swansont

! Moderator Note Threads merged

To add to Genady’s response: you don’t know the exact trajectory, so the photon passes through both slits and interferes with itself

A stepper motor might be (part of) the solution..

No, because each state has an amplitude, and the probabilities of being in the states add to 1.

Do you have an example? Quantum states I can think of will have an energy, and you need to add energy to put a system in the ground state into a superposition of energy eigenstates (unless they’re degenerate)

In my experience yes. I’ve used commercially-available systems that lasted around 1.5 years or so; generally the failure was degradation of the anti-reflection coating, and the laser wouldn’t stay at the desired frequency, but they would still lase. If it’s a homemade system, it’s critical that the electronics prevent the laser from seeing voltage spikes. Those will destroy a laser diode.

! Moderator Note The description of the sandbox: This forum is provided for members to test BB code, learn how to use the various forum functions, and generally get to grips with the system IOW it’s not for discussion. Do not expect anyone to respond.

But you are, apparently, a word-salad smith. A neutron contains mass energy of this amount. But where is the “pure energy” that you claim it contains? How much is there? What is the evidence it exists?

We prefer to discuss only one proposal per topic, so let's focus on your first "theory" What evidence is there of "pure energy"? Has it ever been observed? If not, how would one do so? If this energy can be converted to mass, why haven't we seen a violation of E=mc^2? One glaring problem here is that you have no mathematical framework for your idea, so that nothing can be quantified. That's a fatal shortcoming for a physics proposal.

The length of the day is very different from what it was hundreds of millions of years ago, the length of the year isn’t constant, the orbit of the moon changes, the magnetic poles flip on occasion. Cellular functions can’t be too dependent on such external influences.

! Moderator Note Among the issues here is that this is a series of speculations built on top of each other, and that’s not allowed. There’s not much in the way of actual modeling and evidence, and far too many instances of a claim supported only with a youtube link, which is not permitted. It also suffers from the “wall of text” problem. While this isn’t inherently a rules violation, it’s discouraged, and given the broad spectrum of the material, it is contrary to our preference of one topic per thread.

The phrasing of these articles is that the evidence is quite weak, that they are teasing out small correlations in data. The links you provided don’t present any of the science; they’re just summarizing other studies, so there’s not much science to discuss.

Yes, plausibility is important; the moon would not draw anything up - things don’t fall upwards when the moon is overhead. The earth’s gravity still dominates, even though one can measure a small reduction in the net acceleration toward the earth. There’s also an effect from the change in local mass from tidal effects, so one would expect any effect to be accentuated near coastal regions. Also perhaps see an effect in skydivers and astronauts on the ISS What would aid in a plausibility argument is the calculation of the physics involved, such as a comparison of the moon’s attraction as compared to e.g. wearing a hat. (I know that this has been done to debunk the notion that astrology has some basis in science by comparing the gravitational attraction of planets vs that of the attending staff when a baby is born)

! Moderator Note I think we’ve covered this same ground more times than is necessary. Soapboxing, not science. Locking such threads makes me feel safe

If the moon does not exert a force on the earth, why does the moon orbit the earth? (consider Newton’s third law) In any event, tides present no conflict with Newton’s laws, and nothing is being ignored. The issue is your lack of understanding of physics, which is not going to be fixed by looking at this exercise; there are too many issues to address. (coordinate systems, linear vs rotational physics, action-reaction) Basically, if you think that physics is wrong, it’s invariably your understanding of physics that’s wrong or missing.

Provisionally. I don’t see the relevance. That does not follow, but, again, probably irrelevant. Measurements give some sense of security? I thought you were in the camp that thinks we can’t tell what others perceive or feel. So at best you can claim it gives you a sense of security. And you’re describing the act of measurement, rather than what that measurement is. I think you did not, despite your claim to know everything. You forgot the red and orange of the fire that you’re playing with I thought we were talking about science - the study of how nature behaves.

Fine. Answer my question, then. How is a physical measurement, like how far a kangaroo jumps, a function of human nature?

So, not everything we know is based on human nature, as you had claimed.

You didn’t answer my question, though. Thinking that you understand is a dangerous thing. It needs to be tested, and the empirical evidence is that you do not.

Can you link to the research rather than the paywalled pop-sci summary? At least we’d get an abstract. I can read as far as “a handful of studies have also hinted that” which is not a phrasing that one uses when there’s solid evidence. It sounds like there’s a blip in the data that might not be statistically significant, and perhaps someone has made a plausibility argument

One can do quantitative measurements, like how far a kangaroo can jump, or how long its gestation period is. How are these observations based on human nature? I think these are objective observations.

Wealth isn’t expertise, either

! Moderator Note It was not deleted because we don’t do that. It will not be re-opened. ! Moderator Note No you may not. ! Moderator Note You should not repeat your error of being persistently obtuse as you were in the closed thread

My comment was primarily for the benefit of the OP and any onlookers. Apologies that this was not clear.

A change in position with respect to time is velocity. Toss something up in the air. It's moving up. Later, it moves down. There will be a point in time where it is at the top of its arc, and is motionless. v=0 You can tell if you are accelerating. Even blindfolded. You don't need a point of reference. That's not true of velocity, which is relative to some frame of reference. It's vertical in a cartesian coordinate system, where you have a y axis and an x axis. It depends. g can be a scalar, 9.8 m/s^2, or you can use it as a vector if you acknowledge the direction, toward the center of the earth g is 9.8 m/s^2 If you are not at the surface of the earth, your gravitational acceleration will be different than this value. It's sloppy to use g as a generic acceleration That's why one should use GM/r^2 The problem being that you are posting this for others to read, and it just begs the question of why you would do calculations one way when you can do it in far fewer steps. It's confusing, and points to a lack of understanding of the concepts. Seeing as how often you get a wrong answer, that's not readily apparent Most of the time you use average it's because you've decided to make things more complicated by insisting on using an average value. I might have something to do with your disdain of calculus. Orbit defines a specific set of conditions. Standing on the earth is not an orbit. If you make up your own definitions of terminology you can't communicate with others You miss the point. A circular orbit has more conditions than moving in a circle. Then get the basics right. An orbit means the gravitational force is what is keeping it moving in a circle. Not standing on the ground. The choice of coordinate system seems logical given the problem. You can use x and y if you want, but then there is motion in both the x and y directions, which varies over time. The speed remains the same; v^2 =vx^2 + vy^2 (gosh, that's the equation of a circle!) Simpler and more descriptive. If you are trying t describe a satellite's orbit, vertical only works when it's directly overhead. At any other time "vertical" doesn't work. "free" and "bound" have definitions in physics. You would do well to learn such terminology. A free particle is one that could get infinitely far away without being subject to new force. A bound one cannot. In terms of energy, KE + PE < 0 for a bound particle No, it's not. For a circular orbit, v = sqrt(GM/r) As r gets larger, v decreases The dynamics of it is more complicated than that, actually. You add energy, but this goes into increasing your potential energy (it's negative, but gets smaller in magnitude) and your speed decreases. KE goes down, but PE increases twice as fast. Yes Local means where you are. You can measure it with a spring scale. If you know your mass, the spring scale tells you the apparent weight, and from that you can get the acceleration. If you want to know the rotation (which is another form of acceleration), you can use a Foucalt pendulum. The SI units of acceleration are meters/ second^2 There's no area The m^2 in Joules does not represent an area A satellite is oin circular motion. The center of that circle is the center of the object it's orbiting. Force is tied in with momentum, not momentum squared. The acceleration in circular motion does not change the speed, it only changes the direction (velocity is a vector. Changing the direction of motion requires an acceleration) There is no more, no less. The speed is constant. You don't. It's not a linear (i.e. one-dimensional) system, and the direction is not vertical. The direction is "toward the center of the circle" (i.e., it's radial) If the force is perpendicular to the velocity, it will only change the direction. No work is done, so there is no change in KE