangian, why does not the GEM Lagrangian be the same as the EM Lagrangian for the free field ([itex]\int \frac{1}{4} F^{\mu\nu} F_{\mu\nu}[/itex])? How do Maxwell's equations follow?
  18. Sep 27, 2005 #17


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    Another question. I do not understand how the metric arises. In GR it appears as a dynamic variable in the Lagrangian and is therefore the variable for which the equations of motion are solved. In the GEM Lagrangian the dynamic variable is A and the equations of motion depend on A. However, its covariant derivative is computed according to the metric in spacetime. But that metric itself is a function of the dynamic variable A, am I right? If yes, why don’t you write explicitely the GEM Lagrangian only in terms of A (as the covariant derivative operator is also a function of A)? Sorry if my understanding is completely wrong and if this is trivial, but it is not easy to understand (although you did a very good job explaining it)…
  19. Sep 27, 2005 #18

    There are two questions, so I'll provide two answers in separate posts, this one about Maxwell equations. When J=0, there are no sources, and the Lagrange density applies to a vacuum. For folks reading this, let me clarify the difference between an action [itex]S[/itex] and a Lagrange density [itex]\mathcal{L}[/itex]:
    [tex]\mathrm{the\ action }S=\int\mathrm{the\ Lagrangian\ }\mathrm{L}\mathrm{\ over\ space\ and\ time}[/tex]
    In this case:
    [tex]S_{EM}=\int \frac{1}{4} F^{\mu\nu} F_{\mu\nu}[/tex]
    [tex]\mathcal{L}_{EM}=\frac{1}{4} F^{\mu\nu} F_{\mu\nu}[/tex]
    You probably know this, and I was just using this as a chance to explain some terminology that can be scarier than it should be. What we need to define is the electromagnetic field strength tensor:
    The electromagnetic field strength tensor is what I have been calling the antisymmetric tensor, or the deviation from the average amount of change tensor for the GEM proposal (for the precise reader, they will note I use a mixed tensor in the GEM Lagrangian so I can define a scalar field by taking the trace). This is 6/16ths of the the story, in that there are six terms in the antisymmetric tensor, and there are ten in the symmetric tensor, or [itex](\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu})[/itex].
    Please note the difference between the partial derivatives and the covariant derivatives. A metric is a symmetric object. The change in a symmetric object - the connection - is symmetric. The antisymmetric tensor can have zero information about the change in the metric, that is why it has the partial derivatives. The antisymmetric tensor will be exactly the same no matter how spacetime is curved, although the contraction of the antisymmetric tensor will be changed because tensor contractions depend on metrics.

    Now to your question: How do the Maxwell equations follow? Well, we follow the standard approach: start from the Lagrange density, integrate that over a volume of spacetime to form the action, then vary the action with respect to the 4-potential, and look for a minimum. That is by the way what the Euler-Lagrange equation effectively does. People more skilled with actions than I can look at the action, I think they apply integration by parts, and pull out the field equations:
    0=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}
    This is one way to write the Maxwell equations in a vacuum. One thing about field equations: different Lagrange densities can get to the same field equations. It would be incorrect to presume that the [itex]F^{\mu\nu}[/itex] is a unique, necessary path.

    If you want to see the Maxwell or GEM equations written out in terms of the fields E, B, ..., well, I could do that by making a killer long post, but let me sketch it out instead. [itex]F^{\mu\nu}=E+B[/itex]. One needs to do the same darn thing for the symmetric part. I HATE to make up new terms, that is a crank calling card, but there is a symmetric counterpart to both E and B I write at e and b, along with a field for the diagonal. It is a confusing game to define these and make sure that E, B, e, b, and g together make up the field equation above. It is done explicitly - every darn term written out - in the lecture 3 notes at www.TheStandupPhysicist.com.
  20. Sep 27, 2005 #19
    The answer here is no. To have a viable theory for gravity, you must be able to have an equation whose solution is a dynamic metric consistent with experimental tests of gravity. GR does this is the most straight forward way possible, by taking the Hilbert action and varying the metric field to generate a rank 2 field theory. I cannot do that for the GEM proposal which has rank 1 field equations. Instead, there is a symmetry in the Lagrange density involving the change in the potential and the change in the metric. Here is the differential equation that a metric must solve if the gauge choice is for a constant potential:
    [tex]\rho_{m}=2\partial_{\mu}\Gamma_{\nu}{}^{\:0\mu}A^{ \nu}[/tex]
    I have to rush off to work, but I encourage people to check for themselves that the exponential metric referenced in the first post solves this equation. The answer turns out to be reasonable: since the exponential metric is both diagonal and symmetric, only the [itex]g_{00}[/itex] term makes a contribution, the derivative of the exponential has two parts, the exponential that cancels with [itex]g^{00}[/itex] and the derivative of a 1/R, and Laplace knew the answer to that one.
  21. Sep 28, 2005 #20


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    What I meant was the following. The Lagrangian (density) is a "prescription" for the equations of motion of ALL things in there which are declared "dynamical variable". You can give it a physical interpretation if you want to, but that's not my argument.
    I guess your dynamical variables are:
    - the metric
    - the field A
    - the M and Q current densities, J_M and J_Q.

    right ?

    When you write down the lagrangian, you've now specified ALL equations of motion.
    The metric only appears in the tensor contractions, and has no "dynamical" term of itself.
    But the observation I'm making is that J_M and J_Q only can occur in a very specific combination: J_M - J_Q. So if I now define a new dynamical variable, J_T = J_M - J_Q, and replace J_M and J_Q by this new variable then everything should remain exactly the same in ALL EQUATIONS OF MOTION. So you have a dynamics which cannot in any way distinguish any combination of J_M and J_Q, as long as they result in the same J_T, because the only quantity that appears in the equations of motion is J_T. This comes down to saying that there is no dynamical distinction between "mass" and "charge".

    To give you a simple example of what I mean: let us consider electromagnetism (Maxwell). Now suppose that I define "yellow charge" and" "green charge" respectively described by J_y and J_g. I now say that J_q = J_y + J_g and this is the J_q that goes in the Maxwell equations. Purely based upon the Maxwell equations, only J_q has a dynamical meaning, and for a given J_q (no matter how it is composed of yellow and green charge), I have a certain solution to my dynamical problem (J_q over spacetime, and A over spacetime). So we observe that there is no dynamical distinction between yellow and green charge, that they are INTERCHANGABLE, and that the only physically meaningful quantity is J_q. Yellow charge could be fluctuating in green charge and back at just any rate I wish, it wouldn't change the dynamics. I see exactly the same happening in your theory, where "mass" and "charge" seem to play the same role as yellow and green charge in my toy example. What stops mass from changing into charge and back without your lagrangian "noticing" ?

    I hope I made clear now my difficulty I see with your approach. Maybe it is due to the fact that you do something else but "standard lagrangian dynamics" or that I have missed something else.

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