What is the Flaw of General Relativity Regarding Uniform Gravitational Fields?

[Zanket]

Think again. If the ball was thrown in a gravitational field, at NO POINT in time does it ever have a constant velocity, be it in a uniform or none uniform field. You cannot transform to any inertial frame in which the ball is at rest! On the other hand, the ball that was thrown in an accelerating frame can! There's nothing "negligibly" different here. It's A LOT different.

Now I am completely confused as to what you’re getting at. And I really do want to understand what you’re getting at.

Let the gravitational field of the planet be uniform. In this case, according to the equivalence principle, “it is theoretically impossible to distinguish between gravitational and accelerational forces by experiment” (Encarta). Then how could the situations possibly be a LOT different? No experiment could detect a difference, all else being equal, right? In either situation you can transform to any inertial frame in which the ball is at rest—just be an ant sitting on the ball in either situation. Please elaborate as to how you think an ant sitting on the ball in the planet’s field is not in an inertial frame in which the ball is at rest.

 

[Zanket]

No I didn't - Rising upward would mean acelerating upward.

When I throw an apple upward, it is rising upward but not accelerating upward. Then they don’t mean the same.

The object has upward speed BUT that speed is slowing down due to the 'gravity' the travelers feel. That slowing down is the same as FALLING.

The object is in free fall, but not falling. The dictionary definition of “falling” is “moving downward”. The dictionary definition of “rising” is “ascending”. These terms are not changed in physics. The buoy ascends in their frame, it does not move downward; hence it is rising, not falling. As I said, maybe there's a better term, but by "free-rising" I mean that it is in free fall, rising upward. Do you really think that I should call an object that is ascending a “falling” object? How would I distinguish such an object from one that is “moving downward”?

Of course that's the same as seen on a planet - you just keep calling an intial velocity up as "rising" even when it's slowing down - that does cut it, your just WRONG.

I call it “rising” because it meets the dictionary definition. An object is rising when it’s ascending, whether or not it is slowing down. A rocket accelerating upward, as well as an apple thrown upward, rises.

You need to define all of what the object is really doing - like stopping its "rise" when you reach Andromeda and then your “free rising” object with no change in gravity (that is you don’t turn your engines off at Andromeda and your artificial gravity is taking you back to earth) is now falling on you!

What happens after they reach Andromeda is irrelevant to #3, which deals with objects in free fall that are rising, not falling. I agree that if they don’t turn their engines off at Andromeda then the buoy falls toward them, but that is irrelevant. I don’t see how that fact addresses #4.

I notice you didn’t address that small detail in my comments.
“If you don’t turn your engines off, it will start to fall towards you from that stopped position as you depart Andromeda.”

I agree with this small detail but I didn’t address it because it’s irrelevant, as above.

Care to explain just how your free rising object decides to start falling with no change in gravity??

No, I don’t care to address anything that happens after they arrive at Andromeda, because it’s irrelevant, as above. I want to stick to the case I put forth, which ends when they arrive at Andromeda. The case ends when the object that is rising upward reaches its highest point.

Seems exactly like gravity (uniform tidal negligible if you want) on a planet too me.

The situations are not the same according to GR, even though the equivalence principle demands that they can negligibly differ. In the local frame of an observer on a planet, GR does not predict that objects rising in free fall (e.g. thrown upward) can recede to any given proper distance in an arbitrarily short proper time while causal contact is maintained, as the crew can observe in #3. For example, in the local frame of an observer on a planet (a frame that we agree can be as large as imaginable), a buoy thrown upward cannot recede to a distance of one million proper light years in ten proper years while causal contact is maintained.

How do you think what can happen after the crew arrives at Andromeda, i.e. after the buoy reaches its highest point, resolves that? I don’t get why you even bring that up.

 

[ZapperZ]

Now I am completely confused as to what you’re getting at. And I really do want to understand what you’re getting at.

Let the gravitational field of the planet be uniform. In this case, according to the equivalence principle, “it is theoretically impossible to distinguish between gravitational and accelerational forces by experiment” (Encarta). Then how could the situations possibly be a LOT different? No experiment could detect a difference, all else being equal, right? In either situation you can transform to any inertial frame in which the ball is at rest—just be an ant sitting on the ball in either situation. Please elaborate as to how you think an ant sitting on the ball in the planet’s field is not in an inertial frame in which the ball is at rest.

Put yourself on a spaceship that is far enough away from that planet and in which it is in the same inertial frame as the planet (assuming there's no other bodies around). Now look at the ball being throw vertically up on the planet. You (the one in the "stationary" ship) will see the ball decelerating before it turns around and accelerates back to the ground.

Now tell me what inertial frame can you transform to be stationary with the ball. NONE! The ball is being accelerated by the gravitational field continuously. There is always an "F" acting on the ball!

Compare this with another spaceship which is accelerating "upwards". After the ball is thrown, there's no "F" left acting on it. It moves with a constant velocity. I can easily transform to an inertial frame in which the ball is at rest in that frame.

These two are NOT identical. There's no symmetry between the two. This is the VERY reason why we can tell that it is US that is orbiting around the sun and NOT the sun orbiting around us even when it appears that way just because the sun rises and sets every day.

Zz.

 

[RandallB]

When I throw an apple upward, it is rising upward but not accelerating upward. Then they don’t mean the same.

The object is in free fall, but not falling. The dictionary definition of “falling” is “moving downward”.

O come on Zanket .
We aren’t talking a dictionary world here, this is physics!
You need to decide what you mean by “FREE RISING” that's not in the dictionary so you tell us.
When you hit a pop-up in baseball how is it not “Free-Rising”. Just like your space buoy?
Yes or no is your object slowing down or not, does stop rising when you reach Andromeda or not.

You agree that if they don’t turn their engines off at Andromeda then the buoy falls toward them, but some how that is irrelevant?
Why?
From that point, you must agree then that it “falls” just like the baseball does for the pop-up once it reaches it high point. So, on the ship how do you make a ball do the same thing? How fast and how high would you have to “pop it up” for it to fall just like the buoy will fall if you let the “gravity” continue after reaching Andromeda?
Just how close will it be to the buoy? If you can not solve this problem you certainly cann’t work the one you’ve put forward.

 

[Zanket]

GR does predict this, it just takes a "planet" that is on the verge of becoming a black hole, so that one can get a large enough time dilation factor.

This is a good post. You're getting to the heart of the matter. I'm going to think about this and answer it later, in the meantime responding to other posts.

 

[Zanket]

Put yourself on a spaceship that is far enough away from that planet and in which it is in the same inertial frame as the planet (assuming there's no other bodies around).

You mean a spaceship in free fall that is far enough away that the ship essentially stays at rest with respect to the planet? I’ll assume that for now.

Now look at the ball being throw vertically up on the planet. You (the one in the "stationary" ship) will see the ball decelerating before it turns around and accelerates back to the ground.

Agreed.

Now tell me what inertial frame can you transform to be stationary with the ball. NONE!

I did that already. I transform to the inertial frame of an ant sitting on the ball (well, clinging, since the ant is weightless). Then it’s possible and not NONE, right?

The ball is being accelerated by the gravitational field continuously. There is always an "F" acting on the ball!

That’s the Newtonian viewpoint. It is not gravity we feel, Einstein says, but simply the ground pushing up on our feet. Likewise, in an accelerating rocket, it is not gravity the crew feels, but simply the rocket pushing up on their feet. Gravity is not a force in general relativity. No F acts on the ball in either situation.

Compare this with another spaceship which is accelerating "upwards". After the ball is thrown, there's no "F" left acting on it. It moves with a constant velocity. I can easily transform to an inertial frame in which the ball is at rest in that frame.

In either situation, after the ball is thrown, there's no "F" left acting on it, as above. It moves with a constant velocity in either situation, in the frame of an ant clinging to the ball.

These two are NOT identical. There's no symmetry between the two.

I still don’t see it. How can the ant not be stationary with the ball in both situations? How could the situations possibly be different when, according to the equivalence principle, “it is theoretically impossible to distinguish between gravitational and accelerational forces by experiment” (Encarta). You haven't answered these questions.

This is the VERY reason why we can tell that it is US that is orbiting around the sun and NOT the sun orbiting around us even when it appears that way just because the sun rises and sets every day.

I thought both we and the Sun orbited around the Sun-Earth barycenter, in the absence of other bodies.

 

[ZapperZ]

You mean a spaceship in free fall that is far enough away that the ship essentially stays at rest with respect to the planet? I’ll assume that for now.
Agreed.
I did that already. I transform to the inertial frame of an ant sitting on the ball (well, clinging, since the ant is weightless). Then it’s possible and not NONE, right?

Excuse me? The ball is ACCELERATING. How is it possible to transform to an inertial frame of the ball? Does your dictionary have no definition of an "inertial frame"?

That’s the Newtonian viewpoint. It is not gravity we feel, Einstein says, but simply the ground pushing up on our feet. Likewise, in an accelerating rocket, it is not gravity the crew feels, but simply the rocket pushing up on their feet. Gravity is not a force in general relativity. No F acts on the ball in either situation.

You're contradicting yourself. You just agreed with me that from the point of view of an inertial frame, the ball is accelerating. Now you are telling me there's no "F". So the ball is accelerating all on its own due to... what? Remember, this is from the point of view of an observer in an inertial frame!

But then again, I don't think you know what an inertial frame is judging from your comments above. It makes the rest of your reply (and probably the whole thread) moot.

Zz.

 

[Zanket]

GR does predict this, it just takes a "planet" that is on the verge of becoming a black hole, so that one can get a large enough time dilation factor.

...

This quick aging process would allow objects following a geodesic (what you call free-rising) to reach an arbitrarily large distance in an arbitrarily short amount of your time, while maintaining causal contact, the exact features you want to duplicate as nearly as I can tell from your text. (You will have to move arbitrarily close to the event horizon to achieve this).

GR does not predict this. It predicts that an observable receding object recedes at an observable rate (proper distance over proper time) always less than c. That is the basis of the horizon problem in cosmology based on GR. What you describe, and what I describe in #3, is an observable rate of xc, where x is any factor >= 0. GR predicts that objects receding at a rate where x >= 1 lie beyond a cosmological horizon, where they are unobservable; i.e. there is no causal contact with them.

Now I say, and you seem to agree, that GR must predict the same for both situations (i.e. for both the crew and the observer on the planet). This is demanded by special relativity and the equivalence principle, components of GR. Then GR is inconsistent. When the same is predicted for both situations, the horizon problem can be explained away by gravity.

Note that the time dilation factor is NOT related to the value of acceleration required to hold station, but to the total potential energy of the hovering observer. This may be a subtle point that you are missing.

I think #3 implies this already. The crew can travel from Earth to Andromeda in 20 proper years at a rate of acceleration just over 1 Earth gravity. At this relativity low rate of acceleration they can observe the buoy, once they pass it, to recede at an average observable (not actual) rate of one hundred thousand c. You are correct that the rate of acceleration is immaterial.

 

[Hurkyl]

GR does not predict this. It predicts that an observable receding object recedes at an observable rate (proper distance over proper time) always less than c.

As you've stated this, it's patently false. Remember that any coordinate chart is perfectly valid -- which includes coordinate charts where this distant object has a coordinate velocity millions of times of c.

(And don't forget that light is only guaranteed to travel at c near the origin of "good" coordinate charts. It could be anything at all near the origin of a "bad" coordinate chart, as well as anywhere that isn't at the origin of a "good" coordinate chart)

(I'll confess to not having read the entire thread in depth, so my apologies if you've agreed upon something that makes your statement true)

 

[pervect]

GR does not predict this. It predicts that an observable receding object recedes at an observable rate (proper distance over proper time) always less than c

This is where you are wrong. This is a somewhat confusing topic, but I've dug up a few references to demonstrate this. You might also want to read Hurkyl's post on the same point.

http://casa.colorado.edu/~ajsh/sr/postulate.html

In general relativity, arbitrarily weird coordinate systems are allowed, and light need move neither in straight lines nor at constant velocity with respect to bizarre coordinates (why should it, if the labelling of space and time is totally arbitrary?). However, general relativity asserts the existence of locally inertial frames, and the speed of light is a universal constant in those frames.

What needs to be added to the above is that the clocks and rulers that are used to measure the speed of light must be local clocks and rulers, i.e. clocks and rulers that are physically present in those "locally inertial frames".

Another point that should be obvious but probably should be mentioned is that the local clocks and rulers must be properly normalized / standardized. One should probably imagine lugging along an actual copy of the meter-bar in paris, plus a pair of atomic clocks, and distances being laid out with the bar and times measured with the clock(s). (Since we are talking about _measuring_ the speed of light, we are conceptually reverting to the days before the SI meter was re-defined in terms of the speed of light, so it's appropriate to imagine carrying around a platinum-alloy bar).

GR is perfectly capable of dealing with coordinate systems that are not normalized or standardized in the above or any other manner - it is actually a convention of physicists that such normalization be be done (and be done at the origin of a coordinate system). This is what Hurkyl is talking about when he talks about "good" vs "bad" coordinate systems.

http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html

also addresses this question.

Einstein went on to discover a more general theory of relativity which explained gravity in terms of curved spacetime, and he talked about the speed of light changing in this new theory. In the 1920 book "Relativity: the special and general theory" he wrote: . . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone, it is not clear that he meant the speed will change, but the reference to special relativity suggests that he did mean so. This interpretation is perfectly valid and makes good physical sense, but a more modern interpretation is that the speed of light is constant in general relativity.

The problem here comes from the fact that speed is a coordinate-dependent quantity, and is therefore somewhat ambiguous. To determine speed (distance moved/time taken) you must first choose some standards of distance and time, and different choices can give different answers. This is already true in special relativity: if you measure the speed of light in an accelerating reference frame, the answer will, in general, differ from c.

In special relativity, the speed of light is constant when measured in any inertial frame. In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected).

Again, the clocks and rulers that are used to measure the speed of light in a freely falling frame must be local clocks and rulers if one is to measure the speed of light as 'c'. And they must also be appropriately standardized (no fair using a clock that ticks at a non-standard rate, or a meter-bar that is too short or too long).

Because the tidal forces are neglgible, the clocks will all be running at the same rate, and the rulers will all measure the same distance, in the "locally inertial" region.

It is unfortunate that this important point has not been stressed more in the references I could find on-line.

Our hypothetical black hole observer is both accelerating, and in a region of curved space-time, so he cannot be expected (and will not) measure the speed of light to be globally equal to 'c'.

The black-hole observer will, like the accelerating observer, measure the speed (coordiante speed) of light to be 'c' only at the origin of the coordinate system, and find that it will vary with position.

One can also make the following statement about both observers. The arbitrarily high rate of change of position coordinates with respect to time coordinate of fast-moving free-falling objects will still always be lower than the rate of change of the position coordinate with respect to the time coordinate of a light beam at the same location, as long as the regions are causally connected.

Add:
(This can be deduced from the time independence of the problem, and the fact that light does get from point A to point B. Time independence is an important part of the argument here, without it the above statement would be incorrect).

In this case, the regions are causally connected, as you have already noted. (And the metric can be written in a time-independent manner). Thus the arbitrarily fast coordinate-speed of physical objects will be mirrored by an even faster arbitrarily fast coordinate-speed of light.

Also note that the specific numerical value of this "coordinate velocity" is also totally dependent on the coordinate system chosen, i.e. it is not a coordinate independent quantity.

 

[Zanket]

We aren’t talking a dictionary world here, this is physics!

Agreed. I said that the dictionary definition of “rising” is not changed in physics to “falling.”

You need to decide what you mean by “FREE RISING” that's not in the dictionary so you tell us.
When you hit a pop-up in baseball how is it not “Free-Rising”. Just like your space buoy?
Yes or no is your object slowing down or not, does stop rising when you reach Andromeda or not.

I said that by “free-rising” I mean the object is in free fall, rising upward. So yes to both questions.

You agree that if they don’t turn their engines off at Andromeda then the buoy falls toward them, but some how that is irrelevant?
Why?

Why not? You haven’t shown how it is relevant as far as I can tell. Yes, the buoy’s motion in the crew’s frame is just like that of a pop-up baseball. It rises and then, if they keep their engines running at Andromeda, it falls. I just don’t see how the fall affects #4. Only the rise is needed to make #4 a valid question.

From that point, you must agree then that it “falls” just like the baseball does for the pop-up once it reaches it high point.

Agreed.

So, on the ship how do you make a ball do the same thing? How fast and how high would you have to “pop it up” for it to fall just like the buoy will fall if you let the “gravity” continue after reaching Andromeda?

To make a ball track with the buoy, the crew need only throw up the ball at the moment they pass the buoy, and with the buoy’s relative (to them) velocity. The ball will then remain at rest with respect to the buoy.

Just how close will it be to the buoy?

If you mean how close to buoy’s motion will the ball’s motion be, it will be the same motion. The buoy and the ball will essentially be the same system, a single buoy-ball object.

If you can not solve this problem you certainly cann’t work the one you’ve put forward.

Now that I have solved this, do you have any objections to #1, 2, or 3 in the original post, and if not, can you answer #4?

Can I assume that the reason you brought up the buoy’s fall is because you needed further convincing that the buoy is rising in the crew’s frame just like a pop-up ball could?

 

[Zanket]

Excuse me? The ball is ACCELERATING. How is it possible to transform to an inertial frame of the ball? Does your dictionary have no definition of an "inertial frame"?

Before I answer this and the rest of your post, let’s be clear on something: Are you suggesting that there is no inertial frame possible in a uniform gravitational field of a planet?

 

[RandallB]

If you mean how close to buoy’s motion will the ball’s motion be, it will be the same motion. The buoy and the ball will essentially be the same system, a single buoy-ball object.

Now that I have solved this, do you have any objections to #1, 2, or 3 in the original post, and if not, can you answer #4?

Can I assume that the reason you brought up the buoy’s fall is because you needed further convincing that the buoy is rising in the crew’s frame just like a pop-up ball could?

Why would I have a problem now, as per your solution and explanation in this post “The buoy and the ball will essentially be the same system, a single buoy-ball object.”
Now that the ball on earth, and the buoy in space are moving identically, you’ve shown the two to behave equivalently.

YOU have finally hit on your confirmation that the GR equivalence principal applies just as predicted. And therefore there is not “A Flaw of General Relativity” on this issue.

Congratulations good job – sometimes it just takes working though the details to get it eh – but it feels good when you do, doesn’t it.

 

[Zanket]

As you've stated this, it's patently false. Remember that any coordinate chart is perfectly valid -- which includes coordinate charts where this distant object has a coordinate velocity millions of times of c.

Sure, if the coordinate chart is stretching, stretching the distance. That’s not the case here.

 

[Zanket]

YOU have finally hit on your confirmation that the GR equivalence principal applies just as predicted.

That was the point of #1, 2, and 3. That it should apply.

And therefore there is not “A Flaw of General Relativity” on this issue.

Unfortunately GR does not let the equivalence principle apply to the observer on the planet just as predicted. GR does not predict that the same observation is possible for the observer on the planet. For that observer, a pop-up ball cannot recede to a distance of one million proper light years in ten proper years while causal contact is maintained. For that observer GR predicts that it can recede only less than ten proper light years in ten proper years while causal contact is maintained.

 

[Zanket]

This is where you are wrong. This is a somewhat confusing topic, but I've dug up a few references to demonstrate this. You might also want to read Hurkyl's post on the same point.

I read your whole post. Good info. Unless I missed something, it mostly boils down to Hurkyl’s comment that GR can predict the observation in #3 (in the original post) for the observer on the planet if the coordinate system is expanding, stretching the proper distance to an object thrown upward. But #3 can occur in a static coordinate system. Then #4 remains an open question. Making the two situations as equal as possible includes using the same coordinate system in all respects (like static for both).

Our hypothetical black hole observer is both accelerating, and in a region of curved space-time, so he cannot be expected (and will not) measure the speed of light to be globally equal to 'c'.

The workaround to this issue of measurements being skewed within an accelerated frame is to have the observer fall through a trap door when the object thrown upward reaches its highest point. The spacetime of the experiment is flat (negligibly curved) because we let the tidal force throughout the region enclosing the experiment be negligible (including below the observer, who can be on top of a tower if necessary), and the now free-falling observer can do good measurements and measures the speed of light to be c throughout the region. The observer and the object will remain at rest with respect to each other during the measurement, just like the crew remains at rest with respect to the buoy when they reach Andromeda and shut off their engine. Falling through the trap door is analogous to shutting off the engine.

To show that GR is inconsistent, the observer need measure a proper distance over proper time (between throwing the object up and falling through the trap door) rate >= c. The observer should be able to do that given that the crew can do that (according to the equivalence principle).

An observer with the same acceleration felt as the crew in #3 feels, who throws an object upward at the same velocity as the crew measures for the buoy when they pass it, and waits ten proper years before falling through the trap door, should find that the object is one million proper light years away.

 

[RandallB]

That was the point of #1, 2, and 3. That it should apply.

Unfortunately GR does not let the equivalence principle apply to the observer on the planet just as predicted. GR does not predict that the same observation is possible for the observer on the planet. For that observer, a pop-up ball cannot recede to a distance of one million proper light years in ten proper years while causal contact is maintained. For that observer GR predicts that it can recede only less than ten proper light years in ten proper years while causal contact is maintained.

What are you talking about here ? What do you mean “GR does not predict that the same observation is possible for the observer on the planet.” ?

In your post 61 YOU already have shown that that is exactly predicted to be the same. Your not agreeing with yourself now.

 

[Zanket]

What are you talking about here ? What do you mean “GR does not predict that the same observation is possible for the observer on the planet.” ?

In your post 61 YOU already have shown that that is exactly predicted to be the same. Your not agreeing with yourself now.

It is GR that disagrees with itself. It is inconsistent. GR predicts that the same observation is possible for the observer on the planet, according to its own equivalence principle. But it also predicts that such is impossible. For example, look up "cosmological horizon" to see that GR predicts that causal contact with objects receding at a proper distance over proper time rate >= c is impossible. (In the absence of any stretching of the coordinate system, which stretches proper distances.)

Here is a theory:

Birds can fly.
Birds cannot fly.

If I show you that birds can fly, am I disagreeing with myself or is the theory inconsistent?

 

[RandallB]

It is GR that disagrees with itself. It is inconsistent. GR predicts that the same observation is possible for the observer on the planet, according to its own equivalence principle. But it also predicts that such is impossible. For example, look up "cosmological horizon" to see that GR predicts that causal contact with objects receding at a proper distance over proper time rate >= c is impossible. (In the absence of any stretching of the coordinate system, which stretches proper distances.)

Here is a theory:

Birds can fly.
Birds cannot fly.

If I show you that birds can fly, am I disagreeing with myself or is the theory inconsistent?

Holly Cow what are you doing with all this. Birds??
The only advice I can give you is from Fynnman, and that’s
“SHUT UP AND CALCULATE”

He wasn’t being insulting, just saying stop forcing things to be the way you think. And do the calculations in detail to show what’s really happening.

So find a way to put REAL numbers on your problem.
Defining what you’re looking for has got to be most of your trouble. So here:

Earth “E” to Mid point “M” to Andromada “A” change to any distances easy for you to work with.

On the ship start acceleration “a” at M to slow to a stop at A.
KEEP “a” going to return to M.
On ship at M hit ball at speed v to “stay” at M what is v.
V of ball will be Zero when batter reaches A.
Define height of ball in reference frame of batter while he is at A.

THEN move to the planet, hit a ball up at same starting speed v as batter at M did.
Solve for “g” gravity, required to have ball go up to same height as spaceship batter saw when he reached A.

Then compare “a” to “g”.

If don’t have time to do the math – quit arguing.

 

[JesseM]

Numbered for reference:

1. The equivalence principle tells us that the crew of a rocket, traveling in flat spacetime and where the crew feels a constant acceleration, experiences a uniform gravitational field identical to that experienced locally by an observer on a planet.

Yes, locally--in an arbitrarily small region of space and in an arbitrarily small time-interval.

2. Special relativity http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html that in principle the crew can traverse between any two points A and B in flat spacetime in an arbitrarily short proper time, where the two points are at rest with respect to each other and the rocket accelerates from rest at point A to the halfway point and then decelerates from the halfway point to rest at point B.

Sure.

3. Then the crew can observe free-rising objects (such as an object floating stationary at the halfway point) to recede apparently arbitrarily fast—a million c is not out of the question—while causal contact is maintained since the actual velocity is always less than c.

If you want to talk about what they "observe" as opposed to what they actually see using light signals, then presumably you must be talking about the velocity of the object in some coordinate system where the observer is at rest. The problem is that if you try to create a coordinate system where an observer is at rest throughout an extended period of acceleration, this will not be an inertial coordinate system, and thus it won't be relevant to the equivalence principle, nor is there any restriction on objects traveling faster than c in non-inertial coordinate systems. On the other hand, if you look at the inertial frame where the observer's instantaneous velocity is zero at a particular moment during the acceleration, in this instananeous inertial rest frame the velocity of outside objects will never be greater than c.

Example: Let the rocket travel from Earth to Andromeda, two million light years away as we measure. (Assume that Earth and Andromeda are at rest with respect to each other and the spacetime between them is flat.) Let a buoy float stationary at the halfway point. Let the half of the trip from the buoy take ten proper years as the crew measures. Then during this half the buoy recedes by one million proper light years in ten proper years, an apparent (not actual) velocity of one hundred thousand c. From the crew's perspective the buoy free-rises in a uniform gravitational field.

Only if by "perspective" you mean a non-inertial coordinate system where the crew is at rest throughout the journey. Again, not relevant to the equivalence principle. If you look at the crew's instantaneous inertial frame as they pass various buoys, and restrict it to purely local observations so that this will be relevant to the equivalence principle (ie measuring the velocity of a buoy in the crew's instantaneous inertial rest frame at the moment they pass next to that buoy), then no buoy will be observed to travel faster than c.

4. Then why does general relativity not predict the same possible observation for the observer on the planet?

In formulating this paradox, you seem to have forgotten the part of the definition of the equivalence principle in point 1) that says the equivalence only applies to local observations, and you forgot to specify that it is only an equivalence between what is observed in a gravitational field and what is observed in an inertial frame by the observer moving in flat spacetime, not a non-inertial one. If you keep both of these in mind, you will see there is no paradox.

 

[Zanket]

Holly Cow what are you doing with all this. Birds??
The only advice I can give you is from Fynnman, and that’s
“SHUT UP AND CALCULATE”

The bird theory is a spot-on example of an inconsistent theory. We agree that the equivalence principle demands that the planetary observer should be able to observe what the crew can observe, and you agreed with what the crew observes in #3. The calculation for the cosmological horizon has already been done by cosmologists using GR and they've widely reported their result, which shows that the observation in #3 is impossible for the planetary observer, in which case you must agree that GR is inconsistent.

So I'll assume that “SHUT UP AND CALCULATE” is just your way of giving up.

 

[RandallB]

So I'll assume that “SHUT UP AND CALCULATE” is just your way of giving up.

You've obviously never read Fynnman.

And I take it from your comment you’re not willing to do the math.
So if you’re giving up on yourself, why shouldn’t I give up on you?

Bye

 

[Zanket]

Numbered for reference:

1. The equivalence principle tells us that the crew of a rocket, traveling in flat spacetime and where the crew feels a constant acceleration, experiences a uniform gravitational field identical to that experienced locally by an observer on a planet.

Yes, locally--in an arbitrarily small region of space and in an arbitrarily small time-interval.

Note boldfacing. “Local” here means “of a range throughout which the tidal force is negligible”. That is not necessarily a small region, as suggested by the fact that “small” is an arbitrary amount (what is small to me may be large to you). In principle the spacetime throughout a cube having sides a billion light years long can be flatter than the spacetime throughout any arbitrarily small region you can imagine. Likewise the time interval need not be arbitrarily small. Nevertheless, #4 in the original post applies to any sized region, as long as it’s local.

If you want to talk about what they "observe" as opposed to what they actually see using light signals, then presumably you must be talking about the velocity of the object in some coordinate system where the observer is at rest. The problem is that if you try to create a coordinate system where an observer is at rest throughout an extended period of acceleration, this will not be an inertial coordinate system, and thus it won't be relevant to the equivalence principle, nor is there any restriction on objects traveling faster than c in non-inertial coordinate systems. On the other hand, if you look at the inertial frame where the observer's instantaneous velocity is zero at a particular moment during the acceleration, in this instananeous inertial rest frame the velocity of outside objects will never be greater than c.

There’s no problem here. Yes, I’m talking about observations—measurements, as opposed to what they see. The equivalence principle “holds that forces produced by gravity are in every way equivalent to forces produced by acceleration, so that it is theoretically impossible to distinguish between gravitational and accelerational forces by experiment” (Encarta). Then the principle applies to noninertial frames. The http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html are equations of special relativity for constant acceleration. The equivalence principle demands that these equations apply to any uniform gravitational field, whether it’s one the crew of a rocket experiences or one an observer on the planet experiences locally. (Otherwise an experiment could disprove the principle.) These equations tell us what both an inertial and a noninertial observer would measure.

That said, let the crew shut off their engine at Andromeda. Now they remain at rest with respect to the buoy, which must be one million proper light years away. We had let the spacetime between the Earth and Andromeda be flat. The time elapsed on their clock since passing the buoy was given as ten proper years.

Only if by "perspective" you mean a non-inertial coordinate system where the crew is at rest throughout the journey. Again, not relevant to the equivalence principle. If you look at the crew's instantaneous inertial frame as they pass various buoys, and restrict it to purely local observations so that this will be relevant to the equivalence principle (ie measuring the velocity of a buoy in the crew's instantaneous inertial rest frame at the moment they pass next to that buoy), then no buoy will be observed to travel faster than c.

Agreed that nothing actually moves faster than c, as noted in #3. But apparently the buoy does, as described. The equivalence principle does not apply to only inertial frames. The experimenter trying to “distinguish between gravitational and accelerational forces” can be the crew or an observer on a planet. The equivalence principle refers to a uniform gravitational field and constant acceleration in flat spacetime, in which case the frame of an object in free fall in either situation is inertial. (Like the inertial frame for which t and d apply in the relativistic rocket equations.) Maybe that is why you think the equivalence principle applies only to inertial frames.

In formulating this paradox, you seem to have forgotten the part of the definition of the equivalence principle in point 1) that says the equivalence only applies to local observations, and you forgot to specify that it is only an equivalence between what is observed in a gravitational field and what is observed in an inertial frame by the observer moving in flat spacetime, not a non-inertial one. If you keep both of these in mind, you will see there is no paradox.

Now that I have resolved these seeming problems, what say you?

 

[JesseM]

Note boldfacing. “Local” here means “of a range throughout which the tidal force is negligible”.

I don't think that's correct, every detailed definition of the equivalence principle I have seen says that it only applies in the limit of an arbitrarily small neighborhood of spacetime. There may be other issues besides tidal forces that prevent larger regions from being equivalent to flat spacetime. Can you find any source that defines "local" in the equivalence principle as you do above?

There’s no problem here. Yes, I’m talking about observations—measurements, as opposed to what they see. The equivalence principle “holds that forces produced by gravity are in every way equivalent to forces produced by acceleration, so that it is theoretically impossible to distinguish between gravitational and accelerational forces by experiment” (Encarta). Then the principle applies to noninertial frames.

No, it doesn't. Any more detailed definition will specify that it only applies to inertial frames. For example, from this page:

The equivalence principle can be stated as "At every spacetime point in an arbitrary gravitational field, it is possible to chose a locally inertial coordinate system such that, within a sufficiently small region of the point in question, the laws of nature take the same form as in unaccelerated Cartesian coordinate systems

Or from http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Principle%20of%20Equivalence%20in%20Mathematical%20Form.htm :

General relativity yields the special theory of relativity as an approximation consistent with the Principle of Equivalence. If we focus our attention on a small enough region of spacetime, that region of spacetime can be considered to have no curvature and hence no gravity. Although we cannot transform away the gravitational field globally, we can get closer and closer to an ideal inertial reference frame if we make the laboratory become smaller and smaller in spacetime volume. In a freely falling (non-rotating) laboratory occupying a small region of spacetime, the laws of physics are the laws of special relativity. Hence all special relativity equations can be expected to work in this small segment of spacetime.

The http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html are equations of special relativity for constant acceleration.

You do not need to use noninertial frames to derive the relativistic rocket equations--you can analyze the proper time of a clock that's accelerating from the point of view of an inertial frame, for example. In fact, I'm pretty sure that this is exactly how those relativistic rocket equations were derived, with no use of noninertial coordinate systems.

The equivalence principle demands that these equations apply to any uniform gravitational field, whether it’s one the crew of a rocket experiences or one an observer on the planet experiences locally.

The equivalence principle only demands that an inertial observer's observations of a local accelerating rocket be identical to those of an observer freefalling a gravitational field looking at a rocket at rest in that field.

Agreed that nothing actually moves faster than c, as noted in #3.

That's not what I said. Nothing moves faster than c in any inertial frame, but things do move faster than c in non-inertial frames, and there's no reason to consider inertial frames as representing what "actually" happens while non-inertial frames do not.

But apparently the buoy does, as described.

Again, the distinction between "actually" and "apparently" is physically meaningless, it's just a question of which type of coordinate system you use to analyze the problem.

The equivalence principle does not apply to only inertial frames. The experimenter trying to “distinguish between gravitational and accelerational forces” can be the crew or an observer on a planet.

This is only true to the extent that you can derive this from the fact that the freefall observer must see the same thing locally as the inertial SR observer. For example, if the inertial observer sees a uniformly accelerating experimenter pass by him making measurements, and a freefalling observer passes by an experimenter at rest in a gravitational field making the same type of measurements, the equivalence between the freefall observer's observations and the inertial observer's observations implies that they must see the two experimenters getting the same results on their measurements (note that there is no need to consider non-inertial coordinate systems in seeing why this is true). The fundamental definition of the equivalence principle is always about freefalling observers in curved spacetime vs. inertial ones in flat spacetime, statements about the equivalence of accelerating observers and observers at rest in gravitational fields are just secondary consequences of the basic definition.

 

[Zanket]

And I take it from your comment you’re not willing to do the math.

The math has already been done, for #3 at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html and, for #4, numerous sources about the cosmological horizon. It would be silly to duplicate that here.

 

[JesseM]

The math has already been done, for #3 at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

There is nothing in the mathematics of the relativistic rocket that tells you how external objects would behave from the point of view of a non-inertial coordinate system where the rocket is at rest, so it is not relevant to your #3.

 

[pervect]

The workaround to this issue of measurements being skewed within an accelerated frame is to have the observer fall through a trap door when the object thrown upward reaches its highest point. The spacetime of the experiment is flat (negligibly curved) because we let the tidal force throughout the region enclosing the experiment be negligible (including below the observer, who can be on top of a tower if necessary), and the now free-falling observer can do good measurements and measures the speed of light to be c throughout the region. The observer and the object will remain at rest with respect to each other during the measurement, just like the crew remains at rest with respect to the buoy when they reach Andromeda and shut off their engine. Falling through the trap door is analogous to shutting off the engine.

Let's go back to your example

Example: Let the rocket travel from Earth to Andromeda, two million light years away as we measure. (Assume that Earth and Andromeda are at rest with respect to each other and the spacetime between them is flat.) Let a buoy float stationary at the halfway point. Let the half of the trip from the buoy take ten proper years as the crew measures.

If we shove an observer off the ship at the halfway point, right at the buoy, this observer will measure his velocity relative to the buoy to be some velocity less than 'c'. Let us call this velocity v1. The marooned observer will also measure the velocity of the Earth as v1.

v1 would be my personal pick as the best velocity to use to describe the velocity between the spaceship and the Earth, BTW.

Meanwhile, an observer onboard the ship right at the halfway point will measure his velocity relative to the buoy as v1, and his velocity relative to the marooned observer as zero.

So far so good. All velocities are less than 'c'. But we have not yet talked about the "velocity" of the ship relative to the Earth. The shipboard obsever will have to define a coordinate system before he can measure the coordinate velocity of the Earth.

There is more than one logical way to choose coordinates. A popular choice is to define the instantaneous "distance" of the Earth as that distance measured by an instantaneously co-moving observer i.e. the distance measured by an observer dropped through a trap door.

You appear to be incorrectly viewing these coordinate distances (and velocities) as having the same physical properties that distances (and velocities) have in an inertial frame. This appears to be the root of most of your problems. These coordinates are NOT inertial coordinates, and must not be treated as such. Note that while any single coordinate distance is measured in an inertial frame (some particular trap door frame), the choice of frame varies as a function of time, and thus the distances are not measured consistently in a single inertial frame.

It might be instructive to stop the hand-waving, and work out what happens to coordinate distance for accelerating spaceship, using the formulas at

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

It is easiest to work out the numbers "before turnaround". We can then write, in geometric units, the coordinates of the spaceship in the Earth's coordinate system, and use the Lorentz distance contraction formula to find the apparent distance of the Earth in an inertial frame co-moving with the ship. Doing this we get.

xship = (cosh(g*tau) - 1)/g
vship = tanh(g*tau)
coord-dist = xship*sqrt(1-vship^2) = cosh(g*tau) - 1) / g*cosh(g*tau)

The last formula is, again, the result of Lorentz length - contraction of the distance between the spaceship and the Earth.

We can see from the above that the coord-dist approaches the constant value 1/g, and that the coordinate velocity of the Earth, the rate of change of coord-dist with proper time tau, becomes very low.

It is tricker to work out what happens when the ship starts deaccelerating, but it can be done. (I'm not going to post the formulae at this point). This is where all the numbers greater than 'c' appear. The trick to make the published formulas applicable to the stopping ship is to measure the distance from the ship to its stopping point, and the time as the time until stop. The sci.physics.faq formulas which assume that d=v=0 when t=0 then apply directly.

 

[Zanket]

I don't think that's correct, every detailed definition of the equivalence principle I have seen says that it only applies in the limit of an arbitrarily small neighborhood of spacetime. There may be other issues besides tidal forces that prevent larger regions from being equivalent to flat spacetime. Can you find any source that defines "local" in the equivalence principle as you do above?

How small is “arbitrarily small”? If I give a different specific amount than you do, who is right? How small is “arbitrarily small” is an opinion. Then it’s clear that these definitions are not precise enough. To be unambiguous they need to say how you can determine how small the region must be. Like:

From this site:
We see that gravity is different than other forces. You can make gravity completely disappear in small regions by freely falling. This means that a free fall frame is a perfectly good inertial frame. The only way we can detect the difference is to look for tidal forces which arise if the gravitational field is not perfectly uniform. But for any real gravitational field we can always make the region we consider (our elevator in this case) small enough so we cannot detect the tidal forces.

Now we know how small. The region must be small enough that the tidal force throughout the region is negligible (a better term than “undetectable”). Such a region can be of any size. Then “small” is misleading, and it is better to say that the region must be “of a range throughout which the tidal force is negligible”, which sources call “local”.

There is only one property that determines whether the spacetime of a region of any size is flat, and that is the tidal force. The tidal force and the curvature of spacetime are exactly the same thing expressed in different languages, says Thorne in Black Holes and Time Warps.

When the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html say “we assume that the stars are essentially at rest in [the frame in which the crew started accelerating]”, they are saying that they assume that the spacetime throughout the rocket’s path is flat, which is to say that the tidal force throughout the region is negligible, no matter how large the region is. In other words the rocket travels through a local frame. This flat spacetime, this uniform gravitational field, this local frame of any size, is of course a prerequisite of the equivalence principle in order for equations of special relativity to apply to the whole region.

No, it doesn't. Any more detailed definition will specify that it only applies to inertial frames.

You’re reading something into those quotes that isn’t there. One of the quotes says, “Hence all special relativity equations can be expected to work in this small segment of spacetime.” (Boldface mine.) That includes the relativistic rocket equations, equations of special relativity for constant noninertial acceleration. More on this below.

That's not what I said. Nothing moves faster than c in any inertial frame, but things do move faster than c in non-inertial frames, and there's no reason to consider inertial frames as representing what "actually" happens while non-inertial frames do not.
...
Again, the distinction between "actually" and "apparently" is physically meaningless, it's just a question of which type of coordinate system you use to analyze the problem.

I don’t see any objection here, to what is in the original post. More on this below.

This is only true to the extent that you can derive this from the fact that the freefall observer must see the same thing locally as the inertial SR observer. For example, if the inertial observer sees a uniformly accelerating experimenter pass by him making measurements, and a freefalling observer passes by an experimenter at rest in a gravitational field making the same type of measurements, the equivalence between the freefall observer's observations and the inertial observer's observations implies that they must see the two experimenters getting the same results on their measurements (note that there is no need to consider non-inertial coordinate systems in seeing why this is true). The fundamental definition of the equivalence principle is always about freefalling observers in curved spacetime vs. inertial ones in flat spacetime, statements about the equivalence of accelerating observers and observers at rest in gravitational fields are just secondary consequences of the basic definition.

Now I’m really confused as to what your objection is. It seems that your objection was that #3, what the crew observes, does not apply to #4, what an observer on a planet observes, because their frames are noninertial and you think the equivalence principle applies only to inertial observers. But you acknowledge above that the two noninertial experimenters must get the same results on their measurements according to the equivalence principle. It doesn’t matter whether the principle explicitly says this or whether these are “secondary consequences”. All that matters is that those measurements are observations that must match. Then #3 must apply to #4. So what is your objection that would show that there is no flaw of general relativity?

 

[Zanket]

There is nothing in the mathematics of the relativistic rocket that tells you how external objects would behave from the point of view of a non-inertial coordinate system where the rocket is at rest, so it is not relevant to your #3.

The crew can shut down their engine at Andromeda, and then do the measuring. The relativistic rocket equations predict the measurements. It is given in #2 that they come to rest at Andromeda, so with their engine off there they’d remain at rest with respect to the buoy.

Likewise the observer on the planet can be on top of a tower, and fall through a trap door, analogous to shutting off the engine. Then this observer can do a good measurement relative to a buoy previously thrown upward at the same velocity at which the crew passed the buoy. All else being equal with #3, if the observer waits ten proper years before falling through the trap door, the buoy will be one million proper light years away, as predicted by the relativistic rocket equations. The buoy will remain at rest with respect to the observer as long as the gravitational field remains uniform.

 

[JesseM]

How small is “arbitrarily small”? If I give a different specific amount than you do, who is right?

Arbitrarily small means we are taking a limit as the size of the spacetime region approaches zero. For any size larger than that there will be some finite difference between the observations in curved spacetime and the observations in flat spacetime, but for any threshold of how small you want this difference to be, you can make the size of the region small enough that the difference will be below this threshold.

How small is “arbitrarily small” is an opinion. Then it’s clear that these definitions are not precise enough.

Once you understand that arbitrarily small is a statement about a limit, it becomes precise.

Now we know how small. The region must be small enough that the tidal force throughout the region is negligible (a better term than “undetectable”). Such a region can be of any size. Then “small” is misleading, and it is better to say that the region must be “of a range throughout which the tidal force is negligible”, which sources call “local”.

Again, I'm not sure tidal forces are the only issue--for example, in a sizeable region of curved space there would be the issue of the sum of the angles of a triangle not equalling 180. Perhaps there would be a way of seeing this as a special case of "tidal forces" but I'm not sure.

There is only one property that determines whether the spacetime of a region of any size is flat, and that is the tidal force. The tidal force and the curvature of spacetime are exactly the same thing expressed in different languages, says Thorne in Black Holes and Time Warps.

OK, then it may be that all effects of curved spacetime can be understood in terms of tidal forces. Do you know the page number where Thorne explains this?

When the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html mentions that the twin paradox in flat spacetime can be understood from the traveling twin's point of view if you introduce a uniform gravitational field at the moment of acceleration, but I'm pretty sure the curvature of spacetime is independent of your coordinate system, so if it's flat in the earth-twin's inertial coordinate system it must be flat in the traveling twin's non-inertial coordinate system too, despite the uniform gravitational field seen in this system). The equivalence principle defines an equivalence between local observations in a gravitational field which involves curved spacetime and identical observations made in flat spacetime.

You’re reading something into those quotes that isn’t there.

Care to give your own interpretation of why the first quote refers to "unaccelerated Cartesian coordinate systems" and the second refers to getting "closer and closer to an ideal inertial reference frame"? Those statements seems to be pretty unambiguously saying that the equivalence principle involves an equivalence between local observations of a freefalling observer and observations made by inertial observers in flat spacetime.

For slightly more authoritative sources, I looked up the definitions of the equivalence principle in two of the most popular GR textbooks, Gravitation by Misner, Thorne and Wheeler, and Gravitation and Cosmology by Weinberg. Here's what Gravitation says (p. 386):

in any and every local Lorentz frame, anywhere and anytime in the universe, all the (nongravitational) laws of physics must take on their familiar special-relativistic forms. Equivalently: there is no way, by experiments confined to infinitesimally small regions of spacetime, to distinguish one local Lorentz frame in one region of spacetime from any other local Lorentz frame in the same or any other region.

The technical definition of a "local Lorentz frame" is given on p. 207, but it basically means a coordinate system that is as close as possible to an inertial frame of SR in the neighborhood of a given event--as they put it "A local lorentz frame at a given event p0 is the closest thing there is to a global Lorentz frame at that event". Also note the statement in that quote about "experiments confined to infinitesimally small regions of spacetime", which is another way of saying you're looking at the limit as the size of your region approaches zero.

As for Weinberg's book, it defines the equivalence principle this way (p. 68):

Therefore we formulate the equivalence principle as the statement that at every space-time point in an arbitrary gravitational field it is possible to choose a "locally inertial coordinate system" such that, within a sufficiently small region of the point in question, the laws of nature take the same form as in unaccelerated Cartesian coordinate systems in the absence of gravitation.

So again, note that they are clearly talking about an equivalence between a "locally inertial coordinate system" in a gravitational field and the laws of nature as seen in an unaccelerated coordinate system in flat spacetime, ie an inertial coordinate system. Again, this is what the equivalence principle is fundamentally about, although you can derive the equivalence between local observations of an accelerating observer in flat spacetime and an observer at rest in a gravitational field from this basic statement of the equivalence principle.

One of the quotes says, “Hence all special relativity equations can be expected to work in this small segment of spacetime.” (Boldface mine.) That includes the relativistic rocket equations, equations of special relativity for constant noninertial acceleration.

...equations which can be understood perfectly well from the point of view of an inertial coordinate system, and which can also apply locally if you let the time-intervals become arbitarily small. So, I don't see why you think this would contradict my statement that the equivalence principle is fundamentally drawing an equivalence between local observations made by 1) freefalling observers in a gravitational field and 2) inertial observers in flat spacetime.

That's not what I said. Nothing moves faster than c in any inertial frame, but things do move faster than c in non-inertial frames, and there's no reason to consider inertial frames as representing what "actually" happens while non-inertial frames do not.
...
Again, the distinction between "actually" and "apparently" is physically meaningless, it's just a question of which type of coordinate system you use to analyze the problem.

I don’t see any objection here, to what is in the original post. More on this below.

It wasn't meant to be, I was just objecting to your statements that "nothing actually moves faster than c" and "But apparently the buoy does" in your last post to me--the distinction you were drawing between what "actually" happens vs. what "apparently" happens seems meaningless to me.

This is only true to the extent that you can derive this from the fact that the freefall observer must see the same thing locally as the inertial SR observer. For example, if the inertial observer sees a uniformly accelerating experimenter pass by him making measurements, and a freefalling observer passes by an experimenter at rest in a gravitational field making the same type of measurements, the equivalence between the freefall observer's observations and the inertial observer's observations implies that they must see the two experimenters getting the same results on their measurements (note that there is no need to consider non-inertial coordinate systems in seeing why this is true). The fundamental definition of the equivalence principle is always about freefalling observers in curved spacetime vs. inertial ones in flat spacetime, statements about the equivalence of accelerating observers and observers at rest in gravitational fields are just secondary consequences of the basic definition.

Now I’m really confused as to what your objection is. It seems that your objection was that #3, what the crew observes, does not apply to #4, what an observer on a planet observes, because their frames are noninertial and you think the equivalence principle applies only to inertial observers. But you acknowledge above that the two noninertial experimenters must get the same results on their measurements according to the equivalence principle.

I objected to your vague handwavey statement about what the crew "observes". If you can specify precisely how the accelerating-in-flat-spacetime/at-rest-in-a-gravitational-field observers can set up a physical measuring apparatus to assign coordinates to local events, and show that the accelerating-in-flat-spacetime will find that a local buoy is traveling faster than c according to the coordinates assigned by this device, then I would say that the at-rest-in-a-gravitational-field observer must find precisely the same results according to the equivalence principle. One major problem is that you seem to want to talk about the two observers assigning speeds to distant buoys rather than local ones. But if there is a way of constructing a coordinate system such that even local buoys are measured to move faster than c by the accelerating rocket, then the rocket at rest in a gravitational field would indeed have to measure the same thing.

 

[Zanket]

You appear to be incorrectly viewing these coordinate distances (and velocities) as having the same physical properties that distances (and velocities) have in an inertial frame. This appears to be the root of most of your problems.

This is a non-issue. #3 talks about the distance the buoy recedes (one million proper light years) during time on the crew’s clock (ten proper years). The crew can always do a good measurement of their own clock. As I previously pointed out, they can do a good measurement of the distance to the buoy by shutting down their engine at Andromeda. Then they will be in an inertial frame shared with the buoy, and remain at rest with respect to the buoy. Likewise the observer on the planet can use the trap door to analogously shut down the engine.

To repeat: The observers in either #3 or #4 can measure the final distance in an inertial frame.

It might be instructive to stop the hand-waving, and work out what happens to coordinate distance for accelerating spaceship, using the formulas at

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

I included this link in the original post. We don’t need to do the calculations. The link already includes the result for a trip from Earth to Andromeda like that described in #2. According to the site, such trip takes 28 proper years at an acceleration of 1 Earth gravity. Obviously with a different acceleration it could take 10 proper years to get from the buoy to Andromeda, like in #3. Rather than do a calculation, all you need do is agree that the crew can make the trip from the buoy to Andromeda in an arbitrarily short proper time. (This is not hand-waving; it’s logic.) No matter how long they take on their clock, the buoy will be one million proper light years away when they come to rest at Andromeda, because that distance was a given. They can shut down their engine there and then confirm the distance.

 

[JesseM]

This is a non-issue. #3 talks about the distance the buoy recedes (one million proper light years) during time on the crew’s clock (ten proper years).

And what is this observation supposed to be equivalent to in curved spacetime? If you look at a situation involving a rocket at rest in a gravitational field while a buoy freefalls away from it, then if the buoy moves light-years away, it will experience significant changes in the gravitational field, unless you are imagining that the source of the field is getting arbitrarily large so even objects light-years apart will experience an arbitrarily small difference in gravity. I think this would have basically the same effect as letting the distance that the buoy moves away approach zero in a normal gravitational field...either way the ratio between the distance the buoy moves away from the rocket and the rocket's distance from the center of the planet is approaching zero, and the gravitational field is approaching being perfectly uniform in the region encompassing both the rocket and the buoy, and spacetime is approaching perfectly flat in this region.

If it helps, I think there probably would be an equivalence between 1) a rocket accelerating between two inertial buoys in flat spacetime with no gravitational field and 2) a rocket at rest in a perfectly uniform gravitational field in flat spacetime, with two buoys freefalling past it. But this equivalence doesn't lead to any paradoxes, as far as I can see.

 

[Physics Monkey]

In other words the rocket travels through a local frame. This flat spacetime, this uniform gravitational field, this local frame of any size, is of course a prerequisite of the equivalence principle in order for equations of special relativity to apply to the whole region.

Hi Zanket,

I think you may be a bit confused here because the coordinate system of an accelerated observer cannot be as large as you like. It is limited in size by the magnitude of the acceleration so that the coordinate system breaks down for distances greater than 1/g. I haven't followed the thread closely, but you seem to be using the accelerated frame to draw conclusions about distant objects, a procedure that may not in fact be valid. See for example the excellent discussion in MTW.

 

[Zanket]

I think you may be a bit confused here because the coordinate system of an accelerated observer cannot be as large as you like.

See post 79 and post 81. The coordinate system of an accelerated observer is immaterial here, because the observers (crew and observer on the planet) can measure the distance to the buoy (either passed in the case of the crew, or thrown up in the case of the observer on the planet) in an inertial frame. The distance they would measure is predicted by the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html for any given acceleration a and proper time T.

 

[pervect]

This is a non-issue. #3 talks about the distance the buoy recedes (one million proper light years) during time on the crew’s clock (ten proper years). The crew can always do a good measurement of their own clock. As I previously pointed out, they can do a good measurement of the distance to the buoy by shutting down their engine at Andromeda. Then they will be in an inertial frame shared with the buoy, and remain at rest with respect to the buoy. Likewise the observer on the planet can use the trap door to analogously shut down the engine.

To repeat: The observers in either #3 or #4 can measure the final distance in an inertial frame.

Measuring the distance in an inertial frame is not sufficient to set up an inertial coordinate system unless one always uses the same inertial frame.

This should be quite obvious, but even if it isn't obvious, if you'd do some mathematical analysis or draw some space-time diagrams or read a book on the topic (as Physics Monkey mentions, MTW's treatment is very good, though it does require 4-vectors and the mathematical treatment also uses tensor notation), or listen to several of the people including me who've talked to you on this board about the topic you should be able to understand why this is an incorrect assumption. The fact that making this assumption leads to inconsistencies should be a big tip-off.

Unfortunately it seems that you've done none of these things, and instead chose to blame your own error on "general relativity".

I think I've said about as much as I can say on the topic, so I'm going to take a break from this thread.

 

[ZapperZ]

... and on that note, I believe that we have been MORE than generous to let this thread run THIS long. Considering that this should have been done in the IR forum in the first place, there has been more than enough posting on here.

Zanket: please do not test out various parts of your IR-submitted theory on here. While we did allow this to run its course, if you try this again, it will be deleted without notice.

Zz.