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If Gravity is not a force , then how does this work?

  1. May 23, 2012 #1
    If Gravity is not a "force", then how does this work?

    This has got to be a question with a simple answer. But I can't figure it out.

    I understand that Einstein demonstrated that gravity is not a "force", as was contemplated by Newton. Instead, gravity alters the geometry of spacetime.

    I understand how this affects the path of an object traveling close to a large mass. The object "thinks" it is following a "straight line", but the space in which the object is located is curved by the mass, so that the straight line becomes curved from the perspective of an observer comoving with the large mass. I can visualize that.

    But this I cannot understand:

    If an object is at rest with respect to a large mass, comoving in an inertial frame of reference with respect to the large mass, why does it approach the large mass absent an external force applied to the object?

    Imagine a rock supported on a platform. It is stationary with respect to the earth. If the platform is removed, the rock will move towards the center of gravity of the large mass.

    Why?

    Again, I understand how curved space can alter the trajectory of an object that passes near a large mass. In that case, the straight line the object would otherwise traverse follows the warped space, and the object will veer towards the large mass.

    But if the object starts out stationary with respect tot the large mass, and if there is no force to move the object, why does it fall?


    The object exists and is stationary with respect to the mass within the curved space. But what makes it move with respect the the center of gravity if it was originally stationary?

    How can gravity not be a "force"? Do I simply misunderstand the definition of "force"?

    When Einstein denied that gravity was a "force", did I misunderstand what was said in some fundamental manner?
     
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  3. May 23, 2012 #2

    stevendaryl

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    Re: If Gravity is not a "force", then how does this work?

    That's the same effect as the first one, which you already understand. You have to remember that in relativity, the universe is considered to be 4-dimensional, not 3-dimensional. In the spacetime view, every object has a 4-velocity: it's velocity in the x-direction, in the y-direction, in the z-direction and in the t-direction. A rock sitting on the ground isn't at rest, it's traveling at a hefty speed in the t-direction. The presence of the Earth bends the object's path so that instead of traveling straight in the t-direction, its path bends a little, picking up a small amount of velocity in the z-direction.
     
  4. May 23, 2012 #3

    Nugatory

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    Re: If Gravity is not a "force", then how does this work?

    Two things to consider:
    First, curvature can create the illusion of a force for moving objects. For example, if you and I stand 100 feet apart at the equator and start walking north, we'd discover that there is this force inexorably pushing us closer until we collide at the north pole. This can be a big deal - it's what makes hurricanes spin.

    Second, Einstein's theory of gravitation is based on curvature of four-dimensional space-time, not just three dimensional space. And in four-dimensional space-time you are always in motion... Even when you feel at rest, you're still moving forward in time.
     
  5. May 23, 2012 #4
    Re: If Gravity is not a "force", then how does this work?

    It's not just the curvature of space which matters, it's the curvature of time as well. The initially stationary rock will move towards the Earth because of time curvature. Time curvature isn't exactly an easy thing to visualize, so try to think of it like this:

    Take a spacetime diagram of two objects in flat spacetime, at rest with respect to each other. Their worldlines will be two parallel lines which never intersect in the future.

    Now imagine drawing a spacetime diagram on a curved surface, such as the surface of a ball. Even if the objects are initially stationary with respect to each other (their worldlines are initially parallel), as you extend their "straight" worldlines you will see them eventually intersect.

    http://i.minus.com/ibtnChlQE4HLM5.jpg [Broken]
     
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  6. May 23, 2012 #5
    Re: If Gravity is not a "force", then how does this work?


    Thanks, but again, there must be something fundamental that I misunderstand.

    I posit an object (a rock) comoving with a large mass (like the earth) in all 4 directions. It is sitting on a platform. It exists in a stationary mode with respect to the earth, or comoving in an inertial manner along with the earth.

    When the platform is removed, suddenly the rock is no longer comoving with the earth, but instead, it changes direction and velocity.

    If gravity is not a force, and no other force is applied to the rock, what accounts for its sudden accelleration towards the center of the earth?

    Newton would explain things with some sort of mysterious "force of gravity" which seems to make sense, if one accepts that things such as fores exist, and can affect physical objects.

    Einstein would say that there is no force exerted on on objects by gravity, but instead, that the behavior of objects is determined by their path through curved spacetime.

    But if there is no "path" being followed by the stationary rock (in that it is comoving with the earth in an inertial frame), then why does it suddenly diverge from the former comoving path, and instead, start moving towards the center of the earth?
     
  7. May 23, 2012 #6

    PeterDonis

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    Re: If Gravity is not a "force", then how does this work?

    The fact that, until you removed the platform, there *was* a force acting on the rock: the force of the platform pushing up on it. When you removed the platform, you removed that force, and so the rock took up its "natural" path through spacetime, the one it follows when there are no forces acting on it. That path is the freely falling path, because of the curvature of spacetime around the Earth.

    Another key point here is that the rock being "stationary" while on the platform, and experiencing a "sudden acceleration" when the platform is removed, are frame-dependent statements; that is, they depend on your choice of coordinates. That means they are not good statements of physics in relativity; to properly do relativistic physics, you have to find ways of describing a scenario that are frame-independent, i.e., that hold true regardless of your choice of coordinates.

    Here are some statements about your scenario that *are* frame-independent:

    (1) While sitting on the platform, the rock feels a force--its weight. (In principle we could measure this force by attaching strain gauges to the rock, sitting it on a scale, etc.)

    (2) After the platform is removed, the rock does *not* feel a force--it is in free fall, weightless, and any measurement of the "force on the rock" would read zero. (For example, a scale freely falling with the rock would not register any weight.)

    In General Relativity, these types of statements are used to *define* what a "force" is, precisely because they are frame-independent, whereas statements like "the rock experiences an acceleration when the platform is removed" are not. (There is in fact a frame-independent way to define "acceleration": an object is accelerating if it *feels* acceleration! This is called "proper acceleration". By this definition, the rock is accelerating while on the platform, but not after the platform is removed. The advantage of this definition, like the corresponding definition of "force", is that it is frame-independent.)
     
  8. May 23, 2012 #7
    Re: If Gravity is not a "force", then how does this work?

    Good. You are saying something very similar to the first responder. And what you are saying is something I haven't ever though about closely.

    I am missing something, and that something seems to be related to motion on time.
     
  9. May 23, 2012 #8
    Re: If Gravity is not a "force", then how does this work?


    Ah. So is my misunderstanding related to the impossibility of "visualizing" 4 dimensional frames of reference?

    I can understand the concept of "parallel" lines converging on the surface of a sphere. That is all in 3 dimensions.

    But I'm still struggling with the rock, which seems to suddenly start moving, absent any force, when the platform is removed.

    Must I simply chalk this up to my inability to visualize four dimensional spacetime?

    Would it be correct to say that the rock is not "stationary" when it is sitting on the platform? That it is moving through time, with a force (the platform) altering its trajectory?

    That is mind bending, and I still can't wrap my mind around it.
     
  10. May 23, 2012 #9
    Re: If Gravity is not a "force", then how does this work?


    These are concepts that I thought that I understood in the context of GR.

    Now I see that I do not.

    I need to think about what you have said, and what I thought I understood, and see if I can merge the two.


    Edit:

    To test my new half-understanding, can you tell me if these are true statements?

    On the platform, the rock is not comoving inertially WRT the earth through 4 dimensional spacetime. Instead, the rock has a force applied to it, which is identical to a 1G acceleration from the point of view of the rock.

    The earth is warping and curving both space and time.

    On the platform, the rock is "moving" with respect to the earth - it has a force applied to it by the platform. Such force is exactloy 1G, and is in the direction exactly away from the center of the earth.

    --

    I think I can understand each of the above statements. i need to warp and curve my mind in order to put them all together.

    Damn. I started learning about quantum physics because I thought I had a good handle on Relativity. But I think that I have a ways to go before I really understand.
     
    Last edited: May 23, 2012
  11. May 23, 2012 #10
    Re: If Gravity is not a "force", then how does this work?

    Look at the picture I posted above. Think of it as a graph of position vs. time, with time going vertically. As the objects move forward in time, their paths intersect.
     
  12. May 23, 2012 #11

    stevendaryl

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    Re: If Gravity is not a "force", then how does this work?

    You answered it yourself in your first post. You wrote
    Draw a graph, with the time axis going vertically, and the z-axis going horizontally (ignore x and y, for now). The Earth is at z=0, going straight up. The rock is at z=1 (say), and is also going straight up. So if there were no gravity, these two paths, which start off parallel, would remain parallel forever, never getting closer together. But gravity warps spacetime, so that things follow paths that are curved. The two paths don't continue to go straight up, but curve in toward each other.

    The analogy that people use is lines of longitude on the surface of the Earth. Imagine you have two objects that are both at the equator, one at longitude 0 degrees, and another at longitude 90 degrees East. You send both objects straight north. Although their paths start off parallel, as they proceed north, their paths don't remain parallel; their paths converge. When they get to the North Pole, they will collide.

    So think of the Earth as the object traveling at longitude 0 degrees and the rock as the object traveling at longitude 90 degrees. North in this case represents the time direction, and east-west represents the z-direction. As the objects travel north, their paths bend so that their east-west distance gets smaller, until eventually they collide.

    What do you mean, there is no path being followed by the stationary rock. It's traveling at speed c in the t-direction, just like in the analogy, it's traveling at some speed directly north.
     
  13. May 23, 2012 #12
    Re: If Gravity is not a "force", then how does this work?

    I will do that. I am familiar with the concept of space and time being equivalent in that they are dimensions. I am familiar with the concept of rotating through dimensions.

    But I've never practiced it in my head, and certainly not mathematically.

    I will try to do that.

    And I suppose that there is no answer to why the two objects moving through time converge. Or maybe it all has to do with the curvature of spacetime, and the impossibility of flat spacetime in the presence of a large mass.

    Thanks to all for giving me a handle on things I do not yet understand. This is why I love physics. I need to struggle with it. Most stuff is conceptually easy in comparison.
     
  14. May 23, 2012 #13

    PeterDonis

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    Re: If Gravity is not a "force", then how does this work?

    Incorrect. At least, assuming that by "comoving" you mean "at rest relative to". The rock and the Earth are indeed at rest relative to each other, as you could in principle verify by bouncing light signals between them and verifying that the round-trip travel time is constant (as seen by either the rock or the Earth). It's possible that your concept of "comoving" is conflating things which are the same in flat spacetime (i.e., in the absence of gravity), but not in curved spacetime (i.e., in the presence of gravity).

    Except for the "instead" part, this is correct.

    Yes.

    Incorrect; see above.

    Correct. The fact that you think this goes together with the incorrect statement just above is one reason why I think you are conflating different concepts. Feeling a force does not necessarily imply relative motion, or vice versa.

    One other thing that you may not be taking sufficient note of is that the Earth, in this scenario, is not just an object that sits in the spacetime--it is the *source* of the curvature of the spacetime. That means that you can't treat its "motion" in this scenario the same way you treat the motion of the rock. When I said that the rock and the Earth are at rest relative to each other, you'll note that I was careful to define "at rest relative to each other" in terms of a direct observation--constant round-trip travel time of light signals between them. That means that the small piece of the Earth's surface directly beneath the platform on which the rock sits is accelerated, in the same sense as the rock is--if you could put a scale underneath that piece of the Earth's surface, it would register weight. But the Earth as a whole is not accelerated in the same sense. So you have to be careful in thinking about what is "moving", what is "accelerating", etc.

    Yes (except that 1G is actually a unit of acceleration, not force--but they're basically equivalent).
     
  15. May 23, 2012 #14

    PeterDonis

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    Re: If Gravity is not a "force", then how does this work?

    On re-reading your post I noticed that you said "comoving inertially", not just "comoving". This may be another point of confusion, related to what I said in my last post about not being able to treat the Earth's motion the same as the rock's, and about the Earth being the source of the spacetime curvature. On a scale in which the rock's acceleration is significant, there is no single "inertial frame" that encompasses the entire Earth. (There might be on a scale that was large compared to the Earth but small enough compared to the Earth's orbit about the Sun, so that the spacetime curvature due to the Sun could be neglected. But in such an "inertial frame" the rock's acceleration would be too small to measure.) In a curved spacetime, you can only use inertial frames locally--locally in both space and time, meaning over a small enough piece of spacetime that the effects of spacetime curvature, in both space and time, are negligible. Obviously the Earth itself, as the source of the curvature, can't fit into such a local inertial frame.
     
  16. May 23, 2012 #15

    PeterDonis

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    Re: If Gravity is not a "force", then how does this work?

    Don't give up too easily. :smile: There is. You are already headed in the right direction:

     
  17. May 23, 2012 #16
    Re: If Gravity is not a "force", then how does this work?

    Expanding on the concept is that things move along a "free fall" geodesic when not subject to any forces,
    consider that a man and a marble free falling at the same rate. Since neither is acted on by any force we would expect them to take identical paths.
    Otherwise we would have to believe that Newton's "gravitational force" somehow *exactly* cancels the seemingly unrelated inertial properties of matter.

    Consider John in empty space with rocket motors mounted on the bottoms of his shoes accelerating him at 9.8m/s^2.
    He passes Jane who is floating in space right next to a marble.
    Jane and the marble are just doing what masses do when they are not subject to any forces
    (and this is true to her senses, free falling to earth feels just like floating is empty space).
    John will observe Jane and a marble both accelerating at the same rate. What conclusion should he draw?

    If John conjured up a force that was causing Jane and the marble to
    accelerate (or fall) identically while ignoring the force that he could feel on the bottoms of his own feet, this is essentially Newtonian gravity.
     
    Last edited: May 24, 2012
  18. May 24, 2012 #17
    Re: If Gravity is not a "force", then how does this work?

    That's actually a very very good analogy. You can consider the ground accelerating you "up" at 9.8 m/s/s and someone in free fall would be equivalent to Jane floating in space.
     
  19. May 24, 2012 #18
    Re: If Gravity is not a "force", then how does this work?

    I think I understand the difficulty you are having. In the first case the curvature of space is fairly easy to visualise, but in the second case the object appears to fall straight down and no curvature is apparent. As I understand it, the key concept is that gravity is curvature of spacetime rather than just curvature of space. This is a rather abstract concept and can be difficult to visualise. This diagram may help:

    http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif

    In the above diagram the path of the falling object is a straight line from the top of the house to the bottom of the house, from a purely spatial perspective. The path going from left to right includes a time element and is the straight path in curved spacetime of an object in freefall following a geodesic. Objects that are stationary in the gravity field (e.g. the house) follow curved paths through the spacetime and are subject to proper acceleration. The curvature is result of the relative different rates of clocks at different potentials. I hope that helps.
     
  20. May 24, 2012 #19
    Re: If Gravity is not a "force", then how does this work?

    The rock doesn't experience a sudden acceleration at all. It's the ground flying up towards it, from the rock's reference frame. :smile:
     
  21. May 24, 2012 #20

    A.T.

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    Re: If Gravity is not a "force", then how does this work?

    When you consider only motion along a line, then you can omit 2 space dimensions. Have a look at this:
    http://www.relativitet.se/spacetime1.html
     
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