Equivalence Principle

A light clock installed in the elevator of Einstein which is moving upward @ the rate of 9.8 m/s/s therein free space away from the local attraction of gravity consists of two mirrors, M1 (attached to the east wall) and M2 (attached to the west wall) facing each other at a distance "d" apart.

The beam will follow a straight horizontal path relative to the observer inside the accelerating elevator as he feels a ground condition.

Ok, there is another observer at rest outside the above elevator therein free space shines a flashlight horizontally from west across the elevator toward its far east wall (from west to east) but below or above the light clock.

The beam will follow a curved path downward relative to the observer inside the accelerating elevator.

So which one is right – please

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Following diagram shows the elevator used in the thought experiment by the Einstien for equivalence principle



Elevator at rest on earth: Light beams or pulses emit at A and D at the same time, respectively. Both beams/pulses reach their destination B and C at the same time.

Elevator accelerating @ 9.8 m/s/s in free space: Light beams or pulses emit at A and D at the same time, respectively. Both don't reach their destination B and C at the same time. Pulse on right of observer reaches C faster as the floor of the elevator approaches it than pulse on the left where pulse moves away from the floor of the elevator when the floor tries to approach it (difficult to catch) - Right

So should the distance covered by pulses/ beams in time "t" be the same in both aforementioned cases for equivalency reasons?

Same scenario but with light clock

A light clock installed in the elevator of Einstein consists of two mirrors, M1 (attached to the ceiling) and M2 (attached to the floor) facing each other at a reasonable height "h" apart. Would the bouncing pattern of light beam/ pulse in between M1 and M2 be the same (equivalent) when the elevator is at rest on the ground and moving upward @ the rate of 9.8 m/s/s therein free space away from the local attraction of gravity?

A light clock installed in the elevator of Einstein which is moving upward @ the rate of 9.8 m/s/s therein free space away from the local attraction of gravity consists of two mirrors, M1 (attached to the east wall) and M2 (attached to the west wall) facing each other at a distance "d" apart.

The beam will follow a straight horizontal path relative to the observer inside the accelerating elevator as he feels a ground condition.

Ok, there is another observer at rest outside the above elevator therein free space shines a flashlight horizontally from west across the elevator toward its far east wall (from west to east) but below or above the light clock.

The beam will follow a curved path downward relative to the observer inside the accelerating elevator.

So which one is right – please

The latter.
The light will only follow a horizontal path in an inertial reference frame.
In an accelerating reference frame, the light will follow a curved path (assuming it isn't moving in the direction of the acceleration).

With the equivalence principle, being in free fall is locally equivalent to being in an inertial reference frame, while standing on Earth in its gravitational field is equivalent to being in an accelerating reference frame.

The latter.
The light will only follow a horizontal path in an inertial reference frame.
In an accelerating reference frame, the light will follow a curved path (assuming it isn't moving in the direction of the acceleration).

With the equivalence principle, being in free fall is locally equivalent to being in an inertial reference frame, while standing on Earth in its gravitational field is equivalent to being in an accelerating reference frame.

First of all thanks for your response - My question is more about the equivalency of both frames

If both frames are equivalent then they should have everything common not just 1 or 2 common things b/w them - Right

A person standing in an accelerating reference frame is equivalent to being on Earth in its gravitational field but not relative to its light clocks if installed either horizontally or vertically -

Ok with the standing on the floor of an elevator but why not relative to object (clock inside) so would he still feel standing on earth. 

For those who need more detail

Assume the said elevator is at rest on the ground - A person on the floor of the elevator feels himself on the ground. He sees a light pulse bounces in between the two mirrors M1 and M2 normally (either horizontally or vertically depending upon its installment). Everything is just fine.

Now the said elevator accelerates deep in space away from gravity. A person stands on the floor of the elevator (unaware of acceleration) feels as he stands on earth but he sees a pulse of his light clock doesn't bounce normally.

Here are velocities when we drop thing here on earth

[table 1

Time (seconds)   Velocity (m/s) 0   0 1   9.8 2   19.6 3   29.4 4   39.2 5   49

   
Per second increase in velocity is 9.8 m/s

Acceleration ( a = g= 9.8 m/s/s) is the same while velocity increases with the passage of time “t” therefore imagine if the velocity of elevator is increased from 9.8 m/s to 10,000 m/s or more what would be the falling velocity of an object relative to the observer inside elevator accelerating upward @ 9.8 m/s/s

This means bouncing of a tennis ball either on its floor or vertical walls will not be equivalent to on earth
Similarly, the degree of curve of bending of light downward depends upon the instantaneous speed of the elevator –

The more is the speed of the elevator at which it is accelerating upward the more will be bending of the light beam relative to the said observer
 
Here on earth, the max falling velocity of an object is reached to 200 km/h or sec something like

If both frames are equivalent then they should have everything common not just 1 or 2 common things b/w them - Right
A person standing in an accelerating reference frame is equivalent to being on Earth in its gravitational field but not relative to its light clocks if installed either horizontally or vertically -

Everything should be equivalent, including the effects on clocks and light.
If you have a beam of light travelling initially horizontally, it will curve downwards.
However, the amount it curves down will be quite insignificant.
To an outside observer if you have the light travel 2.9979 km, then it will only take 10 us. In that time the elevator would have accelerated by 98 um/s.
It will have only moved up by 0.49 nm, which is very insignificant.
But that is also how much light would bend down on Earth (although you would need to take not of the change in velocity and thus perform the appropriate relativistic correction to it.

Also note that if the mirrors were perfectly vertical, the light would eventually bend down enough that it misses the mirror. So to actually have the light continually bounce back and forth the mirrors need to be pointed upwards slightly so the light initially goes upwards before apparently curving down (in the accelerating/Earth frame). To the outside observer, this would mean the light continually goes at a more vertical angle as the elevator accelerates.

Another issue that complicates analysis is that time dilation will mean the observer in the elevator and the observer outside will observe the passage of time differently.

Acceleration ( a = g= 9.8 m/s/s) is the same while velocity increases with the passage of time “t” therefore imagine if the velocity of elevator is increased from 9.8 m/s to 10,000 m/s or more what would be the falling velocity of an object relative to the observer inside elevator accelerating upward @ 9.8 m/s/s

It would be the same regardless of when it is released.

If the observer is initially travelling at a velocity of v, along with the ball, and then releases it, then the ball continues to travel at v while the observer continues to accelerate with the velocity increasing linearly.
The velocity at time t of the observer will thus be v + t * 9.8 m/s^2.
We can also determine the velocity of the ball relative to the observer.
That will be v - (v + t * 9.8 m/s^2) = -t * 9.8 m/s^2.
Notice that this is the same as that for an observer on Earth. (noting that the "-" just indicates it is going down).

The same also applies to the bending of light.
However note that shining light which is horizontal to the outside observer when the elevator is already moving is equivalent to shining it down at an angle for the observer in the elevator (and it will still curve down, and the relative velocity of light will complicate it as well).

The size of the box in a thought experiment could be just a normal elevator or 2 m by 29972000 m. it’s just a matter of concept of equivalency.

The light will only follow a horizontal path in an inertial reference frame.
In an accelerating reference frame, the light will follow a curved path (assuming it isn't moving in the direction of the acceleration).

A light pulse bounces perfectly in horizontal line in b/t M1 and M2 of horizontal light clock relative to the inside observer when the elevator is at rest on earth – He doesn’t notice any abnormality in the operation of watch even if observes it closely thereat for one year - Means no bending of light at all – But 

The same light pulse doesn’t bounce perfectly in horizontal line in b/t M1 and M2 of horizontally light clock relative to the observer inside observer when the elevator accelerates @ 9.8 m/s – An observer notice it suddenly

Don’t they contradict each other in equivalency?

Similarly, if a pulse of a vertical light clock (M1 on floor and M2 on the ceiling) is used as a signal for the purpose of stopping the aforementioned elevator - (controlled remotely)

Technically or practically, once the elevator is stopped then it should stop for both observers but theory says it’s not for one of the observers  - Right

Don’t they contradict each other again?

Moreover, there is also a contradiction in following as explained above but you don’t agree - that’s ok

1-   The greater the velocity of the elevator (say 20,000 m/s) the greater the bending observed by an inside observer of elevator accelerating upward @ 9.8 m/s/s.
2-   The greater the velocity of elevator (say 20,000 m/s) the greater the falling velocity of an object observed by an inside observer of elevator accelerating upward @ 9.8 m/s/s.

i may be wrong but the observation of the occupant of the elevator which accelerates upward @ 9.8 m/s/s was just made by the outside observer on his/her behalf

A light pulse bounces perfectly in horizontal line in b/t M1 and M2 of horizontal light clock relative to the inside observer when the elevator is at rest on earth – He doesn’t notice any abnormality in the operation of watch even if observes it closely thereat for one year - Means no bending of light at all –

But it doesn't actually bounce perfectly horizontal.
Instead it would start heading up slightly and bend down to end up going down slightly. Then it bounces to go back up slightly.
You need a very large system in order to notice the bending, or a very rapid acceleration/high gravitational field.

They wont notice any abnormality in their watch unless they have something to compare it to.

Technically or practically, once the elevator is stopped then it should stop for both observers but theory says it’s not for one of the observers  - Right

I'm not sure exactly what you are saying here, if you mean 2 observers in the same elevator, this is to do with relativity breaking the notion of things being simultaneous.

Moreover, there is also a contradiction in following as explained above but you don’t agree - that’s ok
1-   The greater the velocity of the elevator (say 20,000 m/s) the greater the bending observed by an inside observer of elevator accelerating upward @ 9.8 m/s/s.
2-   The greater the velocity of elevator (say 20,000 m/s) the greater the falling velocity of an object observed by an inside observer of elevator accelerating upward @ 9.8 m/s/s.

i may be wrong but the observation of the occupant of the elevator which accelerates upward @ 9.8 m/s/s was just made by the outside observer on his/her behalf

The bending and falling velocity isn't effected by the velocity of the elevator, relative to an observer in the elevator.
As explained before, the falling velocity will always appear to be -a*t for the observer inside the elevator. That is because the velocity increases linearly with time, so the initial velocity cancels out. The way for it to be different is for the outside observer to "drop" it, but that is equivalent to it being thrown downwards at different velocities.
The bending at first might seem like it should be effected, but it is actually the initial direction which is.
i.e. if you have light going horizontal for an outside observer while the elevator is already moving it will be going down at an angle for the observer in the elevator.
This will be the same as if you had the light initially going down at that angle with the elevator starting from rest.

Everything should be equivalent, including the effects on clocks and light.

How about the vertical light clock in the last paragraph of the question. I write it down for you again or you can scroll up.

A light clock installed in the elevator of Einstein consists of two mirrors, M1 (attached to the ceiling) and M2 (attached to the floor) facing each other at a reasonable height "h" apart. Would the bouncing pattern of light beam/ pulse in between M1 and M2 be the same (equivalent) when the elevator is at rest on the ground and moving upward @ the rate of 9.8 m/s/s therein free space away from the local attraction of gravity?

Relative to onboard observer:
Two possibilities:
1- the upward and downward motion of the pulse in b/w m1 and m2 are the same
2- the upward and downward motion of the pulse in b/w m1 and m2 are not the same

which one pattern ( 1 or 2) would you choose in order to make it equivalent to the same vertical clock here on earth at rest and why? 

Do we also need a high rise system here on earth in order to notice the delay in upward motion of a pulse and rapid falling of a pulse?

Use a pulse of above said vertical light clock (M1 on floor and M2 on the ceiling) as a signal for the purpose of stopping the aforementioned elevator - (controlled remotely)

Technically or practically, once the elevator is stopped then it should stop for both observers but theory says it’s not for one of the observers as either a pulse is moving in the same direction of elevator or opposite to it which will make difference in timing of stopping - Right

Don’t they contradict each other again?

Bouncing of a tennis ball on the floor of the elevator

Let the occupant of the elevator release a ball from his hand. The acceleration of the said elevator increases due to the reduction of weight on its floor. 

For an outside observer, the floor of the elevator is catching the ball while the ball stays at its original position. NOTE: the greater the velocity of elevator in its constant acclaration of 9.8 m/s/s the less time it requires to catch the ball

For an inside observer the ball is falling on the floor of the elevator but would he feel an increase in his acceleration as soon as he releases the ball? Similarly would he feel a declaration when the ball hit the floor due to an increase in weight of elevator again? - here you don't agree with above NOTE but thats ok

Would the said elevator decelerate relative to the inside observer if feather and hammer fall on its floor at the same time from the outside environment on its way up.
 
The same phenomenon of declaration and acceleration is applied to the bouncing of ball either on its floor or vertical walls. Assume the elevator is accelerating constantly but at very high speed.
 

But it doesn't actually bounce perfectly horizontal.
Instead, it would start heading up slightly and bend down to end up going down slightly. Then it bounces to go back up slightly.
You need a very large system in order to notice the bending, or a very rapid acceleration/high gravitational field.

There are two possibilities
Relative to inside observer
If the path of a pulse remains straight horizontal then it is a contradiction with curving or bending down of horizontal beam

If the path of a pulse curve down relative to the inside observer then it would eventually miss its target of a mirror. Also, its pattern of bouncing is not similar to the bouncing of pulse on earth as you said. – Again in contradiction with horizontal light clock on ground at rest.

A horizontal beam of light seems to curve down to inside observer of accelerating frame – Such comments are made by the outside thinker here on earth on behalf of an inside observer. An outside thinker thinks that a light beam stays horizontal while elevator along with its occupant goes up with velocity v of constant acceleration, therefore, suggesting on behalf of inside observer who is unaware of his upward acceleration or velocity that a horizontal beam of light would like a curve down for him inside a tiny box but same thinker is unable to observe bending here on earth. Adding velocity of an elevator will, of course, make it straight-line curve but does the inside observer know that he is moving upward relative to the stationary horizontal beam? no

Also, a thinker simply suggests that an inside observer in the accelerating frame would feel a ground condition if he throws a ball upward and then catches it again after falling. Here a thinker, who made comments on behalf of an inside observer, doesn’t add upward velocity (which he added in order to make a horizontal beam curve down) to the velocity of the elevator and subtract downward velocities of a ball from the upward velocity of elevator which will make a huge contradiction to the ground conditions.

Quote

Everything should be equivalent, including the effects on clocks and light.

How about the vertical light clock in the last paragraph of the question. I write it down for you again or you can scroll up.

A light clock installed in the elevator of Einstein consists of two mirrors, M1 (attached to the ceiling) and M2 (attached to the floor) facing each other at a reasonable height "h" apart. Would the bouncing pattern of light beam/ pulse in between M1 and M2 be the same (equivalent) when the elevator is at rest on the ground and moving upward @ the rate of 9.8 m/s/s therein free space away from the local attraction of gravity?

Relative to onboard observer:
Two possibilities:
1- the upward and downward motion of the pulse in b/w m1 and m2 are the same
2- the upward and downward motion of the pulse in b/w m1 and m2 are not the same

which one pattern ( 1 or 2) would you choose in order to make it equivalent to the same vertical clock here on earth at rest and why?
Do we also need a high rise system here on earth in order to notice the delay in upward motion of a pulse and rapid falling of a pulse?

Option 2. That is what should be observed in both cases. But again, it is insignificant.
Note that the math is more complicated in this case, and gets more complex when you throw in relativity. So I will do the times for the outside observer.
If you had a 1 km tall building (reasonable for tall high-rises), then as a first approximation, it would take ~ 3 us for the light to reach the top of the building.
In that time the building would have moved ~0.05 nm, which would take the light an additional 2e-19 s, i.e. basically nothing.

Doing the relativistic correction, you get basically the same number.
For the light path going down, you also get basically the same number, but in the opposite direction (i.e. the light gets there sooner).
However, as the light going up has to go further, more time is spent on the upwards journey and thus the overall effect is a tiny increase in the total time taken.

Technically or practically, once the elevator is stopped then it should stop for both observers but theory says it’s not for one of the observers as either a pulse is moving in the same direction of elevator or opposite to it which will make difference in timing of stopping - Right

Again, not quite.
Practically once the elevator stops, it stops and there is no difference in timing, due to just how insignificant the difference is.
But technically, that is not the case. To start with the elevator can't even stop instantly. Lets assume a simple case of the elevator being thrust upwards by a rocket.
The elevator is then acting somewhat like a spring. The force is compressing it, accelerating it upwards as it does so.
When you remove the thrust, the elevator is still compressed. The bottom stops, but that compression keeps the top accelerating, decompressing as it does so. Eventually the decompression reaches all the way to the top and it has all stopped.

Even if you did manage to have the elevator stop all at once, that would only be the case for a particular reference frame. For other reference frames that will not necessarily be the case.

Let the occupant of the elevator release a ball from his hand. The acceleration of the said elevator increases due to the reduction of weight on its floor. 

For an outside observer, the floor of the elevator is catching the ball while the ball stays at its original position. NOTE: the greater the velocity of elevator in its constant acclaration of 9.8 m/s/s the less time it requires to catch the ball

Is the person just releasing the ball, or are they making its velocity 0 relative to an outside observer?
If the former, then the ball is initially travelling at the same velocity as the elevator and it will take the same amount of time to catch up regardless of the velocity of the elevator.
If the latter, then yes it will reach the floor quicker, but that is because it effectively be thrown at the floor.

For an inside observer the ball is falling on the floor of the elevator but would he feel an increase in his acceleration as soon as he releases the ball? Similarly would he feel a declaration when the ball hit the floor due to an increase in weight of elevator again? - here you don't agree with above NOTE but thats ok

This depends upon a multitude of factors, including what is causing the acceleration. If the acceleration is kept constant then there will be no change. However if it is based upon a constant force, then it will increase when the ball is released and drop when the ball hits the floor.
Also note that the same happens with Earth and gravity. When you drop a ball, not only does the ball accelerate towards Earth, the Earth (and you) accelerate up towards the ball, but again, the effect is insignificant.

If the path of a pulse curve down relative to the inside observer then it would eventually miss its target of a mirror. Also, its pattern of bouncing is not similar to the bouncing of pulse on earth as you said. – Again in contradiction with horizontal light clock on ground at rest.

Not in an ideal environment.
In an ideal environment the light initially goes upwards at an angle of a. During its journey it curves downwards eventually reaching an angle of -a.
It then hits a mirror perpendicular to this and goes back heading upwards at an angle of a, curving down eventually reaching an angle of -a, and bouncing off another mirror.
The light will continue bouncing back and forth between these mirrors.

To the outside observer the light would initially be going upwards at an angle of a, but then get steeper and steeper as it bounces between the mirrors.

does the inside observer know that he is moving upward relative to the stationary horizontal beam? no

Only in the sense of it being too insignificant to notice, or it being equivalent to being stationary in a gravitational field.

Also, a thinker simply suggests that an inside observer in the accelerating frame would feel a ground condition if he throws a ball upward and then catches it again after falling. Here a thinker, who made comments on behalf of an inside observer, doesn’t add upward velocity (which he added in order to make a horizontal beam curve down) to the velocity of the elevator and subtract downward velocities of a ball from the upward velocity of elevator which will make a huge contradiction to the ground conditions.

I'm not sure what you mean by ground condition, but if you mean what it feels like to catch the ball, the velocity of the elevator doesn't make a difference. That is because what matters is the relative velocity. That is because during the collision the velocities need to be matched, i.e. a force is applied to accelerate the objects such that both objects travel with the same velocity.
It would feel the same to catch a ball heading towards you at 1 m/s with you stationary as it would to catch a ball heading away from you at 99 m/s, with you travelling at 100 m/s.

Assuming you have a train system where you live with decent stretches of straight track you can test this with a friend by throwing a ball back and forth as the train is moving.

If you stand at rest in a gravitational field and throw a ball upwards at a speed of v, it will take -2*v/g to come back down, and it will be travelling at a velocity of -v when you catch it and thus has a relative velocity of -v.

If instead you throw it in an elevator in free space accelerating upwards at a rate of a, when this elevator is at a velocity of u, then the position of the ball will follow the equation (u+v)*t, while your hand follows the equation u*t+0.5*a*t^2.
The ball and your hand meet again when these 2 are equal, i.e.:
(u+v)*t = u*t+0.5*a*t^2
u+v=u+0.5*a*t
v=0.5*a*t.
i.e. at a time of 2*v/a.
The velocity of the ball will still be u+v.
The velocity of your hand will be u+a*t=u+a*(2*v/a)=u+2*v.
Thus the velocity of the ball hitting your hand, relative to your hand is -v.
This is the same.

The appropriate analogy for a beam of light curving downwards is throwing the ball horizontally, where it will appear to curve downwards, but unlike light it will typically be noticeable.

As said, it’s just a question about the equivalency of the feeling of an occupant of an elevator when he is at rest here on earth inside elevator and when he is in elevator accelerating upward @ rate 9.8 m/s/s – No one else 

The maximum or terminal velocity of an object (skydiver) if dropped is around 200 km/h with air resistance on it. Here are velocities when we drop thing here on earth - Remember "g =9.8m/s/s" is constant

Time (seconds) : 0   1       2       3       4     5  .....
Velocity (m/s) :   0   9.8  19.6  29.4   39.2  49 ….

This means the velocity of an elevator (which is used for equivalence principle - accelerating upward @ 9.8 m/s/s) is also increasing with time ”t” until it reaches a very high value say 50,000 m/s or more. Remember the elevator doesn’t lose its value of constant acceleration of a=9.8 m/s/s. Therefore if he releases an apple from his hand at such a high velocity

Relative to an occupant of accelerating elevator deep in space:  The falling/ hitting time of an apple decreases due to the increase in velocity of the elevator every moment if fall from the same height.

Relative to occupant of an elevator when he at rest on the ground: The falling/ hitting time of an apple is constant if fall from the same height.

The same is applied to the bouncing of a tennis ball as explained earlier.

So is this not a contradiction?

Option 2. That is what should be observed in both cases. But again, it is insignificant.
Note that the math is more complicated in this case, and gets more complex when you throw in relativity. So I will do the times for the outside observer.

What do you mean by outside observer –

I think you mean he is not on earth but outside? If yes, then irrelevant as said. But still, according to him, the bouncing pattern of pulses on earth will be constant while it is not the same or constant in the accelerating frame of the elevator due to the increase in velocity of elevator every moment. This applies to all clocks installed either horizontally or vertically as well.

Similarly, if you chose option 2 then the same horizontal beam becomes curve down for inside occupant if shines by outside observer toward far wall of elevator while the same beam if used for the horizontal light clock is not curved down but straight? No idea why?

Also note that the same happens with Earth and gravity. When you drop a ball, not only does the ball accelerate towards Earth, the Earth (and you) accelerate up towards the ball, but again, the effect is insignificant.

But we never feel any declaration or acceleration here on earth. Similarly, the force of earth on each object is different if they are different while in elevator there is only one force, which is the mass of elevator and its acceleration. There is no force of an object if put on the floor as it doesn’t produce acceleration other than its resistance. 
So all object on the floor of elevator feel the same force from elevator which is accelerating upward at the rate of 9.8 m/s/s while here on earth exerts force differently on each object if their mass is different i.e. F=Mg, where M is mass of earth while g is acceleration due to gravity of an object at which earth accelerate toward it.

Again, not quite.
Practically once the elevator stops, it stops and there is no difference in timing, due to just how insignificant the difference

Please scroll up to question. You will see the following along with diagram
Elevator accelerating @ 9.8 m/s/s in free space: Light beams or pulses emit at A and D at the same time, respectively. Both don't reach their destination B and C at the same time. Pulse on right of observer reaches C faster as the floor of the elevator approaches it than pulse on the left where pulse moves away from the floor of the elevator when the floor tries to approach it (difficult to catch) - Right

Let a pulse emit at A is used for stopping the elevator as a remote control? Ignoring the pulse emit at D in the above example.

Let a pulse emit at D is used for stopping the elevator as a remote control? Ignoring the pulse emit at A in the above example.

Have you noticed the stopping of the elevator w.r.t the occupant and the outside observer?

The maximum or terminal velocity of an object (skydiver) if dropped is around 200 km/h with air resistance on it. Here are velocities when we drop thing here on earth - Remember "g =9.8m/s/s" is constant

Terminal velocity is a result of air resistance.
It is when the force due to the air pushing the object upwards is the same as the force of gravity pulling it downwards.

This means the velocity of an elevator (which is used for equivalence principle - accelerating upward @ 9.8 m/s/s) is also increasing with time ”t” until it reaches a very high value say 50,000 m/s or more. Remember the elevator doesn’t lose its value of constant acceleration of a=9.8 m/s/s. Therefore if he releases an apple from his hand at such a high velocity

Relative to an occupant of accelerating elevator deep in space:  The falling/ hitting time of an apple decreases due to the increase in velocity of the elevator every moment if fall from the same height.

No, it doesn't.
That is the point I have been trying to make.
This can be shown mathematically.

When the object (apple in this case) is released, it doesn't magically have its velocity change to 0. It continues at whatever velocity it was going at before hand.
So if you have your apple at height h above the elevator floor, initially travelling at velocity v with the elevator, then once released its position will be given as:
y=h+v*t.
The floor of the elevator starts at a position of 0, initially at a velocity of v, but it continues to accelerate at a rate of a. Thus its position will be given as:
y=0+v*t+0.5*a*t^2.

The apple "hits" the floor of the elevator when the 2 positions are equal.
i.e. when:
h+v*t=0+v*t+0.5*a*t^2
This immediately simplifies to:
h=0.5*a*t^2.
t=sqrt(2*h/a)

The term involving the velocity is cancelled out because both the apple and the elevator have that term describing their position.

This is the same as for an observer on Earth, with the apple accelerating at a rate of -a.
Now, the floor remains at 0, and we have the apple at a position of y=h+0+0.5*(-a)*t^2.
So it hits when:
0=h-0.5*a*t^2
0.5*a*t^2=h
t=sqrt(2*h/a)

The exact same as before.

So there is no contradiction.
The time taken for the apple to appear to fall and hit the ground is the same on Earth as it is in the elevator.
The initial velocity of the elevator makes no difference.

The only way for that initial velocity to make a difference is for the apple to have its speed set to 0. But that is not longer equivalent to dropping the apple. That would be throwing the apple down with whatever velocity the elevator has (but in opposite direction).
i.e. if the elevator is travelling upwards at 50 000 m/s, then instead of the observer on Earth just releasing the apple, they would need to throw it downwards at 50 000 m/s.

What do you mean by outside observer –

I was doing it for the case of an outside observer for the elevator in free space. It would be equivalent to some hypothetical outside observer for a person on Earth.
For the observer inside the elevator or on Earth it is more complex due to time dilation.
But time dilation will have the same effect on the person on Earth as it would to the person in the accelerating elevator, so the situations end up equivalent.

Similarly, if you chose option 2 then the same horizontal beam becomes curve down for inside occupant if shines by outside observer toward far wall of elevator while the same beam if used for the horizontal light clock is not curved down but straight? No idea why?

They both curve down.
If you want the light to continue bouncing back and forth between the walls it can't start out horizontal, it need to starts at an angle. The light will follow a parabolic arc, just as a ball will.

But we never feel any declaration or acceleration here on earth.

Because the effect is tiny.
The same would apply if you had a massive elevator the size of Earth accelerating upwards in free space (but without gravity). When you drop something the change in acceleration would not be felt because of how tiny it is.

So all object on the floor of elevator feel the same force from elevator which is accelerating upward at the rate of 9.8 m/s/s

No they don't.
They feel a force based upon their mass, such that they accelerate at the same rate as the elevator.
If they all had the same force then light objects would fly up from the floor while heavy objects would fall through the fall.
The force given will be F=ma.

Also note that to the observer in the elevator, there will be a pseudo-force (or inertial force) acting downwards with a magnitude given by F=m*a.
Again, this has to be dependent on the mass of the object (as all inertial forces/pseudo-forces are). If it wasn't and instead all objects experienced the same pseudoforce, then light objects would accelerate to the floor very rapidly while heavier objects would fall much more slowly.

So again, this is equivalent to an observer on Earth.
The only difference is instead of a, we use g.

Elevator accelerating @ 9.8 m/s/s in free space: Light beams or pulses emit at A and D at the same time

As I said before, with relativity, events occurring simultaneously rely upon a specific reference frame.
So if something occurs "at the same time" there is the question of what frame is this "same time" in?
Because if you use a different frame, it will not necessarily be the same time.

With relativity, even time is relative, both the location in time and the passage of time.

If I understand and recall correctly, if this occurring simultaneously for an outside observer located infinitely far away, then for the observer in the elevator down near A, that beam leaves before the beam leaving D. However due to time dilation and relativity of time, to them the beam hits C before the beam hits B.

Relativity of time can be quite complex and hard to get your head around, even more so when acceleration is thrown in. I struggle with it myself.

Quote

Quote from: E E K on Today at 10:09:09 AM
This means the velocity of an elevator (which is used for equivalence principle - accelerating upward @ 9.8 m/s/s) is also increasing with time ”t” until it reaches a very high value say 50,000 m/s or more. Remember the elevator doesn’t lose its value of constant acceleration of a=9.8 m/s/s. Therefore if he releases an apple from his hand at such a high velocity

Relative to an occupant of accelerating elevator deep in space:  The falling/ hitting time of an apple decreases due to the increase in velocity of the elevator every moment if fall from the same height.
No, it doesn't.
That is the point I have been trying to make.
This can be shown mathematically.

Would object A and B fall at the rate relative to the occupant as shown in the below Fig if your abovementioned statement is true. Object A is an apple (discussed previously). Object B is just above the elevator which means outside the frame of the elevator.

If you want the light to continue bouncing back and forth between the walls it can't start out horizontal, it need to starts at an angle. The light will follow a parabolic arc, just as a ball will.

It seems a pulse follows the same acceleration (at an angle) at which an elevator is accelerating upward in the accelerating frame in order to continue its back and forth bouncing in b/t M1 and M2 - I don't think so its possible.

Similarly, mirrors which are attached to the walls of elevator firmly goes upward with “g” while pulse of the light clock has either vt or ct (upward parabolic arc in your mind for outside observer if I’m not wrong) so eventually a pulse which is bouncing in b/t M1 and M2 will lose it target as vt <0.5 gt^2 while ct >0.5gt^2 (ct is correct) - AGAIN impov

Similarly, if you chose option 2 then the same horizontal beam becomes curve down for inside occupant if shines by outside observer toward far wall of elevator while the same beam if used for the horizontal light clock is not curved down but straight? No idea why?
They both curve down.

if they both curve down then it would be straight horizontal beams for an outside observer.

For the observer inside the elevator or on Earth it is more complex due to time dilation

Use below light clock which is free of time dilation - BTW comment, please

If I understand and recall correctly, if this occurring simultaneously for an outside observer located infinitely far away, then for the observer in the elevator down near A, that beam leaves before the beam leaving D. However due to time dilation and relativity of time, to them the beam hits C before the beam hits B.

Please read the following - it might help what i said earlier
Relativity of Simultaneity (Experiment)
https://en.wikipedia.org/wiki/Talk:Relativity_of_simultaneity/Archive_2#Relativity_of_Simultaneity_(Experiment)

Would object A and B fall at the rate relative to the occupant as shown in the below Fig if your abovementioned statement is true. Object A is an apple (discussed previously). Object B is just above the elevator which means outside the frame of the elevator.

Is object B initially at rest relative to the observer in the elevator (and thus the Apple)?
If so, it will fall and accelerate at the same rate as the apple (but reach a higher speed as it has longer to fall).
If instead it was at rest relative to the outside observer, it would already be moving quite quickly compared to the observer in the elevator.

It seems a pulse follows the same acceleration (at an angle) at which an elevator is accelerating upward in the accelerating frame in order to continue its back and forth bouncing in b/t M1 and M2 - I don't think so its possible.

Similarly, mirrors which are attached to the walls of elevator firmly goes upward with “g” while pulse of the light clock has either vt or ct (upward parabolic arc in your mind for outside observer if I’m not wrong) so eventually a pulse which is bouncing in b/t M1 and M2 will lose it target as vt <0.5 gt^2 while ct >0.5gt^2 (ct is correct) - AGAIN impov

It is the inside observer which sees it following a parabolic arc.
To the outside observer it is initially travelling upwards at a small angle, and as time goes on, that angle increases due to the reflection.
As the mirror isn't vertical in this arrangement, the mirror will cause the light to change angle as it reflects.
This changing angle keeps up with the acceleration, until eventually when the elevator is travelling near the speed of light the light is travelling almost straight up to the outside observer.

if they both curve down then it would be straight horizontal beams for an outside observer.

Yes, assuming they were initially horizontal, otherwise they would just be straight for the outside observer.

Please read the following - it might help what i said earlier
Relativity of Simultaneity (Experiment)
https://en.wikipedia.org/wiki/Talk:Relativity_of_simultaneity/Archive_2#Relativity_of_Simultaneity_(Experiment)

As I said, relativity of time can be quite confusing.
The main article may be more helpful.

Is object B initially at rest relative to the observer in the elevator (and thus the Apple)?
If so, it will fall and accelerate at the same rate as the apple (but reach a higher speed as it has longer to fall).
If instead it was at rest relative to the outside observer, it would already be moving quite quickly compared to the observer in the elevator.

Let object “B” is at rest above the elevator to the outside observer but object “B” seems falling at the rate of 9.8 nm/s/s relative to the inside observer. The occupant in his hand holds an object “A”.

Now B is at height” h” from the floor while A is at a height “h1” from the floor when it is just released by the inside observer. So how would you prove mathematically that both “B” and “A” are falling at the same rate relative to the occupant as you did for an apple previously?

Quote

Quote from: E E K on October 10, 2019, 04:28:39 PM
if they both curve down then it would be straight horizontal beams for an outside observer.

Yes, assuming they were initially horizontal, otherwise they would just be straight for the outside observer.

I think it will clear a lot of things if the outside observer sees the whole event right from the very start when an elevator is at rest and a pulse bounces in b/t M1 and M2 (installed horizontally) to the time when he shines his flashlight toward the far wall of the elevator when it is accelerating upward at the rate 9.8 m/s/s and the occupant stops said elevator via a vertical pulse either going up or going down as asked earlier.

I’m not giving you hard time but could you please explain how would a beam of light and a pulse of light clock look like relative to the outside observer and inside observer from the start of the event to the end when an elevator stops?

As I said, relativity of time can be quite confusing.
The main article may be more helpful.

No it is not. Science may be complex but straight forward, therefore, there is something exists unnatural as when a train in an experiment or aforementioned elevator is stopped then it should stop for both observers – Right. Magic or hocus-pocus is not allowed in science.

Quote

Is object B initially at rest relative to the observer in the elevator (and thus the Apple)?
If so, it will fall and accelerate at the same rate as the apple (but reach a higher speed as it has longer to fall).
If instead it was at rest relative to the outside observer, it would already be moving quite quickly compared to the observer in the elevator.

Let object “B” is at rest above the elevator to the outside observer but object “B” seems falling at the rate of 9.8 nm/s/s relative to the inside observer. The occupant in his hand holds an object “A”.

Now B is at height” h” from the floor while A is at a height “h1” from the floor when it is just released by the inside observer. So how would you prove mathematically that both “B” and “A” are falling at the same rate relative to the occupant as you did for an apple previously?

The general formulas for motion at constant acceleration (which can be 0) are:
a=a,
v=v0+at
y=y0+v0*t+0.5*a*t²

where v0 and y0 are the values for v and y at time t=0. (and a subscript indicates what it is for, e.g. o for object, e for elevator)
We will also take the 0 value for y to be the floor of the elevator at time t=0.

So we have our object (either a or b), at the time of release no longer having any acceleration acting upon it.
This means the y term simplified to y_o=y0_o+v0_o*t.
The floor of the elevator does continue its motion and would have y_e=v0_e*t+0.5*a*t².
Then to find the position of the object relative to the floor of the elevator, we find the difference:
y=y_o - y_e
=y0_o + v0_o*t - ( v0_e*t+0.5*a*t² )
=y0_o + v0_o*t - v0_e*t - 0.5*a*t²
=y0_o + (v0_o - v0_e) *t - 0.5*a*t²

This is now in terms of the velocity of the elevator and object relative to the outside observer. It would be better if we could express it in terms of the velocity of the object relative to the elevator. The initial velocity in this case vi_o will be given by the difference between its velocity and the velocity of the elevator, i.e.:
vi_o = v0_o - v0_e
Substituting that back in to the formula above we end up with:
y=y0_o + vi_o*t - 0.5*a*t²

And now there is no dependence upon the velocity of the elevator at all.
Either object will fall, starting at whatever the initial velocity is, and accelerating at a rate of a.

For object A we have
y=h1 - 0.5 a*t²

For object B we have:
y=h + v*t - 0.5 a*t²
If the initialy velocity of it relative to the inside observer is 0, then this simplifies to
y=h - 0.5 a*t²

Otherwise it is equivalent to being thrown at a velocity of v rather than just being released.

I think it will clear a lot of things if the outside observer sees the whole event right from the very start when an elevator is at rest and a pulse bounces in b/t M1 and M2 (installed horizontally) to the time when he shines his flashlight toward the far wall of the elevator when it is accelerating upward at the rate 9.8 m/s/s and the occupant stops said elevator via a vertical pulse either going up or going down as asked earlier.

I’m not giving you hard time but could you please explain how would a beam of light and a pulse of light clock look like relative to the outside observer and inside observer from the start of the event to the end when an elevator stops?

If the beam of light bouncing back and forth starts horizontal, then once the elevator starts to accelerate the beam of light will:
For the outside observer - continue to bounce back and forth as the mirrors move up and eventually the mirrors will be too high.
For the inside observer -  curve down until it eventually escapes the mirrors.
The above also applies to the flashlight shone horizontally at the far wall, with the outside observer seeing the light just travel off straight, while the inside observer sees it curve down.

On the other hand, if it starts at a slight angle with the elevator accelerating then the beam of light will:
For the outside observer - continue its motion upwards and at every reflection have the angle become steeper. When the elevator stops the light will continue going up and miss the mirrors.
For the inside observer - It will follow a parabolic arc, bouncing between the 2 mirrors. Each bounce has it initially going upwards and curving down to hit the mirror. When it stops, the light will move upwards and miss the mirror rather than curving down.

The stopping of the elevator is the hardest part due to the time dilation involved.
For the outside observer, they see the light shining upwards eventually reaching the top of the elevator, then assuming the elevator is being pulled from the top, they see the elevator stop from the top to the bottom.
For the inside observer, they see the light get to the top faster than the outside observer, but the stopping of the elevator will take longer to propagate back down.

No it is not. Science may be complex but straight forward, therefore, there is something exists unnatural as when a train in an experiment or aforementioned elevator is stopped then it should stop for both observers – Right. Magic or hocus-pocus is not allowed in science.

No, it shouldn't That is a key part of the relativity of time. Things being simultaneous is dependent upon what reference frame it is in.
2 events can be simultaneous in one reference frame but not simultaneous in another.
Lots of things in quantum mechanics and relativity can be quite confusing.

If the beam of light bouncing back and forth starts horizontal, then once the elevator starts to accelerate the beam of light will:
For the outside observer - continue to bounce back and forth as the mirrors move up and eventually the mirrors will be too high.
For the inside observer -  curve down until it eventually escapes the mirrors.
The above also applies to the flashlight shone horizontally at the far wall, with the outside observer seeing the light just travel off straight, while the inside observer sees it curve down.

There you go!

Relative to the inside observer: The clock works perfectly fine when he is at rest on the ground but it stops working for him as soon as the elevator starts accelerating. So no equivalency at all relative to the clock.

I think this is it

The general formulas for motion at constant acceleration (which can be 0) are:
a=a,
v=v0+at
y=y0+v0*t+0.5*a*t2

where v0 and y0 are the values for v and y at time t=0. (and a subscript indicates what it is for, e.g. o for object, e for elevator)
We will also take the 0 value for y to be the floor of the elevator at time t=0.

So we have our object (either a or b), at the time of release no longer having any acceleration acting upon it.
This means the y term simplified to yo=y0o+v0o*t.
The floor of the elevator does continue its motion and would have ye=v0e*t+0.5*a*t2.
Then to find the position of the object relative to the floor of the elevator, we find the difference:
y=yo - ye
=y0o + v0o*t - ( v0e*t+0.5*a*t2 )
=y0o + v0o*t - v0e*t - 0.5*a*t2
=y0o + (v0o - v0e) *t - 0.5*a*t2

This is now in terms of the velocity of the elevator and object relative to the outside observer. It would be better if we could express it in terms of the velocity of the object relative to the elevator. The initial velocity in this case vio will be given by the difference between its velocity and the velocity of the elevator, i.e.:
vio = v0o - v0e
Substituting that back in to the formula above we end up with:
y=y0o + vio*t - 0.5*a*t2

And now there is no dependence upon the velocity of the elevator at all.
Either object will fall, starting at whatever the initial velocity is, and accelerating at a rate of a.

For object A we have
y=h1 - 0.5 a*t2

For object B we have:
y=h + v*t - 0.5 a*t2
If the initialy velocity of it relative to the inside observer is 0, then this simplifies to
y=h - 0.5 a*t2

Otherwise it is equivalent to being thrown at a velocity of v rather than just being released

I didn’t go through it yet but the following calculation is also yours

For Object A:

When the object (apple in this case) is released, it doesn't magically have its velocity change to 0. It continues at whatever velocity it was going at before hand.
So if you have your apple at height h above the elevator floor, initially travelling at velocity v with the elevator, then once released its position will be given as:
y=h+v*t.
The floor of the elevator starts at a position of 0, initially at a velocity of v, but it continues to accelerate at a rate of a. Thus its position will be given as:
y=0+v*t+0.5*a*t^2.

The apple "hits" the floor of the elevator when the 2 positions are equal.
i.e. when:
h+v*t=0+v*t+0.5*a*t^2
This immediately simplifies to:
h=0.5*a*t^2.

For Object B:
Relative to the inside observer: As it was and it is falling at the rate 9.8 m/s/s, therefore, it has zero upward velocity as an apple had in the above case - True. So
y=h.
The floor of the elevator starts at a position of 0, initially at a velocity of v, but it continues to accelerate at a rate of a. Thus its position will be given as:
y=0+v*t+0.5*a*t^2.

Object B "hits" the floor of the elevator when the 2 positions are equal.
i.e. when:
h=0+v*t+0.5*a*t^2
This immediately simplifies to:
h=v*t+0.5*a*t^2


So it means they don't fall at the same rate.

Relative to the inside observer: The clock works perfectly fine when he is at rest on the ground but it stops working for him as soon as the elevator starts accelerating. So no equivalency at all relative to the clock.

The elevator is in free space, not on the ground.
The equivalence still remains.
If instead of having it accelerate you put it right next to a massive planet with gravity, the same thing happens. The light curves down due to gravity.

What isn't equivalent is an observer in an elevator at rest in free space being equivalent to an observer standing on Earth; or an observer in an elevator at rest in free space being equivalent to an observer in an elevator accelerating upwards in free space.

      h of object B=v*t+0.5*a*t^2 ......is not equal to ...... h of object A=0.5*a*t^2.

So it means they don't fall at the same rate.

This comes down to what you mean by rate.
Objects typically don't fall at the same velocity, as they change velocity as they fall.
So that makes it hard for a comparison on that. Instead the more relevant term to use is the acceleration.
And that is equal for the 2.

But even the different velocity doesn't show any problem with equivalency of the 2 frames.
Instead it shows the problem with trying to equate releasing an apple and allowing it to fall with throwing it at the ground.

i.e. object A and object B are not "equivalent" because they don't start at the same velocity relative to our observer.

Object A in the elevator is equivalent to releasing an object on Earth.
Object B in the elevator is equivalent to throwing an object at the ground on Earth.
So the accelerating elevator and Earth remain equivalent.

The elevator is in free space, not on the ground.

I presumed we were on the same page. It’s a thought experiment earth can be disappeared at the timing of starting of acceleration of the elevator.

Anyhow, let both the elevator of the subject along with its occupant and outside observers are at rest in deep free space. 

An outside observer shines a flash of horizontal beam of light through an elevator such that a light beam enters through the left window and leaves via its right window. An occupant who is close to the floor of the elevator doesn’t know the direction of beam of light. He sees a horizontal beam above his head only just before the start of the accelerating of elevator. Now the said elevator starts from rest in free space. When the elevator starts accelerating, a horizontal pulse also emits from one of the mirrors for the operation of the clock (simultaneously).

So how would one explain the bouncing of a light pulse in b/t the two mirrors before and after the acceleration of the elevator relative to the inside and outside observer?

Although the elevator is in a mode of constant acceleration of 9.8 m/s/s but its velocity increases with time "t" therefore would the bending of horizontal beam of light also be constant even if the upward velocity of the elevator is 0.0002 m/s or more or it can even < 0.0002 m/s. And the same is applied to parabolic arc made by the pulse?

Similarly, which way a horizontal beam of light which was shunned from outside would curve down. Please explain if it curves down right then why right and if it curves down left then why left?

Object B in the elevator is equivalent to throwing an object at the ground on Earth.

No, it just falling at rate of 9.8 m/s/s relative to the inside observer. When the said elevator approaches to any objects (at rest to the outside observer unless falling in the opposite direction of the elevator) on its way up it seems, this object in the elevator is equivalent to falling an object on Earth.

Falling velocity, of course, would be different due to differences in height. By rates means acceleration not velocity

An outside observer shines a flash of horizontal beam of light through an elevator such that a light beam enters through the left window and leaves via its right window. An occupant who is close to the floor of the elevator doesn’t know the direction of beam of light. He sees a horizontal beam above his head only just before the start of the accelerating of elevator. Now the said elevator starts from rest in free space. When the elevator starts accelerating, a horizontal pulse also emits from one of the mirrors for the operation of the clock (simultaneously).

So how would one explain the bouncing of a light pulse in b/t the two mirrors before and after the acceleration of the elevator relative to the inside and outside observer?

Before the acceleration it would need to bounce horizontally.
After the acceleration begins it would need to be going upwards at a slight angle to follow a parabolic path.
Starting the acceleration acts the same as switching on gravity.

Although the elevator is in a mode of constant acceleration of 9.8 m/s/s but its velocity increases with time "t" therefore would the bending of horizontal beam of light also be constant even if the upward velocity of the elevator is 0.0002 m/s or more or it can even < 0.0002 m/s. And the same is applied to parabolic arc made by the pulse?

For the inside observer, their velocity is irrelevant. They don't know what their velocity relative to some outside observer.
All they know is their acceleration, so the light would just continue bouncing in parabolic arcs.
The bending of the light, which is a result of their acceleration, remains constant.
For a beam that is horizontal to the outside observer, the bending would still be the same but it wouldn't start horizontal to the inside observer.
For example, if the inside observer is travelling at half the speed of light, then the beam which is horizontal to the outside observer would be going down at an angle of 30 degrees at the start, and then bend from that 30 degrees.

Quote

Object B in the elevator is equivalent to throwing an object at the ground on Earth.

No, it just falling at rate of 9.8 m/s/s relative to the inside observer. When the said elevator approaches to any objects (at rest to the outside observer unless falling in the opposite direction of the elevator) on its way up it seems, this object in the elevator is equivalent to falling an object on Earth.

Falling velocity, of course, would be different due to differences in height. By rates means acceleration not velocity

You could either have it as the initial height is the height when the object was at rest, or have it start to fall at a different velocity.
But yes, an object approaching the elevator, relative to the accelerating elevator, would be equivalent to an object falling on Earth.

Here is the crux

Relative to the inside observer,

Before acceleration
The pulses of light beam shunned from outside and a pulse of light clock is in horizontal motion

After acceleration
The pulses of light beam are bent down but while a pulse of light clock is not but instead bend up when both are initially horizontal when the elevator was at rest.

Why gravity acts differently on a pulse if initially moving horizontally when

1-   it is a part of clock
2-   it is not a part of clock

Similarly, I just want to go a lil bit more in detail about the formation of curve made by single pulse right from the start of its bending down from one wall all the way to its end on the other wall of elevator when the velocity of elevator is very low at initial stage and when the velocity of elevator is very high because it seems to me that the bending is increased with the increase in velocity of elevator if not it should bend at one constant angle.

Here is the crux

Relative to the inside observer,

Before acceleration
The pulses of light beam shunned from outside and a pulse of light clock is in horizontal motion

After acceleration
The pulses of light beam are bent down but while a pulse of light clock is not but instead bend up when both are initially horizontal when the elevator was at rest.

The clock which works while at rest without acceleration will not work when the system starts to accelerate. It would need to be realigned after the acceleration begins such that the light initially travels upwards rather than going horizontally.
It doesn't bend up, it initially goes upwards, bending down such that it hits the mirror and reflects back up.
Both mirrors for the light clock in the accelerating frame need to be angled upwards, however in practice the amount is not detectable for such a low acceleration.

Why gravity acts differently on a pulse if initially moving horizontally when

1-   it is a part of clock
2-   it is not a part of clock

It doesn't.
The light behaves the same regardless of if it is in a clock or not.
The distinction is in the initial direction of the light.

Similarly, I just want to go a lil bit more in detail about the formation of curve made by single pulse right from the start of its bending down from one wall all the way to its end on the other wall of elevator when the velocity of elevator is very low at initial stage and when the velocity of elevator is very high because it seems to me that the bending is increased with the increase in velocity of elevator if not it should bend at one constant angle.

As I said, to the inside observer it follows a parabolic arc.
It starts at some small angle a, heading upwards and to the right at one mirror. When it reaches the other mirror it has been bent downwards such that it is now travelling at an angle of -a.
The mirror is set up such that it is perpendicular to this ray of light.
That means the light reflects off it and heads back in the same direction it came from, now heading upwards and to the left at an angle of a. During this journey it again bends downwards and ends up travelling at an angle of -a when it again hits the mirror, getting reflected straight back along its prior path, so it will again be heading upwards and to the right at an angle of a.

To the outside observer the light is initially heading upwards and to the right at an angle of a. It continues on a straight path, going higher the mirrors (as the mirrors have a smaller upwards velocity) but eventually the mirrors velocity increases above the upwards velocity of the light and they catch up, right when the light gets to the other mirror.
But in this frame the light does hit the mirror perpendicular to its surface. The mirror is still bent upwards and this causes the light to reflect back to the left at an angle higher than a.
This means the light is again travelling upwards faster than the mirror. But again, the mirror's constant acceleration causes it to catch back up at the next reflection.
This process continues with the light travelling upwards at a steeper angle after each reflection.
As the elevator approaches the speed of light the angle approaches 90 degrees, i.e. the light is almost going straight up.

Need help with this

Could you clarify what you need help with, perhaps by asking questions as to exactly what you don't understand?

Quote from: rowetim138 on December 31, 2019, 04:31:10 AM

Need help with this

Could you clarify what you need help with, perhaps by asking questions as to exactly what you don't understand?

Maybe he should click the link in his sig.

The final striking (falling) velocities of object A and B on the floors of elevators are not the same as when the elevator is


1- At rest here on earth and
2- Moving with constant acceleration at the rate of 9.8 m/s/s but with higher velocities deep in space

Isn’t the aforementioned sufficient for non-equivalency before arguing something else that has the same result – Period.

Also, don't forget, the elevator is not moving with constant velocity (inertial frame in which laws of physics are the same) but has constant acceleration.

Thank you and have a wonderful and prosperous new year!

The final striking (falling) velocities of object A and B on the floors of elevators are not the same as when the elevator is


1- At rest here on earth and
2- Moving with constant acceleration at the rate of 9.8 m/s/s but with higher velocities deep in space

Why not?
As long as each object's initial positions, relative to the elevator, are the same "here on Earth" and "Moving with constant acceleration at the rate of 9.8 m/s/s" before they are released.

Isn’t the aforementioned sufficient for non-equivalency before arguing something else that has the same result – Period.

If it were true.  Show your calculations for each case.

Also, don't forget, the elevator is not moving with constant velocity (inertial frame in which laws of physics are the same) but has constant acceleration.

If the elevator were moving with a constant velocity the objects would never fall.
Look what happens in the ISS before and during a "reboost" around 3:20 in:

Space Station Reboost
The high velocity of the ISS has no effect but a very slight acceleration starts the camera moving.

Thank you and have a wonderful and prosperous new year!

And to you.

The final striking (falling) velocities of object A and B on the floors of elevators are not the same as when the elevator is


1- At rest here on earth and
2- Moving with constant acceleration at the rate of 9.8 m/s/s but with higher velocities deep in space

Isn’t the aforementioned sufficient for non-equivalency before arguing something else that has the same result – Period.

No, as you are yet to describe a situation that should be equivalent that isn't.

A and B don't hit the elevator at the same relative velocity, but that is true for both at rest and in the moving elevator.

I have already demonstrated, in a uniform gravitational field, which accelerates objects at a rate of g (noting that g is negative), the time required for the object to hit the ground, assuming it started at rest, is given by sqrt(-2*h/g). Its velocity at any time (t) will be given by g*t, so when it hits the ground it will hit at a velocity of g*sqrt(-2*h*g).

If it starts with some initial velocity, then just project it backwards in time and pretend it started at rest and problem solved.

This is the exact same for the equivalent situation for an elevator accelerating upwards.

For the elevator, it starts at rest relative to the elevator, with both at some initial velocity vref, and the object initially a height h above it, with the elevator accelerating at a rate of a.
At time t, the object will be at a position of h+vref*t.
The floor of the elevator instead will be at a position of vref*t+0.5*a*t^2.
That means the height of the object above the elevator will be (h+vref*t)-(vref*t+0.5*a*t^2)=h-0.5*a*t^2.
This is the same as an object falling on Earth, noting that g=-a.
This gives us the equivalent result as the impact time is sqrt(2*h/a).

The velocity at any given time is vref for the object, and vref+a*t for the elevator, so the relative velocity is -a*t, which is equivalent to g*t, and results in an impact velocity of -a*sqrt(2*h*a).

The initial velocity of the elevator doesn't matter.

Although abovesaid statement is quite clear but still

Compare the striking velocity of falling object A on the floor of an elevator when the elevator is at rest here on ground WITH the striking velocity of the same object A when the elevator is accelerating at 9.8 m/s/s upward but at higher velocities. 

Compare the striking velocity of a falling object B on the floor of an elevator when the elevator is at rest here on ground WITH the striking velocity of the same object B when the elevator is accelerating at 9.8 m/s/s upward but at higher velocities.

(Addendum - The "h" of B is greater (not too greater) than the "h" of A but B hits the floor of an elevator (moving)  earlier than A as B is at rest originally while A is moving with its final velocity when released)

Also, don’t forget,

As the value of “g” of the earth decreases when height h above the surface of earth increases, therefore, the equation of v=gt (where "g" is constant) works only over the short-range of height h where the radius of the earth is considered constant (h doesn’t change significantly).

So imagine the height h of object B where the v= gt works and when it doesn’t work. 

Compare the striking velocity of falling object A on the floor of an elevator when the elevator is at rest here on ground WITH the striking velocity of the same object A when the elevator is accelerating at 9.8 m/s/s upward but at higher velocities.

As shown, the velocity relative to the ground is the same in both cases.
It is given by g*sqrt(-2*h/g)=-a*sqrt(2*h/a), noting that a=-g.

(Addendum - The "h" of B is greater (not too greater) than the "h" of A but B hits the floor of an elevator (moving)  earlier than A as B is at rest originally while A is moving with its final velocity when released)

Have you done the appropriate equivalence for Earth then?
I had already pointed out this problem and what you need to do to actually have it equivalent.
If object B actually starts at rest while the elevator starts moving, then the equivalent situation on Earth is an object which starts heading down towards Earth. It is not an object simply being released.
They will have an initial velocity relative to the floor, and the relative speed of the impact will be the same.

An object released at rest while the elevator is moving is not equivalent to an object at rest relative to the elevator/Earth that is released, and the equivalence principle doesn't say it is.

Also, don’t forget,
As the value of “g” of the earth decreases when height h above the surface of earth increases, therefore, the equation of v=gt (where "g" is constant) works only over the short-range of height h where the radius of the earth is considered constant (h doesn’t change significantly).

The equivalence principle applies locally, i.e. for a region where g does not vary significantly.
At the technical level it only applies over a volume of 0. As an approximation it applies over a larger volume.

But I have seen plenty of FEers reject that g changes with altitude.

Although abovesaid statement is quite clear but still

Compare the striking velocity of falling object A on the floor of an elevator when the elevator is at rest here on ground WITH the striking velocity of the same object A when the elevator is accelerating at 9.8 m/s/s upward but at higher velocities. 
Compare the striking velocity of a falling object B on the floor of an elevator when the elevator is at rest here on ground WITH the striking velocity of the same object B when the elevator is accelerating at 9.8 m/s/s upward but at higher velocities.
(Addendum - The "h" of B is greater (not too greater) than the "h" of A but B hits the floor of an elevator (moving)  earlier than A as B is at rest originally while A is moving with its final velocity when released)

But if both objects, A and B, are at rest relative to the elevator before being released I fail to see why there is any difference?

So give your calculations for each case, either algebraically or with a numerical example, showing what you think this difference should be.

Also, don’t forget,

As the value of “g” of the earth decreases when height h above the surface of earth increases, therefore, the equation of v=gt (where "g" is constant) works only over the short-range of height h where the radius of the earth is considered constant (h doesn’t change significantly).

I haven't forgotten that.

So imagine the height h of object B where the v= gt works and when it doesn’t work.

The elevator thought experiment only compares a region of uniform gravitational field with a region of uniform acceleration.

Einstein’s Experimental Elevator
He formulated this insight in what is known as the “equivalence principle,” which asserts that uniform acceleration is equivalent to the presence of a homogenous, or uniform, gravitational field. Now, of course, since the Earth is spherical, its gravitational field is not, strictly speaking, homogenous or uniform, because its lines of gravitational force diverge. But the equivalence principle still holds for reasonably small regions of space where the divergence is negligible.

Maybe you should read all of it.

In any case, no elevator would be large enough for there to be any significant variation in the magnitude or direction of g so I doubt Einstein saw any need to mention it.

Just wondering why it is so difficult to reach out to the kernel. You can also use your own abovementioned calculation for the following explanation.

ON EARTH

Object B releases at height “h” which is slightly greater than the height of the elevator. The initial velocity of object B is zero but its g is constant. Velocity of B increases when it falls at a rate of 9.8 m/s/s. Note down its striking time and final striking velocity - easy

ELEVATOR - MOVING UPWARD AT THE RATE OF 9.8 M/S/S

Elevator, which is moving upward at the rate of 9.8/s/s, gains velocity with the passage of time. Consider the scenario at higher velocities/speed, not the velocities of falling bodies at ground level where v=gt is applicable

The initial velocity of the elevator is not zero from where the height h starts while approaching object B, which is at rest but located just above its roof level (open).

Although the elevator is moving upward with constant "g" but since its velocities are higher at approaching moments than here on ground condition when B falls from h, therefore, the floor of the elevator hits the object B located at rest deep in space earlier than object B which falls at the same rate from the same height h here on earth.

Similarly, an object A starts moving upward with its final velocity when released in the same moving elevator.

The floor of the moving elevator catches B (located at h slightly greater than the h of A) which is at rest but yet to catch object A which is moving with its final velocity) when moving upward at the rate of “g=9.8 m/s/s”

Just wondering why it is so difficult to reach out to the kernel. You can also use your own abovementioned calculation for the following explanation.

I see words and no calculations.

Remember that, before release, each object is at the same velocity as the elevator.
So at the time of release the elevator is moving at velocity, V₀, and the object is also moving at velocity, V₀.
Once released the object continues moving at velocity, V₀, but the elevator's velocity now increases at g m/s².

The velocity, V₀, just adds to all the velocities but makes no difference to what is seen and felt in the elevator.

That is what is called Galilean invariance or Galilean relativity which states that the laws of motion are the same in all inertial frames.

The elevator itself is not an inertial frame but it can be "placed in" an inertial frame moving at V₀.