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Transformation to local inertial frame

  1. Jan 22, 2014 #1
    I've been working on a problem that I can't seem to get started on. Here is how it is posted:

    Metric of a space is:
    [itex] ds^2 = (1+2\phi^2)dt^2 - (1-2\phi)(dx^2+dy^2+dz^2)[/itex], where [itex] |\phi | << 1 [/itex] everywhere. Given a point [itex](t_0 , x_0 , y_0, z_0)[/itex] find a coordinate transformation to a locally inertial frame to first order in [itex]\phi[/itex]. At what rate does this frame accelerate with respect to the original coordinates (to first order in [itex]\phi[/itex])?

    So far I know that I have to find a transformation that transforms the metric to the Minkowski metric, [itex]\eta_{ab}[/itex] so that [itex]ds^2 = \eta_{ab}dx'^a dx'^b [/itex] but I'm not sure how to get started on this.

    I'm studying this myself so I have no instructor to ask so hopefully someone can point me along the right geodesic in this space of confusion. ;)
     
  2. jcsd
  3. Jan 22, 2014 #2

    Mentz114

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    If you start with an ansatz that the transformation (##{\lambda^a}_\hat{a}##) is diagonal with components ##a,b,c,d## on the diagonal then

    ##{\lambda^a}_\hat{a}{\lambda^b}_\hat{b}g_{ab}= \lambda.g.\lambda^T = diag(a^2g_{00},b^2g_{11},c^2g_{22},d^2g_{33})##
     
  4. Jan 22, 2014 #3
    Thank you for your reply. :) Swift action in here!

    I am not sure I understand your notation, so let me make a quick check to see if I got it right:
    [itex] {\lambda^a}_\hat{a} [/itex]: The matrix that transforms coordinates ##x^a## to what I denoted as primed coordinates ##x'^a##?

    The middle part of your equation: ##\lambda.g.\lambda^T## - what does the . mean? I'm used to ##g## meaning the determinant of the matrix ##g_{ab}##, so does ##\lambda## have the same meaning or it is simply short notation for the ##{\lambda^a}_\hat{a}##-matrix?

    The left side of your equation is the transformation of the entire metric to the new coordinate system? The metric as seen from this new coordinate system (the local inertial frame) should be Minkowskian such that (again in my notation with primes for the local inertial frame) ##g_{a' b'} = \eta_{a' b'}##. Using this and your equation I get

    ##\eta_{a' b'} = diag(a^2g_{00},b^2g_{11},c^2g_{22},d^2g_{33})## from which I have four equations for the variables ##a,b,c,d##.

    From this I get that
    ##a = \frac{1}{\sqrt{1+2\phi}}## and
    ##b=c=d=\frac{1}{\sqrt{1-2\phi}}.##

    I have not used the limitations on ##\phi## yet, so I could do an expansion of these fractions in ##\phi## and keep only the first order terms as requested in the problem.
     
    Last edited: Jan 22, 2014
  5. Jan 22, 2014 #4
    Hm, this leaves us with a global transformation transforming the entire of spacetime to Minkowskian which I take it is not the purpose of the problem.

    We might have to assume (as was assumed in a previous problem in the set I'm working with but not stated explicitly in this problem) that ##\phi = \phi(t,x,y,z)## is a function of the coordinates.
     
  6. Jan 22, 2014 #5

    Mentz114

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    Yes, it does. I don't understand the problem, actually. I can understand finding a local Minkowski space at a point on a parameterised curve, but I've never seen a position vector used in curved space.
     
  7. Jan 22, 2014 #6
    Using a Taylor expansion up to and including first order to keep stuff linear I get

    ##\phi(t,x,y,z) \approx \phi^0 + (t-t_0)\phi^0_{,t} + (x-x_0)\phi^0_{,x} +(y-y_0)\phi^0_{,y} +(z-z_0)\phi^0_{,z}##.

    Putting this into the metric I get

    ##ds^2 \approx (1+2\phi^0)dt^2 - (1-2\phi^0)(dx^2+dy^2+dz^2) +
    2\left[(t-t_0)\phi^0_{,t} + (x-x_0)\phi^0_{,x} +(y-y_0)\phi^0_{,y} +(z-z_0)\phi^0_{,z} \right]dt^2 ##
    ## + 2\left[(t-t_0)\phi^0_{,t} + (x-x_0)\phi^0_{,x} +(y-y_0)\phi^0_{,y} +(z-z_0)\phi^0_{,z} \right](dx^2+dy^2+dz^2)##
    where ##\phi^0## means ##\phi(t_0,x_0,y_0,z_0)##.

    The first part looks like the original metric evaluated at the point ##(t_0,x_0,y_0,z_0)## but the rest of it doesn't look pretty at all. All the coordinates are mixed up so the metric is not diagonal anymore.
     
  8. Jan 22, 2014 #7

    WannabeNewton

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    You're making this much too complicated I'm afraid. The exercise is much simpler than that.

    Let ##\{x^{\mu'}\}## be the coordinates of the local inertial "frame" (more on the use of the word "frame" here after you solve the exercise). At ##p = (t_0,x_0,y_0,z_0)##, the coordinates will transform as ##x^{\mu'} = (\delta^{\mu'}{}{}_{\nu} + \omega^{\mu'}{}{}_{\nu})x^{\nu} + O(\phi^2)## where ##\omega^{\mu'}{}{}_{\nu}\sim \phi##. Now you know that ##\Gamma^{\mu'}_{\nu'\gamma'}(p) = 0## since ##\{x^{\mu'}\}## is a local inertial frame. Take it from there.
     
  9. Jan 22, 2014 #8
    I hate it when I miss something obvious and somehow end up making stuff way more complicated than it needs to be - and I already find relativity quite complicated! :(

    Maybe I've stared at the problem for so long that I miss something here, but I have some questions:
    ##(\delta^{\mu'}{}{}_{\nu} + \omega^{\mu'}{}{}_{\nu})## is the transformation matrix between the two coordinate systems and is what I'm suppose to find. How do I know that ##\omega^{\mu'}{}{}_{\nu}\sim \phi##? I don't recognize the structure of ##(\delta^{\mu'}{}{}_{\nu} + \omega^{\mu'}{}{}_{\nu})x^{\nu} + O(\phi^2)## except that the first term resembles a coordinate transformation with the stuff in () being the transformation matrix, but I guess the general structure is an expansion of something due to the higher-order term we are leaving out.

    The Christoffel symbols in the primed coordinates vanish at point P due to the transformation to a Minkowskian metric. The equation ##\Gamma^{\mu'}_{\nu'\gamma'}(p) = 0## will relate the derivatives of the original metric with respect to the new coordinates evaluated at point P.

    I need more coffee...
     
  10. Jan 22, 2014 #9

    WannabeNewton

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    ##\omega^{\mu'}{}{}_{\nu}## is what you have to find. It contains the first order deviations of the coordinate transformation from the identity matrix, which is what what we would have if the gravitational potential ##\phi## vanished identically. In our case it doesn't vanish but is very small when compared to unity so we can expect the coordinate transformation to be a perturbation of the identity matrix to first order in ##\phi##; this first order perturbation to the identity, in terms of ##\phi##, is represented by ##\omega^{\mu'}{}{}_{\nu}##. In order to find ##\omega^{\mu'}{}{}_{\nu}## you have to cleverly use the fact that ##\Gamma^{\mu'}_{\nu'\gamma'}(p) = 0## to your advantage.
     
  11. Jan 22, 2014 #10

    Bill_K

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    Isn't an infinitesimal coordinate transformation more like x'μ = xμ + ξμ(x)? And I think you'd need to work directly with the components of the metric rather than the Christoffel symbol, since gμν = ημν is a stronger condition than Γσμν = 0.
     
  12. Jan 22, 2014 #11
    Thank you for clearing that up. :) I've spent the last 40 minutes or so trying to grasp how to "cleverly use the fact that ##\Gamma^{\mu'}_{\nu'\gamma'}(p) = 0##" but I have yet to see the light. :(

    So far all I have is the formula for the Christoffel symbols:
    ##\Gamma^{a'}_{b'c'} = \frac{g^{a'd'}}{2}\left( g_{d'b',c'} - g_{b'c',d'} + g_{d'c',b'}\right) = 0## but I'm at a loss on how to move on from here. I figure that the metric seen in the primed system of coordinates should be ##\eta_{ab}## but that only makes the statement ##\Gamma^{a'}_{b'c'} = 0## trivial. I might miss something here.
     
  13. Jan 22, 2014 #12
    I now notice that my original metric is can be written as
    ## \bar{g}_{ab} = \eta_{ab} + h_{ab}## with
    ##h_{ab} = diag(2\phi,-2\phi,-2\phi,-2\phi)## and that the transformation matrix I am after will satisfy
    ##\eta_{ab} = A_a^c A_b^d \bar{g}_{cd}##
     
  14. Jan 22, 2014 #13

    WannabeNewton

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    See here: http://www.mth.uct.ac.za/~peter/Peter_Dunsby/General_Relativity_files/tut5.pdf [Broken]

    ##g_{\mu'\nu'}(p) = \eta_{\mu'\nu'}## does not make the statement ##\Gamma^{\mu'}_{\nu'\gamma'}(p) = 0## trivial because the former condition only holds at a single point, namely ##p##, meaning ##g_{\mu'\nu'}## can deviate from ##\eta_{\mu'\nu'}## in a neighborhood of ##p##; ##\Gamma^{\mu'}_{\nu'\gamma'}(p) = 0## depends on the behavior of ##g_{\mu'\nu'}## in a neighborhood of ##p## so ##g_{\mu'\nu'}(p) = \eta_{\mu'\nu'}## isn't enough to conclude that ##\Gamma^{\mu'}_{\nu'\gamma'}(p) = 0##.
     
    Last edited by a moderator: May 6, 2017
  15. Jan 22, 2014 #14
    Wow, thanks for the link! It helps to know the transformation law for the Christoffel symbols.

    So, by using the fact that ##\Gamma^a_{bc} = \frac{\partial^2 x'^i}{\partial x^b \partial x^c}\frac{\partial x^a}{\partial x'^i}## I was able to find expressions for the two fractions involved. Using the notation from the link you provided I got:
    ##\frac{\partial x^a}{\partial x'^i} = \left(\delta^i_a + L^i_a + L^i_{c,a}x^c \right)^{-1}##
    and
    ##\frac{\partial^2 x'^i}{\partial x^b \partial x^c} = L^i_{b,c} + L^i_{c,b} + L^i_{d,bc}x^d##
    Combining these two I get
    ##\Gamma^a_{bc} = \left(L^i_{b,c} + L^i_{c,b} + L^i_{d,bc}x^d\right)\left(\delta^i_a + L^i_a + L^i_{c,a}x^c\right)^{-1}##

    Now if this was an expansion around the origin ##x^d = x^c = 0## and I could use the symmetry on ##L^i_{b,c}## to get a factor of two and ignore the denumerator because I want ##\phi## to the first order and I could then integrate and get the matrix. However, I still have some terms that involve those pesky ##x^d## and ##x^c## that I am not really sure what to do with.

    Maybe I should start putting in the mentioned Christoffel symbols and see what I end up with...
     
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