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Posts posted by Aethelwulf

  1. There is something called a Planck length. It is much smaller than any length associate with particle


    That's only partially true. In physics, we have something called a Planck Particle, whose Compton wavelength and Schwarzschild radius are equal, and is roughly the size of a planck length. So theoretically speaking, we can have particles which may be equal to a planck length.

  2. I find it hard to imagine living in an electron, if they are truly pointlike. This means they have no dimensions... though... some theories out there treat it with a classical radius... if it had one, this would not be a new dimension, it would just be made up of your normal radial dimensions.

  3. Thanks for the reply, I was referring moreso to when I read on the internet and people give wavelengths to things like tennis balls and actual macroscopic matter. Even though we can't see the wavelike properties(another thing I don't know why) they still have a "wavelength". Does this wavelength no longer exists if said particle has 0 momentum?


    That's right.


    Even you have a wavelength, but it is very small. So it still exists... very technically speaking, you have a wave function which extends way past the milkyway. All macroscopic objects have wave functions, but as I said, they are too small to be visible.

  4. How does this theory, based on Larmors equation account for the mass of the neutron?


    I'll be honest. I don't know because I have not studied these equations for it. If I come across one, which I am looking into as we speak, I will be sure to tell you of one.


    How does this theory, based on Larmors equation account for the mass of the neutron?


    Are you aware of any technique in which the Larmor equation accounts for the mass of any particle?


    As far as my memory recollects, the Larmor equation is


    [math]\omega = \gamma B_0[/math]


    Or do you speak of Larmors formula even? That would make more sense because it has been spoken about in the work, very early on? (I called it Larmor's equation, but it can be a bit confusing as different authors call it by different names.)

  5. From what I understand the wavelength of an object is given by: Planks constant/Momentum.


    Does this mean that when stationary(relatively) that particles have no wavelength and are therefore not waves? If so, why does a particle need to be moving to have a wavelike character?


    I have the feeling there is a simple answer but I would be grateful if it's cleared up, thanks.


    Compton's wavelength?


    You can derive it by dividing through [math]Mc^2[/math] into [math]\hbar c = GM^2[/math]. Photon's can never be at rest , however, as I understand the wavelnegth the energy of a photon can be low enough to have it's wavelength match any particle who is at rest near rest. So the wavelength may be seen perhaps, as the energy of the wave of a particle at near rest which may fit the energy of a photon whose energy is small enough.

  6. Where do the four fundamental forces get their energy from?

    it takes energy to move things, and the forces do it constantly seemingly without an energy source?


    The forces get their energy from the fields which describe them. Gravitational fields for instance, is due to a gravitational energy. Electromagnetic fields store energy as well, as much as the weak and strong nuclear forces. The energy source then is the fields, in which quanta move in.



  7. Wel, it is clear measred by giving power of meter: 10 exp 100 meters is 1 with 100 zeros (1 gugool)


    If you have dimensions, X,Y and Z, it does not matter how much you cut these up to make a meter, a yard ect - they still belong to the known dimensions of space. If you where speaking about a dimesnions which was unseen because it was curled up so small, then you'd have a theory akin to string theory, but from what I am reading, this is not what you are saying.

  8. Abstract


    Mass in our most up-to-date theory was beleived to be a by-product of a Boson and its respective field. Finding this particle, called the Higgs Boson has proven difficult. It seems likely that scientists will have to find a new mechanism for mass - and in this work, we will advocate one of those theories.


    What is Mass?


    So, what do physicists mean when they speak of a mass, or rather, a mass term? For a while, physicists tried to answer this question a number of ways.


    There was such a thing as an electromagnetic mass [1] at one point in physics. The was the idea that the electromagnetic field interacted with a quantum object in such a way that it gave a contribution of mass to a system. This was called the self-energy of a system -- in fact, so mainstream the idea was at one point, it was in fact considered as an explanation for inertial mass itself.


    It is quite curious to mention that there was even an electromagnetic inertia which was experienced by particles which where accelerated, this was explained through what is called the Larmor Equation, but given as an expression here:




    This would in fact describe how much radiation a charged particle would give off as it was accelerated to higher and higher speeds. The faster you made your particle move at, the more power was required to bring it to that speed. This was in fact analogous to inertia which was experienced by particles with a charge: today, our new definition is that inertia is strictly a property of mass. As much as this idea has faided, some scientists still use the equations describing the EM-mass. In a sense, mass has been shown to be an electrical phenomenon derived from first principles.


    There maybe some important clues however that it may still have importance, especially since the current model, the Higgs Boson is now finally faiding into the past as scientists have narrowed the possible energy levels for the Higgs Boson to a very small corner. Some of these clues, may exist in the fact that charged massless bosons don't exist. The Yang Mills Equations once predicted massless charge bosons but the approach has been generally considered today as a false one. But why should this prediction and how it failed important?


    Charge is considered by physicists as an ''intrinsic property'', something which is inherent with particles of mass. When we think about charge, we usually talk about electromagnetic charges [math]e[/math], which will sometimes be found as a coefficient to the electromagnetic potential in equations [math]ieA_{\mu}[/math]. What is important here, is that the electric charge is always found with mass, so maybe electromagnetic mass contributions will arise from the shadows again? Who knows, but what does seem clear that the Higgs Boson is now failing our theory so we will need to revise the equations again. So what is a Higgs Boson? How does it enter our theories?



    Our theory of Gauge Fields [2] where [math]\phi[/math] is a Gauge Boson, has a structure in the equations which invariance. This has been called, Gauge Invariance and this invariance allows physicists to make certain transformations to fields. The laws of physics always remain the same under gauge invariance, indeed, that is what it's all about! It means there is no such thing as an absolute position in physics - the only thing which does count is relative positions. Another feature of the gauge fields is that they retain a symmetry of the theory.


    A symmetry can be understood from the most simplest langrangian term


    [math]\partial\phi^{*} \partial \phi[/math]


    Suppose we had a transformation


    [math]\partial \phi' \rightarrow e^{i\theta}\phi(x)[/math]


    In this transformation, the derivatives of [math]\phi'[/math] are concerned only with [math]\phi[/math]. This transformation will look like


    [math]\partial \phi'(x) = e^{i \theta} \partial \phi(x)[/math]


    [math]\partial \phi'^{*}(x) = e^{-i \theta} \partial \phi^{*}(x)[/math]


    Therefore, when you multiply [math]\partial \phi '[/math] with [math]\partial \phi ^{*}[/math] the [math]e^{i\theta}[/math] and [math]e^{-i\theta}[/math] will cancel out because that is how you would compute them with their conjugates. Thus


    In other words, our field [math]\theta[/math] is constant, and does not require the same derivatives as our boson. In return, we would just get


    [math]\partial \phi^{*}\partial \phi[/math]


    and viola! This was the most simplest demonstration of a symmetry conserving field. An even simpler demonstration would be:


    [math]\frac{d(x +C)}{dt}[/math]


    is in fact simply the same as




    this means that the equations where symmetry existed and what we have in these symmetries are extra constants always remain the same, just as our Gauge example above.


    But what if [math]\theta[/math] also was a function of position [math]\theta (x)[/math]? Well let's see shall we.


    [math]\partial \phi' = (\partial \phi + i \phi \frac{\partial \theta}{\partial x})e^{i \theta}[/math]


    and our conjugate would be


    [math]\partial \phi'^{*} = (\partial \phi^{*} - i \phi^{*} \frac{\partial \theta}{\partial x})e^{-i \theta}[/math]


    Multiplying the two, we need to factorize it


    [math](\partial \phi + i\phi\frac{\partial \theta}{\partial x})(\partial \phi^{*} - i\phi^{*}\frac{\partial \theta}{\partial x})[/math]


    which gives


    [math]=\partial \phi^{*} \partial \phi + i(\phi \partial \phi^{*} - \phi^{*} \partial \phi)\frac{\partial \theta}{\partial x} + \phi^{*} \phi (\frac{\partial \theta}{\partial x})^2[/math]


    The reason, again why this equation turned out to be such a mess was because our [math]\theta[/math]-field now depended on position, so the derives of our boson field also included it.


    This motivated scientists to find a symmetry again in the equations, and to do so required the use of te Covariant Derivative which originally came form the work on fibre bundles.


    To restore symmetry, we need to define the Covariant Derivative as


    [math]A_{\mu}' \rightarrow A_{\mu} - \partial_{\mu} \theta[/math]


    Here, we can see our four-vector potential again [math]A_{\mu}[/math] - you can basically build the electromagnetic fields form this. It's time component in [math]A_0[/math] but that is really not relevant right now. Our Covariant Derivative and the respective conjugate fields are usually denoted as


    [math]D_{\mu} \phi = \partial_{\mu} \phi + iA_{\mu} \phi[/math]




    [math]D_{\mu} \phi^{*} = \partial_{\mu} \phi^{*} - iA_{\mu} \phi^{*}[/math]


    So calculating it all together, we just define the whole thing again as:


    [math]D\phi' = (\partial \phi + i\phi \frac{\partial \theta}{\partial x})e^{i\theta} + i(A_{\mu} - \partial_{\mu} \theta) \phi e^{i \theta}[/math]


    Well, with this, we can see straight away that some terms cancel out. The [math]i\phi[/math] terms cancel, and [math]\frac{\partial \theta}{\partial x}[/math] is in fact the same as [math]\partial_{\mu} \theta[/math]. So what we are really left with is


    [math]D\phi' = D \phi e^{i\theta}[/math]


    and so its conjugate is


    [math]D\phi^{*} = D\phi^{*} e^{-i\theta}[/math]


    Again, the latter terms cancel out when you multiply these two together and so what you end up with is


    [math]D \phi^{*}' D \phi ' = D \phi^{*}D \phi[/math]


    and so by using the Covariant Derivative, we have been able to restore the lost [math]U(1)[/math] symmetry where [math]U(1)[/math] symmetries deal with rotations.


    But when physicists talk about a mass, we don't want to retain symmetry in these fields. In fact, the very presence of a mass term will imply an explicit symmetry breaking. The process of course, is a little more complicated however. It involves another boson, called a Goldstone Boson, which can be thought of as a ground-state photon which lives in the minimum of a Mexican Hat potential. Something which exists in the minimum is the same as saying our system does not contain a mass term and so [math]\phi=0[/math]. In such a potential, we may describe our field as


    [math]\phi = \rho e^{i\alpha}[/math]


    Here, [math]\rho[/math] is a deviation from the ground state. [math]\rho[/math] is in fact our Higgs Boson and [math]\alpha[/math] is our Goldstone Boson. If [math]\alpha[/math] is a frozen (constant) field, then there are no changes in the equations. But if there is a deviation of the Goldstone Boson from the minimum of our potential, then we are saying that it costs energy to do so, and this energy is what we mean by particles like a photon obtaining a mass. In fact, the Goldstone Boson is gobbled up by the Higgs Boson which gives the system we speak about a mass. Our flucuation from the minimum has the identity [math]f \ne 0[/math] where [math]f[/math] plays the role of mass.


    Let's discuss this mass term in terms itself of the electromagnetic field tensor. Such a tensor looks like:


    [math]F_{\mu \nu}F^{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}[/math]


    here, [math]F_{\mu \nu}F^{\mu \nu}[/math] is in fact just [math]F^2[/math] and it makes up the langrangian. [math]F_{\mu \nu}[/math] is an antisymmetric object with respect to swapping its indices. It is like a four dimensional curl.


    It is Gauge Invariant, but checking that, remember our transformation


    [math]A' \rightarrow A - \partial \theta[/math]


    plugging that into the tensor gives:


    [math]F_{\mu \nu}' = \partial_{\mu}A'_{\nu} - \partial_{\nu}A'_{\mu}[/math]


    That is


    [math]\partial_{\nu} \partial_{\mu} A_{\nu} - \partial_{\mu} \partial_{\nu} \theta - \partial_{\mu} \partial_{\nu}A_{\mu} + \partial_{\nu} \partial_{\mu} \theta[/math]


    Since the order of partial differentiation doesn't matter, the [math]-\partial{\mu} \partial_{\nu}\theta[/math] and [math]\partial_{\nu}\partial_{\mu} \theta[/math] cancel and what you are left with is invariance.


    There are some numerical factors left out of this, such as a quarter, but that really isn't all that important in this demonstration. And so, we may have a completely Gauge Invariant term


    [math]D_{\mu}\phi D_{\mu} \phi^{*} - V(\phi^{*}\phi) + F_{\mu \nu} F^{\mu \nu}[/math]


    One should keep in mind that the potential term here [math]V(\phi \phi^{*})[/math] is also manifestly invariant. What becomes interesting however, is the question of what cannot be added to this but still can remain Gauge Invariant. What we cannot add to the electromagntic part ([math]F^2[/math])-part, is this:




    This is just the same as [math]\frac{M^2 A^{2}_{\mu}}{2}[/math]. The reason why, is because a mass-term for a photon would typically look like this. This actually breaks local invariance. In nature there is only one massless spin-1 boson - the photon - then it seems that if we where to use a local gauge group that is less trivial than [math]U(1)[/math], then the symmetry would be broken. It's interesting, that when we speak about symmetry breaking for a photon, we are in fact talking about an electromagnetic term which is squared in the Langrangian.


    It seems, by no far shot, that physicists will not abandon the use of symmetry breaking to describe mass for particles, but it must be noted that our usual, and perhaps, most insightful way out of the problem, the Higgs Boson, may need some drastic changes. As I presented at the beginning of this thread, we have dealt with theories involving electromagnetic theories of mass. If this idea is not recollected, what will we turn to?


    There are some alternatives. These actually go by the specific name of ''Higgless models''. We have actually a reasonable list of different theories which can be read about here:http://en.wikipedia....Higgsless_model .


    But what theory could entertain a gravitational charge best? I have decided that there is a theory which might be able to satisfy the speculations of treating mass as a charge on a system.



    Long before I learned about the Self-Creation Cosmology, I realized that this was an idea I had developed in my own head independantly, which is why it has me very excited.


    The development of Self-Creation Cosmology (SCC) should historically stem from the work of Nordstrom, who developed a relationship for matter and the gravitational field. In his relation, matter actually depended on the gravitational field:


    [math]\Box \phi = 4 \pi \rho[/math]


    In this equation, we can replace the density for the stress-energy tensor


    [math]\Box \phi = 4\pi \lambda T_{\mu \nu}[/math]


    In the (SCC) approach, we may have a trace free solution


    [math]\nabla_{\mu}T_{M \nu}^{\mu} = 4\pi f(\phi) = f(\phi) \Box \phi[/math]


    The is almost identical to an approach made by Ni


    [math]\eta_{\mu \nu} \partial_{\mu}\partial_{\nu} \phi = 4 \pi \rho k(\phi)[/math]


    Where the function [math]k(\phi)[/math] is written slightly different, but which is still related to the metric.


    The density of a particle, I once speculated, might just be seen as


    [math]\frac{1}{4\pi\lambda}\Box \phi = T_{\mu \nu}[/math]


    Essentially, the gravitational field gives rise to the creation of mass. Interestingly, my own argument for a similar model was that a particle coupled to the gravitational field will experience a gravitational charge, similar to how a particle experiences an electromagnetic charge when moving in an Electromagnetic field.


    The Gravitational Charge


    The gravitational charge has already been given with the expression of [math]\sqrt{G} M[/math] by Motz in his paper ''On the Quantization of Mass'' [3] In my own studies, I derived the gravitational charge as the following relationship:


    [math]\sqrt{G}M = \sqrt{E_gr_s}[/math]


    where [math]r_s[/math] is the Schwarzschild radius and [math]E_g[/math] is the gravitational energy. The gravitational energy is interpreted as the intrinsic gravitational energy due to charge, which for a photon would be zero.


    The gravitational charge then, can be thought of as the quantity of inertia inherent within a particle and you may even derive the relationship of the proper density of a particle which will be shown soon. The energy therefore can be given as


    [math]E_g = \frac{GM^2}{r_s}[/math]


    This may even be seen as a potential energy equation


    [math]\frac{GM^2}{r_s} = M\phi[/math]


    The question then is whether mass comes from a potential energy? The prototypical case of color charges in quark condensate also get masses from a potential energy. Is it possible that when we speak of a system getting a mass, is the mass a by-product of potential gravitational energy?


    One thing to keep in mind, is that the Schwartzschild metric describes bodies with mass. Therefore, it might be slightly interesting to realize that when you make [math]r_s \rightarrow 0[/math] in a limit, it may describe the non-presence of a gravitational charge of something,


    [math]\sqrt{G}M = \lim_{r_s \rightarrow 0} \sqrt{E_g r_s} = 0[/math]


    Would be a true statement for massless radiation.


    Photons do not decay spontaneously in space. However, using a catalyst like an electron or maybe even a photon collision, particles with matter can be created


    [math]\gamma \gamma \rightarrow e^{-}e^{+}[/math]


    So it seems it is not entirely unreasonable to believe that such photon collisions create a matter under what would be a special type of energy transition. Given enough energy in KeV, the energy transition will create particles with mass.


    The Planck relationship


    [math]\hbar c = GM^2[/math]


    Can yeild the Compton wave length and the Schwartzschild radius simultaneously by dividing both sides of this equation by [math]Mc^2[/math]


    [math]\frac{\hbar c}{Mc^2} = \frac{GM^2}{Mc^2} = (\frac{\hbar}{Mc} = \frac{GM}{c^2})[/math]


    The proper density of a particle can be given in a relationship as


    [math]8\pi \rho_0 (\frac{G}{c^2})[/math]


    and this can be set equal with the radius or the Compton wave length. The gravitational charge from the energy radius relationship can derive the following equation


    [math]\frac{GM^2}{Mc^2} = r_s[/math]


    which is the gravitational case analogue of the classical electron radius


    [math]\frac{e^2}{Mc^2} = R[/math]


    Increasing the gravitational charge on a system would be effectively increasing the inertial mass quantity of a system. In the Heirarchy problem, we do realize that there are many different masses in the Standard Model. During the energy transition, we may take a Taylor series on the gravitational energy and increase it according to the amount of energy which is contributing to the mass of the system.


    In this work, I have not derived an understanding mathematical how one might be able to approach the idea that mass is a charge as a charged particle would be as it moved through its respective field - in this case, the gravitational field. That is a difficult task but I do have a proposition to head in that direction.


    A mass operator takes into account of the interaction of a particle with it's own field but also including other fields if necessery. It has been shown already that in many senses, mass is an electromagnetic phenomenon from first principles and that charged particles do possess what is called an electromagnetic inertia when being accelerated through spacetime. In fact, we may believe that a portion of the mass of a system might exist as a contribution of electromagnetic phenomena. Then some of the rest of the inertial mass is purely the charge of system moving in it's respective field.


    So it would be a good approach to use a mass operator to describe both the gravitational field contribution of mass and the respective electromagnetic field. I will most likely write up more later, but there is so much in this work it might be best to chew over the first installment first.


    [1] - http://en.wikipedia....romagnetic_mass




    [2] http://www.wiley-vch...7408355_c01.pdf


    [3] http://www.gravityre...d/1971/motz.pdf (The work in which these investigations stemmed from)


    I see this place uses [math] tag latex. I will change it now. Sorry for the delay.


    Has this post been moved, I can't find it now?

  9. People often ask me about when proper mass can't work as good as relativistic mass or even better. I wrote an entire paper to answer this question but people don't read it carefully enough. Probably because they believe that no matter what the paper says they've already made up their mind because the already thought about it carefully alread a long time ago. And that is a good reason. I'd probably do the same thing - Too much reason with no real expectations of changing what they think in any sape or form


    So to take a shot at perhaps clarifying why physicists hold on to the notion of relativist mass. So I've decided to create a challenge for everyone. I have a SR text by Hans C. Ohanian. One of the homework problems is to find the mass density of a magnetic field. That's my challenge to you all. Solve this introductory level SR problem. Let's make it as simple as possible and assume that the magnetic field be uniform. Find the mass density of a the magnetic field. Use whatever definition of mass that you see fit.


    Good luck. :)


    So... you want someone here, to show how one would find the mass density of an electromagnetic field...


    For a non-dispersive material, the energy density of a magnetic field is


    [math]\rho = \frac{B \cdot B}{2\mu}[/math]


    I did find this paper which helps determine a mass density along magnetic field points. http://www-pw.physics.uiowa.edu/~dag/publications/2001_DeterminingTheMassDensityAlongMagneticFieldLinesFromToroidalEigenfrequenciesPolynomialExpansionAppliedToCRRESData_JGR.pdf

  10. Yes.


    But I will add, if there really are an infinite amount of universes (which I don't believe there is any isomorphic to our own), then you can expect even the most radical of universes existing with many dimensions. Theories like string theory, might be purporting to existences outside our own universe.

  11. Currently we know 4 dimensions (3 spatial X-Y-Z and 1 Temporary ) in our Universe.


    I wonder what are the requirements to be met by a new dimension to be considered as such.


    According to string theory or M theory, there are at least 10 or 11 dimensions in the universe.


    In this thread I would like to discuss which might be these new dimensions, but should first be clear what are the requirements to be a New Dimension.


    And also if this dimension is dependent or independent of the other dimensions.


    A pre-requisite to have another dimension, would be a degree of freedom. If you have the necessary space, volume, or whatever you choose, you can create a dimension from it. As far as I am aware, no other requirement is needed.

  12. The short answer: GR and QM are mutually incompatible.


    Good to point out, that being incapable of unifying two theories may not necessarily mean they inherently incompatible. It may be a point of how people are attempting to unify the theories. Every attempt so far has proven difficult - but on the whole does not mean the two theories cannot be unified.


    As a I understand it, string theory is also an attempt develop a general theory to include Gen. Rel. and Quantum theory.


    It's quite an unrealistic approach however. Don't get me wrong, many scientists like Leonard Susskind would bet their life that the world is made of strings... the theory can't convince many however that there really does exist as many dimensions it purports to. Start proving these dimensions exist, all several more of them, and I'm in... otherwise, string theory seems far-fetched.

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