Ok, I have some problem with that last statement. Just because you change the relative velocity of a frame doesn't mean you change the orientation of its axes. Therefore, I don't get how you can say -x is physically the same as x, since both frames keep the same orientation?
It's not that I have a physical reason for it, but it is an assumption and assumptions must be justified. In particular, I am thinking about the following step:
## x' = (x-vt) \gamma(v) ##
## t' = (t-\frac{vx}{c^2}) \gamma(v) ##
Now, to obtain the reverse transformation, one would simply...
In the special theory of relativity, it seems impossible to derive the lorentz transformation without assuming that the lorentz factor is independent of the sign of the relative velocity. For some reason, I can't get my head around why this assumption is so easily made, as if it's trivial. Can...
Well, I suppose for the second method if you have a function and the slope is quite steep locally, then ##x_{n+1}## may be close to ##x_n## even though there is a large change in the function value. This gives the risk of premature convergence. The first method does not have that problem, but...
What I'm wondering is whether the second method is more robust, as it doesn't seem to have that particular problem that I mentioned. Perhaps there are downsides to the second method as well where the first method may perform better; I just haven't figured it out. My overall goal here is to get...
I suppose either would do when you just want the root, but then again I can imagine that if you take a function that comes very close to zero yet doesn't actually cross it it may see it as a false root, say something like x^2 + 0.000001 with a tolerance of like 0.0001 would still lead to a root...
The Newton-Raphson algorithm is well-known:
##x_{n+1} = x_n - \frac{f(x_n)}{f'(x_{n})}##
Looking at a few implementations online, I have encountered two methods for convergence:
1) The first method uses the function value of the last estimate itself, ##f(x_n)## or ##f(x_{n+1})##. Since at...
It's been some years for me, so I don't have a direct answer, but afaik you have to be careful here as to whether or not you are talking about engineering strain or true strain. In the former case, as is also the case for the spring force, L is considered the initial length, and thus is constant...
You may want to think about this:
- Which of Kepler's laws is about the orbital period?
- Does this law say anything about the eccentricity of the orbit?
Finally, I think you may be confusing "maximum radius" (aphelion) with semi major axis.
I only looked at the beginning, but it looks a bit odd to me. For starters, why would there be a tangential force Ft at A? If you assume no slip, then the tangential force there should be zero. If you do assume slip, then you'd need an equation involving the friction coefficient somewhere.
And...
I haven't read through the article, but the figure in the link below looks like it is what I meant. A proof that pressure is isotropic comes from force balance on that tetrahedron
https://www.brighthubengineering.com/naval-architecture/106499-hydrostatic-pressure-and-pascals-law-for-static-fluids/
To get a relationship between the different faces, you'll have to use a tetrahedron instead as there is no way to prove such a relationship with an infinitesimal cube.
It's just a theoretical concept. It's infinitely small. There's not much more than that to it. And yes, if you want to describe it as a charge distribution you'd use a three dimensional dirac delta function for it.