Aethelwulf Posted June 24, 2012 Share Posted June 24, 2012 That helps a lot pmb, thank you. I had just never seen it before. I actually wondered if the derivation had anything to do with momentum over time. Your link is clearing it up for me. 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 Just to let you know then, it must be wrong in your paper... around eq 15 I think. I copied it exactly I believe, except for changing ij = alpha beta. Link to comment Share on other sites More sharing options...

pmb Posted June 24, 2012 Author Share Posted June 24, 2012 (edited) 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. Edited June 24, 2012 by pmb Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Share Posted June 24, 2012 The expression is supposed to be in Newtons, not m/s^{2}. 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. 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] Link to comment Share on other sites More sharing options...

pmb Posted June 24, 2012 Author Share Posted June 24, 2012 (edited) 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. Edited June 24, 2012 by pmb Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Share Posted June 24, 2012 (edited) 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. 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. Oh I know, I was just stating the question I asked because of the 4-acceleration. I was aware your equation was using a 3-acceleration (which is why I asked the question about whether it had dimensions of acceleration). 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.... The definition of canonical momentum corresponding to the momentum operator of quantum mechanics when it interacts with the electromagnetic field is, using the principle of least coupling P=mv + qA'' Where the q we can assume is playing the charge e. Howsoever, it does say right below it ''Non standard terminology is sometimes used for these momenta:^{[18]} P for canonical momentum, Π = mv for kinetic momentum, and q A for potential momentum'' Of course, this is not where I learned to begin saying it was a ''field momentum''. 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... (as for the dimensions thingy) I was wondering whether the christoffel symbol in this case was dimensionless, you have to be careful as well because some of them aren't. 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. Yes, that would be right, but velocity squared doesn't have dimensions of (m/s)^2, that's acceleration. So maybe its just me, but if you christoffel symbol is dimensionless, then in your equation what you have is really mass x velocity squared, which is an energy. Unless I have messed up somewhere. Edited June 24, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

pmb Posted June 24, 2012 Author Share Posted June 24, 2012 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. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Share Posted June 24, 2012 (edited) 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. Edited June 24, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

pmb Posted June 24, 2012 Author Share Posted June 24, 2012 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. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Share Posted June 24, 2012 You're correct of course. Maybe I made a mistake with the units of the Christoffel symols??? I dunno. 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. Link to comment Share on other sites More sharing options...

pmb Posted June 24, 2012 Author Share Posted June 24, 2012 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 http://home.comcast.net/~peter.m.brown/gr/uniform_force.htm 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. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 25, 2012 Share Posted June 25, 2012 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 Link to comment Share on other sites More sharing options...

pmb Posted June 25, 2012 Author Share Posted June 25, 2012 (edited) 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) Edited June 25, 2012 by pmb Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 25, 2012 Share Posted June 25, 2012 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) No problems, take your time and get well! Link to comment Share on other sites More sharing options...

pmb Posted June 25, 2012 Author Share Posted June 25, 2012 No problems, take your time and get well! Thanks. Much appreciated. I can't believe how lousy I've been feeling lately. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 25, 2012 Share Posted June 25, 2012 Pm if you need to talk to someone. Link to comment Share on other sites More sharing options...

pmb Posted June 26, 2012 Author Share Posted June 26, 2012 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 http://home.comcast.net/~peter.m.brown/gr/grav_force.htm http://home.comcast.net/~peter.m.brown/gr/uniform_force.htm http://home.comcast.net/~peter.m.brown/gr/force_falling_particle.htm The last one you might find interesting. The results are consistent with an article I read in the American Journal of Physics Thanks Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 26, 2012 Share Posted June 26, 2012 Certainly no one's buisness than yours... Link to comment Share on other sites More sharing options...

pmb Posted June 26, 2012 Author Share Posted June 26, 2012 Certainly no one's buisness than yours... Call me thick, but I don't understand what that means. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 27, 2012 Share Posted June 27, 2012 Well, no one else needs to know you business than yourself - just basically means it has nothing to do with anyone but your good little self Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted June 27, 2012 Share Posted June 27, 2012 ! Moderator Note Stay on topic please! Link to comment Share on other sites More sharing options...

pmb Posted June 27, 2012 Author Share Posted June 27, 2012 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] Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 27, 2012 Share Posted June 27, 2012 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] That's fine, no probs with that in my eyes Link to comment Share on other sites More sharing options...

pmb Posted June 28, 2012 Author Share Posted June 28, 2012 (edited) No shooting here, only a polite and impersonal remark about the non-existence of gravitational forces in GR. The christoffel symbols are not gravitational fields. The christoffel have no true physical meaning. The definition of "gravitational field", as well as their true physical meaning, is found in Gravitation by Misner, Thorne and Wheeler (MTW) on page 467 One can always find in any given locality a frame of reference in which all "gravitational fields" (all Christoffel symbols; all [math]\Gamma^{\alpha}_{\mu\nu}[/math]) disappear. No [math]\Gamma[/math]'s means "no gravitational fields" and no local gravitational field means no "local gravitational energy-momentum." In the third place, pmb's cannot be derivative with respect to s, unless he is now using another notation than that he used before here when he tried to define kinetic momentum (e.g. in #3). They aren't. The first section of http://home.comcast.net/~peter.m.brown/gr/grav_force.htm defines the [math]v^{\alpha}[/math] as the derivative with respect to coordinate timne, t. Thus it is valid for luxons. In the second place, in the geodesic equation of motion cannot be proper time for massless particles. The end result is derivatives with respect to coordinate time. The expression holds for both luxons and tardons. The m in it is simply [math]m = P^0/c[/math] which is valid for luxons as well as tardyons. ..., [math]m\Gamma_{ij}^{k}v^i v^j[/math] is not a gravitational force because it includes a Christoffel symbol! Gravitation is not a force in general relativity. The summation is over 0, 1, 2, 3. i and j usually denote summation over 1, 2, 3. There is a zero 4-force on a particle in free-fall. However the gravitational force is not a 4-force, it's an inertial force and as such is represented by the Christoffel symbols as is the gravitational field as seen in MTW above The definition of the gravitational force is in Basic Relativity[/b by Richard A. Mould, [i]Springer Verlag[/i], 1994 page 262 Eq. (8.65) [math]G_{\mu} = m\Gamma^{\sigma}_{\mu\tau}v_{\sigma}v^{\tau}[/math] No problems, take your time and get well! I feel a little bit better today. I double checked my expression for the gravitational force with that of Mould's and they are identical. Mould's text was my motivation for my derivation. Edited June 28, 2012 by pmb Link to comment Share on other sites More sharing options...

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