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Author Topic: How is the equivalence theory reconciled with spaghettification  (Read 3325 times)

Online syhprum

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Equivalence theory tells me I should not be able to distinguish between acceleration and gravity but when my tall space ship is sitting on the earth I can measure a difference in G whether my instrument is on the floor or near the ceiling.
Presumably this would not happen when I was away from the Earth and accelerating in space.
« Last Edit: 12/05/2015 19:41:43 by syhprum »


 

Offline chiralSPO

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I believe that equivalence theory only applies (in practice) to very small regions of space.

The theory, itself, I think, states that a uniform gravitational field is equivalent to constant acceleration, which obviously doesn't  apply to any actual gravitational field.
 

Offline Colin2B

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I believe that equivalence theory only applies (in practice) to very small regions of space.

The theory, itself, I think, states that a uniform gravitational field is equivalent to constant acceleration, which obviously doesn't  apply to any actual gravitational field.

It does apply to uniform gravity g, which means a small volume on, say, earth.
 

Offline jeffreyH

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It should also depend upon the relative sizes of the masses involved. Two earth sized masses will not act in exactly the same way as the earth and a tennis ball. The two earth masses will accelerate towards each other at an equivalent rate depending upon initial trajectories. An observer on one of these earths may see theother approaching earth moving at a greater acceleration than a tennis ball falling from the same altitude. Someone else may correct me on this.
 

Offline Colin2B

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An observer on one of these earths may see theother approaching earth moving at a greater acceleration than a tennis ball falling from the same altitude.

They might, but not for long :)

Because the distance in Newton's equation is taken from the centre of each mass the surface of the earths will be closer together than the surface of the tennis ball from earth.
Because the ball is small we can ignore it's mass and use the idea of a gravitational field around the earth, acceleration std g at earths surface. If the tennis ball is a distance from the earths surface equal to the earth's radius then it will only experience an acceleration of 1/4 of g.

This wouldn't be true for another earth with it's centre positioned at the same distance ie both surfaces touching. I don't think we could use the Newton formula, we would need to use calculus to account for the distribution of the masses. But, yes, the acceleration would be greater than for the tennis ball.

An interesting thought however, for the observer on one of the earths they would experience weightlessness just before impact, but not for long  :)



 

Offline jeffreyH

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You should also get a dual atmospheric accretion zone because of the mass equivalence. If you think about it the cancellation of the intersecting fields should decrease the general acceleration effects as the planets approach. So although they will have attained a significant acceleration this should not continue right up to impact. I once did a plot of this and it would tend more and more towards a steady velocity with little acceleration effects the nearer they get. Since when the surfaces are actually touching the field cancels completely at the interface. This is likely why the earth was not totally destroyed by the impact that created the moon.
 

Offline PmbPhy

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Equivalence theory tells me I should not be able to distinguish between acceleration and gravity but when my tall space ship is sitting on the earth I can measure a difference in G whether my instrument is on the floor or near the ceiling.
Presumably this would not happen when I was away from the Earth and accelerating in space.
There are essentially two forms of the equivalence principle that I'm familiar with.

1) A uniform gravitational field is equivalent to a uniformly accelerating frame of reference.

2) A gravitational field is locally equivalent to an accelerating frame of reference.

By local it's meant that that there are restrictions on the spatial size of the frame of reference in which an experiment/observation is done and on the length of time in which the experiment/observation takes place.

I wrote about this in a paper I wrote. See: Einstein's gravitational field by Peter M. Brown (i.e. me) and is located at http://xxx.lanl.gov/abs/physics/0204044
 

Offline PmbPhy

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Quote from: chiralSPO
The theory, itself, I think, states that a uniform gravitational field is equivalent to constant acceleration, which obviously doesn't  apply to any actual gravitational field.
It does apply to fields which can exist in nature or in the presence of bodies which can be constructed. Consider the gravitational field of the Earth. We typically approximate it as a sphere of constant mass density. In that same approximation/idealization we can excavate a spherical cavity out of the Earth somewhere other than the center of the Earth. Perhaps 1,000 m at a depth of 10 km. In such a cavity the field is perfectly uniform. The strength of the field is proportional to the distance between the center of the cavity and the center of the Earth. For a derivation please see:
http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm

Also, the gravitational field of a vacuum domain wall are also uniform.
 

Offline chiralSPO

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It does apply to fields which can exist in nature or in the presence of bodies which can be constructed. Consider the gravitational field of the Earth. We typically approximate it as a sphere of constant mass density. In that same approximation/idealization we can excavate a spherical cavity out of the Earth somewhere other than the center of the Earth. Perhaps 1,000 m at a depth of 10 km. In such a cavity the field is perfectly uniform. The strength of the field is proportional to the distance between the center of the cavity and the center of the Earth. For a derivation please see:
http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm

Thank you for the link, that's an interesting way of generating a very uniform field. Having read the derivation, I understand that the gravitational field in such a cavity is nearly uniform, not perfectly uniform, even if we assume the Earth to be perfectly spherical, of constant mass density, and not near any other gravitational bodies (tidal forces). In other words, I agree with the calculations of your "approximation/idealization," but would still argue that it doesn't hold for "any actual gravitational field."

For that matter, I will also argue that it doesn't hold for any actual accelerating frame of reference either (because there will be some finite, nonzero contribution from massive objects a finite distance away), but maybe I'm just being picky...
 

Offline PmbPhy

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Quote from: chiralSPO
Having read the derivation, I understand that the gravitational field in such a cavity is nearly uniform, not perfectly uniform, ..
I can't imagine why you'd think that. It's a fact of the derivation that if the original object is a sphere with a uniform mass density and you cut out a spherical cavity a distance d from the center of the sphere then the gravitational field inside is perfectly uniform, not nearly uniform. Please provide your argument as to why it would be otherwise.

Quote from: chiralSPO
..even if we assume the Earth to be perfectly spherical, of constant mass density, and not near any other gravitational bodies (tidal forces). In other words, I agree with the calculations of your "approximation/idealization," but would still argue that it doesn't hold for "any actual gravitational field."
Why? I don't understand why you'd make such an assertion and not provide the proof of it in the same post. Why is that?

Quote from: chiralSPO
For that matter, I will also argue that it doesn't hold for any actual accelerating frame of reference either (because there will be some finite, nonzero contribution from massive objects a finite distance away), but maybe I'm just being picky...
Yes. You're going outside the model. Anytime a derivation is provided for something there is a model that is implied. Everything is in accordance with that model. In this case the model is as I stated it. If you post a comment like
Quote
the gravitational field in such a cavity is nearly uniform, not perfectly uniform, ..
you're confusing to the people reading the thread. They don't know why you'd make such an assertion and neither do I. If you mean something like "because there are actually stars somewhere in the universe and the gravitational field of those stars make the field non-uniform" then while correct its a petty point to make and outside the model. Such trivialities only confuse the point being made.

I thought you were aware of how models are used in physics? From that post it's not clear to me that you are.
 

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