Markus Hanke

I am not certain what you mean here. I only mentioned LQG as an example; truth is, at this point in time we simply don't know which of our current hypothesis, if any, will represent a valid model of quantum gravity. No, you are right of course. I should have said that the idea of LQG is that macroscopically it becomes indistinguishable from GR, I did not mean to imply that it is actually the case. You are right that the proof of this is, as per yet, still outstanding. Presently LQG is really just a hypothesis with many ends still be tied together. My apologies for any confusion, I should have been more precise.

LQG is scale dependent; on macroscopic scales it is indistinguishable from deterministic GR, whereas on microscopic scales it is probabilistic in nature. So it incorporates both models. This works because on a macroscopic level probabilistic effects are so small as to be negligible.

Yeah, that's the big question, isn't it. One such attempt is Loop Quantum Gravity; basically what it boils down to is that on large scales it should be indistinguishable from standard GR ( proof of this is pending though ! ), whereas on small scales space-time itself becomes quantized, inducing quantum effects. This would be one way to do it.

It's neither of these. What is needed is an entirely new paradigm, which reduces to GR and QM in their respective energy domains, or implies these as boundary conditions. Consider, just as an example, String theory - the model is mathematically consistent only on a curved space-time background, and if you investigate the constraints on that curvature, you find precisely the GR field equations. At the same time, the energy levels of a String are ( roughly speaking ) its vibrational modes, which are inherently quantized. So there you have it - a model of quantized particle fields, which lives in a space-time governed by GR. P.S. I am not saying that String Theory is a valid model of quantum gravity; we don't know that. I am just using it as an example.

It is incomplete in the sense that it does not incorporate any quantum effects, it is a purely deterministic theory. For example, GR can model the gravitational collapse of a star only so long as quantum effects can be ignored; at the point of the collapse where such effects become important, GR ceases to be a useful model. That is why GR cannot tell us what really happens in the vicinity of a gravitational or cosmological singularity, or even if such singularities really do exist or not. To fully comprehend the physics of such a scenario we would need to have a model which incorporates both relativistic and quantum effects, namely a theory of quantum gravity. This is an area of active and ongoing research.

Division by zero is not defined; writing "1/0 = infinity" is meaningless.

It literally depends on your point of view. For an observer who is located far away outside the BH, an object falling towards the BH would take an infinite amount of the observer's own time to reach the event horizon, while slowly fading away into nothingness. On the other hand, if you were to "piggy back" on the infalling object you would not notice anything special. You would reach - and cross ! - the event horizon in a well defined, finite amount of time as measured on your own watch. Both observers are right, but only in their own frames of reference.

Of course, I completely agree. GR is a deterministic ( classical ) theory, whereas all of quantum physics is based on probabilistic principles. That is one of the reasons why they are so hard to reconcil / unify. We all understand that GR is incomplete in that its domain of applicability is limited.

This is good advice, right here

I disagree. The mathematics are very tedious ( as in time consuming ), but not complicated to understand conceptually. In fact the tensor relation which is the Einstein Field Equations is probably as simple and straightfoward as it gets, on a conceptual level. Turn on the GPS in your car - that's a beautiful demonstration of General Relativity right there. Without taking into account relativistic effects, that GPS would be off by something like 10km each day. A classical model that explains and quantifies the effects of gravity as being a geometric property of space-time, as opposed to mechanical forces between bodies.

It always amazes me how people go on and on, even though they have already been shown wrong. I will never understand this...

Fair enough You see, what is called the "twin paradox" in relativity textbooks & literature is a fairly specific setup with quite specific specs - one of them is the presence of acceleration for one of the twins. Yes, we can find other setups which lead to the same result even without acceleration ( which is what you were referring to ), but I truly believe that calling these alternatives "twin paradox" as well leads to a lot of confusion, as unfortunately happened on this thread. Anyway, I think it is all cleared up now. I am more of a "GR guy", but sometimes going back to the basics is interesting too.

Excellent, if we can all agree on this then we're good I was just starting to get confused as to what the actual argument was, since participants here appeared to be going in different directions, talking about different things.

I'm afraid I don't follow you. I never said anything about "preferred frames". In fact I explicitly stated that the frames are interchangeable without affecting the physical outcome, thereby ruling out that one of them is "preferred". That automatically makes them symmetric. I do not know why you state that on the one hand the motion is relative and symmetric, whereas the "situation" is not. That makes no sense. There is no such thing as "instant acceleration". I must admit that I am starting to wonder what the actual purpose of this thread really is. Of course it is possible to make up the elapsed proper time of an accelerating observer by combining the times of two purely inertial observers. So what ? That does not allow us to state that "acceleration is not important in the twin paradox". The two scenarios are not physically equivalent, in the sense that if you conduct them right next to each other, the participants will always be able to tell which experiment they take part in. So could somebody just state in plain text what the premise of the OP actually is ? I am getting more and more confused. So far as I am concerned acceleration is an integral part of the "twin paradox" scenario as it appears in relativity textbooks, because the point of the exercise is to teach students the physical difference between symmetric and non-symmetric frames. If you eliminate the asymmetry, you eliminate the discrepancy between the two twins after they are brought back together in the same frame at relative rest, defeating the purpose. Likewise, if you eliminate the "bring back together at rest in the same frame" bit, you depart from the original twin paradox scenario. Yes, precisely my point. So, the premise of the thread that "acceleration is not important in the twin paradox" is clearly false. You can consider alternatives to the typical twin paradox scenario to achieve the same outcome, but these are then different scenarios, which are not physically equivalent.

Of course it could, but that is because, in the frame of reference of the alien ship, the distance between earth and sun is length contracted. The pilot of your UFO will not notice anything special happening on his own clock. The point is that the pilot of your alien vessel could consider himself at rest, and argue that it is in fact the sun which is rushing towards him, or the earth receding from him. Because his frame is inertial, in the absence of another outside point of reference he has no way to decide whether it is himself moving, or whether it is the sun/earth that is moving, or some combination of the two. If he was to observe a clock co-moving with the sun or earth, he would also read 1 second on it since it is dilated by exactly the same factor, just like on his own clock is from the other frame. The two frames of reference are freely interchangeable without affecting the physical outcome, so long as all coordinate effects are taken into account; that is what I mean when I say that the clocks cannot disagree, and neither can rulers. Not so in the twin paradox - because the travelling twin measures a local gravitational field in his vessel, his frame is physically distinguishable from the one of the stationary stay-behind twin. They are no longer symmetric, so if you swap frames you will find that the clocks are dilated by different factors. They are not symmetric.

Yeah, I am happy with that Given that, I think it is quite clear that in the scenario of three clocks, all of which are in uniform relative motion with respect to one another, you will never find a "twin paradox". That is really all I was trying to point out. For me, it is most clearly obvious in the proper time integral for each observer.

You mean approximately inertial locally. Globally the entire setup is not inertial, because the clocks are at points of different gravitational potentials. Yes, I can agree to this statement, even though I am not certain what you mean by "inertial forces". That is also the reason why I kept mentioning the metric tensor in my explanations. What they mean is that the "paradox" arises only from the erroneous assumption that both frames are inertial; if that was the case then there shouldn't be any differences in clock readings. Recognizing that one of the frames is not in fact inertial immediately shows that there really isn't a paradox in the first place, because the disagreement in readings is a physical necessity due to the fact that the non-intertial frame "carries" a non-constant metric tensor. This is pretty much what the last two sentences in the link say.

You see, this is where you are being imprecise, and hence run into problems. The twin paradox is not about time dilation as a result of Lorentz transformations between inertial frames. In fact, it is the exact opposite - it is about disagreements between clocks which cannot be related by Lorentz transformations. In other words, it is about relations between frames which are not symmetric. You are correct that there is time dilation involved, but, due to the equivalence principle, it is gravitational time dilation that we are dealing with, not Lorentz time dilation. This is actually made clear in my own post #2 - inertial frames ( as used in SR ) all have constant metric tensors associated with them; the very constancy of the metric tensor in Minkowski space-time explicitly rules out any disagreements between clock readings. In the twin paradox however, the metric tensor in the frame of the travelling twin is not constant; locally, this is equivalent to the presence of a gravitational field. That is the reason why his clock disagrees with the one of the stay-behind twin. You cannot replicate that result with any arrangement of purely inertial frames that you can possible dream up. Here's a pretty good and intuitive overview without complicated maths - have a read through it : http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm

I am surprised that this thread is still going. It is an obvious and well established fact that, if only uniform relative motion in Minkowski space-time is involved, it is impossible to get any disagreements between clock readings. The twin paradox cannot exist under such circumstances, no matter what specific scenario is conjured up. Do you really need the maths again ?

Kudos to you ! I am also self-studied, but it took me a lot longer than 3 years to get to the level where I am, and it is still very much an ongoing process. I am working two full time jobs, and look after a family of six, so the going is tough. But I am curious enough to not give up

Indeed. Repeatedly asserting "inconsistencies" in SR even though it has been mathematically proven that there cannot be any such inconsistencies paints a pretty bad picture.

These three clocks are all in uniform unaccelerated motion relative to each other, so all three of these clocks represents inertial frames, and they are thus perfectly symmetric. It is not possible to get any disagreements in proper times between the clocks. If you look over on the other thread "Paradox in Relativity", you will find the mathematical proof of this there - I have provided it there in a different context, but it holds nonetheless. Also, in order to compare proper times, you have to either bring the clocks to rest in the same frame of reference ( which involves acceleration ), or you calculate the outcomes using relativity of simultaneity. You cannot, however, directly compare clock readings so long as they are in relativistic motion relative to each other. So once again - acceleration is a crucial element in the twin paradox scenario. If it is absent then you will get no disagreements between the clocks.

Can you show why these frames are not symmetric, as you say ?

If there is only uniform relative motion between the three reference frames, then the frames are symmetric, and the clocks can't disagree, because you can exchange any two of them via Lorentz transformations without effecting the physical outcome. Then what is the point of the exercise ? The "twin paradox" is specifically designed to illustrate what happens when frames are not symmetric; if you eliminate acceleration you eliminate the asymmetry. It is a trivial matter to show mathematically that there will not be any "disagreement" between clocks if there is no acceleration. Can the OP elaborate exactly what it is you are trying to show ? That is plain wrong. "Turning around" is equivalent to the presence of acceleration, which of course renders the frames asymmetric, leading to the clocks "disagreeing" at the end of the experiment. That is definitely physically relevant ! Without the turning around symmetry is maintained ( provided there is no other source of acceleration ), and the clocks remain in agreement.

Why make it complicated ? Just model the gravitational field as going radially outward from an isolated point source, and with spherical symmetry ( a physically reasonable model ); this makes the field both irrotational and source-free outside the point source, in other words, it makes it conservative. Mathematically, for a vactor field F, this means that outside the source the following conditions are fulfilled : [math]\displaystyle{\triangledown \cdot \mathbf{F}=0}[/math] and [math]\displaystyle{\triangledown \times \mathbf{F}=0}[/math] So both the divergence and the curl must vanish. In flat 3-dimensional space this automatically implies an inverse square law, as can easily be shown in spherical coordinates from the above differential equations.