Concept of closing speed?

[rotchm]

dono incited me to ask such a question.

In the recent thread entitled "addition of velocities", the op
inquired about closing speeds.

An aggressive debate ensued.

I (and others) maintained that closing speed (of two things)
is simply "v1 ± v2" [depending of the sign conventions].
Or, the rate of change of the separation (distance) between the two things.
[We were discussing trajectories along the x_axis only btw].

Dono however, maintains that the above are not the meaning of closing speed as used here (relativity and kinematics in general). And he reasons that:

"v1+v2 is the closing speed of two objects that STARTED SIMULTANEOSLY. "

" If the two objects don't start simultaneously, one cannot define closing speed."

I am wondering as to what is the consensus here concerning the concept of
"closing speed".

[Dono.]

On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:
> dono incited me to ask such a question.
>
> In the recent thread entitled "addition of velocities", the op
> inquired about closing speeds.
>
> An aggressive debate ensued.
>
> I (and others) maintained that closing speed (of two things)
> is simply "v1 ± v2" [depending of the sign conventions].

Well, you are an idiot, Stephane.
Time has made this only worse.
Closing speed is a convention, used to express the rate at which two objects, starting simultaneously, cover a distance L between them:

v1*t+v2*t=L (see, "t" needs to be the SAME)
t=L/(v1+v2)

By CONVENTION, UNDER the above conditions, u=v1+v2 is called closing speed
The concept is very useful when one of the "objects" is a pulse of light, say v1=c. Then the closing speed of covering a distance "L" is c+v. But cranks, of the type of Thomas Heger and Stephane Baune, call this "relative speed of light". This causes infinitely long debates about light speed not being constant. Most of such debates center around Einstein derivation of the Lorentz transforms in "On the Electrodynamics..." but also in the derivation of the phase difference in the Sagnac experiments where cranks like Stephane Baune will claim that light goes around at c+v in one direction and c-v in the opposite direction. Ergo, the Sagnac effect is "proof" that light speed is "variable".

[rotchm]

On Thursday, September 29, 2022 at 10:44:29 PM UTC-4, Dono. wrote:
> On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:

> Closing speed is a convention, used to express the rate at which two objects,

Finally, you agree to that part (you denied this in the previous thread). Good.

> starting simultaneously,

Simultaneity has nothing to do with it.

> v1*t+v2*t=L

That is not the definition of closing speed.

> (see, "t" needs to be the SAME)

Yes that is what I told you.
Remember, I told you: At time t, the closing speed of two things is the "sum" of their speeds at that time: "v1 + v2".
You denied that.

> t=L/(v1+v2)
>
> By CONVENTION, UNDER the above conditions, u=v1+v2 is called closing speed

Finally we are getting somewhere in this thread...
Yes under the above conditions it is the closing speed.
In the general conditions, closing speed is v1 + v2, including in your above conditions.
No matter when or how to objects initiated their motion, at a given time, they have a closing speed which is
v1 + v2 at that time in question.

In the scenario that initiated this discussion, two bullets were approaching each other with speeds v1 & v2.
The closing speed is v1 + v2, no matter when these bullets were fired. So you finally agree to this!?

> The concept is very useful when one of the "objects" is a pulse of light, say v1=c. Then the closing speed of covering a distance "L" is c+v.

The closing speed between light and an object with speed v wrt a given reference frame is c ± v, independently
when these things initiated their motion.

So it looks like as to not make a fool of yourself in this thread, you finally agree (mostly)
with me & Stan.

[Dono.]

On Thursday, September 29, 2022 at 8:33:39 PM UTC-7, rotchm wrote:
> On Thursday, September 29, 2022 at 10:44:29 PM UTC-4, Dono. wrote:
> > On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:
>
> > Closing speed is a convention, used to express the rate at which two objects,
> Finally, you agree to that part (you denied this in the previous thread). Good.
>
> > starting simultaneously,
>
> Simultaneity has nothing to do with it.
>

That is because you are an imbecile who is incapable to follow how the concept is derived, Stephane.

> > v1*t+v2*t=L
>
> That is not the definition of closing speed.

Actually , it is. But you are too much of an imbecile to be able to follow it.

[rotchm]

Ok, let's see. At time t =0 your definition says
0=L which is the initial seperation, the length of the ship. Mighty good definition you have there!

And at time t= 0.01 say, do the two bullets have a closing speed (wrt ship frame)?

[Dono.]

Idiot,

"t" is the time when the two objects meet. You have outdone yourself , Stephane. Congratulations, bozo!

[rotchm]

You are the bozo here since you did not realize what happened. You cornered yourself again.

You are saying now that closing speed depends on the time at which two objects coincide.

Of course you can use the concept of closing speed to find out which time to object coincide, but that is not the definition of closing speed. The definition of closing speed is independent when the objects were set in motion, and when objects coincide if they do.

For instance, consider two objects leaving each other in separate directions. What is their closing speed at time t?

[Stan Fultoni]

On Thursday, September 29, 2022 at 8:33:39 PM UTC-7, rotchm wrote:

> > starting simultaneously,
>
> Simultaneity has nothing to do with it.

To be clear, two objects with positions x1(t) and x2(t) have the closing speed V(t) = dx1(t)/dt - dx2(t)/dt at time t in terms of S, and in terms of another system S' the closing speed is V'(t') = dx1'(t')/dt' - dx2'(t')/dt' at a given time t'. Obviously the relevant simultaneity (equal t or equal t') is built into this definition, as it is also built into the speed composition formula that lets us weite these as v1-v2 and v1'-v2'. Needless to say, whether or not the objects started simultaneously is irrelevant, as is the distance between them, given only the stipulation that they are both in flight with specified speeds in a specified order.

[Dono.]

On Thursday, September 29, 2022 at 9:19:13 PM UTC-7, rotchm wrote:
> On Friday, September 30, 2022 at 12:12:50 AM UTC-4, Dono. wrote:
> > On Thursday, September 29, 2022 at 9:01:44 PM UTC-7, rotchm wrote:
> > > On Thursday, September 29, 2022 at 11:45:50 PM UTC-4, Dono. wrote:
> > > > On Thursday, September 29, 2022 at 8:33:39 PM UTC-7, rotchm wrote:
> > > > > On Thursday, September 29, 2022 at 10:44:29 PM UTC-4, Dono. wrote:
> > > > > > On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:
> > >
> > > > > > v1*t+v2*t=L
> > > > >
> > > > > That is not the definition of closing speed.
> > > > Actually , it is.
> > > Ok, let's see. At time t =0 your definition says
> > > 0=L which is the initial seperation, the length of the ship. Mighty good definition you have there!
> > >
> > > And at time t= 0.01 say, do the two bullets have a closing speed (wrt ship frame)?
> > Idiot,
> >
> > "t" is the time when the two objects meet. You have outdone yourself , Stephane. Congratulations, bozo!

> You are saying now that closing speed depends on the time at which two objects coincide.
>

Just when I thought you reached bottom, Stephane, you managed to dig yourself even deeper. You are starting to give Ken Seto competition for the village idiot title. Thanks for the entertainment, dumbestfuck!

> For instance, consider two objects leaving each other in separate directions. What is their closing speed at time t?

This is even better than the prior one. You are getting more imbecillic by the post.

[Ken Seto]

On Thursday, September 29, 2022 at 10:35:14 PM UTC-4, rotchm wrote:
> dono incited me to ask such a question.
>
> In the recent thread entitled "addition of velocities", the op
> inquired about closing speeds.
>
> An aggressive debate ensued.
>
> I (and others) maintained that closing speed (of two things)
> is simply "v1 ± v2" [depending of the sign conventions].
> Or, the rate of change of the separation (distance) between the two things.
> [We were discussing trajectories along the x_axis only btw].

In SR closing speed is viewed from an outside observer. As viewed by an observer who receives the incoming light, the closing speed of light is c.

[patdolan]

On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:

The fundamental property of closing velocity consists in the fact that two observers in motion wrt each other will each calculate identical displacements wrt each other during identical time intervals.

[rotchm]

On Friday, September 30, 2022 at 11:00:48 AM UTC-4, patdolan wrote:

> The fundamental property of closing velocity consists in the fact that two observers in motion wrt each other
> will each calculate identical displacements wrt each other during identical time intervals.

Whats the definition of closing speed then?
Whats its "formula" ?
An example ?

[Richard Hachel]

Le 30/09/2022 à 16:27, Ken Seto a écrit :

> the closing speed of light is c.

<http://news2.nemoweb.net/jntp?6bjgLCUEIoCkHZ8jo5ZdMFC_27Y@jntp/Data.Media:1>


R.H.

[rotchm]

Link doesn't work.

[patdolan]

Excellent question, oh man of the miniature biceps. The formulation of closing velocity is an entirely different situation in a Galilean Universe vs. and Einsteinian one. We have been over this before. Add another 10 lbs to your curling bar and see if that helps.

[patdolan]

I shall deign to help you out a bit rotchm, and provide you the formula you crave.

In the case of observer O and observer O', the closing of v must be equivalent to closing velocity v' if the Dolan condition on closing velocity is to hold true

"The fundamental property of closing velocity consists in the fact that two observers in motion wrt each other will each calculate identical displacements wrt each other during identical time intervals."

Therefore, if v == v' then ∆x/∆t == ∆x'/∆t'. This last equality is true enough when it comes to the speed of light, and is the very expression of the second postulate. But it must also be true for ALL closing velocities less than c.

rotchm, use the LTs to solve for v' in terms of v by taking the ratio of ∆x'/∆t'. Then show us what you find. Do this immediately.

[Paul Alsing]

On Friday, September 30, 2022 at 1:35:18 PM UTC-7, patdolan wrote:

> "The fundamental property of closing velocity consists in the fact that two observers in motion wrt each other will each calculate identical displacements wrt each other during identical time intervals."

2 observers in motion wrt each other can *never* measure their closing speed, and what they *do* measure is always less than c.

[rotchm]

On Friday, September 30, 2022 at 4:35:18 PM UTC-4, patdolan wrote:
> On Friday, September 30, 2022 at 11:22:33 AM UTC-7, patdolan wrote:
> > On Friday, September 30, 2022 at 8:25:25 AM UTC-7, rotchm wrote:

> > > Whats the definition of closing speed then?

No answer?

> > > Whats its "formula" ?

No answer?

> > Excellent question, oh man of the miniature biceps. The formulation of closing velocity is an entirely different
> > situation in a Galilean Universe vs. and Einsteinian one.

Really? Is not closing speed of two things the "difference" of their speeds as measured in a chosed frame?
"v1-v2" ?

> In the case of observer O and observer O', the closing of v must

"the closing of v" makes no sense, its gibberish. You mean the closing speed of two "things", not of v.
You definitely need to learn to write coherently in English.

> be equivalent to closing velocity v' if the Dolan condition on closing velocity is to hold true

Off topic. We are not discussing any dolan conditions. We are discussing the notion of closing speed as used in physics, in SR.
Not your physics not somebody else's Theory or anything else.

> "The fundamental property of closing velocity

I asked about closing speed not closing velocity.
Re-read the op.

< rest of irrelevancies snipped>

[rotchm]

Since you talked about closing speeds, can you explain what you mean by such an expression?
That is, according to you, what is the definition of closing speed as used in physics/SR?

[patdolan]

Yes, this is a silly string of words, to be sure. A lot of people in Oregon and California do that a lot.

[patdolan]

The Einstein velocity addition formula makes a mockery of a well defined closing speed for all FoRs other than the co-moving frame of precisely *one* of the objects. And that object must consider itself at rest in it's own FoR while measuring the closing speed of the other object. This is Dolativity 101.

[mitchr...@gmail.com]

Who measured that paul. Have you got to those speeds?

[patdolan]

The me further add to my own brilliant insight that the proper closing speed of an object, described above, is made a mockery of by its coordinate closing speed. But that is a Dolativity 400 level course.

[Paul Alsing]

On Friday, September 30, 2022 at 6:23:14 PM UTC-7, patdolan wrote:
> On Friday, September 30, 2022 at 4:53:48 PM UTC-7, rotchm wrote:
> > On Friday, September 30, 2022 at 6:59:56 PM UTC-4, Paul Alsing wrote:
> > > On Friday, September 30, 2022 at 1:35:18 PM UTC-7, patdolan wrote:
> > >
> > > > "The fundamental property of closing velocity consists in the fact that two observers in motion wrt each other will each calculate identical displacements wrt each other during identical time intervals."

> > > 2 observers in motion wrt each other can *never* measure their closing speed, and what they *do* measure is always less than c.

> > Since you talked about closing speeds, can you explain what you mean by such an expression?

You need to read what Uncle Ben said about this in the first post of this thread...

https://groups.google.com/g/sci.physics.relativity/c/-b3lO0V6RZg

... in this post he says..."Closing speed is the rate of change of the distance between two objects as they both move. Closing speed is certainly a rate of change of a distance, but in the given frame of reference, there is no material object nor even a photon that moves with that speed."

Later he says...

"Consider two light beams aimed at each other, each traveling at speed c. The distance between their light fronts decreases at the rate 2c,
and this is the closing speed."

Pretty simple once you understand that this 2c measurement *must* be observed from a frame that is independent of the frames of either beam. The distance between 2 objects (or beams of light) that are approaching each other can exceed c, but no object itself can exceed c. I just do not understand why this concept is so hard to understand. If both you and I have a top running speed of, say, 20 mph, and we run directly towards each other, then we will be approaching each other at 40 mph, even though neither one of us can run that fast. Same with light beams. Of course, applying the Lorentz transformation, we could show that the 2 of us running towards each other are not actually approaching at 40 mph but something less than that... but at such low speeds it would be unmeasurable... which is why no one bothers to ever calculate approaching speeds using the Lorentz transformations! This concept is also discussed in Uncle Ben's post...

[rotchm]

> > > Since you talked about closing speeds, can you explain what you mean by such an expression?

> ... in this post he says..."Closing speed is the rate of change of the distance between two objects as they both move.

Yes, that's basically it. The closing speed between two entities is the rate of change of the distance (separation) between them, as measured by a given frame (system of coordinates). Its "v1 ± v2" depending of sign conventions.

[Tom Roberts]

On 9/30/22 10:17 PM, rotchm wrote:
>>>> Since you talked about closing speeds, can you explain what
>>>> you mean by such an expression?
>> ... in this post he says..."Closing speed is the rate of change of
>> the distance between two objects as they both move.
>
> Yes, that's basically it. The closing speed between two entities is
> the rate of change of the distance (separation) between them, as
> measured by a given frame (system of coordinates).

Yes, but only for inertial coordinates. The closing speed for a given
pair of objects varies, depending on which inertial frame is used to
calculate/measure it.

> Its "v1 ± v2" depending of sign conventions.

Only when the two objects are moving along a single axis in the inertial
frame used. If that isn't so, the calculation is MUCH more complicated.

Tom Roberts

[patdolan]

Now, Tom Roberts, provide this forum with the proper and coordinate closing speeds, as calulated by observers on each closing object. I will state up front that you won't/can't do it.

[Stan Fultoni]

On Saturday, October 1, 2022 at 10:02:56 AM UTC-7, tjrob137 wrote:

The same kinematics apply to motion along any curve, i.e., within any one-dimensional manifold, taking "distance" to refer to distance along the curve (i.e., distance within that 1D manifold), e.g., to cars moving along a curved road. The rate of change of the distance in this manifold satisfies the same kinematics. Since this is pure kinematics, the key is not inertialness, it is 1-dimensionalness.

> If that isn't so, the calculation is MUCH more complicated.

In a flat three dimensional manifold with inertial coordinates x,y,z,t we typically we are given x(t), y(t), z(t) for each object, and the rate of change of distance is just dL/dt where L = sqrt[(x2-x1)^2 +(y2-y1)^2 +(z2-z1)^2], so it isn't particularly difficult. For any other coordinate system x',y',z',t' we can also determine x'(t'), y'(t'), and z'(t') by direct application of the Lorentz transformation, and compute dL'/dt'. We can also easily transform to an inertial system in terms of which one object is at rest.

[rotchm]

On Saturday, October 1, 2022 at 1:02:56 PM UTC-4, tjrob137 wrote:
> On 9/30/22 10:17 PM, rotchm wrote:

> Yes, but only for inertial coordinates. The closing speed for a given
> pair of objects varies, depending on which inertial frame is used to
> calculate/measure it.

Yes that is what we have been repeating all along this thread.

> > Its "v1 ± v2" depending of sign conventions.
> Only when the two objects are moving along a single axis in the inertial
> frame used.

Yes as we have been repeating all along this thread.

[rotchm]

On Saturday, October 1, 2022 at 1:59:40 PM UTC-4, Stan Fultoni wrote:

> The same kinematics apply to motion along any curve, i.e., within any one-dimensional manifold, taking "distance" to
> refer to distance along the curve (i.e., distance within that 1D manifold), e.g., to cars moving along a curved road.
> The rate of change of the distance in this manifold satisfies the same kinematics. Since this is pure
> kinematics, the key is not inertialness, it is 1-dimensionalness.

Yes, that's more accurate!

[Paul Alsing]

No one can do it, as I already told you. Closing speed *cannot* be observed by any observer on a closing object, by definition. You apparently are clueless about this really simple concept.

[Burt Rotolo]

nonsense, you are both wrong. The motion is always in one dimension, the
point is where you embed this dimension, 2D, 3D, 3D+t, 4D etc. You guys
are talking not knowing what you are talking about.

watch out, not to cross my borderline.

[mitchr...@gmail.com]

How can an atom leave light behind if it does not have its own absolute speed in expanding space?
That is a motion BH that is only temporary.

[patdolan]

Increase in angle subtended by object, divided by proper time. This method will even give you acceleration.

[patdolan]

if it does not, will not
Have its own absolute speed

Expanding space?

That is a motion of Blackness
that is only temporary

[patdolan]

The distance d of an object of cross-section h subtended by an angle 𝜃 is

d = sqrt[ ( 1 - h^2 )/( 2 cos𝜃 ) - ( h^2/4 ) ]

The it's ∆d/∆t and ∆d/∆t^2

[patdolan]

Where did you go, Paul A#2 ? Surely you are not so mortified that you are at a loss for more words to type. Come now, my boy. Admit your lack of mathematical and physical circumspection then continue to struggle. Just like Uncle Bodkin used to do.

[Paul Alsing]

You are just bloviating evasive nonsense, Dolan. Closing speed is a really simple concept that has nothing whatsoever to do with subtended angles, acceleration, or the cross-section of an object. Desperate much? Again, you don't know what you don't know.

[lostgold]

English please

[patdolan]

Gladly. By measuring the change in the size of the angle subtended by a distant object for a given time interval, that object's closing speed can be determined.

[Paul Alsing]

You are assuming that you know the values of d and h... which would be unlikely in the real world. You are changing the initial conditions.

[lostgold]

thanks

[patdolan]

Paul A#2, you foolish fool. Paul A#1 is an excellent algebracist. By that standard, you are actually Paul A#1,001.

You don't need to know the value of h or d in order to determine an accurate closing speed. Assume any value for h. Then ∆d = f(∆cos𝜃) divided by ∆t will give you the same value for closing speed as if you knew d and h.

[Trevor Lange]

On Wednesday, October 5, 2022 at 8:02:33 AM UTC-7, patdolan wrote:
> By measuring the change in the size of the angle subtended by a distant object for a given time interval, that object's closing speed can be determined.

You've overlooked the effect of the finite speed of light. The rays from the object arrive at your eye after a certain delay, and if the object is receding from you, the amount of delay is changing. As a result, the rate of change of the subtended angle q at your eye is not related to the rate of change of the distance L by the simplistic dL/dt = -h/[4sin(q)^2](dq/dt) as it would be if the speed of light was infinite.

To correct for the effect of the finite speed of light (even in Galilean relativity), you need to account for how fast the delay time is changing, and this depends on the relative speed of the receding object... which is what you are trying to determine, so you are chasing your tail. Fortunately, all is not lost, because if you know the "rest color" of the object, you can infer the relative speed using the Doppler effect. Of course, once you have done this, you're done, and there is no point in trying to measure the rate of change of the subtended angle (which would not be practical in most cases anyway).

Up to this point, we're just talking about basic Galilean relativity. To be exact, you would need to use the relativistic Doppler formula, and you would have another reason for not trying to infer the speed from the angles, because you would need to account for relativistic aberration.

Lastly, what you get when you correctly infer the speed of the object using the Doppler effect is the relative speed, meaning the speed of the object in terms of the system of coordinates in which the eye is at rest. This is generally not the closing speed, because the closing speed refers to the sum or difference of the speeds in terms of some arbitrary coordinate system. In Galilean relativity the closing speed and the relative speed are the same, but in special relativity they are not.

[Paul Alsing]

No, your math is fatally flawed...

[Maciej Wozniak]

On Thursday, 6 October 2022 at 06:36:12 UTC+2, Paul Alsing wrote:

No, your math is fatally flawed...

Speaking of math, it's always good to remind that
your idiot guru had to announce its oldest, very
important and very successful part false - as
it didn't want to fit his idiotic postulates.

[patdolan]

Your last point is a good one. As to the previous points, in this forum we always assume that relativistic doppler has been taken into account. But thanks for pointing it out.

My overall effort was more to instruct young Paul A#2 in the multifarious ways that velocity can be ascertained, as he was laboring under another one of his many misconceptions. Your final point did cross my mind. But I considered it way beyond the scope of what the callow Paul A#2 was capable of digesting in a single post. But thanks again.

[patdolan]

On Wednesday, October 5, 2022 at 7:21:51 PM UTC-7, Trevor Lange wrote:

You final point also appears to provide fertile ground for thought on yet another means of falsifying the concept of the absolute relativity of motion. I hope to get back to you on it.

[Tom Roberts]

Consider inertial frame S with coordinates (x,t) -- ignore y and z as
they don't appear in this problem. Object A is located at x>0 and is
moving inertially along the x axis of S with velocity Va, Va<0. Object B
is located at x<0 and is moving inertially along the x axis of S with
velocity Vb, Vb>0. These two objects are clearly approaching each other.

In frame S, the closing speed of these two objects is |Va|+|Vb| = Vb-Va.

(Remember my V's are 3-velocities along the x axis, with
Va<0 and Vb>0.)

In the rest frame of A, object B has velocity Vba such that
Vb = (Va+Vba)/(1+Va*Vba/c^2)
so
Vba = (Vb-Va)/(1-Va*Vb/c^2)
In the rest frame of A, their closing speed is |Vba| = Vba.

In the rest frame of B, object A has velocity Vab such that
Vb = (Va+Vab)/(1+Va*Vab/c^2)
so
Vab = (Va-Vb)/(1-Va*Vb/c^2)
In the rest frame of B, their closing speed is |Vab| = -Vab.

While the expressions are different, |Vab| = |Vba|, and the closing
speeds in the two object's rest frames are the same.

Note that if Va<<c and Vb<<c, the closing speeds in all three frames are
approximately the same, |Va|+|Vb| = Vb-Va.

> I will state up front that you won't/can't do it.

As usual, you are wrong.

Tom Roberts

[Ross A. Finlayson]

On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:
> dono incited me to ask such a question.
>
> In the recent thread entitled "addition of velocities", the op
> inquired about closing speeds.
>
> An aggressive debate ensued.
>
> I (and others) maintained that closing speed (of two things)
> is simply "v1 ± v2" [depending of the sign conventions].
> Or, the rate of change of the separation (distance) between the two things.
> [We were discussing trajectories along the x_axis only btw].
>
> Dono however, maintains that the above are not the meaning of closing speed as used here (relativity and kinematics in general). And he reasons that:
>
> "v1+v2 is the closing speed of two objects that STARTED SIMULTANEOSLY. "
>
> " If the two objects don't start simultaneously, one cannot define closing speed."
>
> I am wondering as to what is the consensus here concerning the concept of
> "closing speed".

The objects meet or pass.

Seems logic dictates "space contraction".

[Ross A. Finlayson]

On Thursday, September 29, 2022 at 7:44:29 PM UTC-7, Dono. wrote:
> On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:
> > dono incited me to ask such a question.
> >
> > In the recent thread entitled "addition of velocities", the op
> > inquired about closing speeds.
> >
> > An aggressive debate ensued.
> >
> > I (and others) maintained that closing speed (of two things)
> > is simply "v1 ± v2" [depending of the sign conventions].

> Well, you are an idiot, Stephane.
> Time has made this only worse.
> Closing speed is a convention, used to express the rate at which two objects, starting simultaneously, cover a distance L between them:
>
> v1*t+v2*t=L (see, "t" needs to be the SAME)
> t=L/(v1+v2)
>
> By CONVENTION, UNDER the above conditions, u=v1+v2 is called closing speed
> The concept is very useful when one of the "objects" is a pulse of light, say v1=c. Then the closing speed of covering a distance "L" is c+v. But cranks, of the type of Thomas Heger and Stephane Baune, call this "relative speed of light". This causes infinitely long debates about light speed not being constant. Most of such debates center around Einstein derivation of the Lorentz transforms in "On the Electrodynamics..." but also in the derivation of the phase difference in the Sagnac experiments where cranks like Stephane Baune will claim that light goes around at c+v in one direction and c-v in the opposite direction. Ergo, the Sagnac effect is "proof" that light speed is "variable".


Or, ..., that it produces higher order objects as waves, ..., that reflect in eventual disposition in
real wave collapse, ..., that c+v and c-v are "imaginary v".

It's called resonance these days some guys just got a Nobel prize for it.

[Ross A. Finlayson]

Figuring the simplest machine's the lever, there's still the ramp, ....

Which provides free work given potential.

Here for example the classical, "is" potential, inverse-squared one way and inverse-cubed the other.

Just like a perfect spherical geometry into Kruskal, or, Sczekeres, Kerr and Schwarzschild making a Chandrasekhar.

[Ross A. Finlayson]

"It's going in a circle", ....

[Ross A. Finlayson]

"Bianchi and Darboux: very close"

Hmm, the Bianchi identities are:

Del R^phi mu = 1/2 Del mu R^0.


Here I'm most interested in it where it's 1/2 instead of 1, that it is covariant differentiation,
that it is not, "re-co-variant", why it would be 1 instead of 1/2 this tensor convention, or
statement as according to formula that the common gradient is one-half one-way.

Del and Grad, here I always write Del and Grad together.

What is this mu here in Ricci tensor, it's associated to the covariant derivative the tensor
analysis, and I think a tensor is after a matroid, here that it's not satisfied, "what was
the tensor connection", that it's only "non-linear", then calling it one half.

It's a convention there are others, and probably it's pretty particular, but mostly
here about the equivalency function its integral is one half, but defines one.


Then there's the right-associator bit, here the connection's shifting out the kernel.

(Of its connection.)

Tensor analysis, usually vectors, ..., "simple tensors".


Tensors are like, "hey thanks for writing all this linear state in a vector space,
yes it's all covariant and what is this covariant derivative of a tensor what?"

Torsion on the kernel, singularities.


I like to write my road to reality in like these layers that are like "Road to Reality", ....

Fundamentally, ....

Yeah, then it's for the twistor theory, there is Penrose and there is the guy,

Rindler, ..., "Road to Reality, II"


"... In1961 Rindler used the Fitzgerald contraction
as the premise of his article "Length contraction paradox".
The thought experiment is now called the "ladder paradox". ...."


Restitutive, dissipative, ....



"... and derives the Einstein tensor, ", ....

[patdolan]

Thank you Tom Roberts for deigning to respond to my challenge. So far we have been making our way up a 5.4 - 5.6 route. Time to find some 5.11 stuff. In your pedestrian implementation of the Einstein velocity addition formula, these are the two most interesting lines that you type:

1) "While the expressions are different, |Vab| = |Vba|, and the closing

speeds in the two object's rest frames are the same."

This result you arrive at is correct of course. But I will demonstrate that it rests on a completely unproved assumption.

2) "Note that if Va<<c and Vb<<c, the closing speeds in all three frames are

approximately the same, |Va|+|Vb| = Vb-Va."

Would you care to speculate on what the ramifications are of closing speeds that are a substantial fraction of c? Say .867c when gamma = 2?

[Ross A. Finlayson]

It's they saiy "there's an non-linear impulse", which adds to zero,
then as it resonates, it adds to 1/2.


That's where, "in a circle, the wave resonance observes itself",
"yeah, radial is infinite and non-linear".

So, it adds half up/down what is input.

[Trevor Lange]

On Friday, October 7, 2022 at 10:38:34 AM UTC-7, patdolan wrote:

> Would you care to speculate on what the ramifications are of closing speeds

> that are a substantial fraction of c? Say .867c.... ?

There's no need to speculate. If, in terms of inertial coordinates in which you are at rest, one object is approaching you at speed 0.86602c from the left, and another is approaching at that speed from the right, their closing speed in terms of this system is 1.73205c and their relative speed is 0.98974c.

[patdolan]

Trevor, you and Tom Roberts just don't seem to get it. Stare at the LTs for a while and contemplate them as the Dolan does. Apprehend that your proper closing speed v is used both to by you AND the one who is closing on you, to convert between your respective proper and coordinate space intervals and time intervals. How strange. How unsupported by the facts. Why would the one closing on you measure the same proper closing speed that you measure on him??? You measure a different coordinate value of space interval for him. You measure a different coordinate value of time interval for him. Those coordinate values of space and time intervals are the constructors of his coordinate closing speed on you. So why in Dolan's name wouldn't you measure a different coordinate value of closing speed for him?

I'll tell you why: You are sooo steeped in classroom relativity that the foregoing is probably gibberish to you. It is to Jan. It is to Dirk. It is to Paul A#1. Bodkin would have at least tried to unravel its meaning. Tom Roberts has re-read the foregoing several times by now and he still can't make sense of it.

[Trevor Lange]

On Saturday, October 8, 2022 at 5:42:18 AM UTC-7, patdolan wrote:
> Your proper closing speed v is used both by you AND the one who is closing

> on you, to convert between your respective proper and coordinate space
> intervals and time intervals.

I think you aren't really meaning to talk about closing speeds, you are talking about what are called relative speeds, meaning the speed of one object in terms of the standard inertial coordinates in which the other object is at rest.

To understand the reciprocity, first consider a simpler case, e.g., a straight line with two sets of markings, called the x coordinates and the x' coordinates, and suppose they are related by x' = x - 3. This relationship can also be written as x = x' + 3. So the x' coordinates are offset from the x coordinates by 3, and the x coordinates are offset from the x' coordinates by -3. This is neither strange nor unsupported by facts. The reciprocity between the offsets is tautological and self-evident.

> Why would the one closing on you measure the same proper closing speed
> that you measure on him???

Just as in the simple example above, consider two systems of coordinates called the x,t coordinates and the x',t' coordinates, and suppose they are related by x' = 2(x - t*sqrt(3)/2) and t' = 2(t - x*sqrt(3)/2). This relationship is can also be written as x = 2(x' + t'*sqrt(3)/2) and t = 2(t' + x'*sqrt(3)/2).

> How strange. How unsupported by the facts.

It is manifestly neither strange nor unsupported by facts. The reciprocity between these systems of coordinates is tautological and self-evident.

> You measure a different coordinate value of space interval for him. You measure
> a different coordinate value of time interval for him.

The relationship between the space and time coordinates for the two systems is tautologically reciprocal and symmetrical, as shown above.

[patdolan]

You are not to be blamed that you are unable to conceive of two observers disagreeing on their relative velocity (of which closing speed is a subspecies).

Trevor measures Tom Roberts' relative velocity along Trevor's x-axis as 0.8660243c. In Trevor's universe (the only universe(unless SR is a many-worlds theory)) Tom's co-moving x-axis which is parallel to Trevor's x-axis contains two meter marks for every one meter mark of Trevor's. This is not an illusion just because the two axes are moving wrt each other. Tom's meters really are only half as long as Trevor's meters. Therefore Tom must measure in his universe--the only universe--two meters of displacement for every one meter of displacement that Trevor measures.

Furthermore, Trevor observers that Tom's clock is ticking at a rate only half as fast as Trevor's clock. This is not an illusion just because Tom's and Trevor's clocks are moving wrt each other. Tom's seconds really are twice as long as Trevor's seconds. Therefore Tom must measure in his universe--the only universe--one second for every two seconds that Trevor measures.

When Tom divides his displacement by his interval of time in his universe--the only universe--it is not an illusion or mathematically artifact that Tom's brain, operating at half the biological speed of Trevor's brain, concludes that his velocity towards Trevor is 3.4681689c or gama^2 times the proper velocity that Trevor calculates for Tom.

Not as tautological as you though now, is it. The equivalence of coordinate and proper relative velocity is ONLY tautological in the Galilean transforms. It is merely assumed in the LTs. Think about it before you fire off a knee-jerk reply.

[mitchr...@gmail.com]

Closing speed applies to light converging at light speed... or diverging.
Light obeys the speed limit but moving with respect to itself
it can go to 2C separation or 2C closing additions.

Mitchell Raemsch

[Trevor Lange]

On Saturday, October 8, 2022 at 12:35:05 PM UTC-7, patdolan wrote:

> Trevor measures Tom Roberts' relative velocity along Trevor's x-axis as 0.8660243c.

> Tom's co-moving x-axis is parallel to Trevor's x-axis...

The x and x' axes may be spatially aligned, but they are not parallel when both space and time are taken into account, because the x axis is at constant t=0, whereas the x' axis is at constant t'=0, and for standard inertial coordinates these are skewed relative to each other (as required by the inertia of energy), just as are the t and t' axes.

> Tom must measure two meters of displacement for every one meter of
> displacement that Trevor measures.

Not at all. For example, suppose Tom is displaced by 0.866 LY in one year in terms of Trevor's inertial coordinates. But Tom is displaced by 0 LY in terms of his own inertial coordinates. These displacements are not related by a factor of 2. This shows that you need to specify precisely what kind of "displacement" you are referring to, and you can't willy-nilly conflate displacements, distances, and lengths. You must actually understand each of these concepts.

Consider, for example, the markings on Trevor's x axis, and ask yourself how far apart they are in terms of Tom's system. You want to know the spatial distance between the markings *at a single instant* of Tom's system (equal t'), but the systems of coordinates have skewed x and x' axes. In terms of S (Trevors's standard inertial coordinates), the spatial distance between markings "1" and "2" at time t is x2(t) - x1(t), and in S' (Tom's system) the distance between those markings "1" and "2" at time t' is x2'(t') - x1'(t').

The crucial thing you are overlooking is that those are two different pairs of events, so your simplistic reasoning fails. In terms of S', the "L meter" markings of x are a distance sqrt(1-v^2)*L apart, and in terms of S the "L meter" markings of x' are a distance sqrt(1-v^2)*L apart. This is all in accord with the equations I provided in the previous message.

> Furthermore, Trevor observers that Tom's clock is ticking at a rate only half as fast
> as Trevor's clock.

To be precise, each clock is ticking at half speed in terms of the standard inertial coordinates in which the other clock is at rest. This too is exactly in accord with the equations I provided in the previous message.

> Therefore Tom must measure one second for every two seconds that Trevor measures.

That's not quite right. As stated above, each clock runs slow in terms of the inertial coordinates in which the other clock is at rest. The relations between those systems of coordinates (given by the equations in the previous message) are perfectly reciprocal.

> Not as tautological as you though now, is it.

It is all as tautological and self-evident as I said, for the reasons explained above.

> The equivalence of coordinate and proper relative velocity is ONLY tautological
> in the Galilean transforms. It is merely assumed in the LTs.

You are making up terms (like "proper relative velocity") without defining them, so your intended meaning can only be guessed at, but it was fully explained in the previous message that the reciprocity of the stated coordinate transformation is tautological and self-evident.

[patdolan]

I have extended properness and coordinancy in a direction that should have been accomplished in 1907. This is how science grows and checks itself.

"You are making up terms (like "proper relative velocity") without defining them, so your intended meaning can only be guessed at, but it was fully explained in the previous message that the reciprocity of the stated coordinate transformation is tautological and self-evident."

The above is your most telling statement regarding your prejudicial mindset. Explain to this forum exactly why there are both proper and coordinate durations in time, proper and coordinate extension in space, but no proper nor coordinate relative velocities.

You, Trevor, are tenaciously holding on to the well trodden paths of an old theory; fearing to explore and expand that theory's own principles, or apply those principles reflexively. Instead you continue to regurgitate and rehearse examples that no longer satisfy. C. S. Pierce categorized four methods of belief. Yours at the moment is the method of tenacity. My advice to you, Trevor, is the same advice that Cool Hand Luke's prison guard gave to him on a similar occasion: "You need to get your mind right."

[patdolan]

Trevor Lange I have come up with a better and time honored way to resolve our differences concerning the uniqueness of proper relative velocity v. Prove to this forum that proper velocity v is always equal to coordinate velocity v' in terms of the LTs. Will you accept this challenge?

[Trevor Lange]

On Saturday, October 8, 2022 at 2:32:48 PM UTC-7, patdolan wrote:
> Prove that proper velocity v is always equal to coordinate velocity v' in terms of the LTs.

As I said, you are making up terms (e.g., proper velocity) and nomenclature (e.g., v') without defining them, so your strings of characters have no definite meaning. If you are asking me to explain to you the reciprocity between relatively moving systems of inertial coortdinates, we covered that before.

As a reminder, first consider a simpler case, e.g., a straight line with two sets of markings, called the x coordinates and the x' coordinates, and suppose they are related by x' = x - 3. This relationship can also be written as x = x' + 3. So the x' coordinates are offset from the x coordinates by 3, and the x coordinates are offset from the x' coordinates by -3.

Likewise, consider two systems of coordinates called the x,t coordinates and the x',t' coordinates, and suppose they are related by x' = 2(x - t*sqrt(3)/2) and t' = 2(t - x*sqrt(3)/2). This relationship is can also be written as x = 2(x' + t'*sqrt(3)/2) and t = 2(t' + x'*sqrt(3)/2). Thus the reciprocity between these systems of coordinates is tautological and self-evident.

[patdolan]

definitions:
c = ∆x_0/∆t_0 = ∆x_0'/∆t_0' second postulate
v = ∆x/∆t
v' = ∆x'/∆t'

prove v === v'

[Trevor Lange]

On Saturday, October 8, 2022 at 3:18:02 PM UTC-7, patdolan wrote:
> > > Prove that proper velocity v is always equal to coordinate velocity v' in terms of the LTs.
> >
> > As I said, you are making up terms (e.g., proper velocity) and nomenclature (e.g., v') without defining them, so your strings of characters have no definite meaning. If you are asking me to explain to you the reciprocity between relatively moving systems of inertial coortdinates, we covered that before.
> >
> > As a reminder, first consider a simpler case, e.g., a straight line with two sets of markings, called the x coordinates and the x' coordinates, and suppose they are related by x' = x - 3. This relationship can also be written as x = x' + 3. So the x' coordinates are offset from the x coordinates by 3, and the x coordinates are offset from the x' coordinates by -3.
> >
> > Likewise, consider two systems of coordinates called the x,t coordinates and the x',t' coordinates, and suppose they are related by x' = 2(x - t*sqrt(3)/2) and t' = 2(t - x*sqrt(3)/2). This relationship is can also be written as x = 2(x' + t'*sqrt(3)/2) and t = 2(t' + x'*sqrt(3)/2). Thus the reciprocity between these systems of coordinates is tautological and self-evident.

Ignoring (again) everything that was just explained to him, Pat Dolan wrote:
> v = ∆x/∆t ..... v' = ∆x'/∆t' ..... prove v === v'

That's can't be (validly) proven, because it is self-evidently false. We covered this before: If, in terms of S, Tom moves ∆x = 0.866 LY in the time ∆t = 1 year then in terms of S' he moves ∆x' = 0 in any non-zero ∆t', so you have v=0.866 and v'=0.

This is all contained in the equations given to you above.

[patdolan]

Whether he's Townes Olsen, Stan Fultoni or Trevor Lange, Legion always balks at proving that the v in the LTs is the same when determining t, t', x or x'

[mitchr...@gmail.com]

Look at a motion black hole...
An atoms movement can leave behind light's constant speed.
That light absolute has to catch up to the fast atom.
Light's absolute competing with the fast atom demonstrates
the atomic motion compares to an absolute.
Does that not show atom has absolute speed compared
to light's faster?

Mitchell Raemsch

[Trevor Lange]

On Saturday, October 8, 2022 at 4:55:26 PM UTC-7, patdolan wrote:
> > Ignoring (again) everything that was just explained to him, Pat Dolan wrote:
> > > v = ∆x/∆t ..... v' = ∆x'/∆t' ..... prove v === v'
> >
> > That's can't be (validly) proven, because it is self-evidently false. We covered this before: If, in terms of S, Tom moves ∆x = 0.866 LY in the time ∆t = 1 year then in terms of S' he moves ∆x' = 0 in any non-zero ∆t', so you have v=0.866 and v'=0.
> >

> Trevor balks at proving that the v in the LTs is the same when determining t, t', x or x'

You've asked for proofs of two different propositions, the first being the proposition that relatively moving systems of inertial coordinates are reciprocally related, which is proven by simple grade school algebra, i.e., the relationship x'=(x-vt)g, t'=(t-vx)g where g=1/sqrt(1-v^2) for any constant v has the unique algebraic inverse x=(x'+vt')g, t=(t'+vx')g. Proof complete. The second proposition you asked to be proven was the absurd assertion that velocities are generally invariant, which isn't even true in Galilean relativity, let alone special relativity. In fact, it is insane. This was patiently explained to you (twice), with an explicit example. You're welcome.

[patdolan]

Legion, stop the constant rephrasing my questions and challenges.

[Paul Alsing]

On Saturday, October 8, 2022 at 12:35:05 PM UTC-7, patdolan wrote:

> You are not to be blamed that you are unable to conceive of two observers disagreeing on their relative velocity (of which closing speed is a subspecies).

See? You are wrong right out of the box. It is NEVER the case that 2 observers observing their relative speeds can EVER determine their closing speed. The very definition of closing speed precludes such a situation. It just cannot be done. Get a clue, for once, you are only making a bigger fool of yourself than you usually do.

Read a dang textbook and do try harder.

[Paul B. Andersen]

Den 08.10.2022 21:35, skrev patdolan:
>
> Trevor measures Tom Roberts' relative velocity along Trevor's x-axis as 0.8660243c. In Trevor's universe (the only universe(unless SR is a many-worlds theory)) Tom's co-moving x-axis which is parallel to Trevor's x-axis contains two meter marks for every one meter mark of Trevor's. This is not an illusion just because the two axes are moving wrt each other. Tom's meters really are only half as long as Trevor's meters. Therefore Tom must measure in his universe--the only universe--two meters of displacement for every one meter of displacement that Trevor measures.
>
> Furthermore, Trevor observers that Tom's clock is ticking at a rate only half as fast as Trevor's clock. This is not an illusion just because Tom's and Trevor's clocks are moving wrt each other. Tom's seconds really are twice as long as Trevor's seconds. Therefore Tom must measure in his universe--the only universe--one second for every two seconds that Trevor measures.
>
> When Tom divides his displacement by his interval of time in his universe--the only universe--it is not an illusion or mathematically artifact that Tom's brain, operating at half the biological speed of Trevor's brain, concludes that his velocity towards Trevor is 3.4681689c or gama^2 times the proper velocity that Trevor calculates for Tom.
>

https://paulba.no/pdf/Mutual_time_dilation.pdf

--
Paul

https://paulba.no/

[Tom Roberts]

On 10/8/22 10:47 PM, Paul Alsing wrote:
> It is NEVER the case that 2 observers observing their relative
> speeds can EVER determine their closing speed. The very definition
> of closing speed precludes such a situation. It just cannot be done.

This is just plain false, and I have no idea why you think it is true.
Any (locally) inertial frame can be used to determine the closing speed
between two objects, including either one of their rest frames (as long
as the object is moving inertially).

[Note that the value of the closing speed depends on which
(locally) inertial frame is used.]

Closing speed is the rate of change of the distance between the two
objects, as measured in the specified (locally) inertial frame. If one
uses the rest frame of either object, the closing speed is equal to the
relative speed of the other object.

Tom Roberts

[Paul Alsing]

I do not follow. 2 vehicles are approaching each other along the same axis. Let's say each is moving at .7c relative to my position in my chair on my balcony looking down on that axis. I can measure the rate of change of the distance between them and come up with a closing speed of 1.4c... but neither one of these guys can make any kind of measurement of the other that will ever exceed c... their measurements are limited by the LT... isn't that correct? Isn't that just basic relativity? Wouldn't this also be true even for much lower speeds? Even at, say, 50 mph each, their closing speed would be 100 mph but technically you could still apply the LT and find that from the frame of either car their perceived relative speed would be less than 100 mph by some incredibly small amount? Like 99.99999999 mph or some such? I am only talking about this one specific scenario, 2 objects moving directly towards each other and the measurements made from several different locations.

Please advise.

[Trevor Lange]

On Sunday, October 9, 2022 at 10:42:38 AM UTC-7, Paul Alsing wrote:
> Two vehicles are approaching each other along the same axis. Let's say
> each is moving at .7c relative to my position in my chair...

Speeds are not expressed "relative to positions", they are expressed in terms of specified coordinate systems, because objects have infinitely many different speeds "relative to your position", depending on the choice of coordinate systems, but they have just one speed in terms of (for example) the standard inertial coordinate system in which you are at rest.

> ...neither one of these guys can make any kind of measurement of the other

> that will ever exceed c... their measurements are limited by the LT... isn't that
> correct?

No, not at all. One of the worst aspects of many popular introductory expositions of special relativity is the conflating of the concepts of "observer" and "coordinate system". These books give young readers the impression that each observer is somehow magically linked to some particular system of coordinates, and any "measurement" performed by an observer somehow magically corresponds to the descriptions of events in terms of the standard inertial coordinate system in which that observer is (momentarily) at rest. Almost all the misunderstandings of special relativity -- on display daily in this newsgroup -- are due to this fundamental (and rather silly) misconception.

The actual situation, epistemologically, is much more sophisticated and frankly beyond your grasp or interest, but suffice it to say that a person need not be at rest in a coordinate system in order to make measurements in terms of that coordinate system. In fact, we do this all the time... you driving around in your car, measuring distances and times in terms of the ECI coordinate system (via GPS), and yet you are not at rest in terms of this system. Likewise we can make measurements in terms of accelerating coordinate system (such as systems rotating with the earth), even though those are not standard inertial coordinate systems. Astronomers make measurements of the solar system in terms of the barycentric system of the CoM, and yet the earth is not at rest in that system, and so on.

Moreover, you are only ever at a single location, so even when you naively think you are making measurements in terms of the standard inertial coordinates in which you are at rest, you are really making indirect inferences about other locations where you are not present. So there is no sense in which a person's measurements are locked to any particular system of coordinates. All measurements are indirect, and we can make measurements in terms of any system of coordinates we like. So, take all the books that conflate observers with coordinate systems and throw them in the trash, and start over trying to learn relativity.

[rotchm]

On Sunday, October 9, 2022 at 1:42:38 PM UTC-4, Paul Alsing wrote:

> 2 vehicles are approaching each other along the same axis. Let's say each is moving at .7c relative to my
> position in my chair on my balcony looking down on that axis.

Thus, relative to your (inertial) coordinate system (CS) [it doesn't matter where you are located at in this CS], the speeds of the two "cars"
are 0.7c each, approaching each other. IOW, you [your CS] measures their speeds top be 0.7c.

> I can measure the rate of change of the distance between them and come up with a closing speed of 1.4c...

Correct.

> but neither one of these guys can make any kind of measurement of the other that will ever exceed c...

If one of the cars [also called an "observer", also called a CS; its associated (inertial) CS] measures the speed of the other (incoming) car,
he will get a value less than c.

> their measurements are limited by the LT... isn't that correct?

Yes.

> Wouldn't this also be true even for much lower speeds? Even at, say, 50 mph each, their closing
> speed would be 100 mph but technically you could still apply the LT and find that from the
> frame of either car their perceived relative speed would be less than 100 mph by some
> incredibly small amount? Like 99.99999999 mph or some such?

Yes.

[Michel Marconi]

rotchm wrote:

> On Sunday, October 9, 2022 at 1:42:38 PM UTC-4, Paul Alsing wrote:
> Thus, relative to your (inertial) coordinate system (CS) [it doesn't
> matter where you are located at in this CS], the speeds of the two
> "cars"
> are 0.7c each, approaching each other. IOW, you [your CS] measures their
> speeds top be 0.7c.
>
>> I can measure the rate of change of the distance between them and come
>> up with a closing speed of 1.4c...
>
> Correct.

what's the speed between them so, the time reach you and the time to reach
each other, you fucking stupid troll. You are a fucking crackpot. And that
yacht, is not yours.

[Maciej Wozniak]

On Sunday, 9 October 2022 at 22:12:19 UTC+2, rotchm wrote:
> On Sunday, October 9, 2022 at 1:42:38 PM UTC-4, Paul Alsing wrote:
>
> > 2 vehicles are approaching each other along the same axis. Let's say each is moving at .7c relative to my
> > position in my chair on my balcony looking down on that axis.
> Thus, relative to your (inertial) coordinate system (CS) [it doesn't matter where you are located at in this CS], the speeds of the two "cars"
> are 0.7c each, approaching each other. IOW, you [your CS] measures their speeds top be 0.7c.

> > I can measure the rate of change of the distance between them and come up with a closing speed of 1.4c...
> Correct.

In the meantimein the real world, however, forbidden
by your bunch of idiots GPS and TAI keep measuring
t'=t in forbidden by your bunch of idiots old seconds.

[Athel Cornish-Bowden]

On 2022-10-10 06:01:24 +0000, Maciej Wozniak said:

>
> [ … ]

>
>
> In the meantimein the real world, however, forbidden
> by your bunch of idiots GPS and TAI keep measuring
> t'=t in forbidden by your bunch of idiots old seconds.

507
--
Athel -- French and British, living mainly in England until 1987.

[patdolan]

Paul A#1, you have not squeezed all of the conclusions possible out of your three event, three clock scenario. For instance, at E1 there are d*gamma x-axis length units between clock A and clock B' on the x-axis as viewed from Frame K'. Therefore at E2 clock B' will have counted that d*gamma x-axis length units have passed it by since E1. According to your own calculation clock A reads d/(v*gamma) at E2. From our vantage point in Frame K' we now construct the coordinate relative velocity of clock A as observed by clock B' at E2 by dividing the Dolan coordinate distance d*gamma by the Paul B. Andersen clock A coordinate time t_2:

Conclusion #5 (d*gamma)/(d/(v*gamma)) = v*gamma^2

Which is exactly the Dolan coordinate relative velocity factor for relative velocity v. The symmetry is obvious.

[patdolan]

Not to belabor the point, but coordinate relative velocity exceeds the speed of light for all proper relative velocities greater that .618c.

[Ross A. Finlayson]

Shut up Paddy!

Tell them what I say!

[Ross A. Finlayson]

On Friday, October 7, 2022 at 11:02:15 AM UTC-7, Ross A. Finlayson wrote:
> On Friday, October 7, 2022 at 10:38:34 AM UTC-7, patdolan wrote:
> > On Thursday, October 6, 2022 at 3:26:16 PM UTC-7, tjrob137 wrote:
> > > On 10/1/22 12:58 PM, patdolan wrote:
> > > > On Saturday, October 1, 2022 at 10:02:56 AM UTC-7, tjrob137 wrote:
> > > >> On 9/30/22 10:17 PM, rotchm wrote:
> > > >>>>>> Since you talked about closing speeds, can you explain
> > > >>>>>> what you mean by such an expression?
> > > >>>> ... in this post he says..."Closing speed is the rate of
> > > >>>> change of the distance between two objects as they both move.
> > > >>>
> > > >>> Yes, that's basically it. The closing speed between two entities
> > > >>> is the rate of change of the distance (separation) between them,
> > > >>> as measured by a given frame (system of coordinates).
> > > >> Yes, but only for inertial coordinates. The closing speed for a
> > > >> given pair of objects varies, depending on which inertial frame is
> > > >> used to calculate/measure it.
> > > >>> Its "v1 ± v2" depending of sign conventions.
> > > >> Only when the two objects are moving along a single axis in the
> > > >> inertial frame used. If that isn't so, the calculation is MUCH
> > > >> more complicated.
> > > >>
> > > >> Tom Roberts
> > > > Now, Tom Roberts, provide this forum with the proper and coordinate
> > > > closing speeds, as calulated by observers on each closing object.
> > > Consider inertial frame S with coordinates (x,t) -- ignore y and z as
> > > they don't appear in this problem. Object A is located at x>0 and is
> > > moving inertially along the x axis of S with velocity Va, Va<0. Object B
> > > is located at x<0 and is moving inertially along the x axis of S with
> > > velocity Vb, Vb>0. These two objects are clearly approaching each other.
> > >
> > > In frame S, the closing speed of these two objects is |Va|+|Vb| = Vb-Va.
> > >
> > > (Remember my V's are 3-velocities along the x axis, with
> > > Va<0 and Vb>0.)
> > >
> > > In the rest frame of A, object B has velocity Vba such that
> > > Vb = (Va+Vba)/(1+Va*Vba/c^2)
> > > so
> > > Vba = (Vb-Va)/(1-Va*Vb/c^2)
> > > In the rest frame of A, their closing speed is |Vba| = Vba.
> > >
> > > In the rest frame of B, object A has velocity Vab such that
> > > Vb = (Va+Vab)/(1+Va*Vab/c^2)
> > > so
> > > Vab = (Va-Vb)/(1-Va*Vb/c^2)
> > > In the rest frame of B, their closing speed is |Vab| = -Vab.
> > >
> > > While the expressions are different, |Vab| = |Vba|, and the closing
> > > speeds in the two object's rest frames are the same.
> > >
> > > Note that if Va<<c and Vb<<c, the closing speeds in all three frames are
> > > approximately the same, |Va|+|Vb| = Vb-Va.
> > > > I will state up front that you won't/can't do it.
> > > As usual, you are wrong.
> > >
> > > Tom Roberts
> > Thank you Tom Roberts for deigning to respond to my challenge. So far we have been making our way up a 5.4 - 5.6 route. Time to find some 5.11 stuff. In your pedestrian implementation of the Einstein velocity addition formula, these are the two most interesting lines that you type:
> >
> > 1) "While the expressions are different, |Vab| = |Vba|, and the closing
> > speeds in the two object's rest frames are the same."
> > This result you arrive at is correct of course. But I will demonstrate that it rests on a completely unproved assumption.
> >
> > 2) "Note that if Va<<c and Vb<<c, the closing speeds in all three frames are
> > approximately the same, |Va|+|Vb| = Vb-Va."
> > Would you care to speculate on what the ramifications are of closing speeds that are a substantial fraction of c? Say .867c when gamma = 2?
> It's they saiy "there's an non-linear impulse", which adds to zero,
> then as it resonates, it adds to 1/2.
>
>
> That's where, "in a circle, the wave resonance observes itself",
> "yeah, radial is infinite and non-linear".
>
> So, it adds half up/down what is input.

Yeah, so non-linear input is atomic.

[patdolan]

Ross!...I don't think there's been a cross word between us since I challenged you to a duel 10 or 15 years ago. What gives? Better yet, meet me here tomorrow at sunrise. Mitch is my second, just like last time. Who is yours?

[Ross A. Finlayson]

You think I duck? No, you duck.

Don't forget to duck!

[Ross A. Finlayson]

On Thursday, October 6, 2022 at 6:57:27 PM UTC-7, Ross A. Finlayson wrote:
> On Thursday, September 29, 2022 at 7:44:29 PM UTC-7, Dono. wrote:
> > On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:
> > > dono incited me to ask such a question.
> > >
> > > In the recent thread entitled "addition of velocities", the op
> > > inquired about closing speeds.
> > >
> > > An aggressive debate ensued.
> > >
> > > I (and others) maintained that closing speed (of two things)
> > > is simply "v1 ± v2" [depending of the sign conventions].
> > Well, you are an idiot, Stephane.
> > Time has made this only worse.
> > Closing speed is a convention, used to express the rate at which two objects, starting simultaneously, cover a distance L between them:
> >
> > v1*t+v2*t=L (see, "t" needs to be the SAME)
> > t=L/(v1+v2)
> >
> > By CONVENTION, UNDER the above conditions, u=v1+v2 is called closing speed
> > The concept is very useful when one of the "objects" is a pulse of light, say v1=c. Then the closing speed of covering a distance "L" is c+v. But cranks, of the type of Thomas Heger and Stephane Baune, call this "relative speed of light". This causes infinitely long debates about light speed not being constant. Most of such debates center around Einstein derivation of the Lorentz transforms in "On the Electrodynamics..." but also in the derivation of the phase difference in the Sagnac experiments where cranks like Stephane Baune will claim that light goes around at c+v in one direction and c-v in the opposite direction. Ergo, the Sagnac effect is "proof" that light speed is "variable".
>
>
> Or, ..., that it produces higher order objects as waves, ..., that reflect in eventual disposition in
> real wave collapse, ..., that c+v and c-v are "imaginary v".
>
> It's called resonance these days some guys just got a Nobel prize for it.

Right, Dono. Tell me I'm right!

[Ross A. Finlayson]

On Thursday, October 6, 2022 at 6:55:41 PM UTC-7, Ross A. Finlayson wrote:
> On Thursday, September 29, 2022 at 7:35:14 PM UTC-7, rotchm wrote:
> > dono incited me to ask such a question.
> >
> > In the recent thread entitled "addition of velocities", the op
> > inquired about closing speeds.
> >
> > An aggressive debate ensued.
> >
> > I (and others) maintained that closing speed (of two things)
> > is simply "v1 ± v2" [depending of the sign conventions].

> > Or, the rate of change of the separation (distance) between the two things.
> > [We were discussing trajectories along the x_axis only btw].
> >
> > Dono however, maintains that the above are not the meaning of closing speed as used here (relativity and kinematics in general). And he reasons that:
> >
> > "v1+v2 is the closing speed of two objects that STARTED SIMULTANEOSLY. "
> >
> > " If the two objects don't start simultaneously, one cannot define closing speed."
> >
> > I am wondering as to what is the consensus here concerning the concept of
> > "closing speed".
>
> The objects meet or pass.
>
> Seems logic dictates "space contraction".


Seems logic dictates!

[patdolan]

Duck...huh? So is it pistols or pillows?

[Ross A. Finlayson]

Nope, it's "die".

[Ross A. Finlayson]

"Like a giant lightning bolt from the center of the universe, ...".

"Like a giant lightning bolt, from the center of the universe, ...".

Straight to the head!

[patdolan]

Ross, you are clearly in crisis tonight. Give us your coordinates and we will send assistance. If you are in Seattle, as I suspect, I will minister to you personally.

[Ross A. Finlayson]

NO, no, no, no, no, no, no, Pat.

You, you are in crisis.

[Ross A. Finlayson]

Clearly I have the advantage from having a logical, mathematical, science.

Which I wrote scholarly thus that it paved me.

[Ross A. Finlayson]

Giant lightning bolt from the center of the universe is not in crisis, ....

[patdolan]

My apologies, Ross. How was I to know, with our keyboards in between us, that last night was a Pascalian "Night of Fire" for you. My best wishes and congratulations, as long as your visitation was not of demonic or chemical origins.

But there can be no doubt that Paul B. Andersen and his cohort are in crisis at this moment.

[Paul B. Andersen]

Den 11.10.2022 04:20, skrev patdolan:
> On Sunday, October 9, 2022 at 1:02:05 AM UTC-7, Paul B. Andersen wrote:
>> https://paulba.no/pdf/Mutual_time_dilation.pdf
>>
>> --
>> Paul
>>
>> https://paulba.no/
>
> Paul A#1, you have not squeezed all of the conclusions possible out of your three event, three clock scenario. For instance, at E1 there are d*gamma x-axis length units between clock A and clock B' on the x-axis as viewed from Frame K'.

This is a convoluted way of saying that vied from frame K',
the distance d⋅γ in K is Lorentz contracted to d.

> Therefore at E2 clock B' will have counted that d*gamma x-axis length units have passed it by since E1. According to your own calculation clock A reads d/(v*gamma) at E2. From our vantage point in Frame K' we now construct the coordinate relative velocity of clock A as observed by clock B' at E2 by dividing the Dolan coordinate distance d*gamma by the Paul B. Andersen clock A coordinate time t_2:
>
> Conclusion #5 (d*gamma)/(d/(v*gamma)) = v*gamma^2

Clock B' can't observe the velocity of clock A.
All B' can observe is that clock A shows d/(v⋅γ) when B' shows d/v.
B' can't calculate a speed from that.

But we, knowing that the position of B was x' = -d
at t'= 0 when A was at x' = 0, can then calculate that
the coordinate relative velocity of clock A in K' is v.

--
Paul

https://paulba.no/