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

Fermi-Walker transport geodesic

  1. Jul 11, 2011 #1
    Hi guys, here's my question:

    An accelerated observer (both in curved or non-curved space) who Fermi Walker transports his own basis vectors set along his world line will have the metric in the minkowsky form [itex]\eta_{\mu\nu}[/itex] at each point of the world line?


    AND if the observer follows a geodesic, the condition of parallel transport [itex]\nabla_{U} V[/itex] along th geodesic is sufficient to keep the metric in the minkiwsky form at each point of the geodesic itself?


    Thanks!
     
  2. jcsd
  3. Jul 11, 2011 #2

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes. Much like the Rindler observer, there will be first order corrections in the metric due to his acceleration, if the acceleration is not zero. i.e. [itex]\partial g_{00} / \partial x[/itex] will be 2a. So, if we call the direction of acceleration "up", above or below the worldline g_00 will vary, but it will have the value of 1 on the wordline itself.

    There will be another first order correction if the observer is rotating. Corrections due to the curvature of space-time will be of second order or higher
     
  4. Jul 12, 2011 #3
    Thanks pervect!!

    But.. does the fact that the first order correction to the metric aren't zero mean that for the accelerated observer spacetime is never (locally) flat (the definition of "flat" I use is that the metric is in the minkiwsky in the 0-order and has null first correction)?

    In that case the accelerated observer wolud disagree on measurement made by a static observer in the point where their world line meet!
     
  5. Jul 12, 2011 #4
    I suppose I am the only one who is confused (not unusual :) ) but an accelerated observer never follows a geodesic. So I do not get the 'AND' you place between the two parts.
     
  6. Jul 12, 2011 #5
    They're two different situation:

    Firstly I wanted to know if an accelerated observer who Fermi Walker transports its own tetrad would have the metric in the minkowsky form at each point of its worldline.

    Then I asked if, for a freely falling observer the condition of parallel transport of his basis vector [itex]\nabla_{U}V=0[/itex] along the geodesic on which it moves is a sufficient condition for the metric to be in the minkowsky form all along the geodesic itself.

    :wink:
     
  7. Jul 12, 2011 #6

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Teddd, could you define your notations U and V? When you refer to the "condition of parallel transport," you give an expression, but you don't set it equal to anything, so it isn't a condition. Do you mean that this expression equals zero?

    [EDIT] OK, I see that the =0 issue is addressed in #5.
     
  8. Jul 12, 2011 #7
    Yeah, I forgot the =0!!!

    I've noticed that before posting the #5...

    By the way, to avoid any confusion: U is the tangent vector to the geodesic and V stands for the vector (defined on the curve) to be parallel transported.

    :tongue:
     
  9. Jul 12, 2011 #8

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm not positive what you're asking here, but taking some typo-interpreting liberties, I'll say that if you have an accelerated observer who fermi-walker transports his basis vectors, the metric in terms of said basis vectors will be Minowskian _on the worldline_ of the observer, by which I mean the product of the basis vectors will be the [itex]\delta^{i}{}_{j}[/itex] as you'd expect, i.e. they'll be orthonormal at any point on the worldline.

    However, the Christoffel symbols, the connection, of such an observer will not be zero if the acceleration is non-zero, because there will be some non-zero first order derivatives of the metric.

    In fact, one can explicitly find that in terms of basis vectors, [tex]\Gamma^{\hat{0}}{}_{\hat{j}\hat{0}} = a^{j}[/tex] on the accelerated worldline, where [itex]a^{j}[/itex] is the acceleration, see for instance MTW pg 330.

    Because you specified fermi-walker transport, there won't be any issue with rotation, and you didn't ask about that so I'll skip talking about it.

    If you go from transporting basis vectors to actual coordinates, you can define the fermi-normal coordinates of an accelerating observer, and find the metric in terms of these coordinates.

    You can find an expression accurate to the second order of the metric in fermi-normal coordinates on pg 332 of MTW in terms of the Riemann.

    If you have a non-accelerating observer, a geodesic observer who parallel transports his basis vectors (and isn't rotating), the Christoffel symbols will be zero as well.
     
  10. Jul 14, 2011 #9
    Ok, help me out with this, becauose I think i've messed up something.

    Let me start from the beginning.

    At each point of the manifold it's possible to choose a basis which makes the spacetime looks flat ([itex]\eta_{\mu\nu}[/itex]) in an infinitesimal neighbour: this change of basis correspond to pick an observer who is freely falling right?
    But does this work also backwards? I mean, only the freely falling observers can have the metric in the minkowsky form (locally)?

    And if that is true, if I parallel transport the basis vector of this freely falling observer along the geodesic on which it moves, will the metric be minkowsky (locally) in each infinitesimal neighbour of every point on the curve (not only between points lying on the curve, as for the Fermi-Walker, here I mean a whole infinitesimal neighbour)?

    This would mean that a freely falling observer has a minkowskian metric (of course at first order) at every point on the curve (right?). Becaouse by parallel transporting those basis vector from a point on the geodesic to the one infintesimally close I'm actually performing an (infinitesimal) lorentz transformation between lorentz observers (right?).

    However any accelerated observer (non freely falling) wolud never have the metric in a minkowsky form, ever.

    (This has not much to do with Fermi-Walker transport: if I should open a new post please let me know.)

    Thanks!
     
  11. Jul 14, 2011 #10

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Let's avoid informal language such as "makes the spacetime look flat" for now,which could be interpreted in different manners, until we get the confusion sorted out.

    At each point on the manifold, there are an infinite number of possible choices of basis vectors. If one restricts oneself to sets of basis vectors that are orthonormal, the product of a basis vector with itself is unity and the product of a basis vector with another basis vector is zero, but there are still an infinite number of such orthonormal basis vectors at any point on the manifold.

    The orthonormality condition is necessary and suficient to say that [itex]\eta^{\mu\nu}[/itex] is diagonal and unity. This might be what you mean when you write [itex]\eta^{\mu\nu}[/itex], the diagonal matrix of the product of basis vectors. I'm not positive, sometimes one sees the same symbol used for the metric, i.e [itex]g^{\mu\nu} = \eta^{\mu\nu} + h^{\mu\nu}[/itex], which is a different meaning of the symbol.

    To define a metric, you need to define coordinates as well as basis vectors. But I don't know if we need to get into this or not, because I'm not positive what you mean when you write [itex]\eta^{\mu\nu}[/itex]. So I'll avoid an extended discussion unless we need one - we might, if you're conflating the basis vectors, coordinates, and coordinate bases.

    Basis vectors are defined at a point, irrespective of the notion of the motion of acceleration of the point.

    If we have a worldline of some observer moving through a point, some specific basis vector at that point represents his notion of proper time. This might help in visualizing the significance of the basis vectors, but you don't need to specify the observer in order to define the basis vector - you might use an observer to help physically interpret the meaning of a set of basis vectors, though.

    To compare basis vectors at different locations, one needs a method of transporting them. Some observers will choose to Fermi-walker transport the basis vectors, others will chose to parallel transport them.

    The transport process defines the Christoffel symbols, the connection coefficients.

    An accelerating observer who uses Fermi-Walker transport will have Christoffel symbols that are non-zero in the direction of his acceleration.

    In terms of metric coefficients, and coordinates, which we haven't really talked about, the metric coefficients can be defined as

    [tex]\Gamma_{abc} = \frac{1}{2}\left( \partial_{a} g_{bc} + \partial_{b} g_{ac} - \partial_{c} g_{ab} \right)[/tex]

    which implies that the partial derivatives of the metric are nonzero for nonzero Christoffel symbols, so the metric coefficients are not constant to first order.

    Since this appears to be related to your question, I suspect we'll actually have to go into the relation between the coordinates and the basis vectors. But I don't want to go into that if I'm getting off-track.

    Another possible point of contention is "flatness". The usual definition of "flatness" is independent of the coordinate choice and is in terms of the Riemann tensor. It's true that many of the complexities of dealing with non-flat space-times are found when using funny coordinate systems in flat space-times, but it's a bit confusing to have a space-time called "flat" or "non-flat" depending on one's coordinate choice, which is why it's usually avoided.

    [add]
    Let me try putting this another way. If you are in a perfectly flat space-time of special relativity, and you aren't accelerating, your basis vectors don't change (as compared to the flat space-time), and your Christoffel symbols are zero.

    But if you are accelerating, your time-basis vector is "boosted" as you travel, your time-basis vector is not constant along your path. And your Christoffel symbols are non-zero, because your basis vectors are changing as a function of time.
     
    Last edited: Jul 14, 2011
  12. Jul 15, 2011 #11
    First of all thanks for your disponibility pervect!!!

    I'm actually talking about the metric here.

    I'm working with some coordintate basis, so defining a coordinate system on the manifold my basis vector are [itex]\vec{e_j}=\frac{\partial}{\partial x_j}[/itex], and with [itex][\vec{e_k},\vec{e_z}]=0\;\;\forall k,z[/itex].


    Ok, here I mean that at any point [itex]P_0[/itex] I can perform a lorentz transformation (thus changing observer) to obtain a new basis set in which the metric, after a Taylor expansion (in [itex]P_0[/itex]) looks like: [tex]g_{\mu\nu}(P_0)=\eta_{\mu\nu}(P_0)+\frac{\partial^2}{\partial x^mx^n}g_{mn}+...[/tex]

    I do not mean that the riemann tenson is zero.


    Of course! But different basis set at a particular point correspond to different observer, i mean both of them are in that particular point of the spacetime but they have different notion of proper time and different velocities (don't they?).


    So, what I wanted to know if that an observer that follows a geodesic (so its timelike vector is parallel to the geodesic, which points on the manifold we will call [itex]x^{geo}[/itex]) ad who parallel transport his basis vector will have always the metric in the form [tex]g_{\mu\nu}(x^{geo})=\eta_{\mu\nu}(x^{geo})+(second \, order \, derivatives \, etc..)[/tex] for every point [itex]x^{geo}[/itex] on the geodesic.


    Feel free to correct all the mistakes i've made!!
     
  13. Jul 15, 2011 #12

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I think you can do some of what you want, but the textbook results aren't quite in the form you present them.

    The first thing to point out is that one has a lot of possible coordinate choices for "the" coordinate system of an obsever. Rather than use the numeric indicies, I'll call the coordinates the more familiar t,x,y,z. And making the metric Minkowskian to any order (ignoring the issue of how it changes away from the worldline) makes the basis vector [itex]\partial / \partial t, \partial / \partial x, \partial / \partial y, \partial / \partial z[/itex] along the worldline.

    But consider non-linear transformations of the sort f' = f(x), where df/dx = 1 near the origin. This won't change any of the basis vectors at the origin, but will represent a different coordinate system - and it will change some of the Christoffel symbols, such as [itex] \Gamma^{x}{}_{xx} [/itex]. So you need to specify a bit more detail to completely describe exactly what coordinate system you want to use.

    The textbooks, (I'm using MTW's "Gravitation", which has a good treatment of your problem, and I'd recommned that you get a hold of it, it's discussed near pg 330, you'll need to know that because it's a huge book and it's hard to find things in it if you do get it) talks about one specific coordinate system, which is the Fermi Normal coordinate system, which specifies things exactly.

    The Fermi normal system is created by taking the set of space-like geodesics through some particular point on an observer's wordline (accelerating or not). Said geodesics are orthogonal to the time-like basis vector of the worldline. To fully specify the coordinates, one maps "distance" to an affine parameterizations of aforseaid spatial geodesics. geodesics.

    Specifically, they map the magnitude of some local vector (x,y,z), which would just be sqrt(x^2+y^2+z^2) as the tangent space is flat. This paramter specifies a particular point on the geodesic due to the affine parameterization. This gets around the ambiguity of the x' = f(x) transformation I mentioned, it specifies the coordinates exactly and unambiguously.

    This coordinate system does allow you to determine the metric near the worldline of the observer, and I think some of it may be equivalent to the way you are thinking about it, but it's not presented quite that way.

    What the textbooks say is you can write the metric to terms that are second order in terms of t,x,y,z. They also say that if you have an observer who is not accelerating, and also not rotating,the fermi-normal coordinate choices makes all the Christoffel symbols vanish, which I think should imply that the first order derivatives of the metric also all vanish.

    This leaves terms like dt^2 (1 - alpha x^2), for instance, that represent local tidal forces. In fact, you can write the metric in Fermi-normal coordinates in terms of the Riemann, some of the terms of said Riemann are of the sort that gives you said tidal forces. So this shouldn't be a surprise, the tidal forces of your Fermi-normal coordinate system are releated to the Riemann tensor.

    YOu'll see some other, purely spacial terms, as well. For instance (1 - alpha x^2) dx^2. These can also be expressed in terms of the Riemann. If you're familiar with the Bel decomposition at all, you can neatly classify the spatial terms as the "topogravitic" part of the tensor, the part of the tensor that describes the spatical curvature of the particular time-slice you've selected. This selection was done (again) by choosing spatial geodesics that were orthogonal to the time basis vector of your observer.

    If you want the exact expression for the line element in terms of the Riemann, MTW gives it, but it's a bit tedious to type in. HOpefully you can get a copy at your local library and check it out and copy it down.
     
    Last edited: Jul 15, 2011
  14. Jul 15, 2011 #13

    atyy

    User Avatar
    Science Advisor

  15. Jul 15, 2011 #14
    Ok, i feel i'm catching it.

    But I'm still confused on one thing, a really basic one!
    I've always assumed that different basis at a particular point on the manifold corrispond to different observer, and different basis vector corresponds to different coordinate system on the manifold.

    Maybe it's a misunderstanding: now I don't mean the coordinate system of the observerm (fermi walker, fermi normal...), but the coordinate system on the manifold itself.



    Allow me a stupid example: take for instance flat-mono dimensional (x,t) space.
    Put on the manifold (a plane sheet) a cartesian like coordinate system: the coordinate basis at any point is [itex]\frac{\partial}{\partial x}\;;\;\frac{\partial}{\partial t}[/itex].
    Now if i make a lorentz transformation i obtain some other basis vector, that correspond to an observer who moves with respect to the first one.

    So in my point of wiew the basis vectors at any point on the manifold (flat or curved) uniqely identify the cinematic status of the observer.


    Am I missing something?
     
  16. Jul 15, 2011 #15

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm guessing you mean kinematic status of the observer, i.e. their velocity.

    What you describe works in flat space-time. But in curved space-time, you run into difficulties. You have one set of basis vectors at one point, and another set of basis vectors at another. Now, at any given point, the basis vectors determine velocity - to be more precise, the time basis vector is the 4-velocity of an observer at that point.

    The problem is, to compare velocities at different points, say point A and point B, you need to transport the basis vectors somehow from A to B. You can do this in curved space-time given a paticular path between A and B by parallel transport - but unlike the flat-space case, the particular path you chose matters, so you can't in general just compare basis vectors (or velocities) at point A and point B.

    That's why you'll see some mathemeticians (Baez for instance) say that in GR, you can't compare velocities, they're in different tangent spaces. Of course, if you have a static space time, you have a notion of static observers, but this is "extra" structure that you don't always have. Without some sort of extra structure like this, you can't unambiguously compare velocities at different points, you need to specify some path-dependent procedure as to how to transport the basis vectors.

    In flat space-time, the path choice turns out not to matter, but in curved space-time, it does.
     
  17. Jul 15, 2011 #16

    atyy

    User Avatar
    Science Advisor

    Usually the cinematic status of an observer is specified by a set of orthonormal vectors (called a "frame" or "orthonormal tetrad"). Coordinate basis vectors are not orthogonal, unless the metric is diagonal in those coordinates. If coordinates are such that the metric is diagonal, then a set of frame vectors can be obtained by "normalizing" the coordinate basis vectors. For a general observer, each frame vector will be a linear combination of coordinate basis vectors. Fermi normal coordinates are constructed by constructing a frame at one point of the worldline, then Fermi-Walker transporting it along then worldline.

    http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity
    http://arxiv.org/abs/1102.0529, section 9.2, 9.3
     
    Last edited: Jul 15, 2011
  18. Jul 16, 2011 #17
    I'm impressed by your disponibility and patience!!
    Please hold on another little bit!

    Pervect: I'm aware that you cannot compare vector defined in different points, they belong to different tangent spaces, and that's why you construct a connection (i.e Christoffel symbols) to "connect" different tangent spaces.
    And obviously different path corresponds to different vector transportation.

    My post was started becaouse my professor stressed out the importance of the possibility of choosing an observer for which (at a particular point) the basis vector make the metric minkowskian up to first order (at that particular point).

    I thought that if the metric is minkowkian to that observer (to first order) this means that i've choosen a freely falling observer, and that the fact that the second order derivatives of the metric aren't zero is beacouse of the curvature of the manifold.

    And becaouse I know that freely falling observers follows geodesics I asked myself if there was a way to mantain the metric minkowskian (always up to first order) for every point on the observer geodesic. And I suppose, and it seems to me that you have confirmed this, that the condition of parallel transport of the basis vector along the geodesic is a sufficient condition to keep the metric in that form for every point of the world line.

    Fermi Walker is the general case of transporting the basis vector, and reduces to the condition above if the observer follows a geodesic (and is non rotating).


    Atyy: According to the wikipedia link i always have to choose coordinate basis that are orthonormal, but why?? (Thanks for the readings!!)
     
  19. Jul 16, 2011 #18

    atyy

    User Avatar
    Science Advisor

    If we want to to represent a family of ideal timelike observers, as the wikipedia link says, then each observer by convention constructs an orthonormal frame to do his local physics in.

    However, the metric in general can be represented using non-orthonormal tetrads, such as in the Newman-Penrose formalism. http://www.scholarpedia.org/article/Spin-coefficient_formalism
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fermi-Walker transport geodesic
  1. Fermi-Walker transport (Replies: 3)

Loading...