Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Local Lorentz frame

  1. Sep 21, 2010 #1
    Blandford & Thorne, Applications of Classical Physics:

    Taylor & Wheeler, Spacetime Physics:

    These definitions seem to be based on the notion of a "physical" or "practical" infinitesimal: a quantity too small to be detected. But how can we measure the accuracy of an imaginary detector? Taylor & Wheeler answer this by saying: you decide. In that case, could we not define a (trivial) global Lorentz frame if we specify zero accuracy, and a Lorentz frame of any other size, from infinite down, by an appropriate choice of accuracy? This seems at odds with the connotation of smallness.

    Blandford & Thorne's clocks and rulers are regarded as in some sense "ideal", yet not ideally accurate. What quality does their idealness consist of if not accuracy? (I take it it's not that they're just a very nice colour and you can check your emails on them.) Is it just that that there's a scale limit on their accuracy, so that, given a finite degree of accuracy of instruments, you can always choose smaller and smaller scales till you find a scale where they detect no curvature--rather than the empty statement that given a degree of curvature, you can always find instruments not accurate enough to detect it!

    I notice that Blandford and Thorne only specify a spatial accuracy at this stage; is that significant?

    Besides curvature, could topology limit the size of a Lorentz frame of a given accuracy?

    The metric tensor field is defined at each point. Its value at each point contains information about curvature. Sometimes the value is derived by an infinitesimal analogue of the Pythagorean formula. If curvature is significant enough that it can't be neglected in such a small region as one point, how can it be neglected in such a large region as a space shuttle? Is it because different degrees of accuracy are being used in these two cases, i.e. this way of talking about the metric tensor field assumes ideal, unlimited accuracy (and if so, how does that mesh with the idea of a differential a linear approximation)?
     
  2. jcsd
  3. Sep 21, 2010 #2

    atyy

    User Avatar
    Science Advisor

    There should be a Taylor expansion around the point somewhere in Blandford and Thorne. That governs the deviation from being perfectly locally lorentz. In principle, locally lorentz means at a point.

    Ideality means the clock measures the proper time along its trajectory.

    Even when restricting to a point, they discuss the need to restrict to phenomena not involving second derivatives, since only the first derivative can be made to disappear at a point.
     
  4. Sep 21, 2010 #3

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I haven't worked through all the details, but this is how I think these things should be done. (I'm sure I would want to change some of this if I did know all the details).

    Suppose that we define a "normal coordinate system" associated with a world line C:[a,b]→M, at a point p in the range of C, as a coordinate system x such that

    1) its 0 axis coincides with the tangent of C at p,
    2) the tangent vectors of the axes at p make up an orthonormal basis of the tangent space at p,
    3) if q is a point on the [itex]\mu[/itex]-axis, then [itex]x^\nu(q)[/itex] is equal to [itex]\delta^\nu_\mu[/itex] times the proper time/distance along the axis from p to q.
    4) [itex]g_{\mu\nu}(p)=\eta_{\mu\nu}[/itex]
    5) [itex]g_{\mu\nu,\rho}(p)=0[/itex].

    Since the definition of a geodesic doesn't involve any second (or higher) derivatives of the metric, any "normal" coordinate system should map all geodesics through p to curves through [itex]0\in\mathbb R^n[/itex] that have zero acceleration at 0. Aren't these exactly the properties that we want a "local Lorentz frame" to have? I expect that it's possible to show that the coordinate system is completely specified by these requirements when spacetime is flat (Edit: Definitely not correct. See #6.), and that there are several possibilities when spacetime is curved. (Riemann, Fermi, etc.)

    GR is a theory of physics, so it can't be defined by Einstein's equation, which is pure mathematics. It's defined by a set of axioms that tell us how to interpret the mathematics as predictions about results of experiments. The axioms have to look something like this:

    0. Motion is represented by curves.
    1. A clock (that's built according to standard specifications) measures the proper time of the curve in spacetime that represents its motion.
    2. A radar device (that's built according to standard specifications, and is moving as described by a congruence of timelike geodesics that fill up a region of spacetime with zero curvature) measures the proper length of the spacelike geodesic from the reflection event to the midpoint of the timelike geodesic from the emission event to the detection event.

    These axioms include a bunch of idealizations. I only mentioned some of them explicitly. These idealizations are particularly annoying in axiom 2, but I doubt that it's possible to come up with a much better axiom. I have never seen an axiom that describes accurate length measurements with a measuring device in an arbitrary state of motion in an arbitrary region of an arbitrary spacetime, and I don't think it can be done.

    It's important to understand the nature of the idealizations in the axioms. The radar device described in axiom 2 will only measure what we want it to measure in very specific circumstances. If we need better accuracy, then we need to make it smaller. How small? Small enough that the measurement result doesn't depend significantly on variations of the size of the measuring device. How much is significant? That depends on the situation, doesn't it?

    So how do we know if our measuring devices are accurate enough? (or at all?) That's a tricky question, and I don't have a complete answer. I think this is a partial answer: If the theory e.g. predicts a specific orbit of Mercury, and we find that Mercury has that orbit to within the desired accuracy, we can conclude a) that if the measuring device is accurate, then the observation says something about the accuracy of the prediction, and b) that if the prediction is accurate, the observation says something about the accuracy of the measuring device. I guess that if we compare the same measuring device to a lot of different predictions, all of the results together can be considered evidence that the measuring device is good enough
     
    Last edited: Sep 22, 2010
  5. Sep 21, 2010 #4

    atyy

    User Avatar
    Science Advisor

    A related to question is: how small does a test particle have to be to be a test particle?

    ie. if the principle of equivalence fails, is it because the EP is false, or because you used a particle that was bigger than what is permissible for a test particle?

    http://arxiv.org/abs/0707.2748

    http://arxiv.org/abs/gr-qc/0309074
     
  6. Sep 21, 2010 #5
    We'd also need an orientation, wouldn't we?
     
  7. Sep 22, 2010 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, and that's not the only thing I forgot. I have only described the assignment of coordinates to point on the axes, and there's no way that that can be sufficient to determine the assignment of coordinates to points that aren't on the axes. Note that the difference between Riemannian and Fermi normal coordinates is what geodesics they use to assign coordinates to other points. Riemannian normal coordinates uses all the geodesics through the point p. Fermi normal coordinates uses, for each q on the 0 axis, all the geodesics through q that are orthogonal to the 0 axis, to assign coordinates to points in the hypersurface [itex]\{r\in M|x^0(r)=x^0(q)\}[/itex].

    Also note that I haven't checked if the way I assigned coordinates to points on the axes agree with Fermi/Riemannian normal coordinates. I might have to rethink that, unless someone else does it for me (wink wink, nudge nudge).
     
  8. Sep 22, 2010 #7

    Dale

    Staff: Mentor

    Yes, if you want trivial results you can make trivializing assumptions.

    Precision and reliability. I.e. they should have no bias, and they should not be affected by any other quantity like temperature or acceleration.
     
  9. Sep 22, 2010 #8
    Hi.

    As for curvature, equivalence principle says that size of inertial frame should be infinitesimal small both in space and in time. If tidal force generated from nearby energy is negligible, we can regard this frame is Lorentz frame practically within certain finite distance and time interval. But due to gravity form your weight, mass of distant stars and all the other energy distribution varying in time and space, the Lorentz or inertia frame must be infinitesimal small in theory.
    Regards.
     
    Last edited: Sep 22, 2010
  10. Sep 22, 2010 #9
    This made me think of two recent posts:

    I haven't really got much idea of what this means yet. All I can do for now is mentally log these statements for future reference. I thought at first the field equations said that curvature came from stress-energy, which manifests at each point, and there is no action at a distance in GR. But apparently there are various kinds of curvature, some curvature due to action at a distance, and some not, and, well, it's all rather confusing. My plan for now is to get to grips with the geometic concepts: what curvature is, what different kinds there are, what they're called, what tides are (another kind of intrinsic curvature not covered by GR, or covered by GR but not due to stress-energy?); how certain charts such as RNC and LLF are defined; play with some coordinate transformations, calculating metric tensors, practice integrating over differently curved manifolds; also get to the bottom of certain elementary things in calculus that are puzzling me; and then turn to what GR has to say about the causes of curvature.
     
  11. Sep 22, 2010 #10

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Tidal forces are components of the Riemann curvature tensor, the abstract geometric entity that represents curvature. So, you can think of the tidal forces as being curvature.

    This may be too much detailed, but If you construct some othronormal frame defined by basis vectors [itex]\hat t[/itex], [itex]\hat x[/itex], [itex]\hat y[/itex], [itex]\hat z[/itex] then the tidal forces will be specifically:

    [itex]R_{\hat t \hat x \hat t \hat x}[/itex], [itex]R_{\hat t \hat y \hat t \hat y}[/itex], [itex]R_{\hat t \hat z \hat t \hat z}[/itex]

    This takes advantage of the fact that a rank N tensor maps N vectors into a scalar, so in our example we are giving the rank-4 Riemann tensor four basis vectors in some specified order, for example [itex]\hat t[/itex],[itex]\hat x[/itex],[itex]\hat t[/itex], [itex]\hat x[/itex], and getting out a scalar the magnitude of the tidal force in that direction.

    The hats are important - they represent the fact that we are working with an orthonormal basis, a "frame field", rather than coordinates (this may also be too technical).
     
  12. Sep 22, 2010 #11
    Thanks Pervect. No need to apologise for being too technical; anything I don't understant now, may be helpful later. I think you're saying, in this context, tides and curvature are two names for the same thing. And from what bcrowell said in the rest of that post I quoted: all curvature, unless otherwise qualified, is intrinsic (when talking in precise mathematical language), and those non-intrinsic qualities which we colloquially call curvature are called topology by mathematicians.

    I notice that the word "frame" sometimes means a chart (as in "reference frame", "local Lorentz frame"), sometimes a basis (as in "frame field"), sometimes a basis field (as when "frame" is used as a short-hand synonym for "frame field"). So thanks for making that distinction clear.

    In relation to surfaces (2d manifolds) embedded in Euclidean 3-space, I've come across the concept of the two principle "curvatures" (not true, intrinsic curvature), and their product, the intrinsic Gaussian curvature. If I understand your post correctly, it takes three numbers to specify the curvature (Gaussian curvature) at a point of a 4d manifold. Is this a general rule, that for an n-dimensional manifold, it takes (n-1) numbers to specify the curvature (tidal forces) at a point?
     
  13. Sep 22, 2010 #12

    JesseM

    User Avatar
    Science Advisor

    Intuitively it seems like it should be "harder" to detect the effects of curvature in a very small region of a given curved spacetime than in a larger region--one would need increasingly accurate detectors to measure tidal effects in smaller and smaller regions. I wonder if there'd be any way to formalize this notion of "detector accuracy" in order to show that in the limit as the size of the region of spacetime goes to zero, the detector accuracy needed to detect a difference from what you'd see in flat spacetime would go to infinity.
     
  14. Sep 22, 2010 #13
    Hi.
    I hereby show you two examples of tidal force.
    #1 Free fall elevator. Its frame of reference seems to be inertia frame. However, floating objects in the same level come close together into the center of the elevator box because the directions of the gravity, i.e. to the center of the earth, are slightly different among the objects.

    #2 Rocket free falling into black hole. Its frame of reference seems to be inertia frame. However, difference of gravity force at the top and at the bottom of the rocket, i.e. tidal force increase as it approaches to BH and the rocket is stretched to spaghetti.

    Regards.
     
  15. Sep 22, 2010 #14

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    The good news is that we do have a simple physical interpretation of curvature as tidal forces for _some_ of the components of the Riemann. The bad news is that it doesn't cover all of the components.

    The Riemann has 4x4x4x4 = 256 components. But they are highly interrelated.
    It turns out it takes 20 independent quantities to specify the Riemann curvature tensor at a point. See for instance http://mathworld.wolfram.com/RiemannTensor.html

    The Riemann can be decomposed into various subparts, unfortunately the texts I have don't cover this very well, and as a consequence I dont fully understand it myself. It's called the Bel decomposition of the Riemann tensor, however - I know that much.

    For what info is online, try the wiki stub:

    http://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=325505109
    and
    http://www.cramster.com/reference/wiki.aspx?wiki_id=37176#Relation_with_Ricci_decomposition

    ((not sure where the last came from or how durable it is))

    The result is that the Riemann is decomposed into three pieces, though - the electrogravitic part, the magnetogravitic part, and the topogravitic part.

    The tidal forces are part of electrogravitic tensor in the Bel decomposition. The full Electrogravitic tensor has six degrees of freedom, however, not just three. I suspect the remaining degrees of freedom are physically interpretable as "tidal torques", but I could be mistaken.

    Curvature can also mimic magnetic forces. These show up as geodesic deviations for moving particles - you could think of them as acting like velocity dependent forces - or as forces on spinning particles. The magnetic analogy applies because the B field is a velocity dependent force - F = v x B by definition, or you can think of how the B field accelerates a spinning charge and not a non-spinning one. The magnetogravitic part adds 8 more according to what I can gather from my online reading.

    Finally, there's something called the "Topogravitic tensor". I'm afraid I don't have much of a clue as to what the physical interpretation of this is. This has the remaining six degrees of freedom.
     
    Last edited by a moderator: Apr 25, 2017
  16. Sep 22, 2010 #15
    Question:

    Suppose we have a non-rigid ball free falling radially in a Schwarzschild metric, how does the ball get deformed (height and width) in terms of r and m?

    Does it matter if the ball travels at exactly the escape velocity (fall from infinity) or not?
     
    Last edited: Sep 22, 2010
  17. Sep 22, 2010 #16
    Either I've misunderstood something, or you're using a totally different definition of inertia(l) frame (i.e. inertial chart) to Taylor & Wheeler and Blandford and Thorne, since theirs is defined by the absence of detectable deviation from an inertial chart on Minkowski space. Although I suppose we'd have a good excuse to neglect gravity in #2, as we'd be dead.
     
  18. Sep 22, 2010 #17
    By "spaghetti effect" the ball is stretched in r direction.
    By "elevator effect" the ball gets pressure in r sphere.
    I am not good in quantity discussion.

    The difference is in quantity not in quality, I think.
    Regards.
     
  19. Sep 22, 2010 #18
    Yes I understand that, but I want to know how to calculate it. Anyone?

    So you think it does make a difference? Two balls radially free falling at the same R in the same Schwarzschild solution with different relative speeds will measure the height and width differently?

    A third question would be how the radar distance relates to the ruler distance in this situation.

    Another question, what in case this ball is in orbit? It seems that the height and width of this ball would not change if it always faces the 'center' of gravity, and thus can we say the ball is truly in an inertial frame as opposed to a ball free falling radially or does Thomas precession spoil the fun? And is this still the case if the orbit is an ellipse and if not how is it different?
     
    Last edited: Sep 22, 2010
  20. Sep 22, 2010 #19
    Yes. As a matter of SR, we need Lorentz transformation for comparison.
    Regards.
     
    Last edited: Sep 22, 2010
  21. Sep 22, 2010 #20
    No, in SR if an observer A on ball X measures the width and height w and h, then the other observer B on ball Y will measure exactly the same w and h assuming the balls are identical.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Local Lorentz frame
  1. Local lorentz (Replies: 1)

  2. Local lorentz frame (Replies: 7)

  3. Locally Lorentz (Replies: 40)

  4. Local Lorentz Frames (Replies: 47)

Loading...