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On page 247 in Gravitation by Misner, Thorne, and Wheeler, they state:

"No metric means no way to quantify length; nevertheless, parallel transport gives a way to compare length!"

Three questions:

  1. By "quantify", do they mean assign a numerical value to the spacetime distance traveled when traveling along a curve in spacetime?

  2. By "compare lengths", do they mean define the length of one spacetime curve in terms of another?

  3. Does having metric mean we are using "the language of coordinates" as they call it? (I think they just mean we can define local coordinates or something...not sure...I am new to GR and Riemannian geometry)

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  • $\begingroup$ Parallel transport means we are taking a rigid stick in one point and we are shifting it to another. In general the length of the stick may change but if the change is independent of the way we do the parallel transport, then we have a physical comparator. Even if the length stays the same, the direction may become dependent on the path (that already happens for parallel transport on a perfectly metrical manifold like a sphere). $\endgroup$
    – CuriousOne
    Commented Dec 30, 2014 at 1:50
  • $\begingroup$ Huh. I knew the direction could change and a sphere is a good example of that. But I didn't realize the length could change as well. Very interesting. So you can have either length or direction or both or neither change depending on the manifold. $\endgroup$
    – Stan Shunpike
    Commented Dec 30, 2014 at 1:54
  • $\begingroup$ I don't remember enough about differential geometry to give you details, but I am pretty sure that there is no requirement for quantities to stay constant during parallel transport in the most general case. $\endgroup$
    – CuriousOne
    Commented Dec 30, 2014 at 2:00

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CuriousOne has, as he/she/it is wont to do, provided a perfectly satisfactory answer as a comment. However just for completeness let's provide the details as an answer.

In general relativity the length of any curve is calculated by integrating the line element $ds$ along it, where $ds$ is given by:

$$ ds^2 = g_{ab}x^ax^b \tag{1} $$

In this equation $g$ is the metric tensor, so without a metric tensor there is no way to calculate the length of a curve. Since the length of an object is just the length of the (straight) curve joining its ends, without a metric tensor we cannot measure the length of objects.

But even in the absence of a metric, we can determine if two objects are the same length by superposing them and seeing if their ends coincide. To do this we have to parallel transport one or both of the objects to sperpose them.

However it should be pointed out that there is no guarantee that parallel transporting an object will preserve its length. This is the case in general relativity, but for example it is not the case in Weyl's proposed theory for unifying gravity and electromagnetism. In the article I link see 7. Problems with Weyl's Theory starting at page seven.

Response to comment:

The raised indices in equation (1) aren't powers. They just identify the components of a four-vector and run from $0$ to $3$.

$$ \vec{x} = \left( x^0, x^1, x^2, x^3 \right) $$

This is the sort of thing you take for granted when you're used to this stuff, and it's easy to forget newcomers to general relativity get confused. Sorry!

The notation I've used was invented by Einstein to make differential geometry a bit less verbose. The $a$ and $b$ are just indices and when an index is repeated it is summed over. So the equation I've written is shorthand for:

$$ ds^2 = \sum\limits_{a = 0}^{3} \sum\limits_{b = 0}^{3} g_{ab}x^ax^b $$

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  • $\begingroup$ I know that the metric tensor $g$ for a flat space-time, is a diagonal 4 by 4 matrix with $-c^2,1,1,1$ as elements. Which gives $ds^2=-c^2dt^2+dx^2+...$ which means $a$ and $b$ exponents in the general formula you've provided are both $=2.$ How does one determine these components? Sorry if it's too naive of a question :( $\endgroup$
    – user929304
    Commented Dec 30, 2014 at 13:03
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    $\begingroup$ @user929304: I've edited my answer to respond to your comment $\endgroup$
    – John Rennie
    Commented Dec 30, 2014 at 14:40
  • $\begingroup$ @JohnRennie How does one integrate using a tensor? As i have mentioned in past questions, I dont understand what the field of tensor calculus encompasses. Are we talking something in the vein of a line integral? $\endgroup$
    – Stan Shunpike
    Commented Dec 31, 2014 at 4:46
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    $\begingroup$ @StanShunpike: you're integrating the scalar $ds$, which is calculated using the metric tensor as described in equation (1). See for example How to calculate spatial distance in space-time? where I go through the calculation, or there must be many such examples hereabouts. $\endgroup$
    – John Rennie
    Commented Dec 31, 2014 at 6:25

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