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John Armstrong Says

May 23rd, 2007 at 8:10 am

Pioneer1, I’ve been over and over your comment and I can’t make heads or tails of it. You keep saying that

it is not true that G is a constant of nature and that they have forgotten that they defined G as a unit

but back in your history you assert

He simply converted Kepler’s constant k then in use in astronomy into British units and called it G, Newton’s universal constant of gravity.

so you really do think G is a constant, and your real quibble is in the system of units.

Let me be clear about this for you: no one here is asserting the constancy of the numeric value.

So, either you’re dreadfully misinformed as to the use of the term “constant”, or as to physics in general. If you’re thinking that “constant” means “always has the same numeric value (in all systems of units) then you’re just wrong there.

On the other hand, if you really think that the physical (not numeric) value of G changes even at Newtonian scales then you really have no idea what you’re talking about.

May 24th, 2007 at 8:19 am

Pioneer1: A unit is a convention. It’s a definition that cannot be “measured”, but is used to give numeric values to other physical observations by comparison. The distance from the floor to the top of my head is perfectly well-defined, but to get a handle on it for calculations it’s convenient to pick a rod and lay copies of that rod end-to-end until they equal that distance. Then the number of copies of the rod is my height in “rod-units”. As we change the rod from an inch to a foot to a meter, the numeric value I get changes, but my height doesn’t.

A constant of nature is a little more complicated. Let’s just go back to the definition of G as we use it now. Maybe that will clear things up.

To an excellent degree of accuracy in daily experience, any two bodies experience an attraction to each other that is proportional to each of their masses and inversely proportional to the square of the distance between them, and depending on no other properties such as their materials. Do you agree or disagree?

Now, this means that when I double the mass of one object (doubling makes sense independently of picking a system of units) or the other I double the force they experience. Similarly, when I double the distance between the objects I quarter the force they experience. All of this holds without ever talking about what system of units I’m using to measure with.

When we pick a system of units we get numbers for all these values. Then we will get a formula: F=Gm_1m_2/r^2. The m_1 and m_2 terms make sure that when we double one mass or the other — doubling its numeric value — we double the numeric value of the force. Similarly, the r^2 in the denoinator ensures the other proportionality relation.

Given the three proportionalities above and given a system of units, the function which takes the numeric values of two masses and a distance and gives back the numeric value of the force experienced is uniquely specified up to a constant multiple. This number (here I’m calling it “G”) doesn’t depend on anything about the system. It can’t depend on the masses or the distance because it would ruin the proportionality. It can’t depend on anything else because then the force would. The only thing it depends on is our choice of units, and it is uniquely specified by that choice.

So, if we decide to use “furlongs” to measure distance, “firkins” to measure mass, and “fortnights” to measure time (and thus “firkin-furlongs per square fortnight” to measure force) we can set up an experiment with two 1-firkin masses placed 1 furlong apart. We then measure a gravitational attraction between the two masses of 0.0005 firkin-furlongs per square fortnight. That is the numerical value of G in the FFF system of units.

Now what you may be confusing is that sometimes we use constant (along with c) to define systems of units. The speed of light shows up as a factor over and over again, so it’s convenient if we choose a system of units in which c has the numerical value 1. So if we start with fortnights as a unit of time and choose the “light-fortnight” ( = 1.8 trillion furlongs) as our unit of distance, c will have the value 1. We can similarly chose a “fortnight-mass” to use as our unit of mass so that G will also have the value 1. We are not changing G here, nor are we using G itself as a unit. We are using the fundamental constant G as a guideline to define a system of units in which G takes the numeric value 1.

Now, you’re also asserting that Kepler’s “k” is just “just the Astronomical Unit”. I take it you’re referring to the length of the semi-major axis of the Earth-Sun system? If so you’re off-base here too. If we use the astronomical unit to measure distances, the solar mass to measure masses, and mean solar days to measure time, then we have a system of units, and thus can get a numerical value for G within this system. Kepler’s constant k is the square root of this numeric value.

Really I don’t see any evidence that you know anything about physics but history. I would fail a high school physics student who couldn’t keep track of units and what they mean. I’m serious: this is extremely basic stuff here, and you’re all over the map on it.

May 27th, 2007 at 5:44 am

Slide 2: What you’re calling “rational unit” and “conventional unit” are really the same thing. Everything else in your misunderstanding seems to flow from this wellspring.

Slide 5: You’re picking two different lengths and trying to use them both as units. This isn’t how a system of units works. It seems to you like you’re picking two different things because of the misunderstanding on Slide 2 (supra).

Basically, it looks like your impression of physicists is that they’ve picked meters and kilometers as units and derived “1000? as a constant. You’re right that that would be simplistic and silly, but that’s very much not how physicists derive G.

I’ll try to rephrase again: G is a very specific property of physical reality, independent of all choices of coordinates. G has no numeric value in and of itself. When we pick a system of units then G has a numeric value within that system and it is very well-understood how that numeric value changes as we change the system of units.

Notice again that I defined G above with no reference whatsoever to units until I needed to derive a numeric value for a particular system of units.

May 27th, 2007 at 2:13 pm

So your “conventional unit” is exactly what I mean by a unit, but your “rational unit” only gives the answers “less than/equal to/greater than”, right?

But then a conventional unit can serve as a rational unit. Just take the numeric value it returns from a measurement and ask if it’s less than, equal to, or greater than 1. To have both floating around is redundant.

Honestly, the slides are — from top to bottom — too terse and confused to pinpoint exactly what’s wrong. I can, however, say with certainty that your two different units are redundant, and what physicists mean by “unit” subsumes both roles. Your problem then stems from having two different things trying to play one role, and the subsequent dissonance.

May 29th, 2007 at 5:28 am

But how do physicists derive G? I believe that G is not derived from measurements but it is defined.

You can believe whatever you want. However, I just told you in a previous comment how G is defined, and even gave an example of how to determine its value in a specific system of units.

If you have a problem with that explanation, say so. If you have a question about that definition, raise it. To just wave your hands past it as if I haven’t said it at all marks you out as being more interested in your own pet theory than the physical truth, and that’s the first sign of a crackpot.

Oh, and by the way: G is k^2, not k. At least get those basic facts straight.

May 29th, 2007 at 10:39 pm

And what you’re ignoring is the fact that at everyday scales some attraction is observed. Bodies do accelerate towards each other. Yes, the Newtonian model is now known to be ontologically lacking, but it is a very good predictor within everyday human scales.

And yes, the given proportionalities are observed. Write down Newton’s law and watch Kepler’s equations fall out. What was once merely a curve fitted to a set of observations is now explained. Double the mass of one object and the other accelerates twice as much. Double the separation and the acceleration falls to a quarter. You can argue philosophy all you want, but at the end of the day nature has the last word, and these proportionalities are borne out in experiment after experiment.

F is a placeholder for R/TT. Placeholders are labels and they don’t vary with other variables.

It seems here (and in the sequel) that you’re confusing dimensions and quantities like you confused units and numbers before. Do you mean “F has dimensions [distance][time]-2??

Assuming that’s the case, it’s not only wrong, but it’s immaterial. Again, the simple fact is that doubling the separation of bodies quarters the force each feels. You’re trying to disprove what I’m saying on the basis of dimensionality by glossing over the fact that the constant of proportionality — Newton’s constant G — itself has dimensions. In fact those dimensions are what tells us how to change the numeric value of G when we change sytems of units!

Yes, force has units. Specifically its dimensionality is [mass][distance][time]-2. Each body’s mass has dimensionality [mass], and the separation has dimensionality [distance]. Thus the observed proportionalities — derived from experimental evidence — show that force is proportional to the product of the masses divided by the separation squared; that is, to a quantity of dimensionality [mass]2[distance]-2. The constant of proportionality, thus, has dimensionality [mass]-1[distance]3[time]-2. This compensates for what you see as a mismatch in dimensions.

Please, stop advancing one after another argument that the high school physics books oversimplify the history of the Cavendish experiment and how it was viewed at the time. Honestly, nobody here is arguing that point. We’re trying to bridge the incredible gaps in your grasp of really basic physics. Now, go pick up one of those high school physics textbooks and actually read the physics instead of just ranting about the unfairness of all the historical notes in them.

May 29th, 2007 at 10:40 pm

Evidently tags are disallowed. Please read all the numbers in my expressions of dimensionalities as superscripts.

May 30th, 2007 at 7:10 pm

Physicists look at A and say it is a constant of nature because it stays invariant under unit transformations. Does this make sense?

Most emphatically no. The numerical value of a constant of nature does not stay the same when we change units. I’ve said time and again that G has different numerical values in different systems of units, and the dimensionality of G (which I had to explain in my last comment) is what tells us how that value changes.

June 2nd, 2007 at 1:33 pm

pioneer, your comment on what constitutes an experiment is garbage from beginning to end. I honestly can’t refute it because there’s nothing sensible there to refute. It’s not even wrong.

June 4th, 2007 at 6:46 pm

I reduce G to k which is a conventional unit

As I’ve already pointed out, G is the same as k^2, not as k. At least get your facts straight.

I show that G was never observed

You rant about the Cavendish experiment. Yes, the standard history there might be somewhat oversimplified. Rest assured, though, that G has been measured plenty of times since then.

I note that G has dimensions

Every quantity in physics has dimensions. This is pure nonsense.

I note that physics does not care if G exists or not.

I don’t see you as having done anything of the sort. To a very good degree of accuracy on everyday human scales such a constant undoubtedly does exist. Its numerical value can be measured as I’ve already indicated. You still have yet to offer a single coherent rebuttal to my description of G in terms of Newtonian mechanics.

All of your nonsense about labels and conventions is just a smokescreen to hide your dazzling ignorance of basic high-school physics. You are unquestionably a crackpot, desperate to gain credibility from being the “outside man”. You cower under a pseudonym and a broken hyperlink, you claim to know better than everyone else, and you refuse to respond to the actuality of physics. Take your medicine show on the road, pioneer. We aren’t buying anything here.

June 4th, 2007 at 9:32 pm

In the hard experimental science of physics there is no definition of what an experiment is… Physicists cannot go wrong by adapting my definition as a starting point

In the science of zoology there is no absolutely precise definition of what a cow is. However, zoologists can most certainly go wrong by adopting a definition of a cow as spherical as a starting point.

Physics (and the philosophy thereof) does indeed have a notion of what consitutes an experiment. You just don’t like it because it doesn’t support your pet rantings, which (again) all stem out of your cries of “wolf” over some oversimplified history.

Cavendish did not intend to measure G, though since his experiments others have used his data to give a value for G. Similarly, astronomers observing the perihelion of Mercury did not intend to measure frame-dragging, but since then their observations have been used as verification of general relativistic effects. You’re right that the standard story is not quite accurate, but everything else you’ve said after that one jumping-off point is wrong.

Oh, and that link? Serious discussions do not look kindly on sock-puppetry. Again, I call on you to unmask and drop this pseudonym farce. And stop patting yourself on the back for being a voice crying in the wilderness. Instead, go and read some actual physics and do some actual experiments. You’re a petulant child, sulking that we won’t play along with your make-believe and it’s high time you grew up.

angryphysicist: I’m sticking with this because these sloppy thinkers are out there all too often without voices to counter them. It’s easy for someone to find their ramblings and be taken in. If we’re committed to the truth, we are also committed to fighting ignorance.

I’m not going to convince pioneer of a damned thing, and I know it. I might, however, serve to dissuade anyone else reading this thread from following him. I might also give other readers ideas of how to respond to this sort of thing other than throwing their hands up in frustration. If nothing else, it forced me to work up my explanation of where G actually comes from (in the Newtonian framework), thus solidifying my own understanding of the matter.

June 5th, 2007 at 6:03 am

What Klein calls, “the atom of the philosophers” was exactly something like G of today. The philosophers defined a unit of the indivisible and labeled it atom. This unit then was turned into a constant of nature.

And again you outdo yourself. Nobody ever, ever used either the term “unit” or the term “constant of nature” to refer to an atom. Again you prove that your uses of the terms have absolutely nothing to do with their use in physics, that you have no clue how they are used in physics, and that you don’t care. Yet you feel privileged to complain about physics.

Go read Alan Sokal’s Fashionable Nonsense and Intellectual Impostures for a lot more of how a real physicist took down a group of poststructuralist crackpots who similarly mangled the use of mathematics and physics terms.

June 5th, 2007 at 6:40 pm

You say you know that G is k^2. However on this entire page this is the first time you’ve said it. You keep saying over and over that G is k. Of course, part of your condition is a pathological inability to admit that you were wrong.

And no, I do not accept that “definition” of G, since

(a) R_0 and T_0 are completely undefined. Of course, this is then held up as evidence against the physicists, completely ignoring the fact that

(b) No physicist uses that purported definition. It’s a complete straw man. This, of course, is because

(c) That page is, again, pure sock-puppetry. You’re basically saying, “look, I agree with me”. You’re putting words into the mouths of supposed physicists and then pointing to those words as evidence against them, all under the pretense that it’s someone other than yourself talking.

Instead, the definition of G is exactly what I explained above here. It’s there for all to read, and it’s you that refuse to read it, despite me pointing you back to it over, and over, and over, and OVER!

Do the experiments to verify the proportionalities that I assert (yes, I have done them myself. It was part of middle-school physics)....

June 5th, 2007 at 6:55 pm

You say you know that G is k^2, but on this entire page, this is the first time you admit to it. You keep saying again and again that “G is k” and I’ve corrected you a number of times. Of course, even when you admit you were wrong you try to spin it in a way that it looks like you aren’t admitting any such thing. Stop lying and just admit it.

Now, as to that “definition”, I most certainly do not agree with that. This is because

(a) The quantities R_0 and T_0 are completely undefined, either on this page or on that one. Of course, this is then used as evidence against physicists, completely ignoring the fact that

(b) No physicist actually uses this purported definition. It’s a pure straw man, which is extremely convenient because

(c) The page is again pure sock-puppetry. You make up these “outside” sources just to point to them and imagine that they support your point.

Instead, the definition of G is exactly what I stated before. It’s on this very page. I’ve referred you back to it over and over and over, and you claim to have read it but it’s clear you’ve done nothing of the sort. It’s here. Now go read it.

Then perform the experiments to verify the proportionalities I assert. Yes, I’ve already done them. It was back when they were part of middle-school physics.

Then for once, just once in your miserable life, take off your blinders and try to hear something without it being filtered through your petty prejudices and silly superstitions.

June 6th, 2007 at 5:41 pm

Pioneer, in my misflagged comment I addressed this. Your paranoia about this single oversimplification in popular and low-level physics books is the root of every other mistake you make.

Physics does not say that Cavendish measured G. Physics says that the data Cavendish’s experiments collected can be used to derive a numerical value for G. Similarly, astronomers observing the perihelion of Mercury were not intending to measure frame-dragging effects, but once Einstein laid out general relativity that experimental data could be used to check against his predictions. Just because the measurement preceded the theory doesn’t mean it can’t be used as a check on the theory.

You’re reifying “physics”, and then putting words in its mouth. This is a straw man, and you’ve even mangled that.

From this point you go on to show your total lack of comprehension of anything physicists actually do have to say. You show prejudicial contempt for the community you claim to try to understand. You ignore all evidence that doesn’t fit into your worldview.

In your previous comment you ask if I agree with your “definition” of G, and if not, why not. I most emphatically reject it because

(a) The quantities R_0 and T_0 are nowhere defined. Not on this page, nor on the page you link to. It’s completely ambiguous. Of course, this doesn’t matter because that purported definition constitutes

(b) a straw-man attack against some imagined “Newtonist” bogey-man. You put words into the mouth of your imagined opponent so you can claim that those who correct you are actually attacking physics itself.

And besides, I already told you how G is defined. I explained it and keep referring you back to it. It’s here in this comment. I don’t care how many times you claim to have read it and then link to something completely unrelated. Actually read it for once.

And then do the experiments to verify the proportionalities I assert. Yes, I’ve already done them back when they were part of a middle-school physics class. Observe the real world around you unfiltered by your blinkered ideologies.

And then read through the logical steps in the derivation. Go ahead and try to find the logical flaw in my argument. The only flaw is the fact that at scales extremely outside those of everyday human experience the proportionalities are ever so slightly off. The derivation I present of G is the one physics actually uses. No actual physicist uses your purported “definition”. If you’re going to attack physics, attack what it actually says and not what you say it says.

Oh, and even if you say you understand that G is k^2 rather than k, this is the first time in this entire thread you’ve said so. I’ve pointed it out before and you’ve only acknowledged your error this once, and only by trying to phrase it so it appears that you’re not admitting any error at all.

June 10th, 2007 at 10:27 am

pioneer: again you elide my explanation above. Go and read it. Don’t just say you have, do it. It’s really all in there, and unless you can find fault with a specific part of my explanation you’re just dodging.

You’ve shown yourself to be more interested in pushing your own agenda than in what actually happens, and now you’re proving it conclusively. You don’t get to ask me again and again what I’ve already answered. Unless you go back to my explanation of G that I even handed you a link to, and you find some actual problem in there (”It doesn’t support my pet theory” doesn’t count), then I consider this conversation over. You’ve proved conclusively that you’re a quack, and a hack who knows absolutely nothing about the subject you rail against. Go tilt against your windmills somewhere else.

June 12th, 2007 at 6:35 pm

There is no ambiguity. Physicists oversimplify the explanation. The more refined statement is as I said: Cavendish made measurements, and after the fact that data was used to give a numerical value for G.

Yes, I’ve already agreed that the standard statement is inaccurate on its face. The remedy, however, is not in throwing out independently-verified experimental evidence. The remedy is to make the statements more accurately reflect the history.

June 18th, 2007

Of course Newton’s “force” doesn’t exist in nature. I don’t know a respectable physicist who would say that it really does. What they say is that Newton’s equations describe a model which gives very accurate predictions at everyday scales.

Then they note that Newtonian mechanics fails in areas where relative velocities are large. In those cases the predictions are no longer so accurate, but we have other models (special relativity) to take up the slack. Similarly, Newtonian mechanics fails in the area of very small actions, and we have yet more models (quantum mechanics) to pick up that slack.

See, physicists don’t hold that Newtonian mechanics is the is-all-end-all the way you pretend they do. They recognize failures of the Newtonian model and devise newer, more accurate models in response.

June 20th, 2007

angry, of course pioneer ignores the fact that when physicists use the term “is” or otherwise speak in semantically realist terms that it’s intended to be expanded into model-theoretic semantics. The whole rant against the straw-man of “Newtonism” is based on this (intentional?) misunderstanding.

So much of the language of physics is shorthand, and pioneer insists on confusing the shorthand for ontological statements.

June 26, 2007

Here we are not talking about Newtonism. We are talking about a specific symbol in physics.

Bullshit, you swing back and forth between the general attacks on the establishment of physics and this particular point whenever it suits you. Whenever we respond to one point you move to the other. It’s a cheap trick, designed to hide the fact that you’re wrong on both counts.