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studiot last won the day on April 21

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    Semi Retired Technical Consultant

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  1. If you must have atoms, rather than electrons, the gadget you want is called a field ion microscope.
  2. I am assuming you want to achieve a straight line of the form y = mx + b, where m and b are constants. Strange though it may seem, this is not actually a 'linear' equation, so it is better not to use that description. 'Linear' has a special meaning in maths. y = mx is linear y= mx + b is what is known as affine. Back to your questions. starting with number 5 Your two variables are length L and frequency. f. the velocity, v and end correction, e are stated to be constants. But there is an inverse relationship between variables L and f. So you need to somehow introduce a new variable by inverting one of these. Can you think of a relationship (another equation) for either L or f that will do this?
  3. Well explain what your rules are and what your objectives are. Interaction of matter waves is an everyday common or garden event in this universe. The point is twofold (1) To have a wave you need a wave equation to be satisfied. There are varying degrees of sophistication of wave equations. (2) To pick out the appropriate solution to said wave equation you also need to apply boundary conditions. In both (1) and (2) we normally use appropriate versions to the situation.
  4. You specified unrelated 'matter waves' If you are using particles (matter waves) generated by the same source, how do you guarantee that they are unrelated? If they are already in separate orbitals, this is guaranteed.
  5. I didn't say they did, but how do you think an electron microscope works?
  6. Why wouldn't I? Or is your question "do I consider an electron to be exclusively a matter wave?"
  7. If the two objects are orbital electrons belonging to different atoms, the two may combine to form new molecular orbitals, one a bonding orbital and the other an antibonding orbital. These new orbitals have new wavefunctions, different to but formed from the original wavefunctions.
  8. You are getting there. The tolerance standards refer to laboratory glassware in general and include glassware (flasks, pipettes etc) that have only one measuring line and no scale. As such they are designed to deliver the rated quantity /- rated tolerance. Since you compared this with engineering practice, consider this. A tape measure with one inch missing from the end will 'measure' a 10 foot wall as 10 feet and 1 inch, from the end. But by measuring from the 1foot marking to the 11 foot marking, the correct length of 10 feet will be obtained. This is because the tape has an error everywhere of 1 inch so using a mesurement by difference this cancels. Length of wall = (11' 0") - (1' 0") = 10 feet. The burette allows this measurement by difference so the resolution and accuracy are not necessarily limited by the overall tolerance. But inEngineering you also have 'limits and fits' tolerance. Suppose you are turning a bar down to 'just fit' through a particular hole. What tolerance would you enforce to turning the bar down? This is equivalent to using a measuring flask where you cannot measure volume by difference. Your tolerance refers to the whole diameter and again you cannot use difference. Remember also that there are other considerations that affect accuracy with a burette and if you try to get too accurate you need to start to consider air bubbles, temperature, liquid density and the scale accuracy of the burette and techniques to ensure that all the liquid you think is transferred actually reaches the receiving vessel. At this point you might move to weight measurement rather than volume.
  9. You need to distinguish between tolerance (which affects accuracy) and reading resolution (which affects precision). Tolerance is a specification of full scale ie a nominal delivery of 50ml will be within the range 49.95mL to 50.05 mL. This is a characteristic of the instrument (in this case a burette or buret in US) This gives a relative tolerance of 0.13% The actual scale can be read more finely than this. This effect is common in many instruments where a scale is read rather than a digital readout obtained. For example mechanical verniers, analogue voltmeters. For both of these latter tolerance is usually given as the relative (of full scale) %. Such a voltmeter would always be specified as 1% (very good) 5% (good) 10%(El cheapo variety). Remember for readings that are made by difference (as in a burette) the tolerance is a systematic error that affects both reading equally and with the same sign so cancels out on subtraction. That leaves reading resolution. This is measured as the smallest scale graduated interval (usually 0.1mL for a 50 mL burette), or some fraction of it. (ASTM E287-02) has half of this at 0.05mL. Vogel suggests this for "all ordinary work", but reading to 0.01 or 0.02 mL with the aid of a lens, "for precision work". Precision work could also entail establishing a calibration curve for use over the whole length of the burette or alternatively repeating the measurement over several different parts of the sscale and averaging.
  10. Timo's posh explanation is just fine but try this simplified one. Looking at the diagram I see that upwards is defined positive. Looking at your equation and the diagram pA - (p+dp) A - ρgAdy So I see that the arrow on the underside is pointing up so pA is positive. But on the top, p and dp point downwards so are negative so (p+dp) is negative on the top dy is positive and g is negative so the third term is also negative. does this help?
  11. Enough pseudomathematics. "more accurately" is a mathematical statement but I see no mathematics. Furthermore the rules here require answers to polite questions and comments from other members
  12. A point is not a frame (hint that is why we have separate words). Besides which the Universe as a whole has no centre. Swansont's reference to a centre was referring to a finite system of bodies not the whole universe.
  13. Well the original question says that the body never experiences a force. But it does not say for how long this period of grace last in the history of the body. I also mentioned momentary conjunctions of the generators of forces acting on the body where the net force is zero. Obviously, in a universe where nothing is moving, no matter how far apart (in space) two bodies are, the 'force field' generated by one will eventually affect the other, though there will be a time lag. The bodies may however be moving so that causal connection comes into play, as Mordred comments. But I would add a rider to his comment that causal connection is not immune from disturbance by a third body which may push something into or out of the connection light cone region.
  14. Thank you for the response, but it is not necessary to consider gravity. Unless the bodies are at exactly the same temperature, there will exist some sort of radiation pressure between them.
  15. I have three points nobody seems to have considered, to add to this discussion. Firstly this idea of a virgin body that has never felt a force. That’s why they might actually be stationary, in other words, no momentum.  If an object has experienced a force to make it move surely it has momentum, wouldn’t it make sense to then choose a frame of reference that hasn’t had a force applied to it? Within the bounds aof all known mathematics this requires a single body in an otherwise empty universe. As soon as you introduce more than one body there will always be some sort of force between them. Secondly zero momentum could be momentary (and frame dependant), but the requirement has always been zero net force, not zero force. Thirdly there is no magic in pair production from radiation.