Since when is pointing out crank nonsense abusive. He started the conversation about black hole physics. The PM system is for private messages and I'm pointing out nonsense 'on the public science forum where the nonsense is being posted'.
You started the conversation about black hole physics and the metric. You haven't made any effort to show I'm wrong in calling your comments about black hole physics and the metric uninformed. That's because you can't. You were happy to continue on in this thread but now you say I'm abusive. The real abuse is you continuing to spam this site with garbage under the pretext that you're an expert. That's what cranks do. You're just dodging any request to show you understand physics in a real knowledgeable way. The reason you think derivations are not important is because you can't do any. That's what cranks do.
But of course he's an expert. He's much more knowledgeable on his crank topics than you are, probably because he came up with them himself. Not to insult you Farsight, but even if you feel that established views are wrong, this conversation is like two people trying to communicate by wrting messages on paper, folding it into paper planes, and hurling those planes into a brick wall between them. Both of you can drag on, not gonna achieve anything.
I already understand the physics so I'm not trying to settle anything about the 'actual' physics. I'm pointing out Farsight doesn't understand the 'actual' physics [yet he claims he does] and should quit posting as an expert. Same thing the PHD's Alphanumeric and Prometheus have been doing for as long as I can remember. That happens in public science forums. So how many 'on topic' posts do you need to answer your question? How long do you remain confused before somebody decides to talk about something else? Whose dragging on? Farsight started the comments about black hole physics in a post to me. New threads don't need to be started to answer posts. Quit whining about your convoluted thread being abused.
oh no, now you're taking it out on me. Too bad I can't suddenly stop being confused. Most of the posts have been off-topic, making it convoluted, and both of you are the ones dragging on. Lol. Well you can argue as much as you want, but its ultimately futile, even if you are always right and Farsight is always wrong,
So, it's about me now? I think you should just focus on Farsight, he's farsighted anyway, can't focus hehe.
If bruce wishes to take you up on that offer than it'll have to be in pseudo or AT because you have a demonstrably flawed/poor/misguided understanding of the material in question. Given you don't know any linear algebra, differential geometry, calculus or tensor algebra you are functionally innumerate in regards to metrics and the Einstein field equations. You talk about coordinate changes but you cannot actually do them yourself, you can only accept what other people say about such procedures and parrot their details or else just make up insufficiently informed arm waving opinions. You have no understanding of any details of any area of physics, particularly electromagnetism, quantum mechanics and relativity. This constant "I understand it" from you is utterly unjustified and repeatedly refuted. If you wish to put forth your take on things in physics you have no experience with then we have places for that on this forum, but they are not this sub-forum. Until such time as you can demonstrate a working understanding of things you claim to understand your work will remain the domain of the pseudo sub-forum. Given it's rejection from every journal you've sent it to, the dismissal of your claims on every forum where actual physicists reside and your inability to answer the most relevant questions about your claims you're going to have to do some serious work over in pseudo before you are allowed to discuss such things here. In the future if you do this "I understand this stuff" and then go on to talk about mathematical constructs, such as the metric, which you are demonstrably innumerate in and therefore cannot understand I'll consider it trolling because it amounts to little more than trolling. You've got a forum to present your ideas, use it rather than derail threads in this forum where people are interested in more than frantic arm waving and delusions of grandeur.
I have to agree with AlphaNumeric. Given the answers to my questions above, it seems clear that Farsight is only interested in deceiving others into believing that he understands this subject but is not interested in engaging the subject matter.
How about you describe what's futile about what I said about black hole physics and the metric? I'm not backing down because I understand the physics. Farsight won't back down because he's a crank.
Is this your first experience posting on a public science forum? I don't care what you think. You should quit trolling me if you want this to end. Alphanumeric has spoken and put 'crank science' where it belongs. Crank science = bs.
Thanks for clarifying that for us. This is the last thing I'll say about the Farsight MO. Whenever he wants to dodge a difficult question requiring some real analysis he says start a new thread and I'll answer it.
Tried to crack a joke, get accused of trollin'. Please Register or Log in to view the hidden image! Well, Farsight considers his stuff "alternative", so as you said, he will never back down. He's just gonna come back with more elaborate methods to promulgate whatever he's already saying. He just sees thing very differently. Through tinted lenses.
Anyway, re angular momentum, this is a nice website: https://motivate.maths.org/content/wonderful-world-gyroscopes/gyroscopic-effect The article is by Hugh Hunt of the Engineering department at the University of Cambridge. Follow the links to find more. When I first saw it I was a bit puzzled by the third picture which refers to moment of inertia as "angular mass". I realised what this meant when I got a gyroscope. When it isn't spinning, you can waggle it easily. However when you spin it up, it resists your attempts to waggle it. Nowadays I've got an electric gyroscope, and it still tickles me when I hand it to somebody and get them to try it out.
I don't know where I've "failed very badly". I should explain it clearly again. I think rpenner will be interested, though he probably knows how to do it already. Here's the crude derivation: \(s_y=f(t) s_x=g(t)\) Then: \(v_y=f'(t) v_x=g'(t)\) \(a_y=f''(t) a_x=g''(t)\) \(r=\sqrt{{s_y}^{2}+{s_x}^{2}}=\sqrt{{f(t)}^{2}+{g(t)}^{2}}\) \(v=\sqrt{{v_y}^{2}+{v_x}^{2}}=\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\) \(a=\sqrt{{a_y}^{2}+{a_x}^{2}}=\sqrt{{f''(t)}^{2}+{g''(t)}^{2}}\) Now we need to find an expression to represent the angle between v and r. At first, I tried using arctan, but its very hard to simplify. So I turned to the sine rule, to be specific, the formula for the area of a triangle, \(\frac{1}{2}bdsin\theta\) Please Register or Log in to view the hidden image! We can express the horizontal and vertical components as triangles. Please Register or Log in to view the hidden image! Now if we want to the use the area formula, we have to scale one of the triangles like so: Please Register or Log in to view the hidden image! We use a scaling factor, k. First we're scaling the horizontal component. \(kg'(t)=g(t)\) \(k=\frac{g(t)}{g'(t)}\) As seen in the 2nd image, θ represents the angle between v and r. The expression \(\frac{1}{2}bdsin\theta\) describes the sum of the areas of the red and blue triangles. \(b=\sqrt{{f(t)}^{2}+{g(t)}^{2}} d=k\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\) \(\frac{1}{2}bdsin\theta=A\) \(A=\frac{1}{2}f(t)g(t)-\frac{1}{2}k^{2}f'(t)g'(t)\) The values of all the time-based functions can be positive or negative. This can produce 4 different combos. What if we're trying to find the area of the red triangle that is not overlapped by the blue one? Please Register or Log in to view the hidden image! \(A=\frac{1}{2}k^{2}f'(t)g'(t)-\frac{1}{2}f(t)g(t)\) That's why we have a minus sign in the above formula. \(\frac{1}{2}bdsin\theta=A k\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}(sin\theta)=k^{2}f'(t)g'(t)-f(t)g(t)\) \(sin\theta=\frac{k^{2}f'(t)g'(t)-f(t)g(t)}{k\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}}\) \(k=\frac{g(t)}{g'(t)}\) \(\frac{k^{2}f'(t)g'(t)-f(t)g(t)}{k\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}}=\frac{[\frac{g(t)}{g'(t)}]^{2}f'(t)g'(t)-f(t)g(t)}{[\frac{g(t)}{g'(t)}]\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}}=\frac{\frac{{g(t)}^{2}f'(t)}{g'(t)}-f(t)g(t)}{[\frac{g(t)}{g'(t)}]\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}}=\frac{\frac{{g(t)}f'(t)}{g'(t)}-f(t)}{[\frac{1}{g'(t)}]\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}}=\frac{f'(t)g(t)-g'(t)f(t)}{\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}}\) \(sin\theta=\frac{f'(t)g(t)-g'(t)f(t)}{\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}}\) \(r=\sqrt{{s_y}^{2}+{s_x}^{2}}=\sqrt{{f(t)}^{2}+{g(t)}^{2}}\) \(v=\sqrt{{v_y}^{2}+{v_x}^{2}}=\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\) \({v}\times{r}=vr(sin\theta)={\sqrt{{f(t)}^{2}+{g(t)}^{2}}}{\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}}\frac{f'(t)g(t) - {g'(t)}f(t)}{\sqrt{{f'(t)}^{2}+{g'(t)}^{2}}\sqrt{{f(t)}^{2}+{g(t)}^{2}}}=f'(t)g(t) - {g'(t)}f(t)\) So finally we get: \(\frac{L}{m}=|f'(t)g(t) - {g'(t)}f(t)|\) Torque Using the same steps as the previous derivation. \(a=\sqrt{{a_y}^{2}+{a_x}^{2}}=\sqrt{{f''(t)}^{2}+{g''(t)}^{2}}\) \(\frac{\tau}{m}={a}\times{r}=ar(sin\beta)=|f''(t)g(t) - {g''(t)}f(t)|\) Is \(\frac{\tau}{m}=\frac{d\frac{L}{m}}{dt}\) ? Assuming mass to be constant: \(\frac{d\frac{L}{m}}{dt}=\frac{d}{dt}f'(t)g(t) - {g'(t)}f(t)\) \(=[f'(t)g'(t)+f''(t)g(t)]-[g'(t)f'(t)+g''(t)f(t)]=f'(t)g'(t)+f''(t)g(t)-g'(t)f'(t)-g''(t)f(t)=f''(t)g(t)-g''(t)f(t)\) \(=f''(t)g(t)-g''(t)f(t)\) When the acceleration is pointing towards the origin: \(\frac{f''(t)}{g''(t)}=\frac{f(t)}{g(t)}\) \(f''(t)g(t)=g''(t)f(t) f''(t)g(t)-g''(t)f(t)=0 \) \(\frac{\tau}{m}=|f''(t)g(t) - {g''(t)}f(t)|=0\) When the acceleration is pointing towards the origin, there is no net torque. Simple, eh? Angular Velocity It seems fairly straightforward that \({v}\times{r}=(r\omega)(r)\) Let's see if its compatible with equations. The angle w.r.t. to the origin, ϕ Assuming basic angles, \(tan\phi=\frac{s_y}{s_x}=\frac{f(t)}{g(t)}\) \(\phi=arctan[\frac{f(t)}{g(t)}]\) \(\omega=\frac{d\phi}{dt}=\frac{d}{dt}arctan[\frac{f(t)}{g(t)}]\) The rules of differentiation: \(\frac{d}{dt}arctan[h(n)]=\frac{h'(n)}{1+h(n)^2}\) \(\frac{d}{dt}arctan[\frac{f(t)}{g(t)}]=\frac{\frac{f'(t)g(t)-g'(t)f(t)}{g(t)^2}}{1+\frac{f(t)^2}{g(t)^2}}=\frac{\frac{f'(t)g(t)-g'(t)f(t)}{g(t)^2}}{\frac{g(t)^2+f(t)^2}{g(t)^2}}=\frac{f'(t)g(t)-g'(t)f(t)}{g(t)^2+f(t)^2}\) \(\omega=\frac{f'(t)g(t)-g'(t)f(t)}{g(t)^2+f(t)^2}\) \(r^{2}=f(t)^2+g(t)^2\) \(r^{2}\omega=[f(t)^2+g(t)^2]\frac{f'(t)g(t)-g'(t)f(t)}{g(t)^2+f(t)^2}=f'(t)g(t)-g'(t)f(t)=\frac{L}{m}\) Yeah, that's about it.
This is not true, in general \(\frac{s'_x}{s_x}=\frac{v_x}{s_x}\) isn't constant, it isn't even time independent , as you are trying to manipulate, so, your \(kg'(t)=g(t)\) is wrong from the start. On the other hand it is remarkable that you put in all this effort to write all this stuff in LateX. You really need to learn how to use vectors, for example, since \(\vec{L}=\vec{r} X \vec{v}\) and \(\vec{T}=\vec{r} X \vec{a}\) and since in general \(\vec{v}\) and \(\vec{a}\) do not have the same direction, you cannot use the same angle \(\theta\) in the expressions of the scalars for angular momentum and torque, as you've tried. Think about the case of circular motion, \(\vec{v}\) and \(\vec{a}\) do not have the same direction, they do not make the same angle with the positional vector \(\vec{r}\). On the other hand, \(\frac{d\vec{L}}{dt}=\frac{d\vec{r}}{dt} X \vec{v}+\vec{r} X \frac{d\vec{v}}{dt}=0+\vec{r}X\vec{a}=\vec{T}\) , true for any motion. One line and the proof is done, you only need to learn how to use vectors.
