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Another Inertial ref frame q :-(

  1. May 30, 2006 #1
    It's not that I don't believe that all i.r.f.'s are on an equal footing, it's just that I can't seem to get that to square with the following thought experiment, undoubtedly 'cause I'm getting lost in my own undies here, and I need help getting a handle on this.

    Suppose you have a universe with just ONE planet in it (so that there's no outside ref points to determine relative motion of any kind).

    Within that universe there would have to be an "absolute" or fundamentally correct reference frame wouldn't there?

    I mean, that planet is either spinning, or it's not. Period. If it's spinning, especially if it's spinning really fast, your weight, standing on it's surface, is going to be lower than you weight would be if it were not spinning, right? If you're going to say it's spinning, spinning in relation to what?

    Using an inertial reference frame that spun with the planet (i.e. the planet is defined as completely motionless) would lead you to an incorrect value for gravitational attraction. The fact that you don't know that the planet is spinning, 'cause you don't have any reference points to compare to, doesn't change the fact that you wind up with the wrong value for "g" does it? I mean wrong really is wrong right? Sooner or later you're going to notice and have to explain coriollis effects. It's not like spinning or not spinning is a subjective condition. You're either spinning, or your not, and those two conditions are distinct with distinct and observable consequences.

    So how can all inertial ref frames stand on equal footing?

    I'm choking on my elastic waste band here. HELP!
  2. jcsd
  3. May 30, 2006 #2


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    Look up "Mach's Principle", the guy has been thinking about exactly that issue in the beginning of the 20th century (or was it the end of the 19th ?).

    In general relativity (or any theory that has general covariance) the problem goes away: ANY coordinate system will do, and what happens to be the metric in your preferred coordinate system will determine what are the paths of non-interacting particles. So no more wondering whether your coordinate system is really an "inertial" one: just pick one (for instance attached to the surface of your one and only planet).
    Now, by doing local experiments everywhere with non-interacting particles, you can determine the metric expressed in this coordinate system.
    If it turns out that it is close to a Schwarzschild solution, then you could say - if you wanted to - that your planet is "not rotating". If it turns out to be another metric, but such that when you apply the coordinate transformation phi --> phi + omega . t, you obtain a metric which looks much more like a Schwarzschild solution, then you MIGHT say as well that "your planet is rotating with rotation speed omega".
    But if you find the last statement a non-physical statement in your space, then you can just as well accept that the metric, expressed in the coordinate system attached to your planet's surface, is what it is, without viewing it necessarily as "rotating".

    The deviation of the empirically measured metric from a Schwarzschild metric will simply include then the variation of effective acceleration from the poles to the equator.
  4. May 30, 2006 #3
    Thanks Vanesch!

    This gives me lots of stuff to look up and read about. Many of the terms you're using go way over my head, but what the heck, learning what terms like "general covariance" and "Schwarzschild solution" mean is half the fun here. If I can't find what I need to pick apart your answer, I'll just ask. But for now I want to see how much I can learn on my own by doin' a little googling and reading.

    Again, thanks. You've given me a great place to start.
    Last edited: May 30, 2006
  5. May 31, 2006 #4


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