How to read the geodesic equation?

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Main Question or Discussion Point

In the formula : ##\frac{d^2 x^\mu}{d\tau^2}=-\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}##

How is the ##x^\mu## understood : a 4-vector or the ##\mu##-st component simply ?

If it is a vector, how to write it in spherical coordinate with extra time dimension ?

Btw: why does one not solve EFE in 3 space timensions and then solve the geodesic equation with ##\tau\rightarrow t## ?
 

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  • #2
PeterDonis
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How is the understood
As a 4-vector function of ##\tau##, the proper time along the worldline. Then ##dx^\mu / d\tau## and ##d^2 x^\mu / d\tau^2## are 4-vectors giving the first and second derivatives of that 4-vector function with respect to ##\tau##. In other words, the equation you wrote is really four equations, one for each ##\mu = (0, 1, 2, 3)##.
 
  • #3
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Yes but the ith component of the second derivative of the vector is not the second derivative of the ith component, i.e. the ##\theta## component of ##\ddot{\vec{r}}## is not ##\ddot{\theta}## for example in spherical coordinates.

If one sees it 4 dimensionally then ##\vec{r}=r\vec{e}_r=r(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)^T## is replaced by for example ##\vec{x}=r(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta\cosh\alpha,\cos\theta\sinh\alpha)^T## ?
 
  • #4
PeterDonis
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the ith component of the second derivative of the vector is not the second derivative of the ith component
No. That is not what is going on here. All four components are functions of the parameter ##\tau##. So the 4-vector ##x^\mu## is just 4 functions of ##\tau##; and the 4-vectors ##d x^\mu / d\tau## and ##d^2 x^\mu / d \tau^2## are just 4-tuples of the first and second derivatives of those 4 functions with respect to ##\tau##.
 
  • #5
PeterDonis
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##\vec{r}=r\vec{e}_r##
No, that is not what the 4-vector ##x^\mu## is. It is just 4 functions of ##\tau##. It is not a vector in the sense you appear to be thinking. In spherical coordinates the 4 functions would just be the 4 spherical coordinates as functions of ##\tau## instead of the 4 Cartesian coordinates as functions of ##\tau##.

Not all texts use the term "4-vector" in this context; perhaps that would be a better idea to avoid confusion. Unfortunately there is no other standard term for the thing that the notation ##x^\mu## denoting 4 functions of ##\tau## refers to. Possibly "4-tuple" would work.
 
  • #6
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I am confused, it it ##\frac{d(x^\mu)}{d\tau}## or ##\left(\frac{dx}{d\tau}\right)^\mu## ? It seems it lacks parentheses.

In fact I want to know what ##x^1## is : is it the 1st component of ##(ct,r,\theta,\phi)## or of ##(ct,r\cos(\phi)\sin(\theta),...)## ?
 
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  • #7
vanhees71
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As a 4-vector function of ##\tau##, the proper time along the worldline. Then ##dx^\mu / d\tau## and ##d^2 x^\mu / d\tau^2## are 4-vectors giving the first and second derivatives of that 4-vector function with respect to ##\tau##. In other words, the equation you wrote is really four equations, one for each ##\mu = (0, 1, 2, 3)##.
It's not four-vectors but four-vector components, and it's
$$u^{\mu} = \mathrm{d}_{\tau} x^{\mu}$$
but
$$a^{\mu} = \mathrm{D}_{\tau} u^{\mu} = \mathrm{d}_{\tau} u^{\mu} + {\Gamma^{\mu}}_{\nu \rho} u^{\nu} u^{\rho},$$
i.e., it's the covariant derivative of the vector components wrt. to the affine parameter ##\tau## (for massive particles usually one chooses ##\tau## to be the proper time, defined by the constraint ##g_{\mu \nu} u^{\mu} u^{\nu}=c^2##.
 
  • #8
DrGreg
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I am confused, it it ##\frac{d(x^\mu)}{d\tau}## or ##\left(\frac{dx}{d\tau}\right)^\mu## ? It seems it lacks parentheses.

In fact I want to know what ##x^1## is : is it the 1st component of ##(ct,r,\theta,\phi)## or of ##(ct,r\cos(\phi)\sin(\theta),...)## ?
If you're working in spherical polar coordinates ##(ct,r,\theta,\phi)## then
$$
\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} = \left(
\frac{\mathrm{d}(ct)} {\mathrm{d}\tau},
\frac{\mathrm{d}r} {\mathrm{d}\tau},
\frac{\mathrm{d}\theta} {\mathrm{d}\tau},
\frac{\mathrm{d}\phi} {\mathrm{d}\tau}
\right) \, ,
$$
so the answer to your question is ##x^1=r##.

If you're working in cartesian coordinates ##(ct,x,y,z)## then
$$
\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} = \left(
\frac{\mathrm{d}(ct)} {\mathrm{d}\tau},
\frac{\mathrm{d}x} {\mathrm{d}\tau},
\frac{\mathrm{d}y} {\mathrm{d}\tau},
\frac{\mathrm{d}z} {\mathrm{d}\tau}
\right) \, .
$$
 
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  • #9
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Let say it is like above. Then I tried to solve EFE in 3 spatial dimension only, with the unknown metric being spherically symmetric :
$$g_{ij}=diag(A(r),B(r),C(r)\sin^2\theta)$$

I got

##C(r)=\sqrt{2\beta+4r}##
##B(r)=\alpha C(r)##
##A(r)=\alpha C'(r)^2/C(r)##

Now is it a way to compute the geodesic for ##\dot{\theta}=\dot{\phi}=0 ## and to make the approximation for weak field (large r) being the Newton equation of motion ?

In doing in 3+0 dimensions then I thought that :

1) timelike loops are avoided
2) there is only one time and not coordinate vs. proper time
3) one can do find the transformation of coordinate from ##\mathbb{R}(3,0)\rightarrow\mathbb{R}(3,1)## s.t. the curved metric is obtained by scalar product of the basis vectors (exterior geometry like for surfaces but for a 3d space embedded in 4-spacetime and not 5-spacetime like in (3,1) metric.


I would like to seek for courses that do that way (if it exists?).
 
  • #10
PeterDonis
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Then I tried to solve EFE in 3 spatial dimension only
This makes no sense. Spacetime is 4-dimensional, not 3-dimensional.

Also, are you asking about the Einstein Field Equation, or the geodesic equation? They're not the same. Your OP in this thread asked about the geodesic equation along a timelike curve; but any timelike curve will have only one point in a spacelike 3-surface, so looking at "3 spatial dimensions" tells you nothing at all about the timelike curve except for a single point on it.

I suggest keeping this thread focused on the geodesic equation. If you want to ask questions about the EFE you should start a separate thread. But first think carefully about what "spacetime is 4-dimensional, not 3-dimensional" means.

In doing in 3+0 dimensions then I thought that :

1) timelike loops are avoided
It's perfectly possible to have a spacelike 3-surface in a 4-dimensional spacetime that has closed timelike curves.

there is only one time
No, there is no time in a spacelike 3-surface.

one can do find the transformation of coordinate from s.t. the curved metric is obtained by scalar product of the basis vectors (exterior geometry like for surfaces but for a 3d space embedded in 4-spacetime and not 5-spacetime like in (3,1) metric.
You are very confused. (3, 1) spacetime is not embedded in anything; it's just a curved 4-dimensional manifold with (3, 1) signature. The curvature of (3, 1) spacetime is intrinsic, not extrinsic. Curvature has nothing to do with transformations of coordinates or scalar products of basis vectors.
 
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  • #11
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If you're working in spherical polar coordinates ##(ct,r,\theta,\phi)## then
$$
\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} = \left(
\frac{\mathrm{d}(ct)} {\mathrm{d}\tau},
\frac{\mathrm{d}r} {\mathrm{d}\tau},
\frac{\mathrm{d}\theta} {\mathrm{d}\tau},
\frac{\mathrm{d}\phi} {\mathrm{d}\tau}
\right) \, ,
$$
so the answer to your question is ##x^1=r##.

If you're working in cartesian coordinates ##(ct,x,y,z)## then
$$
\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} = \left(
\frac{\mathrm{d}(ct)} {\mathrm{d}\tau},
\frac{\mathrm{d}x} {\mathrm{d}\tau},
\frac{\mathrm{d}y} {\mathrm{d}\tau},
\frac{\mathrm{d}z} {\mathrm{d}\tau}
\right) \, .
$$
If this were the case then with the 3dimensional above computed metric we would get :

$$\ddot{\theta}=\frac{1}{2\beta+4r}\dot{\theta}\dot{r}-\sin\theta\cos\theta\dot{\phi}^2/\alpha$$

Would this mean a change of orientation of the ecliptic plane ?(i never heard of such a phenomenon)
 
  • #12
PeterDonis
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with the 3dimensional above computed metric
Which, as I said, makes no sense. You should not be doing anything further with it; it will only mislead you even more than you are already misled.
 
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