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Energy and acceleration

  1. Jan 10, 2012 #1
    This is a dumb question. Thank you for your patience. I understand that acceleration and gravity are one and the same---- as are their effects . Is this to be understood as the earth really accelerating under us , this motion being the force of gravity -- the earth rising up to meet us so to speak making it impossible for us to fall off as its' surface is always coming up under us?
    If this most likely utterly wrong idea were to be really so it would lead to two other questions-- where is the earth's ' supposed acceleration going to or coming from and why---- and what happens to the classic Newtonian, Gallileo match of mass and weight and the earth's gravity in such a case? Thank you.
  2. jcsd
  3. Jan 10, 2012 #2


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    Gravity is not the same as acceleration. It's effects are the same locally as the effects of an accelerating reference frame.
  4. Jan 12, 2012 #3
    Hi joan, welcome to these forums..lots of smart people here [not me].

    The above is entirely correct post..but the meaning may not be clear.....
    [I've never seen the following description anywhere else and have posted it here maybe a dozen times; so far, not one has attempted to refute it...so I still take it as a good intuitive insight of the underlying mathematics.]

    good question,NOT dumb...it took an "Einstein" to make the connection and draw the (limited) proper conclusions from that relationship!!!!

    The earth surely IS accelerating under us...it's rotating about the sun.....but that effect has almost no effect on gravity and is not specifically what Einstein meant. That rotational acceleration is a =v2/r. Such Newtonian (old physics) acceleration does provide an apparent curvature to space. But gravitational curvature is different.

    DrGreg provided me an answer during a related discussion in these forums several years ago:

    [The key is that gravitational curvature IS observer independent and is reflected as curvature of the spacetime manifold ("graph paper" as described below). Frame dependent curvature (observer dependency) is a variable overlay on top of this fixed background curvature.]

    [By the way, Einstein used a gravitational analogy with no 'tidal forces'...you can find what THAT means in wikipedia:

    http://en.wikipedia.org/wiki/Equivalence_principle ]
    Last edited: Jan 12, 2012
  5. Jan 17, 2012 #4
    Dear Naty 1 Thanks very much for replying to me and in such depth. I am a physics illiterate and a math one too so much is wasted on me but I do have a glimmer of understanding from your answer. I think I can at least get the idea that the observance of space time curvature can be from the point of view of two types of observers- two inertials no gravity and one inertial and one accelerated no gravity and there is still no real space time curvature although it looks like there is for the accelerated one due to the paper illusion (?) described (which description is too deep for my understanding really.) And then I see that when you introduce gravity it becomes real and a paper grid is only an approximation.(Ithink.) To tell the truth I cannot fit the rotation of the earth into this great explanation-- too much thinking. I also end up seeing something strange: there is no space time curvature if you are free falling or accelerating only if you somehow "land" and make a dent in space time. And this no matter how heavy you are ? Because you are not heavy you are weightless? What if you threw the moon out of the earth's gravitational field -- would it only free fall and never "land" and bend space time? Sorry and thanks again you have done more than the text books. Verytrulyyours, Joan Pendleton
  6. Jan 17, 2012 #5


    Staff: Mentor

    Joan, let me add my welcome to Naty1's, and also my assurance that the questions you are asking are not dumb. I think you already have a grasp of the fundamental point about gravity, but let me try to suggest some ways of looking at it that may help.

    The key word to add to this is "locally", meaning over a sufficiently small region of spacetime. Basically, this is the "Einstein elevator" thought experiment: if I'm standing on the floor of a windowless room, with no way to see outside, then over a short period of time, I can't tell whether the weight I feel, the force pushing up on me from the floor, is due to the room being a rocket accelerating at 1 g in empty space, or due to the room standing motionless on a planet like Earth whose surface gravity is 1 g. But if I wait long enough, or if the room is large enough, there will be observable differences between these two scenarios. See below.

    You can kind of look at it this way, but you have to be careful doing it, as your further question indicates:

    This question does not really have an answer; the Earth is not "accelerating up" underneath us as a rocket would. It can't possibly be, because it would have to be accelerating in all directions at once in order for people all over the Earth to simultaneously feel a 1 g acceleration towards the center. This is one of the differences I was talking about above; if your "windowless room" is large enough compared to the Earth, you can measure the change in direction of your "acceleration" as you go from one place to another; in a rocket accelerating in empty space, there would be no such change in the direction of the acceleration.

    The equivalence of inertial and gravitational mass (all objects fall with the same acceleration in a gravitational field) is still valid in GR. In fact, this equivalence (which is verified experimentally to a very high accuracy) is the key feature that allows us to interpret gravity as spacetime curvature. See below.

    Correct: in both of these cases, spacetime is flat. Here's how you can tell: use the worldlines of inertial observers (i.e., observers that are in free fall, feeling zero acceleration) as your "grid lines" in the time direction. Then use the standard geometrical test for curvature: take two grid lines that start out parallel, and see if they stay parallel. For inertial observers, this means: take two inertial observers that are at rest with respect to each other at some instant of time. If they both stay at rest with respect to each other for all time, while remaining inertial (feeling no weight), then spacetime is flat. But if they don't, then spacetime is curved. See next comment.

    Given the test I just described, can you see how the presence of gravity equates to spacetime curvature? Suppose I take two objects and drop them off a tall tower at two different heights at some time t = 0. (Assume the tower is evacuated and hollow, so air resistance is absent.) Both objects are in free fall and at rest relative to each other at t = 0 (because I just that instant dropped them). They both fall, but since one is higher than the other, they fall with slighly different accelerations; so they will not remain at rest relative to each other (i.e., even if we switch to a frame of reference in which one stays at rest, say the higher one, the other will not stay at rest in that frame). So the two objects trace out two "grid lines" in spacetime that start out parallel but don't stay parallel--which means spacetime is curved.

    You can also see that the key factor was not just the presence of gravity per se, but the fact that gravity changes strength (and direction--we could run a similar test by dropping two objects from the same height but at different locations, and seeing that they move towards each other) as we go from place to place. In other words, *tidal* gravity is what is equivalent to spacetime curvature.

    I would postpone thinking about this until you are comfortable with the static case, with no rotation (which is what I was assuming in my comments above).

    From the above you should now be able to see that this is not correct; spacetime curvature can be observed solely by looking at the motion of freely falling objects.

    By the equivalence of inertial and gravitational mass, which I mentioned above, yes, the path of a freely falling object is independent of its mass. However, technically this only holds true if the mass is small enough not to measurably affect the gravitational field in which the object is moving. (The term used for this is that the object must be a "test particle", whose own mass is not taken into account in determining the field--in principle *all* massive objects produce gravity, no matter how small, but in practice we can often ignore the contributions of all but one or a small number of masses.)

    The moon and the earth is a case where we can't just consider one object as a "test particle"; the moon's mass measurably affects the field that the earth and the moon move in, and if you somehow threw the moon out of orbit, there would be measurable changes in the field. Also, if you are going to consider throwing the moon out of orbit, you have to include the energy used to do that in your calculations as well.

    Regarding free fall, the moon is in free fall now (as is the earth), because it is moving solely under the influence of gravity. To throw it out of orbit, you would have to subject it to an acceleration somehow, which would mean it would no longer be in free fall while your moon-sized rocket (or whatever device you used) was in operation (which might only be a finite time, after which the moon would presumably be in free fall again). But the moon's gravity would be there the whole time and would produce a measurable spacetime curvature.

    Hope this somewhat long-winded post helps; please feel free to ask further questions.
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