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Why aren't you accelerating when you fall

  1. May 7, 2010 #1
    I get that gravity is acceleration we feel as a result of moving through spacetime. But I don't understand why it is that, when you fall, you are closer to not accelerating than when you sit still.

    Can someone please explain that? thanks!
     
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  3. May 7, 2010 #2

    Borg

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    Acceleration is a change in velocity. There is no acceleration when you are not moving. When you fall, you do accelerate.
     
    Last edited: May 7, 2010
  4. May 7, 2010 #3

    Ich

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    You don't feel any force when you're free falling.
    You feel force when you're on the ground.
    The basic idea of GR is to take these observations at face value. It's the floor that accelerates and takes you with it. Without the floor, nothing accelerates you and you follow your natural, inertial path, called a geodesic.
    Of course, the accelerating floor is a bit difficult to achieve for a spherical body like the earth. That's what "curved spacetime" means, that points that remain stationary wrt each other still can feel acceleration in different directions.
     
  5. May 7, 2010 #4

    Dale

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    As with many relativistic concepts you need to distinguish between two different "flavors" of acceleration.

    One is "proper acceleration" which is the coordinate-independent acceleration measured by an accelerometer. Proper acceleration is 0 for an object in free-fall, and an object sitting still on the ground has a proper acceleration of 9.8 m/s² upwards as measured by an accelerometer.

    The other is "coordinate acceleration" which is the second time derivative of your position in some coordinate system; this is obviously a concept which depends on your choice of coordinates. You can choose a coordinate system where free-falling objects are not accelerating, this is called an inertial coordinate system, or you can choose a coordinate system where free-falling objects are accelerating, this is called a non-inertial coordinate system. In non-inertial coordinate systems you get "fictitious forces" or "inertial forces" like gravity which are required in order to make Newton's laws work in the non-inertial coordinate system.
     
  6. May 7, 2010 #5
    The Einstein elevator thought experiment might give some insight to this. Imagine you are in the elevator far out in space far from any gravitational sources. When you release a mass it tends to stay where you put it or drift at a slow constant velocity relative to you. There is no preferred direction such as up or down. Now imagine you are in the elevator at the top of a tall building on the Earth and the cable is cut. As the elevator free falls, released objects behave in much the same way as the elevator in space, either staying where they are or drifting slowly.

    If on the other hand, when the elevator is on the surface of the Earth, any released objects rapidly accelerate in a preferred direction that we call down. This is pretty much the same as what you would observe if the elevator was in a rapidly accelerating rocket. Stationary on the surface of the Earth is equivalent to being in an accelerating rocket far out in space. In both cases you would feel acceleration and you would measure acceleration of yourself or the elevator using and accelerometer device. On the other hand when you and the elevator are free falling, everything you feel and measure inside the elevator is the same as being in a stationary elevator in space, far from any gravitational sources.
     
  7. May 7, 2010 #6

    Fredrik

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    The concept of "straight lines" in spacetime has to be defined, just as we have to define what a straight line is in a plane. In the plane, we can define a curve to be a "straight line" if the points (x,y) on the curve satisfy ax+by+c=0 where a,b,c are constants. Alternatively, we can define a curve as a function [tex]t\mapsto (x(t),y(t))[/tex], rather than as a set of points (the point set is then called the graph of the curve), and then say that the curve is a straight line if x''(t)=y''(t)=0. In GR (and SR too actually), we use a definition that's very similar to that last one, to define which curves to call "straight lines". Actually the technical term is "geodesics", so I'll use that from now on.

    We then define an object to be "not accelerating" if its motion is represented by a geodesic in spacetime. For massive particles, we then define "proper acceleration" as a measure of how much the particle's motion deviates from geodesic motion.

    GR describes a relationship between the properties of matter in spacetime, and the geometry of spacetime. This is why a massive object like a star will change which curves in that region of spacetime are considered geodesics. If a large enough region is empty, then the projection of a geodesic in spacetime onto the 3-dimensional hypersurface that we consider "space, at time t" would be (approximately) a straight line. But when a star is present, the projection of the geodesics onto space are approximately elliptical instead. There's no way to see "why" this is so without actually doing the calculations.
     
  8. May 7, 2010 #7

    Borg

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    Thank you guys for ignoring my ignorant posting. I wasn't thinking about the question relativistically and I wouldn't have posted if I had. :shy:
    BTW, Fredrik - I've always wanted to say, nice avatar.
     
  9. May 7, 2010 #8
    You don't feel acceleration. You feel your weight, which is the grounds resistance to your acceleration by gravity. In free fall nothing is interferring with your motion.
     
  10. May 7, 2010 #9

    Ich

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    It is your body's resistance to acceleration by the ground.
     
  11. May 7, 2010 #10
    thanks everyone! Most of you have been really helpful.

    What really helped was watching this youtube video. It showed someone in a room out in space, where there was no gravity, so everything was floating. Then the room turned on its thrusters and sped up until the person in the room could walk, just as he would be able to on earth.

    I realized that I would be in a warp in spacetime (I would say falling through a warp in spacetime, but that implies acceleration and it seems if you are in one of the spacetime warps then you are actually not moving) except the earth is blocking me. So it's like the earth is pushing me.

    If you are out in a room in space and it turns on its boosters, then you are accelerating, which is why you can walk normally. In the same way, when I'm sitting here in my chair, earth is pushing me out of the natural path, which is called the geodesic.

    Is that about right? Thanks!
     
  12. May 8, 2010 #11
    Taybot,
    Yep. That rocket analogy is commonly used to describe this. It's one of the better "thought experiments" to help visual this. If you consider a light beam as seen by the rocket, it can also be used to explain why (if the equivalence principle is true), gravity should "bend" the path for light as well.

    Something that has always confused me about this, is that the freefalling / inertial observer would see a table doing positive work on a book that it just sitting on it. Actually, ALL observers will agree the book is experiencing non-zero proper-acceleration. The inertial observer sees the table adding kinetic energy to the book. Where does this energy come from?
     
    Last edited: May 8, 2010
  13. May 8, 2010 #12

    Fredrik

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    What about when you ride an elevator on the way down? Then the normal force from the floor is doing work on all the furniture on every floor you're passing. This increases the potential energy of the furniture. Where does this energy come from? :wink:
     
  14. May 9, 2010 #13
    But the potential energy is not increasing in that example. From the reference frame of the elevator, the furniture is indeed moving upwards, but so is the earth. Assuming the elevator is moving at a constant speed, both the kinetic and gravitational potential energy of the furniture are constant. No? I guess I'm not getting the answer you are hinting at. :(
     
  15. May 9, 2010 #14
    If that's true, why doesn't the earth and the moon collide?
    (without using Newton!)
     
    Last edited: May 9, 2010
  16. May 9, 2010 #15

    Dale

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    The energy comes from the normal force from the table, which is the only force acting on the book.

    I am sure that is not a satisfactory answer to you, because it practically begs the question about where the table got its energy. You can proceed to answer this question by expanding your analysis to include the floor on which the table stands. Where did the floor get its energy? Again, you can answer this by expanding the analysis to include the foundation.

    At each step you have to expand your analysis, and very quickly you get to a point where you are no longer dealing with a local inertial frame, the equivalence principle no longer applies, and you have to deal directly with the curvature of spacetime. Unfortunately, energy in curved spacetimes is a notoriously difficult subject:
    http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/GR/energy_gr.html
     
  17. May 9, 2010 #16

    Dale

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    Because they are not moving towards each other.
     
  18. May 9, 2010 #17

    Fredrik

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    Sorry, I was a little dumb there. I was trying to show that the same "problem" exists in Newtonian physics (i.e. in Galilean spacetime) as well, but the furniture in my example (which we're supposed to think about in Newtonian terms) aren't accelerating, so the sum of the forces on them are zero, and no work is being done.

    To make the point I was trying to make, we have to consider an accelerating coordinate system. So consider the first few seconds of your elevator ride, when the elevator is accelerating (towards the ground). Or if you prefer, imagine that the elevator cable snaps and that you're in free fall. Now we can associate a non-inertial coordinate system with the motion of the elevator. (We're still talking about Newtonian/Galilean physics here). In this coordinate system, the "up" component of the velocity of an arbitrary piece of furniture is increasing, so that object, let's say it's a chair, has a non-zero coordinate acceleration. Force is defined by F=ma, so there's a force acting on the chair. It's often called a "fictitious" force.

    This force is clearly doing work, since one of the position coordinates of the chair is changing as well. W=\int F z'(t) dt. And the kinetic energy is increasing as a result, since the velocity is increasing. So even in Newtonian physics, we have situations where we can ask "Where does the energy come from?". That's the point I wanted to make.

    I wouldn't answer it with "Bah, that's just a fictitious force." I don't think any distinction should be made between "fictitious" forces and "real" forces. They are the same thing. We start by specifying which curves represent non-accelerating motion. Then we define acceleration as a measure of the deviation from non-accelerating motion. Then we define force as the mass times the function that describes how the acceleration depends on the velocity, the position, and the time: x''(t)=f(x'(t),x(t),t)=F(x'(t),x(t),t)/m.

    So what is the answer? I'll leave that as an excercise.

    (That means I have to think about it :smile:).
     
  19. May 9, 2010 #18

    atyy

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    From the fictious potential energy of the fictious gravitational field? (Let's just do Newtonian mechanics here) :tongue2:
     
  20. May 9, 2010 #19
    GR says the earth's mass generates a field that alters the space around it, and the space/field directs objects toward the earth, i.e., they accelerate toward the earth's center.
    The earth's surface is also accelerated toward the center of earth, but the layers beneath it prevent it from moving.
    If the earth's surface were accelerating outward, what would prevent it from moving outward?
     
  21. May 9, 2010 #20
    If you, the table, and the book are in free fall, why is any work done on the book?
    In the space lab, objects remain where you put them, but you can move them in any unrestricted direction with a push. There is no force holding the book to the table, they stay together because all mass gets the same acceleration.
     
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