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I've just been learning about Einstein, relativity, and the equivalence principle in Physics. I'm fascinated with the idea of being inside a free-falling enclosed environment (such as, e.g., rocket, airplane, Einstein elevator, etc.) and weightlessness, but I'm having trouble understanding the difference between being inside something that's free falling and free falling yourself. Obviously in the former is essentially a complete simulation of weightlessness - you can spin around in the air and traverse all three spatial dimensions by throwing stuff away from you or 'paddling.' In this situation, you can effectively push yourself up while inside (say, a shuttle) the falling reference frame.

Set this against free-falling without being in a enclosed frame (like a skydiver before he opens his parachute). In this situation, you still have characteristics of weightlessness (you can stand on your head, stand or sit cross-legged, and spin around) but you can't 'push' yourself up like you can while in a free-falling reference frame (like the shuttle).

Why?

(I feel like I'm missing something very, very obvious here, but I can't place my finger on it.)

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I think when you say "reference frame" you really mean "enclosed environment," is that right? Some parts of your question don't quite make sense as written, e.g. there is no such thing as not being in a reference frame. –  David Z Jan 20 '12 at 21:31
    
@DavidZaslavsky yes, that's right. I guess I didn't really know what else to call it. edited –  Thomas Shields Jan 20 '12 at 21:34

4 Answers 4

up vote 2 down vote accepted

Bear in mind that a skydiver falling before opening his parachute is encountering air resistance, which means he's not really in free fall (in the relativity sense). But let's imagine a hypothetical skydiver falling through empty space. In this case, there is no difference at all between the person in the free-falling shuttle and the skydiver in free fall outside the shuttle.

At first, I wasn't entirely clear about what you meant by

but you can't 'push' yourself up like you can while in a free-falling reference frame (like the shuttle)

but based on what has come up in the comments, I think this explanation would help: when you throw a ball downwards as you're falling, you move up relative to how you would have been moving otherwise. This holds true whether you're in an enclosed environment or not.

The reason it may seem different is that, if you're in a shuttle, when you throw the ball, your motion changes, but the shuttle's doesn't. The shuttle gives you a convenient reference for how your motion would continue if you hadn't thrown the ball. So you now observe a relative motion between yourself and the shuttle, whereas before the throw there was none, and it's obvious that your motion has changed.

It's worth noting that it's not necessary for the environment to be enclosed in order for this to work. If, instead of a shuttle, you had a friend skydiving next to you, and you threw a ball down, you would notice yourself moving up relative to your friend.

On the other hand, if you don't have a shuttle or a friend or anything else falling along with you, there's no other object to serve as a reference for how your motion would have continued. So when you throw the ball down, as a human, it's difficult to notice that your motion has changed. There's no switch from "not moving with respect to X" to "moving with respect to X." Instead, the only convenient object which you have to measure your motion against is the Earth, and you merely change from "falling toward Earth" to "falling slightly slower toward Earth."

Bear in mind that the actual change in your motion is the same as in the other cases, where you had a friend or a shuttle or something falling with you. If you had measured your motion relative to the Earth, instead of relative to the shuttle or friend, then you would see the same effect: you would change from "falling toward Earth" to "falling slightly slower toward Earth." It's just that the limitations of human senses make it hard to detect the difference between those two motions; it's much easier to detect when you start moving relative to something you were previously at rest with respect to.

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so you're saying that if i'm a skydiver (in empty space, hypothetically) and i'm in freefall, if I throw a ball down (in the same direction i'm falling) I will actually move up because of Newton's Laws of Motion? Exactly like i'd move up if I did the same thing inside the shuttle? –  Thomas Shields Jan 21 '12 at 17:49
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Relative to how you would have been moving otherwise, then yes, you would move up. –  David Z Jan 22 '12 at 3:07
    
then that answers my question. Thanks. I guess I was thinking there was some sort of fundamental difference between free-falling inside something vs. outside of it. –  Thomas Shields Jan 22 '12 at 4:33

Here's a completely mathless answer.

Let's simplify the problem. Just think of two billiard balls near each other falling in a vacuum.

If they were originally moving toward each other, they will collide and then move away. They are both still weightless in the sense of having nothing else touching them.

If either one now extends an arm out to one side, spins it around to the other side, and then retracts it, the ball will have rotated in space, which either ball can see by looking at the other.

Now suppose ball A is hollow, and it is big, and ball B is inside it. How does this change things? Not at all. They are still both weightless and can move relative to each other, just as if they were separate.

Now suppose ball A is shrunk to a point so it's invisible, but it can still see ball B. So if ball B does it's arm-twisting trick and rotates, A will see it. If ball B throws out a BB in one direction, causing itself to move in the other direction, A will see it. So if B is all alone, that does not mean it is not twisting or moving, it's just that if there's an A to watch it, A could be effectively "not there", which doesn't mean B isn't doing what it's doing.

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This is a helpful illustration, but it doesn't really answer my question. I'm trying to understand when your reference frame while falling changes from the earth (falling alone) to the shuttle, and why. –  Thomas Shields Jan 21 '12 at 2:18
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@Thomas: You can pick any reference frame you wish - self, shuttle, ball A or ball B. To be weightless, it is only necessary that an object have nothing pushing against it. –  Mike Dunlavey Jan 21 '12 at 3:02

When you say, "push yourself up" what do you exactly mean ? If you mean, you can touch the floor and go up then the answer is clear, in the second case, free falling outside the box, there is nothing to push against. When you are falling you can't move in any direction but down, since there is no force that would push you in any direction other than down.

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a) this should be a comment and b) you can "push up" without something to push off of (for example, but throwing a ball down) –  Thomas Shields Jan 21 '12 at 1:19

I agree with the others. Your question is somewhat ill-posed. Short answer, there is no difference of the type you posed. You have made a difference that makes no difference. Free fall or "free float" (John Archibald Wheeler) means you are on a timelike geodesic in the local geometrodynamical field. The timelike geodesic is inside the local light cone at each point of it and it is the longest proper time connecting any two points on it. Whether or not the spacetime is curved is another issue and people get confused about that. When you are in free fall you are in a Local Inertial Frame (LIF). People really get confused because they think a point on surface of Earth is an "inertial frame." Well it is in Newton's theory, but not in Einstein's. That's a long story.

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a point on surface of Earth is NOT inertial frame, even in Newton's theory as witnessed by the non-inertial (Coriolis) forces. This was known already in the 17th century. en.wikipedia.org/wiki/Coriolis_effect –  magma Jan 24 '12 at 10:36

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