Equivalence principle

Let there is an observer “IO” who holds Newton’s apple in Einstein's elevator. The said elevator is accelerated @ rate of 9.8 m/s/s relative to the outside stationary observer “OO” (not accelerating). It is said “IO” feels a force created by the foregoing accelerated elevator pulling him down. An apple feels the same force which is felt at every point within the elevator and falls (mg) to the bottom of elevator for “IO” if “IO” let it goes. Turning on his flash light towards the far wall, “IO” sees a beam of light bends down. However, “OO” sees the apple is at rest/hovers at the same point in space while the elevator moves upward till its floor hits the apple. And the beam of light doesn’t bend either.

I disagree with all above.

1-   An apple starts moving upward with its final velocity as soon as “IO” leaves it in space. The final velocity of the apple of the subject apple depends upon the instantaneous velocity of the elevator at the moment when the apple was left by “IO”. This final velocity of the apple must be added/subtracted (included in the calculation) to all the velocities which makes the acceleration of elevator before making any conclusion about the topic of “Equivalence Principle” - RIGHT?

2-   How come “IO” feels a force when the direction of the acceleration (a=g) is upward, not downward – This is my old question so please ignore it

The apple doesn't stop.
It continues moving at whatever speed it was going at prior to being released.
The key part is that it doesn't continue accelerating after it is released, until it hits the floor.
This means the person inside sees it accelerate towards the floor.

The apple doesn't stop.
It continues moving at whatever speed it was going at prior to being released.
The key part is that it doesn't continue accelerating after it is released, until it hits the floor.
This means the person inside sees it accelerate towards the floor.

Doesn’t the inside person “IO” sees an apple falling but not at the rate of 9.8 m/s/s as the upward speed/velocity of an apple has to be adjusted with all the velocities of the acceleration of elevator involved before hitting and this will change the order of type of motion of an apple as well.  This can be observed if graph of velocity of an apple and acceleration of elevator is drawn w.r.t time.

Doesn’t the inside person “IO” sees an apple falling but not at the rate of 9.8 m/s/s as the upward speed/velocity of an apple has to be adjusted with all the velocities of the acceleration of elevator involved before hitting and this will change the order of type of motion of an apple as well.  This can be observed if graph of velocity of an apple and acceleration of elevator is drawn w.r.t time.

This depends on if you are reaching a high enough speed for relativistic time dilation and length contraction and so on to kick in and make a significant difference.

e.g. lets say we have out elevator start at rest at time t=0.
We can even have it start at x=0.
With the apple held 1 m above the ground at x=1 m.
And accelerating in the positive x direction at a rate of 10 m/s^2 (easier math). And these are all variables that could be changed.

It is now accelerating upwards till time t=10 s.
That means the base is now at t=500 m, and moving at 100 m/s. The apple is at 501 m, and moving at 100 m/s.
Now the apple is released.
It continues to move at this speed of 100 m/s, while the rest of the elevator continues accelerating upwards.
Then ~0.447213595 s later, at 10.447213595 s, the base of the elevator is now at 545.7213595 m.
The apple, which was at 501 m, and travelling at a speed of 100 m, continues at that speed for the 0.447213595 s, and now is also at 545.7213595 m.
i.e. the apple has hit the floor of the elevator.
For the person inside, they see the apple start at 1 m, and accelerate downwards at a rate of 10 m/s^2 and hit the floor after 0.447213595 s

How about if an apple doesn’t gain any final velocity from the accelerated (say 10 m/s/s) elevator and stays @ its original position as soon as it is released. Doesn’t it accelerate downward @ the rate of 10 m/s/s for the inside person.

How about if an apple doesn’t gain any final velocity from the accelerated (say 10 m/s/s) elevator and stays @ its original position as soon as it is released. Doesn’t it accelerate downward @ the rate of 10 m/s/s for the inside person.

I'm not sure what you mean by that?
Do you mean if when the person releases it, it doesn't have the velocity of the elevator?
In that case, the apple will appear to the opposite of whatever the velocity of the elevator is in an instant.

Doing that, we have the apple stationary at a distance of 501 m. It then remains there. To the observe in the elevator travelling upwards at 100 m/s, that means it appears that when they released the apple it instantly jumped to a speed of 100 m/s downwards, and then continued to accelerate at a rate of 10 m/s^2.
Then at t=10.009995 s, the floor of the elevator reaches 501 m and collides with the apple, and is now travelling at a velocity of 100.09995 seconds
That means to the observer in the elevator, the apple has travelled 1 m in 0.009995 s.

Perhaps this will help:

Jack:

The seen acceleration (10 m/s/s) of an apple can’t be same in both scenario.

1-   When an apple gains its final velocity from the elevator as soon as it is released within the accelerated elevator.
2-   When an apple doesn’t gains its final velocity  as soon as it is released from the accelerated elevator i.e. an apple stays at its original position.

Markjo:

The ball @3:17 in video, doesn’t gain any velocity from the moving elevator (accelerating upward deep in the space) as soon as it is released – Why? - This is my question.
Similarly, @ 2:24, a beam of light would not remain straight but bend in a capsule freefalling on earth while the same beam of light remains straight in a capsule not accelerating deep in space.

The seen acceleration (10 m/s/s) of an apple can’t be same in both scenario.

1-   When an apple gains its final velocity from the elevator as soon as it is released within the accelerated elevator.
2-   When an apple doesn’t gains its final velocity  as soon as it is released from the accelerated elevator i.e. an apple stays at its original position.

The difference is the instant it is released.
Also, it doesn't gain it when it is released, it is gaining all the time it is being held, so when it is released, it is already at that final velocity (until it hits the floor).

The difference for the observer inside is:
1 - They see it start off stationary, and accelerate downwards at a rate of 10 m/s^2.
2 - They see it instantly jump to 100 m/s, and then accelerate downwards at a rate of 10 m/s^2.

So for the observer, in the first case the apple is released with no initial velocity; while in the second case it is released with an apparent initial velocity of 100 m/s.
After that point, regardless of which situation it is in, it accelerates at 10 m/s^2.

The ball @3:17 in video, doesn’t gain any velocity from the moving elevator (accelerating upward deep in the space) as soon as it is released – Why? - This is my question.

The question you should ask is why it should gain velocity and when.
If the ball is not in contact with the elevator, directly or indirectly, why should it accelerate it?
And the simple answer is it shouldn't.
It is when the person inside the elevator, who is standing on the floor, holding the ball, that it is accelerated.
That is because the elevator is accelerating, that in turn pushes on the person accelerating them, which in turn pushes on the ball, accelerating them.
Noting that to the observer in the elevator, it is not accelerating, because they are accelerating with it. Instead it just feels like there is force pulling the ball down.

Similarly, @ 2:24, a beam of light would not remain straight but bend in a capsule freefalling on earth while the same beam of light remains straight in a capsule not accelerating deep in space.

The beam of light would bend to an outside observer. To an observer inside the elevator, the path of the light follows causes it appear to go straight to that observer.

Following is also noticed.

The inside person standing on the floor of accelerated elevator deep in space sees his reflection on the opposite mirror wall shorter than his height due to the bending of light.

If we imagine two identical persons A & B (same height) facing each other but standing at a distance “d” apart on the floor of the accelerated elevator of the subject.

“A” sees his height is greater than “B” and vice versa – RIGHT?

The inside person standing on the floor of accelerated elevator deep in space sees his reflection on the opposite mirror wall shorter than his height due to the bending of light.

If we imagine two identical persons A & B (same height) facing each other but standing at a distance “d” apart on the floor of the accelerated elevator of the subject.

“A” sees his height is greater than “B” and vice versa – RIGHT?

Actually, they would see it higher. But they see everything appear higher. But I'm not sure if it would appear taller or shorter.
This might appear strange and counter intuitive, but it is what happens.

Lets just focus on the light from their eyes, and consider an extreme case.


For the red beam of light, the light goes from their eyes to the wall, initially horizontal but bending down in the process.
It reflects off the wall and continues to bend down, until it gets back to them and hits their shins.
They can't see that light, as it has gone well below their eyes.
However, you can follow this light backwards. Light from their shin, initially going upwards, eventually reaches their eye, appearing level.
So in this extreme example their shins appear to be at eye level.

If instead you focus a different beam of light (the purple one), it goes from their eyes, initially going up at an angle.
It also bends down, but it first has to bend from going up to going level.
Just when it is going level, it hits the wall and now reflects back.
Now on the path back, it bends from level to going down, until it eventually hits their eyes from an upwards angle.

So light bending down, makes things appear higher than they are, because the light needs to start going up more than a straight line to you, and end up curving back down, so approaching from a higher angle.

A light beam bends towards the floor of elevator which is accelerating deep in space for the inside observer i.e. against the direction of force which accelerate the elevator while on the contrary, here on earth, it bends in the direction of the force of gravity of earth. No equivalency again.

Addendum: I may be wrong but the said elevator which is accelerating in any direction just acts as a force for the person who is standing on its floor. According to Newton’s second law of motion, the elevator just accelerates the person in the direction of applied force. Due to the increase in total mass the acceleration of the elevator is reduced.

A light beam bends towards the floor of elevator which is accelerating deep in space for the inside observer i.e. against the direction of force which accelerate the elevator while on the contrary, here on earth, it bends in the direction of the force of gravity of earth. No equivalency again.

Addendum: I may be wrong but the said elevator which is accelerating in any direction just acts as a force for the person who is standing on its floor. According to Newton’s second law of motion, the elevator just accelerates the person in the direction of applied force. Due to the increase in total mass the acceleration of the elevator is reduced.

You misunderstand the equivalency.
A downwards force due to gravity is equivalent to an upwards acceleration.
In both cases, light bends down.

To explain it more:
When you standing on Earth's surface, you have gravity pulling you down, and you have the normal reaction force from you standing on Earth pushing you up.

When you are in an accelerating elevator, you have a force of the elevator pushing you up, with a normal reactionary force from you pushing the elevator down.

The inside observer was floating inside the elevator before the starting of elevator in above case. Anyway,

To explain it more:
When you standing on Earth's surface, you have gravity pulling you down, and you have the normal reaction force from you standing on Earth pushing you up.

When you are in an accelerating elevator, you have a force of the elevator pushing you up, with a normal reactionary force from you pushing the elevator down

if i push the elevator down then i must also accelerate the elevator down with an equal amount - right?

There is no reaction force of "m" in F = ma however this counter force of "m" must be included in the preceding equation - Right?

Further, how about the increase in the acceleration of elevator due to the decrease in total mass as soon as the inside observer release an apple inside elevator?

The inside observer was floating inside the elevator before the starting of elevator in above case.

Yes, just like they would be on Earth in free fall.
And try to think about having the elevator start accelerating. The comparison is between the elevator in free fall to an elevator in deep space; and between the elevator being accelerated to the elevator stationary on Earth.

if i push the elevator down then i must also accelerate the elevator down with an equal amount - right?

That depends on what is controlling the elevator.
Assuming it has a controller that can react fast enough, with a decent feedback look, it will alter the force.

e.g. if it is an elevator of mass M, with an object of mass m inside touching it, then the force applied to the elevator to accelerate it will be F=(M+m)*a.
Then the reactionary force will mean the net force on the elevator is F=M*a.

And if you want to split it up to include the apple as well (with mass n), then you get F=(M+m+n)*a, when the apple is being held, and back to F=(M+m)*a when it is released.

However, this issue can also arise with gravity.
The force due to gravity between 2 objects is F=GMm/r^2.
Importantly, this applies to both objects.
So when you release the apple, as well as the apple accelerating towards Earth, the Earth (taking you with it) accelerates upwards towards the apple. It is just that that change is insignificant.
Likewise, if the mass of the elevator is significant enough, even with a constant force, that is still going to be an insignificant change in the acceleration.

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f i push the elevator down then i must also accelerate the elevator down with an equal amount - right?

That depends on what is controlling the elevator.
Assuming it has a controller that can react fast enough, with a decent feedback look, it will alter the force.

According to the third law of motion, the person attached to the floor of elevator react equally but in opposite direction to the change in each adjusted force. As said earlier, this reactive force of an object of mass m is even missing while formulating F = ma