Markus Hanke

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?

Indeed, you are right, I didn’t quite think this detail through. What’s more, if we want to be dealing with free fall world lines, then the clocks can’t really be at relative rest at the final event, as this would imply acceleration.

That is not the definition I would use, since the possible paths connecting events depend on initial and boundary conditions as well (specifically - whether your test particle starts off at rest, or has initial momentum). It is better to say that the principle implies that any free-fall path connecting two given events must be a geodesic of spacetime. In singly-connected spacetimes, this choice is unique, since for a given set of appropriate boundary conditions, the geodesic equation has precisely one unique solution. Technically you could have spacetimes the topology of which is multiply-connected; in those cases you can then have more than one geodesic connecting the same two events. I don’t know though if it would be possible to have more than one geodesic of the same length connecting the same events. That is an interesting question, but I’d have to think about that first before attempting an answer. I certainly couldn’t think of a physically realisable example right now. Well, the technical definition is that you start with the integral I quoted earlier, hold the start and end points as fixed, and then vary the paths between these points. The principle simply states that the path physically taken is the one that is an extremum of this proper time functional, i.e. the longest one. Note that the principle of extremal ageing is a specific example of a more fundamental principle, the principle of least action. That is the fundamental principle that underlies all field theory frameworks, specifically quantum field theory, and hence the Standard Model. Also GR itself follows from it - applying the principle of least action to the Einstein-Hilbert action gives you the Einstein field equations.

I think it is really important to emphasise that in spacetime, you can only meaningfully compare clocks that start together at rest, and end together at rest, i.e. which connect the same two events. This is what the principle of extremal ageing applies to - if you have a number of different paths which connect the same two events, the principle tells us that that path which represents the longest proper time (i.e. the greatest geometric interval) is physically realised. Schwarzschild spacetime admits a number of Killing vector fields (which is to say it has certain symmetries), so you can extend this principle a little bit in order to compare setups like the one in question here. I think the relevant scenario would be if you have two events A (near a central mass) and B (further from a central mass) which are situated along a radial line with the centre of the mass. How would a test particle move - will it fall from B to A, or would it go from A to B? If you do the maths, you’ll find that - even though the purely spatial distance is the same in both cases -, the world lines A-B and B-A nonetheless have different geometric lengths in spacetime. The length of B-A is longer than the one of A-B, so a clock freely falling from B-A records more proper time than a clock riding a rocket propelling itself from A-B. This is because the path A-B involves proper acceleration in the frame of the clock, which always implies extra time dilation (ref equivalence principle) on top of the background curvature of spacetime; the free-fall path B-A does not. So B-A will accumulate more proper time. Technically speaking, the curve B-A is a geodesic (segment) of Schwarzschild spacetime, whereas A-B isn’t. The presence of proper acceleration always means a shortening of a test particle’s world line. So to make a long story short - test particles tend to remain in free fall, because that is how they accumulate the most proper time on their clocks => principle of extremal ageing. Any external force implies local proper acceleration, which in turn implies time dilation, leading to less proper time recorded between the same set of events. This is why test particles which start off at rest always fall towards a central mass, never away from it - because the two world lines aren’t the same ones, even though the spatial trajectory may appear to be. Obviously, if you start off not at rest, but with some initial momentum, much more complicated dynamics may result. Intuition quickly fails for such scenarios, and it’s really a matter of doing the maths then.

The total proper time accumulated on a clock between two events in spacetime is equivalent to the geometric length of the clock’s world line C that connects these events, hence: \[\tau=\int _{C} ds=\int _{C}\sqrt{g_{\mu \nu } dx^{\mu } dx^{\nu }}\] Just to clarify - I mentioned the comparison to a far away clock only as a pedagogical aid to illustrate a principle; such a comparison is of course not physically precise, because these two clocks would connect a different set of events. Come to think of it, I should not have brought this up at all, as it only confuses the issue further. Please consider it retracted. You can’t really compare such clocks, since there is no good synchronisation convention you could use for this. Not necessarily, because the total proper time generally depends on the path that is taken as well as the metric (see line integral above). For example, in an equatorial orbit around a rotating black hole, you would get different readings depending on whether your orbit the black hole in the direction of its rotation, or in the opposite direction. Apologies, I don’t quite understand what you mean with this...? Possibly they could age the same, but not necessarily - again, because the total time accumulated depends on the metric and the path that is taken in spacetime. So it really depends on how they travel, and what the geometry of the underlying spacetime is like. There is really no generally valid answer to this, it depends on the specific case. I am personally finding that, even though my intuition about all things GR has gotten far more accurate over the years, it is still a slippery slope and sometimes leads me to conclusions that later turn out to be wholly wrong. In cases like the above, really the only way to be sure is to do the math, and evaluate the above integral for the specific case at hand. In very general terms, we can say that the longest possible path through spacetime between two fixed events is always a geodesic of that spacetime. You can take the above integral, fix the start and end points (i.e. decide on two fixed events), and then vary it between all possible paths between these events, finding the extremum - what you will find is that you end up with the geodesic equation. Because this is a differential equation, its solution is determined by initial and boundary conditions - so if you have a test body coming in at high speed, you might get a free fall geodesic that is entirely different from the one you’d get if you start off at rest. The dynamics are really quite complex, especially in non-trivial spacetimes, so intuition often fails here. It’s best to just sit down and calculate. For your scenario with highly elliptic orbits, it is also very important to remember that orbital distance and orbital speed are not good indicators of how long the world line of the test particle is in spacetime. You definitely have to do the maths here to tell the whole story.

Proper time is the geometric length of a test particle’s world line in spacetime - not just in space, and not just in time, but in spacetime. What we find is that in spacetime, the longest possible trajectory between two events is always a geodesic, which is a world line where the test particle does not feel any acceleration at any point. If you start off near a massive object, the only trajectories that avoid you feeling proper acceleration in your own frame are generally those that bring you closer to the massive object (free fall, in the classic sense), since you would need acceleration to do anything else (unless of course you already come in very fast, e.g. in a slingshot manoeuvre). So think about it geometrically - very loosely speaking, world lines near a massive object tend to be longer as compared to similar ones far away, because spacetime there is “stretched out”, particularly in the time direction (that is why e.g. a radar signal passing by a massive planet takes longer to get to its receiver - because its world line through spacetime is longer). Or you can think of it this way - if you sat on a clock somewhere near a massive object, and look back at another reference clock that is somewhere far away, then the far-away clock would appear to go faster. This is due to gravitational time dilation between your own frame, and the far-away frame. So actually, more time is accumulated (very loosely speaking) near a massive object, as compared to anywhere else. Note that this can either be a maximum or a minimum, depending on how you choose the signs in your metric. That’s why it is more generally called the principle of extremal ageing.

A tensor is a multilinear map that maps vectors and 1-forms into other objects. This is true even if the manifold isn’t endowed with a metric, and regardless of the number of dimensions. The vector cross product which you are using here is defined only in 3 dimensions, and is not generally covariant. It also requires the presence of a metric. Furthermore, if you replace a higher rank tensor by a real number, you loose both information and a number of degrees of freedom. Again, you cannot simply replace replace a rank-n tensor field by a scalar field; they are not equivalent objects. Physically speaking, you need to be able to capture all relevant symmetries of the field (in the case of GR, diffeomorphism invariance), which is possible only if it is a rank-2 tensor field. Remember also that, in terms of QFT, the graviton needs to be a massless spin-2 boson, which again means you need a rank-2 tensor field. It appears that you chose your “characteristics” in order to obtain the solution you wanted, not because those choices are motivated mathematically or physically; there is no mention of any boundary conditions. Vanishing Christoffel symbols imply a flat region. Clearly, this would not lead you to the Schwarzschild metric, or any spacetime other than the trivial Minkowski one. I’m afraid it is not. Also, the fact that the EFE is a rank-2 tensor equation has deeper reasons; you can’t just replace it with a scalar relation. No, mostly because these are not generally covariant objects (no diffeomorphism invariance), and do not capture the underlying physics.

As I’ve pointed out earlier, there is no such thing as a “magnetic charge”.