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Drawing lines through spacetime warped by the mass of the earth

  1. Jun 2, 2014 #1
    Been reading laymen's books on general relativity and I'm in need some clarification! ...

    Line 1: The shortest path from a point 9.8 meters above the surface of the earth to another point at the same location in space but 1 second later in time.

    Line 2: The shortest path from a point 9.8 meters above the surface of the earth to another point directly 9.8 meters below it 1 second later in time.

    (Using the earth as the frame of reference)

    Is it true that line 1 is longer than line 2?
  2. jcsd
  3. Jun 2, 2014 #2


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    The short answer to your question as you've asked it is: yes, but maybe not for the reason you think.

    The longer answer requires clarifying some things that you might not have realized while you were framing the question:

    (1) The "shortest path" means a geodesic in spacetime, i.e., the path in spacetime followed by a freely falling object. (Note that this "shortest" path is actually the path of *longest* elapsed proper time between two given endpoints, which is why I am putting "shortest" in quotes.) For line #2, that's probably the path you were imagining: an object free-falls for 1 second straight down, starting from rest. (Btw, the object will actually fall 4.9 meters in 1 second, not 9.8 meters, if it starts from rest.)

    But for line #1, the "shortest path" might not be what you were imagining. I'm guessing that you were imagining a path that "hovers" at 9.8 meters altitude for 1 second; but that is *not* the shortest path between the two points in spacetime that you give (9.8 meters above Earth's surface at time t = 0, and 9.8 meters above Earth's surface at time t = 1 second). The shortest path is the path of an object that starts out moving upward, in free-fall, with just the right velocity to come to rest in half a second, then fall back down so that it is just passing 9.8 meters altitude again in 1 second.

    So we really have three paths involved here: line #1, defined as I just did above; line #2, defined as you did; and line #3, which is what I think you were thinking of when you defined line #1--a path that "hovers" at 9.8 meters altitude above the Earth for 1 second. Of these three, line #1 (as I defined it above) is the longest; then line #3; and line #2 is the shortest. So both line #1 and line #3 are longer than line #2, and the answer to your question is yes (and that's true whether we interpret "line #1" in your question as I think you were thinking--as line #3--or as you actually stated it--as my line #1 above).

    (2) Comparing the lengths of lines #1 and #3 is easy, because they both pass through the same endpoints in spacetime (9.8 meters above the Earth, at 0 and 1 second). But line #2 only shares one endpoint: it starts at the same point in spacetime as the other two, but it ends at a different point (4.9 meters above Earth at 1 second). So when we compare the length of line #2 with the other two, we are implicitly adopting a simultaneity convention: we are assuming that the endpoint I just described for line #2 is in fact simultaneous with the endpoint for #1 and #3, i.e., that "1 second" in all three cases means "1 second according to clocks at rest with respect to the Earth". But with this definition, the only one of the three paths whose length is actually 1 second is line #3; in other words, if we have clocks following all three lines, only the clock following line #3 will show 1 second of elapsed time. The clock following line #1 will show more than 1 second elapsed time, and the clock following line #2 will show less. (This is because "path length" along these paths *is* elapsed time for a clock following the path.)

    But we could adopt different simultaneity conventions. For example, we could adopt the simultaneity convention that is natural to an object following line #2; in other words, we judge events to happen at the same time if they happen at the same time according to an observer following line #2. If we adopt this simultaneity convention, then line 2 is *longer* than the other two lines (line #1 is still longer than line #3, because they share both endpoints, so the comparison of their clocks at each endpoint is an invariant and doesn't depend on our choice of coordinates or simultaneity conventions or anything else).
    Last edited: Jun 2, 2014
  4. Jun 2, 2014 #3


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    Do you mean the shortest path as measured by the space-time interval, which would be the elapsed time (more precisely, the elapsed proper time along the worldline)? Or do you mean the shortest path excluding time and considering only the spatial distance in the chosen frame?
  5. Jun 2, 2014 #4
    Thanks! Yes you were right, I was imagining the "hovering" path - it does start to make more sense now! and I should have remembered that s=ut+0.5at^2 oops!

    I started reading the spacetime entry on wikipedia which helps too. I didn't really appreciate how much "longer" a second was compared to a meter in spacetime - about 300 million times longer? Hopefully I'm not talking nonsense!

    I like the idea of being able to see the warping of spacetime with my eyes, and so felt a bit disappointed before to find out that the distortions of the 3 spacial dimensions due to the mass of the earth are so small that you would never be able to see them directly. But I have realized now (helped by your explanation above) that the motion of a ball if I just toss it straight up into the air and catch it again is a great way to see how warped spacetime really is all around me!
  6. Jun 2, 2014 #5
    pervect, I meant the shortest path including time ...
  7. Jun 2, 2014 #6


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    Actually, as I noted in my previous post, you mean the *longest* path including time; when you include time, the free-falling path has maximal elapsed time, not minimal.
  8. Jun 2, 2014 #7


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    Nope, you've got it right. To be precise, a second is 299,792,458 times longer than a meter in spacetime.

    Yes, exactly. Glad I could help!
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