swansont Posted February 13, 2022 Share Posted February 13, 2022 3 hours ago, rjbeery said: The light ray, B, isn't bending due to gravitational effects; it's bending due to the refraction index being set to the time dilation function. See "n = gamma" in the corner? B is approaching the mass M purely through optical considerations. Why would the index equal gamma? I’m looking for a physics justification. Not something you’re just making up. Link to comment Share on other sites More sharing options...

rjbeery Posted February 14, 2022 Author Share Posted February 14, 2022 (edited) 23 hours ago, swansont said: Why would the index equal gamma? I’m looking for a physics justification. Not something you’re just making up. I have a full response, but I cannot get LATEX to work. Here's the short, ugly version: t_0 * gamma = t_r where gamma >=1, t_0 is the remote observer, and t_r is a clock at radius r above a large mass. For every second of t_r, the remote observer sees "t_0 * gamma" seconds pass on his local clock. Now we take c_0 as the local (relative to the remote observer) velocity of light, and c_r as the velocity that the remote observer measures at radius r above the large mass. Velocity has time on the denominator, so we have c_0 / gamma = c_r where a local light-ray moves one light-second per second, but that same light-ray moves "one over gamma light-seconds per second" at radius r. This should make intuitive sense. If gamma = 2, then the clock at r moves half as fast; therefore light moves half as fast. This gives gamma = c_0 / c_r which is the velocity of light "in a vacuum" over the velocity of light in another medium, which is the precise definition of the index of refraction. Edited February 14, 2022 by rjbeery Link to comment Share on other sites More sharing options...

Ghideon Posted February 15, 2022 Share Posted February 15, 2022 8 hours ago, rjbeery said: Now we take c_0 as the local (relative to the remote observer) velocity of light, and c_r as the velocity that the remote observer measures at radius r above the large mass. Velocity has time on the denominator, so we have I do not follow the physics of your argument, can you please explain? As far as I know gamma* is calculated using an invariant speed of light c. It looks lite you have an observer dependent speed of light, how does that affect gamma? *) Reference: https://en.wikipedia.org/wiki/Lorentz_factor Link to comment Share on other sites More sharing options...

SergUpstart Posted February 15, 2022 Share Posted February 15, 2022 1 hour ago, Ghideon said: I do not follow the physics of your argument, can you please explain? As far as I know gamma* is calculated using an invariant speed of light c. It looks lite you have an observer dependent speed of light, how does that affect gamma? On the contrary, G should be a constant, the speed of light is variable, and if the speed of light is constant, then G should be variable, as a payment for the constant speed of light. Link to comment Share on other sites More sharing options...

swansont Posted February 15, 2022 Share Posted February 15, 2022 13 hours ago, rjbeery said: I have a full response, but I cannot get LATEX to work. Here's the short, ugly version: t_0 * gamma = t_r where gamma >=1, t_0 is the remote observer, and t_r is a clock at radius r above a large mass. For every second of t_r, the remote observer sees "t_0 * gamma" seconds pass on his local clock. In the scenario I brought up r would either be infinite, or is the same for both clocks, because we're discussing a situation where we're ignoring gravity. It's getting increasingly difficult to think that this is just an oversight on your part. So gamma is 1, from your calculation. There is no bending of light in deep space. And yet we can have time dilation Link to comment Share on other sites More sharing options...

rjbeery Posted February 15, 2022 Author Share Posted February 15, 2022 12 hours ago, Ghideon said: I do not follow the physics of your argument, can you please explain? As far as I know gamma* is calculated using an invariant speed of light c. It looks lite you have an observer dependent speed of light, how does that affect gamma? *) Reference: https://en.wikipedia.org/wiki/Lorentz_factor The speed of light is taken to be invariant locally. If time is slowed down by 50% "over there" then the speed of light has also been slowed down "over there". 7 hours ago, swansont said: In the scenario I brought up r would either be infinite, or is the same for both clocks, because we're discussing a situation where we're ignoring gravity. It's getting increasingly difficult to think that this is just an oversight on your part. I'm sorry but I looked at all of your posts following my initial essay post, and I'm not sure which scenario you're referring to. 7 hours ago, swansont said: So gamma is 1, from your calculation. There is no bending of light in deep space. And yet we can have time dilation Yes. Have you looked at the diagram? Gamma is 1 at r = infinity. There is no bending of light in deep space; there is no gravity in deep space; there is no time dilation in deep space. I think perhaps you're mistakenly thinking about the meaning of gamma. Link to comment Share on other sites More sharing options...

swansont Posted February 15, 2022 Share Posted February 15, 2022 1 hour ago, rjbeery said: Yes. Have you looked at the diagram? Gamma is 1 at r = infinity. There is no bending of light in deep space; there is no gravity in deep space; there is no time dilation in deep space. I think perhaps you're mistakenly thinking about the meaning of gamma. If you have a moving clock, it will experience time dilation. Even in deep space. Gamma is speed dependent in special relativity. Link to comment Share on other sites More sharing options...

rjbeery Posted February 15, 2022 Author Share Posted February 15, 2022 7 minutes ago, swansont said: If you have a moving clock, it will experience time dilation. Even in deep space. Gamma is speed dependent in special relativity. I understand now. Yes, you are correct, time dilation (on its own) does not effect a light ray's path because it is the gradient that determines curvature. Even in an accelerated frame, where time dilation would exhibit a gradient, that gradient is in the direction of travel -- so, again, no curvature. We see the same thing when we stick our straw straight-down into the water; the index of refraction changes, but the path of light does not. Link to comment Share on other sites More sharing options...

Markus Hanke Posted February 17, 2022 Share Posted February 17, 2022 (edited) On 2/13/2022 at 6:30 PM, SergUpstart said: Are these 45 degrees visible in the experiment or is it just a theory? Yes, they are apparent in the LIGO data. Searching for additional polarisation states (vector and scalar states in addition to the tensor states) is also under way, as this provides a way to test for modifications to the laws of gravity. On 2/13/2022 at 6:30 PM, SergUpstart said: If any vector is in a plane oriented at an angle of 45 degrees to the direction of wave propagation, then it can be decomposed into longitudinal and transverse (at 90 degrees) components that are modulo equal to each other, is that so? I’m sorry, I don’t quite understand what you mean by this? Gravitational waves are always transverse; their dynamics are also nonlinear, so I doubt it is possible to decompose them in simple ways. On 2/13/2022 at 11:15 PM, rjbeery said: I'm using PE + KE = constant as a shortcut to determine the free-fall kinetic energy as a function of r This is fine - but again requires that such a unique decomposition is possible, which brings us back to the gravitational potential issue. So yes, your approach works - but only given the boundary conditions I mentioned. Another way to look at this is via the action, which is the difference of kinetic and potential energies (Langrangian), integrated over time. In Newtonian gravity, the Lagrangian is just a sum of three terms, so you can easily see the kinetic and potential parts. In GR however the action is of the form \[S=\frac{1}{2\kappa} \int{R \sqrt{-g}}d^4x \] How do you decompose this into T and V parts? You can’t, except under very special circumstances. On 2/13/2022 at 11:15 PM, rjbeery said: GR time dilation is what determines the geodesic. And again, this is only sort-of true under special circumstances. In general, in the geodesic equation, the geodesic depends on all components of the metric tensor in a rather non-trivial way, each one of which may depend on several coordinates. If, however, you introduce the symmetry conditions I mentioned, and in addition assume low velocities and weak fields, then the geodesic depends as an approximation only on initial conditions and g00 - which, in effect, plays the role of a potential. On 2/13/2022 at 11:15 PM, rjbeery said: I'm not refuting GR in any way. Well, you then have to acknowledge that in the general case, the kinematics of test particles depend on more than a single quantity. On 2/14/2022 at 3:41 AM, rjbeery said: gravitational potential may be the same for each ray, and the time dilation may be different How does that work, exactly? Remember that it isn’t position that determines the difference in that example, but orientation (ie motion in direction of rotation, or against it). Edited February 17, 2022 by Markus Hanke Link to comment Share on other sites More sharing options...

rjbeery Posted February 19, 2022 Author Share Posted February 19, 2022 On 2/16/2022 at 7:41 PM, Markus Hanke said: Well, you then have to acknowledge that in the general case, the kinematics of test particles depend on more than a single quantity. It isn't a single quantity, it's a scalar gradient. On 2/16/2022 at 7:41 PM, Markus Hanke said: How does that work, exactly? Remember that it isn’t position that determines the difference in that example, but orientation (ie motion in direction of rotation, or against it). I'm not exactly sure what you're asking here, but frame-dragging is my response. Quote Rotational frame-dragging (the Lense–Thirring effect) appears in the general principle of relativity and similar theories in the vicinity of rotating massive objects. Under the Lense–Thirring effect, the frame of reference in which a clock ticks the fastest is one which is revolving around the object as viewed by a distant observer. This also means that light traveling in the direction of rotation of the object will move past the massive object faster than light moving against the rotation, as seen by a distant observer. It is now the best known frame-dragging effect, partly thanks to the Gravity Probe B experiment. Qualitatively, frame-dragging can be viewed as the gravitational analog of electromagnetic induction. Anyway, as I said, different time dilation effects do not necessitate a change in direction. The straw will continue to appear to move "straight down" into a fluid, regardless of any change in index of refraction, because the surface of the fluid is perpendicular to the movement of the straw. Link to comment Share on other sites More sharing options...

Markus Hanke Posted February 19, 2022 Share Posted February 19, 2022 2 hours ago, rjbeery said: It isn't a single quantity, it's a scalar gradient. Do you mean a scalar gradient - which is a scalar quantity -, or the gradient of a scalar field (which is a vector)? These are very different. 2 hours ago, rjbeery said: I'm not exactly sure what you're asking here, but frame-dragging is my response. Yes, frame dragging is the GR effect. However, note that this arises from off-diagonal terms in the metric tensor - I do not see how you can self-consistently model this using a ‘scalar gradient’ (clarification required as per above) alone. This is why this effect does not (and cannot) exist in Newtonian gravity. Link to comment Share on other sites More sharing options...

rjbeery Posted February 19, 2022 Author Share Posted February 19, 2022 The path of the light ray is determined by the gradient of the scalar time dilation field. It’s a vector. Frame dragging apparently does not affect the radial gradient of time dilation. Note that this if not the same thing as not affecting the time dilation field itself. Link to comment Share on other sites More sharing options...

Markus Hanke Posted February 20, 2022 Share Posted February 20, 2022 5 hours ago, rjbeery said: The path of the light ray is determined by the gradient of the scalar time dilation field. But this isn’t what GR predicts - the geodesic in fact is determined by all components of the metric, plus boundary conditions. I thought you said your idea is meant to replicate the results of GR? Also, what exactly do you mean by “scalar time dilation field”? Time dilation is a relationship between clocks - so assigning a single value to each point in space isn’t enough, you need to also fix your reference clock somehow. 5 hours ago, rjbeery said: Frame dragging apparently does not affect the radial gradient of time dilation. Well then there is a contradiction with your model, because in the real world frame dragging does very much affect the geodesics of light (and massive objects) around rotating objects. Real-world geodesics also depend on more than just the radial coordinate, unless all of the conditions I listed apply. As I mentioned earlier, for rotating objects there will be off-diagonal terms in the metric, making geodesics depend on at least two coordinates (radius and colatitude). I do not see how you propose to have this arise from a single scalar field? The gradient only tells you direction and slope of change at each point, it doesn’t add any extra degrees of freedom. Link to comment Share on other sites More sharing options...

rjbeery Posted February 20, 2022 Author Share Posted February 20, 2022 15 hours ago, Markus Hanke said: But this isn’t what GR predicts - the geodesic in fact is determined by all components of the metric, plus boundary conditions. I thought you said your idea is meant to replicate the results of GR? But in the frame-dragging scenario you provided the geodesics of light rays are identical regardless of direction around the rotating mass. 15 hours ago, Markus Hanke said: Also, what exactly do you mean by “scalar time dilation field”? Time dilation is a relationship between clocks - so assigning a single value to each point in space isn’t enough, you need to also fix your reference clock somehow. The reference clock is the Schwarszchild coordinate time of the infinite observer. 15 hours ago, Markus Hanke said: The gradient only tells you direction and slope of change at each point, it doesn’t add any extra degrees of freedom. I feel like you think my essay is trying to replace GR, but it's literally relying on GR to calculate time dilation and then using the gradient of that to determine geodesics. Geodesics alone to not describe everything about spacetime. Place observer A inside a massive, transparent, hollow sphere, and observer B "at infinity" (with coordinate time). They will each claim to be inertial, with no local gravitational effects. Physics for both observers is identical because their local time dilation gradient is zero, even though the time dilation of observer A is higher than that of B. Link to comment Share on other sites More sharing options...

Markus Hanke Posted February 20, 2022 Share Posted February 20, 2022 (edited) 4 hours ago, rjbeery said: But in the frame-dragging scenario you provided the geodesics of light rays are identical regardless of direction around the rotating mass. I pointed out only that they lie in the same equatorial plane, and pass the body at the same minimum distance. But I also made it clear that they experience different frequency shifts, so the geodesics through spacetime are not identical. In practice, the total deflection angle would likely also be different, despite the same closest approach distance (I’d have to check this first). 4 hours ago, rjbeery said: The reference clock is the Schwarszchild coordinate time of the infinite observer. Which is precisely the first of my conditions - asymptotic flatness. 4 hours ago, rjbeery said: I feel like you think my essay is trying to replace GR, but it's literally relying on GR to calculate time dilation and then using the gradient of that to determine geodesics. Ok, thank you for clarifying, I was indeed confused on this. It makes more sense now. Now, if you demand geodesics to be approximately determined by time dilation and its gradient alone - which is the g00 component of the metric -, that means the other metric components should be negligible within the geodesic equation (which you retain, as you say). This is precisely the other three conditions in my list, plus an extra assumption of low velocity and weak fields (so that g00 dominates over g11 by a factor of c^2). So we have recovered the necessity of my boundary conditions. Your proposal may work, but only if all these conditions are met. The rotating body eg violates spherical symmetry and is not stationary, so the geodesic cannot depend on g00 alone. Each of the other examples I gave violate one or more of these conditions. GR on the other hand makes no assumptions about the metric components - it treats them all equally, and accounts for them all in the geodesic equation. For interior spacetimes (which we haven’t even spoken about yet) these components are all wired up to the various components of the energy-momentum tensor via the field equation, providing a comprehensive account of gravity and its various sources. In such scenarios, tidal effects in both time and space are important, so it goes far beyond time dilation alone. So if you can just acknowledge that your idea is useful in some circumstances, but limited in its domain of applicability, then we can be all good. After all, you can’t replaced a rank-2 tensor with a single scalar field (its 00-component), and expect to not loose any information in the process - that should make intuitive sense, no? Edited February 20, 2022 by Markus Hanke Link to comment Share on other sites More sharing options...

rjbeery Posted February 22, 2022 Author Share Posted February 22, 2022 On 2/20/2022 at 4:11 PM, Markus Hanke said: So if you can just acknowledge that your idea is useful in some circumstances, but limited in its domain of applicability, then we can be all good. After all, you can’t replaced a rank-2 tensor with a single scalar field (its 00-component), and expect to not loose any information in the process - that should make intuitive sense, no? I acknowledge your point and appreciate your input very much. You'll forgive me if I remain skeptical. I need to find a scenario where the time dilation gradient cannot be used to determine the geodesic, and I don't think spinning mass qualifies. If light rays were to curve differently depending upon whether the mass was approached "with or against" the direction of spin then I do think that I would need to concede your point completely. Let me think about it some more. Link to comment Share on other sites More sharing options...

Markus Hanke Posted February 22, 2022 Share Posted February 22, 2022 (edited) 5 hours ago, rjbeery said: I acknowledge your point and appreciate your input very much. You'll forgive me if I remain skeptical. Ok, fair enough. 5 hours ago, rjbeery said: I need to find a scenario where the time dilation gradient cannot be used to determine the geodesic, and I don't think spinning mass qualifies. Well, there are very many other solutions to the field equations where that is the case. For example the FLRW metric - the notion of “time dilation field” doesn’t even make sense here, since this spacetime isn’t asymptotically flat, so no Schwarzschild observer exists at infinity to function as reference clock. You might find the book “Exact Solutions to Einstein’s Field Equations” by Stefani helpful, if you have access to it. It’s a nice survey of known analytic solutions - some very remarkable spacetimes here, which aren’t common knowledge. It’s quite mathematical though. 5 hours ago, rjbeery said: If light rays were to curve differently depending upon whether the mass was approached "with or against" the direction of spin then I do think that I would need to concede your point completely. They most certainly do in spacetime - ie the geodesics differ (there is a difference in frequency shift at least). Whether their purely spatial trajectory differs I’m not 100% certain, but I suspect it does, as the light ray will get “dragged along” by the spinning mass on one side (just as a massive test particle would), so it should experience more deflection when oriented along the direction of rotation. A quick search yields this : https://arxiv.org/pdf/1910.04372.pdf Underneath equation 16, there are plots for “effective potentials” (a mathematical term within the equation of motion); as you can see, these terms differ in the Kerr case between direct and retrograde geodesics - so there is a difference in deflection angles between these cases. The exact expression is given in equation 60, which unfortunately is very complicated, and can only be treated numerically (it’s an elliptic integral); but you find plots of typical cases a bit further down in figures 8 and 9 (dashed is Kerr-direct, dotted is Kerr-retrograde) - proving that the angle is indeed different depending on whether the deflection is direct or retrograde, as I suspected. The difference is in fact a lot larger than I would have suspected. For comparison it also shows the Schwarzschild, Reissner-Nordström and Kerr-Newman cases (we haven’t spoken about electric charge here, but that adds yet another degree of freedom). Edited February 22, 2022 by Markus Hanke 3 Link to comment Share on other sites More sharing options...

studiot Posted February 22, 2022 Share Posted February 22, 2022 2 hours ago, Markus Hanke said: You might find the book “Exact Solutions to Einstein’s Field Equations” by Stefani helpful, if you have access to it. It’s a nice survey of known analytic solutions - some very remarkable spacetimes here, which aren’t common knowledge. It’s quite mathematical though. What an excellent post all round. +1 Link to comment Share on other sites More sharing options...

rjbeery Posted February 22, 2022 Author Share Posted February 22, 2022 11 hours ago, Markus Hanke said: https://arxiv.org/pdf/1910.04372.pdf I think the crux of the light trajectory around a Kerr-Newman black hole is showcased in this diagram. If geodesics were to be strictly determined by time dilation this would imply that the field gradient is no longer strictly radial, but rather manifests as a spiral. Would you agree that this is possible, in theory? I'll investigate. Link to comment Share on other sites More sharing options...

Markus Hanke Posted February 23, 2022 Share Posted February 23, 2022 7 hours ago, rjbeery said: Would you agree that this is possible, in theory? No, because the geodesic depends chiefly on the orientation of the light ray, not its spatial trajectory. You can see this in one of the other examples I gave - the satellite orbiting on the same trajectory in and against the direction of rotation, with different outcomes. So it doesn’t matter what kind of field you define in space, you can’t capture this behaviour. What does it even mean for the gradient to be shaped like a spiral? What kind of scalar field would give rise to such a gradient? But yes, feel free to investigate further. That’s how science is done, after all, and that’s how one learns 👍 Link to comment Share on other sites More sharing options...

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