Recent content by parkner

Nope. The relative quantities in physics are always frame independent, because these are just the elementary relations between the bodies. The reference system is irrelevant here, because the relation is just an operator: a op b, and it must be, and it's always a symmetric operator, due to...

No, because a speed is just |v|, what can't be a negative number.

I wish to remind only: relative velocity is a coordinate independent thing. This is just a relation of the two bodies - a distance derivative. In SR this is also preserved: there is the same v in the both frames: in the stationary and in the moving. In a one dimensional case: v > 0 means a...

I can't agree with this interpretation. The relative velocity is just the 'rate of closure': v = dx/dt, ie. it's a relative distance hange. And in a general case: v = v_r + v_t, where: v_t = wr = df/dt r so there is possible some additional tangential displacement: an angular position...

It adds normally: c - v. For example: we send a signal to the space probe at distance d, and moving away with a speed v. What is a time of the signal journey to reach the probe? Of course, this is not d/c, but just d/(c-v) Thus the signal speed wrt the probe is c-v exactly, not c, what...

This is just an abstract entity - model specific; something like a number in the algebra theory, or a point, line and a vector in the geometry. In the reality we can identify a photon as an event - the detection of impulse, energy, a state change... but this is rather oversimplification. The...

Correct. After all there is no quantisation visible in the black body radiation - it's continuous function. The derivation uses explicitly the condition for the standing waves. This is just the standing waves condition.

This is standard... check the calculations in QM, for example the hydrogen atom, and you should find the cause - explanation of the quantisation. with respect to an oscillator with fundamental frequency denoted as ##\nu## ...

The observed quantisation is always just a consequence of the boundary conditions, and this applies to the atom, the Hall effect, flux quantisation in the superconductivity, tunneling effect, etc.; all the so-called quantum effects without exception. Quantity of energy can be anything - there...

Planck never assume any photons. This is an Einstein's idea, used for photoefect (a simplified version only). The black body radiation is just a sum over all possible wavelengths in the box. There is a simple boundary condition for the standing waves, which gives the quantisation, not any...

Maybe I underappreciated too much these 'estimation' method of r''. This will be an extrapolation. Very little precision... perhaps 2-nd order with three points, but locally, thus globally 1-st only. I plan to compute with 8 order... at last. :) This is probably just implicit ode...

You can't compute in this way, because the r'', which you suggest to compute just geometricaly, will be incorrect - some static version, ie. without any forces (you assume here: v = const, at a position r(t)). This is probably the same, bacause you need to konow acceleration to modify a...

No, i don't konow the r'', because the force is defined as a function of the r'' also. With a standard electric force: F = ke^2/r^2 r^0 and: F = ma, then we can solve: a = F/m = ke^2/mr^2 r^0, simply. But with the Weber force we get: a = F/m = ke^2/ma (1 - r'^2/2c^2 + r''/c^2) r^0; and the...

Maybe I would ask to move this issue to the section of physics... mathematicians probably don't know too much about numerical problems of solving equations of motion. i correct a mistake in my recent post: |r|'' = a_r + v_t^2/r

I need to compute numericaly n-body sys. interacting acording to the Weber force: http://en.wikipedia.org/wiki/Weber_electrodynamics and I have a problem with the acceleration on rhs: r'', because the acceleration is unknown, due to the Newton law: F = ma, and we need just 'a' to do next...