How is the force of gravity transmitted? If it’s through gravitational waves which have wavelengths that are millions of miles long then a rock falling 1 foot would fall between the waves and be unaffected
>>17034975It's not
>>17034984Then what? Gravitons? We can’t detect these despite gravity being so prevalent
>>17034975Through gravitons.
>>17034975Local absorption of the neutrino field.Prove me wrong :^)
>>17034975its transmitted by mass, a wave of water is also a gravitational wave.
>>17035030How is it transmitted by mass? Mass isn’t reaching out and commanding the rock to fall to the ground
>>17035030That's a gravity wave, not a gravitational wave
>>17034975same as magnets
>>17035109We don’t even know how magnets worked
>>17034975>How is the force of gravity transmitted?According to general relativity, gravity is not a force at all. You are not being 'pulled in' by gravity. Rather, you're following a straight line in spacetime ,which is called a geodesic. A massive body creates a spacetime curvature around itself. Objects that are then seemingly "attracted" to the object are actually just following a straight line in this curved spacetime. A satellite orbit may look curved, but it's actually a straight line in spacetime, because it's the spacetime that's curved, not the orbit.A rather simple way to visualize curved spacetime around a massive body is to imagine space continuously flowing into the object over time (pic rel). When you let go of a rock, the rock isn't 'pulled in' by Earth. Rather, the rock is dragged by the flow of space into the Earth. The rock doesn't "feel" this flow. It's, in fact, not moving at all. The rock sits at a single point in space, while the space itself shifts towards the Earth. The rock isn't moving, the space itself is moving and dragging the rock to collide with the Earth.Yes, by this logic, Earth accelerates upwards towards the stationary rock. The earth would in fact explode without spacetime curvature and so would every star.>but how does it actually work???In general relativity metric *is* the gravitational field. Write the Schwarzschild metric (single massive body) in Gullstrand-Painleve coordinates:[eqn]\qquad ds^2 = -c^2t^2 + \left(dr + \sqrt{\frac{2GM}{r}}dt\right)^2 + r^2 d\Omega^2[/eqn]For a free falling rock [math]d\Omega = 0[/math], since a free fall is purely radial. We also set the radial part to zero to make the rock attached to a single point in space.[eqn] \qquad dr + \sqrt{\frac{2GM}{r}}dt = 0,\qquad v = \frac{dr}{dt}=-\sqrt{\frac{2GM}{r}}[/eqn]The speed is v is interpreted as the speed of the flow of space dragging the rock along with it. Then the second derivative would recover the Newton's gravitational law.
>>17034991Not the anon you are replying to, but if you are using general relativity as an explanation of gravity, gravity is an effect, akin to the centrifugal force or other fictitious forces, not a transmitted force.In my opinion, the graviton model is useless. You're trying to describe the deformation of a 4D manifold that describes dimensions with a specific particle transmitted between supposedly the local energy density and the manifold. Utterly ridiculous.If you want to solve this problem, you'll have to abandon GR entirely
