# Equivalence principle

## Main Question or Discussion Point

Hi
I am reading Stephan Hawking's Universe in the Nutshell and there I didnt understand this sentence
"This equivalence didn't work for a spherical earth because people on opposite sides of the world would be getting farther away from each other.Einstein overcome this idea to make spacetime curved"
If you explain it more basic I will be glad
Thanks

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## Answers and Replies

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Nugatory
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He's saying that if we apply the equivalence principle to two people on opposite sides of the earth, they'd have to be accelerating in opposite directions to explain the appearance of gravity pulling them both towards the center of the earth. If they're accelerating in opposite directions, the distance between them should be changing, but it's not - it's constant and equal to the diameter of the earth.

A.T.
"This equivalence didn't work for a spherical earth because people on opposite sides of the world would be getting farther away from each other.Einstein overcome this idea to make spacetime curved"
Locally, the gravity we observe on the surface doesn't require intrinsic curvature, just curve-linear coordinates (equivalent to an accelerating frame of reference) like shown here:

But the above video shows just a small local patch of space-time. To make all those small patches fit together globally, space-time must be curved like shown here:

Unlike the local cone-like patches, this global space-time surface cannot be rolled out flat, without distortion. This is called "intrinsic curvature".

He's saying that if we apply the equivalence principle to two people on opposite sides of the earth, they'd have to be accelerating in opposite directions to explain the appearance of gravity pulling them both towards the center of the earth. If they're accelerating in opposite directions, the distance between them should be changing, but it's not - it's constant and equal to the diameter of the earth.
But electromagnetism from the molecules of the earth are pushing each of them away from one another, thereby offsetting the gravity, so there is no contradiction in any regards.

Same principle as an automobile pushing against something like a rocket sled. If either is more powerful, they move in one direction or another, but if they are balanced they don't accelerate at all.

Locally, the gravity we observe on the surface doesn't require intrinsic curvature, just curve-linear coordinates (equivalent to an accelerating frame of reference) like shown here:

But the above video shows just a small local patch of space-time. To make all those small patches fit together globally, space-time must be curved like shown here:

Unlike the local cone-like patches, this global space-time surface cannot be rolled out flat, without distortion. This is called "intrinsic curvature".

If no forces are acting during the fall, why does a dropped object smash through it's target at the bottom of a fall? No force should have equaled no kinetic energy, no momentum. Why doesn't it bounce off harmlessly, or just "stick" for example?

I don't know about you, but I've accidentally dropped a thing or two on a finger or toe, and always feel just as much, usually a lot more, force, pressure, whatever you want to call it, than simply lifting the object. I realize issues like "Instantaneous Acceleration" modify the amount of damage a given amount of momenta can cause, but the explanation of that video appears to contradict both that notion and common experience as well.

Can you explain this discrepancy?

And for example, in the rubber sheet model of space-time distortion due to mass, the objects don't actually fall toward one another due to the warping of the rubber sheet in the model, but rather the "real" gravity of the Earth. The only reason the model works is because "real" gravity is actually "faking" the modeled gravity, which makes the model invalidated.

WannabeNewton
A.T.'s construction in his video is not even remotely close to the "rubber sheet analogy". The latter is inherently flawed and non-descriptive of general relativity whereas the former is perfectly accurate. This is a question we get very often on this forum.

The short answer is that you are forgetting to include the time dimension; when one models gravity using space-time geometry and space-time curvature, there is curvature in the time dimension that causes an object initially at rest to start freely falling under the influence of gravity.

See this thread for more details: https://www.physicsforums.com/showthread.php?t=726837
and this post for a more mathematical version of what I just said above: https://www.physicsforums.com/showpost.php?p=4597436&postcount=19

Nugatory
Mentor
The short answer is that you are forgetting to include the time dimension; when one models gravity using space-time geometry and space-time curvature, there is curvature in the time dimension that causes an object initially at rest to start freely falling under the influence of gravity.
Which is of course is the key to the second sentence in the original post.

Nugatory
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But electromagnetism from the molecules of the earth are pushing each of them away from one another, thereby offsetting the gravity, so there is no contradiction in any regards.
You are misunderstanding the equivalence principle. The EP basically says that there is no gravity to offset; the force you feel from the molecules of the earth pushing against the soles of your shoes is accelerating you away from your natural inertial trajectory and it is the only force acting on you.

Nugatory
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If no forces are acting during the fall, why does a dropped object smash through it's target at the bottom of a fall? No force should have equaled no kinetic energy, no momentum. Why doesn't it bounce off harmlessly, or just "stick" for example?
"No force" does not mean "no kinetic energy, no momentum". This isn't a relativity thing; if you look at the classical definition of momentum (##p=mv##) and kinetic energy (##E_k=\frac{1}{2}mv^2##) you'll see that they depend on having a non-zero velocity relative to the observer, not a non-zero force. The relativistic definitions are just a bit more complicated, but there's still no force involved in them.

The reason a dropped object smashes through its target is that it's on a collision course with its target and when they collide they hit hard enough to break the target.... Imagine what happens to the windshield of a car if it drives into a weight hanging from an highway overpass, for example. There's no force between the car and the weight, no gravity involved anywhere, but if the car is moving fast enough its windshield will be smashed.

The General Relativity picture of a dropped object is that the object is at rest while the ground underneath it is accelerating upwards towards the object. If the ground is moving fast enough when they meet, something will smash.
We know it's the ground accelerating upwards while the object is just floating free in space because when we stand on the ground we can feel it pushing up against the soles of our shoes; and if we were falling alongside the object we'd be weightless until we and the ground collided.

I can press my hand against a table, either from above or from below, and feel the same pressure as I would were I standing on my hands instead of my feet. The table doesn't magically change the direction of it's natural motion (or non-motion). The experience is the same either way, and I must be in the same reference frame either way. Therefore the explanation does not work, because it is not consistent with observation.

PeterDonis
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If no forces are acting during the fall, why does a dropped object smash through it's target at the bottom of a fall? No force should have equaled no kinetic energy, no momentum.
No force is acting on the free-falling object; but a force *is* acting on the target. In the GR viewpoint, the free-falling object sits at rest and the target is accelerated upward until they smash together. The further away the target is at the start, the more time it has to accelerate, so the harder the smash will be.

PeterDonis
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"No force" does not mean "no kinetic energy, no momentum".
This is true, but I don't think it's the answer to Wade888's question. There *is* a force present; but it acts on the *target*, not on the falling object. See my response to him, just posted.

No force is acting on the free-falling object; but a force *is* acting on the target. In the GR viewpoint, the free-falling object sits at rest and the target is accelerated upward until they smash together. The further away the target is at the start, the more time it has to accelerate, so the harder the smash will be.
Okay, so synchronize two experimenters on exactly opposite points of the Earth. Each drops their ball on their target. Your explanation requires the Earth to expand in both directions simultaneously, in order to "accelerate" the target "upward." Of course, this scenario (not the theory) happens all over the world all the time in asynchronous fashion, but it helps to do the synchronized thought experiment to make the point.

So now I am to understand that the theory of Relativity says that if we lift objects, the Earth shrinks, but if we drop them, the Earth expands, but I drop the object from shoulder height. The Earth cannot be expanding, because I did not sink up to my arm pits in the soil or rock. The explanation is still inconsistent with easily, readily observed reality.

PeterDonis
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Okay, so synchronize two experimenters on exactly opposite points of the Earth.
In which case you're no longer talking about the equivalence principle, since that only applies locally, where "locally" means "over a small enough region of space and time that tidal gravity is negligible". The exact size of such a region depends on how accurate your measurements are, but it's safe to say that experimenters on opposite sides of the Earth are *not* in each other's local regions of spacetime.

So now I am to understand that the theory of Relativity says that if we lift objects, the Earth shrinks, but if we drop them, the Earth expands, but I drop the object from shoulder height.
No, relativity says no such thing. See above.

PeterDonis
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You are misunderstanding the equivalence principle.
No, he's trying to apply it in a situation where it doesn't apply. See my previous post. (The same comment applies to the OP; what Hawking was really saying was that you can't apply the EP over a region large enough to include experiments on both sides of the Earth simultaneously, because tidal gravity is not negligible. Hawking just chose a more colorful way of saying it, as he usually does.)

Nugatory
Mentor
No, he's trying to apply it in a situation where it doesn't apply. See my previous post. (The same comment applies to the OP; what Hawking was really saying was that you can't apply the EP over a region large enough to include experiments on both sides of the Earth simultaneously, because tidal gravity is not negligible. Hawking just chose a more colorful way of saying it, as he usually does.)
You sure? When he talks about the electromagnetic forces from the surface of the earth "offsetting" gravity, I'm hearing a deeper misunderstanding than the one that Hawking was addressing - that description is inconsistent with the equivalence principle even locally, and it's more akin to the Newtonian description that the GR one.

Of course there is always the possibility that I am misunderstanding OP's misunderstanding

PeterDonis
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When he talks about the electromagnetic forces from the surface of the earth "offsetting" gravity
Just to clarify, by "the OP" I meant Quarlep. The electromagnetic force thing was Wade888. Of course, I may be misunderstanding the respective misunderstandings as well.

A.T.
Your explanation requires the Earth to expand in both directions simultaneously, in order to "accelerate" the target "upward."
No, as Hawking explains, this is exactly the issue that space-time curvature solves. In curved-space time you can have two objects experiencing proper acceleration away from each other, without increasing their distance.

proper acceleration = what an accelerometer measures = deviation from free fall = deviation from straight (geodesic) path in space time

PeterDonis
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as Hawking explains, this is exactly the issue that space-time curvature solves. In curved-space time you can have two objects experiencing proper acceleration away from each other, without increasing their distance.
Yes, but it should be noted that, as I said in my previous response to Wade888, if you're talking about spacetime curvature, you're no longer talking about the EP, because spacetime curvature is tidal gravity, and the EP only applies in a small enough patch of spacetime that tidal gravity is negligible.

Dale
Mentor
So now I am to understand that the theory of Relativity says that if we lift objects, the Earth shrinks, but if we drop them, the Earth expands, but I drop the object from shoulder height. The Earth cannot be expanding, because I did not sink up to my arm pits in the soil or rock. The explanation is still inconsistent with easily, readily observed reality.
Expansion is different from acceleration. The surface of the earth is accelerating upwards at 1 g proper acceleration. The earth is not expanding. Those two statements are compatible because the spacetime around the earth is curved.

A.T.
if you're talking about spacetime curvature, you're no longer talking about the EP,
Global curvature allows the EP to work locally everywhere.

1 person
Dale
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Well said!

Nugatory
Mentor
Just to clarify, by "the OP" I meant Quarlep. The electromagnetic force thing was Wade888. Of course, I may be misunderstanding the respective misunderstandings as well.
OK, all is now clear to me Or something

Global curvature allows the EP to work locally everywhere.
Global curvature is a somewhat ill-defined concept for GR's Lorentzian 4-manifolds. The EFE is about the local curvature, and different for each solution, so I fail to see what global curvature you allude to here.

WannabeNewton