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It will be so great if someone help me with this, and try to understand what's my point and what I am trying to say. I have asked a question few days ago, and it was closed because, apparently it wasn't clear. Please be patient with me and tell me if I'm right with this or not, trying to explain all details if I'm not. Thanks to all who will give time to my question, and think about it.

Einstein's equivalence principle says we cannot distinguish between a gravitational field and a uniformly accelerated rocket. I possibly have found a way which someone can know he isn't in earth because he will realize something strange happens in the accelerated rocket or cabin. But first, we have to emphasize some things. As we know, by these equations $F=ma$, $KE=(mv^2)/2$, $\Delta P=Nt$, $P=mv$, increasing mass, means increasing force, kinetic energy and momentum.

I'll give you this example. We have two rockets (A and B) going at 1 g, in both cases the person inside the rocket, drop a ball from, for example, 1,5 m. In both cases everything is the same, the acceleration of the rockets, the distance between the ball and floor, the weight of the ball, the time of the collision, etc...except mass, in B, the rocket has more mass than A. Assuming the floor of the rocket will hit the ball, and assuming it's a perfect elastic collision, in B the ball should go further because there is more force and kinetic energy in the collision (Because there is more mass).

So, that means if we add the enough mass to the rocket ($F=ma$=>with more mass, comes more force, $KE=(mv^2)/2$=>with more mass comes more kinetic energy), we can make the ball surpass the height of the drop in eyes of the inside observer, and the person will know he isn't on earth by just drop a ball and observing it surpass the height.

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  • $\begingroup$ You remind me of the old trope about the engineering prof who challenged his first-year students to find fault with his proof that bumblebees can't fly. His numbers all were correct. His calculations were flawless. The problem was, he was using "laws" that model the performance of a single-engine, fixed-wing aircraft, and that is not how bumblebees actually fly. I think that you are doing something similar with the equations that you cite and the conclusions that you draw from them. E.g., "$F=ma$...[therefore]...increasing mass means increasing force." $\endgroup$ Mar 10 '20 at 2:11
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    $\begingroup$ BTW, you should have fixed your old question instead of deleting it and asking a new one. If you ask too many bad questions the system will ban you from asking questions, even if the bad questions are deleted. $\endgroup$
    – PM 2Ring
    Mar 10 '20 at 3:30
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    $\begingroup$ I've deleted some inappropriate comments and their responses. $\endgroup$
    – David Z
    Mar 10 '20 at 6:21
  • $\begingroup$ I'm not algebraically inclined enough to pen an answer here, but I've got an intuitive hunch that your alleged difference is cancelled out by the fact that when you release the ball, it starts to pull the rocket towards itself (because it has its own gravitational field), and when you increase the relative mass of the ball (relative to the rocket), you increase the amount that the ball's gravitational field (pre-bounce) pulls the rocket forward (thus infinitessimally increasing its acceleration above 1g), which effectively cancels out the "higher bounce" principle based on the same parameters. $\endgroup$
    – Flater
    Mar 10 '20 at 15:25
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It appears that you are thinking of the standard elastic collision formula: $$v_m=\frac{m-M}{m+M}u_m+\frac{2M}{m+M}u_M$$ $$v_M=\frac{2m}{m+M}u_m+\frac{M-m}{m+M}u_M$$ where $m$ is the mass of the ball, $M$ is the mass of the rocket, $u_i$ is the velocity the instant before the collision, and $v_i$ is the velocity the instant after the collision. This formula can be expressed more usefully if we set $k=m/M$ $$v_m=\frac{k-1}{1+k}u_m+\frac{2}{1+k}u_M$$ $$v_M=\frac{2k}{1+k}u_m+\frac{1-k}{1+k}u_M$$

Now, the equivalence principle refers to a uniformly accelerating reference frame, so to implement it in this scenario requires $v_M=u_M$. We can substitute that into the equation above and solve to obtain $k=0$. This means that a uniformly accelerating rocket implicitly assumes that the ball is of negligible mass compared to the rocket.

With $k=0$ the equations above further simplify to $$v_m=-u_m+2u_M$$ $$v_M=u_M$$

Now, for someone on the rocket itself $u_M=0$ so we immediately get $v_m=-u_m$. This is exactly what is observed in a uniform gravitational field when bouncing a perfectly elastic ball on the floor. The equivalence principle therefore holds for bouncing a ball on a uniformly accelerating rocket floor.

Now, suppose that we relax the requirement that the acceleration be uniform. We would no longer be discussing anything related to the equivalence principle, but we could learn how the ball behaves when bounced on the floor of a rocket when $k>0$. In this case, since we are onboard the rocket we still have $u_M=0$. This means $$v_m=\frac{m-M}{m+M}u_m$$ which we plot below:

Bounce velocity as a function of rocket mass

In the limit that the rocket mass becomes very large, the maximum bounce ($v_m=-u_m$) is achieved because the rocket is so massive that it does not move and all of the KE remains in the ball. Note that, as the mass of the rocket increases arbitrarily the speed of the ball does not increase arbitrarily, but reaches the maximum of the original speed (opposite direction). It does change, but it approaches a finite value. Conversely, if the rocket has a small mass, then we will notice some deviation from the "full bounce". The ball will have less energy as some of the ball’s energy goes into moving the smaller rocket.

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  • $\begingroup$ Even if the ball don't reduce the acceleration of the rocket, Don't you think a rocket with, for example, 5000kg going at 1G, will make the ball go further(go at more speed) than another with 500kg going at 1G ( assuming in both cases the balls don't reduce the acceleration)? Because kinetic energy and force are more in the one with 5000kg, than the one with 500kg. $\endgroup$
    – user255961
    Mar 9 '20 at 23:14
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    $\begingroup$ The KE and force is already included in the standard collision formula that I posted. All of those effects are considered, and what I described is the result. You seem to forget that although the more massive rocket brings in more KE it also carries more KE away (in other frames). In the end you get what I posted by considering the mass. $\endgroup$
    – Dale
    Mar 9 '20 at 23:55
  • $\begingroup$ I undestand what you put in your answer, as Einstein said, acceleration will not change no matter what. I'm acctually think that's the standard collision formula doesn't work here, because here, it's different, since v1=u1 as you said that means 0. But what if we use the conservative equation? M1V1 + M2V2 = M1V1' + M2V2'. Using this equiation the result of the final velocity of the ball is 0, which means the ball will just stuck to the floor and won't bounce, because, V1 of M1 = V1' of M1. But here, we assume that the ball will be hitted by the floor and will bounce $\endgroup$
    – user255961
    Mar 10 '20 at 0:57
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    $\begingroup$ Comments are not for discussion. If you need discussion then you should go to a discussion forum, not a question and answer forum. In any case, the conservation of momentum formula is also included in the formula I posted. Both KE and momentum are conserved (that is how the elastic collision formula is derived). Your idea that they somehow don’t apply here is flat out wrong. You should discard it immediately. For now, you really need to focus on learning the very basics. $\endgroup$
    – Dale
    Mar 10 '20 at 2:13
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    $\begingroup$ @Dale i sent you a message $\endgroup$
    – user255961
    Mar 18 '20 at 21:32
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Dale's answer is perfectly correct and provides a nice mathematical demonstrations.

However, I will try to offer an intuitive insight that I think compliments his answer.

In the reference frame of the person standing on the Rocket the floor of the rocket is not moving. A ball left lying on the floor is said (in this frame) to be at rest (even though from the perspective of someone outside the rocket-ship it is accelerating). So long as the ships acceleration is unifrom their is not any "extra" bounce on the ball from the impact, simply because the floor is not moving.

But in practice, in the experiment you describe, the spacecraft's acceleration will NOT be uniform. At the moment the ball hits the floor it exerts a force on the rocket to slow it down. Momentarily, reducing its forward acceleration.

So imagine a very lightweight spacecraft with almost nothing on board, and an extremely heavy ball. In this extreme case as the ball hits the floor it's impact will heavily reduce the ship's acceleration. This "backward lurch" of the rocket itself would look like some "extra bounce" of the ball to someone standing on the rocket.

The entire point of the equivalence principle is that it applies only in a "uniformly accelerating frame". It does NOT apply in a frame which is not uniformly accelerating. In your ball experiment you are using the ball to make the rocket's acceleration non-uniform.

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    $\begingroup$ In a completely negligible order of magnitude, doesn't a ball bouncing on the Earth displace the Earth (relative to its relative mass) just like how it displaces the rocket, effectively making the earth/spaceship examples equivalent? I.e. if a rocket ship had the mass of the Earth, the bounce would be completely indistinguishable from doing it on Earth, even if not in a uniformly accelerating frame? $\endgroup$
    – Flater
    Mar 10 '20 at 15:07
  • $\begingroup$ @Flater I believe that you are correct. $\endgroup$
    – Dast
    Mar 10 '20 at 15:20
  • $\begingroup$ If thrust is not adjusted to maintain 1G, the rocket will exceed 1G by a small amount when it is released, decelerate a bit during the bounce, accelerate after the ball rebounds, and decelerate a bit when caught. $\endgroup$
    – B. Young
    Mar 11 '20 at 23:50
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Einstein's equivalence principle says we cannot distinguish between a gravitational field and a uniformly accelerated rocket.

That is somewhat of a simplification. First, the equivalence principle only applies locally, since a gravitational field will exhibit tidal forces. Second, of COURSE it's possible to distinguish between being on Earth and being on a rocket. The idea that we can't is just silly.

The equivalence principle says that we can't locally distinguish between reference frame of constant acceleration and a gravitational field, not that we can't distinguish objects that create acceleration (such as a rocket) from objects that create gravity (such as the Earth).

increasing mass, means increasing force, kinetic energy and momentum.

No, it doesn't. If mass increases, and acceleration is constant, then force increases.

Assuming the floor of the rocket will hit the ball, and assuming it's a perfect elastic collision, in B the ball should go further because there is more force and kinetic energy in the collision (Because there is more mass).

Mass is not a property of a "collision", it's a property of an object. Speaking of there being more mass "in the collision" obfuscates the issue. The ball has no more mass, so it has no more mass. The rocket has more mass, but it also is accelerating less, so the force is (approximately) the same.

You seem to think that if the rocket has twice the mass, then it hits the ball twice as hard, but that's not how it works. The amount of force the ball experience depends only on the mass of the ball and the acceleration of the ball. The mass of the object that is hitting it has no direct effect.

Now, there is an indirect effect. In an elastic collision, if we keep the mass of the ball constant, then as the mass of the rocket goes to infinity, the magnitude of the final velocity of the ball approaches the magnitude of its original velocity. This is an asymptotic process, and the difference between a rocket that has a thousand times the mass of the ball, versus a rocket that has two thousand time the mass, is negligible.

However, if you still want to know how the equivalence is consistent with even this negligible difference, the key if the phrase reference frame. The Principle doesn't say that the rocket is indistinguishable from a gravitational field, it says that the reference frame is. When the ball hits the rocket, the rocket moves down slightly in the constant acceleration reference frame, which means that while the constant acceleration reference frame is indistinguishable from a gravitational field, the rocket is distinguishable from both the gravitational field and the constant acceleration reference frame.

The thought experiment you propose does not distinguish between an accelerating reference frame and a gravitational field, it distinguishes between a ball hitting an object with relatively small mass in an accelerating reference frame, versus a ball hitting an object with a larger mass in a gravitational field. It's the mass of the object that is hitting, not acceleration versus gravitational field, that is supplying the difference.

we can make the ball surpass the height of the drop in eyes of the inside observer,

No, the height of the ball will asymptotically approach the original height. It will not exceed the original height.

and the person will know he isn't on earth by just drop a ball and observing it surpass the height.

That doesn't make any sense. The Earth has more mass than any rocket ship, so by your logic a ball dropped on Earth should rise higher than on any rocket ship.

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You are wrong in the sense of thinking that because the rocket has more mass the ball will bounce higher: It will always bounce back to the same height. In the equations you normally use to find the height of a bouncing ball it does not matter what causes the acceleration, if the gravitational force, $mg$, or a pseudo-force $ma$, the equations give the same result if a=g. For the equations see here. See for instance, that the equations are oblivious to the "mass of the floor", whatever that means.

UPDATE: In your comment you ask if hitting something with a larger mass will make that move with a larger speed right after the collision. If there is no loss of energy, like in an elastic collision, that is true. However that is not the same situation than in a rocket that accelerates uniformly. In the hitting something example, the hitting object will also change its velocity due to the collision, slowing down, that is, it will decelerate. The larger the mass the less it will decelerate. But the rocket does not decelerate, it keeps accelerating with the same velocity. The equivalent would be a collision in which the large object is given extra energy, as much as it needs to keep its acceleration constant. Part of this energy is transferred to the object inside the rocket so that, from the point of view of someone inside the rocket, the bouncing speed will look the same, regardless of the mass of the large hitting object.

See @Dale answer for a mathematical explanation. And as the equations show, against my initial intuition, it is implicitly assumed that the mass of the rocket is very large compared to the mass of the ball, unless you have some kind of rocket that keeps the acceleration constant.

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  • $\begingroup$ So a ball hitted by something with 100kg, will have the same velocity as if it's hitted by something with 10000kg? In both cases, the time of collision it's the same, the distance it's the same, everything except, as you see, the mass. $\endgroup$
    – user255961
    Mar 9 '20 at 20:37
  • $\begingroup$ I will update the answer. $\endgroup$
    – user65081
    Mar 9 '20 at 20:40
  • $\begingroup$ But if we assume the floor will hit the ball, How can a rocket with more kinetic energy and force, make the ball goes at the same velocity as another one with less kinetic energy and force? Both of them have the same acceleration, that's true, the ball won't reduce the acceleration, okay, we assume that, but i think the one with more kinetic energy and force will transmit more velocity than the one with less kinetic energy and force due to the force in impulse force. $\endgroup$
    – user255961
    Mar 9 '20 at 21:29
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    $\begingroup$ @Yassin-98 the problem with your "counterexample" is that you try to disprove gravitational/accelerational equivalence by first assuming that this equivalence doesn't take place. The ball falling on Earth is affected by the Earth mass. But the ball falling on the rocket is affected by the rocket acceleration. Just as you, standing inside the rocket, would feel the same as standing on the floor, so the ball, bouncing off the floor, would behave identically to its Earthen brothers. $\endgroup$
    – IMil
    Mar 10 '20 at 5:26
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    $\begingroup$ @Yassin-98, There is no rocket in the equivalence principle. It's about the equivalence between a uniformly accelerated frame of reference far from any gravitating body, and a frame of reference that is close to a gravitating body. The rocket is just an illustration: It's not important. If your argument depends on the mass of a rocket, If your argument depends on measuring the fuel flow, If your argument depends in any way on the word "rocket," then you are not talking about the equivalence principle. $\endgroup$ Mar 10 '20 at 12:28
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First off, it should be noted that the increase in mass by relativity is meant for illustration more than reality. What it says that the shape of the kinetic energy curve under special relativity could be understood as being equal to the shape of the kinetic energy curve under Newtonian physics when we assume the mass changes with the velocity. But for serious calculations, it's better not to go the roundabout way of changing mass then doing Newtonian physics then transferring that back, but directly use relativistic kinetic Energy and momentum based on the object's rest mass.

However, in this case, the error you make is more fundamental: You are confusing frames of reference. It is true that a ball lying on the ground in 1g and a ball sitting on the floor of an accelerating rocket have different masses, assuming the ground is stationary and the rocket is not.

But in the situation you described, we don't have such a unified frame of reference. Experiment A happens in its frame of reference of an unmoving ground, and Experiment B happens in its own frame of reference in which the rocket doesn't move. That is basically the prime requirement to even talk about the equivalence principle: That we have some environment around us that we assume as not moving.

What happens when we drop the ball? It is accelerated towards the floor in both cases, with the same magnitude of the force. Compared to our frame of reference, the ball is moving, so it does pick up some relativistic mass. But since we are dropping from the same height, at the same acceleration, that mass increase when it hits the floor is the same in both frames of reference because the velocity is the same.

Now we can use the elastic collision formulas to determine the velocity after the bounce and therefore bounce height. As Dale pointed out in his answer, there is no ratio of ball to ground mass ratio in which the rebound velocity is greater than the initial velocity.

However, as also pointed out, the ratio of initial to final velocity depends on the ratio of ball mass to ground mass. Since the Earth is heavier than the rocket, we expect the ball to bounce slightly higher on Earth. However, if you do the maths with realistic weights, you will see that this difference is so small as to be unmeasurable. The large amount of other factors affecting rebound height (ball elasticity, ground composition, air friction, etc) are orders of magnitude stronger in their effects.

In conclusion, your experiment would be able to tell you the ratio of ball mass to ground mass, and by comparing that to the mass required to create the felt gravity, you could determine whether or not you're in a rocket, but it's a thought experiment because the accuracy of measurement and size of ball required are both unrealistic.

If we look at the experiment from your proposed point of view, with the rocket speeding past, we will get the same result. Because yes, the ball increases in mass by the velocity of the rocket at the time of collision, but so does the rocket. They increase mass by the same ratio, meaning we arrive at the same conclusion as from our rocket-internal frame of reference.

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  • $\begingroup$ "the increase in mass by relativity is meant for illustration more than reality" is a false statement. The increase in mass is real and is manifested by an increased gravitational attraction to moving object. $\endgroup$
    – Paul
    Mar 10 '20 at 14:18
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Your question stems from a muddle on what the equivalence principle asserts.

You must consider two identical rockets, both of whose rocket motors are firing in exactly the same way. Now place one of these rockets a few metres above the surface of the Earth, hovering there, and the other one off in outer space far from any gravitating object.

The equivalence principle asserts that motions inside these two rockets will be described by the same laws (as long as the rockets are small enough that we can ignore the position-dependence of the strength and direction of Earth's gravity).

Whatever physics calculation you apply to motions inside one rocket, including the collisions with the floor, that same calculation will apply to the motions inside the other rocket. To prove this within Newtonian physics one need merely note that the inertial force term that appears inside a uniformly accelerating reference frame produces a force which is everywhere proportional to the mass of the object acted upon. It follows that its effects cannot be distinguished from those of a uniform gravitational field.

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Theoretically, you can not differentiate between gravity and acceleration. So, the equivalence principle as defined, holds good.

However, in practicality, you do not even need two rockets for this. One should be enough as follows -

In order to maintain a constant acceleration, the rocket engine has to know the total mass of the rocket at all the times, so as the fuel is consumed, the total mass changes and engine has to adjust the exhaust in order to maintain constant acceleration. Suppose the engine is able to do so.

Suppose the ball you drop, has mass 1% (as an example) of the total mass of the rocket at the time of drop. When you drop the ball, the engine does not know the ball is dropped and the acceleration increases by 1% till the ball hits the floor and then drops back to original acceleration.

This 1% bump in acceleration is what would help you in many ways to figure out whether it is a rock or a rocket.

But again, theoretically, we are saying constant acceleration, that means we are assuming that the 1% difference will not happen, and there is no way to know.

There are other ways like clock speed at different heights in a gravitational field will be different, but in a rocket, it will be same. These days we have atomic clocks that can measure this difference between few feet within a short amount of time.

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  • $\begingroup$ Clocks at different heights in an accelerating rocket will have time dilation, just like in gravity. In fact, you can use the SR rocket time dilation to derive the gravitational time dilation per the equivalence principle $\endgroup$
    – Dale
    Mar 13 '20 at 2:59
  • $\begingroup$ @Dale: Yes, there will be time dilation in rocket, but if you put two clocks at different heights in the rocket, will they run at different speeds as they do in a gravitational field? On earth for example, a clock at a meter height and a clock at 2 meter height will run at different speeds. Will that be case in rocket too? $\endgroup$
    – kpv
    Mar 13 '20 at 4:22
  • $\begingroup$ “if you put two clocks at different heights in the rocket, will they run at different speeds as they do in a gravitational field” Yes, and by the same amount as in the gravitational field, provided tidal effects are negligible $\endgroup$
    – Dale
    Mar 14 '20 at 0:30
  • $\begingroup$ @Dale: That should not be true unless you assume equivalence principle means that. Is there a proof? But we are discussing the practical ways to investigate the principle itself, so yours becomes a cyclic reasoning. Equivalence principle is not absolute. For example, assuming all other equivalence, how do you explain escape velocity? You shoot upward a marble from a 1g rocket at 12kmh , and it should never fall back to earth, but in case of a rocket, it will! there is no escape velocity in the direction of acceleration for a rocket. however in horizontal direction , any speed will do. $\endgroup$
    – kpv
    Mar 14 '20 at 15:53
  • $\begingroup$ @Dale: 12KM/second $\endgroup$
    – kpv
    Mar 14 '20 at 16:00

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