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Inertial Frame of Reference & Equilibrium

  1. Nov 1, 2009 #1
    Is it safe to say that any object in an inertial reference frame is at a state of equilibrium?

    If so.. is it safe to say the opposite: that any object in a state of equilibrium is in an inertial reference frame?
     
  2. jcsd
  3. Nov 1, 2009 #2

    Vanadium 50

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    No to both. I would argue that a body in a circular orbit is in a state of equilibrium, yet it is not in an inertial frame. This has some dependence on how you define equilibrium, but that's probably a good reason not to use that word.
     
  4. Nov 2, 2009 #3

    Dale

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    I would say yes with the following clarification in bold.
    Also, by "equilibrium" I assume you mean the usual definition used in statics: the sum of all forces is 0 and the sum of all torques is 0.
     
  5. Nov 2, 2009 #4
    That is what I meant, and thank you for your insight.
     
  6. Nov 7, 2009 #5
    I guess I have another question, I'm trying to understand the difference between equilibrium and movement on a geodesic. Let's say for example that a body is at rest on the surface of the earth. Let's also say the earth is not rotating so only gravitational force is applicable. Let's also consider an object that is moving on it's geodesic as it falls towards into a deep canyon on the earth. Let's finally consider a third object which is an object in outer-space far from the reaches of gravity that has no linear momentum yet is rotating (rotational equilibrium). Now we say that acceleration is absolute in the sense that an object that is accelerating will be considered in all frames of reference to be accelerating. We also say that an object that is in free fall on it's geodesic (the one falling into the canyon) is considered to be under no proper acceleration. We also say that the one that is at rest on earth is under no 'coordinate' acceleration. Finally, we say the one in outerspace is in rotational equilibrium. If we take a 4th person who is viewing all of these from a position in outerspace, he will see the object that is rotating as accelerating, he will see the object falling towards the earth as accelerating and he will see the object resting on earth as undergoing no coordinate acceleration.

    If we say that acceleration is absolute and that an object on its geodesic undergoes no proper acceleration, isn't that the object that in all reference frames should be considered to be not accelerating of the three? It seems, however to be the one at rest.

    All three are in states of equilibrium and will require some force to deviate from their paths.
     
  7. Nov 7, 2009 #6

    A.T.

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    You still confuse proper acceleration and coordinate acceleration. Always specify which one you mean, instead just saying "acceleration/accelerating".
     
  8. Nov 7, 2009 #7
    Indeed I am :-) If we say that acceleration is absolute meaning it will show up in all reference frames do we mean coordinate or proper?
     
  9. Nov 7, 2009 #8

    Dale

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    Proper acceleration is the absolute one, it is what is measured by accelerometers. Coordinate acceleration is relative to the coordinate system.

    For example, consider an accelerometer on the edge of a rotating disk in deep space. The accelerometer measures the centripetal acceleration. Considered from an inertial frame this accelerometer reading matches the second time derivative of its coordinates, so the coordinate acceleration and the proper acceleration are equal. However, now consider the disk's rotating reference frame. In this frame the second time derivative of its coordinates is zero, so the coordinate acceleration does not match the measured proper acceleration. To explain this we include a fictitious force, the centrifugal force, which counteracts the centripetal force. Such fictitious forces, being fictitious, are not measurable by accelerometers. Both frames agree on the proper acceleration (the measured acceleration in each case), but they disagree about the coordinate acceleration (equal to the measured acceleration for the inertial frame, equal to 0 for the rotating frame).
     
  10. Nov 7, 2009 #9
    DaleSpam: Are you a teacher? Very impressive. Thank you. You eased a troubled mind.
     
  11. Nov 7, 2009 #10

    Dale

    Staff: Mentor

    Thanks! That is one of the best compliments I could think of. I used to be a physics tutor, but that was a long time ago.
     
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