Gravity-based sensors

[Jedi Master Spock]

It's been a while since I've seen it. When you mentioned it accelerating into the planet, I thought you were taking into account what would happen if kinetic energy stayed the same while mass was reduced.

Oh, so you just forgot it (and I thought you were using it earlier). Too bad, it's interesting and you really should have accounted for it.

It's not really interesting, actually. Not for this topic. It's worth going over when we talk about impulse engines and the typical model of mass lightening as part of the impulse drive.

You used: F=m'g, F=m''a, g=a*m'/m''.

That last equation should be g=a*m''/m'.

Actually, it should be a=g*m'/m''. That was indeed a typo there.

Of course, since inertial mass and gravitational mass are equivalent, you'd just get g=a, and that's not really a surprise, now, is it?

Exactly my point.

An object's own mass will not affect its acceleration due to gravity. And the gravitational constant is applied to the mass that creates the gravity well, not the object's own mass.

It will if you change inertial mass without also changing gravitational mass - as you were trying to claim was happening.

Oh, you don't care to deal with the inherent velocity increase that occurs when mass suddenly drops and kinetic energy stays the same? Interesting.

According to which frame of reference?

Again, what you're suggesting has no relation to what's present in the episode. It's a common description of the mechanism for impulse, but what we see in the episode does not involve the sudden acceleration of the moon - along its current vector or towards the planet - when the warp field is applied to most of the moon.

I've already requested this "trivial" task from you several times, so I'm kind of puzzled as to why it's taking you so long.

It's been done for a long time.

And inertial and gravitational mass are equivalent. That means they're the same.

Thank you for agreeing with me. I'm glad we had this little chat, because earlier you were saying things like:

As far as I'm aware, the gravity well of a mass-reduced object isn't confirmed to have changed.

You then were attempting to further explain this view.

Reducing G locally won't reduce the gravitational mass. You seem to be confused as to what gravitational mass is, and what its relationship to G is.

Not really. Consider the case when an object only interacts 10% as strongly with gravitons (symmetrically, of course).

We can express this as a local change in G by one order of magnitude. We can also - by thinking of the gravitational constant as constant - refer to this as a reduction of the mass to 10% of the original mass. They're two different ways of looking at the same effect.

Now, scientifically, it's bad nomenclature to be mixing both terms. But Geordi is not a scientist; he's an engineer, and engineers are notorious for using scientifically incorrect or inconsistent nomenclature that gets the point across. His line there is a perfectly fine example. It gets across the idea that the moon's mass is lessened gravitationally and inertially.

Since you seem to have figured this out now, I will now proceed with your regularly requested lesson plan.

[Jedi Master Spock]

Hawking radiation would be my best guess as to how a starship can obtain energy from a black hole. The energy obtained via Hawking radiation would be substantial. What makes you think otherwise?

Obviously, mass would need to be continually pumped into the black hole in order to serve as fuel.

A starship requires peak outputs ranging all the way up to the zettawatt range – but should not be idling much higher than the low exawatt range (e.g., 12.75 exawatts). Then you have to wait for a while – a few years to “warm up” to higher powers – and then have to flood the (tiny, see pictures of Romulan power cores) control room with a highly precise stream of thousands of tons of matter to recharge it.

A 12.75 exawatt black hole would, IIRC, be 5,300 tons – and, unfortunately for anyone trying to feed it in a controlled fashion, has a very small event horizon, substantially smaller than an atomic nucleus. You need to be feeding a stream of 140 kg of matter per second to a region a few millimeters in diameter.

That still isn't enough. Remember, it would be entirely possible to set the shadow up such that there are sections with zero graviton flux. Objects starting at any point in the full shadow could all be moved to the same point (one at a time) and then thrown out. Each would receive the same gravitational pull while exiting, yet the math doesn't work out because of the different starting positions. (Or even different entrance points.)

Yes it is. When you are “thrown out” along different paths, you experience different forces as a result.

Then why haven't you done it?

I have. Not explicitly, but I've described exactly what it has to look like in the static case. The precise analytic function invoked doesn't matter (there are an infinite number of possible analytic eddy functions I could use, expressable in terms of the blocked flux, which would satisfy COE).

Here. Since your intuition is failing you... I'll make you a model. I'd make a mesh model if I had more convenient software on hand, but these pictures should describe what I'm talking about.

This is the sort of blockage invoked in the “gravity shield” idea and which you have been talking about: A region of zero gravity, with no fringe effects.

Note the discontinuity in mapping the vector field to the mesh model; this discontinuity represents the “magic energy change” you're talking about. Once you've jumped into that “shadow” of zero gravity that extends out, you have all the potential energy in the world (i.e., zero rather than negative).

Now, it's easy to fix that discontinuity – all you need to do is bend the edges a bit until they meet back up with the original well.

I can take any surface of the form z=f(x,y), and the gradient of f will be a conservative vector field if there are no discontinuities involved in the surface. It's very trivial geometrically that something you can sketch a nice and continuous mesh model for has a line integral of zero along any closed loop on the surface.

Does it matter what any of the actual functions are? Nope. COE is really easy to handle.

Note that – in general – it requires energy to enter the shadow (going “up”) and you gain energy upon leaving it, but as you move away from the blockage, the effects drop off rapidly (1/r^2). Once you get a good distance from the blockage, you don't even notice the effect anymore.

Nope, that doesn't work.

Let's assume that your "outwards eddy" is enough to completely counter the potential energy gained inside the shadow for the wheel.

Now let's make a bigger wheel with the same mass weight such that the weight enters at the same altitude as the original weight but exits the field at a higher altitude.

At which point the eddy has a different value – the eddy drops off as you increase in distance from the blockage – and does not contribute as much delta-v on exit, corresponding precisely to the energy it would have lost traveling from point A to point B in a path that didn't intersect the shadow.

No part of any object, passing through any closed loop, is going to gain any energy.

Oh, another fun thing we could do is push the wheel in a bit such that the eccentric weight is inside the shadow for most of the cycle and only touches the "outwards eddy" for portion of the trip down. The "outwards eddy" thus speeds the descent of the weight while doing nothing to counter trip up. In other words, your explanation is fundamentally flawed and ill-thought out.

No, that doesn't work either. If you brush the edge of something pushing outwards while rotating down, you're going to be sped and slowed by the same amounts. Now you're just getting silly.

That's precisely the problem...

Not a problem at all.

The Gravity Shield PMM (modified such that the weight only enters the "outward eddy" while moving down, as described above) cycles clockwise and gains speed. Does the gravity shield gain or lose energy?

The same Gravity Shield PMM cycles counter-clockwise and loses speed. Does the gravity shield gain or lose energy?

If the answers are different, explain the mechanism for the bidirectional transfer of energy.

If the answers are the same, explain where the unaccounted energy goes to (or is taken from) in the system, and what mechanism allows this.

Your questions are based on the assumption of an incorrectly constructed field.

A neutron inside a field that could have the potential to block neutrons is in an area with "a distinct absence of a field"!? ROFL.

Inside a magnetic envelope is in an area with the distinct absence of a gravitational field, or anything behaving identical to a gravitational field. Do you understand yet why this is relevant?

If you're going to claim an enclosed envelope through which gravity flows around, you need to invoke a strong and most peculiar effect in the edges of the envelope, as well as energy due to the motion of the envelope – which, in turn, invokes messier versions of exactly the same set of problems.

The Crystalline Entity was in visual range. The new vector was pointing directly at it. Your objections are silly and a waste of time, if not on the verge of being offensive.

As “silly” and as much a “waste of time” as your originally bringing up the problem as an objection. If you can establish location of the signal through some flavor of parallax, you can be sure of a source; if not, you cannot be sure of a source.

And just how long does it take before something drops out of warp? (I don't recall anything staying in warp for hours before dropping out.)

You really don't know your Trek then. Try “Encounter at Farpoint Station.” The saucer is detached at warped.

Can it be predicted? If it can, then the massless particles would drop out at a set distance, and thus couldn't reliably be used to measure mass FTL because the area of effect is fixed.

Yes, it can be predicted, at least for larger objects. See also “Encounter at Farpoint Station.” Now, why couldn't that be used for detection of an incoming FTL object? Provide an explanation and we'll talk about it.

Directional graviton detectors require new properties for gravitons to be invented.

Physics may require gravitons to exist. Some “new properties” are to be expected (e.g., the energy of a graviton). Gravitation is an active field of study.

We do not need to resort to directional graviton detectors in order to explain events in Star Trek canon. Therefore, we do not need to invent new properties for gravitons.

We need those either those properties – incidentally, a doppler type effect is more or less required in a relativistic treatment, and directionality would be integral to any particle treatment – or absurd resolution for the sensors to be useful.

As there exist no non-gravity based methods of remote mass measurement without inventing new branches of physics, and we have occasional reference to the detection of graviton fluxes as well as the remote measurement of mass, there need to be useful gravitic sensors.

If the plane has no acceleration relative to earth, then it is at a constant altitude and thus any object in it will be subjected to a normal (for that altitude) level of acceleration relative to the plane. That's a red herring, completely irrelevant.

Not at all. You're assuming a free falling starship completely inappropriately.

And how does one measure force without measuring acceleration, anyway?

Usually one does measure force without measuring acceleration. Stepped on a scale lately?

And how is a spaceship going to measure its current acceleration due to gravity, anyway?

How does a spaceship know its position?

If your instruments aren't precise enough, then you don't receive any meaningful data. Unless you can divine meaning from a stream of zeros.

Of course not.

[SailorSaturn13]

Okay I understand you probably never learned physics... but some statements are "amazing" at least!

Points

Can it be predicted? If it can, then the massless particles would drop out at a set distance,

and thus couldn't reliably be used to measure mass FTL because the area of effect is fixed. If

it can't, then how do they get a measurement when they don't know what percentage of the

actual gravitons they are receiving is?

It can be predicted how big the part of gravitons dropping at any given distance drops out of warp, like : 1% at first LY, 1% of the rest on the second, and so on...

Let's assume that your "outwards eddy" is enough to completely counter the potential energy

gained inside the shadow for the wheel.

Now let's make a bigger wheel with the same mass weight such that the weight enters at the

same altitude as the original weight but exits the field at a higher altitude.

Since the force of gravity is effectively the same on each weight at the bottom and top (the

variance in g is insignificant at this range), then that means the same acceleration would be

felt at entry and exit on the bigger wheel. However, since the wheel should have gained more

potential energy (higher max altitude), the acceleration is no longer enough.

No matter how you look at it, your explanations are full of holes and are not consistent when

actual math is brought to the table.

Oh, another fun thing we could do is push the wheel in a bit such that the eccentric weight is

inside the shadow for most of the cycle and only touches the "outwards eddy" for portion of

the trip down. The "outwards eddy" thus speeds the descent of the weight while doing nothing

to counter trip up. In other words, your explanation is fundamentally flawed and ill-thought

out.

This experiment was thought through with the water model, and this is NOT the problem. The energy change at rand is.

So when ship appears, it sucks out his mass' worth?! How that's not mentioned???

The energy could be in the form of neutrinos.
Where do those neutrinos come from??? ship appears at some point, but you won't find nearly enough neutrinos in the whole system to reduce the mass so.

How did it gain the energy? It didn't move through the field enough to get it from the field.

it does. The field is exactly strong enough to do so.

That doesn't answer the question "How?"

First law of Handwave-Lents - the blocking field transfers all energy needed to change potentials of objects in shadow into the field.

Nor did you answer the second question: Is it possible to suck the blocking field of energy by moving too much mass around in the gravity shadow above it?

Yes. However, due to the graviton diameter point, this is a finite number.

LOL! No, that wouldn't work at all. Water is denser and would slow the movement of the weight down.

which is the problem: left part is "lighter" than right due to weight loss in water so we should , by your reasoning, get a perpetuum mobile.

(And water would pour out at the entrance,

possible to avoid.

the pressure would accelerate the weight away from the entrance.)

NOW this is a good point. well, same happens with gravity shadow.

[Keiran]

It will if you change inertial mass without also changing gravitational mass - as you were trying to claim was happening.

That's a lie. I never claimed that gravitational mass wasn't changed. I said "As far as I'm aware, the gravity well of a mass-reduced object isn't confirmed to have changed." Notice that a gravity well hasn't been confirmed to have changed, not that the gravitational mass was constant.

Please don't strawman my arguments, it's really bad form. And you won't be able to win that way, either.

Again, what you're suggesting has no relation to what's present in the episode. It's a common description of the mechanism for impulse, but what we see in the episode does not involve the sudden acceleration of the moon - along its current vector or towards the planet - when the warp field is applied to most of the moon.

Hence the inertial damping effect I was talking about. (Which was brought up because I thought you were referring to acceleration from another cause.)

t's been done for a long time.

And yet you can't provide it? Interesting. We could clear this all up if you can just provide one. And if it's done, then it's just as simple copy-and-paste and then I can start testing it. It's not like you have anything to fear, right?

Thank you for agreeing with me. I'm glad we had this little chat, because earlier you were saying things like:

As far as I'm aware, the gravity well of a mass-reduced object isn't confirmed to have changed.

You then were attempting to further explain this view.

And what's wrong with that? I didn't say anything about gravitational mass. Are you getting confused again?

Not really. Consider the case when an object only interacts 10% as strongly with gravitons (symmetrically, of course).

We can express this as a local change in G by one order of magnitude. We can also - by thinking of the gravitational constant as constant - refer to this as a reduction of the mass to 10% of the original mass. They're two different ways of looking at the same effect.

That is incredibly broken, mathematically.

Case 1: g = GM/d^2

In this case, reducing G will reduce the gravity well. However, the gravitational mass (and therefore inertial mass) stay the same. The object is not mass-lightened.

Case 2: M = g * d^2 / G

In this case, reducing G will increase the gravitational mass (and therefore inertial mass). The object becomes heavier. Increasing G will reduce the mass, and make it easier to push.

However, changing the local gravitational constant will not change the gravity well.

Your explanation is pure nonsense.

Now, scientifically, it's bad nomenclature to be mixing both terms. But Geordi is not a scientist; he's an engineer, and engineers are notorious for using scientifically incorrect or inconsistent nomenclature that gets the point across. His line there is a perfectly fine example. It gets across the idea that the moon's mass is lessened gravitationally and inertially.

Then what was Q trying to get across? What was his first instinct? "Simple. Change the gravitational constant of the universe." What was that? Reduce G for the universe? "Change the gravitational constant of the universe. And thereby alter the mass of the asteroid..." (emphasis mine).

Oh, change G. Funny... and if you set the equation up correctly, then the mass will reduce, but you gotta increase G. But increasing G is changing it, so it's consistent with what Q said. The gravity well doesn't change, but the mass does.

It appears that Geordi was simply mixing up the direction the gravitational constant would have effectively been changed to locally. It's a pretty easy mistake to make.

Everything else follows from there.

Since you seem to have figured this out now, I will now proceed with your regularly requested lesson plan.

You're too kind. (By the way, that's sarcasm. You were actually kind of mean, all strawmanning me like that.) Your pictures are 404ing tonight, so I'll wait until they're back up before replying.

[Jedi Master Spock]

That's a lie. I never claimed that gravitational mass wasn't changed. I said "As far as I'm aware, the gravity well of a mass-reduced object isn't confirmed to have changed." Notice that a gravity well hasn't been confirmed to have changed, not that the gravitational mass was constant.

Please don't strawman my arguments, it's really bad form. And you won't be able to win that way, either.

Gravity is mutual, you may recall. In the case of the reduction of "gravitational mass," or the local value of G, both the exerted and received forces drop. You start running into complications with regard to CoM otherwise, as I alluded to earlier.

Now, if I have been "strawmanning" you, then you've been "strawmanning" me right back in order to object to things like this:

it loudly and rudely demands a unified reduction of both inertial and gravitational mass.

That is incredibly broken, mathematically.

It's an accurate and quite useful pair of descriptions.

Now, you seem to be hung up on the single-body field equation and some algebraic manipulations thereof to advance the idea that Geordi misspoke. We need not invoke the idea that our expert onscreen is making stupid mistakes like saying "lower" when he meant "raise"; what we have semantically is something quite simpler:

I already told you that when Geordi talked about mass lightening and reducing the gravitational constant, he was referring to two scientifically incompatible (but both highly useful) ways of talking about the same thing. Geordi hasn't made a mistake at all - by the semantic standards of an engineer, at least, if not a logic teacher in the classroom.

We'll call a function that varies with the warp field at a location f(w). Now, take two ways we can rewrite the equation for the vector g:

g=Integral of G(w) multiplied by density and the vector r divided by the scalar r^3 over all space. g shrinks as G shrinks - but pay attention, it's not force per unit mass. It's acceleration, but inertia is no longer fixed to mass in this model. (F=ma needs a warp-field correction factor in this case e.g., F=omega*ma, where omega is the factor by which inertia is reduced).

When you're talking about mass reduction, G becomes a constant again instead of a function. Here, we have:

g=Integral of G times density as a function of warp-altered mass and the vector r divided by the scalar r^3 over all space. g shrinks as m shrinks in the various contributing sources.

In one expression, we have the term G(w)*M in the integral, in the other G*M(w). Both are mathematically equivalent and could be translated to G*M*W(x,y,z), in which case we add a warp factor to all our equations.

Both are valid expressions, just as e=mc^2 and e=gamma*mc^2 are both valid under the appropriate definitions. When Geordi talks about reducing the gravitational constant, he's using the first treatment; when he talks about reducing mass, he's using the second.

The first equation can be held true over all velocities if you're using relativistic mass m(v), and the second holds true with the velocity-invariant rest mass m.

While it is not technically correct to hold both of them as true at once, both methods are an accurate means of description, and fairly intuitive.

Your pictures are 404ing tonight, so I'll wait until they're back up before replying.

They should be working. I can see them in the thread now.

[Keiran]

Okay I understand you probably never learned physics... but some statements are "amazing" at least!

Yes, I took physics and did quite well. Conservation laws were very important in solving many problems. You should learn to understand them some time.

It can be predicted how big the part of gravitons dropping at any given distance drops out of warp, like : 1% at first LY, 1% of the rest on the second, and so on...

The problem is that massless particles don't experience time, so there isn't any obvious reason to allow for a statistical spread like that. What would cause luxons to drop out of warp after different time periods from other luxons?

This experiment was thought through with the water model, and this is NOT the problem. The energy change at rand is.

I can't figure out what you're saying here. Best I can figure is "energy change at random is the problem," but I wasn't talking about any random energy changes, so that still doesn't make sense.

Where do those neutrinos come from??? ship appears at some point, but you won't find nearly enough neutrinos in the whole system to reduce the mass so.

How do you know? Neutrinos react very weakly with matter, so there wouldn't be any big flash or anything. (Heh, and you accuse me of ignorance in physics?) A neutrino burst--or something similar--equivalent to the mass-energy lost would fit our observations and ensures that CoE isn't violated.

it does. The field is exactly strong enough to do so.

[...]

First law of Handwave-Lents - the blocking field transfers all energy needed to change potentials of objects in shadow into the field.

Thank you for that concise exhibit on circular logic. "It does it because it does it."

Hey, I thought you were arguing that CoE doesn't apply in Star Trek?

Nor did you answer the second question: Is it possible to suck the blocking field of energy by moving too much mass around in the gravity shadow above it?

Yes. However, due to the graviton diameter point, this is a finite number.

LOL. Oh, no, a finite number!

(And if you meant infinite, then you answer isn't "Yes" anymore.)

which is the problem: left part is "lighter" than right due to weight loss in water so we should , by your reasoning, get a perpetuum mobile.

There are other factors to consider than the buoyancy of the object in water. The medium is completely different, and work is being done against the gravity field while rising (because there is a gravity field to do work against). The two situations aren't comparable.

possible to avoid.

How? More handwavium?

NOW this is a good point. well, same happens with gravity shadow.

Why? Because you say so?

[SailorSaturn13]

I see I need to explain some basics about gravitons theory. Note: this is ALL speculation, as we don't =know= gravitons exist. But IF they exis (and has properties similar to photons) it works like this:

Gravitons are quantum discretizators of space distortion field. They are particles with no rest mass and can be seen as wave OR particles. Each graviton, like each particle has length (so much for your physics knowledge) which depends on its energy. A body, when alone and resting, DOES NOT emit gravitons - however in universe , no one is truly alone...
Now there are REAL and VIRTUAL gravitons.

Real gravitons are prodused by any body that accelerates, and as long it accelerates. They can interact with other accelerating objects, or be absorbed by matter. However, they DO NOT constitute potential energy. Also important, as mass cannot be negative, such gravitons can only push objects away from, but not towards their source. (The delta impulse is always in the same direction)

The actual between-bodies gravity is enacted by virtual gravitons. The following points are important:
1. A virtual gravoton always happens between TWO bodies (so our graviton blocker WILL "know" there is the other mass around and adjust energy for the field accordingly)
2. During the process of interaction, CoE is timely violated - the system gets more aenergy and then less again. Good meachanism for our field to pump energy in...
3. In order to be able to pull, the graviton MUST be at least as big as the distance between bodies! So a graviton in Earth-Sun interaction have length of over 150000000 kilometers. NOw it is clear that a) a few meters long nullifying chamber will hardly affect such gravitons. and b) as the affecting potentioal decreases with growing graviton length, the loss of potential per meter of distance falls very sharp (fourth degree) and the integral is finite over infinite distance. The latter means that no matter what is after the Gravitonj Stopper, the total energy to change potential will be finite - and not at all that big.

Digest this first....

[Keiran]

A starship requires peak outputs ranging all the way up to the zettawatt range – but should not be idling much higher than the low exawatt range (e.g., 12.75 exawatts). Then you have to wait for a while – a few years to “warm up” to higher powers – and then have to flood the (tiny, see pictures of Romulan power cores) control room with a highly precise stream of thousands of tons of matter to recharge it.

A 12.75 exawatt black hole would, IIRC, be 5,300 tons – and, unfortunately for anyone trying to feed it in a controlled fashion, has a very small event horizon, substantially smaller than an atomic nucleus. You need to be feeding a stream of 140 kg of matter per second to a region a few millimeters in diameter.

So what's your solution? So far you have made these two mutually contradictory claims:

"I could also bring up the issue of Romulans using artificial black holes as power generators with controlled outputs; gravitational blocking and/or local variation of G provide very elegant solutions to a number of problems that otherwise make things quite messy."

and

"Did I mention that if you lighten mass, the entire universe is in the "shadow"?"

You see, a bigger "artificial black hole" made by mass would, well, be very massive. The only way the ship could move around would be to mass-lighten the black hole. But in your version of mass-lightening, the gravity well--the entire point of having a black hole--is also reduced, and so a mass-lightened black hole wouldn't be a viable power source.

Which is it?

Yes it is. When you are “thrown out” along different paths, you experience different forces as a result.

You misunderstand. The exit point is the same, while starting in different areas of "shadow" with 0 graviton flux. They then move to the same location, and the exiting particles would each experience the same force.

I have. Not explicitly, but I've described exactly what it has to look like in the static case. The precise analytic function invoked doesn't matter (there are an infinite number of possible analytic eddy functions I could use, expressable in terms of the blocked flux, which would satisfy COE).

Then do it. The function does matter, because that's what's needed to analyze. Just point out what assumptions you're making.

Being able to point out that one path conserves energy isn't evidence that all possible paths are conservative. That's why an actual function is required.

Now, I could build my own function based on the picture you provided and easily show that it isn't conservative, but then you'd just complain that I didn't build it correctly, right?

Now, it's easy to fix that discontinuity – all you need to do is bend the edges a bit until they meet back up with the original well.

[...]

I can take any surface of the form z=f(x,y), and the gradient of f will be a conservative vector field if there are no discontinuities involved in the surface. It's very trivial geometrically that something you can sketch a nice and continuous mesh model for has a line integral of zero along any closed loop on the surface.

Does it matter what any of the actual functions are? Nope. COE is really easy to handle.

Note that – in general – it requires energy to enter the shadow (going “up”) and you gain energy upon leaving it, but as you move away from the blockage, the effects drop off rapidly (1/r^2). Once you get a good distance from the blockage, you don't even notice the effect anymore.

Yeah, it's really easy to handle when you don't use any math; just use circular logic.

At which point the eddy has a different value – the eddy drops off as you increase in distance from the blockage – and does not contribute as much delta-v on exit, corresponding precisely to the energy it would have lost traveling from point A to point B in a path that didn't intersect the shadow.

No part of any object, passing through any closed loop, is going to gain any energy

Really? 'Cause you have it backwards. The "eddy" would have to increase here. Remember: as the wheel radius increases, the amount of vertical distance it covers while in the eddies is reduced, but the energy gained during the drop outside the shadow is increased. The "eddies" have less of an effect.

No, that doesn't work either. If you brush the edge of something pushing outwards while rotating down, you're going to be sped and slowed by the same amounts. Now you're just getting silly.

The horizontal movement would be canceled, but the vertical acceleration from your "eddies" is not canceled. It'd be really easy to show this to you if you built an actual function that we could use. (After all, you wouldn't accept mine, I'm sure.)

Your questions are based on the assumption of an incorrectly constructed field.

Perhaps you could give me a correctly-constructed field to work with? Because all I'm seeing from you so far is a circular argument: "the field is conservative because I say it's conservative."

Inside a magnetic envelope is in an area with the distinct absence of a gravitational field, or anything behaving identical to a gravitational field. Do you understand yet why this is relevant?

Do you understand that the fact that this field can affect neutrons (and gravitons) is relevant? If it can affect neutrons, then it can affect matter in general. It is obviously not a magnetic field as we know them.

If you're going to claim an enclosed envelope through which gravity flows around, you need to invoke a strong and most peculiar effect in the edges of the envelope, as well as energy due to the motion of the envelope – which, in turn, invokes messier versions of exactly the same set of problems.

Come to think of it, flowing around the envelope may allow for a CoE violation as well. (Situate the aforementioned PMM so that the weight just touches the edge of the "eddy" without entering the envelope, and you've got a waterwheel under a waterfall, so to speak.)

As “silly” and as much a “waste of time” as your originally bringing up the problem as an objection. If you can establish location of the signal through some flavor of parallax, you can be sure of a source; if not, you cannot be sure of a source.

Two different problems. For the CE, you have a new gravitational vector pointing (pulsating, really) towards an object you already know the location of. The required assumptions are minimal.

For the case of dropping into a system with x number of objects and using gravitational sensors as your primary means of mass measurement, you've got a limited number of vectors compared to the gravity wells. Even gathering data over time, the problem isn't guaranteed to converge.

Though I suppose they could just chalk those situations up to "gravimetric interference," but it still doesn't explain detection of starship mass FTL, particularly given how weak a starship's gravity well would be at such distances.

So, again, I must contend that Federation starships use gravimetric sensors as auxiliary sensors, and not primary sensors.

And just how long does it take before something drops out of warp? (I don't recall anything staying in warp for hours before dropping out.)

You really don't know your Trek then. Try “Encounter at Farpoint Station.” The saucer is detached at warped.

You're dodging the question. How long does it stay in warp?

Yes, it can be predicted, at least for larger objects. See also “Encounter at Farpoint Station.” Now, why couldn't that be used for detection of an incoming FTL object? Provide an explanation and we'll talk about it.

Is it a set period of time before dropping out? Or is there variation?

Physics may require gravitons to exist. Some “new properties” are to be expected (e.g., the energy of a graviton). Gravitation is an active field of study.

So you feel you can just make up whatever properties suit your needs?

We need those either those properties – incidentally, a doppler type effect is more or less required in a relativistic treatment, and directionality would be integral to any particle treatment – or absurd resolution for the sensors to be useful.

Black/white fallacy. Or gravity sensors just have limited use, as auxiliary sensors.

As there exist no non-gravity based methods of remote mass measurement without inventing new branches of physics, and we have occasional reference to the detection of graviton fluxes as well as the remote measurement of mass, there need to be useful gravitic sensors.

Federation sensors can detect various types and amounts of materials remotely. Obtaining density and therefore mass from that is trivial. Direct graviton observation is not required for this.

And graviton flux can simply refer to the strength of a detected gravity well. Again, direct graviton observation is not required to accomplish this.

Not at all. You're assuming a free falling starship completely inappropriately.

Perhaps you could explain.

Usually one does measure force without measuring acceleration. Stepped on a scale lately?

A scale technically measures distance.

How does a spaceship know its position?

One possible method is triangulating its position based off of relative locations of known stars. However, in order to get acceleration due to gravity from this, your sensors would have to be able to determine the distance of those stars precise to the subatomic level. Again, we're getting into "absurd" territory here.

[Keiran]

Gravity is mutual, you may recall. In the case of the reduction of "gravitational mass," or the local value of G, both the exerted and received forces drop. You start running into complications with regard to CoM otherwise, as I alluded to earlier.

In this case, changing gravitational mass and the local value of G will not affect each other.

Now, if I have been "strawmanning" you, then you've been "strawmanning" me right back in order to object to things like this:

it loudly and rudely demands a unified reduction of both inertial and gravitational mass.

Now that's just plain dishonest. My reply to that was: "How is that a problem? The "acceleration" would be two vectors canceling out, so the velocity remains the same."

Asking "How is that a problem" is very different from objecting.

Now, you seem to be hung up on the single-body field equation and some algebraic manipulations thereof to advance the idea that Geordi misspoke. We need not invoke the idea that our expert onscreen is making stupid mistakes like saying "lower" when he meant "raise"; what we have semantically is something quite simpler:

Something simpler than a simple mistake? (Note that Q didn't make this mistake. He simply said "change.") This will be good...

I already told you that when Geordi talked about mass lightening and reducing the gravitational constant, he was referring to two scientifically incompatible (but both highly useful) ways of talking about the same thing. Geordi hasn't made a mistake at all - by the semantic standards of an engineer, at least, if not a logic teacher in the classroom.

Combining to incompatible ways of describing something into one idea is a much bigger mistake than a simple misplaced negative.

We'll call a function that varies with the warp field at a location f(w). Now, take two ways we can rewrite the equation for the vector g:

g=Integral of G(w) multiplied by density and the vector r divided by the scalar r^3 over all space. g shrinks as G shrinks - but pay attention, it's not force per unit mass. It's acceleration, but inertia is no longer fixed to mass in this model. (F=ma needs a warp-field correction factor in this case e.g., F=omega*ma, where omega is the factor by which inertia is reduced).

And that's completely irrelevant. If inertial mass and gravitational mass aren't the same in your model, then your model is wrong. As you said, this problem "loudly and rudely demands a unified reduction of both inertial and gravitational mass."

You're also changing things so that the gravitational equation breaks, which also makes your claim immediately suspect.

If G shrinks, then either a decreases or m increases.

When you're talking about mass reduction, G becomes a constant again instead of a function. Here, we have:

g=Integral of G times density as a function of warp-altered mass and the vector r divided by the scalar r^3 over all space. g shrinks as m shrinks in the various contributing sources.

In one expression, we have the term G(w)*M in the integral, in the other G*M(w). Both are mathematically equivalent and could be translated to G*M*W(x,y,z), in which case we add a warp factor to all our equations.

In order for that to work, G must remain constant. If it hasn't changed, then why bother talking about changing it?

This is hardly "simpler."

While it is not technically correct to hold both of them as true at once, both methods are an accurate means of description, and fairly intuitive.

"G has changed, but it hasn't," isn't exactly what anyone would call "intuitive." Why would it change in one equation but not another? That's insanity.

[Keiran]

Now onto the hilarity.

Gravitons are quantum discretizators of space distortion field. They are particles with no rest mass and can be seen as wave OR particles. Each graviton, like each particle has length (so much for your physics knowledge) which depends on its energy.

What are you on about? Photons are treated as point-particles.

Real gravitons are prodused by any body that accelerates, and as long it accelerates. They can interact with other accelerating objects, or be absorbed by matter. However, they DO NOT constitute potential energy. Also important, as mass cannot be negative, such gravitons can only push objects away from, but not towards their source. (The delta impulse is always in the same direction)

Source?

How can they push objects away from their source? Gravitation is an attractive--not repulsive--force.

The actual between-bodies gravity is enacted by virtual gravitons.

Don't forget that virtual particles cannot be detected. Otherwise, they wouldn't be considered "virtual."

1. A virtual gravoton always happens between TWO bodies (so our graviton blocker WILL "know" there is the other mass around and adjust energy for the field accordingly)

In this case, detection of gravitons could be a potential source of causality violations. (The graviton could be detected before it would interact with the source light-years away, thus letting you not only know something is there, but where it will be when the graviton would have reached it!)

3. In order to be able to pull, the graviton MUST be at least as big as the distance between bodies! So a graviton in Earth-Sun interaction have length of over 150000000 kilometers. NOw it is clear that a) a few meters long nullifying chamber will hardly affect such gravitons. and b) as the affecting potentioal decreases with growing graviton length, the loss of potential per meter of distance falls very sharp (fourth degree) and the integral is finite over infinite distance. The latter means that no matter what is after the Gravitonj Stopper, the total energy to change potential will be finite - and not at all that big.

Again, source? Gravitons travel at c, so what you're proposing is that gravitons "grow" to meet the target, and that they do so knowing exactly where the target will be after however many light-years it has to grow. Detecting this could allow a causality violation.

[Jedi Master Spock]

Combining to incompatible ways of describing something into one idea is a much bigger mistake than a simple misplaced negative.

Not at all. I can talk about "mass" referring to both relativistic mass and rest mass in the same extended sentence and it be perfectly meaningful.

Him saying "increase" when he meant "decrease," and not catching himself in the next sentence, suggests that he is screwing up professionally. Criticizing for being perhaps a little ambiguous is a reasonable assumption; saying that he's completely off-base about the sign of the change is not.

As the difference between those two is purely semantic interpretation of Geordi's statement, the explanation that makes more sense is to be preferred.

And that's completely irrelevant. If inertial mass and gravitational mass

Where did I mention inertial mass? It's not m in the equations you cited there. m in that model is a warp-invariant "rest" mass in that model, and varies inertially and gravitationally by the same factor (above, omega).

You're also changing things so that the gravitational equation breaks, which also makes your claim immediately suspect.

It doesn't. It simply changes.

If G shrinks, then either a decreases or m increases.

And in that model, since it is G that is varying and m that is an invariant constant, m remains constant and acceleration decreases. (F=omega*m*a in that model - remember?)

In order for that to work, G must remain constant. If it hasn't changed, then why bother talking about changing it?

Because both models are being used - each in its own part of the line.

if
we wrap a low level warp field
around that moon, we could reduce
its gravitational constant...
make it lighter so we can push
it.

Let's change a few words, preserving the structure:

if we decrease the velocity of that moon, the relativistic correction factor gamma will decrease... we can make it lighter.

So to paraphrase your argument's content:

But E=gamma*mc^2, you say? We can rearrange this like so:

gamma=E/mc^2

So if we decrease gamma, mass increases!

Not so. Mass is simply constant in that equation. (Energy, of course, is not - neither is inertia per unit mass in the first model.)

I can say that increasing gamma increases mass - after all, the relativistic mass is equal to gamma times the rest mass, and that's a kind of mass.

Now, is it ever useful to say that increasing gamma decreases mass? No. Gamma does vary; rest mass does not vary; relativistic mass increases as gamma increases.

Your equivocation does provide a good demonstration of why it's important to know which mass you're talking about, but it's not useful otherwise.

"G has changed, but it hasn't," isn't exactly what anyone would call "intuitive." Why would it change in one equation but not another? That's insanity.

Not at all. It's not really the same "gravitational constant." One's really a constant; one's a variable with respect to warp fields (although still a "constant" for the purposes of calculating the effect of gravity on an object by another object.)

[Jedi Master Spock]

Now onto the hilarity.

We do not condone baiting, teasing, or flaming on SFJ. You may argue with SS13 here all you want, but if you want to make fun of SS13, please do so elsewhere.

[SailorSaturn13]

In this case, detection of gravitons could be a potential source of causality violations. (The graviton could be detected before it would interact with the source light-years away, thus letting you not only know something is there, but where it will be when the graviton would have reached it!)

Problem is: quantum and relativity theories don't mix. For photons, what I said is true AND the virtual photon model ignores relativity. Same for gravitons. Relativity changes things, but I don't know how and don't think it's known yet. This is theory mesh.



So for now, what I said is considered as close it can get. Of course, in reality relativity changes things, but this is beyond question.

How can they push objects away from their source? Gravitation is an attractive--not repulsive--force.

And yet , if you calculate the impulse transfer carried by ANY real particle - including graviton - you see that the impulse transferred forces both bodies away from each other. THEREFORE you need virual gravitons to get an attractive force.

What are you on about? Photons are treated as point-particles.

No. Photons have legth which is equal to their wavelength.
When we talk about detectors, we have a diffrenet model (as probability of particle existing in a point) but
even then it's clear that probability of catching such a graviton are VERY small (and hence, what I wrote about the well stands).

Combining to incompatible ways of describing something into one idea is a much bigger mistake than a simple misplaced negative.

And yet, sometimes we need to speak about particles, and the about field.. those two views are incompatible too!

Then do it. The function does matter, because that's what's needed to analyze. Just point out what assumptions you're making.

Being able to point out that one path conserves energy isn't evidence that all possible paths are conservative. That's why an actual function is required.

Now, I could build my own function based on the picture you provided and easily show that it isn't conservative, but then you'd just complain that I didn't build it correctly, right?

WRONG! Any vectorfield derived as differetial of a potential field WILL have the property that ienergy integral over ANY closed path is zero. Thisd is a mathematical theorem. The analytical function of this field doesn't make difference.
All you need is a well that is continuous.

Virtual particles can't be detected by definition (otherwise they wouldn't be virtual). Therefore, direct graviton detection is impossible.

Remember Hawkins radiation? Any virtual particle can be made real if energy is inserted.

Now, I suppose you could rewrite that as M = a * d^2 / G. However, in this case, reducing the local gravitational constant will increase the object's gravitational mass (and therefore also its inertial mass). Geordi clearly had things backwards. Maybe he misspoke and meant to say he was increasing G?


Now, changing G as the local gravitational constant affects the gravity well generated by the object (a at as set distance d). G outside of the warp field is still the universal gravitational constant, so the object continues to feel the same acceleration from other gravity wells that it did before.

He meant thet reducing gravity constant can be viewed as reducing mass.

Reducing one reduces the other, so there's no dilemma.

except gravity well.

The equation you're looking for is KE = .5m*v^2.

The equation for kinetic energy is (m-m0)c^2, where m - actusl mass, m0 - rest mass, c - light velocity.

Without CoE, analysis of onscreen events is impossible. Anything even remotely realistic has to obey CoE somehow, otherwise we might as well be using cartoon physics.

Well, if we thorow CoE completely, then yes. But if we say it is nearly OK (just like newtonian physics is used all the time in our life) we can. Note that in any closed volume, the gravity well problematic is solvable.