Talk:Weight/Archive 3

Further comments on "ISO definition" section
1. When commenting on the 2006 ISO definition, the article says:


 * "When the reference frame is Earth, this quantity excludes the effect of buoyancy of any fluid in which the body might be immersed."

However, this is not what the definition actually says, at least not as quoted. The definition merely says "The effect of atmospheric buoyancy is excluded in the weight", which says nothing about buoyancy in any other fluid (e.g. objects in water) and also does not specify that this exclusion only applies "when the reference frame is Earth". For example, what about the effects of buoyancy when an object is weighed in an accelerating lift, or even on the surface of another planet with an atmosphere?

2. In the lengthy discursive section beginning "Weight here is the force necessary to put an object in a 'particular reference frame,' ..." and continuing to the end of the section, it becomes increasingly unclear which ISO definition is being discussed, or whether it applies to both. In fact, the flow naturally suggests that the discussion applies specifically to the 1992 wording, but then why would the article spend so much time discussing a superseded definition?

86.181.201.14 (talk) 20:12, 8 June 2011 (UTC)
 * This section is in desperate need of some clean up. It contains way too much editorializing, and too little discussion based on sources.TR 08:50, 9 June 2011 (UTC)

Comments on "Operational definition" section
1. A couple of quibbles about this:


 * "A minor issue with the formulation is that the operational definition, as usually given, does not explicitly exclude the effects of buoyancy. As a result, a floating balloon or an object floating in water might be said to have no weight. However, this is commonly regarded as an instrument-dependent problem, since in theory, an object will always be weighed in a vacuum with the correct instrument."

For something that's supposed to be a definition, the wording "in theory, an object will always be weighed" seems unacceptably vague. Does the operational definition require this, or does it not? Or is the "definition" itself imprecise and open to multiple interpretations? Also, it's not very clear to me what "with the correct instrument" is adding. How could using an "incorrect instrument" cause buoyancy to be present in a vacuum?

2. I think it would be good to put a mention in this section of another significant contributing factor, which is centrifugal force from the Earth's rotation. It is mentioned in the next section, "ISO definition", but not here. Perhaps we could just say something like "In the operational definition, an object's weight is lessened by the effect of the centrifugal force from the Earth's rotation." Is that correct?

86.160.219.84 (talk) 13:40, 8 June 2011 (UTC)


 * Yes, the centrifugal effect also needs to be included in the operational defintion, since "operational definition" just means what your scale tells you. The problem with "operational definition" however, is "WHAT SCALE?" For example, if you're standing on a little sealed bathroom scale on the bottom of a child's playpool in 1 foot of water, it will give you one weight of yourself (an incorrect one). But if you zero a much larger scale and put it under the entire pool, and THEN climb in, it will give you another weight for you-- this time a correct one. That's the problem with all operational definitions-- bouyancy can always be zero'd out if you have a large enough scale, but for baloons and aircraft that scale would have to cover the entire surface of the planet. Weighing in vacuum is merely the alternative to the "giant ultimate scale" (which works even in non-vacuum, since the fluid always transfers the weight somewhere). S  B Harris 20:25, 8 June 2011 (UTC)


 * I added a mention of centrifugal force to that section. 86.161.61.134 (talk) 11:39, 11 June 2011 (UTC)

ISO definition = Operational definition?
1. The lead section is clear that "the International Organization for Standardization (ISO) use an 'operational' definition, defining the weight of an object as the force measured by the operation of weighing it (using a force-sensitive scale, such as a spring scale), in vacuum."

Later, when one reads the ISO definition, it is somewhat offputting to find that it contains no reference whatsoever to the "operation of weighing". Do we want to somehow include the words "equivalent to" in the lead section?

2. The "ISO definition" section contains the following explanatory paragraph:


 * "The definition is dependent on the chosen frame of reference. When the chosen frame is co-moving with the object in question then this definition precisely agrees with the operational definition. If the specified frame is the surface of the Earth, then the definition agrees with the gravitational definition, save for centrifugal effects due to the acceleration of motion at the spot on the surface of the Earth where the measurement is taken."

It's easy to read this as implying that the ISO definition does not agree with the operational definition when the specified frame is the surface of the Earth. That can't be the intention, can it? 86.160.218.185 (talk) 11:31, 9 June 2011 (UTC)
 * The current statement in the lead about the ISO using an operational definition, has no basis in the sources whatsoever. I'm not sure who put it there in the first place.
 * I guess the best way to put it is that the ISO definition agrees with the operational definition if the chosen frame is the rest frame of the measuring apparatus. Normally the operational definition assumes that the measurement apparatus is at rest with respect to the object being weighed. In that sense the ISO definition only agrees with the operational one, when the choosen frame is co-moving with the object.
 * As I mentioned before, this article needs a LOT of work.TR 11:48, 9 June 2011 (UTC)
 * Yeah, unfortunately this article (and also apparent weight, which some want to do away with completely) has been in a poor and unstable state for years. It would be really nice to get it up to scratch. 86.160.218.185 (talk) 13:53, 9 June 2011 (UTC)
 * Oh, BTW, in #2 I really meant "It's easy to read this as implying that the ISO definition does not agree with the operational definition when the specified frame is the surface of the Earth even if the object is at rest on the surface of the Earth ". 86.160.218.185 (talk) 14:03, 9 June 2011 (UTC)


 * Looking at this again, I'm wondering if something has been lost from our quote of the ISO definition. The text "It should be noted that, when the reference frame is Earth..." to me suggests that the term "reference frame" has already appeared in the definition, but, as quoted, it hasn't. Also, in "$$F_g = m g \, $$, where m is mass and g is local acceleration of free fall" it's very easy to understand the "local" acceleration of free-fall to be approx. 9.8 m/s^2 near Earth regardless of how the object being weighed is accelerating. This would mean, for example, that the ISO-defined weight of an object in free-fall near the surface of the Earth is the same as its weight at rest on the surface of the Earth -- not zero as other parts of the article seem to imply. To make it completely unambiguous what "local" means seems to require further careful wording that in our quoted version is absent. (I suppose we are 100% sure that the ISO-defined weight of an object in free-fall is zero?) 86.160.86.247 (talk) 17:40, 9 June 2011 (UTC)


 * In ISO defintions of weight, there is always implied a mass being weighed, and implied also is that you have to be in THAT mass's location AND reference frame, in order to do the weighing! How would you weigh a mass moving past you? A definition that depended on any reference frame you chose to be in, to weigh a given mass (in some other frame!), would be no definition at all, since the answer would vary to anything you like. The ISO mentions the "local acceleration of free fall." By "local" surely they mean not mere physical location of the object being weighed, but also reference frame locaton of the object being weighed. That fixes the problem, and agrees with earlier definitions, which were more complicated, and were couched in terms of the force needed to get the object from the reference frame defined by it (where it was stationary and being weighed) INTO the free fall frame. S  B Harris 19:28, 9 June 2011 (UTC)
 * NOFI, but you are talking out of your ass here. It is absolutely clear that the definition refers to any reference frame you would care to choose. According to the definition, a mass in orbit around the Earth is weightless with respect its co-moving frame, but with respect to the surface of the Earth it not weightless. Anything else is your projection of what you think the definition should be.TR 20:38, 9 June 2011 (UTC)
 * Excuse me? If you think the ISO is going to completely revamp their definition of weight between 1992 and 2006 and not mention that it's been totally changed between, I think it is you who are projecting to your own fantasy. For those complaining about the ISO paywall, note that the Ethiopians have copied all the 2006 revision to the web, so you can see it: . It's really very short, and is no longer than what's quoted in the article now. "Acceleration of local free fall" is not defined, but one presumes the ISO means the same thing they did when they carefully defined it in 1992. They intend "local" to mean the reference frame of mass and scale. They are talking about the acceleration of free fall in THE reference frame where both your mass and your scale are. It's pointless to talk about "weight" (which is the force between mass and scale) in any other context. Indeed, they SAY: If the reference frame is Earth, this quantity comprises not only the local gravitational force but also the local centrifugal force due to the rotation of the Earth. The clear implication of talking about "THE reference frame" is that there is ONE reference frame. The one for both mass and scale. We do not have one for the mass, one for the scale, one for your Aunt Tilly, and different weights given by any given combo. S  B Harris 02:51, 10 June 2011 (UTC)
 * The 1992 version was even more clear that the definition was dependent on an independently specified reference frame.TR 05:38, 10 June 2011 (UTC)
 * Sorry, but you'll have to quote it, because I don't see it. See page 24 of the commentary here on the ISO 31-3 (1992) version: . In science and technology the weight of a body in a particular reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame. You are merely assuming that the "particular reference frame" is not at all connected with the body and can be any reference frame you like. Big assumption. Nowhere is that made clear. In the absense of this, it's easiest to assume that "the weight of a body in a particular reference frame" MEANS "the weight of a (body in a particular reference frame)," which is to say, the reference frame defined by the body, as the one in which the body is at rest. In the only example given, the reference frame is the surface of the Earth, which is rotating, and the body with it. So they've chosen an accelated frame, but one in which the weighed body is at rest. If they meant to weigh bodies in any OTHER frame than the one where they are at rest, they'd have given an example. Nothing of that sort is given or implied. S  B Harris 19:05, 10 June 2011 (UTC)
 * The actual ISO 31 use the wording "in a specified reference frame", strongly suggesting that the reference frame is something that needs to be specified independently. Moreover, reference frames are attached to observers not objects. A "body in a reference frame" is not really something that makes sense physically, although by abuse of terminology you could interpret in the way you want. The most natural interpretation of the wording is as defining the weight of an object respective to the reference frame of some observer. This also more natural with the remark version that weight depends on the reference frame. (Otherwise they would have said it depends on the motion of the body.)TR 10:32, 11 June 2011 (UTC)
 * I suspect the implied "observer" here is the guy with the scale (or the scale itself), since "weight" is a measured quantity. And again, it gets very difficult if you want to talk about the scale being in any relative motion to the mass being weighed, so operationally the two frames are the same. Any "observer" can put themselves in any free-fall frame in that case, and claim ANY mass has zero weight. But what good does it do to have a definition of "weight" that makes Mt. Everest or the great pyramid at Giza weightless to a guy falling by it with a scale in his hand? So what? How does such a definition help technology? S  B Harris 22:53, 11 June 2011 (UTC)
 * Well, such a definition tells designers of safety nets how strong their net needs to be to slow down a 80 kg in free fall. For that purpose it is somewhat pointless to tell the manufacturer that the man is weightless in his own reference frame. (Not a really good example, of course, since for most safety nets somebody falling into it will have reached terminal velocity before reaching the net. My point is that there are situations where it is useful to assign a weight to body from the perspective from another frame then its comoving frame.) Sure you can specify any frame you want to get pretty much any result. In the end the result will only be as meaning full as the frame chosen.TR 23:12, 11 June 2011 (UTC)
 * But you cannot use such a calculation to design a safety net, as the maximal force needed doesn't come from picking out some non-comoving frame and figuring a 'weight'. In particular you need to know that maximal acceleration of your object induced by the net, and that depends on the elasticity of the net and has nothing at all to do with hypothetical weights calculated in various reference frames. At the end, however elastic the net is, the maximal weight will be measured (and felt) when the decelerated person and the net (and supports) are comoving (at rest to each other)-- even for a moment. And that is the same as all our problems where we assume that weighed masses and their scales, are comoving. Yes, if the net fails, the object is never stopped with regard to the net, but in that case, it's not really weighed, either. A force is applied to it, set by the maximal breaking force of the net, and that's it. You can measure THAT from the net support, but it's not a weight-- it's just the force you applied to an object passing by (and would be the same for any object that broke the net, no matter what its mass was, so it's clearly not connected with the mass of the mass). In the case of a mass that is stopped, there will be a moment of time when the force is maximal, and that's the maximum weight the person feels when stopped by the net, and that the the supports feel. At that instant, the net, the net support harness, and person are all co-moving, and there you are. A weight is not simply a force. A weight is the force (even if it is an instantaeous one) needed to put a mass into the inertial frame of your weight scale. This is operational. You can get such a force that varies by jumping up and down on the bathroom scale. That's a varying weight. As in the Vomet Comet, if you average it out, it comes out to be the traditional one. S  B Harris 23:59, 11 June 2011 (UTC)
 * I agree that wasn't the best possible example. Let's do something simpler, an accelerating car (on a straight road in a uniform gravitational field). If you are forced to use the comoving frame of the car, then the weight of the car changes when its acceleration. The weight of the car would not point down, but would in fact be the entire force that the car exerts on the ground (assuming no friction). This a terribly inconvenient way of describing things. From a practical point of view it is more natural to split the force is a weight that is perpendicular to the road (and which is constant), and a traction force that is parallel to the road. This is achieved by taking the rest frame of the road as the frame in which to define the weight of the car. Operationally, this could be achieved by placing a gaint scale under the road and measuring the increase in weight as the car is placed on top of it.
 * This is the main advantage of using a fixed frame of reference to define the weight of an object; the weight becomes independent of the motion of the object.
 * I think, the thing that you are objecting to is that not all frames are equally sensible from an operational point of view. This is certainly true, as is the case in any gauge theory, not all gauges are equally sensible. This doesn't mean that you should always use the same gauge/frame. The comoving frame is useful in many situations (in the above example, if we we're talking about the weight of the driver it would be most natural frame would be the comoving frame of the car.) This doesn't stop you frame defining observable is any frame/gauge, and letting the gauge be picked by the observer.TR 05:59, 13 June 2011 (UTC)
 * We both agree that things inside the accelerated car weigh more-- as in a dragster pulling 5 g's where the driver weighs half a ton and feels it. (And BTW you can't have traction force without friction-- that's where it comes from). Put the comoving scale in the seat and you easily measure that (the seat frame and supports have to take it). As for the whole car you most easily (more on this below) need a comoving scale to look at that weight, and that's not much harder than putting one under the road. In this case, the scale for the whole car could as easily be in the car tie-downs and supports of the accelerating car-hauler used to move the car at the same acceleration it was otherwise going to do for itself. I've thought about this problem and the one above and decided it's very odd and perhaps not permitted when your mass and scale are in different frames (i.e., they have a relative velocity, either perpendicular or horizontal) while the "weighing" takes place. In both cases you (may or may not) get funny forces whose meaning I'm not sure of, and are sort of undefined (since they are tied up with the elasticity of the system, while one part of the test mass is moving, and the other part is not, and there's a elastic tension inside it-- or the same happens inside the scale, say in a deformation of the pan or spring). The reason this is not weight, is that it's very dependent on the makeup on the instrument or the "give" in the thing being weighed (are your knees locked when you bounce on the scale-- surely that shouldn't affect your weight?). I'm actually NOT sure if you can be said to weigh more or less when/while you bounce gently up and down on a spring scale, while keeping contact with it (or hanging from the spring, either). It seems to me that the verb "to weigh" requires, in a sort of unspoken way, that the scale or weighing instrument BE comoving with the mass weighed. And when it is, all this talk of weight of the mass in yet other frames becomes irrelevant and undefined. If you want to define it, how do you do it? You have frame A at rest with your mass, frame B at rest with your scale, frame C the free-fall frame, and now you want to freely pick yet another frame D?? Or is your third frame always the scale frame? And the total formula is WHAT? W = mg where g = A+B-D? S  B Harris 08:30, 13 June 2011 (UTC)
 * First a bit of physics. Objects are never in a frame. Observers can be in a frame, objects simply move. Form the point of view of general relativity even observers are not in a frame, to adequately define a frame you need a whole family of observers that fill (a region of) spacetime. Typically frames are defined by some physical conditions, for example, you can define the frame in which the surface of the Earth is at rest.
 * Second, I don't see this necessity in scales needing to be co-moving to be able to weigh. In the example, I gave above, I see no reason why you shouldn't say that the weight of the moving car (wrt to the frame defined by the road) is constant and independent of its acceleration. In fact, that is a very practical definition if you want to do predictive calculations towards the movement of the car. (Calculating the movement of an object from a frame that is co-moving with the object is incredibly annoying due to backreaction of the frame.)
 * The ISO definition even tells you how to determine the weight in any frame. 1) determine the accelarition of free fall at every point in the frame. This can be derived from the spacetime geometry and the definition of the frame. (please don't ask how to determine the spacetime geometry, and trust me that this can be done. 2) determine the mass of the object. 3)multiply the two to get the weight. There has to be not actual "scale" involved.TR 15:42, 13 June 2011 (UTC)
 * Having looked at the actual full text in, I have to say that I think the ISO definition is remarkably poor. 86.161.61.134 (talk) 11:49, 11 June 2011 (UTC) Hold on though guys, it says "...where m is mass (item 4-1) and g is local acceleration of free fall (ISO 80000-3:2006, item 3-9.2)". So, we should be looking at ISO 80000-3:2006, item 3-9.2 before we pass any judgment or make any speculations, shouldn't we? Sorry, no time to search further right now... 86.161.61.134 (talk) 13:21, 11 June 2011 (UTC)
 * Item 3-9.2 says nothing more than "g is local acceleration of free fall", unfortunately. In the remarks column it just makes a remark about the accepted  "standard value" of g at the Earth's surface.TR 14:10, 11 June 2011 (UTC)
 * In that case I unstrike my remark that the ISO definition of weight is "remarkably poor". It is IMO completely non-obvious that "local acceleration of free fall" is supposed to vary depending on an object's acceleration, and it seems to me that in this section we are hanging an awful lot of text on an interpretation that is mostly guesswork. 86.160.211.189 (talk) 20:08, 11 June 2011 (UTC)

We're stuck with guesswork about the ISO's intended meaning, either way. If we take the "local acceleration of free fall" as the physical location of the object (but not the object rest frame), then what do we pick as the frame from which to measure? If we use the frame at rest WRT to the gravitating source (this would be the frame at the pole, or something, and wouldn't compensate for Earth rotation), we still need to pick a gravity source, and now we're in trouble. What do we do when off Earth-- pick the nearest one? If we're free to pick any coordinate frame we like for a given mass whose weight we're measuring (as has been suggested) so long as the mass-to-be-weighed is at its physical origin, then there are an infinite number of such frames, each giving differenet object weights. Including an infinite subset number of "orbital-type" accelerated (inertial) frames that have one component that cancels the acceleration of free fall at the location of the object on the Earth's surface, so that we find the object's weight is zero in all of them (since they all have the acceleration of "free fall," even though they represent a variety of tangent velocities and inertial paths). Thus, even if an object is sitting still on the ground, minding its own business, we are required to resister its weight as being zero in the frame falling toward the center of the Earth, and also zero in the frame moving horizontally at circular orbital speed (or rather the Earth orbital frame at ground level), and even as zero in the frame coming in hyperbolically at greater-than-Earth-escape, which will pass tangent to the Earth's surface and our mass at 40,000 mph or something. So long as it is an inertial frame (such as an incoming meteoroid would follow). All for the very same object, sitting still, which can now have any weight you like, depending on what frame you pick to look at it from (sort of like it can have any kinetic energy depending on what frame you pick to look at it from). This seems slightly mad to me, as weight is not like kinetic energy, and seems to me to be more of an intrinsically invariant quantity, connected as it is to proper force. I think the opposite of the proper force needed to keep a mass in the frame where it is at rest, is basically what the ISO intended. It's too complicated otherwise. Only if one uses the inertial frame in which the object/mass and weight-scale are at rest, does any definition of "weight" for the mass approach anything that makes for common sense. Put it this way: what's the weight of an astronaut in a rocket? Does this have physical meaning? Sure. Operationally it's what the astronaut sees when he steps on a bathroom scale. It's what he or she FEELS. And of course, it depends on what force the floor is using to move the astronaut away from a free-fall frame (if any). Weight that the scale measures is the reaction force to that. It's not that mysterious. And yes, of course this felt weight will depends on the "object's" (astronaut's) acceleration (which is the same as that of his scale). S B Harris 20:43, 11 June 2011 (UTC)


 * I don't disagree with anything you say. In practice, except maybe in some highly bizarre situation, I think the only sensible definitions of weight are the strictly operational one or the strictly gravitational one -- e.g., in the case of a body in free fall near Earth's surface, weight is either zero or it's the same as at rest on the Earth's surface. Unfortunately I don't think it's obvious, just from the ISO wording, which one they intend. After presenting the ISO definition, the article launches into a discussion, which I'm not saying is wrong, but just not obviously following from the ISO definition. I don't see that in the article we should be guessing at what the ISO definition means and presenting it as fact. If, as you say "We're stuck with guesswork about the ISO's intended meaning" then it seems we're in trouble. Perhaps we should just move that whole discussion to the "Operational definition" section, and present the ISO definition with no commentary and let readers make up their own minds? 86.160.211.189 (talk) 02:32, 12 June 2011 (UTC).


 * As I see it, the ISO definition is intended to unify the operational and gravitational definitions and assign the difference to a different choice of frame. Just putting it in the operational definition section would be wrong.TR 08:20, 12 June 2011 (UTC)


 * "As I see it" seem to be the operative words here. IMO the definition that we quote is so hopeless and vague that interpretation is a matter of opinion. 81.159.108.132 (talk) 11:54, 12 June 2011 (UTC)


 * All this discussion of the ISO definition, yet I cannot find in the article a link or reference to the ISO source.
 * WikiDMc (talk) 09:23, 15 June 2011 (UTC)


 * That may be because ISO standards are esoteric knowledge that is only supposed to be available to the initiated. Most editors have to use bits and pieces from the ISO standards that crop up in more universally available publications. So far I have never had access to ISO standards at an university. Hans Adler 09:53, 15 June 2011 (UTC)


 * That's probably true, but the editor that included the statement ...the International Organization for Standardization (ISO) use an "operational" definition must have had access to some source, even if it was not the original ISO document. That source should be referenced, whether it is freely accessible or not.  If there is no source available then the sentence, which is closely paraphasing a source (When and why to cite sources), should be removed or amended in accordance with WP:CITE.  If ISO cannot be referenced directly, then the policy Say where you got it should be followed.
 * WikiDMc (talk) 12:14, 15 June 2011 (UTC)
 * There is a subtlety here. The "where people got it", is blatant copyright violation of the ISO documents, . Citing that directly, might actually cause trouble. Since that actually is a literal word for word copy of the ISO standard, we might as well pretend that the authors here used the actual ISO standard as a source. (Which is how it is cite in the body of the text, where the relevant portion of the standard is quoted verbatim.)TR 12:47, 15 June 2011 (UTC)
 * Very true. If the Ethiopian Standard is violating copyright, or we strongly suspect it to be, then it should not be cited.  If we are then to treat the ISO publication as the direct reference it should be cited properly as an in-line citation and added to the list of references according to WP:INCITE and/or WP:INTEXT.  Each quote or reference to ISO throughout the article should be given its own in-line citation and, where appropriate, including the relevant page number(s) within the ISO standard (WP:Page numbers).
 * WikiDMc (talk) 15:08, 15 June 2011 (UTC)
 * Be my guest.TR 15:13, 15 June 2011 (UTC)
 * Indeed. I have added an in-line citation to the ISO definition section, so at least the ISO publication now appears in the references list.
 * Before your edit removing the association between ISO and the operational definition, I was going to add a citation to that paragraph but reading ISO I came to the same conclusion as you; there is nothing in ISO to support the claim that ISO defines weight operationally, despite any "implied" operation of weighing (see SBHarris' comments above). That's a long-winded way of saying that I concur with your edit.
 * WikiDMc (talk) 10:03, 16 June 2011 (UTC)

Break

 * We have to also remember that the article is aimed at a general audience. Even if it is clear to physicists what "local" means, it is IMO very unclear to the general reader. And even if the ISO definition feels it does not need to explain "local" and how it might relate to "frame of reference", I think we are obliged to in the article. At the moment there seems to be a kind of unexplained leap of intuition from the actual words of the definition to the interpretation. I believe that many "ordinary readers" with some inkling that falling objects accelerate but little more are very likely to understand "local" to mean "in the vicinity of", i.e. 9.8 m/s^2 near Earth's surface, independent of what the object might be doing. The question of "how would you do the weighing" in another frame, and so on, is not likely to arise in the reader's mind as an objection to this interpretation because, as I mentioned earlier, the ISO definition that we quote makes no mention whatsoever of any actual weighing operation. 86.160.86.247 (talk) 21:06, 9 June 2011 (UTC)
 * Even worse -- unless I'm thoroughly confusing myself -- the article's opening paragraph does use "local" in this completely different sense when it refers to "the magnitude of the local gravitational acceleration". 86.160.86.247 (talk) 21:28, 9 June 2011 (UTC)

Relativity
In the section "Relativity", the article says:
 * "Gravitational force and weight thereby became essentially frame-dependent quantities. This prompted the abandonment of the concept as superfluous in the fundamental sciences such as physics and chemistry (e.g. atomic weight became atomic mass)."

Is there any basis to this claim that the change from "atomic weight" to "atomic mass" had anything specifically to do with ambiguities relating to relativity, as opposed to simply being a correction of a unfortunate historical choice of term? 86.176.208.134 (talk) 02:08, 26 June 2011 (UTC)
 * The source currently quoted, says there is.TR 15:58, 26 June 2011 (UTC)
 * So the term "atomic weight" comes from applying a weight-based, rather than mass-based, system – as opposed to just using the term "weight" to refer to mass as people do in everyday language all the time? I consider this a very surprising claim, and would like to see the source to get a better idea of where it comes from. Unfortunately my university doesn't have a subscription. Can anyone help? It's here. (My homepage with my email address is linked from my user page.) Hans Adler 20:57, 26 June 2011 (UTC)
 * I've remove the contentious example. It was intended (as in the used source) to give an indication of how weight at some point in history weight was used in fundamental chemistry, when people observed that the weight of a substance did not change when it underwent a chemical reaction. Yet the example seems to cause more confusion than it solves.TR 07:55, 27 June 2011 (UTC)
 * Thanks. I would be quite surprised if anyone had ever asked the question, "Do we want the mass here or the resulting force through gravity?", and answered it with "We want the force!" Basically this only makes sense when you are doing statics, and I suspect that historically things worked out as follows:
 * "Atomic weight" defined at a time when people did not distinguish mass and weight and always called it "weight".
 * Distinction between mass and weight becomes clear. Chemists continue to call it "atomic weight", and may be slightly confused whether it's mass or weight. Or just don't care.
 * Chemists continue to call it "atomic weight", and are aware it's a mass.
 * Usage of the term "weight" to refer to mass becomes more and more rare.
 * Chemists decide to update the word in order to prevent confusion.
 * Do you have the source? Does it contradict my theory? Hans Adler 08:46, 27 June 2011 (UTC)
 * More or less. Anyway, you might try the following link. Your institution might have access via that route. TR 10:29, 27 June 2011 (UTC)
 * That didn't work as I wouldn't know what to enter into the login fields, but now that I am physically on the campus and can use a proper browser, the other link suddenly works. I see what you mean by "more or less". It's moot now, but I don't think these vague comments are sufficiently strong to support the claim. On the other hand, some more reading suggests to me that I should switch my first two points. Hans Adler 14:52, 29 June 2011 (UTC)

ISO definition (again)
From the "ISO definition" section:


 * "When the chosen frame is co-moving with the object in question, this definition precisely agrees with the operational definition. If the specified frame is the surface of the Earth, then the definition agrees with the gravitational definition, save for centrifugal effects due to the acceleration of motion at the spot on the surface of the Earth where the measurement is taken."

Since the work cited was apparently published 10 years before the present ISO definition, and moreover the present ISO definition is IIRC significantly different from the previous version, I don't see how the source can directly support any of the article's claims here. I haven't removed it just yet in case it says something relevant in a way that I haven't envisaged (I have no access to it). I still have a problem that the interpretation in the article seems to rest on an assumption that "local acceleration of free fall" must be zero for a free-falling object (for example), yet in the second paragraph of the lead section we use "local" to mean something completely different. And, if this source falls away, we don't seem to have any authority to back up this assumption.

I'm also not sure I agree with the wording "The following points are emphasized ..." given that the first point ("This quantity is dependent on the chosen frame of reference") isn't even directly stated, let alone emphasised. The first and only reference to "frame of reference" is in "It should be noted that, when the reference frame is Earth ...". This sentence seems so obviously to require a previous mention of the term and how it is relevant, and yet there is none.

86.160.218.143 (talk) 01:50, 27 June 2011 (UTC)
 * There is no significant difference in the 1992 and 2006 ISO definitions. The 2006 definition just replaced the very wordy "force needed to be applied to the object to give it the local acceleration of free fall" (or whatever the exact wording was exactly) by a formula summarizing that sentence in a concise way.TR 07:41, 27 June 2011 (UTC)
 * That this indeed the case is also supported by comparing the versions of the "The NIST Guide for the use of the International System of Units" over they years.(ref 12 in the article) They actually kept the explicit wording of the old ISO standard while referencing the new one, indicating that they view them as identical.TR 08:25, 27 June 2011 (UTC)


 * Also, I don't quite see how you come to the conclusion that "the interpretation in the article seems to rest on an assumption that "local acceleration of free fall" must be zero for a free-falling object". The local acceleration of free fall must be zero in free falling frame. (This is basically the equivalence principle in a nutshell.) As a consequence the local acceleration of free fall is zero for free falling object in its comoving frame. (That is actually somewhat of a tautology.)
 * I also don't see how this use of local (as in the value taken in some local frame) is completely different from the one in the lead.TR 07:50, 27 June 2011 (UTC)


 * I don't think being written 10 years apart precludes them from agreeing with eachother.
 * 86.160.218.143, I agree with your view on the, "The following points are emphasized ..." statement. Who is doing the emphasizing?  If it is the wikipedia author, then surely it is WP:OR or WP:POV.  WikiDMc (talk) 07:54, 27 June 2011 (UTC)


 * Once again: In science and technology the weight of a body in a specified reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame. And the amplification: If the reference frame is Earth, this quantity comprises not only the local gravitational force but also the local centrifugal force due to the rotation of the Earth. Clearly the "reference frame" in the second part is DEFINED as the one in which the surface of the Earth is still (accerated reference frame is comoving with Earth's surface). In no other way would "weight" be affected by the centrifugal force of the Earth's rotation. Thus, it seems to me, QED, that the definition assumes that specified refernence frames, whether they can be attached in theory to anything, in the only example they actually GIVE, are co-moving with both the object whose weight we're interest in, and something else supporting it, in the same frame. The idea of weights of masses in specified frames that are NOT comoving with the objects, simply addresses what the weight would be IF the mass WERE comoving with the specified frame. Which would (in turn) be exactly the same as the force (weight) needed to put it into free fall from THAT frame that it isn't in. And that's the operational definition. IOW, you can take an object on the ground and ask what its weight would be, in some other frame not comoving with the ground-- say what would its weight be in a frame if this specified frame were falling down an elevator shaft, and that's the very same question as asking what its weight would be if THE MASS were falling down down the elevator shaft, comoving with the frame that was. Which is asking, what would be the force needed to attach the mass to the frame falling down the shaft, so that they comove. Operational definition all the way, no? We're all talking about the same thing. S  B Harris 08:11, 27 June 2011 (UTC)


 * What are other's thoughts on this statement, "If the specified frame is the surface of the Earth, then the definition agrees with the gravitational definition, save for centrifugal effects ... "?
 * I do not like it. To me, it is saying that ISO = gravitational definition if you ignore ("save for") the difference between gravitational and operational definitions.  It is akin to saying density of seawater = density of fresh water, save for the weight of dissolved minerals.  They are not equal and should not be represented as such.  WikiDMc (talk) 12:05, 27 June 2011 (UTC)
 * In the gravitational definition it is typically unclear whether or not centrifugal effects are included. (Note that centrifugal effects are not "the difference between gravitational and operational definitions".) To also get rid of the centrifugal you could move to a frame that is comoving but not corotating with the Earth. You'd still have tidal contributions from the Sun, but you could also move an (instantanuous) frame that removes those.TR 13:31, 27 June 2011 (UTC)
 * Maybe my understanding isn't as good as it could be. To help me clarify;
 * * Isn't the sun's tidal effect due to the sun's gravity, and therefore affects weight equally under both the gravitational and operational definitions?
 * * If your reference frame is the surface of the Earth, what else causes a difference between operational and gravitational weight besides centrifugal effects?
 * WikiDMc (talk) 13:56, 27 June 2011 (UTC)
 * To your last point. None, in that situation both agree. That is, in the most common situation an object at rest on the surface of the Earth, pretty much all definitions agree. Which is why the distinction often doesn't get that much attention. The problems start when you differ from the standard situation.
 * Tidal forces are a competition between "gravitational" and "centrifugal" forces. Or maybe more accurately, they result when a rigid body moves through a gravitational field with a gradient. In that case only the center of gravity of an object follows a free fall trajectory. The tidal force is the resultant fictitious force on the other points of the body following a non-free fall trajectory.TR 14:40, 27 June 2011 (UTC)
 * Thanks, that does help. Although it raises as many questions in my mind as it answers.  I'll need to think things through before commenting/discussing further.  WikiDMc (talk) 08:02, 28 June 2011 (UTC)
 * Ok, it has sunk in (mostly), but I can't spot a difference between operational and gravitational definitions due to tidal effects, other than centrifugal effects. Tidal effects are, as you say, made up of gravitational and centrifugal components. The gravitational component will affect both the gravitational and operationally defined weight of an object while the centrifugal component only affects the operational weight.  This is a different centrifugal effect than the one due to Earth's spinning on its axis, but it is still a centrifugal effect.
 * Going back to my original point, I think the statement has been levered into the article in order to compare the ISO definition to the operational and gravitational definitions. IMO, it would be far better to do a straight comparison (possibly in a table) rather than say "this is the same as that, if you ignore the differences."
 * The current statement is also for quite a specific case, with reference frame of the Earth's surface (although it is the most common case). Would the following statement be just as accurate, but more general: "In any specified reference frame the definition agrees with the gravitational definition, save effects due to reference frame acceleration." ??
 * WikiDMc (talk) 10:49, 28 June 2011 (UTC) (edited by WikiDMc (talk) 12:29, 28 June 2011 (UTC))
 * The last sentence doesn't really make sense, since according to the equivalence principle any gravitational effect is due to reference frame acceleration.
 * I think you could in principle argue that the old 1901 CGPM gravitational definition really is equivalent to the ISO definition as long as you take "acceleration due to gravity" to be the local acceleration of free fall in the chosen frame, i.e. the quantity that you can obtain from the length and period of a pendulum stationary with respect to the frame. Of course, GR did not exist at the time, so the CGPM did not consider the issue of different accelerated frames and the effect on g, but in the face of GR this is the only way to let that concept make sense.
 * Of course, we are not supposed to be the ones doing the arguing here. Instead we should find a (reliable) source that does the arguing for use.TR 13:20, 28 June 2011 (UTC)
 * Of course, and I would be very interested to see a reliable source that says ISO and gravitational are equivalent, as the article currently does ("save for centrifugal...").
 * "...any gravitational effect is due to reference frame acceleration." Very true.  To make sense the statement would need a qualifier similar to the statement currently in the article, and it would have the same problem; stating two things are the same if you ignore their differences. That is why I don't see how the current statement in the article is any better than 2.0 equals 5.0, save for 3.0.  Although it is correct, it gives the wrong impression (that 2 equals 5, or that ISO is equivalent to the gravitational definition). WikiDMc (talk) 15:20, 28 June 2011 (UTC)
 * It's more like say Pi is equal to 3.14 save for a small correction. The point is that centrifugal correction is relatively small. If you choose the reference frame of the Earths surface an astronaut in orbit is not weightless. (unless he is in a geosynchronous orbit, in which case the centrifugal term actually becomes large.) If you choose an non-rotating frame comoving to the earth, the tidal corrections are even smaller. The errors here basically are a reflection of the ambiguities of the gravitational definition. That is in a relativistic context the gravitational definition is simply I'll defined. It only works in the Newtonian limit. The (relativitic) ISO definition agrees with the gravitational one precisely in that limit.TR 16:08, 28 June 2011 (UTC)
 * I fully understand that the differences are usually small; I'm not arguing the physics behind the definitions but I think the current sentence is poorly worded and gives a misleading impression. My example of 2=5 may have been overly exaggerated but, using your Pi example, the current statement in the article is more akin to "the definition of Pi is 3.14 save for a small correction."  That is not at all how Pi is defined, and it should not be represented in that way.  I think a better statement would be, "Pi differs from 3.14 by a small amount," or, "3.14 is a close approximation to Pi."  It is the current impression of equivalence between two non-equivalent definitions that sits uncomfortably with me, and I think your astronaut example helps to demonstrate why; the difference between Pi and 3.14 is always relatively small but the effect of centrifugal forces can be large and the ISO weight is not always approximately equal to the gravitational weight (even in the reference frame we are discussing).
 * I propose to reword the sentence, "If the specified frame is the surface of the Earth, then the definition agrees with the gravitational definition, save for centrifugal effects due to the acceleration of motion at the spot on the surface of the Earth where the measurement is taken."
 * to...
 * "If the specified frame is the surface of the Earth, the weight according to the ISO and gravitational definitions differ only by the centrifugal effects due to the acceleration of the Earth's surface where the measurement is taken."
 * Although, as a side-note/query (which I am loathe to bring up for fear of distraction from my main point); if the reference frame is the Earth's surface, surely the Earth's surface has zero acceleration in that frame? Or is it referring to acceleration in some other frame and, if so, which frame?
 * WikiDMc (talk) 12:18, 29 June 2011 (UTC)
 * That wording would work for me. Although obviously a wording that is backed by a source would be even better, it is at least better than the current wording. I would actually replace centrifugal effects due to the acceleration of the Earth's surface where the measurement is taken with centrifugal effects due to the rotation of the Earth. This seems clearer. It also answer your question "acceleration with respect to what?" The Earth's surface is accelerating with respect to a non-rotating frame.
 * You could also go with If the specified frame is the inertial frame of the Earth, then the definition agrees with the gravitational definition. The downside is that it relies on how you define inertial frame, in a curved spacetime. That definition could actually turn this statement in a tautology. I would be very uncomfortable with such a formulation without a source.TR 13:48, 29 June 2011 (UTC)

Arbitrary break
Responding to a couple of points made above re my original post in the thread:


 * In the lead section we make the distinction between a "gravitational" definition and an "operational" definition. The "gravitational" definition is said to be based on "local gravitational acceleration". However, it is then said that by the gravitational definition the weight of an object in free fall is the same as if the object were at rest. In other words, "local" means "local to Earth" (assuming we're on or near Earth's surface), not "local to the object's frame of reference". Is this not completely different to the article's interpretation of "local" in the ISO definition?


 * The A. P. French reference appears to be given in support of the article's interpretation of the ISO definition, but is only relevant in this section if the source specifically talks about the ISO definition. If the source merely defines or discusses weight in a way that agrees with the article's interpretation of the ISO definition, but without actually mentioning ISO, then we can't use it to support our interpretation.


 * SBHarris's bolded definition flows better IMO, but my points were about the actual wording of the ISO definition, not some better version that we might be able to come up with.

86.179.119.220 (talk) 13:39, 27 June 2011 (UTC)
 * To your first point. 1)There is no such thing as an object's frame of reference. 2) In both cases local, means the local value in the chosen frame of reference.
 * To your second point, AP French does explicitly refer to the ISO definition. (albeit the 1992 version of that standard, which has a slightly different wording).
 * SBHarris's bolded formulation is the 1992 wording of the ISO definition. (Or the narrow paraphrasing there of in the NIST guide, too lazy to check right now).TR 13:51, 27 June 2011 (UTC)
 * (2) OK, fair enough about the AP French reference, thanks for that. (3) I wonder if we should go back to the 1992 wording?! (1) I am confused about your answer to point 1. As far as I can see, the whole point of the distinction made in the lead section is that in the gravitational definition "local" never changes (assuming objects in vicinity of Earth): we don't/can't choose any frame of reference other than Earth (assuming objects in vicinity of Earth), and g therefore never changes. Otherwise, what is the point? Yet we are saying that in the ISO definition we can choose any frame of reference, and that g changes accordingly. How is this not a completely different definition of "local"? 86.179.119.220 (talk) 14:02, 27 June 2011 (UTC)
 * I'm not sure why you are attaching the difference in meaning to a different use of the word local. The word local is used in (more or less) the same way. There are however different words used around it, these indicate that the gravitational definition uses a fixed reference frame, while the ISO definition doesn't. In both cases the word local simply means, the value at the location of the object (in the used reference frame).TR 14:29, 27 June 2011 (UTC)
 * The way I read it, the two different interpretations of "local" explain the difference you seem to agree exists, and are necessary in order to accommodate it. In the gravitational definition, "local" is always "local to Earth" (for objects in the vicinity of Earth), so is always ~9.8 m/s^2 whatever the object is doing, whether in free-fall, at rest, or anything else. Therefore, an astronaut in Earth orbit has (almost) the same weight as he does standing on the surface of the Earth. In the ISO definition, as we are interpreting it, "local" means "local to any convenient frame of reference". In the case of an astronaut in orbit, "local" can be conveniently chosen to refer to the spaceship (not Earth), so the astronaut's weight is zero from that perspective. The difference hinges on what you consider to be "local". Incidentally, though you say of the gravitational definition "different words used around it [...] indicate that the gravitational definition uses a fixed reference frame", in fact this is not explicitly stated in the lead. Perhaps it should be, so that it's a bit clearer that there is a difference between the "W = mg" of the gravitational definition and the Fg = mg of the ISO definition -- assuming we all agree on these interpretations. 81.159.104.101 (talk) 17:26, 27 June 2011 (UTC)
 * ... continued .... or maybe in the lead we should first say that g is frame-dependent (allowing for definitions like ISO), and then say that in this so-called "gravitational" definition it's always relative to Earth (or nearest large astronomical body??!!), whatever the object is doing. I'm rushing right now, but I hope you can follow the gist of what I'm saying... What do you think? 81.159.104.101 (talk) 17:37, 27 June 2011 (UTC)

In science and technology the weight of a body in a specified reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame. The other editors here have argued that this definition allows one freedom to choose any reference frame whatsoever, and then talk of the weight of a given object in THAT frame. I don't agree, but suppose I cede it for the sake of argument. Then, generally, I think you may still point out that if you choose your free "specified reference frame" as one in which the mass to be weighed is at rest (i.e., its "rest frame") then its weight in THAT frame is the "operational weight." This of course includes the case where this frame is chosen to be the rest frame of the rotating Earth upon which an object sits, which is the condition where we normally weigh things. Is that helpful? S B Harris 00:43, 28 June 2011 (UTC)

Buoyancy
Buoyancy, as has been noted above, is like having somebody stand next to you and pull up on your belt. You weigh less (according to the small scale you stand on) but the other guy weighs more, so some of your weight has just been moved off-scale. If you had a bigger scale platform, your weight would not change. This is not very interesting. It's the same if I put one leg on the ground and one on the scale; I have a smaller "apparent weight" then also. And finally, if I get off the scale completely, my apparent weight I suppose goes to zero. Or does it? Does anybody see my point? Is apparent weight even worth mentioning? S B Harris 03:11, 18 October 2011 (UTC)
 * It is worth mentioning because many textbooks mention it. (Also note, that unlike stepping of a scale, in many real world situations it is impossible to avoid buoyancy affecting weight measurements. It is therefore important to be aware of this effect.)TR 05:56, 18 October 2011 (UTC)
 * At least, however, could we mention that the weight in such buoyancy situations doesn't just disappear somewhere, but is conserved and transfered to a support that the scale cannot see? The only time weight disappears is when gravity or acceleration changes. All the other things that seem to affect weight, are only affecting its distribution and how our instruments are placed to capture it. There's an old physics story about a truck which is overweight due to its load of live chickens-- a problem solved by the driver beating on the side with a bat so all the chickens take off and fly just before the truck crosses the weight scales. Would this work? No, once you understand the principle you know that hovering chickens make the truck weigh the same as when they are perched. Same for a little model airplane flying around inside the truck. Indeed a jet airliner exerts the same weight on the surface of the earth when flying as when on the runway-- it's just spread-out more. What is the "apparent weight" of a flying airplane? Zero? Or some other imaginary number that we imagine is counteracted by "lift"? S  B Harris 18:47, 18 October 2011 (UTC)
 * weight in such buoyancy situations ... is conserved and transfered to a support that the scale cannot see
 * Under the gravitational definition that may be true but under the operational definition the weight cannot be "conserved" because it does not exist other than as a reading on a scale. If the reading changes then, by definition, the (operational) weight also changes. WikiDMc (talk) 02:12, 2 November 2011 (UTC)
 * I hardly think the "operational" definition is meant to apply to scales that mis-read, or are misused, or are somehow fooled. If you stand on a scale with only one leg, or I pull up on your belt to give a wrong number, that doesn't mean what the scale reads is your "operational weight." It means you performed the operation of weighing yourself incorrectly, and therefore your instrument gave you an incorrect number for your weight. Perhaps it gave the correct weight that you placed upon it, but that wasn't YOUR weight-- it was simply A weight. The idea that your weight is not an objective property of you, but is rather whatever the scale says it is, no matter how you choose to use the scale, is bizarre. Can you think of any other property of objects in physics that are so "operationally" defined that the number actually IS taken to be whatever the instrument says it is, no matter how you screw up in getting the instrument to interact with what it is measuring? I certainly can't. What is the point of having an instrument if there no rules on how you must use it? I believe such a thing does not even count as an instrument and whatever you are doing doesn't count as a measurement. You are rather just messing around generating random numbers. S  B Harris 03:31, 2 November 2011 (UTC)
 * Can you think of any other property of objects in physics that are so "operationally" defined
 * In physics; no, I cannot. That is why in physics the operational definition is of limited relevance.  Outside of physics, almost every measured quantity is treated as if the measured value is correct notably including, for this discussion, the weight of items to be traded (without correction for buoyancy due to air).
 * Whether in physics or outside of physics, the examples you give of someone pulling up on your belt, etc, are all examples of a poorly performed weighing procedure. The key to the operational definition is the statement, by the operation of weighing it, which implicitly assumes that the operation of weighing has been performed correctly.  Your definition of "performed correctly" and my definition may differ but I think it's fairly safe to assume that nobody would consider standing on a scale while someone pulls upward on your belt a correct (or comfortable) performance of a weighing procedure.  On the other hand, to weigh an object without removing buoyancy effects would be considered by me, and I assume many, to be an appropriate weighing procedure depending on the intended use of the gained data.  WikiDMc (talk) 04:59, 2 November 2011 (UTC)
 * Well, it wouldn't be a correct procedure if you were weighing precisely (a further correction must be done, as you know from your analytical chem classes), so it's only a matter of precission. Bouyancy removes a couple of ounces from the weight of a person in normal air, and it's only because we don't care about the two ounces that we ignore it, not because we're actually using some "operational definition." If you were weighing a partly inflated balloon, or an object of low density like aerogel, we would be grossly off the mark of correctness. The point is that it's not honest to ignore bouyancy and lump it under "operational definition" when the effects are too small to care about practically, but then change your stance when they are too large to ignore. We require one view for both cases. If you're going to ignore an effect, do it for the right reason (it's too small to care about) not for the wrong reason (pretending you really only care only about what the scale says). Actually you only care what the scale says when what the scales says is nearly right. Otherwise, you ignore the scale reading and and come up with some other method, or else correct the scale reading and don't take it literally. S  B Harris 05:13, 2 November 2011 (UTC)
 * Well, it wouldn't be a correct procedure if you were weighing precisely
 * That depends on the purpose for the weighing and the intended application of the result. If I am designing a rope to suspend a rock submerged under water, then I am interested in the rock's submerged weight.  So I will weigh it submerged in water and I will care only about what the scale says.  It is entirely precise and no corrections are required.  Or I could weigh it in air, or a vacuum, and make an adjustment for how the rock's operational weight will change once submerged in water.  In either case, it is the rock's operational weight when submerged in water that I am interested in, which helps to demonstrate why I think your statement it's only because we don't care about the two ounces that we ignore it is incorrect.  Sometimes we ignore it because we care about the force required to support a person in air, not their mass.
 * IMO, the operational weight of an object is subjective, or at least situation dependent, and whether the operation used to determine the weight is appropriate is also subjective, according to the user of the information. This fits well with the definition of, the force measured by the operation of weighing it without any caveats regarding the method, accuracy or situation of the weighing.
 * Further, I do not know of any source that mentions a vacuum when defining the operational definition, so cannot support wikipedia implying in any way that a vacuum is required to obtain a precise operational weight or that a correction to the force measured is required (for buoyancy). Even reference 7, Galili, does not seem to mention a vacuum even though it is referenced in support of the statement ...in theory, an object will always be weighed in a vacuum...'
 * As for analytical chem, is that not because we are interested in mass rather than weight?
 * WikiDMc (talk) 05:55, 2 November 2011 (UTC) (edited 08:20)

Sensation_of_Weight
i am quite sure that the "the vestibular system, a three-dimensional set of tubes in the inner ear." is used to tell acceleration and balance. as my perception is impaird at the moment, and i cannot tell if the vehicle i am in starts to move slowly unless it hits a bump, and i must rely on my eyes and the pressure under my feet for balance. as for "sensation of weight", i would think you would tell by using your muscles, by perceiving the task requireing more/less effort to lift or move, ie yourself. — Preceding unsigned comment added by 58.169.182.218 (talk) 21:32, 19 December 2011 (UTC)

Weigh as a compound word
I have reverted User:Wendy.krieger's edit for three reasons: WikiDMc (talk) 00:36, 14 October 2011 (UTC)
 * 1) Weigh/weight is not like the examples listed. In the examples, the first word is an adjective while weigh is a transitive verb.
 * 2) In this sense, everyday and legal contexts reflect the true meaning of the word, is incorrect. The true meaning of the word matches the scientific usage.
 * 3) The scientific use is better described as heft (heave/heft), is not only a violation of WP:NPOV and possibly WP:OR but it is incorrect. Heave implies an upward motion of the object while to weigh something is usually a static process.


 * The 'Oxford Dictionary of English Etymology' Edited by C.T.Onions, gives 'weight' as a measure determined by weighing. Since weighing is used to determine mass, then this is what is thus implied.  The old english 'wiht' (ME wight), refers to a thing, animal, entity, matter, in any case, not a force.  One ought not necessarily take scientific usage as being linguistically correct.  (Electrical) displacement, which is used for the vector D, actually refers to what is called polarisation (P), which is a displacement or movement of electric dipole in the direction of the flux.  The ME weight is based on the long/length sequence in that it determines a measure of (rather like modern english -age does).  --Wendy.krieger (talk) 08:54, 15 October 2011 (UTC)


 * While the origin of the word is interesting, and probably deserves some discussion in the article, it should not cloud the issue of the word's current definition which is the force measured by the act of weighing.
 * Any old english definitions will conflate mass and weight because at the time there was no need to distinguish between the two. Now that man is going into space, accelerating at high g-forces etc., a distinction is required and that is why pound and pound-force are two separate and different units of measure, and why weight is strictly defined as a force and not a mass.
 * WikiDMc (talk) 01:00, 17 October 2011 (UTC)


 * Wendy Krieger is correct.
 * In commercial and everyday use, and especially in common parlance, weight is usually used as a synonym for mass. Thus the SI unit of the quantity weight used in this sense is the kilogram (kg) and the verb “to weigh” means “to determine the mass of” or “to have a mass of.” ― NIST Guide to the SI, section 8.3
 * Zyxwv99 (talk) 14:04, 30 March 2012 (UTC)

Too long an introduction for its overall length
Although the overall length is far longer than the lead section, it still strikes me as being overly long. Some of that text should be moved into one of the sections below. Pagen HD (talk) 10:26, 26 January 2012 (UTC)
 * Shortened it. Having trouble getting the weight in bold -- I think maybe there's some wrong somewhere else on the page that's messing up the parser, but I can't seem to get it to work. Nobody Ent 12:23, 26 January 2012 (UTC)

Appreciate the quick response. Pagen HD (talk) 16:57, 26 January 2012 (UTC)

In law and commerce
I recently added a few sentences to the beginning of the lede: ''In everyday English, weight is a synonym for mass. In law and commerce, including product packaging and nutrition labeling, "weight", unless otherwise indicated, means "mass".''

This was soon changed to:

In everyday English, weight and mass are functionally equivalent.

In retrospect, I can see that the first sentence was a mistake, since in everyday English weight can be ambiguous. However, changing it to functionally equivalent to mass doesn't seem to be an improvement, since it seems to exclude the actual equivalent of mass as a possible intended meaning.

My main concern here is with how the word is used in law and commerce, including product packaging, nutrition labeling, medicine, the pharmaceutical industry, etc.

What brought me to this topic is my decades-long interest in the history of customary units of measurement, especially English, e.g., troy weights, apothecary measures, centuries-old weights and measures legislation. On the talk pages of many Wikipedia articles relating to these topics, the mass vs. weight argument has come up repeatedly (usually long before I arrived). Much of the confusion seems to stem from attributing the physics definition of "weight" to situations in law and commerce where it does not apply.

Thus, I would like to see some further discussion on this. Thanks.

From NIST Handbook 130 – 2011 http://www.nist.gov/pml/wmd/pubs/upload/10-iva-packlabreg-11-h130-final.pdf

Uniform Packaging and Labeling Regulation - When used in this regulation, the term “weight” means “mass.” (See paragraph I. in Section I., Introduction, of NIST Handbook 130 for an explanation of these terms.)

From NIST SI Unit rules and style conventions checklist http://physics.nist.gov/cuu/Units/checklist.html

''When the word "weight" is used, the intended meaning is clear. (In science and technology, weight is a force, for which the SI unit is the newton; in commerce and everyday use, weight is usually a synonym for mass, for which the SI unit is the kilogram.)'' Zyxwv99 (talk) 14:05, 6 April 2012 (UTC)
 * This was already near the bottom of the lead, but for clarity & consensus I've changed the lead back (partially), using existing Canada reference. The US reference could be added if desired. Nobody Ent 14:50, 6 April 2012 (UTC)
 * Thanks. Zyxwv99 (talk) 22:43, 6 April 2012 (UTC)

Does an object in free fall actually have any force on it, "due" to gravity??
Newton says yes, Einstein says no. Newton would say "g" makes sense for an object in free fall, but Einstein would say that the g field has disappeared in the view (the reference frame) of an object in free fall. Which is why its weight has gone away, and it is weightless. It isn't that it feels to gravitation, rather in Einstein's view, gravitation no longer exists in the frame of an inertial object. So of course weight disappears as well. The introductory paragraph uses the "gravitational" definition of weight as one possible definition, but it's not clear to me if anybody ever intended that definition to be applied except for objects that are at rest. For objects that are falling or accelerating, clearly some other definition gives forces that better approximate what we feel as weight, and what scales measure as weight. I believe that the attempt to apply the misguided "gravitational definition" of weight in situations where it was never clearly intended to be applied, is misguided! We could shorten the lede of this article considerably if we moved it down to a "historical" or "special cases" or even "Newtonian" section (which already exists). S B Harris 20:09, 6 April 2012 (UTC)
 * If (in the Einsteinian view) there's no gravitation why is the group approaching the fall-frame stationary object at an ever increasing velocity?
 * In any event, I concur the lead was long and estoric and have made a shortening edit. Nobody Ent 23:21, 6 April 2012 (UTC)
 * You cut out too much, as the operational view (which makes us "weigh" more in accelerating cars, or weightless in the space station) is the "operational view", or the Einsteinian view. In the Newtonian view, only gravity changes weight, and all other things that seem to lessen or increase weight, are false effects. Newton would say that a rotating centrifuge (or rotating space-stations of science fiction, ala 2001: A Space Odyssey) only produce the appearance of weight. Einstein would say that it's real weight, any time you can weigh something on a scale (which certainly includes in a rotating space station or centrifuge). Newton would say that you feel the "force" of gravity, and in an elevator falling, gravity is still there as a force, but you don't feel it. Einstein would say that Newton was wrong: you never feel the force of gravity, per se, even standing on the ground-- all you feel is the contact-forces against your feet, that keep you from free fall. For Einstein, the reason you don't feel "weight" in free fall (a falling elevator, vomit comet airplane, or in orbit) is because the push from the floor is gone in all these weightless cases, and you never feel gravity, anyway! So that's really simple. S  B Harris 00:26, 7 April 2012 (UTC)(Refactored 15 July 2012)
 * Is there a reliable source for the above & corresponding edits? Nobody Ent 02:12, 16 July 2012 (UTC)
 * There are lots of teaching sites on the net that make this point. And it's sort of obvious, is it not? If the only force that acts on you is gravity (you are in inertial trajectory), you cannot tell that situation from being weightless out in far space where there is no gravity acting on you at all. Thus, you cannot feel gravity-alone, as it feels just like no-gravity. You have to look out the window to see your acceleration relative to far away things, to infer it. But you don't feel it (absent "microgravity" tidal and gradient effects, which are very small unless you're near neutron stars or star-mass black holes).  S  B Harris 23:15, 16 July 2012 (UTC)

Regarding recent deletion of "in law an commerce"
A user recently deleted sections of this article dealing with the use of the word weight in law in commerce. To understand my concerns about this issue, see above section "In law an commerce." Also, the picture a spring scale certainly illustrates how the word weight is used in physics and engineering, but the balance scale illustrates the original meaning of the how, and how it is still used in law and commerce. Zyxwv99 (talk) 16:04, 15 July 2012 (UTC)
 * Per WP:NOTADICTIONARY, different meanings of a term go into different articles. In this case, this article is about the concept of weight as used in science and engineering. The colloquial meaning of the term is dealt with in the article titled mass. To assist readers to find the correct article when an single term has distinct meanings, we use hatnotes and disambiguation pages.TR 16:55, 15 July 2012 (UTC)
 * In English Law, the pound was described as a "mass or weight". Martinvl (talk) 21:32, 15 July 2012 (UTC)

Pictures of scales
I like the new picture showing two different scales side by side to illustrate the difference between weight and mass. Now if we can just do something about the picture "Weighing grain, from the Babur-namah" which shows what looks like a mass-measuring scale. Zyxwv99 (talk) 22:32, 20 July 2012 (UTC)


 * Well that's asking for the impossible, isn't it? The terms "weight" and "weighing" in English (and "poids" and "peser" in French, etc.) predate the invention of the very scales capable of actually "weighing" anything. Instead for centuries we "weighed" things using a balance scale. There are even English statutes from the Middle Ages specifically defining "weighing" as an action to be done using a balance scale (the same statutes also show that to the extent any confusion of concepts existed then, it was between volume and weight (err, mass), which, if nothing else, at least shows that our predecessors were far more concerned with measuring the quantity of stuff in question than its force of attraction to the ground when they were "weighing" stuff). The logical conclusion (at least to those with an open mind) is that somewhere along the way the term "weight" got redefined from an historical sense of meaning what we now call "mass" to what we now call "weight", thus leading to much of the confusion that exists around the term (if "weight" had been left alone and another term introduced for the gravitational force, we'd not have this confusion at all).


 * The pictures this article has at the moment are of a modern spring scale and a fairly old balance scale and between them include a caption about what happens when each of these scales is brought to the moon, so it's probably just as well time travel doesn't exist as pity the unfortunate 14th century merchant brought forward in time with his scales and weights, asked to use them to measure some objects on Earth before being sent to the moon and asked to do the same thing with the same objects only for him to come to the conclusion that stuff "weighs" the same on the moon as it does on Earth, even though it feels lighter. Reëducation camp for him. D P J (talk) 22:02, 23 December 2016 (UTC)

Error/absence of clarity in intro?
I believe the "engineering version" of weight is the resultant (vertically resolved) force on an object i.e the product of gravity and other forces e.g buoyancy. In this version, an object weighs less in water than in air etc. A helium filled baloon will have negative weight as, though it has (constant) mass, the mass of the air it displaces will be greater than its own. Could the intro be improved to make this clearer? Regarding the weightlessness of freefall, isn't this phenomenom related only to the reference frame? LookingGlass (talk) 03:05, 7 December 2013 (UTC)

The intro is unsatisfactory in its current state because it first states unequivocally that weight is a vector, then goes on to talk about a "rival tradition" in which it is a scalar (the "magnitude" of a force). Later in the article it cites Halliday & Resnick, one of the most widely used physics textbooks, as defining weight as a scalar. So obviously there is not a clear consensus on the vector vs scalar issue, and that should be stated at the beginning. Dark Formal (talk) 02:52, 25 February 2018 (UTC)