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My professor in mechanics course sneakily teach us some basic idea of general relativity. Which one of the basic assumption is particle walks in shortest world line.

I understand shortest path in the Lagrange mechanics means roughly "physics systems tend to be steadiest". but this quite confuses me.

so on the one hand,

  1. what exactly meant by "shortest world line" (my professor told me that we need only just to consider "particle walks in shortest world line" as an assumption)

on the second hand,

  1. Is there another way to see what "particle walks in shortest world line" exactly mean?
Qmechanic
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Shing
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2 Answers2

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Your professor is telling you something that is absolutely fundamental to a proper understanding of relativity. Suppose we draw out the trajectory of some object on a space time graph, we may get something like this:

Spacetime diagram

The path traced out by the object(the blue curve) is called the world line. The length of the world line, $s$, is equal to $c\tau$, where $\tau$ is the elapsed time measured by a clock carried by the observer. this time $\tau$ is called the proper time and is a very important concept in both flavours of relativity. The proper time is an invarient, i.e. all observers who calculate it will get the same answer regardless of what coordinate system they are using.

To see how to calculate the length of the world line, $s$, let's zoom in on the curve:

World line

If we approximate the curve by a series of tiny straight lines of length $ds$ then the length of the curve is just given by summing up all the $ds$s:

$$ s = \int_C ds $$

If you know Pythagoras' theorem then you might guess that the length $ds$ is given by:

$$ ds^2 = dt^2 + dx^2 $$

but this is not the case. This is where the geometry of spacetime differs from Euclidean geometry, because the length $ds$ is given by:

$$ ds^2 = -c^2dt^2 + dx^2 $$

We multiply the time $t$ by the velocity of light $c$ to convert it to a distance (in light-seconds) to make the equation dimensionally consistent.

If we allow motion in all space directions the equation becomes:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 \tag{1} $$

Equation (1) is called the Minkowski metric, and describes the geometry of flat spacetime. It is the equation that defines special relativity, and it is also the vacuum solution of general relativity that has zero ADM mass. So understanding this equation not only means you understand special relativity, but gives you a start on general relativity as well. In general relativity, where spacetime can be curved, we get the more general form of the equation:

$$ ds^2 = \sum_\alpha\sum_\beta g_{\alpha\beta} dx^\alpha dx^\beta $$

where $g$ is the metric tensor. We get the metric tensor by solving Einstein's equation.

To get back to your question, in special and general relativity the path followed by a freely moving object is the one that minimises the length $s$. If the geometry were Euclidean the path wouldn't be very interesting, but that minus sign on the $t$ coordinate means that in relativity minimising the path length gives us all sorts of interesting behaviour like time dilation and length contraction, and of course in general relativity it explains motion in a gravitational field.

I can't emphasise enough how import this simple idea is. Obviously general relativity is a huge and complex area, but just equipped with what I've described above you can start understanding things like how black holes cause time dilation.

Community
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John Rennie
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    $ ds^2 = g_{\alpha \beta } dx^{\alpha } dx^{\beta } \not = g_{\alpha \beta } x^{\alpha } x^{\beta }$ – Alexander May 23 '15 at 06:29
  • @Alexander: oops, thanks :-) Please free to edit obvious typos. – John Rennie May 23 '15 at 06:33
  • Thank you. Are there any points in further inquiry into the absolute fundamental, sometimes, I was wondering why that is so. – Shing May 26 '15 at 02:06
  • @ShingLau: if you're asking why the universe is Lorenztian, i.e. why equation (1) applies, I don't think there is an answer. It just is. – John Rennie May 26 '15 at 05:29
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The shortest world line (geodesic path) is given by the GR coordinate system. GR has $x_0$ (ct), $x_1$ (x), $x_2$ (y), $x_3$ (z). Often in GR, c is set to 1 (c = 1).

Distance squared ($ds^2$) is given by $$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$.

If one second passes, the shortest distance is to stay in the same spatial location, that is $$ds^2 = -dt^2$$ $$ds^2 = -1$$

This is different then Euclidean geometry in the fact that there are 4 dimensions. In Gr we must account for time.

In GR, gravity is the curvature of space-time, a warping of coordinates. A straight beam of light will curve. In GR it is still straight in respect to its coordinate system, but is following a curved coordinate plane.

enter image description here

Jimmy360
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  • Might want to edit your GR coordinates youve missed a few things – Triatticus May 23 '15 at 06:00
  • Might I ask whoever downvoted this why they did and how I can fix it? – Jimmy360 May 23 '15 at 06:01
  • You might want to make the comparison to euclidean distances in your answer (also your last GR coord has the wrong subscript still) I didnt downvote but that might be a good idea as its easy to visualize a shortest distance on a plane – Triatticus May 23 '15 at 06:01
  • You wrote the shortest distance in the traveler's reference frame (this is the only reason for spacial differentials to vanish). Your claim for distance after 1 sec shouldnt be expressed in differentials – Alexander May 23 '15 at 06:05