2161

Physics, Astronomy & Cosmology / Re: Einstein's Clock: What happens if you move towards a clock at light speed?

« on: 29/09/2018 19:19:28 »

If we consider Earth and A to be stationary relative to each other; their clocks can be synced at t=0.

Synced in the frame in which they are stationary, yes.  Not synced in other frames.

If, at t=0, the ship is stationary relative to A and its clock is synced with A’s clock; the ship’s clock will be synced with Earth’s.

However, if we start the scenario with the ship passing A at 0.866c (or any speed); there will be an instant when the clocks on A and the ship can be synchronised (this we call t=0), but at that point the ship’s clock will not be synced with Earth’s.

No, in the frame of the moving ship, the clock at A is not in sync with the clock at B.  So you can say that there is that instant when the ship clock is synced to the departure event at B (synced to an event, not to a clock).  Both clocks are present at that event and they both read zero (are set to zero actually) at that event.

As for the ship parked before departure, it is stationary in B’s frame, not the eventual frame of the moving ship.

2162

Physics, Astronomy & Cosmology / Re: Einstein's Clock: What happens if you move towards a clock at light speed?

« on: 28/09/2018 19:24:42 »

SR can handle acceleration, the only reason to bring in GR is if you are including gravity in the scenario.

Right you are.  I was mistaken to suggest SR doesn't handle acceleration.  Gravity is not included, nor is non-local scales.  Anything within the galaxy is reasonably local enough for SR.

I was keeping it simple to answer the simple question asked in the OP: Does an approaching clock appear to run faster, and yes, it does.

So for example, if the ship is accelerating from point A to point B, with clocks at these points which are synchronized in the inertial frame, then, if the ship leaves point A at t=0 and T=0, and the ship reaches B when the ship clock reads T, then t will be the time at B according to the ship upon its arrival at B.  The time at A (according to the ship) will depend on the ship's velocity with respect to A and B and the proper distance between A and B( relativity of simultaneity). tB will be greater than T, but tA will be less than T.

I take it that tA is what A clock currently reads in ship frame.  So in my simplified example where the ship is already moving when it departs A, T at B is 5, tB is 10, and tA is 2.5.  Yes, I agree with all this given that the acceleration vector is forwardish (scientific term!) the whole way.  If the ship turns around and accelerates the other way for part of the trip, or maybe makes a detour to Arcturus for nachos, it arrives at B with tA possibly more than T, and tA and tB both synced if the ship velocity becomes zero in the inertial frame of A and B.  So tA is less than T only if acceleration is predominantly away from A.

2163

Physics, Astronomy & Cosmology / Re: Einstein's Clock: What happens if you move towards a clock at light speed?

« on: 27/09/2018 20:19:52 »

Good explanation, Halc.  There are a few points/questions that come to mind.

1. The clock, presumably on a craft, travels from A to B.  We identify B as Earth.

2. The distance from A to Earth is (presumably) already known, so we can calculate that a craft travelling at 0.866c will take 10 years in Earth’s RF, to make the journey.

3. Light takes 8.66 years to make the same journey.

4. We, on Earth, have no way of knowing when the craft left A.  Nor do we have any way of synchronising our clock with that on the craft at the time of its departure.

5. On arrival, would the clock on the craft be 5hrs ahead of clocks on Earth?

6. If the answer to 5 is “yes”, why would that be the case if the clocks were not synchronised when the craft left A.

7. If the craft’s clock differs from Earth’s clock, why would it need “to tick 5 years in only 1.34 remaining years”?  Would the difference not remain?

2: Yes, I said that the departure (A) was 8.66 LY away.

4: Well, we're watching.  At year 8.66, we see the ship depart abruptly already at speed.  Yes, until then, we either don't know, or maybe it was a scheduled thing.  We very much can synchronize our clocks, but we need to assume a frame to do it.  We assume that A is stationary relative to Earth, and thus that synchronized clocks have meaning.  The ship frame is different of course.

5: No...   On arrival, the ship clock is 5 years slow.  It reads 5, and the Earth clock reads 10, assuming both Earth and 'A' read 0 at the same time in their mutual frame.

7: The OP asks how that clock would appear as it approaches rapidly.  We see a zero on the ship clock only 1.34 years before it arrives at time 10.  In that 1.34 years, we see it count from 0 to 5 years, and thus 'appears' to run fast when in fact it is dilated by a factor of 2 and only counts 5 years in a journey that actually takes 10 years in Earth's frame.

2164

Physics, Astronomy & Cosmology / Re: Can we feel gravitational attraction from objects at different velocities?

« on: 27/09/2018 05:16:27 »

Do you feel a change in grav. attraction if the moving object changes mass (relativistically).

Reactionless thrust aside, I never saw this point addressed, and I find it interesting.
What is the gravitational formula for masses moving at relativistic speeds?

I think I worked out that the attraction between objects needs to plug in rest-mass into Newton's formula.

My example was the earth/moon system orbiting once a month.  Now consider just that in a frame where they're going at .866 c along the orbital axis.  The planets get squashed into a sort of phulka shape, but still have the same separation.  They mass twice as much, and orbit every 2 months, which is half the acceleration as the system at rest.  F=ma: Double the mass, halve the acceleration.  The force must be the same, so the law of gravitation (F=(G(m₀1•m₀2)/r˛) is computed with rest-mass (m₀), not with mass (m).
Therefore there is no change in grav. attraction if the moving object changes mass (relativistically).

Did I do that correctly?

2165

Physics, Astronomy & Cosmology / Re: Einstein's Clock: What happens if you move towards a clock at light speed?

« on: 26/09/2018 20:26:05 »

Hi,

I watched a video on Youtube about an Einstein thought experiment, in which he realised that if he was travelling away from a clock at light speed the clock would appear to be frozen, but what happens if you travel at the speed of light towards the clock - does the clock appear to speed up?

This is the Doppler effect, the same thing that causes red and blue shift.  The faster a thing moves away from an observer, the slower it appears to go, even without time dilation.  The effect is much stronger than dilation.

So if a clock is coming at you fast (say a big digital clock on the nose of a ship heading this way), the clock will be slower due to dilation, but still appear to run faster due to blue shift.  The clock cannot move at light speed, but it can go as close as you want to it.  If you want it to appear to run 10x fast, there is a speed at which that occurs.

So take a clock coming from 8.66 light years away (Earth frame), moving at .866 c, a 10 year trip in our frame.  Its clock says 0 when it leaves.  At that speed, it will be dilated to half speed and only log 5 years, so it reads 5 years when it gets here.  But it leaves 10 years before it gets here, and we don't see the light from its departure for 8.66 years, leaving us watching the clock appear to tick 5 years in only 1.34 remaining years, so it appears to run about 3.73 the rate of one of our own clocks.
Go even faster, and the clock coming at us appears to run even faster, but it is meaningless to posit a clock moving at actual light speed.

2166

Physics, Astronomy & Cosmology / Re: Why do we have two high tides a day?

« on: 26/09/2018 12:29:16 »

Your basic error comes from same misconception as if somebody said moon (or sun) rotates around earth once every 24 hours ... That is only an apparent movement !

It is apparent movement that I'm talking about.  The tides go up and down twice a day, and that requires movement relative to Earth's surface of thousands of cubic km of water each 6 hours.  In most places, this is not a problem, since the necessary movement is only about 1km or less, and 6 hours is plenty of time for gravity to shove water that far.  In fact, the variance in the Atlantic comes more from resonance than it does from the gravity boost, so a longer day might actually make the tides lower.

So my example of a place that would get much higher tides were the day to be longer is the Mediteranean.  No resonance there, and all the water needed to raise the tides need to fit through Gibraltar, and the gap simply isn't large enough to allow the tide to come in before it starts running out a few hours later.

"Sublunar" bulge is always almost in line with the moon, and its actual cycle is, as moon´s, 28/29 days ... A little more than 7 (rather than 6) days  ahead of next low tide !!

I'm well aware of that, but no water is stationary relative to this bulge.  If it was, it would kill us: New York City and Shanghai wiped out twice daily by a wave of water moving west at over 1000 km/hr.

2167

Physics, Astronomy & Cosmology / Re: Why do we have two high tides a day?

« on: 25/09/2018 19:46:45 »

... the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.

Again: you continue to mix daily earth spinning with moon related tides … Gravity and inertial effects (centrifugal forces) don´t "need" longer days to build full-size tides, because they have plenty of time to deform ocean surface: a little more than a week from low to high tide !!

A little more than 6 hours actually.  A week between spring tide and whatever they call the low between them.  The shoving I spoke of happens every 6 hours.  The river by me runs backwards twice a day due to that shoving.

I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is ,  already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.

This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit.

My bold.  I disagree.  This explanation predicts that tides would be negative if the Earth spun the other way since the inner part has the greatest tangential speed and wants to move outward, and the outer part has the least tangential speed and wants to move inward.  Venus is such a case (it rotates backwards), at it has normal solar tides towards and away from the sun just like Earth.

2168

Physics, Astronomy & Cosmology / Re: Is Wiki right about tidal acceleration?

« on: 23/09/2018 22:15:22 »

The solar tides on Earth do drag the total angular momentum of the Earth/moon system down

Hi Halc,

What you say means that the boost from the moon's offset tides on the earth's orbital speed has to be equivalent to the boost from the earth's offset tides on the orbital speed of the moon.

The comment above refers to the solar tides on Earth, not tides caused by the moon.  That friction slows down the spin of Earth but has no immediate significant effect on the moon.

The lunar tides transfer momentum from Earth to the moon.  Total momentum must be conserved, but not total energy, which gets lost to heat of tidal friction.

Besides, I explained to Janus how a backward boost was lacking in the "tidal acceleration" explanation, and it would still be lacking even if both accelerations were equivalent.

Your comment about backwards boost is mistaken.  Two forward boosts are required to move a satellite to a new higher orbit, as Janus correctly conveyed.  The moon on the other hand gets a small but continuous forward force which slowly adds to its mechanical energy, stealing away Earth’s mechanical energy.

If I understand well, you mean that the moon suffers a tiny boost, waits till it gets at its apogee, and then uses another boost to stay on a circular orbit.
 That would be fine if no other boost would be happening in between and if the moon knew what it was doing, but it is not the case. The boosts happen all along the trajectory, so how does the moon differentiate the first boosts from the second ones.

The force on the moon is continuous, and so the orbit is technically a continuous spiral, never a circle.  The satellites could get to a higher orbit the same way, but rockets are typically not designed to run at low power like that, and the people that paid for it want it there more quickly I suppose.

if we accelerate it continuously as it is the case with the tidal explanation, it will simply never slow down.

The speed of the moon is continuously slowing down due to gradually going into a higher orbit, as is pointed out in the last line of the wiki article you quote in the OP.

2169

Physics, Astronomy & Cosmology / Re: Why do we have two high tides a day?

« on: 23/09/2018 16:41:59 »

As said on one of the NOAA previously linked sites:

Which admitted to being for children and laymen, and just wrong after the 50’s….   Hardly an argument from authority.

”since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”

This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.  They're being sloppy, something they'd not get away with in a publication meant for a higher audience.
Yes, I understand this reasoning.  You are now just repeating it instead of addressing my objections to it in prior posts.  It isn’t completely wrong (the centrifugal force needs to be described in the frame in which it exists), just not proportional to the tides, and most importantly, it doesn’t explain tides in situations where there are no centrifugal forces, such as all these hypothetical situations with multiple moons and such.  That lack of a general explanation makes it wrong I think

At sublunar area, direct moon´s pull is bigger than above mentioned centrifugal force, it prevails, and we have there the more obvious bulge ...

This is the actual cause, and it works without computing the difference with centrifugal forces to see which is prevailing.  It is a general solution that works even in non-revolving situations.

Not sure if all such accelerated motion can be considered centrifugal since the acceleration is more typically inward instead of outward, and 'centrifugal' is typically used only for the tendency towards outward.

I still have to post here my ideas about the generally badly used issue of inertial and non-inertial reference frames ...

Other frames are valid, but the laws of physics are not the same as in inertial frames.  Centrifugal force is a valid force in rotating frames, which is contrary to what I first said.  Your use of it in descriptions of the scenario in inertial frames seems like bad use of frames.
The moon and Earth are stationary in the rotating frame of the barycenter, and gravity prevents the natural tendency for both objects to accelerate away due to centrifugal force.

I´m procrastinating in that realm, because I know it is tough to convey.
But please note that the term "centrifugal acceleration" doesn´t exist

It does. In a rotating frame, an object at rest  begins to move if not impeded.  Sounds like acceleration to me.  The acceleration is initially outward, but quick bends until it is mostly lateral, but always accelerating.

... "Centripetal force" is the name we give to any force that is bending a moving object trajectory.

In an inertial frame yes, and then only sometimes. The thrust of a rocket turning left isn't usually described as centripetal. In a rotating frame, that centripetal force might be holding an object stationary against its unimpeded tendency to start moving.

Centrifugal force is a controversial term, but it is always an inertial reaction of the affected body to the fact that it is being obliged to change the direction of its speed ...
But if the object moved outward, that would actually mean that it´s got somehow free, because both centripetal and centrifugal forces had ceased to exist, at least partially.

That bolded part I didn’t understand.  Maybe if you specify a frame it would make sense, but it sounds like you’re just describing an object moving forever in a line in an inertial frame, which yes, it does if no forces act upon it, per Newton’s first law.  I hadn’t seen that law described as something “getting free”.

And regarding:
"...In a uniform gravitational field (say at a sufficient distance from the galactic black hole ... you have the same inertial forces at work, but undetectable tides ..."
 please kindly note that in that case centrifugal forces would also be very, very tiny, proportional to v˛/r, being "r" enormous !! 

No, the centrifugal forces would be identical to what the moon generates since they have to counter the gravitational pull of that black hole at the exact distance where the force was the same.  The acceleration of Earth in such a situation would be the same as in a pure Earth/moon system, countered by the same inertial resistance that you’ve been labeling centrifugal force.  I chose that scenario specifically for its identical gravitational forces.

If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not.

@PmbPhy wouldn’t make that mistake. I think by now you will have realised that he was talking about non inertial, rotating frames.

I have since corrected my statement above.  Yes, in a rotating frame reference, the two forces exist and can cancel, holding the object stationary.

I think the whole subject of centrifugal and what are called fictitious forces is badly taught. NASA tends to refer to all 3 as inertial forces and includes G force in linear acceleration as a 4th.

What are the other two?  Centrifugal is an inertial force in a rotating frame. Coriolis accelerates things into curved trajectories. G force is an inertial force in an accelerating reference frame. There are inertial effects involved anytime some mass is accelerated in an inertial frame. That doesn't mean that inertial effects are all centrifugal.

Tides happen or not in both cases, so tides are not a function of either. That was my point. Tides are a function of the stresses and strain on an object due to non-uniform gravitational field, and are a function of only that.

”since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”

This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.  They're being sloppy, something they'd not get away with in a publication meant for a higher audience.

I think they are considering the earth to be a rigid body, so any force at the centre of mass will be transferred to near and far sides.

They seem to be making an explicit reference to forces on specific points, not about where those forces get transferred due to the rigidity of the body. It was a nit, not really detracting from the point being made by NOAA.

Although the water isn’t rigid it is incompressible and would tend to find it’s own level so might be considered to act as the earth does. Whether it’s a reasonable assumption is up for debate, but I don’t think the article was intended to be completely rigorous.

Sure. For one, the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.

2170

Physics, Astronomy & Cosmology / Re: Does gravity have an infinite range?

« on: 23/09/2018 16:18:27 »

Thanks for some thought-provoking responses.

Sorry for the reply delay, but I'm still figuring out this site, and there seem to be no notifications about replies to my posts if I turn off the emails.

There is nowhere where you have zero gravitational potential energy.

 GPE = mgh.  If h is the separation between two bodies; could h be infinite?  If so, what would be the GPE?

I wasn't talking about a specific body.  I'm talking about the general GPE that is mostly due to the most nearby objects.  The most remote places anywhere are still only a billion or two light years from several of the closest superclusters.  You cannot get away from them any further and raise your GPE all the way to zero.

…….. certainly galaxy superclusters have an effect on each other….

What is this effect, and how is it measured?

Gravity attracts, what else?  Maybe the effect offers resistance to the rate at which the expansion of space would otherwise proceed.  The effect doesn't really change the trajectory of the superclusters significantly since they pretty much cancel each other out, being reasonably well distributed in all directions.

BTW. I agree that interpretation is an important factor, but if “yes” and “no” are both correct some clear interpretation is needed.

Well, I said 'no, gravity is limited' in relation to the event horizon.  Gravity waves propagate at light speed, and so gravitons cannot ever reach an object beyond the event horizon and effect it there.
Gravity itself does not propagate, and so I cannot claim that mass here has zero effect on stuff beyond the event horizon.  I can claim it has no net effect since any pull from our direction is balanced by equal pull from the opposite direction.

2171

Physics, Astronomy & Cosmology / Re: Why do we have two high tides a day?

« on: 22/09/2018 15:16:10 »

#316 Halc
When I read carefully your post, I´ll try and refute what found necessary ...

You should perhaps quote the parts of mine that you are refuting, because I pretty much agree with your whole post except the last bit.
I looked up centrifugal force and realized it is not a fictional force. In an inertial frame, there is no such force.  But it is a term legitimately used in rotating frames, where none of Newton's 3 laws apply.  An object at rest tends to accelerate, and the force that causes this acceleration is centrifugal.  No reaction stands opposite this action.  Moving objects do not follow straight lines.  Not sure if all such accelerated motion can be considered centrifugal since the acceleration is more typically inward instead of outward, and 'centrifugal' is typically used only for the tendency towards outward.  If not, I don't know the name of the reactionless force that appears to accelerate objects inward in such rotating frames.

"This is the reason for the equatorial bulge of the Earth due to its own axial rotation[/b]"
Please kindly analyze his explanation, and say if you agree or doesn´t ...

This part I is fine.  He doesn't really discuss tides, but it is mentioned once in the 'stress' section.  Yes, stress causes tides.  Such stresses can kill you if it is strong enough.

I took issue with his blocky diagram above in the 'stress' section, which shows two stacked blocks in a gravitational field and a weight force preventing the whole setup from accelerating from the gravity.  It labels the spring as tension, despite the arrows depicting the spring pushing on the two blocks, which would compress the spring.  The arrows are correct, but text incorrectly labels this as tension.  If the arrows represent tension T, they need to point the other way, and the tension would be negative.  So while I understand the concept of stress and strain (I took civil engineering courses), the description of it there needs a bit of review.

He certainly doesn´t use the term "centrifugal force", but the root of his argument is on the REAL effects of INERTIA ...

Yes.  In an inertial frame, the equator bulge occurs from a combination of at least 3 effects: You weigh less (and so does the water) there because you're further from the typical particles that make up the planet, and because part of the gravitational force is used for acceleration and need not be resisted by the force of weight from below.  The third part is due to the fact that the typical particle is concentrated more below and less to the sides, and this component makes you weigh a bit more, but nowhere near enough to overcome the negative effects of the other two.
The second component I listed is the inertial one.

If that happens as an inertial reaction of the water (and the man´s mass) when "forced" to follow the equatorial curved path, why wouldn´t happen something similar when the water has to revolve, following during some 28 days a circumference (considering only earth-moon "dance") of a radius equal to the distance between earth CM and earth-moon barycenter ??

Inertial effects are indeed involved, but not directly like that.  The far tide has far more acceleration about the barycenter (which represents the inertial frame of a two-body model), but both tides are the same magnitude, which they would be even if the moon was further away putting the barycenter above the surface.  The direct inertial explanation of the tide would push the water down, not upward, since it is now rotating on the same side of the barycenter.  Inertial of the water on the sides would tend to carry it to the far side.

So no, I cannot agree with this wording.  In a uniform gravitational field (say at a sufficient distance from the galactic black hole to have a gravitational force on earth similar to that of our moon (2e20N), you have the same inertial forces at work, but undetectable tides.  It is the non-uniformity of the gravitational field, not inertia, that causes the stress, and  tides which are strain resulting from that stress.

You mean the CM can be used to track Earth’s motion as if it were a point mass.  Indeed, it doesn’t work when the discussion is about more local effects like tidal forces.  It also doesn’t work in general.  I can have two pairs of all identical mass objects with identical difference in separation by CM, but one is a pair of spheres and the other is a pair of long barbells arranged perpendicularly.  The spheres will have more gravitational attraction than the crossed barbells despite the same mass and distanace between their respective centers of mass, because most of the mass is nowhere near that center like it is with the spheres.  So point taken (??).  You can’t treat objects as point masses when the difference matters.

Sorry, but that is utterly absurd ...
A not uniforme distribution of masses, even two o more separated objects, can be treated (for certain purposes) as a whole, with a common center of mass, useful for its dynamic analysis ...
E.g., the very stuff we are talking about: earth-moon system.

Earth and moon are hardly shaped like barbells.  If all objects can be treated as point masses, then I should weigh more and more as I dig myself into a hole.

It has a common CM (the barycenter), which is what follows the elliptical orbit around the sun.

Well, to be fully anal about it, and in an ideal situation with no other planets, we would orbit the CM of the solar system, not the sun itself.  The two points are almost the same thing if we discard the other planets.

It´s quite reasonable to analyze the complex dynamic of moon-related tide formation in different stages: mentioned orbiting of common CM, earth-moon revolving/rotation around the barycenter, gravitational and inertial forces on the earth as a whole (where earth CM has an important roll), and then the detail of forces acting on each area of our planet (internal force changes due to earth-moon dynamic included) ...
Those last ones are mainly in the moon´s CM-barycenter-earth´s CM direction, both "moonwards" and outwards (or "moonfugal" or "centrifugal"...)

Complicated but reasonable to analyze the dynamics this way, sure.  Some of them assume a rotating frame of reference, and it is reasonable to analyze the dynamics in that frame.

And they are the direct cause of tides ... certainly generated by moon varying pull, but also the inertial effects on all and each particle of the earth.

The tides are directly proportional to the varying pull, and not directly proportional to the inertial effects.  So the latter is a related cause, but not a direct cause.  As I said, it is reasonable to analyze it in such terms, but it is more complicated due to this non-proportional relationship.  The inertial effects from the sun are far greater (about 175x) than those from the moon, yet produces a significantly smaller tidal effect.

As I´ve said many times, earth particles only can "feel" gravity at their respective location, and pushes and/or pulls from contiguous matter. And they always have their tendency not to change their velocity vector (inertia), what causes inertial effects ...

Yep

What they can´t "feel" is either the so called "differential gravity" or the gravity on locations many km away, let alone they can "subtract" different gravitational pulls to react to that difference !![/b]

Right.  No individual particle can detect a gravity gradient.  It requires pairs of separated particles.  What the particles might 'feel' is the changing forces put on them by their immediate neighbors, which in turn is effectively them 'feeling' the differential gravity on locations many km away.  This is similar to the way you feel a train go by 100m away.  The stresses are communicated between particles at different locations.

2172

Physics, Astronomy & Cosmology / Re: Why do we have two high tides a day?

« on: 21/09/2018 21:27:26 »

#309 Halc
Thank you and welcome.

And thank you for the welcome. I’ll be glad when I become human and I get more than a postage-stamp size window in which to compose my responses.  As it is, I am forced to do it in an offline wordprocessor.

But today, I´ll only tackle what I consider ESSENTIAL for any possible understanding: NOT to have completely opposite ideas regarding INERTIA, that can manifest itself in different ways, one of them as “centrifugal force", real force (or, at least, causing quite real effects ...)

I suppose it is a matter of interpretation if it is real or not.  I kind of go with force being the F=ma in Newton’s 2nd law, whereas inertial effects (centrifugal being among them) are the ‘m’ in that equation, not the F.  It isn’t causing any real acceleration of anything in the direction of the supposed force.

When I first entered here (more than three years ago !!), I used to approach this issue considering earth´s CM revolving “uses” 100% moon´s gravity pull there to cause the required centripetal acceleration for that movement. No “spare” pull (which, if existing, could produce other effects) there …
But on closer to moon hemisphere there is an excess of moon´s pull. Any given material particle there can´t “use” that excess of pull to get a bigger acceleration, because the rest of the planet “forces” them to move together, with same acceleration … That imbalance generates internal stresses in both senses (between contiguous parts opposite forces are exerted on each other, due to Newton´s 3rd Motion Law).

You mean the CM can be used to track Earth’s motion as if it were a point mass.  Indeed, it doesn’t work when the discussion is about more local effects like tidal forces.  It also doesn’t work in general.  I can have two pairs of all identical mass objects with identical difference in separation by CM, but one is a pair of spheres and the other is a pair of long barbells arranged perpendicularly.  The spheres will have more gravitational attraction than the crossed barbells despite the same mass and distanace between their respective centers of mass, because most of the mass is nowhere near that center like it is with the spheres.  So point taken.  You can’t treat objects as point masses when the difference matters.
You describe tidal forces nicely there.  Yes, they put stresses on things, enough to tear them apart if they get within the Roche limit.

Initially I didn´t agree, and insisted: only previously mentioned outward internal forces could be considered “centrifugal”, to fully satisfy Newton´s 2nd Motion Law …

And since nothing is accelerating outward, this would be consistent with an outward centrifugal force not being real.  Sorry, just thinking out loud when you haven’t got to the point yet.

In any case, both ways are equivalent, because deducting required centripetal force vector from moon´s pull on a given earth particle (the force the particle “interchanges” with the rest of the planet), gives same result than adding the “arguably” called centrifugal force vector to moon´s pull …

Still agree…  As I said, I suspect the difference is interpretational.

But it is necessary to tackle the “apparently” insolvable fact that earth´s CM is actually accelerating …
It is rather tough to explain through posts here. I did have in mind a couple of ways. With your post, yet another. In chronological order:
A) One in relation to the erroneously (to me) used concept of “non-ínertial reference system" …

That seems unnecessary if you get rid of the sun and everything else and just have Earth and moon in the inertial frame of their mutual CM.  Keeping the sun and its frame lets you see that the ‘centrifugal’ tide is actually accelerating away from the moon at full moon.

B) Another way (though kind of equivalent) in relation to the scenarios used by D.C., imagining earth not revolving …

  So they wouldn’t be called tides anymore because the distortion would be permanent and not noticed.  This is assuming Earth rotating in sync with the month, not stopped altogether, which would have the tides occurring twice a month.  The day is lengthening, but it will never be longer than the lunar month.

C) Last one, in relation to what you say: “… If the two forces (moon´s pull and centrifugal force) where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not".
ANOTHER DAY I´ll try and do it.

Sorry, but I never figured out from these three ways what this insolvable fact is concerning earth’s CM accelerating.  My comment quoted in C there was just another way of pointing out why I don’t consider inertia to be a force.

Anyhow I´ll copy and paste something I´ve already said a couple of times (way A), needing to elaborate on).
PmbPhy had said:
"In particular, the centrifugal force is an inertial force which is observed in rotating (non-inertial) frames”.
I replied:
"I´m afraid that is rather confusing ... What do you mean with "... "which is observed in …"?
Does it mean it is a real force "observed" by us? Or, in that "particular" case, is it rather a kind of tricky tool WE apply to bring a "fictitious" situation (an accelerating frame of reference) back to reality (with all REAL effects which have disappeared in that frame of reference)?"

Centrifugal force seems to be the force that holds the water in the pail being swung around, similar to how gravity holds water in the same pail hanging from my arm.  But both are wrong.  It is the pull of my arm that holds the water in the pail in both cases.  Without that, the water would drift out of the pail, even in the gravity field.  Water isn’t held with pails on the space station, despite no lack of gravity up there.

2173

Physics, Astronomy & Cosmology / Re: Does gravity have an infinite range?

« on: 20/09/2018 20:47:34 »

I think some of the experts might be right, but are interpreting the question differently.
So the universe has no spatial limit, but does gravity of the Milky Way effect something beyond the Hubble Sphere?  It one way it cannot (Eurlng).  The galaxy doesn't specifically exist to a thing that distant, which yes, is an interpretive statement.  On the other hand, there is a general density of matter everywhere, impossible to escape at any distance (Brewer).  In relation to that very distant place, there is definitely something somewhere around here, and it contributes to the overall gravity well of even the most remote places.  There is nowhere where you have zero gravitational potential energy.  Potential energy is negative everywhere.

Anyway, certainly galaxy superclusters have an effect on each other, but that is for the most part balanced in all directions, so they don't accelerate towards each other. Spatial expansion gives the greater effect and the clusters are all separating.

As for your 3, yes, spacetime is pretty dang flat in the nowhere that is between clusters.  There is gravity, but again, it is balanced, so no net effect.

2174

Physics, Astronomy & Cosmology / Re: Why do we have two high tides a day?

« on: 20/09/2018 13:21:04 »

      BY THE WAY, on a link posted by PmbPhy some days ago it is said:
...
… for a particle at the earth´s center, the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite.

This illustrates the danger of considering centrifugal force to be a force.  If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not.  It is accelerating due to the one and only force acting upon it here: gravity.

Yes, the tides can sometimes be explained in these term, but by eliminating the curved trajectory of Earth by putting it further out, or by putting it between two moons as David Cooper has done, the centrifugal explanation falls apart.  There is still gravity, but no centrifugal force, and yet the tide remains.

Let's do another version which makes things even more stark. Let's give the Earth two moons which always stay on opposite sides.

Or Earth centered between equal binary stars, unstable, but it gets rid of any possible acceleration altogether.  Tides totally without anything that can be considered centrifugal.

If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!

David’s examples very much did remove that rectilinear accelerated movement, and no compression of the atmosphere on the far side of Earth results.

Your "second" moon, together with real one, are getting what I said on #211: the earth does´t move at all, without the necessity of any "megaman" ...
And what happens is what I said then, BUT TWO OPPOSITE-WAY TIMES:
"... outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)"

There is no compressive effect on the far side due to either moon.  The attraction of the far side is less than the attraction on the Earth below it.  That equates to tensile stress, and these stresses from each moon add to a double magnitude tide, not cancel out to something lower than what we see with one moon.

What happens is that on each earth side the increase of sea level due to the pull from closer moon, is bigger than the DECREASE of sea level caused by the farther moon !!

No, they’re both increases.  This seems to predict a partially cancellation of tides.

I see several references to post 211:

If, somehow, we "remove" the revolving, both earth and moon will accelerate towards each other … Though only a transient movement, bulges would remain (and increase) until the collision.

You seem to be envisioning removal of orbital speed, not the revolving of Earth and the moon.  If you remove the revolving, you get two tides per month, and we would get to see what is currently the far side of the moon.

OK. Therefore, revolving was not causing them[/i] (YOUR deduction).
IN A QUITE SIMILAR WAY, we could reason as follows:
If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)

On this I agree.  If a mysterious force grabbed only the rocky part of Earth and pushed it, but not its water away from the moon, then yes, the water would naturally slosh to the side away from the unnatural acceleration caused by this selective force.  But no such megaman exists.  The sun does it, but it does not decline to act on the water as well.  The megaman example does not reflect what might actually keep Earth from accelerating in the various examples above.

Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!

I’m obviously in the camp that says the differential gravity very much does cause the far bulge.

2175

Physics, Astronomy & Cosmology / Re: Why has action at a distance not been disproved?

« on: 19/09/2018 18:07:07 »

QM requires that when the state of one member of an entangled set is observed, the states of all other members is fixed, in that instant. This is exactly what QM requires.

OK, I see what you're saying.  My comment was that it didn't require the two measurement events to be simultaneous, just separated in a space-like way.

QM doesn't require what you say.  See PmbPhy's reponse for that.
I suppose that if an interpretation asserts what you said above, there must be some preferred ordering of events so that the statement makes coherent sense.  Not sure which interpretation says this.
Copenhagen is essentially an epistemological interpretation, not a metaphysical one.  That means if I make a measurement here, I instantly know what the other measurement is/was/will be.
Bohm says there are hidden variables that determine the outcome of both measurements long before they are taken, so the act of measuring doesn't change anything.

To others:  Please inform me if I am misrepresenting these interpretations.  I'm not really up on my Bohmian mechanics.

Interpretations with locality (MWI, relational, to name two) do not posit any spooky-action.  I've a preference for relational-QM myself.

2176

Physics, Astronomy & Cosmology / Re: Why has action at a distance not been disproved?

« on: 19/09/2018 17:17:56 »

  The action-at-a-distance interpretation of entanglement requires simultaneity.
And, 110 years of testing Relativity has proven non-simultaneity.
So, doesn't that rule out AAAD as a possible theory?

PmbPhy relates what QM actually says.
There is no single action-at-a-distance interpretation, but interpretations that hold to locality deny it.
None of them require simultaneity.  The two measurments are simply ambiguously ordered depending on frame of choice.

The fact that none of the interpretations has been disproven is that they all make the same predictions.  Some interpretations have fallen away when they were shown to conflict with empirical measurements.

2177

Question of the Week / Re: QotW - 18.09.04 - Is it possible to terraform the Moon?

« on: 18/09/2018 17:36:26 »

To overcome the above-mentioned problems with living on our moon, you could create underground spaces and use solar panels on the surface to supply electrical power.

Is it really considered terraforming if you simply make a slightly larger life-support system than the one you brought with you?  The definition of the word implies it is done on a planetary scale.

This is my first post attempt, so I'm doing this partly to learn how.  My first try got an error concerning the letter are not those I typed.  Subsequent attempts error out that I've already posted this, but the post doesn't show up.
I take it I need to prove I'm not a robot, or at least until I get some level of seniority or something?