Spacetime Physics (2nd ed.) problem 2-2: Bathroom scale and a Trampoline

[benorin]

Summary:: Strap a weight-measuring bathroom scale to your feet and jump on a trampoline: weight measurements at different points of each jump? What is the longest part of the cycle you are in the free-float frame?

I am studying Spacetime Physics 2nd ed. by Taylor and Wheeler at the suggestion of one of you fine physicists (thanks!). I need help with

Problem 2-2 (pg. 45) [paraphrased]:

Strap a weight-measuring bathroom scale to your feet and jump on a trampoline: (i) weight measurements at different points of each jump? (ii) When does the weight-measurement read zero? (iii) What is the longest part of the cycle you are in the free-float frame?

Answers so far: (i) I am impressed to note my cousin (non-science grad) pointed out to me that such a weight-measuring scale would be completely uncalibrated rather quickly ;) I wish I had thought of that. On the way up whilst still in contact with trampoline (I cheated and thought of Newton of course, but college physics courses were taken like about two decades ago so help me a bit please) there is a net acceleration upward meaning the force of the trampoline pushing up is stronger than your weight under gravity so the scale will register more than your rest weight being pressed on top and bottom (assuming that the bathroom scale works on compression); zero after leaving contact with the trampoline and thus airborne; on the way down whilst in contact with the trampoline this is the same as on the way up (net acceleration is upward) only instead of becoming airborne you cycle to stationary for and instant because the initial velocity is downward; (ii) whilst airborne; (iii) I'm lost here because there is no physical container that defines the boundary of the frame, there was typically a house or a railcar in the text and I'm not certain what my test particles would behave like: since the potential free-float frame begins just after losing contact with the trampoline and ends just before making contact again with the trampoline, do the test particles travel along the same parabola as me? It totally makes sense if the test particles are inside me though (and my skin is the boundary of the frame) lol

 

[Nugatory]

I'm lost here because there is no physical container that defines the boundary of the frame, there was typically a house or a railcar in the text...do the test particles travel along the same parabola as me? It totally makes sense if the test particles are inside me though (and my skin is the boundary of the frame)

No boundary is needed. One way of seeing this is to start with a situation you’re comfortable with - inside a closed railcar - and then imagine that we remove the walls one at a time until there’s no “inside”. You wouldn’t expect the motion of any test particle to be affected as long as it remains reasonably close.

 

[benorin]

But would the test particle in my would be inertial frame be also following the trajectory of, in this case me, were it not in some sort of vehicle that launched with me off the trampoline? In the text they give example of railcars and houses and I also found one without a physical vehicle/container of falling from a height under gravity but in this most closest example gravity explains why test particles follow the trajectory of the frame. In the current "after I've parted contact with the trampoline and before I make contact again" frame were I too instantiate a test particle at rest within the frame would it simply fall immediately under the influence of gravity down to the trampoline whilst I continued upward (and hence not be at rest?--not sure if that makes sense) or would it, even though it could never have been influenced by the trampoline because of the spacetime boundaries of the frame, nevertheless follow my trajectory at truly be at rest within the frame? Sorry for the paragraph, and thank you for helping me!

 

[DEvens]

In order to get reasonable answers you need to make some assumptions about the trampoline. Let's suppose it behaves like an ideal spring. That is, it pushes with a force that is proportional to the magnitude of displacement.

F = -K x

So x will be negative when you are below the rest point. So it pushes up when you are below the rest position of the trampoline. Were you able to stick to the trampoline it would pull down when you were above the rest point. So it means the maximum trampoline force is when it's at the bottom.

Also, neglect the trampoline's mass. Treat it as not rising above the rest point, and let it restore to the rest point arbitrarily fast so it always keeps up with you on the rise. That's an evergreen problem avoided.

But wait! When you are just below the rest point and rising, what is your acceleration? The force of the trampoline is very small, less than gravity. So just before you launch your acceleration is *not* upwards. Your acceleration is only upwards for the portion of the bounce when you are low enough so that the magnitude of Kx is larger than g. When you are above this value of x then your acceleration is downward.

So you need to calculate: How far down does the trampoline get pushed before you stop? And how far up do you then bounce at the top? And how long does it take from rising off the trampoline to returning? And what will the scale read at the bottom?

 

[Nugatory]

In the current "after I've parted contact with the trampoline and before I make contact again" frame were I too instantiate a test particle at rest within the frame would it simply fall immediately under the influence of gravity down to the trampoline whilst I continued upward (and hence not be at rest?--not sure if that makes sense) or would it, even though it could never have been influenced by the trampoline because of the spacetime boundaries of the frame, nevertheless follow my trajectory at truly be at rest within the frame? Sorry for the paragraph, and thank you for helping me!

If the particle starts at rest relative to you and is in free fall, then it will remain at rest relative to you. You can see this just by writing your height and the height of the particle as functions of time and then subtracting one from the other to get the distance between you and the particle - it will be constant. That’s enough to show that if we choose to use coordinates in which you are at rest (that is, position coordinate does not change with time - just declare your position to be the origin of the coordinate system) the particle will be also be at rest.

With a bit more work, you can also show that if the particle starts out moving relative to you, its speed relative to you will be constant (try writing its position using coordinates in which you are at rest, differentiate once with respect to $t$ to get the velocity, note that it is constant, differentiate the velocity to get the acceleration, it will be zero). In fact, the first result (if at rest relative to you , it stays that way) is just the $v=0$ case of the second... and this second result is nothing more or less than Newton’s first law, indicating that these coordinates define an inertial frame.

Two interesting things about the frame in which you are at rest while in free fall:
First, using the coordinates assigned by the frame in which you are at rest, the surface of the Earth is not at rest, it is accelerating towards you at 1G.

Second, the frame is only inertial if you and the test particle are sufficiently close. For an extreme example, imagine that the test particle is on the other side of the earth; now you and it are falling in opposite directions, accelerating towards one another instead of maintaining a fixed relative position. However, no matter how close you are, there will always be some small deviation from perfectly inertial behavior. If you and the particle are at different heights you’ll be at different distances from the center of the Earth so the gravitational force will be slightly different; you and the test particle will accelerate away from one another. If you and the test particle are at the same height but separated, your paths won’t be exactly parallel because they intersect at the center of the earth; you and the test particle will accelerate towards one another.

This is the answer to your original question about the boundary of the frame. There’s no hard border or container wall. Instead the inertial frame extends as only as far the deviations from ideal inertial behavior (they are properly called “tidal effects” and ocean tides are an example) are negligible for the problem we’re working on. For your trampoline bouncer problem, we’re working at a scale of only a meter or so and these effects can be completely ignored.

 

[haruspex]

In the real world, the scale itself has weight, and some of that is above the internal spring. In free fall it would have a negative reading.