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I'm having trouble understanding how the units of distance are handled when we think of (non-Cartesian) coordinates on a spacetime manifold. All sources I know of seem to brush this question under the rug.

My main thought-process/confusion comes down to the following:

  • Since coordinates determine positions, I would think that the values of the coordinates would come with units of length. So if we have a point $p\in M$ and a local coordinate system $(U, (x^{\mu}))$ about $p$, we would have $p = (cx^{0}, x^{1}, x^{2}, x^{3})$ with something like $$ cx^{0} = 2\text{ m},\; x^{1} = -1.5\text{ m},\; x^{2} = 4\text{ m},\; x^{3} = 10\text{ m} $$ as an example.
  • However, we're thinking about coordinates in general, so they're arbitrary and I can imagine rescaling them any way I want. This line of reasoning ends up conflicting with the previous bullet point, because if I could rescale them, then there is no meaning to the units of meters that are invoked here. By this reasoning, coordinate values must be unitless.

The waters here become more muddied when we try to think about velocity vectors of curves. If we assume the values of coordinates are unitless, then I can take any curve $q:[0, 1]\rightarrow M$ with $q(t) = (q^{0}(t), \ldots, q^{3}(t))$, and then take the derivative to get $$\dot{q}(t) = \dot{q}^{\mu}(t) \dfrac{\partial}{\partial x^{\mu}}$$ as a velocity vector of $q(t)$. Now if the components $q^{\mu}(t)$ are unitless, then $\dot{q}^{\mu}(t)$ must have units of inverse seconds $\text{1/s}$. This makes very little sense, because nowhere else in physics do we say that the units of velocity are inverse seconds instead of meters/seconds. My worries come down to the following questions:

  1. If I wanted to includes units in my calculations in GR, where do the units of position go? Is the first bullet point correct or the second (or neither)?

  2. Is there a consistent system of units that resolves the dilemmas that I mentioned?

An answer to any one of the questions would be very helpful.


Edit: I thought I'd put an idea out there before I forget it. Perhaps we can consider the coordinates as unitless, but then assign length units to the components of the metric, so we'd have something like $g_{00} = 3\;\text{m}^2, \; g_{10} = 2\;\text{m}^2$, etc as an example. This would make sense because any calculation of a physically meaningful quantity involving distance would invoke the metric tensor.

Edit 2: This post gives some units of various objects in the case where the coordinates have a length unit to them.

Edit 3: Based off of J. Murray's reply, I gather that what's going on is that there are two conventions. One convention is that you leave the coordinates unitless, but then take the components of the metric with units of length squared (or radians squared, etc). The other convection is that you attach units to the coordinates and leave the components of the metric as unitless. As long as you are consistent and keep track of units, either convention works.

This is sort of similar to have some people choose $(+, -, -, -)$ for the Minkowski metric signature and some people choose $(-, +, +, +)$. Either one works as long as you're consistent and keep track of the signs.

Edit 4: I think my problem here is that we haven't defined what a meter would even mean in the context of curvilinear coordinates (especially if the coordinates gridlines aren't geodesics). I'm temped to say they are just mere labels or notational aid, but that can't be true. If this were just a matter of arbitrary labeling, then I could take coordinates with meter units, and then resize the coordinates so that 1 meter actually meant 1 yard. Clearly that's absurd, but I haven't seen anyone explain exactly what's wrong with doing that.

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    $\begingroup$ Related. $\endgroup$
    – J. Murray
    May 21 at 22:23
  • $\begingroup$ @J.Murray I see. That clears up my confusion regarding tangent vectors. Regardless of whether my first or second bullet point holds, they will have units T^-1 either way. So now what's left is to understand how position coordinates are handled. It seems like units of position are redundant if your coordinate system already sets the scale, and yet we always invoke them when we're working in Cartesian coordinates. I cannot make sense of them in the general setting, even though I feel as though they should be applicable. $\endgroup$ May 21 at 22:37
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    $\begingroup$ Already in simple Euclidean geometry, you can use polar coordinates, which don't all have the same units, and everything works out fine. $\endgroup$
    – Javier
    May 21 at 23:02
  • $\begingroup$ Tip: Let's not have posts look like revision histories $\endgroup$
    – Qmechanic
    May 22 at 13:05
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You can use whatever units you'd like in your coordinates. However, the infinitesimal spacetime interval $$\mathrm ds^2 = g_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu$$ has dimensions of length squared. As a result, whatever choice of units or dimensions you make in your coordinates are reflected in the components of the metric. For example, if you choose Cartesian coordinates expressed in meters, then the metric components $g_{\mu\nu}= \mathrm{diag}(-1,1,1,1)$ are dimensionless. If you choose polar coordinates $(r,\theta,\phi)$, then $g_{\mu\nu} = \mathrm{diag}\big(-1,1,r^2, r^2\sin(\theta)\big)$, and the latter two components have dimensions of length squared in accordance with $(\theta,\phi)$ being dimensionless.


I gather that what's going on is that there are two conventions. One convention is that you leave the coordinates unitless, but then take the components of the metric with units of length squared (or radians squared, etc). The other convection is that you attach units to the coordinates and leave the components of the metric as unitless.

It's not a binary choice of convention, it's just a choice of units. In ordinary cartesian coordinates, all of the coordinates have dimensions of length and the corresponding metric components are dimensionless. You could divide all of these coordinates by a characteristic length $L$ if you wanted, in which case all of your new coordinates would be dimensionless and your metric components would be scaled by $L^2$. You could make your time component $t$ instead of $ct$, in which case it would have dimensions of time and the metric component $g_{00}$ would gain a factor of $c^2$.

There are plenty of coordinate choices which are mixed as well. In spherical polar coordinates, some coordinates have dimensions of length ($ct$ and $r$) and some coordinates are dimensionless ($\theta$ and $\phi$).

At the end of the day, a choice of coordinates is simply a way to assign a collection of numerical labels to a spacetime event. You can do this in just about any way you can imagine as long as choice of labels is continuous and unambiguous.

I'm temped to say they are just mere labels or notational aid, but that can't be true. If this were just a matter of arbitrary labeling, then I could take coordinates with meter units, and then resize the coordinates so that 1 meter actually meant 1 yard. Clearly that's absurd, but I haven't seen anyone explain exactly what's wrong with doing that.

There's nothing wrong with that at all. However, if you do that then your metric tensor components are going to be rescaled accordingly, so physical distances (e.g. $\mathrm ds = \sqrt{g_{\mu\nu} dx^\mu dx^\nu}$ ) are unchanged.

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  • $\begingroup$ Ok, this is a helpful post. Would I be correct in saying that whether any particular coordinate comes with a unit or not is a matter of choice? If $x^{\mu}$ has a length unit, then $g_{\mu\mu}$ will be unitless. If $x^{\mu}$ has no unit, then $g_{\mu\mu}$ will have units of length squared. Whether I adopt the former convention or the latter is up to me, because both conventions are equally valid (as long as you keep track of units). I'm imagining that in polar coordinates $(r, \theta, \phi)$ I could choose to make $r$ unitless and $\theta, \phi$ come with units of radians if I wanted to. $\endgroup$ May 22 at 2:54
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    $\begingroup$ @MaximalIdeal I would agree with all of that, except for the fact that the angles I wrote are already in radians (note the $\sin(\theta)$ in the last metric component). $\endgroup$
    – J. Murray
    May 22 at 2:57
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    $\begingroup$ @MaximalIdeal I've updated my answer to address your additional edits. $\endgroup$
    – J. Murray
    May 22 at 23:37

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