# Can gravity induce any jerk?

The equivalence principle states that constant acceleration inside a closed elevator is indistinguishable from gravitational force. But prior to its reaching constant acceleration, any elevator has positive jerk (the derivative of acceleration); and it also has negative jerk when it stops.

This leads to the following: suppose you are in a closed elevator in free fall. Say you are standing parallel to the $z$-axis. You experience a downward directed force $F(t)$ consistent with a normal elevator, $0\le t\le 1$. Say

1. $F(0)=F(1)=0$
2. $F'(t)>0$ for $0<t<\epsilon$
3. $F(t)>0; F'(t)=0$ for $\epsilon<t<1-\epsilon$
4. $F'(t)<0$ for $1-\epsilon<t<1$

Can you infer that you are in an elevator that is being externally propelled, or could the accelerations you observe be due to the gravitational influence of (moving) external bodies?

You can assume the jerk is continuous, but it is also possibly interesting to consider the discontinuous case: we can think of $F$ as a step function (taking $\epsilon=0$). At first I thought discontinuous $F'$ could not induced from gravity, but maybe if one imagines a huge burst of energy traveling at the speed of light and moving past the elevator, the gravity of the energy would result in a discontinuous jerk? (I have no idea if this is correct).

But even for continuous jerks, are there limitations on what could be induced by gravity? Is there an "equivalence principle for jerk"?

Also, if you want you can just eliminate conditions (3) and (4) above to make the problem simpler.

• Comment to the title question (v2): No need to go to GR. Already a free-falling particle in Newtonian gravity has non-zero jerk since the gravitational acceleration depends on altitude. – Qmechanic Jun 5 '17 at 5:57
• It's true that a free-falling particle experiences a jerk. However, (a) for some jerks, the particle might need to fall at superluminal speed. Moreover, (b), if the elevator "cable" is rigid and $F$ is a step function, the jerk is infinite and the speed for gravitation-induced jerk would be infinite too. These two issues, (a) and (b) are what I am asking: for which $F$ is there a gravitational equivalent? – kdog Jun 5 '17 at 9:50
• @kdog In the real world, there is no such thing as "a rigid body," and no such thing as a mathematically precise step function. If you get clear in your mind whether your questions are really about mathematics or physics, most of the confusion will disapear. Mathematically, the answers are straightforward. Physically, the questions may be meaningless and therefore unanswerable. – alephzero Jun 5 '17 at 17:14
• @alephzero as I explained in the original question, consideration of the rigid body formulation is purely optional as something "possibly interesting to consider". The main question is what jerks are physically realizable due to gravitation (that is without superluminal velocities for one thing). – kdog Jun 8 '17 at 6:11