Split Thread The validity of classical physics (split from: DWFTTW)

[humber]

Honestly. Humber. If you live near a ballooning site, for chrissake go and ASK a balloon pilot if what you say is right!

$1000 bucks US says you aren't!

Balloons pass overhead quite often. This one was at night, and surprisingly fast. I see them, but I do not assume that the operators of a commercial enterprise are physicists, let alone knowledgeable.
It this why you seem to take your instruments at face value?

Many devices can gain lateral velocity by using energy gained from the gravitational field.
I submit for your approval; The Roller Coaster.
Balloons can do the same. If not why not?
You need more than money to deny that.

 

[humber]

No, that's not what I pointed out. I pointed out that Spork had forgotten to change the explanation to match the diagram. Now he's done so: the diagrams and the explanations below them now make complete sense.

No. The objects are said to be accelerating? And? In this case, the conditions are such that the masses and elevator are one mass. No explanation is given as to how the force is transmitted to those objects without reaction from the elevator. Making the text quack like the ducks, makes no 'sense'.

 

[Michael C]

As an addendum to Spork's diagrams, it's worth pointing out that it is only possible to tell the difference between the two situations if the accelerometers are accurate enough to distinguish the difference "delta" and the difference in the angles of the vectors at right and left. However accurate the accelerometer, there is always a degree of uncertainty. First lesson in classical mechanics: a measurement has no meaning without knowledge of the degree of uncertainty.

The equivalence principle says this: it is always possible to define a small enough elevator and a short enough period of observation that the two situations will be indistinguishable.

 

[humber]

I'm almost certain that you can't. But with my recent history of getting simple things wrong you

Legendary.

should almost certainly get a second opinion.

No 'almost'

It seems to me that if you could to this, it would be over a bounded region that had no dimension - but that takes us back to the fact that gravity and acceleration are already indistinguishable in that zero-dimensional region.
But it's a curious question. I'll (probably) try and think of a way to attempt it.

What does a gyroscope say?

 

[spork]

As an addendum to Spork's diagrams, it's worth pointing out that it is only possible to tell the difference between the two situations if the accelerometers are accurate enough to distinguish the difference "delta" and the difference in the angles of the vectors at right and left.

Yes, thanks. Indirectly that is what I was trying to point out. The two cases are the same if not for the gradients (angles and deltas). And in an actual elevator in the proximity of Earth, those angles and deltas are VERY small.

 

[humber]

Yes, thanks. Indirectly that is what I was trying to point out. The two cases are the same if not for the gradients (angles and deltas). And in an actual elevator in the proximity of Earth, those angles and deltas are VERY small.

In layman's terms, that's known as a 'difference'.

 

[Subduction Zone]

If you hollow out a large depression in a planetary body, can you create a situation where the gravitational field gradient is indistinguishable from that of acceleration over some bounded region such as inside an elevator car?

Theoretically you don't have to go nearly that much work, just drill a big tunnel straight down. The square/cube relationship of radius and mass means that as you go deeper the force of gravity goes down. I just got up so I am not quite ready to work out the formula for a uniform sphere. On the Earth it is a bit more complicated than that. The Earth is not a uniform sphere, the inner and outer core are much more dense than the mantle. So on the Earth the force of gravity actually goes up the deeper you get until you hit the boundary of the outer core. Then after that it will go to zero as you approach the center of the Earth.

ETA: I told you it was early, I forgot to add my quote of Dan

 

[humber]

As an addendum to Spork's diagrams, it's worth pointing out that it is only possible to tell the difference between the two situations if the accelerometers are accurate enough to distinguish the difference "delta" and the difference in the angles of the vectors at right and left. However accurate the accelerometer, there is always a degree of uncertainty.

(a) First lesson in classical mechanics: a measurement has no meaning without knowledge of the degree of uncertainty.

(b) The equivalence principle says this: it is always possible to define a small enough elevator and a short enough period of observation that the two situations will be indistinguishable.

'define' contradicts (a)

 

[Michael C]

Balloons pass overhead quite often. This one was at night, and surprisingly fast. I see them, but I do not assume that the operators of a commercial enterprise are physicists, let alone knowledgeable.
It this why you seem to take your instruments at face value?

Did you measure the speed of the wind at the altitude of the balloon? Can you prove that the balloon wasn't moving at the same speed as the wind?

Many devices can gain lateral velocity by using energy gained from the gravitational field.
I submit for your approval; The Roller Coaster.
Balloons can do the same. If not why not?
You need more than money to deny that.

The roller coaster accelerates laterally by pushing against the inclined track, which, as it should, pushes back. Balloons don't have any tracks to push against.

 

[humber]

Did you measure the speed of the wind at the altitude of the balloon? Can you prove that the balloon wasn't moving at the same speed as the wind?

I need prove nothing about that example to argue the general case.

The roller coaster accelerates laterally by pushing against the inclined track, which, as it should, pushes back. Balloons don't have any tracks to push against.

A balloon or other airborne object, may use drag, or even thrust, to provide that "push".

'define' contradicts (a)

 

[spork]

Theoretically you don't have to go nearly that much work, just drill a big tunnel straight down. The square/cube relationship of radius and mass means that as you go deeper the force of gravity goes down.

Yes, but I think the objective is to create a finite region in which the gravitational field is identical to linear acceleration. In other words accelerometers scattered about this region would all read acceleration of exactly the same direction and magnitude. Are you suggesting this could be done by drilling a cylindrical hole in a homogeneous sphere?

On another note, has anyone noticed that humber seems to be thrashing even more than usual? It's as if he's incredibly disturbed that the adults are having a conversation rather than focusing all attention on him.

 

[Michael C]

'define' contradicts (a)

Not at all: once the degree of uncertainty is known, it is always possible to define elevator size and observation time so that the differences between the two cases are smaller than the degree of uncertainty.

 

[sol invictus]

If you hollow out a large depression in a planetary body, can you create a situation where the gravitational field gradient is indistinguishable from that of acceleration over some bounded region such as inside an elevator car?

Cheap example: spherical planet, core hollowed out in a perfectly spherical cavity concentric with the center. Then there is precisely zero gravitational force anywhere inside, which is equivalent to zero acceleration.

As for non-zero acceleration, you can approximate that "gravitational field" with arbitrary accuracy in various ways. Basically any time you're close to a very large body, so that the surface can be approximated as planar, the gravitational field is very close to pure acceleration (a famous example is the region near the horizon of a large black hole). But it's not truly indistinguishable - just can be made parametrically close. Then again it's hard to make perfect spheres, so maybe it's just as good.

 

[mender]

The force driving the chute come from the car in your case.

Okay, humber, I think I see what the problem is. I'm trying to get a statement from you on two specific scenarios, and you are commenting on the first one but changing the second one. I may not be asking clearly enough.

I can make this simpler:

1) A parachute is tied to the ground in a 30 ft/s wind. For an arbitrary parachute size, the tension on the tether is say 100 lbs. What will the tension be in a 60 ft/s wind?

2) A parachute is attached to a 100 lb load and is falling through the air at a steady 30 ft/s. What will the steady state speed be if the 100 lb load is replaced with a 200 lb load?

Can you please just respond to those two questions?

 

[humber]

Not at all: once the degree of uncertainty is known, it is always possible to define elevator size and observation time so that the differences between the two cases are smaller than the degree of uncertainty.

That bounds the uncertainty, but still leaves an uncertainty. Not definitive enough to make your claim stick.

 

[humber]

Cheap example: spherical planet, core hollowed out in a perfectly spherical cavity concentric with the center. Then there is precisely zero gravitational force anywhere inside, which is equivalent to zero
acceleration.

A Bargain Basement example. There is perhaps no net force in any direction, but the force is still there. Mass likes gravity.

As for non-zero acceleration, you can approximate that "gravitational field" with arbitrary accuracy in various ways. Basically any time you're close to a very large body, so that the surface can be approximated as planar, the gravitational field is very close to pure acceleration (a famous example is the region near the horizon of a large black hole). But it's not truly indistinguishable - just can be made parametrically close. Then again it's hard to make perfect spheres, so maybe it's just as good.

Those pesky 'differences' again.

 

[Subduction Zone]

Yes, but I think the objective is to create a finite region in which the gravitational field is identical to linear acceleration. In other words accelerometers scattered about this region would all read acceleration of exactly the same direction and magnitude. Are you suggesting this could be done by drilling a cylindrical hole in a homogeneous sphere?

On another note, has anyone noticed that humber seems to be thrashing even more than usual? It's as if he's incredibly disturbed that the adults are having a conversation rather than focusing all attention on him.

Well, now I am not sure what Dan O was asking. And the answer to your question is no spork. If you were off center slightly you would have the the very small side deflection that you show in your illustration. And after I did the work to find the force of gravity in a uniform sphere as you approach the center, it was fairly trivial after all. After doing the work I have to list it:

The acceleration that a falling body would undergo would be for a uniform sphere of density p (I don't have a "rho" character) g=4(pi)pr/3.

Dang it I don't have a pi character either. Regardless you can see that the gravity goes down at a linear rate as r approaches zero.

 

[spork]

humber has actually made a statement worth responding to.

A Bargain Basement example. There is perhaps no net force in any direction, but the force is still there.

Sorry, you can't make a case for that. Since the gravity on every particle is 0 there are not even internal forces causing stress or strain on the object.

 

[Michael C]

That bounds the uncertainty, but still leaves an uncertainty. Not definitive enough to make your claim stick.

That's not my claim. It's Einstein's principle of equivalence. The uncertainty can be made arbitrarily small. Since you haven't got as far as understanding Newton, I don't have any hope of you understanding general relativity, so I'll leave it there.

 

[humber]

Okay, humber, I think I see what the problem is. I'm trying to get a statement from you on two specific scenarios, and you are commenting on the first one but changing the second one. I may not be asking clearly enough.

I can make this simpler:

1) A parachute is tied to the ground in a 30 ft/s wind. For an arbitrary parachute size, the tension on the tether is say 100 lbs. What will the tension be in a 60 ft/s wind?

That is the static force in both cases. If you simplfy the situation as I think you are suggesting, then the answer is 200lbs

2) A parachute is attached to a 100 lb load and is falling through the air at a steady 30 ft/s. What will the state state speed be if the 100 lb load was replaced with 200 lbs?

How could it fall, Mender? With 100lb's each way?
No. The parachute is moving, so the static force is no longer applicable. The drag, and therefore the velocity, can be calculated using the formula you know, but with gravity as the input force.
There is no 'drive' from the chute, only drag that opposes motion due to gravity.

Can you please just respond to those two questions?