Let's say i'm in Einstein's elevator

[rocca123]

let's say I'm in einstein's elevator
or let's say I'm in a box with no measuring possibilities, no sensors to help me
a box that's in a spaceship that's traveling with an acceleration of 9.8m/s^2
would i feel the same way as if i were on earth, on the ground, in a box?
or, between inertial mass and gravitational mass - would i feel any difference?

 

[ghwellsjr]

Yes, you would feel exactly the same, at least until your fuel ran out which would be rather quickly, but then, we're playing pretend, aren't we?

 

[Passionflower]

let's say I'm in einstein's elevator
or let's say I'm in a box with no measuring possibilities, no sensors to help me
a box that's in a spaceship that's traveling with an acceleration of 9.8m/s^2
would i feel the same way as if i were on earth, on the ground, in a box?
or, between inertial mass and gravitational mass - would i feel any difference?

No difference, that is the whole idea behind this gedanken experiment.

Of course if you had very precise measuring instruments you would be able to find out.

 

[ghwellsjr]

No difference, that is the whole idea behind this gedanken experiment.

Of course if you had very precise measuring instruments you would be able to find out.

Then it can't be "No difference". What and how would you find out with very precise measuring instruments?

 

[Gigasoft]

In Einstein's elevator, the field would be uniform. On a planet, it's not. With decreasing distance to the planet, the non-uniformity becomes more and more apparent.

 

[HallsofIvy]

I think Passionflower misspoke. Locally no measuring instruments will show any difference. But "in the large", either in time (after your fuel runs out!) or in space (where you find no difference in force as you climb upwards) there will be a difference.

Oh, wait. That is what Passionflower meant- the better your measuring instruments are, the smaller "locally" becomes- because you can better measure slight differences.

 

[Passionflower]

Then it can't be "No difference". What and how would you find out with very precise measuring instruments?

Without very precise instruments, e.g. only using our senses, we would most certainly not be able to sense any difference as the differentials are too small.

Assuming the elevator has at least the height the size of a human body we could, at least in principle, determine whether we are in an elevator or on Earth.

Some differences:

• The difference in proper acceleration between the top and bottom of the elevator would differ in the two cases.
• A lightbeam going from the middle left to the right of the elevator would have a different height mark in the two cases.

That is what Passionflower meant- the better your measuring instruments are, the smaller "locally" becomes- because you can better measure slight differences.

Yes that is what I meant.

The human senses are not exact enough to be able to measure any difference and currently no instruments are exact enough either.

 

[jubair.pk]

My opinion is different.

Firs of all, let me make the scenario clear. The acceleration of the elevator is assumed to be 'uniform'. Of course, in reality, uniform acceleration for ever is not possible. But, let us assume that, our elevator is in a uniform acceleration and it is going be in this state for ever.

In this case, no physical experiment can differenciate whether you are on Earth (under gravity) or accelerating at the rate 'g'. No matter, how precise your instrumentis are, it is NOT POSSIBLE to differenciate.

Another interpretation of this is: Suppose you are freely falling towards earth, (ie, you are accelerating towards Earth at rate of 'g') no experiment can differenciate that you are falling towards Earth or you are somewhere in space at rest (or with a unifrom velocity). All the laws of physics for inertial frame of reference (ie, non-accelerating frame of reference) is applicable to you.

This is one of the fundamental principle of General Relativity. This principle is known as "The principle of Equivalence"

Please refer: http://csep10.phys.utk.edu/astr162/lect/cosmology/equivalence.html
http://www.suite101.com/content/the-equivalence-principle-a43525
http://en.wikipedia.org/wiki/Equivalence_principle

Your opinions are welcome.

 

[D H]

My opinion is different.

You don't quite understand the equivalence principle then. You referred to the wikipedia article. Here is the lead paragraph from that article (emphasis mine):

In the physics of general relativity, the equivalence principle refers ... to Albert Einstein's assertion that the gravitational "force" as experienced locally while standing on a massive body (such as the Earth) is actually the same as the pseudo-force experienced by an observer in a non-inertial (accelerated) frame of reference.​

That word locally makes all the difference in the world. Strictly speaking, the equivalence principle (this aspect of the equivalence principle) applies only to a uniform gravitational field. There is of course no such thing as a uniform gravitational field. Locally, however, every gravitational field looks like a uniform gravitational field. Inside Einstein's elevator the deviation from a uniform gravitational field are presumably so small as to be non-measurable, and hence the equivalence principle still applies to within the limits of scientific measurement.

Of course if you improve the quality of the sensors those deviations can indeed become measurable. The solution is to make the elevator smaller.

 

[jubair.pk]

Yes, DH is correct. Earth's curvature does have impact on gravitational field. I should have used word "uniform magnetic field" instead of "earth". I apologize for the mislead.

To understand the basics, this kind of assumptions (like, uniform gravitational field over an infinite plane surface) helps. Later, when it comes to apply to the real world, we should consider other parameters like, curvature of earth.

-Jubair

 

[Passionflower]

As I said before currently there is no way to test the difference between the 'Einstein elevator' between Earth and space, but that does not prevent us from making the calculation just for fun.

So let's attempt to calculate, hopefully without mistakes, the difference.

First a few assumptions:

- We ignore the rotation of the Earth.
- The Earth is a perfect sphere and the density is evenly distributed.
- The Schwarzschild radius of the Earth is exactly: 0.00887005622 meters.
- The Schwarzschild r value of the Earth's surface is exactly: 6375416.
- The elevator cage is Born rigid and is 2.5 meters high.​

So let's start by calculating the acceleration on the surface of the Earth.
We use:

<br /> {\frac {0.5\,r_{{{\it schw}}}{c}^{2}}{<br /> {r_{{{\it surface}}}}^{2}\sqrt {1-{\frac {r_{{{\it schw}}}}{r_{{{\it surface}}}}}}}}<br />

We get: 9.806651166 meters/second²

Now we have to calculate the Schwarzschild r value of the ceiling of the elevator because the Schwarzschild r is not a measure of distance, however it is so close to 6375416 + 2.5 that we simply take that value.

Thus the acceleration at the ceiling becomes: 9.806643473 meters/second².

Which gives us a difference of: 0.000007693 meters/second².

So now let's take the same elevator but now in space. We match the acceleration at the floor with the surface acceleration of the Earth, so we have: 9.806651166 meters/second². And now we calculate the acceleration at the ceiling. We use:

<br /> {\frac {{c}^{2}\alpha_{{{\it floor}}}}{{c}^{2}+h\alpha_{{{\it floor}}}}}<br />

Which gives us: 9.806651166 (the difference is 7.*10^-9)

So the acceleration at the elevator ceiling on Earth is 9.806643473 while in space it is 9.806651166 a difference of 0.000007686 meters/second².

What would be more fun is to attempt to calculate it by using the Kerr solution, of course the results would be extremely close as the rotation of the Earth is too small to be significant. Anyone willing to take a stab at that one?