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Apologies in advance if some of the questions below seem overly simple.

In an introductory GR book, I find the following expression for the autoparallel of the affine connection (the upper bound of the summation operators should be 3, for brevity's sake I reduce it to 1; also I leave the summation operators explicit because the Einstein convention still freaks me out a little):

$$\ddot{x}^{\mu} +\sum_{\rho=0}^{1}\sum_{\sigma=0}^{1} \Gamma_{\rho \sigma}^{\mu}\dot{x}^{\rho}\dot{x}^{\sigma}~=~0\qquad\qquad\qquad Eq. 1$$

Where

$$\Gamma_{\rho \sigma}^{\mu}~=~\sum_{\nu=0}^1\frac{\partial x^{\mu}}{\partial x^{\nu}}\frac{\partial^2 x^{\nu}}{\partial x^{\rho} \partial x^{\sigma}} \qquad\qquad\qquad Eq. 2$$

So Eq. 1 written out more explicitly would be:

$$\ddot{x}^{\mu} +\left(\frac{\partial x^\mu}{\partial x^0}\frac{\partial^2 x^{0}}{\partial x^{0}\partial x^{0}}+\frac{\partial x^{\mu}}{\partial x^{1}}\frac{\partial^2 x^{1}}{\partial x^{0}\partial x^{0}}\right)\dot{x}^{0}\dot{x}^{0} $$ $$+ \left(\frac{\partial x^\mu}{\partial x^0}\frac{\partial^2 x^{0}}{\partial x^{1}\partial x^{1}}+\frac{\partial x^{\mu}}{\partial x^{1}}\frac{\partial^2 x^{1}}{\partial x^{1}\partial x^{1}}\right)\dot{x}^{1}\dot{x}^{1}~=~0.$$

1) Did I write this out correctly?
2) Particularly, am I right in assuming the off-diagonal terms (eg. $\frac{\partial x^\mu}{\partial x^\nu}\frac{\partial^2 x^{\nu}}{\partial x^{0}\partial x^{1}}\dot{x}^{0}\dot{x}^{1}$) are all zero, since the dimensions are orthogonal to each other?
3) Could I have saved space by writing $\partial_\nu x^\mu$ instead of $\frac{\partial x^\mu}{\partial x^\nu}$ and $\partial_{\rho \sigma}x^\nu$ instead of $\frac{\partial^2 x^{\nu}}{\partial x^{\rho}\partial x^{\sigma}}$?

Then, about fifteen pages later, I find the following expression for the geodesic:

$$\ddot{x}^{\mu} +\sum_{\rho=0}^{1}\sum_{\sigma=0}^{1}\Gamma_{\rho \sigma}^{\mu} \dot{x}^{\rho}\dot{x}^{\sigma}~=~\frac{d \ln{L}}{d \tau}\dot{x}^\mu \qquad\qquad\qquad Eq. 3$$

Where

$$\Gamma_{\rho \sigma}^{\mu}~=~\frac{1}{2}g^{\mu \nu}\left(\frac{\partial g_{\nu \sigma}}{\partial x^\rho}+\frac{\partial g_{\nu \rho}}{\partial x^\sigma}-\frac{\partial g_{\rho \sigma}}{\partial x^\nu} \right)\qquad\qquad\qquad Eq. 4$$

($L$ is the Lagrangian and $\tau$ is the parameter, but this is not really important for my present questions)

4) I am having trouble seeing how Eq. 2 and Eq. 4 are the same. Would appreciate it if someone could point me in the right direction.

5) Eqs. 1 and 3 are not single equations but 4 by 4 systems of equations, right?

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1) No one I think would bother to tell you whether you can add variables correctly. 2) Understand exactly what the Christoffel coefficient is and the derivatives inside it. 3) Yes. 4) Know what the metric is. 5) You tell me... what's $\mu$? –  Chris Gerig Dec 28 '11 at 3:04
    
@Chris: of course your input is appreciated but if you're answering the question, it's best to post it as an answer, not as a comment. –  David Z Dec 28 '11 at 3:37
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3 Answers 3

up vote 2 down vote accepted

The reason for your confusion, and the answer to question 4, is that the first equation you give (although you copied it missing crucial primes on the indices or on the variables that serve to distinguish between the x and x' coordinate labels, as Lubos Motl pointed out) is the geodesic equation for a flat space-time in nonlinear coordinates, while the second form of the equation tells you the geodesics in a curved space-time with an arbitrary metric.

These are not a 4 by 4 system of equations, but a system of only 4 equations, since there is only one free (not summed over) index. Please do not be freaked out by the Einstein convention--- it is simple to add the summations explicitly as you did, and if you know how to do this correctly, you are not confused, since there is nothing more to it. The Einstein convention is the only simple way of keeping track of covariant/contravariant quantities, tensor products, and the covariant linear maps allowed between them.

It is good to begin in two dimensions, as you did. The explicit equations are (writing out the sum in full):

$$ \ddot{x}^0 + \Gamma^0_{00} \dot{x}^0 \dot{x}^0 + \Gamma^0_{01} \dot{x}^0 \dot{x}^1 + \Gamma^0_{10} \dot{x}^1\dot{x}^0 + \Gamma^0_{11} \dot{x}^1\dot{x}^1 = F^0 $$ $$ \ddot{x}^1 + \Gamma^1_{00} \dot{x}^0 \dot{x}^0 + \Gamma^1_{01} \dot{x}^0 \dot{x}^1 + \Gamma^1_{10} \dot{x}^1\dot{x}^0 + \Gamma^1_{11} \dot{x}^1\dot{x}^1 = F^1 $$

Where $F^0, F^1$ are an additional non-gravitational force, which I will omit (because it doesn't matter for the gravitational force, you understand that separately--- this extra force is in the equation in the nonhomogenous case).

These are two equations, not a 2 by 2 system of equations (question 5), and in 4 dimensions they are 4 equations, not a 4 by 4 system (although they can be a "4 by 4 system" if you mean that you need to solve 4 linear equations to find the accelerations in a new coordinate, see below). Also, note that the cross terms appear, they are not zero, and this answers question 2, although also note that the two cross terms are equal (because the $\Gamma$ is symmetric in the lower two indices for the usual metric connection), so that you can add the middle two terms in both equations, but you don't want to do this, so that the explicit Einstein contractions in the equation are made obvious.

The answer to question 3 is yes, you can use this convention, with two caveats: * make sure you know what the derivative means in all cases, in the case you give first, it is differentiating one coordinate with respect to another coordinate (see below). * Make sure that you don't confuse the quantity ${\partial y^\mu \over \partial x^\nu} = \partial_\nu y^\mu = y^\mu_{,\nu} $ with the inverse quantity ${\partial x^\mu \over \partial y^\nu} = M^\mu_\nu $ where $M^\mu_\nu$ is the inverse matrix to $y^\mu_{,\nu}$.

It is easy to lose sight of the meaning of the equations in the soup of symbols and formal relations, and this is no good. To keep the meaning of everything straight, you need a few examples in the back of your head, and these are useful anyway.

Explicit example

Consider 2d spacetime in the regular x,t rectangular coordinantes, and define the new coordinates

$$t'=t$$ $$ x'= x-t^2$$

These coordinates are useless, the center of the x'-coordinate is accelerating to the right with a constant Galilean acceleration, but they are sufficiently not special to explain equation 1. The inverse map is simple

$$ t = t' $$ $$ x = x' + t'^2 $$

In the original rectangular system of coordinates, the geodesic equation is

$$\ddot{x} = 0$$ $$\ddot{t} = 0$$

You can find the new equations just by substitution (but be sure to understand what this means--- you are substituting the x,t coordinate values as a function of the proper $\tau$ in terms of the x',t' coordinates of the same trajectory).

$${d^2\over d\tau^2} (x' + t'^2) = \ddot{x}' + 2t' \ddot{t}' + 2 \dot{t}'^2 = 0$$ $$\ddot{t'} = 0$$

So you see that to solve for the t' and x' accelerations, you need to invert the coefficients of the second derivative terms, which form the Jacobian matrix of the coordinate transformation, and the coefficients of the first derivative squared terms are the second derivatives of the coordinate transformation. The result is silly n this case:

$$ \ddot{x}' = - 2 \dot{t}'^2 $$ $$ \ddot{t}' = 0 $$

Which solves to

$$ t = a \tau $$ $$ x= b \tau - a^2 \tau^2 $$

which is as expected, it is the transformation of a straight line.

It is easy to work out the general case from this.

General Linear/quadratic transformations

If you look near a point x,t, you can define the general quadratic change of variables

$$ x' = ax + bt + P x^2 + 2Qxt + R t^2$$ $$ t' = cx + dt + S x^2 + 2Txt + U t^2 $$

With constants a,b,c,d P,Q,R,S,T,U. You should study this, because it is the general case! Any coordinate transformation is linear plus quadratic to second order near a point, and the cubic terms are not important for determining the transformation law of the connection coefficients $\Gamma$.

You write this more formally in symbol-soup as

$$ x^{\mu'} = {\partial x^{\mu'} \over \partial x^\mu} x^\mu + {1\over 2} {\partial ^2 x^{\mu'} \over \partial x^{\alpha} \partial x^{\beta} } x^{\alpha} x^{\beta} $$

Where the quantities ${\partial x^{\mu'} \over \partial x^\mu}$ are just the (a,b;c,d) coefficients, and the second derivative quantities are $A,C,B,P,Q,R$. Remember that these are just some numbers at every point. This thing is valid to second order near x=0 (which maps to x'=0 under the coordinate change--- this can be arranged just by additively shifting the x' coordinate).

You can formally write the inverse map too (remember that here and above, primed indices or primed coordinate refer to the new coordinates, while unprimed indices or unprimed coordinates are the old ones).

$$ x^{\mu} = {\partial x^{\mu} \over \partial x^{\nu'} } x^{\nu'} + {1\over 2} {\partial ^2 x^{\mu} \over \partial x^{\nu'} \partial x^{\sigma'}} x^{\nu'} x^{\sigma'} $$

Then you can solve for the new form of the geodesic equation by taking the second $\tau$ derivative along a curve $x^{\mu}(\tau)$, setting it to zero, and inverting the linear coefficients multiplying the second derivatives to solve for the second derivatives, as above (except easier, because you only need to do it near the one point x=0).

$$ {\partial x^\mu \over \partial x^{\mu'} } \ddot{x}^{\mu'} + {\partial^2 {x^\mu}\over \partial x^{\alpha'} \partial x^{\beta'} } \dot{x}^{\alpha'} \dot{x}^{\beta'}$$

If you solve this "system of equations" (I don't like this distinction--- 4 equations are 4 equations, whether the leading linear coefficients are explicitly diagonal or not) for the accelerations using the formal inverse matrix to ${\partial x^\mu \over \partial x^{\mu'}}$, you get the form of the Christoffel symbols in the new coordinates, as your book has it.

The meaning of these formal equations is only fully clarified with appropriate examples, which you should provide for yourself.

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But if the cross terms appear, doesn't that imply that the dimensions are not orthogonal to each other? And if the dimensions are not orthogonal, then why bother with the square root in the original Lagrangian (from which the geodesic is derived) since in that case $\mathcal{L}=\sqrt{g_{\mu \nu}\dot{x}_\mu\dot{x}_\nu}=g_{\mu \nu}\dot{x}_\mu\dot{x}_\nu$? Thanks –  ben Jan 1 '12 at 15:21
    
@ben: The cross terms in the metric , those say that the different coordinate directions are not orthogonal. I don't understand the rest of the question. What is the problem with the square root? The length is the square root even when the directions are not orthogonal, an the second equality you write is impossible--- something with a square root is never equal to something without. –  Ron Maimon Jan 1 '12 at 16:47
    
If I understand correctly, (in 2 dimensions) $g_{\mu\nu}\dot{x}_\mu\dot{x}_\nu=\left( \frac{ds}{d\tau}\right )^2=\left(\frac{ds}{dx_0}\dot{x}_0+\frac{ds}{dx_1}\dot{x}_1\right )^2=\left(\frac{ds}{dx_0}\dot{x}_0 \right )^2+2\frac{ds}{dx_0}\frac{ds}{dx_1}\dot{x}_0\dot{x}_1+\left(\frac{ds}{dx_1}\dot{‌​x}_1 \right )^2$ (Continued below...) –  ben Jan 1 '12 at 18:29
    
(continued) My main concern is that by taking the square root you sort of "undo" the metric tensor: $\sqrt{g_{\mu\nu}\dot{x}_\mu\dot{x}_\nu}=\sqrt{\left( \frac{ds}{d\tau}\right )^2}=\sqrt{\left(\frac{ds}{dx_0}\dot{x}_0+\frac{ds}{dx_1}\dot{x}_1\right )^2}=\frac{ds}{dx_0}\dot{x}_0+\frac{ds}{dx_1}\dot{x}_1$ –  ben Jan 1 '12 at 18:30
    
@ben: your equation for $({ds\over d\tau})^2$ makes no sense. The quantity s is not a function of $x_0$ and $x_1$. You are writing gibberish. The quantity under the square root in your second comment is meaningless, and this is not the source of the metric tensor. –  Ron Maimon Jan 2 '12 at 8:25
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Equation (2) is in general not correct as it is written in the question(v5). Recall that the Christoffel symbols $\Gamma^{\lambda}_{\mu\nu}$ do not transform as a tensor under a general coordinate transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ from an $x$ chart to a $y$ chart. If the Christoffel symbols $\Gamma^{(x)\lambda}_{\mu\nu}=0$ vanish in the $x$ chart, then the Christoffel symbols $\Gamma^{(y)\lambda}_{\mu\nu}$ in the $y$ chart read

$$ \Gamma^{(y)\lambda}_{\mu\nu} ~=~ \frac{\partial y^{\lambda}}{\partial x^{\sigma}} \frac{\partial^2 x^{\sigma}}{\partial y^{\mu} \partial y^{\nu}}. \qquad (2^{\prime})$$

I suspect that eq. $(2^{\prime})$ is the correct version of eq. $(2)$; see also related comments by Ron Maimon and Luboš Motl.

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Ah yes, it is as you and Ron Maimon have pointed out. I left out a bar over the x's (where you have $x^\sigma$). Thanks. –  ben Dec 29 '11 at 15:46
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1) No, you didn't understand the equation correctly. The "dot" above a variable means $\partial / \partial \tau$, a partial derivative with respect to a parameter labeling points along the paths: the equation you wrote is the equation for a geodesic. It contains no derivatives with respect to $x^\mu$ whatsoever: it only differentiates the functions $x^\mu(\tau)$ of a single variable $\tau$ with respect to $\tau$.

2) No, you can't make any "diagonalization" assumption here because of the previous point. The derivatives $\partial_\mu x^\nu$ would be $\delta^\nu_\mu$ i.e. diagonal. But no such things appear in the geodesic equation. If they did appear, one could radically simplify the expressions. None of the terms in the geodesic equation is "diagonal".

3) No, you can't write $\partial_\mu x^\nu$ as the Kronecker delta function in this equation because nothing of the sort appears in the equation in the first place. Also, if it did appear somewhere, the second derivatives $\partial_\lambda \partial_\mu x^\nu$ would be not just diagonal: they would all universally vanish (derivatives of a constant are equal to zero).

4) No, equations 2 and 4 aren't "the same". They describe two equations involving the same object but an object in maths and physics doesn't have to enter just one equation. For example, $13=18-5$ and $13=39/3$ are two equations that aren't "the same" even though they both involve $13$. Equation 2 is a geodesic equation for a path while equation 4 is a formula for the connection in terms of the metric tensor and its derivatives. Moreover, your equation 2 is completely nonsensical because of your misunderstandings in all the points above. More generally, you can't extract all the ($4^3$) components of $\Gamma_{\lambda\mu}^\nu$ out of a single geodesic equation (which only has $4$ components) because the lower indices are being summed over.

5) No, equations 1 and 3 are not systems of $4 \times 4$ equations. They're systems of $D$ equations where $D$ is the space or spacetime dimension. In the large spacetime around us, $D=4$ but not $D=4\times 4$. However, your formulae with the explicit summing over indices $\sum_{\mu=0}^1$ indicate that your spacetime dimension is $D=2$ (the indices have two possible values, namely $0$ and $1$). So it's neither sixteen, nor four, it's two. Alternatively, you have completely misunderstood what the values of the indices are as well and they should have been summed over 4 possible values, not 2. Hard to tell. At any rate, your statement isn't right.

Apologies that I didn't find a single thing that you did or understood correctly but you were really far from this outcome.

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Regarding 1) I understand what the dotted x's are about, and as far as I can tell I am not differentiating w.r.t. $x^\mu$. Also, according to my book, Eq. 1 is almost, but not quite, the geodesic: "If the rhs of [Eq. 3] were zero, then this equation would be formally identical to [Eq. 1] of an autoparallel of an affine connection." But anyways yes I am very lost, especially if the metric tensor, as you suggest, is not a diagonal matrix. –  ben Dec 28 '11 at 13:33
    
And it is confusing that you say Eq. 2 is completely nonsensical since I copied it verbatim from the book. Although perhaps I have left out some crucial context by which it would make more sense to you. –  ben Dec 28 '11 at 13:40
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@ben: equation 2 gives the Christoffel symbols after a coordinate transformation away from a flat space in rectangular coordinates, where the symbols are zero. These coordinate changes make the geodesic equation only superficially nontrivial, the solutions are still straight lines at constant velocity. –  Ron Maimon Dec 28 '11 at 16:37
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Dear @Ben, eqn 2, as you wrote it, is nonsensical because the right hand side is tautologically zero - second derivatives of $x$ with respect to another component of the same $x$. Maybe you omitted some primes for a different coordinate system? When you wrote that you "rewrite equation 1 more explicitly", you suddenly lose all derivatives with respect to $\tau$ and produce lots of partial derivatives with respect to $x$. In my opinion, this means that you don't understand that the dot means differentiation with respect to $\tau$. –  Luboš Motl Dec 29 '11 at 13:21
    
See my comment to Qmechanic's answer. Yes I "left out some primes," or in this case bars. –  ben Dec 29 '11 at 15:47
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