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Mark_Friedenbach comments on What are your contrarian views? - Less Wrong Discussion

10 Post author: Metus 15 September 2014 09:17AM

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Comment author: [deleted] 15 September 2014 04:16:37PM 3 points [-]

the Relativity isn't coherent either

Care to explain this one?

Comment author: Thomas 15 September 2014 04:28:16PM *  1 point [-]

Yes.

A big pie, rotating in the sky, should have apparently shorter circumference than a non-rotating one, and both with the same radii.

I can't swallow this. Not because it is weird, but because it is inconsistent.

Comment author: The_Duck 15 September 2014 11:13:16PM 6 points [-]

A big pie, rotating in the sky, should have apparently shorter circumference than a non-rotating one, and both with the same radii.

I can't swallow this. Not because it is weird, but because it is inconsistent.

There is no inconsistency. In one case you are measuring the circumference with moving rulers, while in the other case you are measuring the circumference with stationary rulers. It's not inconsistent for these two different measurements to give different results.

Comment author: Thomas 15 September 2014 11:27:50PM 0 points [-]

No. I am measuring from here, from the centre with some measuring device.

First I measure stationary pie, then I measure the rotating one. Those red-white stripes are either constant, either they shrink. If they are shrinking, they should multiply as well. If they are not shrinking, what happened with Mr. Lorentz's contraction?

Comment author: Jiro 16 September 2014 02:02:57AM 1 point [-]

If you measure a wheel with a ruler, and the wheel is moving relative to the ruler, then your measurement assumes that both ends of a piece of the wheel line up with both ends of a piece of the ruler at the same time. Whether these events happen at the same time, and therefore whether this is a measurement of the wheel, are different depending on the frame of reference.

Comment author: Alejandro1 15 September 2014 05:40:34PM 2 points [-]

Why is it inconsistent?

Comment author: shminux 15 September 2014 06:28:33PM *  5 points [-]

Special Relativity + some basic mechanics leads to an apparent contradiction in the expected measurements, which is only resolved by introducing a curved space(time). So this would be a failure of self-consistency: the same theory leads to two different results for the same experiment.

However, the two measurements of ostensibly the same thing are done by different observers, so there is no requirement that they should agree. Introducing curved space for the rotating disk shows how to calculate distances consistently.

Comment author: [deleted] 17 September 2014 04:52:14PM 1 point [-]

which is only resolved by introducing a curved space(time).

The spacetime itself is flat (if the mass of the pie is negligible), but the spacelike slices are curved because you're slicing it in a weird way.

Comment author: DanielLC 15 September 2014 10:27:44PM 1 point [-]

The problem is that it's inconsistent with solid-body physics?

Solid-body physics is an approximation. This isn't hard to show. Just bend something.

Consider the model of masses connected by springs. This is consistent with special relativity, and can be used to model solid-body physics. In fact, it's a more accurate model of reality than solid-body physics.

Comment author: shminux 15 September 2014 11:03:01PM *  1 point [-]

No, that's not the issue. The problem is that no flat-space configuration works.

Comment author: Thomas 15 September 2014 06:21:43PM 1 point [-]

I have two photos of two different pies, one of rotating one and one of not rotating. Photos are indistinguishable, I can't tell which is which.

On the other hand, both pies have one-to-one correspondence with photos an one should be slightly deformed on the edge.

Even if it is, on the photo can't be. The photo is perfectly Euclidean. I have measured no Lorentz contraction.

Comment author: DanielLC 15 September 2014 10:41:37PM 4 points [-]

Just to clarify, is the spinning pie a set of particles in the same relative position as with a still pie, but rotating around the origin? Is it a set of masses connected by springs that has reached equilibrium (none of the springs are stretching or compressing) and the whole system is spinning? Is the pie a solid body?

What exactly we're looking at depends on which of the first two you picked. If you picked the third, it is contradictory with special relativity, but there's a lot more evidence for special relativity than there is for the existence of a solid body. Granted, a sufficiently rigid body will still be inconsistent with special relativity, but all that means is that there's a maximum possible rigidity. Large objects are held together by photons, so we wouldn't expect sound to travel through them faster than light.

Comment author: Thomas 15 September 2014 11:18:59PM 0 points [-]

The spinning set of particles is a toroidal with let say 1 million light years across - the big R. and with the small r of just 1 centimetre. It is painted red and white, differently each metre.

The whole composition starts to slowly rotate on the signal from the centre. And slowly, very slowly accelerate to reach the speed of 0.1 c in a several million years.

Now, do we see any Lorentzian contraction due to the SR, or not due to the GR?

(Small rockets powered by radioactive decay are more than enough to compensate for the acceleration and for the centrifugal force. Both incredibly small. This is the reason why we have choose such a big scale.)

Comment author: DanielLC 15 September 2014 11:34:40PM 2 points [-]

I'm going to assume mass is small enough not to take GR into effect.

From the point of view of a particle on the toroid, the band it's in will extend to about 1.005 meters long. Due to Lorentz contraction, from the point of reference of someone in the center, it will appear one meter long.

Comment author: Thomas 15 September 2014 11:39:59PM 0 points [-]

The question is ONLY for the central observer. At first he sees 1 m long stripes, but when the whole thing reaches the speed of 0.1 c, how long is each stripe?

Comment author: DanielLC 15 September 2014 11:54:49PM 3 points [-]

One meter.

I just want to clarify. I'm assuming the particles are not connected, or are elastic enough that stretching them by a factor of 1.005 isn't a huge deal. If you tried that with solid glass, it would probably shatter.

Come to think of it, this looks like a more complicated form of Bell's spaceship paradox.

Comment author: [deleted] 17 September 2014 04:57:08PM 1 point [-]

I think you're right, but you're interpreting “sees” literally I'm not 100% sure of that, because of light aberration (the Terrell-Penrose effect).

Comment author: gjm 15 September 2014 08:05:19PM 3 points [-]

In other news, the earth is really flat because photographs of the earth are flat.

Comment author: Slider 15 September 2014 07:41:24PM 1 point [-]

Place red and white equilength rulers on the edge of the cylinder. The rotating cylinder will have more and shorther rulers. Thus the photos are not the same. Even better have the cylinder slowly pulse in different colors. The edges will pulse more slowly thus not being in synch with the center.

Related phenomenon is that moving ladders fit into garages that stationary ones would not.

Comment author: Thomas 15 September 2014 08:52:26PM 0 points [-]

Place red and white equilength rulers on the edge of the cylinder. The rotating cylinder will have more and shorther rulers.

They will multiply as the orbital speed increases? Say that Arab numerals are written on the rulers. Say that they are 77 at the beginning. Will this system know when to engage the number 78?

Or will there be two 57 at first? Or how is it going to be?

Comment author: Slider 15 September 2014 09:19:47PM 3 points [-]

I was thinking of already spun cylinder and then adding the sticks by accelerating them to place.

If you had the same sticks already in place the stick would feel a stretch. If they resist this stretch they will pull apart so there will be bigger gaps between them. For separate measuring sticks they have no tensile strenght in the gaps between them. However if you had a measuring rope with continous tensile strenght and at a beginning / end point where the start would be fixed but new rope could freely be pulled from the end point you would see the numbers increase (much like waist measurements when getting fatter). However the purpoted cylinder has maximum tensile strenght anywhere continously. Thus that strenght would actually work against the rotating force making it resist rotation. a non-rigid body will rupture and start to look like a star.

So no there would not be duplicate sticks but yes the rope would know to engage number 78.

If you would fill up a rotating cylinder with sticks and spin it down the stick would press against each other crushing to a smaller lenght. A measuring rope with a small pull to accept loose rope would reel in. A non-rigid body slowing down would spit-out material in bursts that might come resemble volcanoes.

Comment author: Jiro 15 September 2014 07:46:50PM *  0 points [-]

Saying that a moving ladder "fits" means that the start of the ladder is in the garage at the same time that the end of the ladder is. If the ladder is moving and contracted because of relativity, these two events are not simultaneous in all reference frames. Thus, you cannot definitely say that the moving ladder fits--whether it fits depends on your reference frame. (In another reference frame you would see the ladder longer than the garage, but you would also see the start of the ladder pass out of the garage before the end of the ladder passes into it.)

Comment author: Slider 15 September 2014 08:07:39PM *  1 point [-]

Why have that definition of "fit"? I could eqaully well say that fitting means that there is a reference frame that has a time where the ladder is completely inside.

If you had the carage loop back so that the end would be glued to the start you could still spin the ladder inside it. From the point of the ladder it would appear to need to pass the garage multiple times to oene fit ladder lenght but from the outside it would appear as if the ladder fits within one loop completely. With either perspective the one garage space enough to contain the ladder without collisions. In this way it most definetly fits. Usually garages are thought to be space-limited but not time limited. Thus the eating of the time-dimension is a perfectly valid way of staying within the spatial limits.

edit: actually there is a godo reazson to priviledge the rest frame oft he garage as the one that count as ragardst to fitting as then all of the fitting happens within its space and time.

Comment author: DanielLC 15 September 2014 10:49:03PM 1 point [-]

Why have that definition of "fit"? I could eqaully well say that fitting means that there is a reference frame that has a time where the ladder is completely inside.

In that case, the ladder fits.

From the point of the ladder

Each rung of the ladder has a distinct reference frame. "From the point of the ladder" is meaningless.

Comment author: Slider 16 September 2014 09:03:28AM 0 points [-]

If the ladder point of view is ildefined so is the garage point of view as the front and back of the garage have distinct reference frames. Any inertial reference frame is equally good. The ladder is not accelerating thus inertial. In the sense that we can talk of any frame as more than a single event or world line the ladder frame is perfectly good.

Comment author: DanielLC 16 September 2014 03:54:30PM 2 points [-]

In the normal example, where the ladder is straight and moving forward, it has only one reference frame. Strictly speaking, each rung has a different reference frame, but they differ only by translation.

From what I understand, you modified it to a circular ladder spinning in a circular garage. In this case, each rung is moving in a different direction, and therefore at a different velocity. Thus, each rung has its own reference frame.

Comment author: Alejandro1 15 September 2014 07:41:56PM 0 points [-]

If the rotating pie is a pie that when nonrotating had the same radius as the other one, when it rotates it has a slightly larger radius (and circumference) because of centrifugal forces. This effect completely dominates over any relativistic one.

Comment author: Thomas 15 September 2014 08:14:56PM 0 points [-]

The centrifugal force can be arbitrary small. Say that we have only the outer rim of the pie, but as large as a galaxy. The centrifugal force at the half of speed of light is just negligible. Far less than all the everyday centrifugal forces we deal with.

Now say, that the rim has a zero around velocity at first and we are watching it from the centrer. Gradually, say in a million years time, it accelerates to a relativistic speed. The forces associated are a millionth of Newton per kilogram of mass. No big deal.

The problem is only this - where's the Lorentz contraction?

As long as we have only one spaceship orbiting the Galaxy, we can imagine this Lorentzian shrinking. In the case of that many, that they are all around, we can't.

Comment author: DanielLC 15 September 2014 10:35:01PM 2 points [-]

If you have a large number of spaceships, each will notice the spaceship in front of it getting closer, and the circle of spaceships forming into an ellipse.

At least, that's assuming the spaceships have some kind of tachyon sensor to see where all the other ships are from the point of reference of the ship looking, or something like that. If they're using light to tell where all of the other ships are, then there's a few optical effects that will appear.

Comment author: Thomas 15 September 2014 10:59:27PM 0 points [-]

The question is what the stationary observer from the centre sees? When the galactic carousel goes around him. With the speed even quite moderate, for the observer has precise instruments to measure the Lorentzian contraction, if there is any.

At first, there is none, because the carousel isn't moving. But slowly, in many million years when it accelerate to say 0.1 c, what does the central observes sees? Contraction or no contraction?

Comment author: DanielLC 15 September 2014 11:26:53PM 3 points [-]

He will see each spaceship contract. The distance between the centers of the spaceships will remain the same.

Comment author: Thomas 15 September 2014 11:32:28PM 0 points [-]

But no, those ships are just like those French TGV's. A whole composition of cars and you can't say where one ends and another begins.

It's like a snake, eating its tail!

Comment author: [deleted] 16 September 2014 07:45:39AM 1 point [-]

The problem is only this - where's the Lorentz contraction?

Each piece of the ring is longer as measured by an inertial observer comoving with it than as measured by a stationary one (i.e. one comoving with the centre of the ring). But note that there's no inertial observer that's comoving with all pieces of the ring at the same time, and if you add the length of each piece as measured by an observer comoving with it what you're measuring is not a closed curve, it's a helix in spacetime. (I will draw a diagram when I have time if I remember to.)

Comment author: Thomas 16 September 2014 07:50:23AM 0 points [-]

The inertial observer in the centre of the carousel measures those torus segments when they are stationary.

Then, after a million years of a small acceleration of the torus and NOT the central observer, the observer should see segments contracted.

Right?

Comment author: [deleted] 16 September 2014 12:19:37PM *  2 points [-]

During the million years of small acceleration, the torus will have to stretch (i.e. each atom's distance from its neighbours, as measured in its own instantaneous inertial frame will increase) and/or break.

Specifying that you do it very slowly doesn't change anything -- suppose you and I are holding the two ends of a rope on the Arctic Circle, and we go south to the Equator each along a meridian; in order for us to do that, the rope will have to stretch or break even if we walk one millimetre per century.

Comment author: Thomas 16 September 2014 12:42:55PM -1 points [-]

I don't see any reason this very big torus should break.

Forces are really tiny, for R is 10^21 m and velocity is about 10^8 m/s. That gives you 10^-5 N per kg of centrifugal force. Which can be counterbalanced by a small (radioactive) rocket or something on every meter.

Almost any other relativistic device from literature would easily break long before this one.

If breaking was a problem.

Comment author: [deleted] 15 September 2014 09:39:30PM *  -1 points [-]

There's a reason it's called special relativity. It only works in special cases. Eucludian geometry and Newtonian mechanics are inconsistent, btw. Special relativity solves these inconsistencies in the special contexts where they originally came up (predicting the Lorentz contraction and time dilation which is experimentally observed). It wasn't until the curved space of general relativity was discovered that we had a fully consistent model.

And yes, curved space of general relativity fully explains the rotating disc in a way that is self-consistent in in agreement with observed results (as proven by Gravity Probe B, among other things).

Comment author: DanielLC 15 September 2014 10:52:30PM 3 points [-]

Special relativity is consistent. It just isn't completely accurate.

It's inconsistent with solid-body physics, but that's due to the oversimplifications inherent in solid-body physics, not the ones inherent in special relativity.

Trying to fit solid-body physics into general relativity is even worse. With special relativity, it works fine as long as it doesn't rotate or accelerate. Under general relativity, it can only exist on flat space-time, which basically means that nothing in the universe can have any mass whatsoever, including the object in question.

Comment author: [deleted] 16 September 2014 12:34:49AM -2 points [-]

Twin paradox.

Comment author: DanielLC 16 September 2014 01:03:43AM 1 point [-]

What about the twin paradox?

Comment author: Azathoth123 16 September 2014 12:35:20AM 1 point [-]

You need GR if you want to treat talk about the rotating reference frame of the disk. Otherwise SR is fine.

Comment author: [deleted] 16 September 2014 07:52:05AM 0 points [-]

“claim[ing] that special relativity can't handle acceleration at all ... is like saying that Cartesian coordinates can't handle circles”

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

But then again, the question whether the study of flat spacetime using non-inertial reference frames counts as SR depends on what you mean by SR. If you mean the limit of GR as G approaches 0, then it totally does.

Comment author: Thomas 15 September 2014 10:06:06PM 0 points [-]

curved space of general relativity fully explains the rotating disc in a way that is self-consistent

Is it any Lorentz contraction visible in the case of around the galaxy rim?

Are all the Lorentzian shrinks just cancelled out?

I'd really like to know that.

Comment author: The_Duck 15 September 2014 11:10:46PM *  0 points [-]

You don't need GR for a rotating disk; you only need GR when there is gravity.

Comment author: [deleted] 16 September 2014 12:34:15AM -1 points [-]

Rotation drags spacetime.

Comment author: Azathoth123 16 September 2014 01:27:54AM 4 points [-]

Only if the rotating object is sufficiently massive.

Comment author: [deleted] 16 September 2014 03:19:15AM 0 points [-]

Only if the rotating object has any mass at all.

Comment author: DanielLC 16 September 2014 04:45:11AM 3 points [-]

For a rotating object of sufficiently small mass, the mass can be ignored, and reasonably accurate results can be found with special relativity.

Comment author: [deleted] 16 September 2014 05:11:15AM *  0 points [-]

I don't disagree. This discussion was philosophical in the pejorative sense, being about absolutely exact results, not reasonable approximations.

Comment author: DanielLC 16 September 2014 05:30:07AM 2 points [-]

The OP was claiming that special relativity was incoherent, not just that it wasn't absolutely exact.

If you want absolutely exact results, you'll need a theory of everything. There are quantum effects messing with spacetime.