Yes - the gravitational force decreases with distance from the mass; and, as sjastro pointed out, there will also be tidal variations in a gravitational field.
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Yes - the gravitational force decreases with distance from the mass; and, as sjastro pointed out, there will also be tidal variations in a gravitational field.So you're saying that if I have two gravity sensors, one a the top of the room and one near the floor, if they both read the same I am on an accelerating spacecraft, and if they read different, I am in a gravitational field?
Yes - the gravitational force decreases with distance from the mass; and, as sjastro pointed out, there will also be tidal variations in a gravitational field.
Yes, but only in the very brief period until the structure has become uniformly compressed by the acceleration; i.e. not long enough to notice.Of course, since there is no such thing as a truly rigid body, if the spacecraft accelerates, won't there be a greater acceleration at the bottom because it is being pushed by the engine, and the top will have less acceleration because it is being squished a tiny amount?
I see you have confused yourself by referring to an equation which has a big G in it instead of a little g.Gravity = acceleration?
No.
F=G([m1*m2]/D^2)
Yes, but only in the very brief period until the structure has become uniformly compressed by the acceleration; i.e. not long enough to notice.
Sure - after a brief interval where the engines are accelerating and the far end of the ship hasn't yet been affected, the entire structure becomes uniformly compressed and then it all accelerates at the same rate.But if the rocket is continually firing, won't it be constantly subject to the deformation?
No worries - it's all good practice!Sorry, I just woke up, and this is half me being silly and half brain addled by lack of coffee.
Sure - after a brief interval where the engines are accelerating and the far end of the ship hasn't yet been affected, the entire structure becomes uniformly compressed and then it all accelerates at the same rate.
If this were not the case, e.g. imagine a spaceship made of cheesecake, the crunchy base with the engines on would catch up with the soft pointy end and the spaceship would flatten out until it was uniformly compressed (you'd need a crunchy base much wider than the rest of the spaceship, to hold the flattened front-end, otherwise bits of cheesecake would fall off and get left behind).
The same end result would probably happen in gravity, but because the gravitational attraction reduces towards the pointy end (by the inverse square law), it would weigh less and so collapse more slowly (in principle). IOW, a rock weighs less at the top of Everest than it does at ground level.So why wouldn't the same thing happen in gravity?
The same end result would probably happen in gravity, but because the gravitational attraction reduces towards the pointy end (by the inverse square law), it would weigh less and so collapse more slowly (in principle). IOW, a rock weighs less at the top of Everest than it does at ground level.
No. Once uniformly compressed, all parts of the structure transmit the acceleration uniformly. If that wasn't the case, the engines would eventually catch up with the nose and the spaceship would be squashed flat.And the tip of the rocket, being farther from the engine, would experience the force less as well.
Ah, no. The correct logic is that the conservation of energy laws are only applicable where the coefficients of the metric are not time-dependent.
In GR, space and time are dynamic, they can change (as is empirically well-established).
But over sufficiently small spatial and temporal scales (i.e. everyday human scales), space & time can be considered to be static and energy conserved; i.e. energy is conserved in the limit.
No. Once uniformly compressed, all parts of the structure transmit the acceleration uniformly. If that wasn't the case, the engines would eventually catch up with the nose and the spaceship would be squashed flat.
No.Translation: Space expansion/contraction violations conservation of energy laws (just like I said).
I'm sorry, I was assuming constant acceleration (as under gravity, which was the original comparison).But once that happens and the rocket is travelling at a steady velocity, it's no longer accelerating. If the rocket then fires again, would the compression increase? Thus, wouldn't the rocket ship continue squishing as the rocket fires?
But once that happens and the rocket is travelling at a steady velocity, it's no longer accelerating. If the rocket then fires again, would the compression increase? Thus, wouldn't the rocket ship continue squishing as the rocket fires?
I could never bring myself to watch more than a minute or two, even though it was kinda ok in the first 2 minutes. Is it about the Von Neumann–Wigner interpretation - Wikipedia ?This is an awesome presentation, which compiles different experiments in Quantum Physics; as it challenges the way that many see the world.
One can hypothesise an infinite number of possible universes... However, the one we observe is expanding:A GR based cosmology theory which *does not* rely upon "space expansion" as a cause of redshift doesn't automatically violate any conservation of energy laws. It's therefore not the fault of GR theory that the LCDM violates such laws. Period.
One can hypothesise an infinite number of possible universes... However, the one we observe is expanding:
"The success of Big Bang Nucleosynthesis depends on the fact that we understand how fast the universe was expanding in the first three minutes, which in turn depends on how fast the energy density is changing. And that energy density is almost all radiation, so the fact that energy is not conserved in an expanding universe is absolutely central to getting the predictions of primordial nucleosynthesis correct." - Sean Carroll, among other things, author of Spacetime and Geometry, a graduate-level textbook on general relativity.
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