Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Le Repteux on 12/10/2018 15:54:44
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NOTE: THIS TOPIC HAS BEEN SPLIT OFF FROM https://www.thenakedscientists.com/forum/index.php?topic=49715.0 BECAUSE THAT TOPIC IS BEING CONFUSED BY DISCUSSION ON CENTRIFUGAL FORCE.
I think any further discussion of rotating frame analysis should be done in a separate thread as it is disjointing the discussion between @rmolnav and @David Cooper
Not if it is rmolnav that is confusing reference frames, and idea that your reference frame concern just gave me.
As I said before, it doesn’t matter if we know we are rotating, it makes no difference; in fact it is essential we know so we can calculate the centrifugal force.
What if we simply called that force rotational? This way, the expression "centrifugal force" could be reserved to a force that adds to the rotational one, what can be done by throwing stones towards the center of a merry go round when we are on its edge for instance. We may also call centrifugal the force applied on the ground by two teams playing tug-of-war for instance. Saying that, I realize that throwing stones towards the moon from the earth's surface would move the earth away a bit, and that rmolnav's centrifugal force doesn't. Anyway, the force we feel in an accelerating car is not centrifugal, and the one we feel on a merry go round is of the same kind, so why call it centrifugal? Just because we feel we are going to be ejected outward if the door opens while the car is turning? Aren't we confusing frames when we think that way? If we know the physics, we know we are not going to move outwards, we know we will simply go on moving in the same direction and speed we were going when the door opened, we know it is the car that will be moving away. If we don't chose the right frame to analyze that situation, we will simply end up with epicycles. If we are in a car that is making circles, and we think that the ejected passenger suffered the same outward force we are feeling, we will think it is making epicycles while moving away. I think that's what rmolnav is doing, except that he knows that the passenger will be moving straight line, so he swithches reference frames in his head in the middle of his explanation. By the way, the same frame switching is used in the main stream explanation of the Twins paradox, while it would be so much clearer to admit that we simply know which Twin has accelerated. We know we are accelerating, so why tergiversate? There is no such thing as a rotating relative reference frame, rotating frames are unfortunately absolute.
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Anyway, the force we feel in an accelerating car is not centrifugal, and the one we feel on a merry go round is of the same kind, so why call it centrifugal? Just because we feel we are going to be ejected outward if the door opens while the car is turning? Aren't we confusing frames when we think that way? If we know the physics, we know we are not going to move outwards, we know we will simply go on moving in the same direction and speed we were going when the door opened, we know it is the car that will be moving away.
... but you don´t actually "know the physics", the SCIENCE called physics. You only "know" what your not too educated mind deduces from what you see, or think to see !!
" ...the force we feel in an accelerating car is "of the same kind" as "the one we feel on a merry go round" ONLY as far as its root cause is concerned, because both are due to INERTIA.
But, as I´ve already said several times, inertia can manifest itself in different ways, frequently as a "tendency" to oppose to any acceleration the considered object may be given ...
In your first case the acceleration given to the car is rectilinear, and the inertial "FORCE" (different than "movement" , terms that you often mix up ...) is with same direction as the car acceleration, but backwards ... NO ROTATION, and NO centrifugal force whatsoever.
But "on a merry go round", the acceleration given to us is "centripetal", towards the axis of rotation ... Therefore, the inherent inertial force is opposite to the one we are being given, that is, in a radial outward direction: centrifugal force !!
And quite another thing is that, if we step off the MGR, we will follow the tangent ... But NOT because the previous inertial force had that direction, as you seem to think ... That is due to the fact that, at the very moment we cease to be on the MGR, centripetal acceleration suddenly disappears, and logically so does that inertial manifestation, the centrifugal force.
And, logically (1st Newton Motion Law), since that moment we move on with the linear speed we had ...
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But "on a merry go round", the acceleration given to us is "centripetal", towards the axis of rotation ... Therefore, the inherent inertial force is opposite to the one we are being given, that is, in a radial outward direction: centrifugal force !!
Resistance to acceleration always opposes the force that produces the acceleration, and it is only accidental that the acceleration is centripetal in the case of rotation. During rotation, nothing pulls outward and nothing moves outward, the acceleration is inward and the force too.
And quite another thing is that, if we step off the MGR, we will follow the tangent ... But NOT because the previous inertial force had that direction, as you seem to think ...
What I said is "we know we will simply go on moving in the same direction and speed we were going when the door opened", and I add that such a tangential motion is perpendicular to the centripetal force, so it cannot interfere with it.
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Quote from: rmolnav on 12/10/2018 19:05:02
But "on a merry go round", the acceleration given to us is "centripetal", towards the axis of rotation ... Therefore, the inherent inertial force is opposite to the one we are being given, that is, in a radial outward direction: centrifugal force !!
Resistance to acceleration always opposes the force that produces the acceleration, and it is only accidental that the acceleration is centripetal in the case of rotation. During rotation, nothing pulls outward and nothing moves outward, the acceleration is inward and the force too.
You not only have basic misconceptions in Dynamics (as a SCIENCE), but also in LOGICS !
If "Resistance to acceleration always opposes the force that produces the acceleration" as you rightly say, during rotation that resistance has to "oppose" centripetal force, "the force that produces the acceleration" (the centripetal acceleration) ... Basic logics !
But then you say: "... it is only accidental that the acceleration is centripetal in the case of rotation".
ONLY ACCIDENTAL ? Did it happen "to be there", just as an "accident", when the considered object was rotating ?
I used to think it is D.C. the only one which finds the concept of "centripetal force" a "grey area" ... But now I can tell your lack of education on Dynamics also reaches that rather unbelievable level !! What I said is "we know we will simply go on moving in the same direction and speed we were going when the door opened", and I add that such a tangential motion is perpendicular to the centripetal force, so it cannot interfere with it.
"Such a tangential motion ... cannot interfere with it (the centripetal force)", but not due to their perpendicularity whatsoever !! They cannot "interfere" with each other because they DON´T EXIST" simultaneously ... The tangential motion "is born" when the passenger gets "free" (when the door opens), and in that very moment the centripetal force (which was previously being exerted on the passenger) "dies" ...
I´ve already said that several times, but it seems you also has D.C.´s habit of discussing what I say without reading my posts with any minimum care and interest ...
And it is not just something I say: there are lot of sites on Dynamics where you can find it is BASIC physics ... Some of them might certainly be not 100% right, but not on such basic fact. Forces need to act on an object some NOT NULL time to change its velocity vector in any way ...
Therefore, as centripetal force (previously acting on the passenger) instantaneously disappears when the door opens, it cannot have any effect on the "tangential" movement whatsoever, and the passenger carries on with the linear speed he had got at that very instant (Newton´s 1st Motion Law) ...
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What if we simply called that force rotational? This way, the expression "centrifugal force" could be reserved to a force that adds to the rotational one, what can be done by throwing stones towards the center of a merry go round when we are on its edge for instance.
Firstly, we have a term ‘centrifugal’ which is defined and pairs well with centripetal, calling it rotational could be confused with torque.
Secondly, throwing stones at the centre of a roundabout doesn’t cause a rotation. Throw them tangentially. That’s why it’s important to get your frames right.
We may also call centrifugal the force applied on the ground by two teams playing tug-of-war for instance.
No, these are just ordinary linear forces and reactions, no need for a special term.
Anyway, the force we feel in an accelerating car is not centrifugal, and the one we feel on a merry go round is of the same kind, so why call it centrifugal? Just because we feel we are going to be ejected outward if the door opens while the car is turning? Aren't we confusing frames when we think that way?
If we know the physics, we know we are not going to move outwards, we know we will simply go on moving in the same direction and speed we were going when the door opened, we know it is the car that will be moving away.
The force we feel in an accelerating car is not centrifugal because centrifugal is reserved for rotating systems. However, the force is still a frame dependant one because the inertial observer doesn’t think it exists, they see the car seat pushing the passenger forward.
As I’ve said before, it doesn’t matter what we ‘know’ it’s where we are measuring from. If we keep the consistency of frames that’s ok, it’s just a convenient way of making some calculations easier.
If we don't chose the right frame to analyze that situation, we will simply end up with epicycles. If we are in a car that is making circles, and we think that the ejected passenger suffered the same outward force we are feeling, we will think it is making epicycles while moving away.
As I’ve said before, we don’t end up with epicycles, they are a construction of a mind that doesn’t realise it is rotating. We do get retrograde motion, which we understand when we know we are rotating and so we don’t need to invent epicycles.
Sometimes it can be very useful to understand the motion relative to the rotating frame so we can analyse the view from that frame and display what we actually see from that frame.
By the way, don’t confuse ordinary centrifugal force with reactive centrifugal force, the latter exists in the inertial frame as well.
There is no such thing as a rotating relative reference frame,
Of course there is, you use them all the time.
Extreme example. Let’s take a vinyl turntable to the geographic pole. Before we switch it on it isn't rotating. Well, of course it is, relative to the fixed stars, but relative to earth it isn’t. Set it rotating in the reverse direction and speed to the earth and it will be rotating relative to earth, but not to the fixed stars. You make these same relative judgements all the time.
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There is no such thing as a rotating relative reference frame,
Of course there is, you use them all the time.
Extreme example. Let’s take a vinyl turntable to the geographic pole. Before we switch it on it isn't rotating. Well, of course it is, relative to the fixed stars, but relative to earth it isn’t. Set it rotating in the reverse direction and speed to the earth and it will be rotating relative to earth, but not to the fixed stars. You make these same relative judgements all the time.
We refer to the stars to measure a faster rotational motion because we think it is a better reference frame, so we know that referring to the galaxies would be better, and to the universe as a whole even better. What that means is that there exists an absolute reference frame somewhere even if we can't localize it. If we rotate a Sagnac interferometer at the pole at the same speed than the turntable, we will measure no rotation. A satellite that is not moving with regard to the stars will fall on the earth even if it is rotating with regard to it. That's hard to admit from a relativistic viewpoint since it may mean that inertial motion is not relative, but rotation is unfortunately not relative. If we are alone in the universe, we can know if we are rotating or not, so it means that the space in which we are is absolute, as it was the case for ether before it was discarded. We won't detect that ether if we are not rotating, but we will if we are. A Sagnac interferometer will give us our exact angular speed with regard to ether providing it is sensible enough for the speed at which we are rotating.
Of course we won't detect our linear motion even if we are actually moving through ether, but it's only because the way light travels between the particles of a MM interferometer prevents us from detecting it. Particles are exchanging light the same way light travels between the mirrors of the interferometer, and they need to stay at the right bonding energy even while moving through ether, so they contract and the interferometer contracts and the observers contract with all their measurement tools so the motion is impossible to detect. If we had a faster than light device, we could measure it, but we don't. On the contrary, we don't need a faster than light device to measure rotation because light makes a roundtrip in the interferometer, the same kind of roundtrip that permits us to know which one of the twins is going to get younger. The only way for this twin to get younger is to travel more absolute distance through ether than his twin has traveled, and as my simulation of the twins paradox shows (http://www.motionsimulations.com/Twins%20paradox), the faster the light clock travels that distance, the slower it tics even if it contracts, and it is so only because light travels more absolute distance between the mirrors.
That roundtrip stuff is probably the reason why resistance to acceleration is observable. In my simulation on acceleration (http://www.motionsimulations.com/Acceleration%20with%20two%20particles), the left particle doesn't accelerate all the time because it has to wait till the photon has made a roundtrip before increasing its speed. Even if the force is still pushing it forward, it doesn't accelerate immediately. If particles would accelerate immediately, we wouldn't feel a force when we accelerate a body. Notice that all my simulations depend on a speed and a direction of light that is constant, and that I need a propagation medium that is at rest with regard to the screen to simulate that. In the case of my twins paradox simulation, I don't have to change reference frames in the middle of the simulation since I know which twin is accelerating. In the case of my simulation on acceleration, I know that the particles are getting speed because I know they are accelerating, and as the display at the left shows, I can also know how much speed they got since the beginning of the acceleration.
We still have a problem though, we don't know at what speed and in what direction the particles were going through ether before being accelerated, which is precisely why the resultant inertial motion is relative, not because there is no ether, or no absolute reference frame. If I need an absolute frame to move my particles, then I can't see how nature could avoid it since it is nature that I'm trying to simulate. Relativity was built out of light and sources of light where the speed and the direction of light was constant, and it is precisely how my simulations work.
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Just commenting on some random inaccuracies and other observations I see in here.
If we rotate a Sagnac interferometer at the pole at the same speed than the turntable, we will measure no rotation.
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That's hard to admit from a relativistic viewpoint since it may mean that inertial motion is not relative, but rotation is unfortunately not relative.
Rotating frames are absolute, yes. Such a device detects absolute rotation, or lack of it, just like Newton's bucket. Nobody claims otherwise, but as Colin2B points out: rotational reference frames are quite useful. Paper maps of cities use a rotating frame of reference. They'd be pretty useless in an inertial or accelerating frame.
Centrifugal force exists in rotating frames. It does not exist in inertial ones. If you do your computations using a rotating frame, you need to know the mathematics that apply. Newton's laws of motion do not apply to rotating frames.
A satellite that is not moving with regard to the stars will fall on the earth even if it is rotating with regard to it.
Actually, Earth will take off like a shot and leave this stationary satellite behind, rotating or not. The satellite will not fall to Earth. This is not 'actual stationary', just stationary relative to some chosen star, all of which have different velocities. Stationary to a selected average is about as good as you can do, and that is still relative to your selection, not an absolute one.
A Sagnac interferometer will give us our exact angular speed
A bit of over-fancy device to do that. Two rocks will suffice nicely.
The only way for this twin to get younger is to travel more absolute distance through ether than his twin has traveled,
Not so. I can get the twin taking the longer path to age more. It just isn't a function of distance travelled, even in an absolute reference system like you describe.
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Not so. I can get the twin taking the longer path to age more. It just isn't a function of distance traveled, even in an absolute reference system like you describe.
Hi Halc,
To me, what you say means that the twin that stays at home could eventually age less than the traveling one, and my simulation shows the contrary, so can you elaborate a bit.
Actually, Earth will take off like a shot and leave this stationary satellite behind, rotating or not. The satellite will not fall to Earth. This is not 'actual stationary', just stationary relative to some chosen star, all of which have different velocities. Stationary to a selected average is about as good as you can do, and that is still relative to your selection, not an absolute one.
There is an orbital speed at which a body will fall directly to the earth, and I think it is quite close to the speed it will have if it stops on the line between the center of the earth and any fixed star. If it's not what you understood, that's what I meant, otherwise I don't understand what you mean again.
Centrifugal force exists in rotating frames. It does not exist in inertial ones. If you do your computations using a rotating frame, you need to know the mathematics that apply. Newton's laws of motion do not apply to rotating frames.
I invite you to take a look at that post (https://www.thenakedscientists.com/forum/index.php?topic=49715.msg556608#msg556608), and to tell me what you think of it.
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Not so. I can get the twin taking the longer path to age more. It just isn't a function of distance traveled, even in an absolute reference system like you describe.
Hi Halc,
To me, what you say means that the twin that stays at home could eventually age less than the traveling one, and my simulation shows the contrary, so can you elaborate a bit.
You said 'longer path', not 'traveling one'. Maybe they're both traveling, especially if 'home' isn't a stationary place. If a set of twins both travel from point A to point B, the one that take the longer path to do it does not necessarily age less. It's just not a function of path length.
There is an orbital speed at which a body will fall directly to the earth, and I think it is quite close to the speed it will have if it stops on the line between the center of the earth and any fixed star.
Again, you said something else. You said 'not moving with regard to the stars', which I took to mean 'not moving relative to the stars'. A satellite on the line you describe will be almost exactly stationary relative to Earth and a very non-stationary relative to your selected star. If that is what you mean by 'with regard to the stars' then yes. An object nearby Earth and stationary relative to it will fall to Earth, much like any rock I drop. You said this is 'hard to admit from a relativistic viewpoint'. I don't see how. Einstein did not deny that rocks drop to the ground if dropped.
I invite you to take a look at that post (https://www.thenakedscientists.com/forum/index.php?topic=49715.msg556608#msg556608), and to tell me what you think of it.
Seems pretty confused. It speaks of tides being stronger at the center of Earth, whatever you mean by that.
The post also mixes reference frames, talking about the equatorial bulge accelerating, an inertial frame reference, and then centrifugal force lower down, which is a rotating frame reference. Pick one or the other to describe your dynamics, but to mix them like that is going to produce nonsense.
The equatorial bulge can be described in inertial terms or in rotational terms, but not both in the same description.
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You said 'longer path', not 'traveling one'. Maybe they're both traveling, especially if 'home' isn't a stationary place. If a set of twins both travel from point A to point B, the one that take the longer path to do it does not necessarily age less. It's just not a function of path length.
I was referring to my simulation of the original Twins paradox mind experiment where only one of the twins travels. If I added to it a third twin that travels in a different direction and at a different speed, then I would get the same result than relativistic calculations.
You said this is 'hard to admit from a relativistic viewpoint'.
What is hard to admit is that rotation is not relative, and you agreed with me on that point, so it means that you are probably not a mainstream thinker.
The equatorial bulge can be described in inertial terms or in rotational terms, but not both in the same description.
That discussion is about whether a centrifugal force is needed to explain the tides. David Cooper thinks it is not, and I agree with him, and Rmolnav thinks it is. So when you introduced centrifugal force in our discussion, I referred you to my post where I explain that even the equatorial bulge depends on differential gravitation, which is the argument David uses to show that the tidal bulges don't need a centrifugal force to build up. A centrifugal force would throw an object directly away from the rotation center, and it is not what we observe: all the objects leave any rotating system at 90 degree from the hypothetical centrifugal force, and the only force that can do that is centripetal. It is due to trying to accelerate a body that resists being accelerated. The is no centrifugal force when a car accelerates, so there is no centrifugal force either when a body accelerates towards the center of rotation. The passenger in the car is not going to be thrown away from the car if we stop the acceleration, and the object that is rotating neither if we let it go. Wiki says "Centrifugal force is an outward force apparent in a rotating reference frame", and it adds "It does not exist when a system is described relative to an inertial frame of reference". If it's only apparent, it simply means that it doesn't exist either. If rotation is absolute, then the resultant force is absolute, and in this case, the only available one is centripetal.
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Good then, but relativistic computations don't always have the twin that travels more age less. You gave an example where he did, but there are examples to the contrary. If your simulation is accurate, it would confirm this if you ran such a scenario.
If you mean that the numbers will get different if the traveling twin is traveling the same distance at a different speed, then you're right, and my simulation will show it.
You introduced it. Look at the thread title.
That title is not mine, it's from Colin2B that split the thread where I referred you to create this one. The expression "rotational force" is mine though.
Tides exist even in absence of any centrifugal or centripetal forces, so explanations using these terms does not predict the observed tides. During half-moon, the tides are lowest where the centripetal/centrifugal forces are greatest. Again, the explanation using these terms predicts the wrong thing. That's what David is trying to show.
You got it wrong. Rmolnav uses a centrifugal force, and David a differential gravitation. As far as the tides are concerned, David says exactly what you say, and I do too, but in the post I referred you to, I add that the equatorial bulge is also due to differential gravitation.
If I drop a rock, it accelerates to the ground via an apparent force of gravity. Apparent means that the object visibly accelerates.
No need to add the word apparent if the acceleration is real. Inertial motion is apparent, but not the gravitational one. A force that can be measured directly is absolute, and the resulting acceleration is absolute too.
In a rotating reference frame, a stationary rock accelerates away from the center. That is real acceleration that acts on the rock. Any force that causes real motion like that is a real force.
A rock that rotates on the edge of a rotating disc is not stationary, it rotates at the same speed the edge is rotating, an if we let it go, it does not accelerate away from the center, it moves at a tangent to the edge, thus at right angle to the so-called centrifugal force, and at the speed at which the edge was rotating.
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In that frame, yes it does.
It doesn't accelerate since it doesn't increase its speed.
It is in the rotating reference frame of the disk, and centrifugal force is only meaningful in such a frame.
If you and me and David are right about centrifugal force having no influence on tides, then using that force to explain rotation is misleading. It leads to guys like Rmolnav believing that the tides depend on it.
You're claiming gravity is less at the equator than at the poles?
No, I simply claim that the gravity force is not the same at different heights inside the bulge, and I compare that difference to the one that produces the two tidal bulges, which are also at different heights from the moon. I we stop the rotation in the two cases, the two kinds of bulge both fall towards the center of the force, and it is only because the equatorial bulge is stopped from free falling that the bulge disappears. Preventing the earth to fall on the moon while it is stopped from orbiting it would also crunch the tides.
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It doesn't accelerate since it doesn't increase its speed.
In a rotating frame of reference, objects at rest begin to move due to centrifugal force. That is an increase of speed. In other frames of reference, there is no such thing as centrifugal force.
Terminology correction: Acceleration is not defined as "increase of speed". It is a rate of change of velocity per unit time.
If you and me and David are right about centrifugal force having no influence on tides, then using that force to explain rotation is misleading.
Nobody ever said anything about centrifugal force explaining rotation.
No, I simply claim that the gravity force is not the same at different heights inside the bulge,
This doesn't work. Gravity would be slightly less as much on the poles as on the equator.
I have an ocean around me, and I build a 21 km tower of water somewhere which you say stays up because gravity happens to be a tiny bit weaker near the top. Such towers of water collapse. But if I spin the water ball, the 21 km tower does indeed stay there.
You're not running any numbers. Try computing the weight of water 21 km deep and try to balance that with the gravitational difference. Turns out gravity would have to be zero above sea level for that to work. Any pull at all and the water would level out.
I we stop the rotation in the two cases, the two kinds of bulge both fall towards the center of the force,
Not so with tides. They stay there even if nothing is orbiting. David and I both found examples of how that might be set up.
Preventing the earth to fall on the moon while it is stopped from orbiting it would also crunch the tides.
That would be application of an additional force.
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The reason we don't call it "rotational force" is because the force required to sustain rotation, or generated by a rotating engine, is the force element of torque.
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What is hard to admit is that rotation is not relative, and you agreed with me on that point, so it means that you are probably not a mainstream thinker.
@Halc is a mainstream thinker, but terminology is getting confused here.
In mainstream, rotation is considered absolute, meaning you don’t need an external reference to tell something is rotating. However, this does not mean there are no relative rotating frames.
Halc gave one example, I gave the example of a turntable either rotating or stationary relative to earth surface.
Think also geostationary orbit. Stationary relative to the earth (geo).
Even if you have an absolute reference you can still have frames relative to it or to each other. There are quite a few defined for astronomy, space navigation etc.
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In a rotating frame of reference, objects at rest begin to move due to centrifugal force. That is an increase of speed.
That is an instantaneous increase of speed, which is not physical. If we let go a ball out of a merry go round, and we look at it getting away from the center of rotation, the ball seems to start moving instantly, whereas in reality, it only goes on moving at a tangent at the same speed the merry go round is rotating.
In other frames of reference, there is no such thing as centrifugal force.
There is no real measurable centrifugal force in any reference frame. Thinking that a force pulls the ball outward is like thinking the sun is rotating around the earth: it is simply wrong and it leads nowhere.
I have an ocean around me, and I build a 21 km tower of water somewhere which you say stays up because gravity happens to be a tiny bit weaker near the top. Such towers of water collapse.
I was talking about the gravity that gets weaker inside the earth, not outside. At the center, there is no more gravity. If we stop the earth from rotating, the inner parts of the equatorial bulge will fall less rapidly than the surface because gravity is stronger at the surface, which is the inverse of what happens with tides if we stop the earth/moon system from rotating. Nevertheless, both phenomenon are similar since in both cases, the force is only gravitational.
Not so with tides. They stay there even if nothing is orbiting.
They stay there while the earth falls on the moon, and the equatorial bulge does the same thing at the beginning: it falls towards the center of the earth. The difference is that there is no void inside the earth, so the bulge is rapidly stopped from falling, whereas the earth can fall towards the moon for a while before it is stopped from falling. I asked David what he thought about that, and here is what he answered me (https://www.thenakedscientists.com/forum/index.php?topic=49715.msg556701#msg556701).
Rotation is not relative in the sense straight motion is relative. We can know we are rotating whereas we can't know we are moving straight line. In that sense, rotation is absolute, and it is misleading to say it is relative. Until we find a new way to explain motion, we should study separately the two kinds of motion. We should also admit that straight motion is no more relative if acceleration and roundtrips are involved, because that's exactly what happens during rotation. Relativity would then only concern cases where we cannot determine which observer is moving because relativistic phenomenon prevent us to determine it. In the case of the atmospheric muons for instance, the earth could very well actually be moving at half the speed of light in a particular direction and we would be unable to know it, so in reality, the muons would not be traveling at the same speed all around the earth and we would be unable to measure it. If the laser signal reflected at the moon was instead sent from the moon, how could we know with precision the distance it has traveled? It might as well take more or less time for a particular alignment's direction and we wouldn't be able to know it. Again, the signal is useful only because it makes a roundtrip, because then, we can use the meantime and split it in two to get the right distance. That's precisely how my simulation of the twins paradox works (http://motionsimulations.com/Twins%20paradox): for all the tics to be comparable, it is the mean time the photon takes between the mirrors that counts for a tic.
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The sun rotates around me every day. It isn’t wrong to say that, and in most day to day contexts, it makes practical sense. Rotating frames of reference are probably used more often than inertial ones, as is appropriate for beings that live in a rotating environment.
It makes sense for those that don't know better, but not for those who do. If we want to make any progress in understanding motion, we can't rely on people that don't even know the basics, and if we want them to understand the basics of rotation, for the sake of knowledge, let's not use the reference frame principle.
That is just a different frame of reference. Both of them are equally real.
It is precisely not equally real because, if we know the basics, we can discover we are rotating.
That makes no sense. Physical would have waited a moment?
In real life, it takes a while to accelerate a body from 0 speed to any speed, and in the case we are discussing, it takes no while at all. The ball immediately starts getting directly away from the center at a certain speed without going through intermediate speeds,as if it was already getting away at that speed before we let it go. It was indeed going at a certain speed, but not directly away. There is visibly no force pushing on the ball, but since we feel one ourselves, we have to invent one that comes from outside the merry go round to move the ball away. That's what @rmolnav is doing on the thread about tides (https://www.thenakedscientists.com/forum/index.php?topic=49715.msg557265#msg557265), and he might not be doing that if he had learned the right way in the beginning. He thinks that centrifugal force is real because he learned it that way and because some scientists he refers us to also think that way. If the "centrifugal force" wording had never been invented, he wouldn't think that way, and we wouldn't have this discussion. I suspect Colin2B, responsible for this thread's name, to be unsure whether centrifugal force is real or not.
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So if I go out and discover a new species, the new species is not real???
Geocentrism seemed real before we discovered it was not, so I think we can now say that it was not as real as heliocentrism.
How long does it take to accelerate a stationary body in real life? How much less time does it take for a stationary object to begin moving due to centrifugal force?
To increase our speed from 0 to 1m/s at an acceleration of 1m/sec², it takes one second, but since a ball is already moving at a constant speed on a merry go round before we let it go, it starts moving at that constant speed immediately, there is no acceleration, so there is no acceleration either seen from the center, it simply starts moving away at a constant speed without having accelerated.
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To increase our speed from 0 to 1m/s at an acceleration of 1m/sec², it takes one second, but since a ball is already moving at a constant speed on a merry go round before we let it go, it starts moving at that constant speed immediately, there is no acceleration, so there is no acceleration either seen from the center, it simply starts moving away at a constant speed without having accelerated.
In a rotating frame of reference, the ball on the merry go round is stationary until you let go of it. Only in such a frame does centrifugal force exist, and the ball accelerates from a stop due to that force.
From any point anywhere in the rotating frame, the ball is stationary until you let go of it. At that moment, it begins to accelerate from a stop, same as me dropping a ball on the floor of my Kitchen from a meter up.
Once in motion, the path bends to one side due to Coriolis force, which doesn't happen in an accelerating linear frame, so at that point one can tell the difference between a linear frame and a rotating one.
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In a rotating frame of reference, the ball on the merry go round is stationary until you let go of it. Only in such a frame does centrifugal force exist, and the ball accelerates from a stop due to that force.
Sorry, I didn't realize that, from the center's viewpoint, the ball seems to accelerate, while seen from the place where it was freed, it is only moving at constant speed. At the moment it is freed, its motion is perpendicular to the line of sight of the observer at the center, so it has no speed with regard to him, and the speed increases while it is getting away, as if it was pulled away. No need to wait for the Coriolis effect to take place then before realizing we are rotating, that kind of force pulling from a center is already fictitious.
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Yes, and the acceleration curve is asymptotic, with time, the speed of the ball gets near the speed it had on the merry go round, but never reaches it. I still maintain the importance of acknowledging that acceleration is absolute though, so that changing reference frames in this case does not give a symmetrical result. For instance in my simulation of the twins paradox (http://motionsimulations.com/Twins%20paradox), I need to decide which clock is accelerating, otherwise I would get a symmetrical result. I can't change reference frames in between either, thus decide that the earthbound clock is moving away in the first half, and decide that the traveling clock gets back to it in the second one. Since I move my clocks with regard to the screen, it wouldn't produce the right numbers. Both clocks would be moving the same distance at the same speed, so they would both show the same number.