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Reference frames, reference particles, coordinate systems and all that

  1. May 25, 2013 #1
    Previously, before getting into relativity, I've always thought of a 'reference frame' of basically an "observer carrying a coordinate system" - where I thought of an observer as anything which could record information of positions and velocities of particles etc. Now, however, I'm reading a book about SR and GR (Øyvind Grøn - Einstein's general theory of relativity) where these concepts are defined precisely in a way that at least is new to me. The definitions are as follows

    Coordinate system: "A coordinate system covering a region of spacetime, is a continuum of four variables ##\{x^\mu\}## that uniquely label every event in the region."

    Reference frame: "Is a continuum of non-crossing time like or light like curves in spacetime. We may thus think of a reference frame as a continuum of particles, called reference particles or observers. These reference particles need not move freely."

    Comoving Coordinate system: "A comoving coordinate system in a reference frame is defined by the requirement that the reference particles have constant spatial coordinates."

    Thus, by these definitions, it seems like one can think of a reference frame of being constructed out of particles. So, I wonder, what is the usefulness, in relativity, of thinking of reference frames in terms of particles moving along world-lines? Is there somehow a necessity, in relativity, to take this view?

    I also note that particle worldlines never can be space like, so while coordinates can label events which lie on space like curves, the curves of a 'reference frame' can not, according to the above definition.

    Any insight on these concepts would be appreciated.
     
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  3. May 25, 2013 #2

    ghwellsjr

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    Just remember this one salient fact: whatever "observers" or "particles" or rulers or clocks you want to employ to help you establish the concept of coordinates for an Inertial Reference Frame (IRF), when you use the Lorentz Transformation process to create the coordinate system for a new IRF moving with respect to the first one, it will have none of the "observers" or "particles" or rulers or clocks available to help you make sense in the new IRF. All these "helps" that are used to explain what a coordinate system is should eventually be abstracted away so that you can think of an IRF and its associated coordinate system with total freedom to populate it with anything (observers, objects and clocks) in any coordinate positions at any coordinate times and with any motions and accelerations you desire.
     
  4. May 25, 2013 #3

    WannabeNewton

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    His definition of reference frame is a weird one. Nowadays, what he calls a reference frame, would usually be called a time-like or null-like congruence. Let ##U\subseteq M## be an open subset of space-time; a time-like congruence on ##U## is a family of time-like curves in ##U## such that each ##p\in U## is contained in exactly one curve in the family. The tangent vectors to the curves at every event then define a vector field ##\xi^{a}## on ##U## and you can picture the integral curves of this vector field as being the world-lines of a family of observers defined on ##U## if you wish (so that ##\xi^{a}## is the 4-velocity field of the observers). I have no idea why the author calls these reference frames but it may be older terminology because reference frames (more precisely, local reference frames) in GR have a very clear definition in terms of frame fields (also called tetrad fields): http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity, http://physics.stackexchange.com/qu...frame-of-reference-mean-in-terms-of-manifolds, http://casa.colorado.edu/~ajsh/phys5770_08/grtetrad.pdf [Broken]

    A coordinate system ##\{x^{\mu}\}## should just be seen as coming out of a smooth chart ##(U,\varphi)## because for any ##p\in U##, ##\varphi(p)\in \mathbb{R}^{4}## can be broken down as the 4-tuple ##\varphi(p) = (x^{0}(p),...,x^{3}(p))## where the ##x^{\mu}## are functions ##x^{\mu}:U\rightarrow \mathbb{R}##.

    EDIT: I don't know of any standard GR text, other than MTW, that goes into much if any detail on local Lorentz frames and frame fields in GR so hopefully someone can give you reference to a more accessible text that does or just list the relevant section(s) in MTW.
     
    Last edited by a moderator: May 6, 2017
  5. May 26, 2013 #4
    Here are some posts I kept for my notes from other discussions...I don't have the thread links,
    but you can try searching as well...

    A few basic thoughts: [a synopsis]
    "A reference frame comprises an origin and a set of axes. The essential idea is that coordinates do not exist a priori in nature, but are only artifices used in describing nature, and hence should play no role in the formulation of fundamental physical laws.
    Coordinates don't have direct physical meaning, so neither do transformations between them.
    All observables in GR are defined in terms of coordinate independent quantities (not just covariant quantities)."

    edit: and a reminder: light doesn't have a rest frame in relativity..

    PAllen:

    from Wikipedia:




    Nugatory:

    Ben Neihoff:
    A frame, on the other hand, is a local system of measuring rods and clocks. There is no apriori reason this should have anything to do with the underlying coordinate system. A frame is just a collection of vectors at a point, of known lengths and angles, against which you can make standard measurements of other vectors at that point.


    DrGreg:


    The following isn't intended to be rigorous but a simplified argument:

    1. An inertial particle is defined to be one that moves freely without being influenced by any other forces
    2. It's an experimentally observed fact that (in regions where the tidal effects of gravity are negligible) inertial particles move at constant velocity relative to each other. Physics can't really explain why this should be, it's just the way our universe works.
    3. We use inertial particles to define inertial frames.
    4. Proper acceleration of a particle is acceleration relative to the inertial frame in which it is momentarily at rest. Equivalently it's what an accelerometer measures. Equivalently (for those who understand the terminology) it's the norm of the particle's 4-acceleration vector. It follows from this that inertial particles are those that that have zero proper acceleration.
    5. When people talk about "acceleration", you need to be clear whether they mean "proper acceleration", or "relative acceleration" = "coordinate acceleration". Every object has a unique proper acceleration (which any observer can calculate, and all get the same answer, and which the object can measure directly with an attached accelerometer without reference to any frames), but can have multiple coordinate accelerations dependent on who is calculating them.
     
    Last edited: May 26, 2013
  6. May 26, 2013 #5

    pervect

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    This definition of a reference frame may not be the "usual" one, but it's rather helpful in understanding Ehrenfest's paradox, in my opinion.

    Trying to find an actual set of points, an actual spatial slice, to describe the "geometry of the disk" is difficult and perhaps even impossible, due to the clock synchronization issue. See the previous argument, where I was trying for a realization (which I'm not totally happy with) and other posters objected that what I was doing was impossible.

    Using the notion of a reference frame as written here, though, one is lead to the idea of finding the distance between worldlines, the "worldlines" of what Gron calls the reference frame (and what I would agree is usually called a congruence).

    I'm not sure if anyone has described in detail the process of finding the distance, but actually, a pair of worldlines which maintains a constant distance is the very best representation of a ruler. A ruler has two ends, both of which exist at multiple instants of time - i.e. each end of the ruler is a worldline. And a ruler has a constant length (physically). So all we really need to do to pin down the mathematics behind this physical idea of a ruler "having a constant length" in detail.

    I've got several thought on how to do this, but I'm afraid I don't know of anyone who has approached this in quite the same manner as I do.

    As far as my thoughts go:

    Take any point on one of the worldlines. Find the space-like hypersurface that's orthogonal to the time-like worldline. Measure the distance to the closest worldline in this spatial hypersurface. Declare this to be the "distance between worldlines" at some particular instant of proper time. Demand, for a ruler, that this "distance between worldlines" be independent of proper time.

    This should be interpreted in the Born sense, as we demand for Born rigidity that the distance to "nearby objects" be constant.

    I *think* this winds up ultimately with the notion of a Born-rigid rotating frame (i.e. the congruence of time like worldlines) as being generated by time-like Killing vectors.

    See http://arxiv.org/pdf/1004.1935.pdf

    In the end I wind up with a notion of geometry of the disk not being a geometry of points of some particular plane-like set at some instant in time, but a geometry of the distance between worldlines.

    If we "smash down" the worldlines to points, I think we get a quotient space. (This may be what the authors who mention quotient spaces are trying to say, unfortunately the treatments I've read omit the details of how the quotient spaces were to be constructed).
     
  7. May 26, 2013 #6

    Mentz114

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    What you've said in this bit reminds me of the expansion scalar ##\mathcal{\Theta}## which is a velocity field between an observer ( a point ) on a worldline of the congruence and a nearby worldline. I think this scalar gives the extent that rulers will get shorter or longer. In fact the rate of change of ##\Theta## wrt to proper time ##d\Theta/d\tau## can be found from Raychaudhuri's equation.
     
    Last edited: May 26, 2013
  8. May 26, 2013 #7

    WannabeNewton

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    Consider, for simplicity, a time-like geodesic congruence and let ##\xi^{a}## be the tangent field to the congruence. Define the quantity ##B_{ab} = \nabla_{b}\xi_{a}##. This quantity can be used to define the shear, expansion, and twist of the congruence but it by itself has a physical interpretation. Consider a separation vector ##\zeta ^{a}## from a geodesic in the congruence to a nearby geodesic. ##\zeta^{a}## will get Lie transported along the congruence i.e. ##\mathcal{L}_{\xi}\zeta^{b} = \xi^{a}\nabla_{a}\zeta^{b} - \zeta^{a}\nabla_{a}\xi^{b} = 0##. Then, ##\nabla_{a}\zeta^{b} = B^{b}{}{}_{a}\zeta^{a}## meaning ##B_{ab}## measures the extent to which the separation vector fails to be parallel transported along the congruence. This means that the geodesic with which ##\zeta^{a}## originates will see nearby geodesics rotate and stretch etc.

    For each ##p\in M## consider the subspace ##\Sigma \subseteq T_pM## perpendicular to ##\xi^{a}##. The tensor ##h_{ab} = g_{ab} + \xi_{a}\xi_{b}## can be used to project any ##v \in T_p M## onto ##\Sigma## via ##h^{a}{}{}_{b} = g^{ac}h_{cb} ## as you can easily check for yourself. Now we define the following quantities: the expansion ##\theta = B^{ab}h_{ab}##, the shear ##\sigma_{ab} = B_{(ab)} - \frac{1}{3}\theta h_{ab}##, and the twist ##\omega_{ab} = B_{[ab]}##.

    It is pretty clear what ##\omega_{ab}## does: it just measures the rotation of nearby geodesics surrounding a reference geodesic, all in the congruence. What ##\theta## does is not measure stretching but rather the average expansion of nearby geodesics away (or towards) from a reference geodesic in the congruence. It is in fact ##\sigma_{ab}## that measures stretching. In particular, it will measure by how much an initial sphere gets stretched (or squeezed) towards an ellipse as it flows along the congruence.
     
    Last edited: May 26, 2013
  9. May 27, 2013 #8

    Mentz114

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    Good answer, thank you. Your definitions are recognisably what I'm used to ( see e.g. Abreu&Visser, arXiv:1012.4806v1). I'm still struggling to imagine what ##\Theta## and ##\sigma## do, exactly.

    I guess the key is the averaging in ##\Theta##, whereas ##\sigma_{ij}## gives directional information.

    [Edit]
    According to Raychaudhuri, the shear and rotation contribution to ##d\theta/d\tau## can cancel like this ,##\sigma_{ij}\sigma^{ij}-\omega_{ij}\omega^{ij}=0##
     
    Last edited: May 27, 2013
  10. May 27, 2013 #9

    WannabeNewton

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    I can explain it, following Eric Poisson, in a visually intuitive, non-rigorous way using the special case of a two-dimensional (finitely sized) sheet (e.g. a thin sheet of rubber). Consider again a small separation vector ##\zeta^{a}##, along the sheet, from a reference point ##O## on the sheet and recall from above that the time evolution equation for ##\zeta^{a}## is ##\frac{\mathrm{d} \zeta^{a}}{\mathrm{d} t} = B^{a}{}{}_{b}\zeta^{b}##. Consider two infinitesimally separated times ##t_{1},t_{2}## so that ##\zeta^{a}(t_2) = \zeta^{a}(t_1) + \Delta \zeta^{a}## where ##\Delta \zeta^{a} = B^{a}{}{}_{b}(t_{1})\zeta^{b}(t_1)\Delta t## (I am dropping all second order terms because the time interval is taken to be extremely small).

    Let's say the separation vector is taken to be, for simplicity, ##\zeta^{a} = r_0(\cos\varphi,\sin\varphi)## so that if we sweep out ##\varphi## to get a separation vector pointing in every direction from our reference point ##O##, we get a circle of radius ##r_0##.

    Consider first the case where ##B_{ab}## is proportional to the identity matrix (so that we don't have to care about a symmetric or anti-symmetric part and can just focus on expansion). Recalling that the expansion is given equivalently by ##\theta = B^{a}{}{}_{a}##, we can write the matrix as ##B^{a}{}{}_{b} = \begin{pmatrix}
    \frac{1}{2}\theta & 0\\
    0& \frac{1}{2}\theta
    \end{pmatrix}##. Then, ##\Delta \zeta^{a} = (\frac{1}{2}\theta \Delta t) r_{0}(\cos\varphi,\sin\varphi)## i.e. the radius of the circle changes by ##\Delta r = (\frac{1}{2}\theta \Delta t) r_{0}## meaning the area of the circle changes by ##\Delta A = \pi r_{0}^{2}\theta \Delta t = A_{0}\theta \Delta t## giving ##\theta = \frac{1}{A_0}\frac{\Delta A}{\Delta t}##. If we take the limit as ##\Delta t## goes to zero then expansion is giving us the rate of change of the area with respect to time of the circle, per unit area. This can be pictured as the circle either contracting towards or expanding away from our reference point ##O##. For higher dimensions you can instead picture volumes containing spheres and visualize the spheres expanding away from or contracting towards a reference point.

    Now consider the case where ##B_{ab}## is symmetric and trace-free (so that we can concentrate only on the shear). Then, ##B^{a}{}{}_{b} = \begin{pmatrix}
    \sigma_{+} & \sigma_{x} \\
    \sigma_{x} & -\sigma_{+}
    \end{pmatrix}## (the notation will make sense in a second). For the same circle setup, we now have ##\Delta \zeta^{a} = r_0\Delta t(\sigma_{+}\cos\varphi + \sigma_{x}\sin\varphi, -\sigma_{+}\sin\varphi + \sigma_{x}\cos\varphi)## so that adding the change in radius to the original radius gives us a curve ##r = r_{0}(1 + \sigma_{+}\Delta t \cos2\varphi + \sigma_{x}\Delta t \sin2\varphi)##. If ##\sigma_{x} = 0## then this is the parametric equation for an ellipse whose major axis is aligned with the horizontal whereas if ##\sigma_{+} = 0## then we get an ellipse whose major axis is aligned with ##\frac{\pi}{4}##. That is, the shear has distorted the circle into an ellipse (without affecting the area of the circle). This is exactly what happens, for example, when a gravitational wave passes by a circular ring of test particles. You can, just as before, then abstract this visually to higher dimensions. I hope this helped.
     
  11. May 27, 2013 #10

    Mentz114

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    That looks good and meaty. I can't absorb it all right now but I look forward to digesting it later. Great, thanks again.
     
  12. May 27, 2013 #11

    WannabeNewton

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    Let me add something extra Mentz, that I think you will like. Note first that ##\xi^{c}\nabla_{c} B_{ab} = \xi^{c}\nabla_{c}\nabla_{b}\xi_{a} = \xi^{c}\nabla_{b}\nabla_{c}\xi_{a} + R_{cba}{}{}^{d}\xi_{d}\xi^{c}## where I have used the fact that ##\nabla_{c}\nabla_{b}\xi_{a} - \nabla_{b}\nabla_{c}\xi_{a} = R_{cba}{}{}^{d}\xi_{d}##. Since ##\nabla_{b}(\xi^{c}\nabla_{c}\xi_{a}) = \xi^{c}\nabla_{b}\nabla_{c}\xi_{a} + \nabla_{b}\xi^{c}\nabla_{c}\xi_{a}## by the Leibniz rule, we can rewrite the above as ## \xi^{c}\nabla_{c} B_{ab} = \nabla_{b}(\xi^{c}\nabla_{c}\xi_{a}) - \nabla_{b}\xi^{c}\nabla_{c}\xi_{a} + R_{cba}{}{}^{d}\xi_{d}\xi^{c} = -B^{c}{}{}_{b}B_{ac} + R_{cba}{}{}^{d}\xi_{d}\xi^{c}##.

    Now take the trace-free, symmetric part of ## \xi^{c}\nabla_{c} B_{ab}## in vacuum, and consider the case of vanishing expansion and rotation (just for the sake of writing down a simpler expression), to get ##\xi^{c}\nabla_{c}\sigma_{ab} = C_{cbad}\xi^{c}\xi^{d} - \sigma_{ac}\sigma^{c}_{b} + \frac{1}{3}h_{ab}\sigma_{cd}\sigma^{cd}## where ##C_{cbad}## is the Weyl tensor. This gives a very nice geometric interpretation of the Weyl tensor as the part of the Riemann curvature that mediates the rate of change of the shear of a geodesic congruence in vacuum.

    You can do a very similar thing with the expansion of a geodesic congruence and see that its rate of change is mediated by the Ricci tensor (lending to why you might have seen geometric interpretations of the Ricci tensor as the part of the Riemann curvature that affects how spheres expand or contract about a point).

    Sorry for dropping all this information on you all at once but knowing you, this is right up your alley :)
     
  13. May 27, 2013 #12

    Mentz114

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    For the moment, going back to post#9. It has helped me realise what the parts of ##B_{ab} = \nabla_{b}\xi_{a}## signify. The traceless antisymmetric part generates rotations, the symmetric part generates boosts and the trace is a scaling factor. All of these in the local frame at some point, of course. I'm still digesting the details of #9.

    The post about the Weyl tensor will have to wait ... looks tasty.
     
  14. May 27, 2013 #13

    WannabeNewton

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    I'd be very interested in learning how you would do an analysis of all this in a local Lorentz frame at a point. From the thread I started recently, you could probably tell that my handle on local Lorentz frames is poor (one of the main disadvantages of learning GR solely from Wald and Carroll!). If you could possibly start a new thread on it (so that this one doesn't veer too far off from the OP) or send me a PM, or even link to a paper that talks about shear,expansion etc. in terms of local Lorentz frames, I would be very much in debt to you. I really need to get a firm handle on local Lorentz frames. Cheers!
     
  15. May 27, 2013 #14

    pervect

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    WanabeNewton has already answered better than I could on the main point - you need shear free as well as expansion free.

    The reference I cited earlier pretty much says exactly that as well - they say straight out that a shear-free and expansion free flow (I interpret "flow" in this context to mean congruence, what Gron is callling a reference frame) is a Born rigid rotational flow.

    I am physically interpreting a Born rigid rotational flow as the worldlines of the ends of the "rulers" ala my previous post - the ones that maintain a constant distance. (But perhaps the Born rigid rotational flow is the velocity field of the congruence rather than the congruence itself - I'm not positive of the definitions here. A relatively minor point, until it confuses someone.)

    As an aside, we see that Born rigid rotating disks do exist (as can be argued from their flows existing), but they can't stop rotating without loosing their Born-rigidity, much as a non-rotating disk can't start to rotate without doing the same.

    [correction]
    The other interesting result from the paper I cited (the main point of it, in fact) is that the Born-rigid flow turns out to be generated by Killing vector. If we have a disk rotating about the ##\theta## axis, (the metric is ##-dt^2 + dr^2 + r^2 d\theta^2 + dz^2##), and let ##\alpha## be some constant, the velocity field
    ##\partial_t +\alpha \, \partial_{\theta}## generates our Born rigid flow as the flow is the orbit of the above field.

    The linear sum of the two Killing vectors is another Killing vector.

    This velocity field may be more recongnizable as the 3-velocity, whose component in the ##\hat{\theta}## direction is ## r \omega##. Of course, for the vector field to remain time-like, ##r \omega## must be less than the speed of light.
     
    Last edited: May 27, 2013
  16. May 27, 2013 #15

    atyy

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    As WannabeNewton and Pervect say, Gron's "reference frame" is a more usually called a "congruence". The concept is useful in saying things like "the universe is expanding". A famous exposition of GR using conguences of timelike test particles is Baez and Bunn's The Meaning of Einstein's Equation, whose mathy counterpart is the Raychaudhuri equation.
     
  17. May 27, 2013 #16

    Mentz114

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    I've always assumed that the same logic applies if ##B## is calculated in the frame basis, using the tetrad covariant derivative. I've seen this expression ##B_{\hat{a}\hat{b}} = \nabla_{\hat{b}}\xi_{\hat{a}}## where the hats indicate that this is done in the frame basis. The usual decomposition of ##B_{\hat{a}\hat{b}}## gives the rotation, shear and expansion scalar ( which will not change, naturally) in the local frame. Obviously (?) the components of ##\sigma_{ab}## and ##\omega_{ab}## can be different in the frame basis from their holonomic basis counterparts.

    I'll see if I can find any references, but I hope in the meantime that my assumption above holds true.

    I've done the decomposition for the circular geodesic congruence in the Schwarzschild vacuum in the holonomic and the frame basis if an actual example would be helpful.
     
    Last edited: May 27, 2013
  18. May 27, 2013 #17

    Mentz114

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    Thanks, Pervect. I've downloaded the paper you cited and look forward to reading it.
     
  19. May 27, 2013 #18

    WannabeNewton

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    Yeah it's really going to the local Lorentz frame at an event on an arbitrary worldline that I have trouble coming to grips with physically. I'll definitely take a look at the decomposition you mentioned, I'm sure it will help. I assume it's the one linked in your blog post? Thanks again!
     
  20. May 27, 2013 #19

    Mentz114

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    The one in my blog is the frame field for all the Schwarzschild time-like geodesics there is no comparison with the holonomic results. I'll PM you something more concise.
     
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