Aethelwulf Posted September 4, 2012 Report Posted September 4, 2012 (edited) Coupling of the Gravielectric and Gravimagnetic Fields with the Gravitational Charge Abstract The purpose of this paper is to investigate further the question of what a gravitational charge is and how it relates to the fundamental physics. It is a continuation from my previous work (1). In this work we shall see how the Coriolis effect for particles gives rise to the gravimagnetic coupling to fields. I will investigate this from the approach made by Motz and Sciarma when discussing this coupling in terms of a cross product between the field and the particles respective velocity. The Coriolis Field The Coriolis field is a gravimagnetic field. Rotating objects (or even those with intrinsic angular momentum) couple to this field. Now going back to Motz paper, which is referenced on page 1, post 1, but here it is again: http://www.gravityresearchfoundation.../1971/motz.pdf Motz explains how there is a coupling of matter to the gravimagnetic field through the effort of a cross product, which he explains is fully discussed by Sciarma. Now I went to some effort to find Sciarma's paper as well: http://articles.adsabs.harvard.edu/f...NRAS.113...34S It's free which is good. Motz explains how a moving charge [math]\sqrt{G}M[/math] with velocity [math]v[/math] is not only coupled to Newtonian gravitational fields and gravielectric fields, but also to the Coriolis Field, which is the gravimagnetic field which he has defined as [math]\frac{2\omega c}{\sqrt{G}}[/math]. This seems to have come from (my guess) that the Coriolis acceleration is [math]a = 2(\omega \times v)[/math] and the force is [math]F = -2M\omega \times v[/math] so you must get his quantity by dividing the force by the gravitational charge [math]\frac{F}{\sqrt{G}M} = \frac{2\omega v}{\sqrt{G}}[/math] Such a coupling of the charge to gravimagnetic field is achieved, as explained by Motz [math]\sqrt{G}M \frac{v}{c} \times \frac{2\omega c}{\sqrt{G}}[/math] according to Motz. Now, I will explain something else. The cross product of the terms [math]v \times \omega[/math] actually give rise to a matrix determinent, which I am not going to write out but I hope you can take my word for it. The Lorentz force is of course [math]evB[/math] and to refresh our minds, the Coriolis force is [math]-2M \omega v[/math] (where we are omitting the cross products). What is interesting is if you set them equal, [math]evB = -2M\omega v[/math] (setting these two quantities equal with each other should not be a surprise, since the Coriolis force is a type of gravimagnetic field (1)) cancel the linear velocities and divide the gravitational charge on both sides you get [math]B = -\frac{2 \omega}{\sqrt{G}}[/math] Now, according to Sciarma, the gravimagnetic field is in general not zero, in fact one could argue that for all fundamental particles, it is never zero due to an intrinsic angular momentum (and for the fact we have never discovered a spin-zero particle before, or as Sciarma put it, any rotating charge distribution). In fact, Sciarma notes: [math]v \times B = 2v \times \omega[/math] Understanding the Work Further We may do a cross product on the term I derived [math]-\frac{2 \omega}{\sqrt{G}} \times v = 2v \times \omega[/math] Normally if we just concentrated on the right hand side expression, we may see that if the elementary charge was coupled to this, we would obtain the force [math]e(-\frac{2 \omega}{\sqrt{G}} \times v)[/math] What is perhaps interesting is the common relationship this has with the usual approach made in gravielectromagnetic equations when converting the electromagnetic equations to their respective gravitational analogues. In GEM (gravitoelectromagnetic) equations, the force equations are the standard force equation [math]F = ev \times B[/math] and the GEM analogue [math]F = mv \times B_g[/math] Going back to [math]-\frac{2 \omega}{\sqrt{G}} \times v = 2v \times \omega[/math] Notice on the left we have potentially the force term, as explained before, if the charge had been coupled to it. What we have on the right hand side is the acceleration, thus the force on the right is obtained if one multiplies the mass. So what we essentially have is something similar to the force equations [math]F = ev \times B[/math] [math]F = mv \times B_g[/math] A Different Mathematical way to View the Coupling to Gravimagnetic field I begin with this equation [math]\frac{GM^2}{\hbar} = (\Omega \times r_s)[/math] The derivation is quite long, so I won't be writing it out for the purpose of getting through this as quick as possible. One can take my equation [math]B = -\frac{2\omega}{\sqrt{G}}[/math] and do the cross product with the term on the right of the equation just submitted. We have [math]-\frac{2\omega}{\sqrt{G}} \times (\Omega \times r_s) = \frac{\omega \times v}{\sqrt{G}}[/math] Where on the left, we can see Motz' term he defines as the gravimagnetic field. One can obtain this by saying [math]ev \times B = -2M \omega \times v[/math] and divide the gravitational charge on both sides we end up [math]v \times B = \frac{-2\omega \times v}{\sqrt{G}}[/math] and thus by substituting all the respective terms together, one can end up with this relationship: [math]-\frac{2\omega}{\sqrt{G}} \times (\Omega \times r_s) = B \times v[/math] Interestingly, [math](\Omega \times r_s)[/math] is just a rotational velocity. It certainly seems appropriate to consider rotating bodies coupling to such gravimagnetic fields. [math]\Omega[/math] is actually perpendicular to the radius component [math]\Omega \perp r_s[/math]. However, where we have speculated a rotational velocity, this only applies to systems which are not pointlike but rather sphere's. Sphere's of course will have rotational velocities. Conclusions I think the approach made by Motz warrants further investigation from as many independant sources as possible. This investigation could lead to a new understanding of how gravitational forces play a role at the fundamental level and will perhaps itself help pave a better understanding of the unification of such phenomenon. (1) - http://arxiv.org/ftp/arxiv/papers/1109/1109.3624.pdf Edited September 4, 2012 by Aethelwulf Quote
Aethelwulf Posted September 4, 2012 Author Report Posted September 4, 2012 (edited) Hmmm... three equations are finding it difficult to translate (just testing something) [math]F = e(v \times B)[/math] [math]F = ev \times B[/math] Edit/Sorted. Edited September 4, 2012 by Aethelwulf Quote
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.