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Markus Hanke

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Markus Hanke last won the day on December 31 2020

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About Markus Hanke

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  1. Quite possibly so! To be honest, I am still struggling to understand what it is the OP is actually trying to do. It also seems he has abandoned this thread, just like all the other ones he opened before this. Indeed - very well summarised! +1
  2. Yes, you can indeed - the result is the hyperbolic transformation I gave earlier. This just doesn’t seem to be what the OP was doing or getting at - and if it is, then the conclusion he reaches is of course meaningless (“in this article we see that the particle cannot accelerate”). It is, however, correct and fully consistent (albeit trivial) with the absence of proper acceleration in an inertial frame. Hence my reading.
  3. 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.
  4. 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).
  5. 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”?
  6. 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.
  7. 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.
  8. 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.
  9. 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.
  10. 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.).
  11. 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.
  12. 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.
  13. 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.
  14. 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.
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