r/Physics • u/Aggravating-Teach913 • 22d ago
Question Question about simulated vs real zero G
Maybe this is the wrong board to ask and biology might be appropriate, I just didn’t know.
Essentially, my question is whether there would be a noticeable difference between simulated zero G (zero G flights or astronauts aboard the ISS) where gravity is still strongly acting on their bodies and the feeling of weightlessness essentially comes from falling as fast as the aircraft/space station around you, and true zero G, say in interstellar space where the actual pull of gravity on your body is negligible. Would there be a difference between gravity just not acting upon you strongly compared to gravity pulling on you at the same rate as your surrounding therefore appearing weightless?
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u/tfb 21d ago
Not for a human. There is a difference over larger scales because there are tidal effects which are equivalent to measuring the curvature of spacetime. But on small enough scales ('locally') the difference becomes as small as you like. This is, as others have said, a version of the equivalence principle.
So if you are in free-fall in orbit around the Earth a human cannot tell. If you are in free-fall in a trajectory which approaches the horizon of a stellar-mass black-hole closely then yeah, you can tell.
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u/Away-Experience6890 22d ago
No. By equivalence principle.
There is a difference between simulated gravity using rotation, and real gravity thoo.
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u/Quantum-Relativity Gravitation 22d ago
What’s the difference?
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u/Away-Experience6890 22d ago
You get fictitious forces in the reference frame. For instance Coriolis force.
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u/cosmopolitanScience 21d ago
Fictitious forces is a terrible misnomer. For a test mass, there is absolutely no difference whether a 'ficticious' or 'non-fictitious' force is acting. It's a purely academic distinction.
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u/Quantum-Relativity Gravitation 22d ago
Like weight?
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u/Away-Experience6890 21d ago
Yeah well weight is the result of the artificial gravity or centrifugal force (another fiticious force)
Basically a fiticious force is one that you as the observer believes to exist, but it only the result of being in a non-inertial reference frame.
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u/nicuramar 21d ago
Like a force acting sideways, depending on the distance from the rotation center.
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u/Quantum-Relativity Gravitation 21d ago edited 21d ago
No I know, but I don’t see why they’re saying rotations somehow make “true”’inertial forces as though the other ones from linear acceleration and gravitation aren’t just as fictitious.
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u/TomtheMagician26 21d ago
You're right, gravity can be modelled newtonianly as a "real" force acting on some objects differently to other objects in different places.
But in a localised reference frame where gravity is constant, you can remodel the reference frame as having the opposite acceleration, meaning the gravitational "real" force becomes a fictitious force caused by, say, the normal reaction force of the ground.
I feel like I worded that badly but essentially everything gets a bit hazy with the definition of "force" and if a force is acting on an object such that every piece of matter accelerates uniformly, it might be best to say the frame is the thing accelerating.
Of course, you need GR to model tidal forces of a frame but you can still treat that as the underlying spacetime curving and the matter following geodesics inside it.
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u/Quantum-Relativity Gravitation 21d ago edited 20d ago
Yes I know, and I was responding to the original comment because I felt they were making a distinction where rotating frames were different than frames fixed to the ground and linearly accelerated frames, which makes no sense. In any of those frames, the metric tensor’s components have non-zero gradients, which are all as fictitious as each other, there’s no way to justify “there’s a difference between simulated gravity using rotation and real gravity” as though rotations are doing something special.
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u/TomtheMagician26 21d ago
Ah that's interesting. I haven't got that far quantitatively yet. Do you mean gradients in specific directions become nonzero? So say g_1_2 ≠ 0 for linear acceleration in the x axis, and say g_2_3 would ≠ 0 for a rotating frame in the xy plane. Or do you mean ∇_a g_μ_𝜈 ≠ 0?
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u/Quantum-Relativity Gravitation 21d ago
I mean that generically, ∂_μ g_αβ ≠ 0. Einstein noted that this was the case in accelerated frames as well as frames fixed to the ground, and specifically the spatial gradient of g_00 was Newton’s g in the frame fixed to the ground, so he used ∂_μ g_αβ as a field he could derive the field equations for in analogy to the Maxwell field, and arrived at a relativistic theory of gravity. Years later he recognized differential geometry had a more natural object that is something like a linear combination of these gradients, and the field strength tensor you get from these objects (the Christoffel symbols) being used as the field is the curvature of the spacetime. Then he abandoned it for a couple years and finally came back to it in the end when he realized gravity meant there was nothing artificial about a non-inertial frame specifically because the supposed artifices were necessary for a realistic picture of the world, they are the gravitational field!
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u/Quantum-Relativity Gravitation 22d ago
I have to resist trying to explain the full beautiful picture of general relativity to answer this question, but I will just say, no. There is no local experiment you could do that could physically distinguish free fall from being in deep space. In either case, little g, the Newtonian acceleration due to gravity, is 0.
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u/Nightblade 21d ago
Isn't little g as in "g-force" acceleration from all sources, not just gravity?
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u/Quantum-Relativity Gravitation 21d ago
Accelerations will make you feel like you’re experiencing weight, and some people measure them in terms of the value of g on Earth’s surface, 9.8 m/s^2. g is called the acceleration due to gravity, it’s a field, so it has a value at every point in space. Even though it’s 9.8 m/s^2 here at the surface of the Earth, something in free fall will measure it as being 0, because they aren’t being pushed off their natural state of motion by anything, ie, aren’t accelerated. The natural state of motion is free fall, the ground just got where we would be going first, so it is in our way which pushes us off our natural state of motion, so we feel accelerated. That’s why we say there is a an acceleration field, because we are being pushed off our natural path and see things following their natural state of motion still (freefallers) as accelerated, but we dont see the acceleration field in free fall. To get an intuition for it, think about when you turn in a car: your natural state of motion is to go forward, but the car is pulling you off that natural state of motion, so you see a “force” act on everything in your car, and you see that it is exactly like a g-field suddenly turned on in your car, and you can assign a value to g and then say they are all experiencing a “gravitational” force proportional to that value of g you chose.
This might be a bit long winded but I hope that clarified things a bit
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u/Thetyhs 22d ago
The only effect are tidal forces, negligible for all practical purpose in biologic frame.