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Time measured by a clock if the metric is not Minkowski

  1. May 25, 2015 #1
    Starting from the general expression of the metric in coordinates, what is the time misured by a clock in a non inertial reference sistem?
     
  2. jcsd
  3. May 25, 2015 #2
    In general relativity i can t say that the time measured by the clock is simply the differential dt in the expression of the metric, because it changes under diffeomorfisms and this means that it is not measurable.
     
  4. May 25, 2015 #3

    wabbit

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    The clock measures proper time, which is in essence the metric itself ##\tau=\int ds=\int\sqrt{g_{ij}dx^idx^j}## (using c=1 units).
     
  5. May 25, 2015 #4

    bcrowell

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    (This is for the +--- signature; for -+++, there would be an extra minus sign inside the square root.)

    You don't need a frame of reference. General relativity doesn't even have frames of reference except locally. The expression wabbit gave is valid for any coordinates whatsoever.
     
  6. May 26, 2015 #5
    Just adding to what wabbit said:
    You can also parametrize the total elapsed time in terms of an affine parameter ##\lambda## , in which case the integral becomes $$τ= \int \sqrt{-g_{\mu ν} \frac {dx^{\mu}}{d \lambda} \frac {dx^{ν}}{d \lambda} } d \lambda $$ (-+++ signature)
     
  7. May 26, 2015 #6
    I still have problem in defining the time in a curved space-time.
    What i read from wabbit is the expression of the proper time in a general coordinates, but this is only the time measured by a free falling observer, what i need is a more general definition of time, not only for free falling observers.
    Anyway if that expression is true, what rapresents the differential of dx-mu (dt, dx, dy, dz) related to the clock in that reference system (locally)?
    I need an operative way to the define the time of the clock starting from that expression: if i can t say that the dx mu is physical, what do they rapresent?
     
  8. May 26, 2015 #7

    pervect

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    I'm not sure what was unsatisfactory about the previous answers - I can only think that the tensor notation may be throwing you.

    Let's start with a reviewe of special relativity.

    In special relativity, for any clock (in free fall or not) we can write the proper time, which we will denote as ##\tau##, as a function of coordinate space and time (we will use t,x,y, and z for our coordinates) as follows.

    ##d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2##

    It's a differential equation, hence the dx, dy, dz, dt. To get the total proper time along a curve we just need to integrate ##\int dt## along the curve.

    In general relativity, let us write the analogous formula for a diagonal metric. For any clock, in free fall or not, we write:

    ##d\tau^2 = g_{00}dt^2 - g_{11}dx^2 - g_{22}dy^2 - g_{33}dz^2##

    So it's the same in GR as it is in SR, except that we've introduced the metric coefficcients ##g_{ii}##

    We will now introduce the tensor notation that may (or may not) have been an issue, while still considering only a diagonal metric. We replace dt, dx, dy, dz with numbered coordinates for notational convenience, so that ##dt = dx^0, dx=dx^1, dy=dx^2, dz=dx^3##.

    ##d\tau^2 = g_{00} (dx^0)^2 - g_{11} (dx^1)^2 - g_{22} (dx^2)^2 - g_{33} (dx^3)^ 2##

    Now we write the general metric without restricting it to a diagonal metric:

    ##d\tau^2 = \sum \sum g_{ij} dx^i dx^j##

    The summation is considered to be over the range i,j = 0..3. We usually omit the summation signs, they are implied along with the range

    We also have relax the restrictions on what sort of coordinates we allow in GR - in SR, it is understood that a unit change in coordinate represents a unit change in time or distance. In GR, there is no such requirement on the coordinates.
     
  9. May 26, 2015 #8

    Ibix

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    Apologies - my grasp of the maths of this is not complete, so my notation may be flawed. However, perhaps another way to look at it: [itex]dx^\mu/d\lambda[/itex] represents a (small) change in position along some path parameterised by [itex]\lambda[/itex]. It doesn't mean an inertial movement - at least, not necessarily. To insist that it is an inertial movement requires that the [itex]dx^\mu/d\lambda[/itex] also satisfy the geodesic equation.

    Wabbit's equation is valid for any time-like path (taking into account comments about the metric signature). Only by adding further constraints does it become about inertial motion only.
     
  10. May 26, 2015 #9

    Nugatory

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    Wabbit's (and PWiz's equivalent) integral is not limited to free-falling observers. It works just fine for any clock moving on any time-like worldline, whether free-fall or not.

    [Rats! Beaten by Ibix!]
     
  11. May 26, 2015 #10

    wabbit

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    As I understand it, it goes the other way round : the clock time is a physical quantity, and the metric is the tensor that allows to compute any clock's proper time in any coordinate system (the existence of such a tensor is not obvious at all, but in GR it does exist).

    While the ## x^{\mu} ## are generic coordinate functions with no meaning, ## g_{\mu\nu} ## is defined so as to express in these coordinates a physically meaningful quantity, the line interval, which is not a lot more than another name for proper time along a clock's worldline (inertial or not, as others have stressed).
     
  12. May 26, 2015 #11

    Ibix

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    Great minds think alike. And I do, too. :wink:
     
  13. May 26, 2015 #12
    I don't want to repeat what everyone else here is saying, but just to be sure: ##g_{\mu ν}## is not equivalent to the diagonalized Minkowski metric tensor (which is frequently represented by ##η_{\mu ν}##).

    ##g_{\mu ν}## is the metric tensor specific to the situation, and it reduces to the Minkowski metric when the Riemann tensor is 0 ("flat" spacetime), but can take more complicated (even non-diagonalized) forms when there is non-zero intrinsic spacetime curvature. The relationship between the differential elements ##dx^{\mu}## is held in the metric tensor, and the equations posted by me and Wabbit will hold in all scenarios involving timelike paths, provided you use the metric tensor ##g_{\mu ν}## which corresponds to that scenario (it "takes care" of everything else).
     
  14. May 26, 2015 #13

    George Jones

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    I am not sure what you mean. There isn't a conflict between diffeomorphisms and proper time.
     
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