Understanding Christoffel symbol in GR context

[JD_PM]

TL;DR Summary

This thread is aimed at gaining insight on the Christoffel symbol in GR context.

What I would like to discuss is the following:

1) Is there a Physics reasoning on why for curvilinear coordinate systems not all Christoffel symbol components vanish? Honestly, this fact goes against my logic.

2) Could you give me an extra exercise that gives me more insight on Christoffel symbols? This technique has really worked for me in PF. See below for evidence.

I was reading chapter 3 of Carroll's book up to page 100, where he mentions that the coefficients of Christoffel symbol in flat spacetime vanish in cartesian coordinates but not in curvilinear coordinate systems.

I was thinking on why this happens. My logic tells me that these coefficients should also vanish while using other coordinate system as the spacetime remains flat anyway.

I wanted to check that the coefficients indeed do not vanish using curvilinear coordinate systems and I picked as an example spherical coordinates.

The metric in spherical coordinates is

$$ds^2 = dr^2 + r^2 d \theta^2 + r^2 \sin^2 \theta d \phi^2$$

We also know that the metric tensor in spherical coordinates is

$$g_{\mu \nu} = \begin{pmatrix}1&0&0\\0&r^2&0\\0&0&r^2 \sin^2\theta\\ \end{pmatrix}$$

The definition of the Christoffel symbol is

$$\Gamma_{\mu \nu}^{\sigma} = \frac 1 2 g^{\sigma \rho}(\partial_{\mu}g_{\nu \rho} + \partial_{\nu}g_{\rho \mu} - \partial_{\rho}g_{\mu \nu})$$

OK so based on the Christoffel symbol definition we see that it is certainly possible to get non-zero terms iff one of the indices appears twice! Actually I see 6 independent terms (by independent I mean that there are more nontrivial components that can be obtained out of these 6 due to the lower symmetry of the Christoffel symbol: $\Gamma_{\mu \nu}^{\sigma} = \Gamma_{\nu \mu}^{\sigma}$).

For the sake of completeness I computed them:

$$\Gamma_{\theta \theta}^{r} = -r$$

$$\Gamma_{\phi \phi}^{r} = -r \sin^2 \theta$$

$$\Gamma_{\phi \phi}^{\theta} = -\sin \theta \cos \theta$$

$$\Gamma_{\theta r}^{\theta} = 1/r$$

$$\Gamma_{\phi \theta}^{\phi} = \cot \theta$$

$$\Gamma_{\phi r}^{\phi} = 1/r$$What I would like to discuss is the following:

1) Is there a Physics reasoning on why for curvilinear coordinate systems not all Christoffel symbol components vanish? Honestly, this fact goes against my logic.

2) Could you give me an extra exercise that gives me more insight on Christoffel symbols? This technique has really worked for me in PF. Evidence:

1) Orodruin's #39 here https://www.physicsforums.com/threa...rem-to-get-a-constant-of-motion.982721/page-2

2) Orodruin's #19 here https://www.physicsforums.com/threa...leaves-the-given-lagrangian-invariant.984601/

3) PeterDonis #2 here https://www.physicsforums.com/threa...-timelike-vectors-and-inertial-frames.984948/

Thank you very much.

JD.

 

[Nugatory]

My logic tells me that these coefficients should also vanish while using other coordinate system as the spacetime remains flat anyway.

Your logic would be fine if you were applying to the curvature tensor invariants, which are of course the same no matter what coordinate system we use (that's what "invaroant" means) and are zero in flat spacetime (mre or less by definition).

But the Christoffel symbols are something different. For a somewhat handwaving but very intuitive explanation, try page nine of https://preposterousuniverse.com/wp-content/uploads/2015/08/grtinypdf.pdf.

 

[PeterDonis]

1) Is there a Physics reasoning on why for curvilinear coordinate systems not all Christoffel symbol components vanish? Honestly, this fact goes against my logic.

The short answer is no, there isn't a physics reason.

The somewhat longer answer is that Christoffel symbols are not a property of the geometry; they are a property of the coordinates you choose. The coordinates you choose are not a matter of physics, so there isn't a physics reason why the Christoffel symbols don't vanish in curvilinear coordinates in flat spacetime.

Could you give me an extra exercise that gives me more insight on Christoffel symbols?

The simplest case is probably polar coordinates on the Euclidean 2-plane. Compute the Christoffel symbols for this case: you should find that only one of them is nonzero. Then look at the "grid lines" for polar coordinates, and note how only one set of them is curved. (Since the plane itself is flat, "curved" is easy to see visually.) Think about how that one set of curved grid lines corresponds to the one nonzero Christoffel symbol--or, looking at it the other way, how the one nonzero Christoffel symbol is telling you about the one set of grid lines for these coordinates that are curved.

In short, think of the Christoffel symbols as telling you how the coordinate "grid lines" are curved, not how the underlying geometry is curved.

 

[pervect]

Summary:: This thread is aimed at gaining insight on the Christoffel symbol in GR context.

What I would like to discuss is the following:

1) Is there a Physics reasoning on why for curvilinear coordinate systems not all Christoffel symbol components vanish? Honestly, this fact goes against my logic.

2) Could you give me an extra exercise that gives me more insight on Christoffel symbols? This technique has really worked for me in PF. See below for evidence.

For an exercise, I would suggest something in a non-GR context. This would be calculating the Christoffel symbols in polar coordinates on a flat 2-d space (no time, just a 2d space), a plane, in polar coordinates, and understanding why they are non-zero in polar coordinates and zero in cartesian coordinates.

The "physical" motivation would be to consider that on a plane, the concept of two vectors being parallel should be independent of whether one uses cartesian coordinates or polar coordinates.

To move on to GR, at some point you'll need to learn about parallel transport. Perhaps you already have learned about parallel transport, though your question makes me think you may not have, or what you have learned needs more work. If you're up to it, feel free to revisit the problem in terms of parallel transport, the GR concept, rather than the non-GR concept o "parallel" that I suggested above as being more likely to be intuitive to you.

In terms of parallel transport, the point is that on a flat space, when you parallel transport a vector around a closed curve, it comes back unchanged. This gives meaning to the idea of "parallel" on a flat space.

To gain some insight into non-flat spaces, you can try to apply the mathematics you are learning in the context of spherical trig - replace the flat plane with the curved surface of a sphere, which is the 2d surface of a 3d globe. But that would be the next step, after you've solved your current issue.

For GR, you'll eventually want to learn about non-flat space-times, not just not flat space. But you can start with dealing with just flat space. It may or may not make things easier for you to eliminate time, but there is a reasonably high possibility that you might have less intuition about the Lorentzian geometyr of flat space-times than the Euclidean geometry of flat space in the absence of time.

 

[Orodruin]

In flat space, you can make an easy mental geometrical image of what the Christoffel symbols mean:
Assuming that you have a set of coordinate basis vector fields $\vec E_a$, the Christoffel symbol $\Gamma_{ab}^c$ is the $c$-component of the derivative in the $x^a$ direction of the vector field $\vec E_b$, or in other words
$$
\partial_a \vec E_b = \Gamma_{ab}^c \vec E_c.
$$
The reason the Christoffel symbols are non-zero in curvilinear coordinate systems is that the coordinate basis changes between points, i.e., what is the radial direction depends on where you are.

 

[JD_PM]

The simplest case is probably polar coordinates on the Euclidean 2-plane. Compute the Christoffel symbols for this case

Nice, let's compute it.

The metric in polar coordinates is

$$ds^2 = dr^2 + r^2 d \theta^2$$

The metric tensor is

$$g_{\mu \nu} = \begin{pmatrix}1&0\\0&r^2\\ \end{pmatrix}$$

you should find that only one of them is nonzero.

Mmm I get two though...

$$\Gamma_{\theta \theta}^{r} = -r$$

$$\Gamma_{\theta r}^{\theta} = 1/r$$

 

[JD_PM]

To move on to GR, at some point you'll need to learn about parallel transport.

I did not get into parallel transport yet but I will do it soon ;)

 

[JD_PM]

In flat space, you can make an easy mental geometrical image of what the Christoffel symbols mean:
Assuming that you have a set of coordinate basis vector fields $\vec E_a$, the Christoffel symbol $\Gamma_{ab}^c$ is the $c$-component of the derivative in the $x^a$ direction of the vector field $\vec E_b$, or in other words
$$
\partial_a \vec E_b = \Gamma_{ab}^c \vec E_c.
$$
The reason the Christoffel symbols are non-zero in curvilinear coordinate systems is that the coordinate basis changes between points, i.e., what is the radial direction depends on where you are.

Could you please give an example? :)

 

[vanhees71]

Nice, let's compute it.

The metric in polar coordinates is

$$ds^2 = dr^2 + r^2 d \theta^2$$

The metric tensor is

$$g_{\mu \nu} = \begin{pmatrix}1&0\\0&r^2\\ \end{pmatrix}$$
Mmm I get two though...

$$\Gamma_{\theta \theta}^{r} = -r$$

$$\Gamma_{\theta r}^{\theta} = 1/r$$

That's correct.

The most simple way to get the Christoffels is to use the Hamilton principle to get the geodesics in terms of an affine parameter. For that you simply take as Lagrangian
$$L=\frac{1}{2} g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}.$$
It's easy to see that the Euler-Lagrange equations give
$$g_{\mu \nu} \ddot{q}^{\nu} + \partial_{\alpha} g_{\mu \nu} \dot{q}^{\alpha} \dot{q}^{\nu}-\frac{1}{2} \partial_{\mu} g_{\alpha \nu} \dot{q}^{\alpha} \dot{q}^{\nu},$$
which you can rewrite as
$$g_{\mu \nu} \ddot{q}^{\nu} + \frac{1}{2} \left (\partial_{\nu} g_{\mu \alpha} + \partial_{\alpha} g_{\mu \nu}- \partial_{\mu} g_{\alpha \nu} \right )\dot{q}^{\alpha} \dot{q}^{\nu}.$$
That gives the Christoffel symbols with lower indices
$$\Gamma_{\mu \alpha \nu}=\frac{1}{2} \left (\partial_{\nu} g_{\mu \alpha} + \partial_{\alpha} g_{\mu \nu}- \partial_{\mu} g_{\alpha \nu} \right ).$$
The usual Christoffel symbols are then given by contraction with $g^{\mu \beta}$:
$${\Gamma^{\beta}}_{\alpha \nu} = \frac{1}{2} g^{\mu \beta}\left (\partial_{\nu} g_{\mu \alpha} + \partial_{\alpha} g_{\mu \nu}- \partial_{\mu} g_{\alpha \nu} \right ).$$
With them the geodesics equations read
$$\ddot{q}^{\mu} + {\Gamma^{\mu}}_{\alpha \beta} \dot{q}^{\alpha} \dot{q}^{\beta}=0.$$
For the polar coordinates you have
$$L=\frac{1}{2} (\dot{r}^2 + r^2 \dot{\theta}^2)$$
The Euler-Lagrange equations read
$$\ddot{r}-r \dot{\theta}^2=0, \quad \ddot{\theta} + 2 \frac{1}{r} \dot{r} \dot{\theta}=0,$$
from which you immediately read off that
$${\Gamma^{r}}_{\theta \theta}=-r, \quad {\Gamma^{\theta}}_{r \theta} = {\Gamma^{\theta}}_{\theta r}=\frac{1}{r}.$$

 

[PAllen]

Please think hard on @Orodruin ‘s answer. It is the essence of the issue - I was going to post the same, but saw he beat me to it.

 

[JD_PM]

In flat space, you can make an easy mental geometrical image of what the Christoffel symbols mean:
Assuming that you have a set of coordinate basis vector fields $\vec E_a$, the Christoffel symbol $\Gamma_{ab}^c$ is the $c$-component of the derivative in the $x^a$ direction of the vector field $\vec E_b$, or in other words
$$
\partial_a \vec E_b = \Gamma_{ab}^c \vec E_c.
$$
The reason the Christoffel symbols are non-zero in curvilinear coordinate systems is that the coordinate basis changes between points, i.e., what is the radial direction depends on where you are.

Could you please give an example? :)

Please let me explain myself; I asked for an example because, honestly, I do not completely see what you mean here. I think I would see it better with an example.

Thanks.

 

[PAllen]

Please let me explain myself; I asked for an example because, honestly, I do not completely see what you mean here. I think I would see it better with an example.

Thanks.

well, @Orodruin did give an example, but I will try to make it very explicit.

First, you said you haven't dealt with parallel transport yet. However, in the simple case of a flat Euclidean plane, the concept is really simple: moving a vector around without changing its direction or magnitude. This notion doesn't depend on any coordinates placed on the plane.

Now consider cartesian coordinates. The x direction and the y direction are 'basis vectors'. They exist at every point. Cartesian coordinates have the property that the basis vectors at one point, parallel transported to another point, are the same as cartesian basis vectors at that point. More simply, the x and y directions don't change from point to point.

As a result, if an arbitrary vector is moved from one point to another, its projection on the that point's basis vectors don't change.

Now consider polar coordinates, superimposed on Cartesian coordinates. Along the x axis, the radial direction is in the x direction and the tangential direction corresponds to the y direction. However, along the y axis, the radial direction is the y direction, and the tangential is the negative x direction. So If you move, e.g. the vector (x,y) = (1,2) from somewhere along the x-axis to the y axis, its representation in cartesian coordinates doesn't change, but in polar coordinates it changes from (r,θ) = (1,2) on the x-axis to (r,θ)=(2,-1) when moved to the y axis. This is necessary to prevent its direction from changing.

The connection coefficients tell you how the representation of a (parallel transported) vector changes from point to point as the basis vectors change direction. They vanish only if the basis vectors don't change from point to point.

NOTE: all of this discussion is restricted to the flat space situation. I have deliberately glossed over additioanl issues that arise in curved space because you specifically asked about non-Cartesian coordinates in flat space.

 

[PeterDonis]

I get two though...

Yes, you're right, I misspoke.

 

[Orodruin]

Ok, here is an explicit example:

In polar coordinates the (non-normalised) basis vectors are given by
$$
\vec E_r = \cos(\phi) \vec e_1 + \sin(\phi) \vec e_2, \quad \vec E_\phi = r[-\sin(\phi)\vec e_1 + \cos(\phi)\vec e_2].
$$
Clearly, both of these vectors depend on position (i.e., on the values of $r$ and $\phi$). If we differentiate $\vec E_r$ with respect to $\phi$, we find
$$
\partial_\phi \vec E_r = -\sin(\phi) \vec e_1 + \cos(\phi) \vec e_2 = \frac{1}{r} \vec E_\phi.
$$
By the relation I quoted above, this is equal to $\Gamma_{\phi r}^c \vec E_c$ and thus $\Gamma_{\phi r}^r = 0$ and $\Gamma_{\phi r}^\phi = 1/r$.

 

[Ibix]

Sometimes a diagram is useful. I apologise for the hand-drawn nature of this, but hopefully it's clear enough. It's more or less a restatement of @PAllen's last post.

Here are two vectors at points on a Euclidean plane, one red and one blue. I've added the vector $x^a$ (which I realize I accidentally wrote $x_a$ in the diagram, but I'm not going to try to fix now) which points from one point to the other.

The red and blue vectors are in some sense the same - they point in the same direction and have the same length. And if you use Cartesian coordinates and the matching basis vectors they are indeed both expressed the same way. In the diagram below I've shown Cartesian basis vectors, and it's clear that the red and blue vectors have the same expression in this basis - something like $(L/\sqrt 2,L/\sqrt 2)$ where $L$ is the magnitude of the vectors.

However, this is not the case in all bases. Below, I've added a polar coordinate system and basis - radial lines to the tails of the vectors are shown in grey, and the basis vectors again in black.

In this system, the basis vectors $\vec E_r$ and $\vec E_\phi$ are not the same at different positions, so vectors that are the same are not written the same way. The red vector is something like $(0,L)$ while the blue vector is more like $(\sqrt 3L/2,-L/2)$ - even though they are still the same vectors I drew in the first diagram. There is no such difference in a Cartesian basis, so the Christoffel symbols are zero. But in the polar basis the components do differ and the Christoffel symbols are non-zero.

As others have pointed out, this somewhat glosses over issues raised in curved spaces.

What the Christoffel symbol $\Gamma^c_{ab}(r)$ does is tell you how the $b$ components of two identical vectors differ in your chosen coordinates and basis if they are located at some point $r$ and another point a differential distance away in the $x^a$ direction.

 

[Orodruin]

So with all of this, let us try to connect back to curved spaces. Unlike in an affine space (such as Euclidean space), there is no longer an inherent notion of what it means for a vector field to change from point to point. In fact, there is generally not even a way to compare vectors at different points since they belong to different tangent spaces.

This is where we need a connection, a way of defining what we mean when we say that a vector field changes. In the general case, this is not uniquely defined between points, only along curves. We define a connection $\nabla$ such that it has a number of properties that we expect from a directional derivative. In short, what is necessary to define a connection is to know how it acts on a set of basis vector fields $E_a$, i.e., to fully define the connection we need to know what $\nabla_{E_a}E_b$ is for all combinations of a and b. Using the typical short-hand $\nabla_{E_a} = \nabla_a$, we know that $\nabla_a E_b$ is what should be roughly interpreted as ”rate of change in $E_b$ in the $E_a$ direction”, where ”rate of change” is defined by the connection. Since this is a vector field, we can expand it in the $E_a$ basis and call the expansion coefficients $\Gamma_{ab}^c$, i.e.,
$$
\nabla_a E_b = \Gamma_{ab}^c E_c.
$$
These gammas uniquely define how the connection acts on any vector (or indeed tensor) field and are called ”connection coefficients”. In general, the connection is not unique and we can specify any connection coefficients we like unless we put further constraints.

The two most common constraints (and the constraints assumed in most GR formulations) are:
- The connection should be metric compatible.
- The connection should be torsion free.
The first of these conditions you can only put when you have a manifold that has a metric tensor and in essence requires that if you have two vector fields that do not ”change” (as defined by the connection) along a curve, then their inner product as defined by the metric is constant along that curve. The second condition is a bit less intuitive but in essence says that (assuming the $E_a$ are a coordinate basis) $\nabla_a E_b = \nabla_b E_a$. The connection satisfying both these conditions is unique and is called the Levi-Civita connection. Its connection coefficients are the Christoffel symbols.

Again connecting back to Euclidean space (or any affine space), there is then a natural way to define what it means for a vector to ”change” due to the natural identification of the tangent spaces at different points. Using this, the connection will be exactly the Levi-Civita connection of the standard metric and the Christoffel symbols will be zero in all Cartesian (and affine) coordinate systems.

Note: The definition of a connection does not require a metric. However, defining a connection to be metric compatible of course does.

 

[pervect]

In flat space, you can make an easy mental geometrical image of what the Christoffel symbols mean:
Assuming that you have a set of coordinate basis vector fields $\vec E_a$, the Christoffel symbol $\Gamma_{ab}^c$ is the $c$-component of the derivative in the $x^a$ direction of the vector field $\vec E_b$, or in other words
$$
\partial_a \vec E_b = \Gamma_{ab}^c \vec E_c.
$$
The reason the Christoffel symbols are non-zero in curvilinear coordinate systems is that the coordinate basis changes between points, i.e., what is the radial direction depends on where you are.

Please let me explain myself; I asked for an example because, honestly, I do not completely see what you mean here. I think I would see it better with an example.

Thanks.

So let's look at an example. In particular, let's look at $\Gamma^r{}_{\theta\theta}$. We've just worked out what the value of $\Gamma^r{}_{\theta\theta}$ is, it's equal to -r. See vanhees' post.

from which you immediately read off that
$${\Gamma^{r}}_{\theta \theta}=-r, \quad {\Gamma^{\theta}}_{r \theta} = {\Gamma^{\theta}}_{\theta r}=\frac{1}{r}.$$

So, it's saying that the basis vector $\vec{e}_{\theta}$, when we move it in the $\theta$ directions, needs to add a bit of the vector $-\vec{e}_r$.

This is because $\partial_{\theta} \, \vec{e}_{\theta}$ is equal to $\Gamma^r{}_{\theta \theta} \vec{e}_{r}+ \Gamma^{\theta}{}_{\theta \theta} \vec{e}_{\theta}$ = -$r \, \vec{e}_{\theta} + 0$

A diagram might help. We draw $e_{\theta}$ at theta=0, and again with $\theta > 0$. We see that this is in fact correct.

One thing that will most likely confuse you is that $e_{\theta}$ isn't in general a unit length vector. To see this, look at the dot product $\vec{e}_\theta \cdot \vec{e}_\theta = r^2$. However, in the diagram above, you can consider that we are drawing $\vec{e}_\theta$ at r=1, so in this particular diagram it is unit length.

It is hopefuly obvious what direction $e_{\theta}$ points in, but you also have to get it's length right when you use a coordinate basis. Orthonormal basis are more physically intuitive, but you get stuck with the coordinate basis in order to use some very useful formulae for covariant derivatives, a topic that's coming up shortly7 in your course, I'm sure.

 

[George Keeling]

Going right back

The definition of the Christoffel symbol is$$\Gamma_{\mu \nu}^{\sigma} = \frac 1 2 g^{\sigma \rho}(\partial_{\mu}g_{\nu \rho} + \partial_{\nu}g_{\rho \mu} - \partial_{\rho}g_{\mu \nu})$$

Carroll did not say that. At his 3.5 he said the connection coefficients $\Gamma_{\mu \nu}^{\sigma}$ are defined to correct the failing in the partial derivative, to give the covariant derivative which is a tensor. He then goes on to show the that the coefficients must be given by that formula. (On the way at his 3.10 for the coordinate transformation equation the sign is wrong. Here's why.)

I understand your instinct that "coefficients should also vanish" and your instinct will be correct when you apply it to the Riemann tensor which you will meet in section 3.6. It is defined to measure the curvature of the metric and turns out to be$$R_{\ \ \sigma\mu\nu}^\rho=\partial_\mu\Gamma_{\nu\sigma}^\rho-\partial_\nu\Gamma_{\mu\sigma}^\rho+\Gamma_{\mu\lambda}^\rho\Gamma_{\nu\sigma}^\lambda-\Gamma_{\nu\lambda}^\rho\Gamma_{\mu\sigma}^\lambda$$Normally this is ghastly to calculate. But in two dimensions it has only one independent component (Carroll 3.137), so in plane polar coordinates where you calculated the Christoffel symbols, you can just calculate one component of the Riemann tensor and it vanishes (e.g. $R_{\ \ rrr}^r=0$ ). So they all vanish in this alternate flat coordinate system as you expected the Christoffel symbols to.
PS I am referring to an edition of Carroll "Second Impression 2017".

 

[Orodruin]

I understand your instinct that "coefficients should also vanish" and your instinct will be correct when you apply it to the Riemann tensor

Or indeed the components of any tensor. The point is that the connection coefficients are not the components of a tensor. Indeed, as @George Keeling points out here, they correct for the failure of partial derivatives to be tensors.

you can just calculate one component of the Riemann tensor and it vanishes (e.g. Rr rrr=0R rrrr=0R_{\ \ rrr}^r=0 )

Note that $R^r_{rrr} = 0$ is not sufficient to draw the conclusion that the Riemann tensor is zero. If all indices are the same, the component must be zero regardless of whether the space is curved or not (due to the symmetries of the tensor) so this tells you nothing. What you need to check in the two-dimensional case is, for example, $R_{1212}$ (note: first index lowered).

For example, for the usual metric on the sphere, $R^\theta_{\theta\theta\theta}=0$ but the curvature is non-zero. If you check the components that are not trivially zero, they will be different from zero.

 

[Orodruin]

It is defined to measure the curvature of the metric and turns out to be

Rρ σμν=∂μΓρνσ−∂νΓρμσ+ΓρμλΓλνσ−ΓρνλΓλμσR σμνρ=∂μΓνσρ−∂νΓμσρ+ΓμλρΓνσλ−ΓνλρΓμσλ​

R_{\ \ \sigma\mu\nu}^\rho=\partial_\mu\Gamma_{\nu\sigma}^\rho-\partial_\nu\Gamma_{\mu\sigma}^\rho+\Gamma_{\mu\lambda}^\rho\Gamma_{\nu\sigma}^\lambda-\Gamma_{\nu\lambda}^\rho\Gamma_{\mu\sigma}^\lambdaNormally this is ghastly to calculate.

I think it is also worth pointing out that using formulas like this or the expression for the Christoffel symbols in terms of the metric is typically way more cumbersome than necessary. The formulas are mostly useful for computer computations where the computer keeps track of all the indices for you (I have my own Mathematica code for this to double check answers when I correct GR exams ...). I generally find it much easier to use other methods such as computing the Christoffel symbols from variational calculus (see #9 by @vanhees71 ) and the curvature components from $R^d_{abc}\partial_d = (\nabla_b\nabla_c-\nabla_c \nabla_b)\partial_a$. Of course, the actual computations will be the same, but it is just much easier to keep track of things.