Markus Hanke

You can explain it purely geometrically - the world line of one of the twins is a geodesic of spacetime, the other one is not. That’s all there is to it. Physically speaking though, a world line not being a geodesic implies the presence of acceleration at some point; but you don’t need to explicitly cast it in those terms.

Gravity - and hence also gravitational waves - are a geometric properties of spacetime. It has neither elasticity nor density, nor any other material property. It is not a material of any kind. Nonetheless it does have degrees of freedom, but these are of a geometric nature.

SR is well capable of handling accelerated frames as well; it’s just that the relationships between such frames are more complicated than simple Lorentz transforms. Where SR breaks down is when gravity is involved, i.e. when the region of spacetime in question is not flat. All relativistic effects are relationships between frames; you would never observe any such effects, if you looked only at a single inertial frame.

A variation in c, and a variation in a particle’s world line in spacetime, are not equivalent. There is no force field. An accelerometer attached to a test particle in free fall will read exactly zero at all times, so gravity is not a force.

Why are they not diffeomorphic? I am not a mathematician, so it is very possible that I am misunderstanding something. However, it seems clear that all diffeomorphisms are also isomorphisms (but not vice versa); furthermore, the definition of a diffeomorphism I am familiar with is that of an invertible function that maps one manifold into another (1-to-1) such that the function is smooth and differentiable throughout its domain. Why does this not apply to the cylinder and flat sheet example? That is true for extrinsic curvature, but not for intrinsic curvature. Concluding that the higher dimension does not exist is not trivial. There is nothing that rules out per se that such a dimension exists. However, one can look at the physics in such a universe - it is likely that at least some of the laws of physics in such a world are “sensitive” to the presence of the extra dimensions. For example, there are fundamental reasons why in a universe with 3 spatial dimensions we would expect to find inverse square laws of various kinds; if we were to empirically find (e.g.) inverse cube laws instead, then that could potentially be due to the presence of an extra spatial dimension. So there may be ways. However, it is also possible that all laws of physics are confined to within the embedded manifold; in such a world, detecting the extra dimensions might be very difficult. I don’t know if it would be impossible - that’s a good question. A test particle in free fall (gravity only) experiences no forces - an accelerometer falling in the same frame as the test particle will read exactly zero at all times. So clearly, gravity is not a force.

The metrics are not the same, but they are diffeomorphisms of one another. The crucial difference is that, on the surface of a cylinder, geodesics in one direction are closed curves - you can walk all around the cylinder, and end up again at the same spot where you started. On a flat sheet, there are no closed geodesics, they all extend to infinity, so no matter in what direction you walk, you will never get back to where you started. So based on this observation, they could probably figure out that they are indeed on a surface with the topology of a cylinder. But that alone does not allow them to conclude anything about whether or not that surface is embedded in a higher dimensional space. The crucial point however is that, in both cases these world lines are locally straight everywhere. You can smoothly deform one “world” into the other while exactly preserving all path lengths and angles.

If you can somehow get hold of a copy of Misner/Thorne/Wheeler, I cannot recommend it highly enough! He talks about all these things in detail, it’s a truly excellent text, and considered the gold standard for a good reason. Sadly, it’s difficult to obtain and very expensive. I was lucky to find an old and damaged copy (still legible though) on eBay some years ago. As others have pointed out, for all intents and purposes we are in a vacuum here. The deviation from “true” vacuum is negligibly small. This isn’t the right way to look at it, spacetime is not a medium. c does not change, the only thing that changes if you vary gravity is the path that an electromagnetic wave travels through spacetime (which is always a null geodesic).

To be honest, I am having a hard time thinking of a physical scenario where you’d have both maxima and minima on the same world line, with said world line still remaining a free fall geodesic. I cannot think of any physically possible background spacetime where that could be the case. That nonewithstanding, let’s say for argument’s sake that we are in such a region of spacetime. You can then break up the world line into curve segments, and vary the path lengths of each section locally. So, you perform the variation only for nearby world lines in each section. You would then have sections that are maxima, and sections that are minima locally. I am unsure what this implies about the global status of such a geodesic - then again, this would be a very unusual kind of background spacetime I think.

The easiest and most straightforward way to do this would be to simply compare the geometric lengths of the two observers’ world lines in spacetime. This way the explanation is a purely geometric one, and makes no direct reference to forces, accelerations, or frames.

Under certain circumstances you can use either one to describe the same thing, but that is not always the case. As a counterexample, consider the (2D) surface of a (3D) cylinder - it is extrinsically curved, but intrinsically flat. So clearly, in this case these two descriptions are not equivalent. In the case of GR, spacetime is not thought to be embedded into any higher dimensional space, so it has only intrinsic curvature. For arguments sake, it is possible to construct a mathematical model that embeds spacetime into something higher-dimensional, and then use extrinsic curvature to capture all the same information. The problem with this is that the embedding would have to have a lot of dimensions; I can’t actually remember the exact number, but I think it was 48. So I fail to really see the advantages in this, as it makes most of the maths very much more complicated than it already is in standard GR. In general terms, any Riemann manifold can be embedded into a higher-dimensional Euclidean space in such a way that paths lengths are preserved. This is called the Nash embedding theorem.

Gravitational waves are affected by the background curvature of spacetime, and you can get many of the same effects that light would be subjected to, such as deflection, frequency shift etc. However, the actual dynamics of gravitational waves are potentially much more complicated than those of light, because gravity is non-linear, unlike electromagnetism; this means that such waves also interact with each other and with themselves. Can you construct an arrangement that acts like a lens for gravitational waves? Yes, you certainly could, but depending on what exactly it is you are trying to achieve, this may be a very complex problem (both mathematically and practically). This is a very complex topic, and truthfully speaking I have not done (or even seen) the exact maths of how this comes about. The general idea is that you start with a wave pulse (i.e. a more or less sharply defined packet of wave fronts) and send this through a region of spacetime that has substantial background curvature due to the presence of some gravitational source, e.g. a black hole. What happens then is that the wave pulse, as it travels through this region, backscatters off the background curvature, which leads to its shape and polarisation to change in some very specific manner. The deformation is such that a “tail” is produced behind the travelling pulse, and that wave tail propagates at less than the speed of light. My understanding (someone more expert at this particular detail please correct me if I am wrong) is that the wave tail travels at below c due to its own gravitational self-interaction - which is a non-linear process.

It doesn’t. Electromagnetic waves propagate exactly at c everywhere locally, they are never slowed down. The dynamics of gravitational waves are non-linear, and hence different from EM radiation. Static gravity does not propagate at all, it is a geometric property of the entire spacetime, so nothing needs to “leave” the event horizon. It is only changes in the gravitational field that propagate as waves - and even then, these changes need to be of a specific nature (they need to have a quadrupole or higher multipole moment) to be the source of gravitational radiation. For the hypothetical case of there being a source of gravitational radiation below the event horizon of a black hole, this radiation would not cross the horizon, same as any other test particle. Good point!

No, it just means that they weren’t emitted simultaneously - which is what one would expect, since these two forms of radiation are the result of different physical processes. This is inconsistent with the basic principles of GR, as well as with the specific mathematics of gravitational waves. The opposite is in fact the case - since the dynamics of gravitational waves are non-linear (unlike e.g. EM waves), they interact both with other gravitational waves as well as with themselves. In this manner you get a number of effects that are exclusive to gravitational waves, and some of these actually propagate slower than the speed of light (specifically so-called “wave tails”). However, a free wave in otherwise empty space must propagate at exactly c.

Yes, you are right of course. This is why one can go and ask someone knowledgeable in the subject matter - or learn the skills needed to do it themselves.

Well, I purposely did not specify the nature of the dimensions on my simplistic 2D manifold, and the choice of coordinates was arbitrary. My main goal was to show where and when boundary conditions come into this. Note in this context that you can parametrise the path in any way you want, it does not need to be via the time coordinate. For example, the null geodesics of photons cannot be parametrised via proper time (because by definition ds=0), so one chooses what is called an affine parameter \(\lambda\) instead. I cannot see any issue with this, tbh. As an analogy, think back to our high school calculus times - there is a simple prescription to find the extrema of any given curve, using derivatives. This same prescription will equally return both maxima and minima of our function (and there is a separate procedure to distinguish one from the other). It’s really not an issue. An extremum can be either a maximum or a minimum, and either be local or global. Yes, this is the main point.

QFT stands for “quantum field theory”; it’s a mathematical framework that describes the dynamics of quantum fields. It’s not a function. I do not mean to be rude, but little of what you write makes any sense - it seems obvious that you have not actually studied things like GR and QFT, and thus are not familiar with these models. It is not generally very wise to comment on things one does not understand.

1. A QFT is not a function. 2. “Derivative of QFT” is a meaningless term. 3. In any case, derivatives don’t measure the area under a curve. 4. “Curve of the theory” is a meaningless term. Yes, that is what I was getting at.

Violation of Bell’s inequalities does not imply such as thing as “action at a distance” - which is in itself a meaningless concept. Quantum entanglement is simply a statistical correlation between measurement outcomes; there is no causative “action” involved. I’m afraid this is completely meaningless.

Yes, one needs to be careful to compare like with like; the principle does not really say anything about non-geodesic paths. Do you mean different geodesics connecting the same pair of events? I own a copy of Misner/Thorne/Wheeler, it’s the text from which I learned a large junk of my GR knowledge. Feel free to ask away! I had a quick look at page 316, but couldn’t spot anything that contradicts this thread. It gives the integral I wrote earlier (up to a sign), and points out that geodesics are extrema (as opposed to always being maxima) of that variation, as I did earlier as well. Suppose we want to determine free fall world lines on a flat 2D manifold (like a piece of flat paper). For simplicity, let’s choose a simple Cartesian x-y coordinate system on our manifold, like the ones we all used in high school; the world lines we are looking for can then be written in the form \(y(x)\). I’ll be very sloppy with notation here, since I only want to show the basic blueprint of how this is done. First, we apply the principle of extremal ageing. It tells us that free fall world lines - regardless of what kind of manifold we are on - must be geodesics of that manifold; so in other words, the principle tells us that, among all possible world lines, the ones representing free fall must be geodesics of the manifold. That means that proper acceleration must vanish at all points of such paths - the principle can thus be mathematically stated as \[a^{\mu}=0\] Since we are in 2D, the index \(\mu \) can take the values (0,1), so we have a system of two equations. But since they are not independent, in the sense that the world line we are looking for is parametrised as \(y(x)\), we need only one of these equations to find our solution. We remember also that acceleration is the second covariant derivative wrt time, which leaves us with the equation \[\frac{d^{2} y^{0}}{dx^{2}} +\Gamma ^{0}_{\mu \nu }\frac{dy^{\mu }}{dx}\frac{dy^{\nu }}{dx} =0\] We know that we are on a flat manifold, so all metric coefficients are constants, meaning the Christoffel symbol in the above equation vanishes, leaving us with simply \[\frac{d^{2} y}{dx^{2}} =0\] Integrate once: \[\frac{dy}{dx} =A=const.\] Integrate again: \[y=Ax+C\] Of course, we recognise this immediately as the equation of a straight line. Which is what one would expect - the geodesics on a flat 2D manifold are straight lines. But note that the above isn’t any specific line - it’s actually a 2-parameter family of lines, which is parametrised by two constants A and C. So what the principle of extremal ageing does is reduce the set of all possible world lines down to the set of just those world lines that are geodesics. But there are still infinitely many of them, corresponding to the infinitely many possible choices of A and C. Here is where boundary conditions come in - to reduce the set of all geodesics down to one specific, unique geodesic, we impose boundary conditions. In this case we have two undetermined constants, so we have to supply two boundary conditions to uniquely determine a geodesic. For example, that could be two events such as (0,1) and (2,2), which yields the geodesic \[y( x) =\frac{1}{2} x+1\] So there you go, this is the general workflow. In curved GR spacetimes, the specific maths become much more complicated, since the Christoffel symbols don’t vanish, and you generally have more than one equation remaining in the system - but the general outline is still the same. Is this helpful in any way?

I’d just like to point out that this conservation law works only if the spacetime patch in question exhibits time-translation symmetry, i.e. if it admits a time-like Killing vector field. In other words, the metric of said spacetime patch needs to be stationary for there to be a global notion of energy-momentum conservation. This is approximately true for earth-based particle accelerators, of course. It is not, however, true for more general regions of curved spacetime. This just as a side note.

The above two statements appear to contradict one another - which one do you mean to say ages more? Generally speaking, the fixed stone is subject to constant proper acceleration, so it will experience a time dilation commensurate to that. Yes, indeed. Yes, also correct. Yes, absolutely right! Specifying two events A and B is one possible example of initial/boundary conditions. I personally think you understand the principle of extremal ageing very well In fact, you have a much better grasp on world lines, geodesics and extremal ageing than many other amateurs would, at least that is the impression I got on this thread. There was only some slight confusion on the role initial/boundary conditions play in this, but that is understandable. If you like I could work through an actual example for you - something extremely simple, such as geodesics on a flat 2D manifold. Obviously this is a completely trivial example, but it would nevertheless show where and how initial/boundary condition come into this. Let me know if you think that would be of benefit for you!

Both of these are true statements, and actually express the same thing. The first statement reflects the formulation of the principle in terms of the geodesic equations \[a^{\mu } =0\] whereas the second statement expresses it in terms of the variational principle (i.e. variation of the integral I quoted earlier, with fixed start and end points). Since variation of the proper time functional yields the above differential equations, these statements are physically and mathematically equivalent. The principle of extremal ageing applies always, everywhere, and in all cases where free fall takes place; it basically tells us that any free fall world line that is physically realised must be a geodesic of the underlying spacetime. A geodesic is a curve which parallel-transports its own tangent vector, which physically means it is a trajectory on which proper acceleration is exactly zero at all instances. So, we write down the equation above. This is a system of partial differential equations, since acceleration is the second covariant derivative of the position function wrt whatever quantity it has been parametrised with. In order to find a specific, unique solution to the equations, we must impose initial/boundary conditions; that is the nature of a differential equation. So, I am talking about initial/boundary conditions, because these are needed to select a specific geodesic, amongst all geodesics within the region of spacetime in question. This does not at all limit the applicability of the principle in any way - it’s just the procedure we need to follow to mathematically determine a unique, specific geodesic. In terms of calculus, we are dealing with segments of curves; since world lines have to be smooth and differentiable everywhere, we need to ensure that this remains the case at the start and end points of the segment as well; this physically corresponds to our boundary conditions, and selects the specific geodesic.

The conserved quantity is energy-momentum, not just energy (which is always a frame-dependent quantity); specifically, it is the energy-momentum tensor. If receiver, emitter, and the electromagnetic field itself are all located in a patch of Minkowski spacetime, then energy-momentum is always conserved: \[\int _{V} \partial _{\mu } T^{\mu \nu } dV=0\] But the same is not necessarily true if the spacetime in question is non-Minkowski (but it can still hold in specific cases). I understand your concerns, and you are absolutely correct in that they are not within the same frame of reference. However, the universe as a whole is described by a known solution of the field equations (FLRW metric), so we know how the two frames are mathematically related. So even if the usual conservation laws do not technically hold, we can still calculate what happens to light that travels to us from a distant galaxy. The result is the known redshift relation. This is again one of those cases when I would simply sit down and employ the mathematical machinery, rather than trying to figure out the situation by intuition based on local laws.

It should be noted here also that in curved space-times, conservation of energy-momentum is a purely local conservation law; it does not necessarily apply globally across regions of non-Minkowski geometry, unless the spacetime in question has the specific symmetries that give rise to such conservation laws.

“Rest” is indeed frame dependent, but proper acceleration (as opposed to coordinate acceleration) is not - it’s something that all observers agree on. I think you may have misunderstood what I was attempting to say, possibly I didn’t make my thoughts clear enough. Of course the principle always applies - it is a fundamental principle of physics, derived from the principle of least action, and there are no exceptions to its validity in the classical world. However, calculating a specific free-fall geodesic is an operation that must depend on initial and boundary conditions, since the underlying equations are differential equations. You cannot obtain a specific, unique solution to a differential equation without imposing some form of initial/boundary condition. So this is how different specific geodesics of the same spacetime are distinguished - they are just different solutions to the same equation (i.e. the same principle of extremal ageing). So you can have different free fall world lines around the same central mass, all of which naturally adhere to the principle of extremal ageing; but they also represent different sets of physical boundary conditions, which you’d need in order to physically realise such trajectories. So the uniqueness of solutions to the geodesic equation is preserved. Hence, the geometry of spacetime along with the principle of extremal ageing determine its geodesic structure, and boundary conditions pick out the specific geodesic that is being realised. Does this makes more sense now?