joigus

Oh, boy. Thanks, dear islander. +1

I may have been not quite precise in my statement. Galilean or not is not coded in the force law though. It's almost all in the (mass)x(acceleration) of Newton's 3rd law. But not all... To be more precise, the force law should comply with certain constraints to be compatible with the whole Galilean group (isotropic, translation-invariant). So any force law that's of the form, \[ \boldsymbol{F}_{ij}=\boldsymbol{F}_{ij}\left(\left\Vert \boldsymbol{r}_{i}-\boldsymbol{r}_{j}\right\Vert \right) \] for particles i and j, would be compatible with the full Galilean group, \[ \boldsymbol{r}'=R\boldsymbol{r}-\boldsymbol{v}t+\boldsymbol{b} \] \[ t'=t \] where \( \boldsymbol{b} \) is a fixed translation, \( \boldsymbol{v} \) is a fixed velocity (Galilean boost), and \( R \) is a fixed rotation matrix, \[ RR^{t}=I \] \[ \det R=+1 \] \( I \) being the identity matrix. The matrix \( R \) codes a possible change of orientation between frames. There are velocity-dependent forces (magnetic) that can be made compatible with special relativity, but not with Galilean relativity. The whole shebang: https://en.wikipedia.org/wiki/Galilean_transformation#Galilean_group

Yes. For example, a model of gravity that tried to investigate deviations from Newton's gravity only in the power law, like, \[ \boldsymbol{F}_{12}=-G\frac{m_{1}m_{2}}{\left.r_{12}\right.^{2+\varepsilon}}\boldsymbol{u}_{12} \] for some small \( \varepsilon \) would be "Galilean," since it complies with, for (t',x') inertial observers vs (t,x) inertial observers. It would be perfectly Galilean, although not Newtonian, theory of gravity. I would have problems in reconciling well-trusted conservation laws, rotational invariance, etc. though.

Thanks. I subscribed to the YT channel some time ago. I'm hoping to have Kuhn interview Neil Turok, but no luck so far.

Thanks for your comments. Yes, I'm getting similar formulae for rotating satellites in free fall. I haven't thought of "proper" vs "coordinate" accelerations, but I think that's because the generality of the proper-time expression already accounts for that. Because the line element is 1st-order, quadratic in coordinate differences, when substituting general trajectories, the acceleration terms relevant are only quadratic in velocities, rather than 2nd-order time derivatives... I think you can pull off a precise-enough discussion without entering into effects of rotating Earth --Kerr metric--, as the angular-momentum term for the Kerr metric is J/Mc --with J spin angular momentum of Earth, really small--, but it would be interesting to generalise to that case for further precision. The thing that got me intrigued is @Genady's comment that consideration of exact expression for radial escape velocity falling from spatial infinity gives twice the correction that I was thinking about, corresponding to clock falling from finite --even if very far away-- distance. I think I'm on my way to understanding this discrepancy. Maybe I can comment on it later if anyone's interested.

Vagueness, vagueness everywhere, nor any prop to think! --Rime of the Ancient Physicist

"Galilean" makes reference to the space-time symmetry group of the equations of motion. All of pre-relativistic physics is "Galilean". If I come up with a model to describe certain interactions tomorrow that assumes absolute time as part of the mathematical features, it will be Galilean. You're absolutely right. Einstein's relativity and Galilean relativity are joined at the hip. One could say that, in a sense, Galilean relativity is but a particular case of Einstein's relativity with c=infinity. This forces instant interactions. If interactions were not instantaneous, inertial observers --in a Galilean world-- would be capable of telling their absolute state of motion.

And he didn't measure anything. Big mistake.

Yes, a thing of the past. Aristotle thought that F=mv instead of F=ma, and that gravity was the natural tendency of things from the sub-celestial world to go back to their place. So... a thing it is. Or it was.

Galilean gravity is Newtonian gravity. What @geordief is proposing is redoing GR with the Galilean group as the locally inertial symmetry group (how different observers co-relate their observations). That's Galilean GR. That's certainly possible. And as Markus pointed out, you simply wouldn't have gravitational waves, horizons, and other things. Singularities would be avoidable, as you can always assume particles to be non-pointlike. The Einstein tensor would have fewer degrees of freedom. But geometry of the distorsion of space sections of space time would be described by the inverse-square law.

Flat GR does not exactly give Newtonian gravity + Newton's equations + Galilean group (as its ST-symmetry). It produces Newtonian gravity as a theory of the potential + Lorentz-group relativity, which is a strange beast I have never met. In order to obtain Newtonian gravity + Newton's equations + Galilean group, you must substitute the Minkowski metric by the Euclidean metric, while isolating time in so-called simultaneity fibers that don't mix with space under transformations between inertial observers. IOW, time must become absolute. So I suppose my answer is yes, Galilean gravity is a thing. Galilean group gives you the symmetry group of Newton's equations of motion, while Newton's law of gravitation relates sources of gravitational field with the gravitational field everywhere.

According to whom? "Explode at the same instant" is a frame-dependent occurrence. As @MigL and @Markus Hanke pointed out, it's important to distinguish between frame-dependent and frame-independent.

Yes, I had to go back to @Genady's comment, and then further back to OP. He meant clock C. If you use the radial escape velocity relation you do get twice the rs/r correction. Interesting... Thanks both. More discussion tomorrow. I'm tired and it's very late here.

I'm with you a 100 %. No, it's not. Or the word "general" means something completely different from what I thought.

OK. We're not talking about black holes. We're talking about the Earth. We're talking satellites orbiting the Earth. Schwarzschild's solution is not just about BH's. It's what any static, spherical distribution of mass looks like far enough away from it. We're talking more like somewhere between 6000 and 7000 Km from a horizon that, BTW, doesn't exist for the Earth. Initial velocities can, only too obviously, take any value we want, not related to r. The formula that you're using clearly comes from some assumption of initial conditions that's not compatible with the case we're discussing, and in particular implies radial approach. It's only too obviously not valid in general. Certainly not valid for rockets moving in an arbitrary way, for example. Make some substitutions and it will be transparent to you. You can't take any velocity-radius dependence, or whatever other similar dependence used for a very particular problem from an article or book --in the case you took it from it's clearly a radial free fall--, and plug it into a completely differently motivated physical problem. The method I used is described in such classics as Lightman's Problem Book in Relativity and Gravitation. It's clearly thought out and reasoned from the start. You may doubt its validity only because it doesn't allow you to separate in your mind the "kinematic" effects and the 1/r effect. But it's known to work, because it's derived from first principles and the approximations I have justified at every step. Essentially: Weak fields (but not so weak that all GR effects disappear) and slow velocities, which is an essential part of the discussion. I understand that @mistermack's clocks are moving much more slowly than the speed of light.

Are you not happy with the fact that \( 2\times10^{-22} \) is much less than \( 2.4\times10^{-9} \)? Your algebra might be correct, but inspection of Schwarzschild's metric tells you how small objects move somewhere near an approximately spherical, static source of gravitation. I know my approximation to be correct to 1st order in rs/r, as well as being the usual one. So I think I'm right. Centrifugal terms are quadratic in velocities, so negligibly small in comparison with r-dependent term. c2 is already accounted for in rs. For Earth it gives about .88 cm which is much more sizeable than so-called "kinematic" corrections. I'll keep thinking about it as soon as I can though. Where do you get this from?

That's essentially what I was proposing. That along with ignoring r times angle-dot for approximantion, as speeds are much smaller than c.

I see. Thanks.

OK. I've got a little bit to digest for the time being. I do believe you can get away with the rs/r term in 1st order and ignore kinematical corrections. Reason being that rs/r is of order \( 2.4\times10^{-9} \), while \( v/c \) is of order \( 1.4\times10^{-11} \). First corrections to "kinematic" terms are square of the latter, so order \( 2\times10^{-22} \) while first corrections for the r-dependent term are just of order (1st order in rs/r for sqrt of g00 and grr) as said before. Mmmm. @Mordred. I think I understand your main argument on rotation/acceleration, but... I wouldn't invoke Cartesian coordinates either. It's all more transparent in spherical coordinates. I'm confused by isotropic dust in the discussion too... What are you deploying on me man? I did make a rough estimation for rotating A and B clocks and didn't find that much of a difference if speeds are safely orders below c, but I could be wrong. I picked circular paths with fixed radii and used a 3rd-Kepler kind of approximation. GR corrections to Kepler's law don't seem to change scenario significantly, I think. They differ only by a sqrt(g00)... I'll keep an eye on this for further comments. I always try to keep as pedestrian as humanly possible. So far I stand by my estimation. Maybe later...

I see how you would think that. It sounds very common-sensical. But that's not how it works. Keep in mind that any exact solution of Einstein's field equations already has any possible kinematical effects included. It's a package deal. Most of these solutions have a structure, \[ f\left(r\right)dt^{2}-\frac{1}{f\left(r\right)}dr^{2}-r^{2}d\Omega^{2} \] The \( f\left( r \right) \) and \( 1/f\left( r \right) \) play the role of "local contraction and dilation gamma factors" so to speak. Because GR is obtained from --among other things-- a demand that it satisfies SR locally, it is guaranteed to take care of that. \( \left(1-\frac{r_{\textrm{s}}}{r}\right)^{-1} \) plays the role of a gamma factor of sorts, while \( \left(1-\frac{r_{\textrm{s}}}{2r}\right) \) plays the role of an inverse gamma factor of sorts, if you will. It's true that the SR metric is different, but GR is under no obligation to follow SR's intuitions so closely. The equation that you wrote is off by a square in the differentials, so it should be, \[ d\tau^{2}\simeq\left(1-\frac{r_{\textrm{s}}}{r}\right)dt^{2} \] Which, after taking the square root becomes, \[ d\tau\simeq\left(1-\frac{r_{\textrm{s}}}{r}\right)^{1/2}dt \] And, only after Taylor expanding and keeping first-order terms, \[ d\tau\simeq\left(1-\frac{r_{\textrm{s}}}{2r}\right)dt \] which is the approximation I used.

For Earth's escape velocity the v/c parameter (the beta relativistic parameter) is about 1.4x10-11. For the other velocities involved it's much smaller. The rs/r term in Schwarzschild metric though is of order 2.4x10-9. rs being the Schwarzschild radius of the Earth, and r the actual radius of the Earth. In order to tackle these problems, you'd better not start thinking in terms of combined effects of r-dependent time dilation plus kinematical time dilation. Otherwise, you get confused, make a mess, and probably get the answer wrong. As @Markus Hanke often says, GR is highly non-linear, so it's not a matter of this effect plus that effect. I'm sure @Mordred agrees on this particular point. What you do is write the Schwarzschild metric and do all your calculations of proper times from there for different trajectories --elapsed time from different POV's, as Markus suggests. The metric really gives you everything you need for small clocks either falling or accelerating, etc. in the "background" field. You just plug in the trajectories \( r\left( t \right) \), \( \theta \left( t \right) \), and \( \phi \left( t \right) \). It's for finer approximations that things might get hairy. I've tried to do that for this case, and here's what I get: \[ d\tau^{2}=\left(1-\frac{r_{\textrm{s}}}{r}\right)dt^{2}-\frac{1}{c^{2}}\left(1-\frac{r_{\textrm{s}}}{r}\right)^{-1}dr^{2}-\frac{1}{c^{2}}r^{2}\sin^{2}\theta d\theta^{2}-\frac{1}{c^{2}}r^{2}d\phi^{2} \] You can further assume that all trajectories are equatorial, so, \[ d\tau^{2}\simeq\left[\left(1-\frac{r_{\textrm{s}}}{r}\right)-\left(1+\frac{r_{\textrm{s}}}{r}\right)\left(\frac{\dot{r}}{c}\right)^{2}-\left(\frac{r\dot{\phi}}{c}\right)^{2}\right]dt^{2} \] For small velocities \( \dot{r},r\dot{\theta},r\dot{\phi}\ll c \),and after Taylor-expanding the square roots, you get, \[ d\tau\simeq\left(1-\frac{r_{\textrm{s}}}{2r}\right)dt \] For clocks A, B, and C, we get proper times from 0 to a certain time standard assymptotic time T: \[ \tau_{A}\simeq\int_{0}^{T}dt\left(1-\frac{r_{\textrm{s}}}{2r_{A}}\right)=\left(1-\frac{r_{\textrm{s}}}{2r_{A}}\right)T \] \[ \tau_{B}\simeq\int_{0}^{T}dt\left(1-\frac{r_{\textrm{s}}}{2r_{B}}\right)=\left(1-\frac{r_{\textrm{s}}}{2r_{B}}\right)T \] \[ \tau_{C}\simeq\int_{0}^{T}dt\left(1-\frac{r_{\textrm{s}}}{2r_{C}\left(t\right)}\right) \] Remembering that, \[ r_{A}<r_{B} \] and rC monotonically goes from rB to rA, \[ r_{C}\left(0\right)=r_{B} \] \[ r_{C}\left(T\right)=r_{A} \] So that, \[ \tau_{A}<\tau_{C}<\tau_{B} \] So the proper time for observer A, who is at sea level on the Earth is shorter –proper, so from his own POV– than it appears to be from the POV of an observer at distance rA, and both shorter still than time elapsed from POV of an asymptotic observer, which is T: \[ \tau_{\infty}=\lim_{r\rightarrow\infty}\left(1-\frac{r_{\textrm{s}}}{2r}\right)T=T \] Which means that those far-away inertial observers see B slowing down, C slowing down even more, and A running the slowest of all. For clocks moving in any funny arbitrary way this would not be true. I think this approximation is good enough for our purposes, as the kinematic terms are at least a couple of orders of magnitude smaller than the r-dependent term. That's what I get from my analysis, anyway. The method I know to be correct. Please do tell me if you think I overlooked something important.

Quarks don't want to get pulled apart...

That physical intuition makes sense to me. Something like that is to be expected. But a proton is uud only on the average. I don't think it would be anything as simple as a dielectric.

Same distance, same "force," meaning same Feynman diagram to all levels in QED (except for sign) if you assume the proton to be point-like. If you ramp-up the collision energy though you would eventually find EM form factors in the proton, but no form factors in the electron, so... You must not assume, as a matter of course, a question to have the most possible conceivable context, I think. You must narrow down the possibilities by assuming a natural context. This could be said of any question.

Well, yes. But if I'm allowed to play with distances, then any force can be made stronger, or weaker, as any other. I'm assuming same distance.