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

Currently understand by parallel transport

  1. Aug 10, 2010 #1
    This is what I currently understand by parallel transport. The definitions I've read don't talk about it in quite this way, as a vector field, but I think this is equivalent to them:

    Given a tangent vector V0 at some point P0, construct a vector field along an oriented curve P(?), where ? stands for any appropriate parameter, such that (1) the value of the vector field at P0 is V0, (2) there exists some parameter t for which dV(P(t))/dt = 0 at every point along the curve. This vector field is the parallel transport of V0, and a parameter that meets the condition in 2 is called an affine parameter because it can be transformed into any other affine parameter by an affine transformation, that is, a function whose rule takes the form or a linear equation a + bt. Proper time is one example of an affine parameter.

    Now, I've just come across the idea of Fermi-Walker transport in Robert Littlejohn's notes on Thomas precession, linked to from the Wikipedia article on that subject. Fermi-Walker transport is defined by this equation:

    [tex]\frac{\mathrm{d} s}{\mathrm{d} \tau}=-\left ( s \cdot \frac{\mathrm{d} u}{\mathrm{d} \tau}\right )u[/tex]

    which he says is a "parallel transport equation, of which there are many kinds in physics."

    This seems to be a different and broader definition of parallel transport from the one I'd previously encountered. Is one definition more prevalent than the other, or are they really the same in some sense.* If they are different, would people who use the narrower definition say rather: "There are many kinds of transport (of a vector) in physics, including parallel transport and Fermi-Walker transport"? In the dialect where Fermi-Walker transport is a kind of parallel transport, are there any non-parallel transports, and what special name (if any) is given to a transport whose derivative wrt an affine parameter is zero everywhere along the curve?

    *I see the Wikipedia article Fermi-Walker transport mentions a thing called a Fermi derivative which it says is zero for a Fermi-Walker transport. It says this is in connection with a Riemannian manifold, but the examples it gives seem to be from general relativity. Unfortunately it doesn't define the parameter it calls s, with respect to which this derivative is taken [ http://en.wikipedia.org/wiki/Fermi-Walker_transport ], whether this is arc length or any parameter or an affine parameter. Does each kind of parallel transport, in the broader sense, have its own specially defined derivative which must be equal to zero?
     
  2. jcsd
  3. Aug 10, 2010 #2

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Transport

    I've wondered about the same thing Rasalhague is asking, and have puzzled over the same WP articles. Here are some notes I have that record what little I understand about this topic -- which is not very much.

    Fermi transport/Fermi coordinates

    -done by Fermi, 1922, as a young student; "Sopra i fenomeni che avvengono in vinicinanza di una linea ovaria" Atti R. Accad. Lincei Rend. Cl. Sci. Fis. Mat. Nat. , 31 (1922) pp. 21–51

    -Fermi showed it could be done in a locally Euclidean space; was later generalized to a locally Minkowski space

    -is a generalization of Riemann-normal coords

    -define local coordinate axes with gyroscopes, and then let them free-fall

    -all Christoffel coefficients along the geodesic vanish, so whereas Riemann-normal coords make the metric flat at one point, Fermi coordinates make it flat along an entire geodesic

    -a test body's ang. mom. remains constant when measured in Fermi coordinates (obviously, because basis is defined by gyroscopes)

    -describes a nonrotating observer

    Fermi-Walker transport

    -a further generalization; doesn't have to be along a geodesic
     
  4. Aug 10, 2010 #3
    Re: Transport

    Exactly. They are not that much different but one is able to be derived from the other if the time-like curve that we suppose as a background curve to define Fermi-Walker transport is a geodesic. Usually FW transport is useful for conceiving the dynamics of the spin 4-vector [tex]\textbf{S}[/tex]. This is because the transport has a differential operator [tex]F_{u}[/tex] acting on a vector field [tex]\textbf{X}[/tex] along a time-like curve [tex]\xi(\tau)[/tex] whose tangent vector [tex]u^{\nu}[/tex] satisfies [tex]u^{\nu}u_{\nu}=c^2[/tex] with [tex]\tau[/tex] being the curve parameter (proper time) that if we replace this vector field by the spin vector, it yields

    [tex]F_{u}\textbf{S}=0,[/tex]

    where

    [tex]F_{u}=\frac{D{X^{\mu}}}{d\tau}+c^{-2}X_{\nu}\left(\frac{d{X^{\mu}}}{d\tau}\frac{D}{d\tau}\left[\frac{d{X^{\nu}}}{d\tau}\right]-\frac{d{X^{\nu}}}{d\tau}\frac{D}{d\tau}\left [\frac{d{X^{\mu}}}{d\tau}\right]\right).[/tex]

    When setting [tex]X^{\mu}=u^{\mu}=dX^{\mu}/d{\tau}[/tex], this is identically zero as

    [tex]\frac{D}{d\tau}\left (\frac{d{X_{\nu}}}{d\tau}\frac{d{X^{\nu}}}{d\tau}\right)=2\frac{d{X_{\nu}}}{d\tau}\frac{D}{d\tau}\left (\frac{d{X^{\nu}}}{d\tau}\right)=0,[/tex]

    following the fact that [tex]d{X^{\nu}}d{X_{\nu}}=c^2d{\tau}^2.[/tex] So that FW transport of the tangent vector is zero, i.e. [tex]F_u{\textbf{U}}=0.[/tex] Now the definition of FW transport is seen to be as follows: A vector field [tex]\textbf{X}[/tex] is said to be FW-transported along [tex]\xi[/tex] if [tex]F_u{\textbf{X}}=0.[/tex] It is obvious that on a time-like geodesic, [tex]F_{u}=\nabla_{u}[/tex] and thus FW and P transports coincide. Nice to be noted, [tex]F_{u}[/tex] is the simplest linear differential operator that reduces to P transport along a geodesic.

    I hope this helps.

    AB
     
    Last edited: Aug 10, 2010
  5. Aug 10, 2010 #4

    Mentz114

    User Avatar
    Gold Member

    Re: Transport

    Sure does. Thank you.
     
  6. Aug 10, 2010 #5
    Re: Transport

    You're welcome. There is one other interesting thing related to FW transport. Consider all quantities defined in my early post. Now let [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] be two vectors FW-transported along [tex]\xi(\tau).[/tex] The purpose is to measure the change in the scalar quantity [tex]A_{\mu}B^{\mu}[/tex] along the curve using FW operator, [tex]F_{u}[/tex]. First we compute the covariant derivative

    [tex]\frac{D}{d\tau}\left (A_{\mu}B^{\mu}\right)=B_{\mu}\frac{D A^{\mu}}{d\tau}+A_{\mu}\frac{D B^{\mu}}{d\tau}.[/tex]

    From the definition of [tex]F_{u}[/tex], we find

    [tex]\frac{D A^{\mu}}{d\tau}=(F_{u}A)^{\mu}-A_{\alpha}K^{\alpha \mu},[/tex]
    [tex]\frac{D B^{\mu}}{d\tau}=(F_{u}B)^{\mu}-B_{\alpha}K^{\alpha \mu},[/tex]

    where [tex]K^{\alpha \mu}[/tex] is an antisymmetric tensor,

    [tex]K^{\alpha \mu}=u^{\alpha}\frac{D u^{\mu}}{d\tau}+u^{\mu}\frac{D u^{\alpha}}{d\tau}.[/tex]

    Now it is really an easy exercise to get

    [tex]\frac{D}{d\tau}\left (A_{\mu}B^{\mu}\right)=A_{\mu}(F_{u}B)^{\mu}+B_{\mu}(F_{u}A)^{\mu}-A_{\alpha}B_{\mu}K^{(\alpha \mu)}.[/tex]

    Since the symmetric part of an antisymmetric tensor is always zero, [tex]K^{(\alpha \mu)}.=0;[/tex] obtaining

    [tex]\frac{D}{d\tau}\left (A_{\mu}B^{\mu}\right)=A_{\mu}(F_{u}B)^{\mu}+B_{\mu}(F_{u}A)^{\mu}.[/tex]

    This means that the inner product of two vectors remains constant if each of the vectors undergoes the FW transport along a curve. With having this done, we've given away three properties of FW:

    1- [tex]F_{u}=\nabla_u[/tex] if [tex]\xi[/tex] is a geodesic.

    2- [tex]F_{u}\textbf {U}=0[/tex].

    3- Along [tex]\xi[/tex], if [tex]F_{u}\textbf {A}=F_{u}\textbf {B}=0[/tex] then we can conclude that [tex] A^{\mu}B_{\mu}[/tex] is constant.

    AB
     
  7. Aug 10, 2010 #6
    Re: Transport

    (EDIT: I'll leave this post for the record, but it'll probably make more sense to skip straight to my next one, as I think I figured some of this out eventually.)

    Okay, I can see that Richard Littlejohn's equation defining Fermi-Walker transport

    [tex]\frac{\mathrm{d} s}{\mathrm{d} \tau}=-\left ( s \cdot \frac{\mathrm{d} u}{\mathrm{d} \tau} \right ) u[/tex]

    = 0 when the curve is a geodesic because the 4-acceleration will be 0. When you wrote

    [tex]F_{u}=\frac{D{X^{\mu}}}{d\tau}+X_{\nu}\left(\frac{d{X^{\mu}}}{d\tau}\frac{D}{d\tau}\left[\frac{d{X^{\nu}}}{d\tau}\right]-\frac{d{X^{\nu}}}{d\tau}\frac{D}{d\tau}\left [\frac{d{X^{\mu}}}{d\tau}\right]\right)[/tex]

    (I'll drop powers of c for now) did you mean [itex]F_u(X^{\mu}) = ...[/itex]? If so, for the case you mention where [itex]X^\mu=u^\mu=dX^\mu/d \tau[/itex], I get,

    [tex]F_u(u^\mu)=\frac{D{u^{\mu}}}{d\tau}+u_{\nu}\left(u^\mu\frac{D u^\nu}{d\tau}-u^\nu\frac{D u^\mu}{d\tau}\right)[/tex]

    or, using RL's notation b for proper acceleration:

    [tex]b^\mu+(u \cdot b)u^\mu-(u \cdot u)b^\mu=(u \cdot b)u^\mu=0[/tex]

    But under what circumstances would [itex]u^\mu=du^\mu/d \tau[/itex]? This seems to mean: the mu'th component of 4-velocity is equal to its own derivative wrt proper time (at every point along the curve). What would that say about the motion of a particle? Probably I'm misinterpreting this notation, since you say "So that FW transport of the tangent vector is zero..." (in general, presumably), whereas the mu'th component of 4-velocity isn't always equal to its own derivative wrt proper time.

    Would another way of writing the definition of the Fermi derivative be

    [tex]F_us=\frac{\mathrm{d} s}{\mathrm{d} \tau}+(s \cdot b)u[/tex]

    so we can say that a vector field s is FW transported iff

    [tex]\frac{\mathrm{d} s}{\mathrm{d} \tau}+(s \cdot b)u=0[/tex]

    (with minus instead of plus if the opposite sign convention is used, and leaving the door open to the possibility of a more general definition allowing other parameters, analogous to the definitions of the covariant derivative and geodesic motion.) Or is this only a special case?

    In RL's example, s is an everywhere spacelike vector field; is the Fermi derivative defined for any kind of differentiable vector field s? Can u be replaced by any other differentiable vector field defined along a curve, or does it always have to be proper velocity?
     
    Last edited: Aug 10, 2010
  8. Aug 10, 2010 #7
    Re: Transport

    Altabeh, now that I've looked at your post #5, I think I might be able to answer some of my questions. Whether rightly remains to be seen... First, is there a typo in #3, perhaps some of those Xs were meant to be the Greek letter [itex]\xi[/itex], indicating position on the curve, and others upper case X, standing for a general differentiable vector field defined along the curve? And in your definition of K in #5, did you mean to write minus before one of the terms? When both terms have the same sign, K seems to be symmetric rather than antisymmetric. With a minus in there, your easy exercise becomes much easier ;-)

    Comparing your two posts, it looks as if your general definition should be, in the notation of #3,

    [tex](F_u \textbf{X})^\mu = \frac{D X^\mu}{d \tau}+X_\alpha\left ( \frac{\mathrm{d} \xi^\mu}{\mathrm{d} \tau} \frac{D}{d \tau} \left [ \frac{\mathrm{d} \xi^\alpha}{\mathrm{d} \tau} \right ]- \frac{\mathrm{d}\xi^\alpha}{\mathrm{d}\tau} \frac{D}{d \tau} \left [ \frac{\mathrm{d} \xi^\mu}{\mathrm{d} \tau} \right ]\right )[/tex]

    or, in the notation of #5,

    [tex](F_u \textbf{A})^\mu = \frac{D A^\mu}{d \tau}+A_\alpha\left ( u^\mu \frac{Du^\alpha}{d \tau} - u^\alpha \frac{Du^\mu}{d \tau} \right )[/tex]

    that is

    [tex]F_u \textbf{A} = \frac{d \textbf{A}}{d \tau}+(\textbf{A}\cdot \textbf{b})\textbf{u}-(\textbf{A}\cdot \textbf{u})\textbf{b}[/tex]

    where [itex]\textbf{A}[/itex] is any spacetime vector field defined along the curve with proper velocity (4-velocity) [itex]\textbf{u}[/itex] and proper acceleration (4-acceleration) [itex]\textbf{b}[/itex]. Then RL's notes on Thomas precession do deal with a special case, the case where [itex]\textbf{A} \cdot \textbf{u} = 0[/itex] at some point, and therefore all points along the curve.

    That gives us three kinds of transport: dA/dT = 0 (parallel transport), dA/dT + (A.b)u = 0 (maybe we could call it spin transport), and dA/dT + (A.b)u - (A.u)b = 0 (FW transport). As RL says there are many other kinds, I assume this doesn't exhaust the list. Is there a completely general equation defining transport per se, or are there just lots of equations like this defining different kinds, perhaps each with its own differential operator?
     
    Last edited: Aug 11, 2010
  9. Aug 11, 2010 #8
    Re: Transport

    What? That was a huge typo I made! I must have written [tex]X^{\mu}=u^{\mu}=dx^{\mu}/d\tau.[/tex] [tex]X^{\mu}[/tex] is not the coordinates used to describe the curve but a vector field along the curve.

    I'm really sorry for such an inattentive post. Yes this was the case.

    In case your [tex]s[/tex] is a curve parameter, no! The operator doesn't act on a curve parameter but a vector field. For the moment, another way would be in the bold-vector form

    [tex]F_u\textbf{X}= \nabla_u\textbf{X}+c^{-2}\left[ (\textbf{X},\nabla_u\textbf{U})\textbf{U}-(\textbf{X},\textbf{U})\nabla_u\textbf{U} \right].[/tex]
    So this illustrates that [tex]s[/tex] is not a scalar parameter but a vector!? What kind of vector? If it is a vector along a curve with a parameter [tex]\tau[/tex], then there might be a specific constraint that reduces the main equation to this one; otherwise this is nothing but nonsense.

    I have to read RL's paper to continue here. So wait for my reply.

    AB
     
  10. Aug 11, 2010 #9
    Re: Transport

    Good, that's what I'd worked out by the time I wrote my post #7. I was still rather confused in #6, so don't pay too much attention to that; maybe I should have deleted it after all.

    Sorry, I should have explained more. There are so many different notation conventions! By s, if I've understood correctly, he means a 4-vector field defined along a timelike curve with the constraint that its value at each point on the curve is orthogonal to the 4-velocity of the curve at that point. So it turns out his equation is a special case of FW transport as you defined it. Your equation reduces to his when the vector field you called X in #3 is everywhere orthogonal to u because then their dot product is always zero, so the last term disappears.

    On p. 5, RL writes:

    By "conventional rest frame", I think he means an orthonormal basis for the tangent space at some point on the curve chosen so that its timelike basis vector is the 4-velocity at that point (the derivative of the curve wrt proper time), and its spacelike basis vectors (and thus its spatial orientation) are chosen by some arbitrary convention.
     
    Last edited: Aug 11, 2010
  11. Aug 13, 2010 #10
    Re: Transport

    Okay, so provisionally (if my post #7 was right), I think we've sorted out the general definition of FW transport, thanks to Altabeh.

    A linguistic detour paragraph. I notice that people talk about a vector field being transported, apparently meaning that some kind of field generated by a formula involving one or more derivatives (the exact details depending on what sort of trasport this is), whose value at each point depends on the value of the original vector field at that point, has values equal to zero everywhere that it's defined. So a proper velocity field might be said to be parallel transported along a geodesic, rather than (as I'd been thinking of it) an individual vector at some point being parallel transported, i.e. a parallel transport field being generated from that vector, which can then be compared to the proper velocity field and the question asked: are the values of the proper velocity field everywhere equal to the values of the parallel transport field of one (and hence any) value of the proper velocity field. Oh well, I expect I'll get used to the terminology as I read more.

    Returning to the special case of FW transport that RL discusses, the creation of a vector field along a timelike curve, with nonzero acceleration, whose value at each point is orthogonal to the proper velocity of the curve so that

    [tex]\frac{\mathrm{d} s}{\mathrm{d} \tau}=-\left ( s \cdot \frac{\mathrm{d} u}{\mathrm{d} \tau}\right )u[/tex]

    he talks about the values of this vector field as being derived by a "sequence of infinitesimal boosts". Boost can mean a coordinate boost or a boost of the vector itself. I think here it must mean boosts of the vectors themselves, since the values of the proper velocity field differ along the curve. I've been told that infinitesimal means a linear approximation, but I can't see how it can mean that here; how can a physical effect be due to an arbitrary human choice of how to approximate it? I'm going to try to put in my own words what I've learnt and mislearnt, and maybe that will help focus on which bits are confusing me. I've filled in the gaps with some guesses, which may be wildly off the mark, so consider every statement here a question!

    RL seems to be talking about a recipe for generating a vector field, given some initial value.

    (1) s is a continuous vector field, but pretend for now that it's made up of discrete values at separate points along the curve. Suppose we know the value of s at some point. Parameterise the curve by proper time, T, so that that value of s at this point is s(0), and has a zero time component in the instantaneous co-moving inertial rest frame, ICMIF, associated with this point on the curve.

    NOTE: RL says nothing explicitly about this discrete step, but I'm guessing it's implied in his formulas such as s(T + Delta T), since it makes as much sense to say the vector field is generated by the value of this as Delta T goes to 0, as it does to define addition by the rule: x + y = x as y goes to zero. And if Delta T is taken as not zero, it raises the question of what is the "next" real number after some value if T.

    (2) For some mysterious reason (said by various sources to be due to the relativity of simultaneity, but why would a 4-vector, which exists independently of arbitrary coordinate representations, care about the relativity of anything?), it happens that the value of s at every point is orthogonal to the proper velocity of the curve. So, from the starting value of s, which we'll express as (0,s), its coordinates in an ICMIF defined by the inital proper velocity and some arbitrary choice or spatial orientation, we'll use the following iterative rule to generate the rest of the (discrete) values of s: let s(Ti+Ti+1) be obtained by the matrix equation L s(Ti), where L is the pure boost

    [tex]\begin{bmatrix}
    \gamma & \gamma \textbf{v}^T\\
    \gamma \textbf{v} & I+\frac{\gamma -1}{v^2}\textbf{v}\textbf{v}^T
    \end{bmatrix}[/tex]

    with v the relative velocity (3-velocity) of u(Ti+1) in an ICMIF defined by u(Ti) and spacelike basis vectors obtained from the initial spacelike basis vectors by the same interative rule. And I is the 3x3 identity matrix. I am the 3x3 identity matrix ;-)

    Q. Am I right in thinking that the coordinates of s(Ti) in the ICMIF so defined corresponding to u(Ti) will be the same as the coordinates of s(Tj) in the ICMIF so defined corresponding to u(Tj) for all values of i and j from 1 to n, the number of values in this discrete vector field?

    EDIT: A. Must be. Reading on, I see RL deals with this in eq. (5.20). This only concerns spatial coordinates, but we already know the time coordinate is 0.

    Q. For finite n, there will never be any Thomas precession for movement in a complete circle at constant speed?

    (3) Find the limit of the vector fields generated by this process as the interval between neighboring values of T goes to zero and n goes to infinity. This will be the actual continuous vector field s, representing the spin or angular momentum of a gyroscope along a timelike curve with nonzero acceleration. This limiting process results in a vector field of spin vectors whose value at the end of a curve representing motion in a complete circle with constant velocity can somehow be different from the value at the beginning, and will thus have different coordinates in the same inertial reference frame. I don't understand how they can differ, but I can believe that weirdness could creep in where infinities are involved. To find out by how much they differ, integrate... something.
     

    Attached Files:

    • FWT.jpg
      FWT.jpg
      File size:
      8.3 KB
      Views:
      117
    Last edited: Aug 13, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook