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Deriving the metric

  1. Sep 27, 2014 #1
    Hello,

    I have a simple question about deriving $$g_{ij}
    \frac{\partial x^i}{\partial t}\frac{\partial x^j}{\partial t}$$
    with respect to time t.

    I have noticed that the first term after derivation turns out to be$$ \frac{\partial g_{ij}}{\partial x^k} \frac{\partial x^k}{\partial t}\frac{\partial x^i}{\partial t}\frac{\partial x^j}{\partial t}$$

    Does $$ g_{ij} $$ depend on time? If so, why can't we just write $$ \frac{\partial g_{ij}}{\partial t} $$ Why did we plug in the $$x^k$$?
    Thanks!
     
    Last edited: Sep 27, 2014
  2. jcsd
  3. Sep 27, 2014 #2

    Orodruin

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    Generally ##g_{ij}## is a function of all space-time coordinates. Along a world line, the ##t## coordinate is changing just as well as the spatial coordinates. It thus makes a difference if you are differentiating ##g## along a world line or just with respect to the time coordinate. The existence of the coordinate derivatives seem to indicate that you do want to differentiate with respect to a world line in which case the total derivative follows the chain rule:
    $$
    \frac{d}{dt} = \frac{dx^i}{dt}\frac{\partial}{\partial x^i} = \frac{\partial}{\partial t} + \vec v \cdot \vec\nabla.
    $$
     
  4. Sep 27, 2014 #3
    Thank you very much for your reply. You mean that what was supposed to be written is $$\tau$$ instead of t? And thus in this case, I should use the chain rule?
     
  5. Sep 27, 2014 #4

    Orodruin

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    Well, you can also parameterise the world line with the coordinate ##t## (assuming that there is only one point on the world line for every ##t##). It may be that that is simply what is assumed about the parameterisation.
     
  6. Sep 27, 2014 #5

    stevendaryl

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    There are several different types of derivatives used in relativity, and I think your question might be mixing up two different types. If you have a field, which is a quantity (real number, or a vector or a tensor) defined at every point in spacetime, you use partial derivatives or covariant derivatives. If you have a spacetime path [itex]\mathcal{P}(s)[/itex] and you have some quantity defined at each point along the path (where s is the path parameter), then you don't use partial derivatives, since the quantity only depends on one parameter, [itex]s[/itex]. You use a total derivative, [itex]\frac{d}{ds}[/itex]. In nonrelativistic physics, the path variable is typically the time [itex]t[/itex], but in relativistic physics, it is often proper time [itex]\tau[/itex] instead.

    You write [itex]\frac{\partial x^i}{\partial t}[/itex]. But what is [itex]x^i[/itex]? If you mean that it is a coordinate, then what does it mean to take a partial derivative with respect to [itex]t[/itex]? Partial derivatives only make sense for fields--quantities defined at each point in spacetime.

    What I'm guessing that you mean is that you have a path parametrized by [itex]t[/itex] (Do you really mean the time [itex]t[/itex], or do you mean the proper time [itex]\tau[/itex]?). So [itex]x^i(t)[/itex] is the coordinate describing the location along the path at time [itex]t[/itex]. In that case, you don't actually want a partial derivative, you want an ordinary derivative, [itex]\frac{dx^i}{dt}[/itex].

    The other question about what you wrote is what the index [itex]i[/itex] or [itex]j[/itex] means in [itex]x^i[/itex] and [itex]g_{ij}[/itex]. In the mathematics of relativity, people usually combine spatial coordinates and the time coordinate [itex]t[/itex] into [itex]x^\mu[/itex], where, [itex]\mu[/itex] can be 0, 1, 2, or 3, and where [itex](x^0, x^1, x^2, x^3) = (t, x, y, z)[/itex].

    So a more typical quantity that you might encounter in relativity is

    [itex]g_{\mu \nu} \frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}[/itex]

    This expression involves both fields ([itex]g_{\mu \nu}[/itex] is a tensor field) and path quantities ([itex]x^\mu(\tau)[/itex] is a function of a path variable, the proper time [itex]\tau[/itex]). It doesn't actually make sense to take a derivative of this quantity unless you make sense of it as a path quantity, or as a field. The way to make sense of it as a path quantity is to understand that [itex]g_{\mu \nu}[/itex] is supposed evaluated at the point [itex]x^\mu(\tau)[/itex]. So you really have:

    [itex]g_{\mu \nu}(x^\lambda(\tau)) \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}[/itex]

    Now, everything is a function of [itex]\tau[/itex] and we can take a derivative with respect to [itex]\tau[/itex]. To take a derivative of [itex]g_{\mu\nu}(x^\lambda(\tau))[/itex] with respect to [itex]\tau[/itex], you have to use the chain rule:

    [itex]\frac{d}{d\tau} g_{\mu\nu}(x^\lambda(\tau)) = (\frac{\partial}{\partial x^\lambda} g_{\mu \nu}) \frac{d x^\lambda}{d\tau}[/itex]
     
    Last edited: Sep 27, 2014
  7. Sep 27, 2014 #6
    Thank you very much for making this clearer!
     
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