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Posts posted by pmb

  1. Stay on topic please!

    Back on topic - The m that appears in the force equation is defined as [math]m = m_0\frac{dt}{d\tau}[/math]. It is rightly called the passive gravitational mass of the particle. This is just the value m in [math]P^{\mu} = (mc, p_x, p_y, p_z)[/math]

  2. Well I figure, to have an ordered set of events you need to be able to speak about time. In the beginning, there was no kind of time we can deal with in a relativistic sense. We are traditionally told, that everything diverged from a ''single point'' without dimensions. If this is the case, then the kind of time we often think about, the conventional space-time doesn't really hold. So whilst we may speak of ''first instances'' there was no kind of geometry to actually speak about time. Geometry only appears a little later when the universe has cooled down sufficiently to allow geometrogenesis. Which is actually a topic I believe Wheeler introduced, you may have heard of it?

    Isee what you're saying. I had forgotten about that notion. Thanks for reminding me. No. I never heard of geometrogenesis. What us it?

  3. I gave a specific example. Read the thread.

    You err when you claim that I didn't read the thread. Iread it and can't find such an example. Instead of my guessing why not simply tell me he post number and make this conversation shorter than it need be?


    The only thing I can find is this

    First you use [math]\nabla (\frac{\partial }{\partial t}\Psi) = \frac{\partial }{\partial t}(\nabla\Psi)[/math] but latter you use [math]\frac{\partial }{\partial t}(\nabla \Psi) = (\frac{\partial }{\partial t}\nabla) \Psi[/math] which is incorrect.

    One can tell merely by inspection that this expression is correct.


    [math]\nabla \Psi = \frac{\partial \Psi}{\partial x}\hat{e}_x + \frac{\partial \Psi}{\partial y}\hat{e}_y + \frac{\partial \Psi}{\partial z}\hat{e}_z[/math]


    [math]\frac{\partial}{\partial t}(\nabla \Psi) = \frac{\partial \Psi}{\partial t \partial x}\hat{e}_x + \frac{\partial \Psi}{\partial t \partial y}\hat{e}_y + \frac{\partial \Psi}{\partial t \partial z}\hat{e}_z[/math]


    [math]\frac{\partial}{\partial t}(\nabla \Psi) = (\frac{\partial}{\partial t \partial x}\hat{e}_x + \frac{\partial}{\partial t \partial y}\hat{e}_y + \frac{\partial}{\partial t \partial z}\hat{e}_z)\Psi[/math]


    [math]\frac{\partial}{\partial t}(\nabla \Psi) = (\frac{\partial}{\partial t}(\frac{\partial}{\partial x}\hat{e}_x + \frac{\partial}{\partial y}\hat{e}_y + \frac{\partial}{\partial z}\hat{e}_z))\Psi[/math]


    [math]\frac{\partial}{\partial t}(\nabla \Psi) = (\frac{\partial}{\partial t} \nabla)\Psi[/math]


    which proves that the expression is correct.


    I never said or insinuate that.

    Then all you had to say was either yes or no. It was unclear in your previous posts. That's why I asked. Since you didn't directly answer the question with either yes or no then I'll assume the answer is no, you're not saying that Hermitian operators don't follow the associative law. Then that's good since they do.


    But you aren't. You're quantizing Hooke's law.

    What does it mean to "quantize" an equation/law? Do you mean to cast the classical law/equation into an expressuion containing operators?

  4. Nice quote thank you.


    I agree strongly with that statement. I have even remarked myself in the past, that the beginning of the universe is devoid of such order that we associate to things today. I will expand on this if you want.

    Yes. Please do.

  5. Err there is a lowest energy. It is called the Zero Point Energy and always corresponds to


    [math]\frac{1}{2} \hbar \omega[/math]


    Of any quantum of energy.


    This is why we say, ''the lowest energy a photon can have resides at the zero point energy scale''.

    Zero point energy only pertains to bound states.

  6. Isn't the quantum a smaller measure of energy than a photon?


    Can't photons be different multiples of quanta?

    Photons are quanta of light. As such there is nothing which restricts either the energy or momentum of a photon to a multiple of a fundamental quantity. You can always find a photon at any energy that you desire. A simple way is through the Dopler Efect, i.e. by merely changig your frame of refrence to a new one moving relative to your initial frame. In general the photons energy will be anything you desire merely by carefull selection of the new frame of refernce.


    In modern use the term quanta is a small finite unit of something. A quantum of charge is the value of the charge of an electron. In the case of charge, any charge is a multiple of the value f an elctrons harg. A quantum of ligt is a photon, i.e. a quantum of electromagnetic radiation. Therefore in modern lingo photon and qantum are not the same thing.

  7. I agree... It cannot tell you what happened.



    I won't argue.


    If any of these pre-bang situations are your topic, I will argue them however... maybe these cyclic theories?

    The world is not cyclic, they certainly happened once and right now.



    ... and nothing is isotropic to these actions.

    Peebles has some interesting comments on the Big Bang. From Principles of Physical Cosmology, page 6

    If there were an instant, at a "big bang," when our universe started to expanding, it is not in our cosmology as now accepted, because no one has thought of a way to adduce objective physical evidence that such an event really happened.

  8. Hi pmb


    Thanks for those links. I had a good look at them and realised how much I didn't know that I didn't know! blink.gifbiggrin.gif

    But having said that it did start to cement some of what I had read on these forums. I think just seeing things written again and again helps make sense of it and it starts sinking in (that's the hope anyway!)

    It's a great place to learn things and to get help from others who can help.


    Cheers again!

    That's awesome. There are a lot of smart people here whom you'll find very helpful in your journey into physics. They're here to help so make full use of them. Best wishes.

  9. In Nuclear energy, a few grams of mass generatea huge energy.

    ...Therefore, in a reversible process, a huge energy is needed tocreate just few grams of mass. Hence, in order to generate the whole mass of theuniverse in a split of a second an infinite and irrational energy is needed. Therefore, is it a feasible process???

    The zero-energy universe hypothesis states that the total amount of energy in the universe is exactly zero. The total energy from matter is positive. The total energy of the gravitational field is negative. The sum is zero. See details at http://en.wikipedia.org/wiki/Zero-energy_universe


    [/font][/size]3. Overcomingthe force of gravity (escape velocity) – The requested escape velocity fromblack hole is so high that even the light can't escape. In the universe thereare billions of black holes. If we placed all of them in one singular point andwe also add the whole mass of stars in the universe, than the estimated therequired escape velocity would be so high that nothing could escape. Therefore,even if there was the energy that created the whole mass in the universe in a fractionof a second, than the requested escape velocity would be so high that notingcould escape. Anyhow, the requested escape velocity should be million or even billontimes faster than the speed of light. Is it feasible???

    No. In order to do what you say you'd have to move stars and black holes around. It takes energy to do that. There may not be enough energy available in the universe to do that. There are also other laws that you'd have to violate to ger that done like conservation of momentum, conservation of the center of mass etc.,


    4. Mass travel at a speed higher than the speedof light –..

    That's impossible. You may be confusing the faster than light expansion of space as moving particles faster than the speed of light.


    5. Is it possible that the steady state theory is correct? Just as Fred Hoyle have stated: " In steady state views, new matter is continuously created as the universe expands"

    No. It is not possible. Observation tells us that its not true.


    No you can't talk about time before the big bang.

    The standard model cannot tell you what happened before the big bang. That doesn't mean you can't ask about what happened before it. There are other theories which allow one to ask such questions. One such theory is the Pre-Big Bang Scenario. E.g. see http://arxiv.org/abs/hep-th/9907067

  10. Pm if you need to talk to someone.

    Thanks Aethelwulf. I don't run into a lot of nice people like you in these forums. Let it be known that your presence is greatly appreciated. I PMd you.


    I think I know partly what is wrong with me. I have to have gastric surgery. I was given omeprazole to help me in the mean time. I ran out last week so that might account for my stomach aches and nausea. I'm going back on it today.


    I'd appreciate your help on my web page if you have the time. I was wondering if you could read these carefully





    The last one you might find interesting. The results are consistent with an article I read in the American Journal of Physics


  11. No your relation is not generally true.

    Please provide a counter example.


    Associativity means [math]\hat{A} (\hat{B}\hat{C}) = (\hat{A} \hat{B})\hat{C}[/math]

    What kind of operators are you talking about? Hermitian? If so and given this statement and the comment above, are you saying that Hermitian operators don't follow the associative rule?

  12. Thankyou very much swansont you are the only person to have answered this question for me.

    Call me crazy, mind you I have never taken a physics class, but I think it theoretically eventually can reach zero.

    So maybe when i do start college in the fall i will eventually see why I'm wrong. I most likely am wrong, for i have no grounds to say otherwise and have no idea what those equations you showed me means.

    I just don't think something physical can infinite.


    The meaning of F = GMm/r^2 is as follows. First off his "G" is not what you used in your first post. It's a constant of proportionality. The M is the mass of a point sized object and m is also the mass of a point sized object. The r is the distance between the two point particles. In our case we can treat the Eath as a partilce because the gravitational field around the Earth is as if its due to a point sized object,so long as you stay outside the body of the Earth. Let m be the mass of your body. Compared to the size of the Earth your body can be considered a point object. The force between you and the Earth is then given by F = GMm/r^2.


    The same thing holds between any two objects which are far enough from each other so that the shape of the body doesn't come into play. E.g. if you and a truck are 50 feet a part then the force between the truck anbd you are F = GMm/r^2 where now M = mass of truck and m = mass of your body.

  13. How do you work out the space-time curvature according to general relativity? Iv seen the special relativity equations, but never the general

    It depends on the information which you have in front of you. If you have the metric tensor you can calculate the spacetime curvature straight from that. If you're given the stress-energy-momentum tensor T then you plug that into Einstein's equation and then solve for the metric tensor, Once you have the metric tensor you can calculate the curvature tensor.

  14. "In 1926 the chemist Gilbert N. Lewis (first) coined the name photon for these particles, and after 1927, when Arthur H. Compton won the Nobel Prize for his scattering studies, most scientists accepted the validity that quanta (quantum) of light have an independent existence, and Lewis' term photon for light quanta was accepted." (parenthesis added)

    Yes. The term was what was accepted. However the concept as he defined it was rejected. Gilbert thought of his photons as existing inside atoms and moved from atom to atom as a conserved quantity. The photon we know today is not a conserved quantity. They can be created and destroyed.


    As that wiki article explains

    Gilbert Lewis, who published a speculative theory in which photons were "uncreatable and indestructible".[6] Although Lewis' theory was never accepted as it was contradicted by many experiments, his new name, photon, was adopted immediately by most physicists

  15. I appreciate you have taken time into this. However, I still see a complication.


    let us assume that the christoffel symbol used does indeed have dimensions of acceleration, that means in your force equation, you have an extra factor of velocity squared to account for. Take into consideration your equation


    [math]f^k = m\Gamma^{k}_{\alpha \beta} v^{\alpha}v^{\beta}[/math]


    So if the connection was dimensionless, you would have the problem I showed before. If it has dimensions of acceleration, then what you'd have is


    force = mass * acceleration * velocity squared

    First, please note that the correct form is (note: I prefer G over f)


    [math]G_k = m\Gamma^{\alpha}_{k\beta} v_{\alpha} v^{\beta}[/math]


    I think that there is an error somewhere but I can't see it right now. I've been sick lately so my energy level is down. I'll figure it out when I get better.


    I that web page I lost the c^2 going from Eq. (7) to Eq. (10)

  16. A statement can sometimes be destroyed by an ugly fact --- I don't know what you did wrong, but not the end of the world.

    I'm not sure I actually did anything wrong. There is no way to determine uniuqly the dimensions of a Christoffel symbol. When I calculated them for a uniform gravitational field they turned out to be unitless. This will change on both the metric and the particular Christoffel symbol. In the case of a uniform gravitational field the two non-vanishing Christoffel symbols are


    [math]\Gamma ^0_{03} = \frac{1}{1 + gz/c^2}[/math]


    [math]\Gamma ^3_{09} = g(1 + gz/c^2)[/math]


    Notice that the [math]\Gamma ^0_{03}[/math] has no units while [math]\Gamma ^3_{09}[/math] has units of acceleration.


    If you were to follow the derivation at



    you'd see that, in the case of a particle in a uniform gravitational field, the units are that of force and since its expressed as G = -mg then the units are that of mass times acceleration. Notice that in this case there is no velocity dependance of the gravitational force. That only happens when the gravitoelectric of gravitomagnetic forces are present.

  17. Maybe you can clear the dimensions problem up for me as well. I think you may have a made a mistake. I think you defined velocity squared as (m/s)^2 but that's acceleration I thought.


    If it was acceleration, that would make sense, since kg*m/s^2 = force. But your equation doesn't use acceleration, it has a velocity squared term in it.

    You're correct of course. Maybe I made a mistake with the units of the Christoffel symols??? I dunno.


    I can't sit up straight any longer so I have to go for now. I'll recheck this and get back to you.

  18. I think the reason I called it field momentum is because that is what I have written down as. Just having a look at wiki, it describes a type of field momentum


    ''Electric and magnetic fields possess momentum regardless of whether they are static or they change in time....

    wiki is wrong. If you have an E field without a B field or a B field without an E field then the momentum density is zero. Only when there is both an E field and B field present can there be field momentum. As I said, use caution when useing wiki.


    No doubt this was a terminology my lecturer used for pi. But there is some clarification needed with the wiki article as well, since it defines a field momentum then described non-standard terminology..

    I think its safe to say that long as you define your terms where its possible for people to misunderstand then you're all set. I perfer to call the p in p = mv the momentum and refer to


    [math] P_i = \frac{\partial L}{\partial v_i}[/math]


    as canonical momentum. Although you have to be careful here. A lot of people don't recognize that term. People will often refer to it as generalized momentum. And of course, as you know, in QM p always means canonical momentum. QM texts are very clear about this. They always want to make sure that the p means canonical momentum and not ordinary momentum.

  19. Well I said what I said because of the four acceleration


    [math]a^x = \frac{du^x}{d\tau} = \Gamma^x_{ij} v^i v^j[/math]

    That expression is incorrect. 3-acceleration is defined as


    [math]a^x = \frac{du^x}{dt}[/math]


    Using MKS units the units of the Christofel symbols unitless. The m has units of mass (in kg) and the units of the product of those velocities are velocity squared, i.e. (m/s)^2. The means that the units of G are kg*(m/s)^2 which is force. It seems to me that the units do indeed check out.


    I don't call it kinetic momentum, I tend to call it a field momentum.

    Why? First of the value can be the same independant of whether the momentum density of the field is zero. And when there is just a magnetic field present then the momentum density of the field will be zero. The charged particle can undergo an acceleration yielding a changing momentum and the canonical momentum will be identical to the ordinary momentum.



    For these reasons I don't see why you'd connect the (e/c)A term with field momentum, especiall since the field might not even have any momentum. This is different than the article talke about since he was referring to two interacting charges whose joint field has a non-zero momentum density to it.

  20. Well, his paper defined it as the gravitational 3-force. Don't shoot the messenger, this is why I asked the questions I did.

    There are two kinds of forces in relativity, and in mechanics as a general rule. One type of force is anything for which the 4-force on a particle is non-zero. That means that when you changes the loccoordinate system to a locally inertial one then the particle is accelerating in tht frame. An inertial force is one that only exists in non-inertial frames. I.e. when the 4-acceleration is zero and the particles 3-acceleration is non-zero then its said that there is an inertial force on the particle.


    Regarding intepretation we need to seek out what is being taught in GR, i.e. look it up in the GR texts and see how its being taught. Not all will teach it the same way of course. Refering to the geodesic equation D'Invernos text states on page 130

    The additional terms involving [tex]\Gamma^{a}_{bc}[/tex] which appears are precisely the inertial force terms we encoutered before. The the equivalence principle requires that the gravitational forces, as well as the inertial forces, should be given by an appropriate [tex]\Gamma^{a}_{bc}[/tex].


    Refering to inertial forces D'Inverno states

    We shall adopt the attitude that if you judge them by their effects then they are very real forces

    The derivation In my web page was motivated by Mould's definition. I believe mine was clearer. But The definitions are identical


    I would say, should it be surprising that a force can have a christoffel symbol in it?

    Not at all. You should expect it.


    Here it is. You where right, it is not partial derivatives, but it does use proper time.

    To see how the partial derivatives come into play I've worked out an example using a uniform graivtational field to show that G = -m grad Phi. Please see Eq. (21) at http://home.comcast.net/~peter.m.brown/gr/grav_force.htm


    The partial derivatives connect the gravitational potentials [tex]g_{\alpha\beta}[/tex] to the Chistoffel symbols.


    Warning: Use caution when using wikipedia. Anyboldy can go in and make changes. You know nothing about who is saying what there. It's useful but still prone to a lot of errors.

  21. Pmb, where was the thread you posted your work on mass? Was it this one, I had been looking for it because I had a question.

    I don't recall the thread. The paper I wrote about the concept of mass is found at



    In your paper, you define the gravitational charge component in a gravitational 3-force equation. I believe it had the form


    [math]f^k = m\Gamma_{ij}^{k}v^i v^j[/math]


    I just wanted to know, how one would derive this equation.

    What you have as a superscript should be a subscript. The derivation is at



    The same derivation can be found in Basic Relativity by Richard A. Mould. The author was motivated by Moller's text so I assume you can also find that derivation there as well.


    Presumably the connection is playing your usual role of the gravitational field yes?

    Yes. To be precise it was Einstein who made this identification, not I.


    I see there is a dependance on the velocity. The [math]\Gamma_{ij}^{k}v^i v^j[/math] part looks a little bit like it comes from the geodesic equation of motion which involves usually a term of second derivatives.

    All forces in relavity are velocity dependant, e.g. like the Lorentz force on a charged partilce.


    The equation I had in mind was this one...So I am just wondering how the force equation ''gets'' the terms it has. I'm especially interested in the dependence of the velocity and how this comes about.

    You'll see when you follow the derivation in the URL I posted.


    This adds up to the same dimensions as you find in that force equation. But then I start thinking about the dimensions of the equation. [math]m\Gamma^{\mu}_{ij} v^i v^j[/math] How does this equate to a 3-force? If we just take it at face value, the part described as


    [math]\Gamma^{\mu}_{ij} v^i v^j[/math]


    would have to have dimensions of acceleration.

    The expression is supposed to be in Newtons, not m/s2. I'm sure that if you were to write the expression out in all its detail and I'm sure that you the dimensions are correct. Meanwill I'd double check it for myself when I have the time.

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