Interstellar -the movie, planet with slower time

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The discussion centers on the physics of time dilation as depicted in the movie "Interstellar," specifically regarding a planet where one hour equals seven years on Earth due to gravitational time dilation. Participants analyze how this extreme time difference would affect physical processes, suggesting that the planet would be superheated due to the inability of heat to dissipate effectively. They also explore the implications of observing the universe from such a planet, noting that radiation would be significantly blueshifted and that the environment would be hostile due to high-energy particles. The conversation touches on the complexities of relativistic motion and the effects of orbiting near a black hole, leading to questions about energy conservation and the nature of time dilation in different frames of reference. Overall, the discussion highlights the intriguing yet challenging aspects of applying real physics to science fiction narratives.
  • #61
Ken G said:
Of course, but the point is, "being in free fall" is not some magical state that automatically determines the redshift of everything you will see-- that's why I mentioned situations where two free-fallers can cross paths at relativistic relative speeds.

Indeed, but unless one of those observers has a relativistic rocket on hand that's not going to be the case. It certainly won't be the case for Miller's planet and the shell of fire.

Yes, and the answer is, "redshifted and blueshifted like mad, depending on which direction you look."

If you're hovering. If you're falling free, the answer is you'll see no shift compared to when you started at infinity. Neither Miller's planet nor the disk are hovering. They are in free fall, and they did not arrive with absurdly high relativistic base velocities.

The movie depicted no change in frequency looking at the Gargantua from any orbit around Gargantua, and even on an approach orbit.

Not sure what an "approach orbit" is, but why would you free falling observer to see Gargantua's light red or blue shift depending on the height of the orbit?

You mean, it was not a problem because the Endeavor was not parked an AU out, or it was not a problem because they chose not to worry about how the Ranger could cover an AU in a few hours without sustaining deadly g forces?

Upwards 7 AU, actually. And they did worry about how the Ranger could arrive without sustaining deadly g forces. Gravitational slingshots around sufficiently massive objects. This is a supermassive black hole system. Think more in terms of the center of the galaxy rather than a star system. There's at least one neutron star in the vicinity. Space could be littered with other stars, intermediate black holes, and the like.
 
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  • #62
Pete Cortez said:
Indeed, but unless one of those observers has a relativistic rocket on hand that's not going to be the case. It certainly won't be the case for Miller's planet and the shell of fire.
I'm afraid I don't know what you are saying here, are you claiming there is something magical about a circular orbit, compared with all the other possible orbits that would cross a circular orbit at relativistic speeds, such that the circular orbit is the one that does not see redshifts and blueshifts? Because that's just not going to be right.
If you're hovering. If you're falling free, the answer is you'll see no shift compared to when you started at infinity.
Are you now assuming that your "free fall" orbit includes infinity? Because circular orbits don't. Miller's planet will cross at a relativistic speed any orbit that involves falling free from infinity, so you cannot have it both ways here-- one or the other will see dramatic Doppler shifts. I suspect both will, but I'm sure the circular orbits will.
Neither Miller's planet nor the disk are hovering. They are in free fall, and they did not arrive with absurdly high relativistic base velocities.
Again, you need to stop imagining that "free fall" is some special state of existence. Various different free-fall orbits will have highly relativistic relative speeds at the points where they cross.
Not sure what an "approach orbit" is, but why would you free falling observer to see Gargantua's light red or blue shift depending on the height of the orbit?
By "approach orbit", I just meant "not the circular orbit of Miller's planet." It is perfectly obvious that observers falling free from infinity will see the light from an accretion disk that is either redshifted or blueshifted. For example, when they start their free-fall at infinity, they will see substantial gravitational redshift, and essentially no blueshift. When they arrive, in their free-fall, at the disk, they will have significant relative speed to the disk, so will see blueshifts. All these shifts will also depend on direction. A difficult calculation, I would never attempt it-- but I didn't just do a nearly-petabyte ray-tracing calculation either. But once the decision was made to alter the appearance of Gargantua for dramatic reasons, I suppose it became less of a priority to make anything about it look right, other than the gravitational lensing.
Upwards 7 AU, actually. And they did worry about how the Ranger could arrive without sustaining deadly g forces. Gravitational slingshots around sufficiently massive objects. This is a supermassive black hole system. Think more in terms of the center of the galaxy rather than a star system. There's at least one neutron star in the vicinity. Space could be littered with other stars, intermediate black holes, and the like.
Well, I won't beat them up about this, because sci-fi always needs to do impossible things to have a story, but note that the Ranger did not have any particular launch window for getting back to the Endeavor, so we can hardly claim they are going to be using gravitational assists from random orbiting flotsam. Still, if their picture is that the Endeavor is orbiting some 7 AU away, and they are managing to dart around the relativistic orbits of the various planets by gravitational assists, perhaps by dropping into the ergosphere and drawing on the Penrose effect, then I'm not going to sweat it-- that's their picture, and that's fine. But the light should still look Doppler shifted.
 
  • #63
KenG: you don't need a particular launch window for getting back to the endeavor, miller´s planet rotates a lot faster due to time dilation than the neutron star. The perfect time windows may be every 10 mins or less.

Kip Thorne said that he would like to have another black hole to make this kind of maneuver, because with the neuntron star the tidal forces will be too strong. But Nolan did not want 2 black holes because that would confuse the audience. He is right.

So? much talk about the red/blue shift, but I was left hanging with my questions about wormhole orbits in both side and gravitational effects crossing side to side.

I guess any gravitational effect from the other side it will be severely reduced, to picture this I imagine a wave over the water and a small tube over the surface connecting to a different pool, the wave which form from the exit of the tube it will be very small, also moving in all directions.
 
  • #64
AngelLestat said:
Ok, what I want to said:

If we have a laser close to a black hole sending 1 TW by second, how much energy will get a receiver by second far away from the black hole if the laser not spread?

The time dilation between these 2 is 2400000, happy?

I have to ask, where are you getting 2400000 from?
 
  • #65
Ken G said:
I'm afraid I don't know what you are saying here, are you claiming there is something magical about a circular orbit, compared with all the other possible orbits that would cross a circular orbit at relativistic speeds, such that the circular orbit is the one that does not see redshifts and blueshifts? Because that's just not going to be right.

Didn't mention circular orbits there. Statement applies to any freefall motion.

Are you now assuming that your "free fall" orbit includes infinity?

I'm assuming an elliptical orbit with extremely distant apsides crossing a circular orbit.

Because circular orbits don't. Miller's planet will cross at a relativistic speed any orbit that involves falling free from infinity, so you cannot have it both ways here-- one or the other will see dramatic Doppler shifts. I suspect both will, but I'm sure the circular orbits will.

I'm pretty sure it's not that simple. An object with a highly elliptical orbit intercepting Miller's is still subject to immmense frame-dragging tending it towards prograde revolution around Gargantua. I would be surprised if closing velocity for intercepting orbits is relativistic.

Again, you need to stop imagining that "free fall" is some special state of existence. Various different free-fall orbits will have highly relativistic relative speeds at the points where they cross.

By what mechanism? Freefall paths are constrained by the curvature and rotation of space-time around Gargantua alone.

By "approach orbit", I just meant "not the circular orbit of Miller's planet." It is perfectly obvious that observers falling free from infinity will see the light from an accretion disk that is either redshifted or blueshifted.

It's obvious that hovering observers, not free falling ones, will see redshifts and blueshifts.
 
  • #66
Pete Cortez said:
An object with a highly elliptical orbit intercepting Miller's is still subject to immmense frame-dragging tending it towards prograde revolution around Gargantua. I would be surprised if closing velocity for intercepting orbits is relativistic.
And that is the crux of our disagreement-- I'd be surprised if the closing velocity is not highly relativistic between those orbits. So we simply need to resolve that issue.
By what mechanism? Freefall paths are constrained by the curvature and rotation of space-time around Gargantua alone.
They are constrained by a lot more than that, they are constrained by the initial conditions that cause them to be such different orbits.
It's obvious that hovering observers, not free falling ones, will see redshifts and blueshifts.
I agree it's more obvious that hovering orbits will see shifts, but I think it is still the default expectation that very different orbits will also see very different shifts. The easiest way to resolve that would probably be just to look at the various possible photon orbits, since if it is possible for photon orbits to cross, it's obvious there must be significantly relativistic closing velocities among free-fall orbits.
 

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