  # pmb

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379

## Everything posted by pmb

1. If you're attempting to say that given two operators P and F when inserted into Newton's secind law yields $\frac{d \hat{p}}{d t} = \hat{F}$ then you're wrong.
2. Back on topic - The m that appears in the force equation is defined as $m = m_0\frac{dt}{d\tau}$. It is rightly called the passive gravitational mass of the particle. This is just the value m in $P^{\mu} = (mc, p_x, p_y, p_z)$
3. Isee what you're saying. I had forgotten about that notion. Thanks for reminding me. No. I never heard of geometrogenesis. What us it?
4. 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 One can tell merely by inspection that this expression is correct. $\nabla \Psi = \frac{\partial \Psi}{\partial x}\hat{e}_x + \frac{\partial \Psi}{\partial y}\hat{e}_y + \frac{\partial \Psi}{\partial z}\hat{e}_z$ $\frac{\partial}{\partial t}(\nabla \Psi) = \frac{\partial \Psi}{\partial t \partial x}\hat{e}_x + \frac{\partia 5. Zero point energy only pertains to bound states. 6. 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 electr 7. Peebles has some interesting comments on the Big Bang. From Principles of Physical Cosmology, page 6 8. 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. 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 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 10. Call me thick, but I don't understand what that means. 11. 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 http://home.comcast.net/~peter.m.brown/gr/grav_force.htm http://home.comcast.net/~peter.m.br 12. Please provide a counter example. 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? 13. 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 y 14. 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. 15. Thanks. Much appreciated. I can't believe how lousy I've been feeling lately. 16. 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 17. 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}$ 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)
18. 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 $\Gamma ^0_{03} = \frac{1}{1 + gz/c^2}$ $\Gamma ^3_{09} = g(1 + gz/c^2)$ Notice that the $\Gamma ^0_{03}$ has no units while $\Gamma ^3_{09}$ has units of acceleration.
19. 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.
20. 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. 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 $P_i = \frac{\partial L}{\partial v_i}$ as canonical momentum. Although you have to be careful here. A lot of people don't recognize that
21. That expression is incorrect. 3-acceleration is defined as $a^x = \frac{du^x}{dt}$ 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. 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
22. 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 th
23. I don't recall the thread. The paper I wrote about the concept of mass is found at http://arxiv.org/abs/0709.0687 What you have as a superscript should be a subscript. The derivation is at http://home.comcast.net/~peter.m.brown/gr/grav_force.htm 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. Yes. To be precise it was Einstein who made this identification, not I. All forces in relavity are velocity dependant, e.g. like the Lorentz force on a
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