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I know my question doesn't make sense all ready. Suppose I have points $(0,0,0,0)$ and $(1,1,0,0)$. The distance traveled is zero since $ds^2 = 0$ in this case. And light has moved from one point to another. And since these two points are distinct, it has traveled on the manifold ¿presumably? along a geodesic.

But I don't quite follow what this means. How can $ds^2$ be zero and I can move somewhere? If $ds^2$ doesn't measure spacetime distance, how can I say, for instance, the sun is 0.00001581 light years away from the earth?

Stan Shunpike
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  • $\mathrm{d}s^2$ does not represent spatial displacement, or really any sort of distance you could imagine with your Euclidean intuition. – ACuriousMind Aug 11 '15 at 23:39
  • I got that. I am not sure how I communicated something that indicated otherwise, but I dont doubt that I did – Stan Shunpike Aug 11 '15 at 23:45
  • You say "How can $\mathrm{d}s^2$ be zero and I move somewhere?" - if you know that $\mathrm{d}s^2$ doesn't represent spatial distance (or only does that when the time coordinates of two events are equal), then where do you see the issue with something moving and the $\mathrm{d}s^2$ being zero? – ACuriousMind Aug 11 '15 at 23:48
  • I meant "move" on the manifold, not spatially. Do you have a better term for moving between points? I don't think I have ever really come to terms with the fact that spacetime necessitates a semi-Riemannian manifold IIRC. I dont get how distance can be zero and yet you move somewhere. – Stan Shunpike Aug 12 '15 at 00:35
  • Maybe try to stop using the word distance and use the term interval always instead. –  Aug 12 '15 at 00:38
  • Hmm....but for a Riemannian manifold they use the word distance. – Stan Shunpike Aug 12 '15 at 00:38
  • On a 4 D mainifold in physics? Or a math description? Anyway, it works for me, after a long time trying to use physical intuition, I just gave up trying, never looked back:) –  Aug 12 '15 at 00:42
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    That's because the Riemannian case cannot have $\mathrm{d}s^2 = 0$, and there your intuition works! I'm sorry, but it seems to me you have not understood what a metric tensor is, and that $\mathrm{d}s^2$, in the Lorentzian case, is not actual "distance", spatial, in spacetime, or otherwise, because two points which are connected by a curve of length $0$ are not necessarily equal. – ACuriousMind Aug 12 '15 at 00:42
  • All that $ds^2=0$ tells you is that either the two points are the same or that the motion between them is light-like. It does not mean absence of motion. (Light travels at c, which is not zero.) – David Hammen Aug 12 '15 at 00:45
  • @ACuriousMind it seems you have diagnosed the problem. I must revisit what a metric tensor is. And the rest of your comment makes clear that considering it a distance makes no sense. I never thought of it like that somehow – Stan Shunpike Aug 12 '15 at 00:54

2 Answers2

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A Minkowski diagram is showing by its light cone that null geodesics are linking different points in spacetime. In your example, the space interval is 1, but the spacetime interval is 0. The null geodesic of a photon includes a point of emission and a point of absorption, each one is a different point in spacetime.

The spacetime interval of two points on the worldline of a mass particle is diminishing if the mass particle is approaching c. In the same way the spacetime interval is zero for photons moving on a lightlike worldline with v=c.

It might be important for you to emphasize once more that the point of emission and the point of absorption remain two distinct points, only their spacetime interval is zero.

It is interesting to know that this fact seems to provide a classical explanation of a thought experiment of two entangled photons moving at c, see this question.

Community
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Moonraker
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Remember that $ds^2$ is calculated using the Minkowski metric. As @ACuriousMind is saying, you need to understand what a metric means. A metric is basically a prescription for how distances are measured in the space. So, for example, a metric for a sphere would be useful for finding the distance between two points on Earth given their lattitudes and longitudes. As long as all the coefficients in a metric are positive (like the metric for a sphere) then distances between two points are always positive and your normal intuition about the meaning of distance will work.

But the Minkowski metric has a term (the $dt^2$ term in many/most people's sign convention) with a negative coefficient. So "distances" can be positive, negative or zero. This is a signal to you that the meaning of "distance" here isn't what you are used to. Since we are talking about space-time (not space or time) "distances" do not refer to distances in space.

This is reflected in the language we use. If $ds^2 < 0$ (in the usual sign convention) then it refers to a "time-like interval". That is, there is some reference frame in which the two events separated by the interval occur at the same spatial location, and so ds can be interpreted as a time interval (in that reference frame). Conversely, if $ds^2 > 0$ then the interval is "space-like" and the two events do not occur at the same place in any reference frame. Instead, there is some reference frame where the two events are simultaneous and ds can be thought of as a distance in space.

So, the intermediate case is where the two events are neither simultaneous nor at the same location in any frame. Light leaving one event could arrive at the other event. These events lie on each others "null cones". Being intermediate between space-like ($ds^2 > 0$) and time-like ($ds^2 < 0$) they correspond to $ds^2 = 0$ even though neither $dt^2$ nor $dx^2+dy^2+dz^2$ is zero.

gleedadswell
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