# Author: alan

## (1000) James G Bridgeman’s answer to Do the effects of special relativity apply to non inertial frames of reference? If I accelerated to a constant velocity of 99% c would I continue to experience time dilation for as long as I maintained that speed or only when accelerating? – Quora

## (1000) Greg Hansen’s answer to Is the theory of relativity valid in a non-inertial frame? – Quora

## The Relation Between Acceleration and Time Dilation in Special Relativity

http://www2.math.uconn.edu/~bridgeman/posts/acceleration.pdf

## The twin paradox: the role of acceleration

https://arxiv.org/pdf/1807.02148.pdf

## Orbits as Conic Sections

The answer to the question is, I think, Yes.

In fact I believe that most of the proofs in Newton’s Principia were expressed in an essentially geometric form rather than making use of the calculus. (He may well have used calculus in his preliminary thinking and in other less formal applications but apparently preferred to avoid it in his most formal presentations).

A possible geometric approach to the conic sections result might be to look at the changes in position, [math]\vec{r}[/math], and velocity, [math]\vec{v}=\dot{\vec{r}}[/math], of the test particle relative to the planet [math]P[/math], and to just show directly that the effect of an inverse square force law giving a centripetal acceleration of magnitude [math]\frac{\mu}{r^{2}}[/math] (so giving [math]\ddot{\vec{r}}=-\frac{\mu\vec{r}}{r^{3}}[/math]) ensures that they satisfy some geometric property of conic sections (rather than following the usual modern approach of deriving the parametric or polar equations of the full trajectory).

There are various geometric conditions that could be used, but the one that seems most promising (because it applies to all of the conic sections in a similar way) is the extension of the Focus and Directrix definition of a parabola. This uses a point [math]P[/math] (the focus), a line [math]D[/math] (the directrix), **and **a number [math]e[/math] (the eccentricity), with the requirement that for any point on the curve, its distance from [math]P[/math] is [math]e[/math] times the distance from [math]D[/math]. The case of [math]e=1[/math] is of course the parabola, the cases [math]1>e>0[/math] give ellipses, [math]e>1[/math] gives hyperbolas, and the limiting case of [math]e=0[/math] gives just a point if the line [math]D[/math] is fixed but can be a circle if [math]D[/math] goes off to infinity at the same time as [math]e[/math] goes to zero.

An alternative version of this, which avoids treating the circle as an awkward limiting case, is to specify instead of [math]e[/math] and [math]D[/math] an “eccentricity vector”, [math]\vec{e}[/math], of length [math]e[/math], and to require that the component of [math]\vec{r}[/math] in the [math]\vec{e}[/math] direction plus [math]\frac{1}{e}[/math] times the length, [math]r[/math], of [math]\vec{r}[/math] remains constant. This would ensure that, for [math]e\neq 0[/math], the set of points at distance [math]\frac{r}{e}[/math] from the particle in the direction of [math]\vec{e}[/math] is a line perpendicular to [math]\vec{e}[/math] at a distance [math]\frac{r}{e}[/math] from the particle (and so [math]r_p + r_p/e[/math] from the focus, where [math]r_p [/math] is the minimum “periapsis” distance of the path from the focus). When [math]e\neq 0[/math], this line then serves as the directrix [math]D[/math] in the previous definition. But this version also makes perfectly good sense (without any mention of a directrix) even when [math]e=0[/math] (in which case the direction of [math]\vec{e}[/math] does not need to be defined).

Apparently [math]\vec{e}=(\frac{v^2}{\mu}-\frac{1}{r})\vec{r}-\frac{\vec{r}\cdot\vec{v}}{\mu}\vec{v}[/math] does the job. But I haven’t checked whether this is easy to see on traditional geometric grounds or from the picture, let alone whether there is a nice intuitive way to come up with it. (I may add more to this answer if I can get anywhere with that.)

OK. I’ve made *some* progress. So here goes!

At each stage I’ll start by using modern notation and tools to motivate a proposition and then maybe come back later and try to give or suggest a more geometric proof.

Let’s start by noting that for *any* central force, directed always at point [math]P[/math], there is no torque about [math]P[/math] so we expect the angular momentum to be constant. Or in terms of vector geometry, if [math]\ddot{\vec{r}}||\vec{r}[/math] then [math]\frac{d}{dt}(\vec{r}\times\dot{\vec{r}})=\dot{\vec{r}}\times\dot{\vec{r}}+\vec{r}\times\ddot{\vec{r}}=\vec{0}[/math]. This leads immediately to Kepler’s area law, but of greater interest to us right now is that it provides an invariant vector [math]\vec{h}=\vec{r}\times\dot{\vec{r}}[/math] – albeit one that is perpendicular to rather than in the plane of our expected orbit.

Before going on, you might object that, by using vectors and derivatives, this is not a geometric argument in the intended sense of Euclidean geometry. But really the use of vectors is just an alternative way of expressing geometric facts and we can’t expect to completely avoid calculus in a problem whose statement is basically about acceleration. So for now I’ll just continue in the same spirit, but I may come back later to see if we can phrase the arguments in a more classically Euclidean form.

To get a vector in the orbital plane we could now just take the cross product of [math]\vec{h}[/math] with anything in that plane, such as either [math]\vec{r}[/math] or [math]\dot{\vec{r}}[/math] for example. Of course we have no reason to expect either of these to be invariant, but let’s see what they are.

Now the case of [math]\vec{r}\times\vec{h}[/math] gives [math]\frac{d}{dt}(\vec{r}\times\vec{h})=\dot{\vec{r}}\times\vec{h}+\vec{r}\times\dot{\vec{h}}[/math] and for any central force we have [math]\dot{\vec{h}}=\vec{0}[/math] which just brings us back to the other case with no opportunity to make use of the inverse square law.

But for the other case we have [math]\frac{d}{dt}(\dot{\vec{r}}\times\vec{h})=\ddot{\vec{r}}\times\vec{h}+\dot{\vec{r}}\times\dot{\vec{h}}[/math] and using the inverse square law and the fact that [math]\dot{\vec{h}}=\vec{0}[/math] gives [math]\begin{align}\frac{d}{dt}(\dot{\vec{r}}\times\vec{h}) &=-\frac{\mu}{|r|^3}\vec{r}\times\vec{h}+\dot{\vec{r}}\times\vec{0} \\&= -\frac{\mu}{|r|^3}\vec{r}\times(\vec{r}\times\dot{\vec{r}})=-\mu\frac{(\vec{r}\cdot\dot{\vec{r}})\vec{r}-(\vec{r}\cdot\vec{r})\dot{\vec{r}}}{(\vec{r}\cdot\vec{r})^{3/2}} \\&= \mu[-(\vec{r}\cdot\vec{r})^{-3/2}(\vec{r}\cdot\dot{\vec{r}})\vec{r}+(\vec{r}\cdot\vec{r})^{-1/2}\dot{\vec{r}}] \\&= \mu[-\frac{1}{2}(\vec{r}\cdot\vec{r})^{-3/2}(\vec{r}\cdot\dot{\vec{r}}+\dot{\vec{r}}\cdot\vec{r})\vec{r}+(\vec{r}\cdot\vec{r})^{-1/2}\dot{\vec{r}}] \\&= \mu\frac{d}{dt}\frac{\vec{r}}{(\vec{r}\cdot\vec{r})^{1/2}} \end{align}[/math]

So [math]\vec{L}=(\dot{\vec{r}}\times\vec{h})-\mu\frac{\vec{r}}{(\vec{r}\cdot\vec{r})^{1/2}}=(\vec{v}\times\vec{h})-\mu\frac{\vec{r}}{r}[/math] is constant.

Furthermore,

[math]\begin{align}\vec{r}\cdot\vec{L}&=\vec{r}\cdot(\vec{v}\times\vec{h}-\mu\frac{\vec{r}}{r})=\vec{r}\cdot(\vec{v}\times\vec{h})-\mu\frac{\vec{r}\cdot\vec{r}}{r}\\&=(\vec{r}\times\vec{v})\cdot\vec{h}-\mu r=\vec{h}\cdot\vec{h}-\mu r=h^2-\mu r \end{align}[/math]

So [math]\vec{r}\cdot\vec{L}+\mu r=h^2[/math], which is also constant.

Thus [math]\vec{e}=\frac{\vec{L}}{\mu}[/math] is a constant vector for which [math]\vec{r}\cdot\vec{e}+|\vec{r}|[/math] is also constant, and dividing by [math]e=|\vec{e}|[/math], we see that the projection, [math]\vec{r}\cdot\frac{\vec{e}}{e}[/math] of [math]\vec{r}[/math] in the direction of [math]\vec{e}[/math] when added to [math]\frac{1}{e}[/math] times the length of [math]\vec{r}[/math] also gives a constant (which is exactly the condition we used to define the eccentricity vector).

Since [math]\vec{e}=\frac{\vec{L}}{\mu}=\frac{\vec{v}\times(\vec{r}\times\vec{v})}{\mu}-\frac{\vec{r}}{r}=\frac{(\vec{v}\cdot\vec{v})\vec{r}+(\vec{v}\cdot\vec{r})\vec{v}}{\mu}-\frac{\vec{r}}{r}=(\frac{v^2}{\mu}-\frac{1}{r})\vec{r}-\frac{\vec{r}\cdot\vec{v}}{\mu}\vec{v}[/math], this does match the definition proposed above.

Also, [math]\vec{r}\cdot\vec{e}=\frac{h^2}{\mu}-r[/math], and if [math]\phi[/math] is the angle between [math]\vec{e}[/math] and [math]\vec{r}[/math] then [math]\vec{r}\cdot\vec{e}=re\cos{\phi}[/math], so [math]re\cos{\phi}=\frac{h^2}{\mu}-r[/math], and [math]r=\frac{h^2/\mu}{1+e\cos{\phi}}[/math] is the polar equation for the orbit relative to origin [math]P[/math] and axis direction [math]\vec{e}[/math].

The case [math]e=0[/math] gives [math]r=\frac{h^2}{\mu}[/math] which is constant so we have a circle. Either using [math]h=v r[/math], or just equating the centripetal and gravitational accelerations, we then get [math]v=\sqrt{\mu/r}[/math], and so the orbital period is given by [math] T=\frac{2\pi r}{\sqrt{\mu/r}}=\frac{2\pi}{\mu}r^{3/2}[/math].

Extras:

[math]\begin{align}\frac{d}{dt}(\vec{r}\cdot\vec{L}) &=\dot{\vec{r}}\cdot\vec{L}+\vec{r}\cdot\dot{\vec{L}}=\dot{\vec{r}}\cdot[(\dot{\vec{r}}\times\vec{h})-\mu\frac{\vec{r}}{(\vec{r}\cdot\vec{r})^{1/2}}]+\vec{0} \\&=-\mu(\vec{r}\cdot\vec{r})^{-1/2}\vec{r}\cdot\vec{r’}=-\mu\frac{d(\vec{r}\cdot\vec{r})^{1/2}}{dt}=-\mu\frac{d|\vec{r}|}{dt}\end{align}[/math]So [math]\frac{d}{dt}(\vec{r}\cdot{\vec{L}}+\mu|\vec{r}|)=0[/math].

[math]\vec{e}=\vec{0}[/math] so the constancy of [math]\vec{r}\cdot\vec{e}+|\vec{r}|[/math] just gives [math]|\vec{r}|= constant[/math] which is a circle. And the condition that [math](\dot{\vec{r}}\times\vec{h})-\mu\frac{\vec{r}}{|\vec{r}|}=\mu\vec{e}=\vec{0}[/math] tells us that [math]\dot{\vec{r}}\times(\vec{r}\times\dot{\vec{r}})=\mu\frac{\vec{r}}{|\vec{r}|}[/math] and so (since [math]\dot{\vec{r}}\times(\vec{r}\times\dot{\vec{r}})=(\dot{\vec{r}}\cdot\dot{\vec{r}})\vec{r}-(\dot{\vec{r}}\cdot\vec{r})\dot{\vec{r}}[/math], and [math]|\vec{r}|= constant[/math] gives [math](\dot{\vec{r}}\cdot\vec{r})=0[/math]), we find that [math](\dot{\vec{r}}\cdot\dot{\vec{r}})\vec{r}=\mu\frac{\vec{r}}{|\vec{r}|}[/math]## fromQuora: Isn’t a ‘probability wave’ simply a statistical function and not a real wave? Does it no more ‘collapse’ than me turning over a card and saying that the probability ‘wave’ of a particular deal has collapsed?

Well, as other answers have noted, the wave function of quantum mechanics is not a probability wave as its values are complex and it is only the *squared* amplitude that gives a probability density. But the process of “collapse” involved in a quantum measurement does involve something like your playing card analogy.

There are actually two stages in the measurement and observation process. One is the interaction with an incompletely known measurement apparatus which reduces or eliminates the prospects for future interference and basically turns the previously pure state of the isolated system (considered as a subsystem of the larger world) into a statistical mixture. And the second is the collapse of that statistical mixture by observation – with the result that, from the observer’s point of view, of the many possible alternatives only one is actually true.

And if this makes it seem to you that the “state” of the system actually depends on the observer then you are on the right track. (But it is nothing special about the “consciousness” of the observer that is relevant here. Almost any localized system could play the same role relative to the rest of the universe.)

Any configuration history of any physical system can be considered as “seeing” the rest of the universe in a “relative state” which “collapses” when the configuration history in question passes a point beyond which the configuration includes information about that particular measurement value.

## Why does the Boltzmann distribution of gas particles not collapse to equal velocities (thermal equilibrium)? I would like help explaining it to high school chemistry students. – Quora

Despite some rather dismissive responses from other experts I think that this is a very good question which deserves a real answer.

It is indeed reasonable for a student to ask why the equal distribution of energy between degrees of freedom in subsystems that appears to result from equipartition and heat flow does not extend down to the level of individual molecules in a gas. And the answer does not require a complete calculus-based derivation of the Maxwell-Boltzmann distribution.

All that is required is some clarification of the distinction betwee microstates and macrostates, where the former are specified in complete detail but the latter always involve some unknown details which have to be dealt with by probabilities.

For a “gas” of just two identical spherical atoms confined by perfectly elastic collisions with the walls of a massive container, the actual energy distribution of the atoms never changes (because during a collision between two atoms conservation of energy and momentum requires that both [math]{v_1}^{2}+ {v_2}^{2}[/math] and [math]v_1+v_2[/math] remain unchanged which allows only that [math]v_1[/math] and [math]v_2[/math] either stay the same or exchange values). But if the interactions with the walls allow for energy transfers to or from the atoms then the distribution may change over time. And if the transfers are small and random in such a way that we end up not knowing the individual particle energies but do know that there is negligible accumulated change in the total energy of the system, then it seems reasonable to make the **assumption **that all of the microstates (ie actual [math]v_1[/math] and [math]v_2[/math] values) corresponding to the macrostate defined just by total energy are equally likely.

For a simpler analysis (avoiding the need to discuss a continuous range of energy values – which *would *require calculus) let’s just imagine that the two atoms are distinguishable, that each of them can have energy values corresponding to whole numbers 0,1,2,3… , and that the macrostate is defined by having a total energy of 2. Then there are just three equally likely microstates with the actual values of [math](e_1,e_2)[/math] being (0,2), (1,1), or (2,0). In this situation the case of equal energies is actually less likely than that of their being unequal, but if the atoms were indistinguishable then the cases (0,2) and (2,0) would be the same and the probabilities would be equal. (And the fact that identical particles really are indistinguishable does make an experimentally significant difference in statistical mechanics.)

For the case of N distinguishable atoms in the macrostate with total energy E, the possible microstates are [math](e_1,e_2,…e_N)[/math] with [math]e_1+e_2+…+e_N=E[/math].

With [math]N=3[/math] and [math]E=6[/math], this allows [math](e_1,e_2,e_3)[/math] =(6,0,0),(5,1,0),(5,0,1),(4,2,0),(4,1,1),(4,0,2),(3,3,0),(3,0,3),(3,2,1),(3,1,2),(2,4,0),(2,0,4),(2,3,1),(2,1,3),(2,2,2),(1,5,0),(1,0,5),(1,4,1),(1,1,4),(1,3,2),(1,2,3),(0,6,0),(0,0,6),(0,5,1),(0,1,5),(0,4,2),(0,2,4),(0,3,3). For each particle, [math]p[/math], this includes one microstate with [math]e_p=6[/math], two with [math]e_p=5[/math], three with [math]e_p=4[/math], four with [math]e_p=3[/math], five with [math]e_p=2[/math], six with [math]e_p=1[/math], and seven with [math]e_p=6[/math], (and in only one of the 28 cases are all three energies equal).

When we combine two such systems with different temperatures (amounts of energy per degree of freedom) then at first the distribution of energies will have two peaks (with one corresponding to each of the combined populations). But assuming that the interactions with the external world allow (perhaps indirect) energy exchange in the usual way, then eventually we’ll reach a situation in which the probability of a microstate depends only on its total energy and every particle has the same probability distribution for its possible energy values.

And even for the case in which the two combined systems each consists of a single particle, the evolution towards a situation in which the two particles both have the same energy *distribution* does not lead to one in which they each actually have the same energy.

(This simple model was not intended to be realistic but just to show why the process of approaching thermal equilibrium does not lead to all of the molecules having the same velocity.)

## Another question on quantum entanglement from a non-physicist : QuantumPhysics

## index – QuantumPhysics

Source: *index – QuantumPhysics*