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I Why can't Gravitational Accelerations vanish everywhere?

  1. Jun 14, 2015 #1
    In attempt to describe the consequences of the Equivalence Principle, this is almost said:
    When there are gravitational accelerations present, as for example in the
    gravitational field of the earth, the space cannot be the flat Minkowski space. Indeed,
    in the Minkowski space we can have
    $$\Gamma^{\lambda}_{\mu\nu}=0$$
    everywhere. This should then be interpreted as meaning that the sum of the inertial
    and the gravitational acceleration could be made equal to zero everywhere. This does,
    however, not correspond to our experience about gravitational accelerations: When
    gravitational accelerations exist, it is not possible to make them vanish everywhere
    .
    We can only make them vanish at one point, or approximately in a small region, by the
    use of an appropriate coordinate system.
    Therefore, when a gravitational field is
    present, the space will be necessarily a curved Riemannian space. The gravitational
    field will then appear as the expression of the fact that we are in a curved riemannian
    space and no longer in the flat Minkowski space.

    For the sentence in bold, what experience that tells us so about gravitational accelerations (that we can not make them vanish everywhere except in a small region)? How do we know that?
     
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  3. Jun 14, 2015 #2

    jedishrfu

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    I imagine a simple pendulum bob would show that as it would point toward the center of the earth hence in different directions in different places.
     
  4. Jun 14, 2015 #3

    WannabeNewton

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    Take two particles separated by some small distance and drop them to the floor; if the height of the fall is large enough you will see them converge towards each other. This means there is tidal gravity i.e. the gravitational field is not uniform hence we cannot transform it away everywhere by going to a freely falling frame. We can only do so locally in a freely falling frame that exists for a small period of time in a small region of space.
     
  5. Jun 14, 2015 #4
    I see, I still have one question then. In the question above, it is also mentioned that " This should then be interpreted as meaning that the sum of the inertial
    and the gravitational acceleration could be made equal to zero everywhere. " Why sum? How was this interpreted? I mean interpretation is something that follows from what preceded, I couldn't relate this sentence I just quoted to what preceded.
     
  6. Jun 14, 2015 #5

    WannabeNewton

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  7. Jan 14, 2016 #6
    @WannabeNewton I am also interested in the question asked, which is why sum? I didn't understand how the link you provided settles this?
     
  8. Jan 14, 2016 #7

    PeterDonis

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    The interpretation offered is not correct. "Inertial acceleration" and "gravitational acceleration" are not the same thing and you can't add them together or use one to offset the other. More precisely, you can add them mathematically, but the sum has no physical meaning.

    What the OP calls "inertial acceleration" is acceleration you actually measure with an accelerometer. It is a direct observable. (The more usual name for it in GR is "proper acceleration", or just "acceleration", since it is the only kind of acceleration that is an invariant, independent of coordinates.)

    What the OP calls "gravitational acceleration" is a coordinate-dependent phenomenon; you can make it vanish, at least locally, by choosing appropriate coordinates. (It is a special case of what, in GR, is called "coordinate acceleration".) So even if the "sum" of inertial and gravitational acceleration, calculated mathematically without regard for the physical meaning of the two, happens to be zero in one system of coordinates, you can make it change just by changing to another system of coordinates. But changing coordinates doesn't change any physics, so this "sum" can't have any physical meaning.

    A better way of interpreting the fact noted in the OP would be to say that, in flat Minkowski spacetime, we can find a system of coordinates such that objects in free fall, i.e., with zero proper acceleration, have zero coordinate acceleration everywhere; i.e., you can make "gravitational acceleration" vanish everywhere, globally, by choosing appropriate coordinates. In a curved spacetime, we can't do this globally; we can only do it locally, in a small patch of spacetime. In any global system of coordinates in a curved spacetime, freely falling objects--objects with zero proper acceleration--will have nonzero coordinate acceleration in at least some regions.
     
  9. Jan 16, 2016 #8
    @PeterDonis thanks very much for your answer. I understand your argument very nicely if I considered e.g. when you said " you can make "gravitational acceleration" vanish everywhere, globally, by choosing appropriate coordinates." is correct. However, I can't tell you that I know what you mean by "appropriate" coordinates here. I would appreciate an example or something if this is not difficult to give.
     
  10. Jan 16, 2016 #9

    PeterDonis

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    "Appropriate" coordinates are just standard inertial coordinates in flat spacetime: ##t##, ##x##, ##y##, ##z##, with the metric ##\eta_{\mu \nu} = - dt^2 + dx^2 + dy^2 + dz^2##.
     
  11. Jan 16, 2016 #10

    David Lewis

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    PeterDonis wrote: " What the OP calls inertial acceleration is acceleration you actually measure with an accelerometer. "

    David Lewis wrote: I suppose an accelerometer will also measure gravitational acceleration, and that it's unable to distinguish between inertial and gravitational acceleration.
     
  12. Jan 16, 2016 #11

    Ibix

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    No - this is the point Peter is making. If you are in a space-station, or dropped out of a hovering rocket over an airless planet, you are in free-fall. Your accelerometer (a bathroom scale, essentially) will read zero. You will be able to detect tidal effects with sufficiently precise instrumentation, but no acceleration.

    However, if you are sitting in a chair on earth or in the rocket under power then your accelerometer will have a non-zero reading. This is because the earth or the rocket floor is accelerating you out of your free-fall path.
     
  13. Jan 16, 2016 #12

    David Lewis

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    In free fall (near the earth's surface) your accelerometer will initially display the sum of inertial acceleration (9.81 m/s^2) and gravitational acceleration (-9.81 m/s^2).
     
  14. Jan 16, 2016 #13

    Ibix

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    No. It will display your proper acceleration - zero.

    What force is causing the "inertial acceleration" you are talking about?
     
  15. Jan 16, 2016 #14

    PeterDonis

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    You suppose incorrectly. You need to think more carefully about the definitions of the terms you are using.

    Why do you think "inertial acceleration" will be +9.81 m/s^2? What definition of "inertial acceleration" are you using? On the definition I gave, "inertial acceleration" in free fall is always zero; that includes the case you are describing. You are correct that "gravitational acceleration" in your scenario is - 9.81 m/s^2, but as I've already pointed out, "gravitational acceleration" is just a coordinate effect which can be made to vanish, locally, by an appropriate choice of coordinates; it's not what an accelerometer measures.
     
  16. Jan 16, 2016 #15
    @PeterDonis , I ve read some place that when we are in Minkowski space we have##\frac{d^2}{d\tau^2}x(\tau)=0## and then we we do general coordinate transformation, we get ##\frac{d^2}{d\tau^2}x^{\mu}+\Gamma^{\mu}_{\nu\lambda}\frac{dx^{ni}}{d\tau}\frac{dx^{\lambda}}{d\tau}=0##, the author calls the second term "inertial acceleration" does this contradict with what you said?
     
  17. Jan 16, 2016 #16

    PeterDonis

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    What place?
     
  18. Jan 16, 2016 #17

    PeterDonis

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    Can you give a chapter/page reference?
     
  19. Jan 16, 2016 #18
  20. Jan 16, 2016 #19

    PeterDonis

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    Ok, it looks like he is using the term "inertial acceleration" here to refer to what I am calling "coordinate acceleration", and what the OP appears to refer to as "gravitational acceleration". His terminology is confusing, however, since he uses the term "gravitational acceleration" to refer to the same thing. (The OP appears to be quoting from the same Papapetrou book, just a little further on from the passage you quote.) Basically what he seems to be saying is that this thing--the term involving ##\Gamma^{\mu}_{\nu \lambda}## in his equation 18.3--can be called either "inertial acceleration" or "gravitational acceleration", or a combination of both, depending on how we're looking at it.

    This provides a good example of why ordinary language is the wrong tool to use when discussing physics. When you look at the math, it's obvious that the term involving ##\Gamma## is coordinate-dependent. That's true whether spacetime is flat or curved, and whether you think of the term as "inertial acceleration" or "gravitational acceleration" or whatever. This coordinate dependence is the reason why the usual term for this thing, if you have to use ordinary language to describe it, is "coordinate acceleration".

    But coordinate-dependent things are the wrong things to focus on when trying to understand the actual physics. What Papapetrou appears to be doing in the quoted passages is to try to match up ordinary Newtonian intuitions about non-inertial frames and "gravitational fields" with the correct relativistic math as best he can. IMO it would be better just to say straight up that Newtonian intuitions are wrong and that trying to understand GR in terms of them is not going to work well.

    Furthermore, when you look at the math, you can also see that the term involving ##\Gamma##, in and of itself, tells you nothing about the actual physical observable in question: the reading on an accelerometer carried by the object, i.e., the proper acceleration (which is what I called "inertial acceleration" in earlier posts--but it's clear now that that's not how Papapetrou is using that term, so I'll avoid it from now on). The proper acceleration is given by the sum of both terms in equation 18.3 of Papapetrou, i.e., it is given by

    $$
    \frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} \frac{dx^\lambda}{ds}
    $$

    If this sum is zero, the object is in free fall, feeling zero acceleration. If this sum is nonzero, the object is feeling acceleration, the magnitude of which is given by the sum. And this is true regardless of the coordinates we choose.
     
  21. Jan 16, 2016 #20
    @PeterDonis I am confused. I have to say. Then what happens to your answer here:
    How can this answer (that answered my question about why sum and OP's question what experience does this tell us) be translated after knowing that the book we're using means different things?
    Also, notice there that the author says gravitational=inertial acceleration after using Principle of Equivalence, so would this mean that what you said previously can still hold? Now that I am lost a little, I would appreciate if you could translate your previous answers and I will read them as if I am reading those for the first time. Thanks Pete.
     
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