Markus Hanke

Well, they are useful to represent the perspective of a stationary observer at infinity. The crucial point to realise is that that is the only situation where measurements of space and time made in Schwarzschild coordinates actually coincide with what physically happens, since such measurements are always purely local in curved spacetimes. Anywhere else other than for a stationary observer at infinity, Schwarzschild coordinates are only a bookkeeping device, but they do not necessarily reflect what actually happens there, locally speaking. They also don’t cover the entirety of the spacetime. Many students of GR either do not understand this, or refuse to acknowledge it, since it goes against Newtonian intuition. The unfortunate result is all manner of misconceptions and misunderstandings. So you have good reason to mistrust Schwarzschild coordinates - they can be useful in certain circumstances, but they are also dangerous if not understood correctly. I think the only reason why pretty much every GR text uses them is because they are algebraically simple.

Yes - this is trivially true, since a vanishing Riemann tensor means you are on a flat manifold. More generally, if a tensor vanishes then so do all of its contractions - this is true for any tensor. Not really. It establishes only that its contractions vanish if Riemann itself vanishes. The reverse is not true, however - the vanishing of the Ricci tensor and/or scalar do not necessarily imply that Riemann is zero. Ricci flatness is a necessary but not a sufficient condition for the absence of Riemann curvature; to make it a sufficient condition, you need to demand the vanishing of Weyl curvature as well. The Ricci tensor is the trace of Riemann, whereas the Weyl tensor encodes the trace-free part of Riemann (the decomposition isn’t exactly trivial, though).

Ok. As a little tip - conventionally, writing (t,x,y,z) will imply Cartesian coordinates to most readers; if you want to indicate a general coordinate basis, it is better to use the notation \({x^0,x^1,x^2,x^3}\), as it is less ambiguous. Ok, that’s an important difference. I’m not sure what you mean by “relative sizes”?

As I said previously, my response was based upon my own understanding of the OP. If it completely missed the point, then it is the OP’s job to clarify things. Unfortunately this poster seems to be in the habit of opening threads, and then abandoning them after a couple of comments are made. Yes that’s pretty much it. Accelerated motion involves a little more than simply allowing gamma to vary.

I’m sorry, but I can’t make heads nor tails of this at all. What kind of coordinates (t,x,y,z) are you using here? Are these Cartesian coordinates, or Schwarzschild coordinates, or something else entirely? And what is your reasoning behind the metric (1)? This form of metric is incompatible with both Schwarzschild coordinates and Cartesian coordinates in a Schwarzschild spacetime, so please explain how you arrived at this ansatz.

I am uncertain what it is really is that the OP is trying to show, so I can’t guarantee this. What I presented is my own understanding of what he has posted. Yes of course, but that isn’t what the OP has been doing. That is kind of my point. He’s essentially using an inertial coordinate system to show that there is no acceleration - which is trivially true. Ok, but then, what is his point? He starts with a metric in Cartesian coordinates, then manipulates it using relations that imply an inertial observer, and ends up with the conclusion that there is no acceleration...? I’m not sure I understand your question. Both charts cover the same spacetime, so the difference is merely one of coordinate basis. Rindler is what you naturally get when you apply the proper transformations that arise in a frame with non-zero constant acceleration (since such frames are not related by Lorentz transformations): \[ct\rightarrow \ \left(\frac{c^{2}}{a} +x’\right) sinh\left(\frac{at’}{c}\right) \] \[x\rightarrow \ \left(\frac{c^{2}}{a} +x’\right) cosh\left(\frac{at’}{c}\right) -\frac{c^{2}}{a} \] The advantage here is that the world line of such a uniformly accelerated particle is one of constant x (in this coordinate basis!) as t “ticks along”, so in some sense a particle in this coordinate frame is “at rest”, even though it is uniformly accelerating. It’s the most natural choice for uniformly accelerated motion. Very simply put, the line element \(ds^2=...\) does exactly what it says on the tin - it is an infinitesimal section (element) of a curve in spacetime. To obtain the entire geometric length of a curve C, you integrate: \[L=\int _{C} ds\] To describe a uniformly accelerated particle, you can do one of two things: 1. Keep the line element in Minkowski coordinates, but make C a hyperbola to describe the motion 2. Use the line element in Rindler coordinates, which makes C a “straight line” in these coordinates Of course, the result will be the same (since this integral is an invariant), so both are valid, but the computational effort differs - it turns out that option (2) is very much easier to do. It’s simply the more natural choice - in much the same way as (e.g.) spherical coordinates are the natural choice to describe the surface of a sphere, rather than Cartesian coordinates.

It’s how I understood it, based on the fact that the metric given is of a form that would generally be used by an inertial observer, so it is natural to assume that these are Minkowski coordinates, and not hyperbolic ones. The geodesics calculated from his metric ansatz are straight lines, not hyperbolas - unless the coordinate basis is not Cartesian, but the OP never indicated that. This is also consistent with the OP’s conclusion: “[...] we see that the particle cannot accelerate”, which is of course trivially true, based on that metric.

Oops, sorry

Yes, proper acceleration (unlike coordinate acceleration) is a quantity that everyone agrees on, so the coordinate basis does not matter. I was referring to the fact that, if an accelerometer co-moving with an observer measures something other than zero, then that observer cannot be inertial by definition. I understood the OPs comment to say that a particle can be in a state of proper acceleration, yet still be inertial - hence my comment. If you want to use transformation laws, then those can be found on this page (scroll down to the relations shown in red and blue). As mentioned before, accelerated particles will undergo hyperbolic motion in Minkowski spacetime, so the coordinate transformations use hyperbolic functions and their inverses. Alternatively, and that’s what I have been suggesting above, you can just adopt an appropriate coordinate chart and metric from the beginning. The Rindler metric is the most commonly used, but it is valid only for constant accelerations; for arbitrary accelerations you can use Fermi coordinates (e.g.).

Not necessarily true. A region of spacetime being special relativistic means only that the Riemann tensor vanishes everywhere within that region, so that spacetime is flat. It does not mean that all frames necessarily need to be inertial, or that there is only uniform relative motion. You can have non-inertial frames in flat spacetime, so SR is perfectly capable of handling acceleration. Not true. If a frame is inertial, then by definition there is no proper acceleration. There can, however, be coordinate acceleration, but that is an artefact of how we choose to label events, and not something that a physical accelerometer would measure. This is incompatible with (1) and (2) in your last post. An accelerated particle undergoes hyperbolic motion in Minkowski spacetime, so you can’t just make the gamma factor variable, but retain a Minkowski coordinate chart (with proper time) and its corresponding metric. I suggest what you should do is start with the correct metric ansatz - i.e. a Rindler or Fermi metric -, and work through it again, and see what result you then get.

The combination of coordinate chart and metric you are using as an ansatz along with your constituent relation \[\gamma=\frac{dt}{d\tau}\] already implies inertial motion from the beginning, so the end result is hardly surprising. If you want to allow for the possibility of uniformly accelerated motion, you need to use a Rindler chart; and if you want to consider arbitrarily accelerated frames, you need to use a Fermi chart to cover your spacetime, along with the appropriate constituent relationships between the components of the velocity and acceleration vectors in each chart.

Of course...that’s why I was saying I was going off on tangents But you have to admit that oftentimes tangents are fun. There’s only one gauge condition in spacetime, because there’s only one potential field and one electromagnetic field. But once we decouple this into E and B fields, the gauge condition also decouples into a constraint on the temporal part of A (which is the scalar potential), and a constraint on the spatial part of A (which is the Newtonian vector potential). The U(1) symmetry applies to the electromagnetic field in spacetime, but not to the E and B fields in Euclidean 3-space. I think the conclusion is the same though, because of this: consider the Lorentz gauge (e.g.) \[\partial_{\mu}A^{\mu}=0\] This is a second order partial differential equation, so to obtain a unique solution, you need to supply precisely two boundary conditions. Since by definition in general field theory (omitting any constants) \[dA=F_{\mu \nu } dx^{\mu } \land dx^{\nu }\] the A’s should be fully determined, given Maxwell’s equations (which now essentially become relations between the components of A). So the boundary conditions above should be the only physical DOFs in the theory. I have not researched this, I’m just pulling this out of my hat as I go along, so it’s possible - likely even - that I’m missing something. This is strictly my own two cents. EDIT: Sorry, the above is of course nonsense, this isn’t a second order PDE at all. What I meant to say is that this conditions leads to the constraint equation for A, which is \[\square A^{\alpha } =\mu _{0} J^{\alpha }\] This requires exactly two boundary conditions for a unique solution.

Ok, point taken Actually, I feel a bit outside of my comfort zone here, since I was never particularly interested in EM theory. However, I am tempted to argue that in actual fact EM fields can have no more than 2 physical DOFs. Given conservation of sources, the EM field arises uniquely from a vector potential and a scalar potential - that’s 3+1=4 “free” quantities. But, both of these potentials are invariant under suitable gauge transformations, so in each instance we can arbitrarily pick a gauge, which eliminates one DOF each - leaving only two physical DOFs in the end. This seems to also be consistent with the fact that EM waves have exactly two possible polarisation states at a 90 degree angle. But perhaps I’m getting this all wrong, I’ve never really thought about this in any detail.

I have studied GR in some detail, so I am aware of all these possible scenarios; by personality I also tend to be a “natural worrier” who easily gets anxious even over minor things and life events. In addition, I am an Aspie too. Yet I feel no sense of depression, worry or anxiety over the possibility of a cyclical cosmology. Furthermore, you need to remember that the observational evidence we have at the moment is much more consistent with other global topologies, and not a cyclical universe. I believe you when you say that you yourself might find such an idea depressing, but remember that this does not imply that others necessarily relate to the concept in the same way. Most of us here understand the notion of a cyclical cosmology well enough, but don’t find it depressing. I find that people often tend to miss the salient point of Schopenhauer’s philosophy - he did not advocate despair, depression, or absolute nihilism. The main point he was trying to make was about acceptance. It is strictly necessary to fully understand and acknowledge the inherent limitations of the human condition - such as the impossibility to permanently satisfy desires and craving, and the futility of constant strife towards some ideal utopia -, but then it is also necessary to accept them for what they are, and thus arrive at a position of peaceful coexistence with those limitations. Philosophical pessimism does not imply despair and meaninglessness. And of course, philosophical pessimism is only one possible life philosophy, which is by no means shared by everyone.

I think it is better to look at it as a 4-dimensional geometric object in spacetime (the electromagnetic field), which is mathematically described by a rank-2 object (the EM field tensor and its dual). The electric and magnetic parts of the field aren’t independent, they are actually the same object seen from different vantage points, so the 6D description is a little problematic, as the basis vectors of that state space wouldn’t all be linearly independent. But that’s just me going off on tangents again 😄

This kind of depends on what exactly you mean by “seen to stop”; in particular it depends on which clock you use, as measurements of time are always purely local in curved spacetimes. As measured on a far-away, stationary clock (Schwarzschild time), the in-falling test particle never reaches the horizon, as that would require an infinite amount of Schwarzschild time. As you correctly say, it will just be “seen” from this vantage point to move more and more slowly as it approaches the EH, and become more and more red-shifted. On the other hand, for a clock co-moving along with the freely falling test particle (Gullstrand-Painleve time), nothing special happens at all - it will record a finite and well defined amount of proper time to reach the horizon, and also a finite and well defined amount of proper time to the “crunch” at the centre. The horizon itself won’t even be easily noticeable by such an in-falling observer; it’s locally as smooth and regular as any other region outside the BH. It’s only when they fire their thrusters in order to get back out to “far-away” that they will notice something being amiss, because no matter how much radial thrust is applied, they cannot stop themselves from falling further in; they can’t even remain stationary wrt to some external reference.

Oh the memories! I got this in college (don’t you guys say anything...!), and was one of the unlucky ones who get very severely ill with it, to the point where hospital attendance was required. And this was despite the fact that I was perfectly healthy, fit, and athletic at the time, and rarely if ever got sick at all. It was nasty and painful, and took a long time (several months) to fully clear. Point being, even very common pathogens shouldn’t be trivialised - even when in perfect health, you can catch a bad dose of something “common”. Common does not imply harmless.

You need only look at the very first equation: a=0. When you are in free fall, you experience no acceleration and thus no forces act on you - and yet you are under the influence of gravity. That’s why gravity isn’t fundamentally a force in the Newtonian sense. That’s all there is to it. The other equations are simply different ways to write that same statement; so there’s more than one way to look at it, but ultimately it always comes back to the geometry of spacetime, which is the fundamental “object” underlying all of this. No, because both matter and anti-matter have positive energy density (roughly speaking, technicalities aside for now) - so gravitationally they behave the exact same. To get repulsive gravity, you would need to create a region of negative energy density - this is called exotic matter (which isn’t the same as anti-matter), and there is no evidence that such a thing exists. No, the photon is its own anti-particle. It’s kind of difficult to explain why this is so (and must be so) in a non-mathematical way; just suffice to say here that the fact can be shown mathematically. They are not. Neutrinos are much more closely related to electrons than to photons, and they form their own class of particles. There are three types of neutrinos and three types of anti-neutrinos, and (as being fermions) they have no direct relationship to photons. The Standard Model is in excellent agreement with experiment and observation, so unfortunately its level of complexity is necessary.

Yes, but that is only because you are in a non-inertial frame. The important point here isn't the table itself, but the fact that its presence prevents the test particle from remaining in its natural state of motion, being free fall. To put it differently, gravity "acts" on freely falling test particles (by determining their trajectories) even though a co-moving accelerometer reads exactly zero everywhere and at all times - meaning gravity isn't a force in the Newtonian sense. There are four main ways to look at this: 1. Kinematics. An accelerometer in free fall reads zero everywhere, so the equation of free fall motion is simply: \[a^{\mu}=0\] If we set up a local coordinate system, we can denote the position vector of our test particle as \(x^{\mu}(\tau)\), and the above then becomes \[x{^{\mu }}{_{||\tau \tau }} =0\] wherein the || denotes covariant differentiation, since we are in a curved spacetime. The solutions of this equation of motion are geodesics of spacetime, being free fall world lines. No need to reference the concept of "force" anywhere here - which would be difficult, since a=0 implies F=0. 2. Geometry. Rewrite the above equation as \[\nabla _{u}\vec{u} =0\] This is the same equation as above, just written in terms of 4-velocity instead of acceleration. The geometric meaning of this is that free fall world lines are curves in spacetime that parallel-transport their own tangent vectors. Again, the concept of "force" does not come into this at all, it is purely geometric. 3. Pseudo-forces. Start with the equation in (1), and write out the covariant derivative fully in component form: \[\frac{d^{2} x^{\mu }}{d\tau ^{2}} =-\Gamma ^{\mu }_{\alpha \beta }\frac{dx^{\alpha }}{d\tau }\frac{dx^{\beta }}{d\tau }\] Again, this is the same equation, only written out fully. The left hand side is a Newtonian acceleration, the right hand side can be interpreted as pseudo-forces, which originate from the fact that the background spacetime is not flat. This is a valid way to look at this for massive particles; however, the right hand side of the equation taken on its own is not covariant, so it depends on the observer, and one can always locally eliminate the Christoffel symbol by going into the rest frame of the falling particle. Of course, this interpretation actually fails in the case of photons, since the very concept of Newtonian "pseudo-forces" is meaningless for massless particles. 4. Least action. Free fall world lines are those for which total proper time is an extremum, i.e. they are the longest (or shortest, depending on sign convention) possible world lines between two given events: \[\tau =\int _{C} ds=\int ^{B}_{A}\sqrt{-g_{\mu \nu } dx^{\mu } dx^{\nu }} =\text{extremum}\] This again does not require any notion of force. If you look at all of the above ways to consider the geodesic equation (which is what this is), then you will find that the common factor in all of them is not force, but geometry. I have parametrised the test particle's world line by proper time here, but everything remains valid if you replace this by a more general affine parameter, in order to include massless test particles as well. Do note though that the "pseudo-forces" way to look at it is highly problematic in the case of massless particles. The point I am trying to make is that it is best to look at gravity as a geometric property of spacetime, since that is the most general description that works for any test particle, and any observer, and does not require any extra concepts.

The natural state of motion for all bodies is free fall - this is the state where no forces act on the body, i.e. it is a state in which a co-moving accelerometer reads exactly zero. In this scenario, the weight is prevented from freely falling by the surface of the table, so it is the table which exerts a force on it, not the other way around.

The source of gravity in GR isn't mass, it's the energy-momentum tensor. All forms of energy (including, but not limited to, mass) contribute to the local geometry of spacetime, and all test particles are affected by that, not just massive ones. The equivalence principle applies only locally in small enough reference frames; on larger scales, gravity is tidal in nature. You can do this locally in a small enough region for massive test particles, but it is really much better to take gravity for what it actually is - geodesic deviation. This concept does not rely on any restrictions of scale, and applies to all cases, not just massive test particles. In my opinion, resorting to local definitions of pseudo-forces unnecessarily confuses things, it is much better and simpler to just stick to a purely geometric picture.

This makes no sense, because gravity in General Relativity is not a force, and does not obey any kind of simple inverse square law. General Relativity is a model of gravity, it has nothing to say about the dynamics of electrons within atoms. Again, gravity is not a force - even though its description can be approximated by Newtonian mechanics in the weak-field, low velocity regime.

Can you provide a reference to look at?

! Moderator Note I’m moving this to Speculations for now, as that is the correct forum section for personal theories. Why would accelerated expansion be an issue? It’s a natural geometric property of this type of spacetime, and thus fully consistent with the gravitational field equations. Gravity is a geometric property of spacetime; to be more exact, it is geodesic deviation, i.e. the failure of initially parallel geodesics to remain parallel. Using the mathematical tools of cosmology, it is possible to construct a universe that - starting from a Big Bang - first expands, then slows, stops, and re-contracts to end up in a Big Crunch again, only for the cycle to repeat over and over again. The problem is that this is not consistent with what we actually observe in the real world. We already have a very detailed model of (classical) gravity, being General Relativity, which works extremely well - what you seem to propose is not very consistent with what we already know about gravity.

This, and almost everything else in that post, is most certainly not mainstream, whatever you might think.