Defining proper distance in GR.

1. Sep 20, 2013

center o bass

At page 234 in Landau and Lifshitz' Classical Theory of Fields the proper time element is defined through the line element by $ds^2 = c^2 d\tau^2$, then for a stationary observer, setting $dx^i=0$ for $i=1,2,3$. One then obtains the relation
$$c^2 d\tau^2 = g_{00}(dx^{0})^2.$$

He then goes on to define the spatial proper distance $dl^2$ and notes that in SR $dl^2$ can be defined as the spatial distance between two events infinitesimally separated, but with the same coordinate time, i.e. by setting $dx^0 = 0$ in the line-element. He claims that this is not possible in GR because the proper time in a gravitational field may have a different dependence on $x^0$ at different points in space.
He then goes on to define $dl^2$ by sending and reflecting back a light ray from/to a stationary observer.

I remember naively doing exactly this when I started learning about GR, but getting the wrong answer. I never really understood why it was not allowed and I'm not sure if I understand what the author is saying here, so I wondered if someone could elaborate on why we can not define the proper distance in this way as one did in SR.

PS: I know it's become common to completely ignore the notion of a line-element and just work
with tensors, i.e. the metric tensor. But I would like to understand it in both ways.

2. Sep 20, 2013

PAllen

You can define a distance with the line element. If you have two events with spacelike separation that you consider to be simultaneous, you can define the proper distance between them as the integrated line element (with the right sign) along them for the geodesic connecting them. If they are 'near by', this will be unique. In general, in GR, it will not always be unique - there can be multiple geodesics. You can adopt a convention of smallest length.

However, the real issue becomes relating this to measurements. What you typically have in physical scenario is two world lines. So to execute the above, you have to decide which pairs of events on the world lines you consider simultaneous. Radar distance solves this by jointly providing a simultaneity choice and distance measurement that can be readily carried out of substantial distances, by one of the world lines.

For reasonably nearby world lines, a 'ruler distance' is typically defined, and presumed to predict the result of using a reasonable implementation of a ruler. The definition is to find a spacelike geodesic 4-orthogonal to the tangent of one of the world lines at some event that intersects the other world line. Then, the proper length of this geodesic segment is the ruler distance between them per the first world line at the chosen time along the first world line.

3. Sep 20, 2013

center o bass

The author seems to be defining distance and spatial geometry by what you call "radar distance". He then obtains that

$$dl^2 = (g_{ij} - \frac{g_{0i}g_{0j}}{g_{00}}) dx^i dx^j \equiv \gamma_{ij}dx^i dx^j$$

where he calls $\gamma_{ij}$ the spatial metric. I guess my question is what is wrong with the following argument?:

The spacetime interval between two infinitesimally close events are

$$ds^2 = g_{00}(dx^0)^2 + 2g_{i0}dx^0 dx^i + g_{ij} dx^i dx^j$$

and for a stationary observer ($dx^0 = 0$), $ds^2 = dl^2$ so that

$$dl^2 = 2g_{i0}dx^0 dx^i + g_{ij} dx^i dx^j.$$

4. Sep 20, 2013

PAllen

The problem is that what you propose is physically meaningless. A line of constant coordinate time would typically have no meaning for an observer following world line of constant coordinate position. Thus, the only physically plausible alternative to radar distance that I know of is the ruler distance definition I gave.

5. Sep 20, 2013

center o bass

I think I agree; it would only make sense if the worldline was parametrized by $x^0$. But then on the other hand does it make sense to set $dx^i = 0$ to find a relation between the proper-time interval and $dx^0$ like Landau did?
Can you check out the derivation of the proper length in Landau's book? You can find it here

http://archive.org/details/TheClassicalTheoryOfFields

at page 234. If so would you say that the derivation is sensible?

6. Sep 20, 2013

PAllen

No, it would only make sense if the line of constant coordinate time was a geodesic, and it was 4-orthogonal to the world line of constant coordinate position. Neither of these is likely to be true.

Proper time is different because it gives the time read by a clock following the constant coordinate position world line, always. No assumptions or restriction, as long as it is actually a world line (timelike path).

His derivation looks fine to me. He is only defining it for an 'infinitesimal' distance, in which case it is not hard to show that there is no difference between radar and ruler distance. He describes the issues I allude to for integrating it, but makes no suggestion other than for case of a static metric.

7. Sep 20, 2013

center o bass

If the worldline of an observer is (affinely) parametrized by $x^0$, then $x^0$ is proportional to that observers proper time. Setting $dx^0 = 0$ would then be equivalent with setting the proper-time interval to zero, which is physically meaningfull. Why would it have to be a geodesic and four orthogonal to constant coordinate worldlines?

I thought it was just the time read by a clock following a worldline in general. Why does it have to be a constant position-coordinate world line? Can not the observer use coordinates for which the "spatial coordinates" need not be constant along the world line? Wouldn't that correspond to a coordinate system with a coordinate basis for which $\vec{e_0} \cdot \vec{e_i} \neq 0$ which is allowed?

8. Sep 20, 2013

WannabeNewton

Yes this is true. And in the comoving coordinates associated with the observer described by the worldline (in which the observer is of course at constant spatial coordinates-naturally the origin of the comoving coordinates), the proper time read by said clock is just the coordinate time of said coordinates, by construction.

9. Sep 20, 2013

PAllen

It would have to be a geodesic if you wanted to integrate along it - else it would have no meaning as a distance. If you are looking only at infinitesimals, all you need is 4-orthogonal. That says it s consistent with local Lorentz simultaneity. In general coordinates in GR a spatial coordinate will not be 4-orthogonal to the time coordinate at some point. You can tell easily: if the metric at that point has no off diagonal elements, then you meet the locally orthogonal condition.
I was answering as to why don't have to worry about anything except g00 to talk about the rate of proper time along the time coordinate, while for distance you have to worry about othogonality, even for infinitesimals.

Last edited: Sep 20, 2013
10. Sep 27, 2013

center o bass

But exactly what in the argument is wrong and why?; If the worldline of an observer is parametrized by $x^0$ then for that observer, $dx^0 = 0$ is equivalent with $d\tau = 0$ regardless of whether the spatial coordinates are time-orthogonal or not. Thus the spacetime interval between two infinitesimaly separated events which for the observer in question happens at the same proper time is

$$ds^2 = g_{00} \times 0^2 + 2 g_{i0} dx^i \times 0 + g_{ij}dx^i dx^j$$

and this we call proper length.. But it's not.

Last edited: Sep 27, 2013
11. Sep 27, 2013

PAllen

The issue is proper length for who? At least infinitesimally, the formula you give will be proper length for 'someone' but not, in general, for the observer who follows a line of constant coordinate position. For this observer, their local simultaneity will have to be 4-orthogonal to the world line of constant coordinate position. This will not be a line of constant time coordinate unless the metric at that point has no off diagonal elements.

12. Sep 27, 2013

WannabeNewton

To add on to PAllen, let's say we have an observer with 4-velocity $\xi^{\mu}$ then at any event $p$ on the observer's worldline, there exists locally a space-like hypersurface which is orthogonal to $\xi^{\mu}$ at $p$. Operationally, it is constructed by sending out space-like geodesics in every direction emanating from $p$ that are orthogonal to $\xi^{\mu}$ at $p$; the images of these geodesics, when restricted to a sufficiently small open subset, will sweep out a space-like hypersurface orthogonal to $\xi^{\mu}$ at $p$. This defines the local simultaneity slice at $p$ relative to this observer (all we're doing is using the exponential map). Now if you were in the coordinates comoving with the observer then a surface of constant $x^0$ need not correspond to such a local simultaneity slice if the representation $g_{\mu\nu}$ of the metric tensor in the comoving coordinates has terms $g_{0i}$ because then the surfaces of constant $x^0$ are not orthogonal to $\xi^{\mu}$ and do not translate over to a local simultaneity slice.

Take for example the observers at rest in the Kerr chart on Kerr space-time; this chart by itself defines comoving coordinates for these static observers following orbits of the time-like killing field $\xi^{\mu} = (\partial_{t})^{\mu}$. However $\xi^{\mu}$ is not orthogonal to the surfaces of constant $t$; the surfaces of constant $t$ do not represent local simultaneity slices for these observers. They are however orthogonal to the vector field $\nabla^{\mu}t$ but the observers following orbits of $\nabla^{\mu}t$ are not at rest in the Kerr chart (they will be orbiting the central body).

Last edited: Sep 27, 2013