elfmotat

I really don't know why you insist on doing this. I know this is irrelevant nonsense. You (probably) know this is irrelevant nonsense. Just about everybody knows it's irrelevant nonsense, yet you post it anyway. What satisfaction do you get out of constantly reaffirming your ignorance to the rest of the community?

Just use rest mass in all of your equations.

Just as a little aside, nobody uses the concept of relativistic mass anymore. It's not very useful and tends to confuse people.

I recommend you watch the videos here: https://www.khanacademy.org/science/physics . Start with 1-D motion and work your way down. You need to understand the fundamentals of physics, and you're going to have to set some time aside to learn them.

Honestly Relative, this is by far the most relaxed science board I've come across that isn't run by cranks. Most sites do not allow speculative posts at all (honestly I'm a bit surprised this one still does given the nonsense that goes on in the speculations section). Most sites require questions to be of the form that they can be answered concretely. This site gives you a soapbox for whatever strange thoughts you might conjure up, and the rules for topics in the mainstream science sections are incredibly relaxed. If you can find another site with knowledgeable people that puts up with even half of the BS that this site does, I'd be surprised.

Sounds like you lean towards the "shut up and calculate" approach .

I play the drums, so this half makes me wish I went to school in the UK.

To add to what Strange said, it's called an "interpretation" for a reason. If it were testable there would be no need for interpretation!

Not really, no. This discussion has been entirely non-relativistic so far.

No problem. Yes, I just put V in function notation to make it clear that it's a function of r. [math]V_B - V_A = V(r_B) - V(r_A)[/math].

Do you have any experience with vector calculus? I suggest looking into "line integrals." I think I remember Khan Academy having some decent videos on the subject. So in general you can examine the difference between two points by taking some general path from A to B. The path doesn't have to be a straight line, it could be some curvy path. [math]\hat{r}[/math] is the unit vector in the r-direction. What [math]\hat{r} \cdot d \vec{l}[/math] is telling you is that integral only depends on differences in r. In other words the integral only depends on integrating the r-component of any path from A to B. If you're familiar with spherical coordinates, For some general curve in spherical coordinate, [math]d \vec{l}[/math] will contain r, theta, and phi components: [math]d \vec{l} = dr ~ \hat{r} + rd \theta ~ \hat{\theta} + rsin \theta d \phi ~ \hat{\phi}[/math] from here it's easy to see that [math]\hat{r} \cdot d \vec{l}=dr[/math]. I put it in this notation to make it explicit that V only depends on r. They mean the same thing.

Ah, okay. So what they're doing is looking at the potential difference between two points in space for a static point charge at the origin. They define the potential difference between two points as: [math]V_B-V_A=-\int_{A}^{B}{\vec{E}\cdot d \vec{l}}[/math] For a point charge at the origin we have: [math]\vec{E} = \frac{kq}{r^2} \hat{r}[/math] [math]V_B-V_A=-kq \int_{A}^{B}{\frac{1}{r^2} \hat{r} \cdot \vec{dl}}[/math] Since [math]d \vec{l}[/math] is a small change in the path from A to B, we have: [math]\hat{r} \cdot d\vec{l} = dr[/math]. Therefore: [math]V(r_B)-V(r_A)=-kq \int_A^B \frac{dr}{r^2}[/math]

Well that equation isn't true in general, so you must have some additional information about the form of the electric field - namely that it's static - to justify setting the electric field to kq/r2. I can't really tell you why they're doing this without some context.

Using applied Quantum Funk Theory, I have come to the conclusion that this is what you seek: https://www.youtube.com/watch?v=yi8811qAWfc

Before I got into physics I used to watch the usual pop-sci shows and documentaries, and I read the usual pop-sci books. So as you might imagine, I had quite a few analogies and simplified descriptions floating around in my head. Once I really started to learn physics I realized how very little I actually knew. All of the analogies and simplifications were actually very counter-productive for me; not only did I have to re-learn the basic concepts, but I had to un-learn all of the analogies to prevent myself from getting completely confused. So maybe I'm biased, but I've always felt that being candid about what words and equations mean is the best course of action. Giving the OP, who is asking a very direct question, a less than honest answer would be a disservice to him. He doesn't understand? Great! Now he has something to learn! Feeding him half-truths to satiate his curiosity is bad - now he's not as curious, and he has a bad model in his head.

It's part of it, but not the whole thing. If two events are separated in space by [math]\Delta x[/math] and happen simultaneously, then in a reference frame moving at [math]v[/math] there is a time [math]\Delta t'= \frac{\gamma v}{c^2} \Delta x[/math] between them. This is just nonsense.

"More obscure" meaning "specific," or "not very general." Diodes are not the general case for electrons, they are a specific case. And, in this thread, probably not very relevant. (I've also never seen them described by electron "clouds" either, so if you have a source for what you're talking about it would be appreciated.)

Agreed. But the OP asked a question regarding whether the "electron cloud" is a "tangible entity," or just a way of visualizing things. I don't think he's looking for analogies. That's a rather rude and accusatory response. Any "condescension" you interpreted was most likely projection. Do you have some sort of problem with me? Yes we have indeed been talking about QM, which shouldn't be surprising. I'm not particularly impressed that the term "cloud" may be well-defined in some more obscure context, because I don't think it's particularly relevant to the OP.

Yes: write your message in red ink, then accelerate it near the speed of light toward its recipient. The light from the message will be blue-shifted, and will in fact be blue at high enough speeds.

Clouds are probably a bad way of thinking about it. As soon as we start introducing ambiguous words into the discussion there are bound to be misconceptions. If everyone sticks to discussion of states, wave-functions, amplitudes, and probabilities, then everything is well-defined and there is likely to be a concrete answer to whatever OP may ask. Once we introduce ambiguous terms like "cloud," inevitably we'll start getting nonsensical questions like, "can electron clouds precipitate(?)," or, "how do electron clouds condense?" Everyone's better off if we stick to the well-defined terminology.

There's no "cloud." The electron has a wave function which satisfies the Schrodinger equation, and the square of the wave function tells you the probability of finding an electron in any given region of space. No clouds necessary .

But it's not. The radius is only that same at the boundaries. Between (0,0) and (1,1) the radius is never the same. So the integral is not the same. Nothing is the same.

The integrals still aren't the same for those boundaries. For y=x2 I get ~5.3, and for y=x10 I get ~7.5.

This is incorrect. It looks like a mish-mash of the formula for revolving around the x and y axes. What you should have is: [math]A_x = \int 2 \pi y \sqrt{1+ \left ( \frac{dy}{dx} \right )^2} \, dx[/math] [math]A_y = \int 2 \pi x \sqrt{1+ \left ( \frac{dx}{dy} \right )^2} \, dy[/math] where Ax and Ay are the surface areas from revolving around x and y respectively. How is the radius the same? [math]y^{1/2} \neq y^{1/10}[/math]

I still don't understand what the problem is. Let's work out the surface obtained by revolving your two example functions around the y-axis, y=x2 and y=x10, over the interval y=[0,2]. So we have: [math]A_1 = \int_0^2 2 \pi x \sqrt{1+\left ( \frac{dx}{dy} \right )^2} dy = 2 \pi \int_0^2 y^{1/2} (1+y^{-1}/4)^{1/2} dy \approx 13.61[/math] [math]A_2 = \int_0^2 2 \pi x \sqrt{1+\left ( \frac{dx}{dy} \right )^2} dy = 2 \pi \int_0^2 y^{1/10} (1+y^{-9/5}/100)^{1/2} dy \approx 14.01[/math] What here do you take issue with?