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I am getting slightly confused regarding the formal definitions of different four vectors in General Relativity. Many texts on relativity begin with four vectors and dynamics in Minkowski space, and often don't make clear definitions hold in curved space as well. Specifically:

  1. The four velocity seems to be defined as $u^\mu = \frac{dx^\mu}{d\tau}$ (for example in this post in GR. Why is this not defined using the covariant derivative, as in $$u^\mu = \frac{Dx^\mu}{D\tau} = \frac{dx^a}{d\tau}\nabla _a x^\mu = \frac{dx^a}{d\tau}\delta_a^\mu + \Gamma ^\mu _ {ac}x^a\frac{dx^c}{d\tau}~?$$ Is $u^\mu = \frac{dx^\mu}{d\tau}$ tangent to the worldline and the second definition is not in curved space? Why is this so? I don't see why we would use one definition over another (i.e. what we want the definition to correspond to)

  2. On the other hand, the four-acceleration is $a^\mu = \frac{Du^\mu}{D\tau}$ with the covariant derivative?

  3. And I believe the definitions of the four momentum and four force as $p^\mu = mu^\mu$ and $f^\mu = ma^\mu$ are consistent with $\frac{Dp\mu}{D\tau} = f^\mu$ from the definition of $a^\mu$ but not with $\frac{dp\mu}{d\tau} = f^\mu$. Again, I don;t see why one of these is preferrable to the other, or what they correspond to.

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    $\begingroup$ Covariant derivatives are applied to tensors to get new tensors. $x^\mu$ is not a tensor so you shouldn't be taking covariant derivatives. $\endgroup$ – jacob1729 Jan 21 at 19:36
  • $\begingroup$ @jacob1729 $x^\mu$ is a tensor. Any four vector satisfies the tensor transformation law (it is a rank 1, type (1,0) tensor). $\endgroup$ – 21joanna12 Jan 21 at 21:11
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    $\begingroup$ A bunch of comments removed. Don't argue in the comments, people. If someone's said something that you can't reply to politely, flag it for moderator attention and move on to a different part of the internet for a while. $\endgroup$ – rob Jan 24 at 16:16
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I have never seen definitions for four-velocity in terms of covariant derivative. Don't even know what it means (what is your source).

It would seem that you are confusing notation.

So let us separate coordinates from functions. Lets say that position on the manifold ($\mathcal{M}$) is given by four coordinates $x^{(i)}$. Also lets say that we have a world-line which is given by four equations ($\tau$ is proper time):

$x^{(0)}-A\left(\tau\right)=0$

$x^{(1)}-B\left(\tau\right)=0$

$x^{(2)}-C\left(\tau\right)=0$

$x^{(3)}-D\left(\tau\right)=0$

With functions $A: \mathbb{R}\to\mathbb{R}$, etc. The four-velocity of this world-line is then given by:

$u^0=\dot{A}$

$u^1=\dot{B}$

$u^2=\dot{C}$

$u^3=\dot{D}$

How would you define four-velocity in terms of covariant derivative here, bearing in mind that $A,B,C,D$ are maps from reals into reals.

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First of all, $x^\mu$ is not a vector, so it does not make sense to write $\nabla_\nu x^\mu$. For example, consider spherical coordinates, there you have $x^0 = t$, $x^1 = r$, $x^2 = \vartheta$, $x^3 = \varphi$. The 4-tuple $(t,r,\vartheta,\varphi)$ surely does not satisfy any property that would allow you to call it a vector such as its transformation with a Jacobian matrix. In flat space-time, there are Cartesian coordinates in which the 4-tuple $x^\mu_{\rm C}$ formally has the same transformation properties as a vector with respect to the linear transformation not shifting the origin. However, 1) it does not behave as a vector with respect to any other coordinate transformation, and 2) there is generally no Cartesian system of coordinates in General relativity anyway.

As for $d x^\mu/d\tau$. One has to understand that tangents to parametrized curves are what defines the differential-geometric term vector! In other words, when we require something to "transform as a vector", we require that it "transforms as $d x^\mu/d \tau$". There is really no more fundamental definition of something being covariant, it has to vary the same way as $d x^\mu/d \tau$ (and its duals, tensor products...) so that we can write meaningful equations involving dynamics of bodies. It then does not come as a large surprise that it does not make sense to "covariantize" what already is already the definition of covariance.

Now we have the space of vectors $d x^\mu/ d\tau$ at every point. When computing $d x^\mu/d \tau$, we simply took an ordinary derivative of the 4-tuple $x^\mu (\tau)$. However, it turns out that if we take another derivative, the 4-tuple $d^2 x^\mu/d \tau^2$ does not transform as $d x^\mu/d \tau$ and these then cannot be written into a single equation. What is more, $d^2 x^\mu/d \tau^2$ does not even transform with a linear matrix, and the equations we would be able to write with it would be some weird nonlinear laws. It turns out this is true for an ordinary derivative of any tensor. We thus must introduce the notion of a covariant derivative of a tensor, so that $D V^\mu/d \tau$ transforms the same as $d x^\mu/ d\tau$ (and any $D T^{\mu...}_{\nu...}/d \tau$ as tensor + dual products of $d x^\mu/ d\tau$).

This answers your questions about $p^\mu = m u^\mu$. $p^\mu$ is a vector, and if we want to make a derivative of it and put it in an equation with another vector, we must use the covariant derivative.

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Coordinates are not tensors. Under a general coordinate transformation, a diffeomorphism, they do not obey the required property. Derivatives of scalars and differential elements do obey the correct relationship. One should treat the coordinates as an n-tuple of scalar functions. As pointed out in the other answer, each coordinate is a map from R-->R (or at least for a subset of each).

The transformation of coordinates would look like x^mu(x^nu'(lambda)), for arbitrary affine parameter, lambda. That's as far as it goes. Differentials will however transform like a tensor.

dx^mu = {d(x^mu)/d(x^nu')} * dx^nu'

And for this reason velocity is a tensor quantity.

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