Interstellar -the movie, planet with slower time

[Pete Cortez]

But that isn't correct. It is just as wrong to claim that "freely falling" observers don't see blueshifts and redshifts as it is to take the naive answer we considered above, that the local time dilation factor is the answer to everything. I didn't read that whole thread, because I quickly encountered errors like this, by what poster it doesn't matter: "From the viewpoint of an observer at rest in the gravitational field, the freely falling frame is accelerating downward. Suppose a photon is emitted upward towards a freely falling observer some distance above, who is at rest in the gravitational field at the instant the photon is emitted. By the time the photon reaches the observer, it will have redshifted, but the observer will have picked up just enough downward velocity so that when the observer receives the photon, there will be a Doppler blueshift that exactly cancels the gravitational redshift."

Of course that is not even close to right, when the free-faller first starts to fall, there is no blueshift at all, but there is certainly redshift, and if the freely falling observer falls all the way to the place where the static emitter is emitting the light, there will obviously be a substantial blueshift that does not "cancel out" because at that point there's no gravitational redshift any more. So the real answer is, it's very hard to make generalizations in relativity, unless one does the calculations.

The problem with your statement is static emitter. Gargantua's disk is not stationary with respect to the gravitational field. It's orbiting--free-falling--around Gargantua. Just as Miller's planet is. Just as, to an absurd degree, the Ranger is when it slingshots in and out of proximity of Miller's planet.

I think you misinterpret my meaning. I was not saying Thorne didn't enjoy the science, or that he was troubled by his own mistakes, I meant that he must have been quite frustrated over all the people (such as the thread on here that characterized the science as "stupid") making incorrect criticisms of what he did because they didn't do the work he did to make it plausible. Much like a coach having to listen to criticisms of their decisions, coming from people who were not aware of the machinations of the game that actually went into that decision.

No, didn't misinterpret your meaning. I'm saying I find it unlikely that Thorne didn't enjoy the criticism, which by all evidence he cheerfully took up in writing his book and also in interviews after the first critiques came out. Thorne knows very well that we're seeing, for the first time, depictions of these exotic objects that approach what we might see in reality. It must be exciting to be one of the first people who get to cross that uncanny valley, and I imagine he quite enjoys drawing a map for others to find their way as well.

That is clearly not right, as two free-falling observers can be at the same place and time-- and have a significant Doppler shift relative to each other. So we know they will not see the same things.

This assumes two observers have built up a massive difference in velocity with respect to one another--which means at least one of the observers was not free falling from infinity to the present.

 

[hankaaron]

Except the farm equipment of course.
Perhaps, as someone pointed out, we shouldn't be taking every single word of the characters as Gospel.

The criticism of the film isn't being driven by a few flawed plot points. The film is stacked with a load of them.

 

[DaveC426913]

The criticism of the film isn't being driven by a few flawed plot points. The film is stacked with a load of them.

Not sure what the take-away from that is: that a summary judgement has been made, and it's above scrutiny?

You're listing flawed plot points that aren't flawed plot points. That doesn't help your case.

In general, the film was very fast-paced. The direction of events in any given scene might have been explained in just a couple of words (for example, Endurance's orbit).In a film almost 3 hours long, this must be the case.

While the movie may be tough to follow, that's not the same as plot holes.

 

[Pete Cortez]

The absence of Doppler shift, the idea that Miller's planet could be so time dilated yet not be inside the accretion disk and not be much closer to Gargantua...

See above reply. The disk is not stationary relative to the gravitational field.

...the idea that the Endeavor could orbit anywhere near Miller's planet and still experience the 60,000 factor time dilation for the guy left behind (or maybe I'm guilty of the same problem, and the factor was much less than 60,000 for him-- I didn't bother to check this, it seemed like they were using that same factor)

Endeavor occupies a parking orbit far enough out that its proper time is only modestly compressed compared to that at infinity. Not too far out, because the Ranger has to be able to reach Miller's planet in about two hours (as measured by the expedition on the Ranger).

...and the idea that a human in a spacesuit pushing books off a shelf would be the best way for an advanced civilization to communicate back in time to humanity...

worst, the question is what alternatives were available. As advanced as an ancient civilization may be, the first thing they need to deal with is the problem of a receiver.

seem like the main ones that could have been replaced with something more plausible.

One of the beautiful things about the film is that it's so terse in most of these areas you can either 1) imagine a more plausible off-screen explanation for what you see than what you might come up with on first viewing or 2) conjure up reasons why a depicted event is also an eminently plausible one.

But even these issues aren't really so bad in my view, because they are central to the drama of the story and the impact of the visuals. Basically, it's a movie, not a science textbook, but at least it is a movie that brings common people into contact with relativity in a reasonable way. The only thing that bothers me is the failure to include Doppler shifts in that relativity lesson, it just seems like a missed opportunity that could have been handled within the same overall story.

With gravitational redshift, two reasons. First, you don't want to fry the cast. Two, you can come up with a plausible configuration for Gargantua's system that allows you to avoid frying your cast. This isn't a movie about characters dying in extreme radiation environments, after all. It's a movie about how characters survive falling into a black hole.

 

[Pete Cortez]

The criticism of the film isn't being driven by a few flawed plot points. The film is stacked with a load of them.

None of which you've actually pointed out.

 

[DaveC426913]

One of the beautiful things about the film is that it's so terse in most of these areas you can either 1) imagine a more plausible off-screen explanation for what you see than what you might come up with on first viewing or 2) conjure up reasons why a depicted event is also an eminently plausible one.

Well said.

 

[AngelLestat]

About the gravity anomalies, I guess they can be caused due to the gravitational waves from gragantua crossing the wormhole which were enoght strong to be detected by ligo?
That´s how they find the wormhome after all.

What I don't know is how the size of the wormhole may reduce the gravitational waves comming from the other side.
We know that the wormhole is orbiting saturn, from the other side is orbiting gargantua (not sure how far), so this mean that there is no gravitational forces going from side to side?

Some pictures to help with our reasoning:

Extra image about gargantua disk.
http://i.space.com/images/i/000/043/505/i02/interstellar-wormhole-travel-141107c-02.jpg?1415391304

 

[Ken G]

See above reply. The disk is not stationary relative to the gravitational field.

Haven't I already demonstrated that argument is not correct? The disk will still appear with different colors on different sides. It will appear that way to a hovering observer, it will appear that way (in a different way) to an orbiting observer who is at the depicted distances, and to someone falling in. All of those different situations require detailed calculations and all will give different results, and none will look like what the movie did a detailed calculation to show us. That is the thing that disappoints me, I would have liked to see what the science says it should have looked like in these various frames. It was not an artist's conception after all, they did a very detailed calculation-- they just decided not to calculate a color. Perhaps it would have made the calculation too difficult, I wouldn't mind if they would just say "we didn't have the flops for that."

Endeavor occupies a parking orbit far enough out that its proper time is only modestly compressed compared to that at infinity. Not too far out, because the Ranger has to be able to reach Miller's planet in about two hours (as measured by the expedition on the Ranger).

I know that, that's what I was questioning. The scale of Gargantua is an AU, so for the guy left behind to have his time act like Earth, he would need to have been a very long way out, at least an AU or so. Maybe the Ranger has a whopping great engine, but g forces are still a problem there. I don't think my expectation there is wrong, and I'm sure Kip Thorne knew it too, so it's curious that was not mentioned in the companion book if indeed it was not. It was probably a problem either way-- if you want to "park" the Endeavor, you have to be a huge distance away from Miller's planet, but then the Ranger has to climb out of the black hole by itself. Probably they should have used the ergosphere to maneuver the Endeavor down close to Miller's planet. Even better, just don't go to Miller's planet at all-- that was an obvious plot device, never really made sense but it's a movie.

 

[Pete Cortez]

Haven't I already demonstrated that argument is not correct? The disk will still appear with different colors on different sides. It will appear that way to a hovering observer, it will appear that way (in a different way) to an orbiting observer who is at the depicted distances, and to someone falling in.

We agree in part. There is a difference between a hovering observer and a free falling one. However, an orbiting observer is also in free fall. He's just falling around the mass rather than into it. The question, remember, is what does the rest of the universe look like to said observer.

All of those different situations require detailed calculations and all will give different results, and none will look like what the movie did a detailed calculation to show us. That is the thing that disappoints me, I would have liked to see what the science says it should have looked like.

Can you explain what you expect an orbiting free faller to see and how it deviates from the movie?

I know that, that's what I was questioning. The scale of Gargantua is an AU, so for the guy left behind to have his time act like Earth, he would need to have been a very long way out, it makes no sense that he should be in a very different time state than those who went down to the planet. I don't think my expectation there is wrong, and I'm sure Kip Thorne knew it too, so it's curious that was not mentioned in the companion book if indeed it was not.

I have not done the calculations myself, but according to Thorne this is not a problem:

How did I come up with these locations? I use the parking orbit as an illustration here and discuss the others later. In the movie, Cooper describes the parking orbit this way: “So we track a wider orbit of Gargantua, parallel with Miller’s planet but a little further out.” And he wants it to be far enough from Gargantua to be “out of the time shift,” that is, far enough from Gargantua that the slowing of time compared to Earth is very modest. This motivated my choice of five Gargantua radii (yellow circle in Figure 6.3). The time for the Ranger to travel from this parking orbit to Miller’s planet, two and a half hours, reinforced my choice.

 

[Ken G]

We agree in part. There is a difference between a hovering observer and a free falling one. However, an orbiting observer is also in free fall. He's just falling around the mass rather than into it.

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.

The question, remember, is what does the rest of the universe look like to said observer.

Yes, and the answer is, "redshifted and blueshifted like mad, depending on which direction you look." Which is exactly what the movie decided not to portray, and I wish they had not made that choice, because I think the general population could have handled Doppler shifts just as easily as time dilation effects. The problem might have been, the Doppler shifts would have made so much of it X-rays or infrared that it wouldn't be visible at all.

Can you explain what you expect an orbiting free faller to see and how it deviates from the movie?

The movie depicted no change in frequency or intensity looking at the Gargantua from any orbit around Gargantua, and even on an approach orbit. That's clearly wrong, there should have been spectacular Doppler shifts and intensity differences. Hard to calculate, but ray-tracing was no easy matter either, why not track the frequency and intensity factors along each of those rays as well? You have to assume an input spectrum, but what they chose was rather golden. A nice effect, but no spectrum is completely flat, they could have chosen something and Doppler shifted it correctly as they did their ray tracing.

I have not done the calculations myself, but according to Thorne this is not a problem:

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?

 

[Pete Cortez]

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.

 

[Ken G]

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.

 

[AngelLestat]

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.

 

[Pete Cortez]

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?

 

[Pete Cortez]

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.

 

[Ken G]

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.