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Spin precession Killing fields

  1. Jun 22, 2014 #1

    WannabeNewton

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    Hello. I was going through the problems in the angular momentum chapter of the text "Problem Book in Relativity and Gravitation"-Lightman et al in preparation for a project and I came upon part (c) of problem 11.10. It basically starts with a family of observers with 4-velocity ##u^{\mu} = (-\xi_{\gamma}\xi^{\gamma})^{-1/2}\xi^{\mu}##, where ##\xi^{\mu}## is a time-like Killing field, and a Lorentz frame ##\{e_a \}## Lie transported along ##\xi^{\mu}##, and asks to show that the angular velocity of comoving gyroscopes relative to this frame is given by ##\omega_{\text{gyro}}^{\alpha} = -\frac{1}{2}(\xi_{\gamma}\xi^{\gamma})^{-1}\epsilon^{\alpha\beta\mu\nu}\xi_{\beta}\nabla_{\mu}\xi_{\nu}##.

    It's the requirement that ##\xi## be a Killing field that I would like to discuss. Assume that Lie transport ##\mathcal{L}_{\xi}e_a =0## holds (which is physically equivalent to the condition that spatial axes constructed by any observer in the family remain fixed to the same neighboring observers in the family for all of his proper time); denote by ##e^{\mu}{}{}_a## the coordinate components of ##e_a##. Under what conditions will ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## hold? Well ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = \mathcal{L}_{\xi}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = e^{\mu}{}{}_a e^{\nu}{}{}_b\mathcal{L}_{\xi}g_{\mu\nu}## so ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## if and only if ##\xi^{\mu}## is a Killing field given that ##\mathcal{L}_{\xi}e_a = 0##. I bring this up because for this physical setting we require ##\mathcal{L}_{\xi}e_a = 0## so on top of this if we also want ##\{e_a \}## to be a Lorentz frame then the Lie transport along ##\xi^{\mu}## has to preserve the condition ##g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b = \eta_{ab}## and as we've seen, under said Lie transport, the necessary and sufficient condition for this is that ##\xi^{\mu}## be a Killing field.

    So we've established that we need ##\xi^{\mu}## to be a Killing field at minimum in order for the desired physical setup to remain consistent. But is that all we need ##\xi^{\mu}## to be a Killing field for? The reason I ask is, Lightman et al provide their own (rather long) calculation in the solutions showing that ##\omega_{\text{gyro}}^{\alpha} = -\frac{1}{2}(\xi_{\gamma}\xi^{\gamma})^{-1}\epsilon^{\alpha\beta\mu\nu}\xi_{\beta}\nabla_{\mu}\xi_{\nu}## wherein, on top of making use of a handful of various relations from relativistic kinematics, they also explicitly make use of the fact that ##\xi^{\mu}## is a Killing field.

    To me however this seems completely unnecessary so I was hoping someone could check my (rather short) calculation, which makes no use of the fact that ##\xi^{\mu}## is a Killing field, and see if it has any fundamental errors in it:

    We start with the relation ##u^{\beta}\nabla_{\beta}e^{\alpha}{}{}_a = -\Omega^{\alpha}{}{}_{\beta}e^{\beta}{}{}_a## where ##\Omega_{\alpha\beta} = a_{\alpha}u_{\beta} - u_{\alpha}a_{\beta} + \epsilon_{\alpha\beta\mu\nu}u^{\mu}\omega^{\nu}## (see MTW exercise 6.8); here ##\omega^{\alpha}## is the precession angular velocity of ##\{e_i \}## relative to comoving gyroscopes, which is (minus) the quantity we're after.

    Using Lie transport ##\xi^{\beta}\nabla_{\beta}e^{\alpha}{}{}_a =e^{\beta}{}{}_a \nabla_{\beta}\xi^{\alpha}##, and defining ##\varphi \equiv (-\xi_{\gamma}\xi^{\gamma})^{-1/2}## we then have ##\varphi e^{\beta}{}{}_a\nabla_{\beta}\xi^{\alpha} = -\Omega^{\alpha}{}{}_{\beta}e^{\beta}{}{}_a##; contracting both sides with ##e_{\gamma}{}{}^a## and using ##e_{\gamma}{}{}^a e^{\beta}{}{}_a = \delta_{\gamma}^{\beta}## (see Carroll appendix J) we get [tex]\varphi \nabla_{\beta}\xi_{\alpha} = - \Omega_{\alpha\beta} = -a_{\alpha}u_{\beta} + u_{\alpha}a_{\beta} - \epsilon_{\alpha\beta\mu\nu}u^{\mu}\omega^{\nu}[/tex] hence [tex]\varphi \epsilon^{\alpha\beta\delta \lambda}\nabla_{\beta}\xi_{\alpha} = -\varphi\epsilon^{\delta\lambda \alpha\beta}\nabla_{\alpha}\xi_{\beta} = 2\epsilon^{\alpha\beta\delta\lambda}u_{\alpha}a_{\beta} + 2u^{\delta}\omega^{\lambda} - 2u^{\lambda}\omega^{\delta}[/tex] where I've used ##\epsilon^{\alpha\beta\delta \lambda}\epsilon_{\alpha\beta\mu\nu} = -4\delta^{[\delta}_{\mu}\delta^{\lambda]}_{\nu}##.

    Contracting both sides with ##u_{\lambda}## yields ##-\varphi^2\epsilon^{\delta\lambda \alpha\beta}\xi_{\lambda}\nabla_{\alpha}\xi_{\beta} = 2\omega^{\delta}## since ##\epsilon^{\alpha\beta\delta\lambda}u_{\lambda}u_{\alpha}a_{\beta} = \omega^{\lambda}u_{\lambda} = 0## and ##u^{\lambda}u_{\lambda} = -1##.
    This yields ##\omega^{\delta} = \frac{1}{2}(\xi^{\gamma}\xi_{\gamma})^{-1}\epsilon^{\alpha\beta\mu\nu}\xi_{\beta}\nabla_{\mu}\xi_{\nu}## as desired.

    Are there any flaws in the solution? Thanks in advance.
     
    Last edited: Jun 22, 2014
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  3. Jun 22, 2014 #2

    Matterwave

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    Looking over your solution carefully, I personally don't see a flaw in it, other than the mismatch of indices at the end (you should have an ##\omega^\alpha## instead of an ##\omega^\delta##).

    The part that I didn't quite follow was the statement:

    $$\epsilon^{\alpha\beta\delta\lambda}u_\lambda u_\alpha a_\beta=\omega^\lambda u_\lambda = 0$$

    Can you explain this conclusion?
     
  4. Jun 23, 2014 #3

    WannabeNewton

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    Oops haha.

    Sorry, I wrote it in a confusing way. What I meant was ##\epsilon^{\alpha\beta\delta\lambda}u_\lambda u_\alpha a_\beta=0## and ##\omega^\lambda u_\lambda = 0##; the first expression is zero since it's a contraction of two antisymmetric indices from ##\epsilon^{\alpha\beta\delta\lambda}## with two symmetric indices from ##u_\lambda u_\alpha## and the second expression is zero by definition of ##\omega^{\lambda}## as a 4-rotation.

    Thanks for checking it over! If there are no flaws in it then I guess the solution provided by Lightman et al for the problem was simply more complicated than need be.
     
  5. Jun 23, 2014 #4

    Matterwave

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    Ah...I was trying to figure out how the first expression reduced to the second one for a while...:P

    Yea, I don't see any errors in your derivation. Maybe someone else will check too.
     
  6. Jun 23, 2014 #5

    WannabeNewton

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    Thanks! One thing that was unsettling to me was the equality ##\varphi \nabla_{\beta}\xi_{\alpha} = -\Omega_{\alpha\beta}## because ##\Omega_{\alpha\beta}## is antisymmetric so the equality would only make sense if ##\nabla_{\beta}\xi_{\alpha}## was antisymmetric as well, which is equivalent to it being a Killing field. So it seemed like I made a mistake somewhere in going from ##\varphi\xi^{\beta}\nabla_{\beta}e^{\alpha}{}{}_a = -\Omega^{\alpha}{}{}_{\beta}e^{\beta}{}{}_a## to ##\varphi \nabla_{\beta}\xi_{\alpha} = -\Omega_{\alpha\beta}## because nowhere in between did I explicitly require ##\xi^{\mu}## to be a Killing field.

    However I think it's ok because in going from one expression to the other I made use of ##\mathcal{L}_{\xi}e_a = 0##; furthermore, since the transport equation ##\varphi\xi^{\beta}\nabla_{\beta}e^{\alpha}{}{}_a = -\Omega^{\alpha}{}{}_{\beta}e^{\beta}{}{}_a## preserves ##g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b = \eta_{ab}## i.e. ##\xi^{\beta}\nabla_{\beta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0##, as is easy to verify***, and since ##\xi^{\beta}\nabla_{\beta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) =\mathcal{L}_{\xi}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b)##, we necessarily have ##\mathcal{L}_{\xi}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## but as I noted in post #1, ##\mathcal{L}_{\xi}e_a = 0## combined with ##\mathcal{L}_{\xi}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## necessarily requires ##\xi^{\mu}## to be a Killing field so the equality ##\varphi \nabla_{\beta}\xi_{\alpha} = -\Omega_{\alpha\beta}## seems consistent. Does that sound alright to you?


    ***
    ##\varphi \xi^{\beta}\nabla_{\beta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = \varphi g_{\mu\nu}(e^{\nu}{}{}_b\xi^{\beta}\nabla_{\beta}e^{\mu}{}{}_a + e^{\mu}{}{}_a\xi^{\beta}\nabla_{\beta}e^{\nu}{}{}_b) \\= -e^{\mu}{}{}_b\Omega_{\mu\beta}e^{\beta}{}{}_a -e^{\mu}{}{}_a\Omega_{\mu\beta}e^{\beta}{}{}_b = -\Omega_{\mu\beta}e^{\beta}{}{}_a e^{\mu}{}{}_b + \Omega_{\mu\beta}e^{\beta}{}{}_a e^{\mu}{}{}_b = 0##
     
  7. Jun 23, 2014 #6

    Matterwave

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    I am not entirely familiar with the requirement that ##\mathcal{L}_\xi e_a=0## and ##\mathcal{L}_\xi (g_{\mu\nu}e^\mu_{}{}_a e^\nu_{}{}_b)=0## imply that ##\xi## is a killing vector (meaning that you can only Lie drag a vierbien along Killing fields?), but if this is so, then your derivation uses implicitly this fact when you assumed that you can Lie-transport your tetrad while also keeping that tetrad a frame field (when you imposed ##e_{\gamma}{}{}^a e^{\beta}{}{}_a = \delta_{\gamma}^{\beta}##). So I think your requirement that ##\xi## is a Killing field is built in to your derivation.

    That's the way I see it anyways.
     
    Last edited: Jun 23, 2014
  8. Jun 23, 2014 #7

    WannabeNewton

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    I showed that in post #1, near the top. ##0 = \mathcal{L}_\xi (g_{\mu\nu}e^\mu_{}{}_a e^\nu_{}{}_b)= e^\mu_{}{}_a e^\nu_{}{}_b(\mathcal{L}_{\xi}g)_{\mu\nu} = (\mathcal{L}_{\xi}g)_{ab} = 0##. But if ##\mathcal{L}_{\xi}g## vanishes in one frame then it must vanish in all so that tells us that ##\xi## is a Killing field.
     
  9. Jun 23, 2014 #8

    Matterwave

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    Yes, I did see that. I just intuitively never thought about it until now. :P

    But it does seem legit to me.

    Thinking about this in a physical way it makes sense. If your field is not a Killing field, then the metric along it will not be constant. This means that your tetrad cannot be Lie dragged along this field since a tetrad vector must have unit norm.

    To confirm, you explicitly used this fact when you Lie dragged the tetrad ##\{e_a\}## along ##\xi## meaning you assumed that ##[e_a,\xi]=0,\quad\forall a##.
     
    Last edited: Jun 23, 2014
  10. Jun 24, 2014 #9

    WannabeNewton

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    It's actually even more restrictive than that. For example, consider why it isn't enough for ##\xi^{\mu}## to simply be Born-rigid in order to satisfy ##\mathcal{L}_{\xi}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## when we have ##\mathcal{L}_{\xi}e_a =0##. Born-rigidity means that ##\mathcal{L}_{\xi}h =0## where ##h_{\mu\nu} = g_{\mu\nu} + \xi_{\mu}\xi_{\nu}##; intuitively this means of course that the geometry of the infinitesimal Einstein simultaneity surface of ##\xi^{\mu}## is invariant under its flow i.e. the "spatial geometry" relative to ##\xi^{\mu}## is invariant under its flow. Furthermore ##\mathcal{L}_{\xi}e_a =0## means that connecting vectors between neighboring observers following orbits of ##\xi^{\mu}## will always remain pointing at the same neighbors.

    We see then that if Born-rigidity and Lie transport hold, ##\mathcal{L}_{\xi}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = \mathcal{L}_{\xi}(\xi_{\mu}\xi_{\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = \mathcal{L}_{\xi}(e^{\mu}{}{}_a (e_b)^{||}_{\mu})## where ##(e_b)^{||}## is the projection of ##e_b## onto ##\xi^{\mu}##. So the issue here is, even if a vector is initially in the infinitesimal simultaneity surface of ##\xi^{\mu}## at an initial event, Born-rigidity does not guarantee that this vector will remain in that surface under the flow of ##\xi^{\mu}##; in the above the "angle" between ##e_a## and ##(e_b)^{||}## (which is parallel to ##\xi^{\mu}##) will not remain constant in general which prevents ##\{e_a \}## from being a Lorentz frame because the axes of a Lorentz frame cannot start precessing relative to one another. So it's not so much an issue of length but rather the "angle" from the infinitesimal simultaneity surface, which a Born-rigid vector field cannot preserve but a Killing field can.

    Anyways, thanks again!
     
  11. Jun 24, 2014 #10

    Matterwave

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    Hmmm, definitely food for thought. :D
     
  12. Jun 25, 2014 #11

    WannabeNewton

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    Actually I don't even think the solution in Lightman et al is correct. They start with the relation ##\frac{d}{d\hat{t}}S^{\hat{i}} = \epsilon^{\hat{i}\hat{j}\hat{k}}\omega_{\hat{j}}S_{\hat{k}}## in the Lorentz frame ##\{e_\hat{\alpha} \}## (which is ##\{e_a \}## in our notation) and they then write this covariantly as ##\frac{d}{d\tau}S^{\hat{\beta}} = \epsilon^{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}}u_{\hat{ \alpha }}\omega_{\hat{\gamma}}S_{\hat{\delta}}##.

    But this expression makes no sense to me because it requires that ##u_{\hat{\beta}}\frac{d}{d\tau}S^{\hat{\beta}} = -\frac{d}{d\tau}S^{\hat{0}} = 0## in the Lorentz frame which is certainly not true in general (https://www.physicsforums.com/showpost.php?p=4783148&postcount=6). See the attachment. What do you think?
     

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  13. Jun 25, 2014 #12

    George Jones

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    Along a worldline, what derivative restricts to ##d/d\tau##, the covariant derivative or the Fermi derivative, i.e., along a worldline does ##d/d\tau = \nabla \mathbf{u}##, or does ##d/d\tau = F_\mathbf{u}##?
     
  14. Jun 25, 2014 #13

    WannabeNewton

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    Isn't ##\frac{d}{d\tau} = \nabla_u## always true for the covariant derivative along a world-line? For example in p.49 of Straumann one has in the local rest frame ##\nabla_u S = (\frac{dS^0}{dt},0)## and in the local rest frame ##\frac{dS^0}{dt} = \frac{dS^0}{d\tau}## so presumably ##\frac{d S}{d\tau} = \nabla_u S## even when speaking of Fermi-Walker transport. On p.53 of the same book, the expression ##\frac{dS^i}{d\tau} = \epsilon^{ijk}S_j \omega_k## is given and given the above notation this would mean ##\nabla_u S^i = \epsilon^{ijk}S_j \omega_k##.

    Also in p.171 of MTW, the authors write, for Fermi-Walker transport of a vector ##v^{\mu}##, the expression ##\frac{d v^{\mu}}{d\tau} = (u^{\mu}a^{\nu} - u^{\nu}a^{\mu})v_{\nu}## so presumably again ##\nabla_u v^{\mu} = \frac{d v^{\mu}}{d\tau}## even when speaking of Fermi-Walker transport.
     
  15. Jun 25, 2014 #14

    George Jones

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    That's what I thought. If this is true, then, in index-free notation, you have show using symmetry/antisymmetry ##\mathbf{u} \cdot \left( \nabla_\mathbf{u} \mathbf{S} \right) = 0##, but the last equation from your book image gives ##\mathbf{u} \cdot \left( \nabla_\mathbf{u} \mathbf{S} \right) = - \mathbf{S} \cdot \mathbf{a}##.

    If, however, ##d/d\tau = F_\mathbf{u}##, then

    $$\mathbf{u} \cdot \left( F_\mathbf{u} \mathbf{S} \right) = \mathbf{u} \cdot \left( \nabla_\mathbf{u} \mathbf{S} \right)_\bot = 0.$$
     
  16. Jun 25, 2014 #15

    WannabeNewton

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    Yeah but directly above that they write ##\frac{d}{d\hat{t}}S^{\hat{\beta}} = \epsilon^{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}}u_{\alpha} \omega_{\hat{\gamma}} S_{\hat{\delta}}## and since ##\hat{t} \equiv \tau## in the local rest frame of the spinning particle, this would require ##u_{\hat{\beta}}\frac{d}{d\tau}S^{\hat{\beta}} = 0## which is certainly not true so isn't their claim that ##\frac{d}{d\hat{t}}S^{\hat{\beta}} = \epsilon^{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}}u_{\alpha} \omega_{\hat{\gamma}} S_{\hat{\delta}}## wrong?
     
  17. Jun 25, 2014 #16

    George Jones

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    Aha!

    What happens when ##\hat{\beta} = \hat{0}##?
     
  18. Jun 25, 2014 #17

    WannabeNewton

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    The RHS vanishes since ##\epsilon^{\hat{0}\hat{\alpha}\hat{\gamma}\hat{\delta}}u_{\hat{\alpha}} = -\epsilon^{\hat{0}\hat{0}\hat{\gamma}\hat{\delta}} = 0##, which would imply ##\frac{d}{d\hat{t}}S^{\hat{0}} = 0## but that's also not true in general :frown:
     
  19. Jun 25, 2014 #18

    George Jones

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    I am going to use the notation of the book.

    This is consistent with ##\mathbf{S}## being a purely spatial vector, i.e., if ##S^{\hat{0}} = 0## at any instant. then it stays zero.

    More after I get home.
     
  20. Jun 25, 2014 #19

    WannabeNewton

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    Thanks George but I don't think ##\frac{d}{d\tau}S^{\hat{0}} = 0## is true. What's true, as far as the spin being a purely spatial vector goes, is ##\frac{d}{d\tau}(S_{\hat{\alpha}}u^\hat{{\alpha}}) = 0## so that if ##S_{\hat{\alpha}}u^\hat{{\alpha}} =0## initially on the world-line then ##S_{\hat{\alpha}}u^\hat{{\alpha}}=0## everywhere along the world-line, but this doesn't necessarily mean ##\frac{d}{d\tau}S^{\hat{0}} = 0## in the local rest frame (https://www.physicsforums.com/showpost.php?p=4783148&postcount=6).

    In fact, if when you get home you have access to Straumann, check out p.49. Straumann works in the local rest frame and derives the relation ##\frac{dS^0}{dt} = \langle S, a \rangle## where all quantities defined are with respect to the local rest frame (so ##t \equiv \tau## in Straumann's notation) so certainly ##\frac{dS^0}{dt}= 0## is not true in general.
     
  21. Jun 25, 2014 #20

    George Jones

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    At home now. Left Straumann in my office.

    Okay, I want you to do something very difficult. I want you to pretend that the abstract index doesn't exist. I want to follow the problem book's notational conventions and interpretations. Isn't the problem book's expression the derivative of a component, not the component of a derivative, i.e., ##\frac{d}{d\tau}\left(S^\hat{\beta} \right)##, not ##\left(\frac{d \mathbf{S}}{d\tau}\right)^\hat{\beta}##?

    Set ##\mathbf{A} = \frac{d}{d\tau}\left(S^\hat{\beta} \right) \mathbf{e}_\hat{\beta}##. Then ##u_\hat{\mu} A^\hat{\mu} = 0##, but ##u_\hat{\beta} \left(\frac{d \mathbf{S}}{d\tau}\right)^\hat{\beta} \neq 0##.

    Straumann takes ##S^\hat{0} = 0##, and thus ##\frac{d}{d\tau}\left(S^\hat{0} \right) = 0##.
     
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