addition of velocities

[Mark-T]

There's something not made clear in the texts.

A stationary observer watches a passing rocket, with
a large window. He sees two bullets approaching
each other, fired from front and rear.

He measures each bullet's velocity, relative to himself,
as given by the addition formula. What does he see as
their closing speed?

Replace one bullet by a light beam. Now what does he see
as their closing speed?


Mark

[Dono.]

Hint: in the frame of the rocket the closing speed is (v1+v2). The rocket moves with speed V wrt the observer. Apply the speed composition.
For extra points: what happens when v1 or v2 is equal to "c"?

[Richard Hachel]

Le 25/09/2022 à 02:17, Mark-T a écrit :

> There's something not made clear in the texts.

When you read a book on special relativity, you actually realize that
things ring false.

But not everything rings true.

As for the general addition formula of relativistic velocities, for
example, we have a perfect equation that students should memorize, as one
memorizes remarkable identities, the Pythagorean theorem, or Thales
equalities. .

There is no problem there.

The problems are elsewhere, and they are not so difficult to cure, if one
wants to, one day, show something other than hatred, arrogance,
compunction as soon as the others speak.

I put here this magnificent equation, which you simply have to admire for
what it is, that is to say one of the most beautiful equations in physics.

<http://news2.nemoweb.net/jntp?UPlsOyrKLIZ_E-LPBGxez4H6-YA@jntp/Data.Media:1>

> Mark

R.H.

[Dono.]

On Sunday, September 25, 2022 at 1:03:53 AM UTC-7, Richard Hachel wrote:
> Le 25/09/2022 à 02:17, Mark-T a écrit :
>
> > There's something not made clear in the texts.
> When you read a book on special relativity, you actually realize that

> dr. Hachel is an insane imbecile

Yep

[Richard Hachel]

Le 25/09/2022 à 16:11, "Dono." a écrit :

> insane imbecile

Oh! a terrorist!

R.H.

[Sylvia Else]

Which text does not make this clear?

Sylvia.

[Mark-T]

On September 24, 2022 at 5:24:29 PM UTC-7, Dono. wrote:
>> A stationary observer watches a passing rocket, with
>> a large window. He sees two bullets approaching
>> each other, fired from front and rear.
>> He measures each bullet's velocity, relative to himself,
>> as given by the addition formula. What does he see as
>> their closing speed?
>> Replace one bullet by a light beam. Now what does he see
>> as their closing speed?
>

> Hint: in the frame of the rocket the closing speed is (v1+v2). The rocket moves with speed V wrt the observer.
> Apply the speed composition.

The transformed velocities are:
(v1)' = V + v1/(1 + (V(v1)/c²)))
(v2)' = V + v2/(1 + (V(v2)/c²)))

So the observer sees the closing speed:
(v1)' + (v2)'

I'm not sure, but a good guess gets full credit on the exam.

> For extra points: what happens when v1 or v2 is equal to "c"?

That's what I want to know.

Mark

[rotchm]

On Monday, September 26, 2022 at 1:29:49 PM UTC-4, Mark-T wrote:

> (v1)' = V + v1/(1 + (V(v1)/c²)))
> (v2)' = V + v2/(1 + (V(v2)/c²)))
>
> So the observer sees the closing speed:
> (v1)' + (v2)'

> > For extra points: what happens when v1 or v2 is equal to "c"?
> That's what I want to know.

Replace v1 v1 by c.
What do you get then for v1' ?

[Dono.]

On Monday, September 26, 2022 at 10:29:49 AM UTC-7, Mark-T wrote:
> On September 24, 2022 at 5:24:29 PM UTC-7, Dono. wrote:
> >> A stationary observer watches a passing rocket, with
> >> a large window. He sees two bullets approaching
> >> each other, fired from front and rear.
> >> He measures each bullet's velocity, relative to himself,
> >> as given by the addition formula. What does he see as
> >> their closing speed?
> >> Replace one bullet by a light beam. Now what does he see
> >> as their closing speed?
> >
> > Hint: in the frame of the rocket the closing speed is (v1+v2). The rocket moves with speed V wrt the observer.
> > Apply the speed composition.
> The transformed velocities are:
> (v1)' = V + v1/(1 + (V(v1)/c²)))
> (v2)' = V + v2/(1 + (V(v2)/c²)))
>
> So the observer sees the closing speed:
> (v1)' + (v2)'
>

No, he doesn't, this is a rookie mistake. Your mistakeS are:
The two objects do not start simultaneously in the observer frame, you have failed to account for that.
One object "chases" the rocket, the other "runs against" the rocket, you failed to account for that.

> I'm not sure, but a good guess gets full credit on the exam.
> > For extra points: what happens when v1 or v2 is equal to "c"?
> That's what I want to know.
>

You need to do the calculations the way I showed you. ONLY in the frame of the rocket is the closing speed the sum of the two speeds. Why is that?

[Dono.]

You can't do that, Stephane, you can't use v1'. You still haven't learned relativity.

[Mark-T]

On September 26, 2022 at 11:40:56 AM UTC-7, Dono. wrote:
>> >> A stationary observer watches a passing rocket, with
>> >> a large window. He sees two bullets approaching
>> >> each other, fired from front and rear.
>> >> He measures each bullet's velocity, relative to himself,
>> >> as given by the addition formula. What does he see as
>> >> their closing speed?
>> >> Replace one bullet by a light beam. Now what does he see
>> >> as their closing speed?
>
>> > Hint: in the frame of the rocket the closing speed is (v1+v2). The rocket moves with speed V wrt the observer.
>> > Apply the speed composition.
>
>> The transformed velocities are:

>> (v1)' = (V + v1)/(1 + (V(v1)/c²))
>> (v2)' = (V + v2)/(1 + (V(v2)/c²))

>> So the observer sees the closing speed:
>> (v1)' + (v2)'
>
> No, he doesn't, this is a rookie mistake. Your mistakeS are:
> The two objects do not start simultaneously in the observer frame, you have failed
> to account for that.

What difference does that make?

> One object "chases" the rocket, the other "runs against" the rocket, you
> failed to account for that.

It's true, from the observer's viewpoint, the rear bullet, moving
with the ship, looks faster in the +x direction.

However, both bullets are carried equally in the rocket. When we
consider only their velocity relative to each other, that fact cancels out.

So what's their relative velocity, i.e. the closing speed?

>>> For extra points: what happens when v1 or v2 is equal to "c"?
>
>> That's what I want to know.
>
> You need to do the calculations the way I showed you.

You didn't show anything.

> ONLY in the frame of the rocket is the closing speed the sum of the two speeds. Why is that?

Dunno

Mark

[Dono.]

On Monday, September 26, 2022 at 3:00:48 PM UTC-7, Mark-T wrote:
> On September 26, 2022 at 11:40:56 AM UTC-7, Dono. wrote:
> >> >> A stationary observer watches a passing rocket, with
> >> >> a large window. He sees two bullets approaching
> >> >> each other, fired from front and rear.
> >> >> He measures each bullet's velocity, relative to himself,
> >> >> as given by the addition formula. What does he see as
> >> >> their closing speed?
> >> >> Replace one bullet by a light beam. Now what does he see
> >> >> as their closing speed?
> >
> >> > Hint: in the frame of the rocket the closing speed is (v1+v2). The rocket moves with speed V wrt the observer.
> >> > Apply the speed composition.
> >
> >> The transformed velocities are:
> >> (v1)' = (V + v1)/(1 + (V(v1)/c²))
> >> (v2)' = (V + v2)/(1 + (V(v2)/c²))
> >> So the observer sees the closing speed:
> >> (v1)' + (v2)'
> >
> > No, he doesn't, this is a rookie mistake. Your mistakeS are:
> > The two objects do not start simultaneously in the observer frame, you have failed
> > to account for that.
> What difference does that make?

It makes a huge difference since in the frame of the observer the two objects travel DIFFERENT amounts of time.
In the frame of the rocket, the two objects travel the SAME amount of time, making the calculations really easy:

v_1*t+v_2*t=L_rocket
t=L_rocket/(v_1+v_2)
Hence, the closing speed , in the frame of the rocket is the trivial v=L_rocket/t=v_1+v_2
In the frame of the observer, the calculations are much more complicated.

> > One object "chases" the rocket, the other "runs against" the rocket, you
> > failed to account for that.
> It's true, from the observer's viewpoint, the rear bullet, moving
> with the ship, looks faster in the +x direction.
>
> However, both bullets are carried equally in the rocket. When we
> consider only their velocity relative to each other, that fact cancels out.
>
> So what's their relative velocity, i.e. the closing speed?
> >>> For extra points: what happens when v1 or v2 is equal to "c"?
> >
> >> That's what I want to know.
> >
> > You need to do the calculations the way I showed you.
> You didn't show anything.

You only needed to follow the hint, I wasn't going to do your homework for you.

> > ONLY in the frame of the rocket is the closing speed the sum of the two speeds. Why is that?
> Dunno
>

Because speed is frame variant, so you cannot expect that the closing speed is v'_1+v'_2

[Stan Fultoni]

On Monday, September 26, 2022 at 10:29:49 AM UTC-7, Mark-T wrote:

> A rocket is moving at speed V in the positive x direction of an inertial
> coordinate system S, and inside the rocket there are two bullets
> fired from rear and front, with velocities v1' and v2' respectively in
> terms of the inertial coordinate system S' in which the rocket is at rest.
> Note that v1' is positive (moving in the positive x' direction) and v2' is
> negative (moving in the negative x' direction).
>
> What is the closing speed of the bullets in terms of S', and what
> is the closing speed in terms of S?

Using units so c=1, the closing speed in terms of S' (in which the rocket is at rest) is simply v1' - v2'. The closing speed in terms of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].

> Replace one bullet by a light pulse. Now what is the closing speed
> of the bullet and pulse in terms of S?

Trivially, if the foward bullet is replaced with a light pulse, the closing speed is (1 - v2')(1-V)/(1+Vv2') = 1 - (V+v2')/(1+Vv2'), and if the rearward pulse is replaced by the pulse, the closing speed is (v1' - 1)(1-V)/(1+Vv1') = (V+v1')/(1+Vv1') - 1.

[rotchm]

On Monday, September 26, 2022 at 2:42:37 PM UTC-4, Dono. wrote:
> On Monday, September 26, 2022 at 10:46:58 AM UTC-7, rotchm wrote:
> > On Monday, September 26, 2022 at 1:29:49 PM UTC-4, Mark-T wrote:
> >
> > > (v1)' = V + v1/(1 + (V(v1)/c²)))

> > Replace v1 v1 by c.
> > What do you get then for v1' ?
> You can't do that, Stephane, you can't use v1'. You still haven't learned relativity.

I did not read his op/scenario, so I have no idea what his v's represent , but nonetheless,
in the relativistic speed composition formula, you can typically replace one of the speeds by the SoL c, even by
speeds greater than c. This is what I was/am saying.

[rotchm]

On Saturday, September 24, 2022 at 8:17:16 PM UTC-4, Mark-T wrote:
> There's something not made clear in the texts.

Nor in your question. See below.

> A stationary observer watches a passing rocket, with
> a large window.

The observer or rocket has the window?

> He sees two bullets approaching
> each other, fired from front and rear.

Front and rear of the rocket? [I will assume so].

> He measures each bullet's velocity, relative to himself,
> as given by the addition formula.

? He does not measure "as given by the addition formula". He measures with his own instruments, "grid of rulers & clocks".

> What does he see as
> their closing speed?

"see" or "measures" ?
You are using ambiguous terminology, the same ambiguous terminology that plagues many SR documentation.
You meant "measure".

Assuming your scenario is:

In the typical config, relative to inertial me [frame S], an inertial rocket [frame S'] of velocity u, fires two bullets.
I measure the velocities of these bullets [wrt me] as v1 & v2 resp.

Is this your scenario?

[Dono.]

On Monday, September 26, 2022 at 7:00:43 PM UTC-7, rotchm wrote:
> On Monday, September 26, 2022 at 2:42:37 PM UTC-4, Dono. wrote:
> > On Monday, September 26, 2022 at 10:46:58 AM UTC-7, rotchm wrote:
> > > On Monday, September 26, 2022 at 1:29:49 PM UTC-4, Mark-T wrote:
> > >
> > > > (v1)' = V + v1/(1 + (V(v1)/c²)))
> > > Replace v1 v1 by c.
> > > What do you get then for v1' ?
> > You can't do that, Stephane, you can't use v1'. You still haven't learned relativity.
> I did not read his op/scenario, so I have no idea what his v's represent ,

So, you had no idea what you were doing. Obvious.

[Dono.]

No, it isn't.
Besides, the problem has been answered , by both me and Stan.

[rotchm]

On Monday, September 26, 2022 at 10:42:44 PM UTC-4, Dono. wrote:
> On Monday, September 26, 2022 at 7:11:32 PM UTC-7, rotchm wrote:


> Besides, the problem has been answered , by both me and Stan.

You did provide an answer, but was/is wrong:

You initially said: "in the frame of the rocket the closing speed is (v1+v2)."
But there are no v1 & v2 in the op.
Basically just said "it's closing speed is its closing speed". That does not help the op.

Then you said "The two objects do not start simultaneously in the observer frame, you have failed to account for that.".
But as the op noticed, that has nothing to do with the problem; it is not needed.

But I see where your posts are going...

I will let the op converse with me....

[Dono.]

On Monday, September 26, 2022 at 8:30:27 PM UTC-7, rotchm wrote:
> On Monday, September 26, 2022 at 10:42:44 PM UTC-4, Dono. wrote:
> > On Monday, September 26, 2022 at 7:11:32 PM UTC-7, rotchm wrote:
>
>
> > Besides, the problem has been answered , by both me and Stan.
> You did provide an answer, but was/is wrong:
>
> You initially said: "in the frame of the rocket the closing speed is (v1+v2)."
> But there are no v1 & v2 in the op.

I defined them as the speeds of the two bullets wrt the rocket. You are still the same old fart idiot, Stephane.

> Basically just said "it's closing speed is its closing speed". That does not help the op.
>
> Then you said "The two objects do not start simultaneously in the observer frame, you have failed to account for that.".
> But as the op noticed, that has nothing to do with the problem; it is not needed.
>

Actually, if one tries to solve the problem like the op, IT IS. You are the same old fart idiot, Stephane.

> But I see where your posts are going...
>
> I will let the op converse with me....

The solution has already been provided, Stephane. But, keep up the entertainment, dig yourself deeper.

[Dono.]

The above can't be right, the closing speed in frame S is not the difference of the Lorentz transforms of the speeds in frame S'. It is the Lorentz transform of the closing speed in frame S'. The reason is that closing speed calculation as a sum/difference presupposes that the two objects start simultaneously. This is true in frame S' but not true in frame S.

[rotchm]

On Monday, September 26, 2022 at 9:35:34 PM UTC-4, Stan Fultoni wrote:
> On Monday, September 26, 2022 at 10:29:49 AM UTC-7, Mark-T wrote:
> > A rocket is moving at speed V in the positive x direction of an inertial
> > coordinate system S, and inside the rocket there are two bullets
> > fired from rear and front, with velocities v1' and v2' respectively in
> > terms of the inertial coordinate system S' in which the rocket is at rest.
> > Note that v1' is positive (moving in the positive x' direction) and v2' is
> > negative (moving in the negative x' direction).

For some reason, I do not find this post of Mark-T.
But this version is well described; its not ambiguous as the op.
[I assume its Stan's interpretation of the op].

> > What is the closing speed of the bullets in terms of S', and what
> > is the closing speed in terms of S?
>
> Using units so c=1, the closing speed in terms of S' (in which the rocket is at rest) is simply v1' - v2'.

I agree.

The closing speed in S is v1 - v2.
With v1 = (V+v1')/(1+Vv1') and v2 = (V+v2')/(1+Vv2') we get the closing speed in S ...

> The closing speed in terms of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].

Agreed.

Mark: Recall that for you, your v1' is positive and your v2' is negative. Use their correct signs in the above.

> > Replace one bullet by a light pulse. Now what is the closing speed
> > of the bullet and pulse in terms of S?
>
> Trivially, if the foward bullet is replaced with a light pulse, the closing speed is
> (1 - v2')(1-V)/(1+Vv2') = 1 - (V+v2')/(1+Vv2'),

Agreed. using v1' = +1 ....

> and if the rearward pulse is replaced by the pulse,

> the closing speed is (v1' - 1)(1-V)/(1+Vv1') =...

Disagree. You jumped a step...too fast [pun intended].

Using v2' = -1 ...
Thus, we get that the required closing speed is (v1' + 1)(1+V)/(1+Vv1').
[Indeed, just consider the case here where v1' = +1. The closing speed should be "2", as in my result above].



[Mark: note that 'closing speed' has a few different definitions or applicable formulas. And "speed" typically references the magnitude (but not applicable to 'closing speeds'). Speed & velocity are not the same thing. And 'closing speed' may be defined as the rate of change
of the distance between them; thus can be negative (if approaching) or positive (receding). All this can greatly confuse the required signs...be very careful, know your context and used words!!]

[rotchm]

On Monday, September 26, 2022 at 11:59:34 PM UTC-4, Dono. wrote:
> On Monday, September 26, 2022 at 8:30:27 PM UTC-7, rotchm wrote:

> > You initially said: "in the frame of the rocket the closing speed is (v1+v2)."
> > But there are no v1 & v2 in the op.
> I defined them as the speeds of the two bullets wrt the rocket.

That was not specified (but one could assume so).

> > Then you said "The two objects do not start simultaneously in the observer frame, you have failed to account for that.".
> > But as the op noticed, that has nothing to do with the problem; it is not needed.
> >
> Actually, if one tries to solve the problem like the op, IT IS.

No it is not. The op is clearly interested in discussing it via the composition of speeds only.

> The solution has already been provided,

Not by you.

By Stan, yes [although with an acceptable error/oversight].
And by me with clarifications...

Where do you fit in all this again?

[rotchm]

On Tuesday, September 27, 2022 at 12:18:26 AM UTC-4, Dono. wrote:
> On Monday, September 26, 2022 at 6:35:34 PM UTC-7, Stan Fultoni wrote:

> > ...

> The above can't be right, the closing speed in frame S is not the difference of the Lorentz transforms
> of the speeds in frame S'.

You perhaps want to think that through again...

> It is the Lorentz transform of the closing speed in frame S'.

Oh dear....here we go again...
Stan, I leave him with you...

> The reason is that closing speed calculation as a sum/difference presupposes that the two objects
> start simultaneously. This is true in frame S' but not true in frame S.

Just for posterity ...

[Dono.]

Actually, Stan replicated the errors made by Mark-T (a very rare case).
You, on the other hand, being the total idiot when it comes to relativity, swallowed the wrong solution by Stan/Mark-T line, hook and sinker. Keep up the entertainment, Stephane.

[Dono.]

Basically, it fits under the fact that you do not understand the basic notion of closing speed and how it is defined. Keep up the entertainment, Stephane.

[Dono.]

On Monday, September 26, 2022 at 9:50:58 PM UTC-7, rotchm wrote:
> On Monday, September 26, 2022 at 9:35:34 PM UTC-4, Stan Fultoni wrote:
> > On Monday, September 26, 2022 at 10:29:49 AM UTC-7, Mark-T wrote:
> > > A rocket is moving at speed V in the positive x direction of an inertial
> > > coordinate system S, and inside the rocket there are two bullets
> > > fired from rear and front, with velocities v1' and v2' respectively in
> > > terms of the inertial coordinate system S' in which the rocket is at rest.
> > > Note that v1' is positive (moving in the positive x' direction) and v2' is
> > > negative (moving in the negative x' direction).
> For some reason, I do not find this post of Mark-T.
> But this version is well described; its not ambiguous as the op.
> [I assume its Stan's interpretation of the op].
> > > What is the closing speed of the bullets in terms of S', and what
> > > is the closing speed in terms of S?
> >
> > Using units so c=1, the closing speed in terms of S' (in which the rocket is at rest) is simply v1' - v2'.
> I agree.
>
> The closing speed in S is v1 - v2.
> With v1 = (V+v1')/(1+Vv1') and v2 = (V+v2')/(1+Vv2') we get the closing speed in S ...
> > The closing speed in terms of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].
> Agreed.

Nope, you are repeating Mark-T rookie mistakes. Turns out that you have no clue what closing speed is. Keep up the entertainment , Stephane. Keep puffing yourself up.

[Stan Fultoni]

On Monday, September 26, 2022 at 9:50:58 PM UTC-7, rotchm wrote:

> > Note that v1' is positive (moving in the positive x' direction) and v2' is

> > negative (moving in the negative x' direction). The closing speed in terms

> > of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].
>
> Agreed.
>

> > Trivially, if the foward bullet is replaced with a light pulse, the closing speed is
> > (1 - v2')(1-V)/(1+Vv2') = 1 - (V+v2')/(1+Vv2'),
>
> Agreed. using v1' = +1 ....
>
> > and if the rearward pulse is replaced by the pulse,
> > the closing speed is (v1' - 1)(1-V)/(1+Vv1') =...
>

> Disagree. You jumped a step... Using v2' = -1 ...

> Thus, we get that the required closing speed is (v1' + 1)(1+V)/(1+Vv1').

Right, v2' is -1, not +1, so the closing speed is trivially (1+v1')(1+V)/(1+Vv1') = 1 + (V+v1')/(1+Vv1').

Dono wrote:
> The above can't be right, the closing speed in frame S is not the difference

> of the Lorentz transform[ation]s of the speeds in frame S'. It is the Lorentz

> transform of the closing speed in frame S'.

By definition the "closing speed" between two particles in terms of inertial coordinate system S is the difference v1 - v2 between the speeds v1 and v2 in terms of S of those particles, just as the closing speed in terms of S' is v1' - v2'.

> The reason is that closing speed calculation as a sum/difference
> presupposes that the two objects start simultaneously.

We stipulate that the particles both been fired, i.e., in terms of S we consider a time t at which both bullets are in flight with their specified speeds. Granted, in terms of S the guns were not fired simultaneously, so there is a period of time (values of t) when the forward bullet is in flight and the rearward hasn't been fired yet, but the question is asking for the closing speed when both bullets are in flight, and this doesn't depend on when they were fired.

[Dono.]

On Tuesday, September 27, 2022 at 6:44:43 AM UTC-7, Stan Fultoni wrote:

> Dono wrote:
> > The above can't be right, the closing speed in frame S is not the difference

> > of the Lorentz transforms of the speeds in frame S'. It is the Lorentz

> > transform of the closing speed in frame S'.
> By definition the "closing speed" between two particles in terms of inertial coordinate system S is the difference v1 - v2 between the speeds v1 and v2 in terms of S of those particles, just as the closing speed in terms of S' is v1' - v2'.

No, it isn't "by definition", it is "by derivation from base principles". I explained that in a previous post.
It goes like this:

Two particles A and B are fired SIMULTANEOUSLY towards each other with respective speeds v_A, v_B. The distance between the two particles is L. Then:

v_A*t+v_B*t=L
t=L/(v_A+v_B)

Therefore we can introduce (in THIS PARTICULAR situation) the notion of closing speed :
u=v_A+v_B

If the particles are fired at different times, then things get more complicated:

v_A*t_A+v_B*t_B=L


As an aside, you keep writing the "separation speed" v1-v2. "Closing speed" is v1+v2.

[Mark-T]

On September 26, 2022 at 6:35:34 PM UTC-7, Stan Fultoni wrote:
>> A rocket is moving at speed V in the positive x direction of an inertial
>> coordinate system S, and inside the rocket there are two bullets
>> fired from rear and front, with velocities v1' and v2' respectively in
>> terms of the inertial coordinate system S' in which the rocket is at rest.
>> Note that v1' is positive (moving in the positive x' direction) and v2' is
>> negative (moving in the negative x' direction).

A more precise edit of my somewhat loose note.

>> What is the closing speed of the bullets in terms of S', and what
>> is the closing speed in terms of S?
>
> Using units so c=1, the closing speed in terms of S' (in which the rocket is at rest)
> is simply v1' - v2'. The closing speed in terms of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].

So you transform each velocity, then simply add them?
That's what I figured, but it isn't so obvious.

>> Replace one bullet by a light pulse. Now what is the closing speed
>> of the bullet and pulse in terms of S?
>
> Trivially, if the foward bullet is replaced with a light pulse, the closing speed is
> (1 - v2')(1-V)/(1+Vv2') = 1 - (V+v2')/(1+Vv2'), and if the rearward pulse is replaced
> by the pulse, the closing speed is (v1' - 1)(1-V)/(1+Vv1') = (V+v1')/(1+Vv1') - 1.

So v1' becomes 1 (= c), treated the same as the bullet?
Not so trivial, since light is special in Einstein relativity.

Mark

[Mark-T]

On September 26, 2022 at 7:11:32 PM UTC-7, rotchm wrote:
>> A stationary observer watches a passing rocket, with a large window.
>
> The observer or rocket has the window?

The rocket has a window.

>> He sees two bullets approaching
>> each other, fired from front and rear.

> Front and rear of the rocket?

Right.

>> He measures each bullet's velocity, relative to himself,
>> as given by the addition formula.
>
> ? He does not measure "as given by the addition formula". He measures with his
> own instruments, "grid of rulers & clocks".


>> What does he see as their closing speed?
>
> "see" or "measures" ?
> You are using ambiguous terminology, the same ambiguous terminology that
> plagues many SR documentation. You meant "measure".

I'll bite. What's the difference between see and measure?

> Assuming your scenario is:
> In the typical config, relative to inertial me [frame S], an inertial rocket [frame S']
> of velocity u, fires two bullets.
> I measure the velocities of these bullets [wrt me] as v1 & v2 resp.
> Is this your scenario?

Using Dono's notation, the rocket is traveling at V, and the velocities
within the rocket are v1 and v2.


Mark

[rotchm]

On Tuesday, September 27, 2022 at 2:04:34 PM UTC-4, Mark-T wrote:
> On September 26, 2022 at 6:35:34 PM UTC-7, Stan Fultoni wrote:

> > Using units so c=1, the closing speed in terms of S' (in which the rocket is at rest)
> > is simply v1' - v2'. The closing speed in terms of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].
> So you transform each velocity, then simply add them?

Yes add or subtract, (depending on your precice meaning/signs)

> That's what I figured, but it isn't so obvious.

Closing speed is v1-v2. In another frame S' its v1' - v2'. In frame S'' its v1'' - v2'' etc.
Its always the same formula/concept here.

> >> Replace one bullet by a light pulse. Now what is the closing speed
> >> of the bullet and pulse in terms of S?

> So v1' becomes 1 (= c), treated the same as the bullet?

Yes. The speeds in the LT's or speed compo of speeds formula, you can put any speed in there; even speeds greater than c (
(these would be "propagation" of non-causal effects, effects rutinely used in physics). For some combos of speeds, the denimitator may be zero, in which case you need to correctly understand and apply correctly the formulas)

> Not so trivial, since light is special in Einstein relativity.

Not that special. Its a limiting speed of matter, objects and information.

[Stan Fultoni]

On Tuesday, September 27, 2022 at 8:04:57 AM UTC-7, Dono. wrote:
> As an aside, you keep writing the "separation speed" v1-v2. "Closing speed" is v1+v2.

That's just a sign convention. I explicitly specified that the velocities are defined as positive in the positive x' direction, so in this example the value of v2' is negative. In other words, if v1 is 1000 ft/sec, and v2 is -1000 ft/sec, the closing speed in terms of these coordinates is v1 - v2 = 2000 ft/sec.

> > By definition the "closing speed" between two particles in terms of inertial coordinate system S is the difference v1 - v2 between the speeds v1 and v2 in terms of S of those particles, just as the closing speed in terms of S' is v1' - v2'.
>

> No, it isn't "by definition"...

The definition of closing speed in terms of S for objects with velocities v1 and v2 in terms of S is just v1 - v2, noting the sign conventions.

> Two particles A and B are fired SIMULTANEOUSLY towards each other with respective speeds v_A, v_B. The distance between the two particles is L.

What matters is not the distance L=x1-x2 but the rate of change of distance, dL/dt = dx1/dt - dx2/dt = v1 - v2. The question stipulates that it wants the closing speed when the bullets are both in flight and approaching each other, so it is unambiguous.

[rotchm]

On Tuesday, September 27, 2022 at 2:17:06 PM UTC-4, Mark-T wrote:
> On September 26, 2022 at 7:11:32 PM UTC-7, rotchm wrote:


> > "see" or "measures" ?
> > You are using ambiguous terminology, the same ambiguous terminology that
> > plagues many SR documentation. You meant "measure".
> I'll bite. What's the difference between see and measure?

> Using Dono's notation,

EEccch... You will soon notice that one should stay away from dono's discussions.
I adopted Stan's notaion. See my followups therein.

> I'll bite. What's the difference between see and measure?

Often, "see" has the conotation of "photographable" scene, where light flight time is important; visual effcts.

See:

https://en.wikipedia.org/wiki/Terrell_rotation
https://arxiv.org/ftp/physics/papers/0106/0106049.pdf

For instance, in the above article, although the length of the (moving) ruler is measured to be
0.5 units, say, its picture (photo, eyes) would indicate 1.1 (or some other values depending on the speed, where its endpoints are etc).
Best to avoid the word "see" in relativity. Try to adopt the word "measures".

See the references in the above links too, and Google Terrell effect etc.
Good reading!

[Mark-T]

On September 25, 2022 at 6:16:23 PM UTC-7, Sylvia Else wrote:
>> There's something not made clear in the texts.

>> A stationary observer watches a passing rocket, with

>> a large window. He sees two bullets approaching

>> each other, fired from front and rear.

>> He measures each bullet's velocity, relative to himself,

>> as given by the addition formula. What does he see as
>> their closing speed?
>
> Which text does not make this clear?


http://www.modelmechanics.org

Mark

[Stan Fultoni]

On Tuesday, September 27, 2022 at 11:04:34 AM UTC-7, Mark-T wrote:
> > Using units so c=1, the closing speed in terms of S' (in which the rocket is at rest)
> > is simply v1' - v2'. The closing speed in terms of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].
>
> So you transform each velocity, then simply add them?

Well, with the stated sign convention, you subtract them.

> > Trivially, if the foward bullet is replaced with a light pulse, the closing speed is
> > (1 - v2')(1-V)/(1+Vv2') = 1 - (V+v2')/(1+Vv2'), and if the rearward pulse is replaced

> > by the pulse, the closing speed is trivially (1+v1')(1+V)/(1+Vv1') = 1 + (V+v1')/(1+Vv1').

>
> So v1' becomes 1 (= c), treated the same as the bullet?

Yes.

> Not so trivial, since light is special in Einstein relativity.

Light follows the same kinematics as material particles in special relativity. There is nothing special or exceptional about light in that sense. The velocity composition formula applies to everything. It's a common newbie misconception to think that relativistic effects apply only to light and not to material objects.

[Dono.]

On Tuesday, September 27, 2022 at 11:31:44 AM UTC-7, Stan Fultoni wrote:
> On Tuesday, September 27, 2022 at 8:04:57 AM UTC-7, Dono. wrote:
> > As an aside, you keep writing the "separation speed" v1-v2. "Closing speed" is v1+v2.
> That's just a sign convention. I explicitly specified that the velocities are defined as positive in the positive x' direction, so in this example the value of v2' is negative. In other words, if v1 is 1000 ft/sec, and v2 is -1000 ft/sec, the closing speed in terms of these coordinates is v1 - v2 = 2000 ft/sec.
> > > By definition the "closing speed" between two particles in terms of inertial coordinate system S is the difference v1 - v2 between the speeds v1 and v2 in terms of S of those particles, just as the closing speed in terms of S' is v1' - v2'.
> >
> > No, it isn't "by definition"...
>
> The definition of closing speed in terms of S for objects with velocities v1 and v2 in terms of S is just v1 - v2, noting the sign conventions.

We are not communicating, I explained clearly that "closing speed" is not a convention, it is derived. I do not see any point continuing.

> > Two particles A and B are fired SIMULTANEOUSLY towards each other with respective speeds v_A, v_B. The distance between the two particles is L.
> What matters is not the distance L=x1-x2 but the rate of change of distance, dL/dt = dx1/dt - dx2/dt = v1 - v2. The question stipulates that it wants the closing speed when the bullets are both in flight and approaching each other, so it is unambiguous.

There are many examples of how closing speed is derived, you can find a lot of them in the analysis of the Sagnac experiment as well as in the famous Einstein paper "On the Electrodynamics....". None of them uses the convention dL/dt = dx1/dt - dx2/dt.

[Dono.]

This is written by the resident crank , Ken Seto,
You can safely ignore "rotchm" as well, he is a reformed crank (but not totally) pretending to be an "instructor".

[Mark-T]

On September 27, 2022 at 6:44:43 AM UTC-7, Stan Fultoni wrote:
>> > Note that v1' is positive (moving in the positive x' direction) and v2' is
>> > negative (moving in the negative x' direction). The closing speed in terms
>> > of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].

>> > Trivially, if the foward bullet is replaced with a light pulse, the closing speed is
>> > (1 - v2')(1-V)/(1+Vv2') = 1 - (V+v2')/(1+Vv2'),
>

>> The above can't be right, the closing speed in frame S is not the difference
>> of the Lorentz transform[ation]s of the speeds in frame S'. It is the Lorentz
>> transform of the closing speed in frame S'.
>> By definition the "closing speed" between two particles in terms of inertial
> coordinate system S is the difference v1 - v2 between the speeds v1 and v2
> in terms of S of those particles, just as the closing speed in terms of S' is v1' - v2'.

Another thought.... replace the space between the bullets
by a shrinking rod... length contraction, hmmm....

So, is relativistic length contraction germane to this question
of closing speed?


Mark

[rotchm]

On Tuesday, September 27, 2022 at 2:36:13 PM UTC-4, rotchm wrote:
> On Tuesday, September 27, 2022 at 2:17:06 PM UTC-4, Mark-T wrote:

> > I'll bite. What's the difference between see and measure?
> Often, "see" has the conotation of "photographable" scene, where light flight time is important; visual effcts.
>
> See:
>
> https://en.wikipedia.org/wiki/Terrell_rotation
> https://arxiv.org/ftp/physics/papers/0106/0106049.pdf

And btw, these effects are not restricted to SR. These effects also occur in plain ol' highscool physics.
If you consider the time of flight of the light (delays), we would see objects distorted. We dont notice this since the speeds involved of the moving objets are way too small. And our eyes,brains,cameras have too slow timing systems...

[Dono.]

If you insist on doing the calculations in the observer frame, yes. Actually things get very messy in that frame.On the other hand, if you calculate the closing speed in the rocket frame and you transform it in the observer frame, things are very clean.

[Dono.]

On Tuesday, September 27, 2022 at 11:53:42 AM UTC-7, rotchm wrote:

> > https://arxiv.org/ftp/physics/papers/0106/0106049.pdf
Eric Baird is a hard core crank. But, then again, so are you.

[Dono.]

On Tuesday, September 27, 2022 at 11:31:44 AM UTC-7, Stan Fultoni wrote:

> What matters is not the distance L=x1-x2 but the rate of change of distance, dL/dt = dx1/dt - dx2/dt = v1 - v2. The question stipulates that it wants the closing speed when the bullets are both in flight and approaching each other, so it is unambiguous.

Ok,

I see your problem , you are confusing "closing speed" with "relative velocity". What you are defining is "relative velocity", not "closing speed". In general , I agree with your posts but in tis instance, I disagree with your reasoning.

[Stan Fultoni]

On Tuesday, September 27, 2022 at 12:22:35 PM UTC-7, Dono. wrote:
> > What matters is not the distance L=x1-x2 but the rate of change of distance, dL/dt = dx1/dt - dx2/dt = v1 - v2. The question stipulates that it wants the closing speed when the bullets are both in flight and approaching each other, so it is unambiguous.
>
> Ok, I see your problem , you are confusing "closing speed" with "relative velocity". What you are defining is "relative velocity", not "closing speed". In general , I agree with your posts but in tis instance, I disagree with your reasoning.

In the relativity literature, the expression "relative velocity" of two objects is defined as the velocity of one object in terms of the inertial coordinate system in which the other object is at rest. The expression "closing speed" in terms of S is defined as the difference between their velocities in terms of S. These are just the definitions of those expressions.

[Dono.]

On Tuesday, September 27, 2022 at 12:36:21 PM UTC-7, Stan Fultoni wrote:
> On Tuesday, September 27, 2022 at 12:22:35 PM UTC-7, Dono. wrote:
> > > What matters is not the distance L=x1-x2 but the rate of change of distance, dL/dt = dx1/dt - dx2/dt = v1 - v2. The question stipulates that it wants the closing speed when the bullets are both in flight and approaching each other, so it is unambiguous.
> >
> > Ok, I see your problem , you are confusing "closing speed" with "relative velocity". What you are defining is "relative velocity", not "closing speed". In general , I agree with your posts but in tis instance, I disagree with your reasoning.
> In the relativity literature, the expression "relative velocity" of two objects is defined as the velocity of one object in terms of the inertial coordinate system in which the other object is at rest.

Agreed.

The expression "closing speed" in terms of S is defined as the difference between their velocities in terms of S. These are just the definitions of those expressions.

Disagree. If that were true, you wouldn't have any difficulty in providing the theory of the Sagnac experiment using your definition dL/dt = dx1/dt - dx2/dt = v1 - v2 for closing speed.

[Richard Hachel]

Le 27/09/2022 à 22:28, "Dono." a écrit :
> On Tuesday, September 27, 2022 at 12:36:21 PM UTC-7, Stan Fultoni wrote:

>> In the relativity literature, the expression "relative velocity" of two objects
>> is defined as the velocity of one object in terms of the inertial coordinate system
>> in which the other object is at rest.
>
> Agreed.

Oui, bon, on va pas y passer la nuit.

LOL.

R.H.

[Stan Fultoni]

On Tuesday, September 27, 2022 at 1:28:19 PM UTC-7, Dono. wrote:
> The expression "closing speed" in terms of S is defined as the difference between their velocities in terms of S. These are just the definitions of those expressions.

> If that were true, you wouldn't have any difficulty in providing the theory of the Sagnac experiment

There is no difficulty accounting for the Sagnac effect. The definitions of terms don't affect physical facts.

[Dono.]

I would be very interested in you deriving the equations. Please show me.

[rotchm]

On Tuesday, September 27, 2022 at 2:53:19 PM UTC-4, Mark-T wrote:


> by a shrinking rod... length contraction, hmmm....
>
> So, is relativistic length contraction germane to this question
> of closing speed?

Relativistic length contraction & closing speed are different concepts.
But you can "germane" then in the sense that you are thinking:
Let a rod shrink from the tips of the approaching bullets all wrt S'.
Now just translate that (the LTs) to frame S.

But the basic idea/math for all these problems is to view the scenarios via the relevant events.
For instance, in your bullet scenario, wrt S' you state that the eqs of motion of the bullets are:

x1'(t') = v1' t'
x2'(t') = L' - v2' t'.

Where L' is the proper length of the rocket (initial separation of the bullets).

IOW, you have the set of events (t', v1' t') and (t', L' - v2' t') as described in S'.
Just use the LT's to translate these two sets to frame S .
Then you will be able to find their eqs of motion in S.

Same idea for a rod, moving or changing shape or whatever.

Note that length [of an object] means the distance (separation, x2-x1 say) between its endpoints, taken *simultaneously* in the chosen frame.
IOW, separation, distance, position, length... may have slightly different meanings. One must be very clear when presenting scenarios and questions.

[Stan Fultoni]

On Tuesday, September 27, 2022 at 4:41:49 PM UTC-7, Dono. wrote:
> > > > The expression "closing speed" in terms of S is defined as the difference between their velocities in terms of S. These are just the definitions of those expressions.
> > > If that were true, you wouldn't have any difficulty in providing the theory of the Sagnac experiment
> > There is no difficulty accounting for the Sagnac effect. The definitions of terms don't affect physical facts.
>
> I would be very interested in you deriving the equations. Please show me.

Is there something in particular you don't understand about the elementary derivation of the Sagnac effect, and, in your mind, is this somehow related to the standard definition of the term "closing speed"? Remember, the name we choose to give to a certain quantity has no effect on physics. The definition of the term "closing speed" between two objects in terms of S is v1 - v2 where v1 and v2 are the speeds of those objects in terms of S, with the stated sign conventions. This is not controversial, it is a simple definition, and changing the name to something else obviously wouldn' affect any physical phenomena.

[Stan Fultoni]

On Tuesday, September 27, 2022 at 11:53:19 AM UTC-7, Mark-T wrote:
> >> > Note that v1' is positive (moving in the positive x' direction) and v2' is
> >> > negative (moving in the negative x' direction). The closing speed in terms
> >> > of S is (v1' - v2')(1-V^2)/[(1+Vv1')(1+Vv2')].

> >> > Trivially, if the foward-going bullet is replaced with a light pulse, the closing

> >> > speed is (1 - v2')(1-V)/(1+Vv2') = 1 - (V+v2')/(1+Vv2'),
> >>

> >> By definition the "closing speed" between two particles in terms of inertial
> > coordinate system S is the difference v1 - v2 between the speeds v1 and v2
> > in terms of S of those particles, just as the closing speed in terms of S' is v1' - v2'.
>

> So, is relativistic length contraction germane to this question
> of closing speed?

Sure. The closing speed is a coordinate-dependent quantity (in general), since speeds are related by the relativistic speed composition formula, which is based on the Lorentz transformation, which entails time dilation, length contraction and the skew of simultaneity.

[Richard Hachel]

Le 28/09/2022 à 07:41, Stan Fultoni a écrit :
>> So, is relativistic length contraction germane to this question
>> of closing speed?
>
> Sure. The closing speed is a coordinate-dependent quantity (in general), since
> speeds are related by the relativistic speed composition formula, which is based
> on the Lorentz transformation, which entails time dilation, length contraction and
> the skew of simultaneity.

You often say interesting things.

Not always true in my opinion.

But interesting.

It is a pity that you like to dialogue but not to discuss.

Otherwise, you are talking here about the addition formula of relativistic
velocities.

I would like to know if you think, for example, that mine is correct.

<http://news2.nemoweb.net/jntp?hO0ULDqb9Nkv4s4ddJNiLNHsOvM@jntp/Data.Media:1>

It would still be a good start to a dialogue.

So far, I have seen two camps. The first one who tells me that the formula
is wrong, and that I'm an idiot. The second who tells me that it is right,
but that I copied it.

Good day to all.

<http://news2.nemoweb.net/?DataID=hO0ULDqb9Nkv4s4ddJNiLNHsOvM@jntp>


R.H.

[rotchm]

On Wednesday, September 28, 2022 at 6:51:48 AM UTC-4, Richard Hachel wrote:
> Le 28/09/2022 à 07:41, Stan Fultoni a écrit :

> I would like to know if you think, for example, that mine is correct.
> <http://news2.nemoweb.net/jntp?hO0ULDqb9Nkv4s4ddJNiLNHsOvM@jntp/Data.Media:1>

Link doesn't work

> It would still be a good start to a dialogue.

Since its a new question, start it in a new thread, DuH!
You are not really bright, aren't you!?

[Stan Fultoni]

On Wednesday, September 28, 2022 at 3:51:48 AM UTC-7, Richard Hachel wrote:
> You are talking here about the addition formula of relativistic
> velocities. I would like to know if you think that mine is correct.

Given two standard and aligned systems S and S' of inertial coordinates, such that the spatial origin of S' is moving at speed v in the positive x direction in terms of S, an object that is moving at speed u in terms of S' is moving at speed (u+v)/(1+uv) in terms of S. Agreed?

[Richard Hachel]

Bon, allez, je vois qu'il se fout de ma gueule.

Je lui donne la formule générale d'addition des vitesses relativistes,
il me demande si sa formule
de simple addition longitudinale est correcte.

Vous voulez que je fasse quoi?

HELP!

R.H.

[Stan Fultoni]

On Wednesday, September 28, 2022 at 7:08:39 AM UTC-7, Richard Hachel wrote:
>> You are talking here about the addition formula of relativistic
> >> velocities. I would like to know if you think that mine is correct.
> >
> > Given two standard and aligned systems S and S' of inertial coordinates, such
> > that the spatial origin of S' is moving at speed v in the positive x direction in
> > terms of S, an object that is moving at speed u in terms of S' is moving at speed
> > (u+v)/(1+uv) in terms of S. Agreed?
>
> Bon, allez, je vois qu'il se fout de ma gueule.

Can you answer the question? Do you agree with the one-dimensional composition? If so, we can go on to explain the three-dimensional composition, but if not, there is no point in going on to three dimensions if you don't even understand one dimensions. (The 3 dimensional composition follows trivially from the one-dimensional plus ordinary spatial rotations.)

[Dono.]

On Tuesday, September 27, 2022 at 10:34:38 PM UTC-7, Stan Fultoni wrote:
> On Tuesday, September 27, 2022 at 4:41:49 PM UTC-7, Dono. wrote:
> > > > > The expression "closing speed" in terms of S is defined as the difference between their velocities in terms of S. These are just the definitions of those expressions.
> > > > If that were true, you wouldn't have any difficulty in providing the theory of the Sagnac experiment
> > > There is no difficulty accounting for the Sagnac effect. The definitions of terms don't affect physical facts.
> >
> > I would be very interested in you deriving the equations. Please show me.
> Is there something in particular you don't understand about the elementary derivation of the Sagnac effect, and, in your mind, is this somehow related to the standard definition of the term "closing speed"?

Please don't talk down to me. I understand perfectly the Sagnac effect, this is why I challenged you to use the definition of relative speed in order to derive it. Please show your work.

[Richard Hachel]

It's all in my equation.

<http://news2.nemoweb.net/jntp?CZk34JjbclO9U28Elhpf3cj6MTU@jntp/Data.Media:1>

Low speeds, relativistic speeds, large angles, small angles, the limit of
a possible observable speed.

What more needs to be shown or proven?

If you want logitudinal addition, you replace µ with 0°

It's still not difficult!


<http://news2.nemoweb.net/?DataID=CZk34JjbclO9U28Elhpf3cj6MTU@jntp>

R.H.

[rotchm]

On Wednesday, September 28, 2022 at 10:26:23 AM UTC-4, Dono. wrote:
> On Tuesday, September 27, 2022 at 10:34:38 PM UTC-7, Stan Fultoni wrote:

> > Is there something in particular you don't understand about the elementary derivation of the Sagnac
>> effect, and, in your mind, is this somehow related to the standard definition of the term "closing speed"?

> Please don't talk down to me.

LOL, look Who's Talking! If someone on this news group talks down to others, it is you!

> I understand perfectly the Sagnac effect,

Doubtful. You have shown your past ignorance. And from this thread, we see that you haven't changed.

[Dono.]

On Wednesday, September 28, 2022 at 7:39:48 AM UTC-7, rotchm wrote:
> On Wednesday, September 28, 2022 at 10:26:23 AM UTC-4, Dono. wrote:
> > On Tuesday, September 27, 2022 at 10:34:38 PM UTC-7, Stan Fultoni wrote:
>
> > > Is there something in particular you don't understand about the elementary derivation of the Sagnac
> >> effect, and, in your mind, is this somehow related to the standard definition of the term "closing speed"?
>
> > Please don't talk down to me.

Stephane,

You need to stop puffing yourself up, you are the same pretender crank you have always been.

[Stan Fultoni]

On Wednesday, September 28, 2022 at 7:33:03 AM UTC-7, Richard Hachel wrote:
> >> > Given two standard and aligned systems S and S' of inertial coordinates, such
> >> > that the spatial origin of S' is moving at speed v in the positive x direction
> >> > in terms of S, an object that is moving at speed u in terms of S' is moving at
> >> > speed (u+v)/(1+uv) in terms of S. Agreed?

> >> > The 3 dimensional composition follows trivially from the one-dimensional
> >> > plus ordinary spatial rotations.
>

> It's all in my equation.

Your equation? What is "your equation"? (I don't click on links in usenet groups. If you type "your equation" I'll be happy to tell you if it is correct or not.)

[Stan Fultoni]

On Wednesday, September 28, 2022 at 7:26:23 AM UTC-7, Dono. wrote:
> I understand perfectly the Sagnac effect, this is why I challenged you to use
> the definition of relative speed in order to derive it.

If you understand the Sagnac effect, you know it isn't derived from a definition, it is a real physical effect. In fact, if you understand any physics at all, you know it doesn't matter what names we choose to give to various quantities, or even if we give them names at all, as long as we are clear what we mean. For example, given two objects with speed v1 and v2 in terms of a standard inertial coordinate system, the quantity v1 - v2 is commonly called "closing speed", but we could just as well call that quantity something else (Fred, Nancy,...) and it would not affect any derivation of any physical effect.

[Dono.]

On Wednesday, September 28, 2022 at 11:43:31 AM UTC-7, Stan Fultoni wrote:
> On Wednesday, September 28, 2022 at 7:26:23 AM UTC-7, Dono. wrote:
> > I understand perfectly the Sagnac effect, this is why I challenged you to use
> > the definition of relative speed in order to derive it.
> If you understand the Sagnac effect, you know it isn't derived from a definition, it is a real physical effect.

You need to stop squirming: all experiments have a set of underlying formulas that predict their outcome. So does Sagnac. So, stop weaseling and derive the expression of the phase difference between the counterpropagating signals starting from your dL/dt=dx1/dt-dx2/dt

>. For example, given two objects with speed v1 and v2 in terms of a standard inertial coordinate system, the quantity v1 - v2 is commonly called "closing speed", but we could just as well call that quantity something else (Fred, Nancy,...) and it would not affect any derivation of any physical effect.

You need to stop squirming: all experiments have a set of underlying formulas that predict their outcome. So does Sagnac. So, stop weaseling and derive the expression of the phase difference between the counterpropagating signals starting from your dL/dt=dx1/dt-dx2/dt.
Or admit the obvious: you can't.

[Stan Fultoni]

On Wednesday, September 28, 2022 at 2:18:57 PM UTC-7, Dono. wrote:
> > If you understand the Sagnac effect, you know it isn't derived from a definition, it is a real physical effect.
>

> Please show me how to derive the expression of the phase difference between the
> counterpropagating signals...

There two subjects here. First, there is the subject raised in the OP, asking about closing speed. Your attempt to answer the OP revealed that you misunderstand the concept of "closing speed". Remedial Tutorial on Closing Speed: For any given system S, the closing speed between two objects moving at speeds v1 and v2 in terms of S is simply v1-v2 with suitable sign conventions.

Second, on an unrelated subject, you've confessed that you don't understand the Sagnac effect, and you've pleaded for someone to explain it to you. Remedial Tutorial on Sagnac Effect: For any incremental segment of a rigid planar loop, translating incrementally at velocity v, the difference in optical path lengths and hence transit times (to the first order) in the two directions is twice the dot product v*ds, and by Green's theorem the integral of this around the closed loop rotating about any fixed point in the plane with angular speed w is 4Aw. This is the time difference to traverse the loop (at speed 1) in opposite directions.

Now that you've been taught the meaning of "closing speed", and how to derive the Sagnac effect, you should be able to see that your contention that the concept of closing speed is incompatible with the Sagnac effect was simply insane. You're welcome.

[Dono.]

On Wednesday, September 28, 2022 at 4:30:51 PM UTC-7, Stan Fultoni wrote:
> On Wednesday, September 28, 2022 at 2:18:57 PM UTC-7, Dono. wrote:
> > > If you understand the Sagnac effect, you know it isn't derived from a definition, it is a real physical effect.
> >
> > Please show me how to derive the expression of the phase difference between the
> > counterpropagating signals...
>
> There two subjects here. First, there is the subject raised in the OP, asking about closing speed. Your attempt to answer the OP revealed that you misunderstand the concept of "closing speed".

The person who doesn't understand closing speed turns out to be you

> Second, on an unrelated subject, you've confessed that you don't understand the Sagnac effect, and you've pleaded for someone to explain it to you.

I understand the Sagnac effect perfectly. I asked you to use the concept of relative velocity to derive the phase difference. After much squirming, it turns out that you have no clue how to. Turns out that you are full of shit.

>For any incremental segment of a rigid planar loop, translating incrementally at velocity v, the difference in optical path lengths and hence transit times (to the first order) in the two directions is twice the dot product v*ds, and by Green's theorem the integral of this around the closed loop rotating about any fixed point in the plane with angular speed w is 4Aw. This is the time difference to traverse the loop (at speed 1) in opposite directions.

I expected the above, you couldn't use your inapplicable definition : dL/dt=dx1/dt-dx2/dt so you resorted to something completely different. Which doesn't even give the correct result. To deflate you pompous gassbag, here is the correct derivation, using the correct meaning of "closing speed producing the correct result:


c*t1+w*R*t1=2Pi*R
c*t2=2Pi*R+w*R*t2

t2-t1=2Pi*R(1/(c-w*R)-1/(c+w*R))=4A*w/(c^2-w^2*R^2)

[Stan Fultoni]

On Wednesday, September 28, 2022 at 5:06:31 PM UTC-7, Dono. wrote:
> > > Please show me how to derive the expression of the phase difference between the
> > > counterpropagating signals...
> >

> > There are two subjects here. First, there is the subject raised in the OP, asking about closing speed. Your attempt to answer the OP revealed that you misunderstand the concept of "closing speed". Remedial Tutorial on Closing Speed: For any given system S, the closing speed between two objects moving at speeds v1 and v2 in terms of S is simply v1-v2 with suitable sign conventions.
>
> [Thanks for teaching me this.]

You're welcome.

> > Second, on an unrelated subject, you've confessed that you don't understand the Sagnac effect, and you've pleaded for someone to explain it to you. Remedial Tutorial on the Sagnac Effect: For any incremental segment of a rigid planar loop, translating incrementally at velocity v, the difference in optical path lengths and hence transit times (to the first order) in the two directions is twice the dot product v*ds, and by Green's theorem the integral of this around the closed loop rotating about any fixed point in the plane with angular speed w is 4Aw. This is the time difference to traverse the loop (at speed 1) in opposite directions.
>
> [Thanks for teaching me this too. It is far more general and powerful than the pre-school method that was previously all I could recite.]

Sure. No problem.

> I asked you to use the concept of relative velocity to derive the phase difference.

No, again, relative velocity between two objects is defined as the speed of one object in terms of the inertial coordinates in which the other is at rest. That is not (repeat: not) the same as closing speed. You claimed that if the terms closing speed refers to v1-v2 then there would be no Sagnac effect. As shown, your claim is utterly insane.

> You couldn't use your inapplicable definition : dL/dt=dx1/dt-dx2/dt

Again, the quantity dL/dt = v1 - v2 is called the closing speed, and the Sagnac effect is (to first order) 4Aw, as explained above. Now that you've been taught both of these things, it should be clear to you that your fantasy that they conflict with each other was utterly insane.

> Here is the correct derivation, using the correct meaning of "closing speed

Again, physical effects do not depend on definitions. And, in fact, you did not mention "closing speed" in your sophomoric little "derivation" (which is vastly less general than what I just taught you), so you again are exhibiting symptoms of mental illness, insisting that the term "closing speed" is essential to your "derivation", and then forgetting to mention it in your "derivation". LOL. And if you had mentioned it, you would merely have been confirming what I just taught you. Duh.

To reiterate, for any given system S, the closing speed between two objects moving at speeds v1 and v2 in terms of S is simply v1-v2 with suitable sign conventions. Hopefully you understand this now, and in the bargain you've learned the grown-up derivation of the Sagnac effect. You're welcome.

[Dono.]

On Wednesday, September 28, 2022 at 4:30:51 PM UTC-7, Stan Fultoni wrote:
> the difference in optical path lengths and hence transit times (to the first order) in the two directions is twice the dot product v*ds, and by Green's theorem the integral of this around the closed loop rotating about any fixed point in the plane with angular speed w is 4Aw.

I have anticipated that you will cheat, v*ds doesn't have dimensions of speed (neither has the integral) , so, you are not using your inane definition of "closing speed". Of course you aren't, it is unusable for the purpose, this is precisely why I gave you the exercise. I never anticipated that you are THAT dishonest. Now, that I taught you the correct way of deriving the phase difference and the correct answer (not yours), you are welcome, pompous gassbag.

[Dono.]

On Wednesday, September 28, 2022 at 5:48:38 PM UTC-7, Stan Fultoni wrote:
> insisting that the term "closing speed" is essential to your "derivation", and then forgetting to mention it in your "derivation".

Pompous gassbag

c*t1+w*R*t1=2Pi*R
c*t2=2Pi*R+w*R*t2

ARE the basic equations of closing speed. You just reached a new low.

[Stan Fultoni]

On Wednesday, September 28, 2022 at 7:26:08 PM UTC-7, Dono. wrote:
> > the difference in optical path lengths and hence transit times (to the first order) in the two directions is twice the dot product v*ds, and by Green's theorem the integral of this around the closed loop rotating about any fixed point in the plane with angular speed w is 4Aw.
>

> v*ds doesn't have dimensions of speed ...

Right, it has units of time, as does Aw, as they should. Remember, we are using units so c=1, so distances and times both have units of seconds. (Another remedial tutorial.)

> so, you are not using your inane definition of "closing speed".

As I explained to you from the start, physical phenomena are not derived from definitions, no matter how much your diseased brains wants to believe they are. The term "closing speed" is an arbitrary name for obvious sums and differences of velocities in terms of a given system of coordinates. Any given derivation may consider such sums and differences, regardless of what they are called.

> c*t1+w*R*t1=2Pi*R and c*t2=2Pi*R+w*R*t2 ARE the basic equations of
> closing speed.

Those expressions are just referring to quantities of the form v1 - v2 with the appropriate sign conventions, i.e., you are using the very definition of closing speed that the other part of your diseased brain is vociferously denying. Yet your whole (insane) claim was that the Sagnac effect relies on the denial of the simple arithmetic of these sums and differences. Everything you are saying is completely and utterly irrational.

[Stan Fultoni]

On Wednesday, September 28, 2022 at 7:33:03 AM UTC-7, Richard Hachel wrote:
> > >> You are talking here about the addition formula of relativistic
> >> >> velocities. I would like to know if you think that mine is correct.
> >> >
> >> > Given two standard and aligned systems S and S' of inertial coordinates, such
> >> > that the spatial origin of S' is moving at speed v in the positive x direction
> >> in
> >> > terms of S, an object that is moving at speed u in terms of S' is moving at
> >> speed
> >> > (u+v)/(1+uv) in terms of S. Agreed?

> > Can you answer the question? Do you agree with the one-dimensional composition?
> > If so, we can go on to explain the three-dimensional composition, but if not,
> > there is no point in going on to three dimensions if you don't even understand one
> > dimensions. (The 3 dimensional composition follows trivially from the
> > one-dimensional plus ordinary spatial rotations.)
>
> It's all in my equation.

And yet, when I invite you to post your question, you run away. Again, if you type "your equation" I'll be happy to tell you if it is correct or not. My guess is that, since you refer to a single equation, you are just expressing the magnitude as a function of the angle. Yawn. You copy the equations of special relativity, claim them as your own, misinterpret the symbols, and then deny relativity. Completely irrational.

[rotchm]

On Wednesday, September 28, 2022 at 10:59:12 PM UTC-4, Stan Fultoni wrote:
> On Wednesday, September 28, 2022 at 7:26:08 PM UTC-7, Dono. wrote:

> Everything you are saying is completely and utterly irrational.

Welcome to dono-land,
where, solutions to differential equations do not satisfy the differential equation in question,
where, simultaneous events in one (inertial) frame are simultaneous in another if generated by a horizontal falling rod,
where 0=1,
etc...

[Dono.]

On Wednesday, September 28, 2022 at 7:59:12 PM UTC-7, Stan Fultoni wrote:
> On Wednesday, September 28, 2022 at 7:26:08 PM UTC-7, Dono. wrote:
> > > the difference in optical path lengths and hence transit times (to the first order) in the two directions is twice the dot product v*ds, and by Green's theorem the integral of this around the closed loop rotating about any fixed point in the plane with angular speed w is 4Aw.
> >
> > v*ds doesn't have dimensions of speed ...
>
> Right, it has units of time,

...meaning that you are not using your inane definition of "closing speed". Meaning that you cheated and you got caught.

>
> > c*t1+w*R*t1=2Pi*R and c*t2=2Pi*R+w*R*t2 ARE the basic equations of
> > closing speed.
>
> Those expressions are just referring to quantities of the form v1 - v2 with the appropriate sign conventions,

No, dumbfuck

They represent the way two entities cover one expanse of space.

[Dono.]

On Wednesday, September 28, 2022 at 8:10:47 PM UTC-7, rotchm wrote:

>

> where 0=1,
> etc...

Stephane,

You are rushing to kiss ass to Stan. Congratulations!

[rotchm]

On Wednesday, September 28, 2022 at 11:17:17 PM UTC-4, Dono. wrote:
> On Wednesday, September 28, 2022 at 8:10:47 PM UTC-7, rotchm wrote:

> Stephane,

<sigh>. if you bothered to do your homework, you would know that stephane and selleri are together now.
Hey, but I still can be david... (I hope idiot ken doesn't see this ;) ).

[Dono.]

On Wednesday, September 28, 2022 at 8:31:26 PM UTC-7, rotchm wrote:
> On Wednesday, September 28, 2022 at 11:17:17 PM UTC-4, Dono. wrote:
> > On Wednesday, September 28, 2022 at 8:10:47 PM UTC-7, rotchm wrote:
> > Stephane,
>
> <sigh>. if you bothered to do your homework, you would know that stephane and selleri are together now.

But you are Stephane Baune, the crank that used to publish crank stuff in Apeiron

[rotchm]

From beyond the grave, directly to your screen.
Good one. I didn't know you were into paranormal activity...boooooh

[Stan Fultoni]

On Wednesday, September 28, 2022 at 8:16:18 PM UTC-7, Dono. wrote:
> > > > the difference in optical path lengths and hence transit times (to the first order) in the two directions is twice the dot product v*ds, and by Green's theorem the integral of this around the closed loop rotating about any fixed point in the plane with angular speed w is 4Aw.
> > >
> > > v*ds doesn't have dimensions of speed ...
> >
> > Right, it has units of time,
>
> ...meaning that you are not using your inane definition of "closing speed".

Again, physical effects are not derived from definitions. The definition of the term "closing speed" is v1 - v2 as explained to you before. This is simply a fact, and it explains why your response to the OP was erroneous.

On the other subject you raised, the simplistic Sagnac derivation for a circular ring with central rotation axis that you recited (obviously without comprehension) actually makes explicit use of quantities of the form v1-v2, so your example actually disproves your own insane claims. And you seem oblivious to this. Then, to top it off, you complain because you can't understand the fully general derivation I provided to you, at your request (you're welcome). The crowning incoherence is your claim that the general derivation of the Sagnac effect proves that people cannot refer to quantities of the form v1-v2 as closing speeds. Sheesh. You are completely irrational.

[Dono.]

On Wednesday, September 28, 2022 at 9:55:09 PM UTC-7, Stan Fultoni wrote:
> On Wednesday, September 28, 2022 at 8:16:18 PM UTC-7, Dono. wrote:
> > > > > the difference in optical path lengths and hence transit times (to the first order) in the two directions is twice the dot product v*ds, and by Green's theorem the integral of this around the closed loop rotating about any fixed point in the plane with angular speed w is 4Aw.
> > > >
> > > > v*ds doesn't have dimensions of speed ...
> > >
> > > Right, it has units of time,
> >
> > ...meaning that you are not using your inane definition of "closing speed".
> Again, physical effects are not derived from definitions.

I never said it was, dumbfuck, I asked you to make use of your inane approach. Then, I showed you how it is done using the correct definition of closing speed.

>The definition of the term "closing speed" is v1 - v2 as explained to you before.

What you are explaining is the useless "relative speed". You couldn't use it for the derivation of the phase difference , so you cheated and you got caught.

> On the other subject you raised, the simplistic Sagnac derivation for a circular ring with central rotation axis that you recited (obviously without comprehension) actually makes explicit use of quantities of the form v1-v2,

No, it doesn't, it makes use of the correct definition of closing speed, something that I have showed you time and again . Yet it doesn't go thru your thick skull.

> Then, to top it off, you complain because you can't understand the fully general derivation I provided to you,

Now, you are outright lying, I am pretty familiar with the derivation. This is the cheat I anticipated that you will try to pull. And you got caught.

> The crowning incoherence is your claim that the general derivation of the Sagnac effect proves that people cannot refer to quantities of the form v1-v2 as closing speeds.

No, dumbestfuck, what I claimed is that you COULD NOT USE v1-v2 in the derivation of the phase difference, so you tried to pull a fast one and got caught. Keep digging yourself deeper, do you want me to hand you a shovel or a pick axe or both?

[Dono.]

On Wednesday, September 28, 2022 at 9:55:09 PM UTC-7, Stan Fultoni wrote:
> The definition of the term "closing speed" is v1 - v2 as explained to you before. This is simply a fact, and it explains why your response to the OP was erroneous.

I missed the above, just when one thought that you couldn't get any lower: I told the OP that the most straightforward way is to calculate the closing speed u=v1+v2 in the rocket frame and to transform it into the observer frame.

[Stan Fultoni]

On Wednesday, September 28, 2022 at 10:06:54 PM UTC-7, Dono. wrote:
> > Again, physical effects are not derived from definitions.

> I never said it was..

Yes you did. You said if v1-v2 were called the closing speed, there would be no Sagnac effect. (Scroll back up to your message if you've forgotten.)

> I showed you how it is done using the correct definition of closing speed.

The derivation you recited (obviously without comprehension) makes explicit use of quantities of the form v1 - v2. It's amazing that you don't even realize this.

> What you are explaining is the useless "relative speed".

Again, the term "relative speed" of two objects is the speed of one object in terms of the inertial coordinates in which the other is at rest. This was explained to you before. Remember?

> You couldn't use it for the derivation of the phase difference...

Again, you pleaded with me for a derivation of the Sagnac effect and I provided it to you. (You're welcome.) The bizarre notion that this (or any other) derivation somehow implies that quantities of the form v1-v2 can or cannot be called closing speeds is entirely yours... and entirely insane.

> No, it doesn't, it makes use of the correct definition of closing speed,

Again, the correct definition of closing speed in terms of system S for two objects with speeds v1 and v2 in terms of S is v1-v2 with suitable sign conventions. Remember? And these quantities appear prominently in the very derivation of Sagnac that you yourself have parrotted (obviously without comprehension).

[Volney]

Will that even work? If I shine two beams of light at each other in a
rocket, I get a closing speed of 2c. How can I transform that into
another frame?

[Richard Hachel]

Le 29/09/2022 à 05:06, Stan Fultoni a écrit :

> Completely irrational.

No.

That's not what I said.

I said that the theory of relativity was a true theory, but that being
very badly explained, and sometimes giving dramatically false results,
such as the distance from the earth to the subject, when the Langevin
rocket turned around (36 ly and not 7.2 ly), or as the proper time
calculated in the Tau Ceti problem (4.776 years and not 3.139 years), it
deserves to be re-written.

I didn't say it was all wrong.

For example, the general relativistic velocity addition equation that
physicists use is mathematically correct.

But it's very poorly written.

I prefer mine, more aesthetic.

The beautiful being the splendor of the true.

<http://news2.nemoweb.net/jntp?qxMzMATAkgqx5ak4avwU9Pq93s8@jntp/Data.Media:1>

R.H.



--
"Mais ne nous trompons pas.
Il n'y a pas que de la violence avec des armes : il y a des situations de
violence."
Abbé Pierre
₀₀₀
<http://news2.nemoweb.net/?DataID=qxMzMATAkgqx5ak4avwU9Pq93s8@jntp>

[rotchm]

On Thursday, September 29, 2022 at 1:18:02 AM UTC-4, Dono. wrote:
> On Wednesday, September 28, 2022 at 9:55:09 PM UTC-7, Stan Fultoni wrote:
> > The definition of the term "closing speed" is v1 - v2 as explained to you before. This is simply a fact, and it
>> explains why your response to the OP was erroneous.

>...I told the OP that the most straightforward way is to calculate the closing speed u=v1+v2 in the rocket

> frame and to transform it into the observer frame.

Yes you said to transform u=v1+v2 to the obs frame.
But there were no v1 nor v2 in the op.
And the correct notation is not v1 nor v2; its v1' & v2'.
And transforming your quantity (v1+v2) [should be (v1'+v2')] to the obs frame does not yield the closing speed in the obs frame.

Can you state here the definition (or formulas) of closing speed?
Perhaps this will clear things up.

[Stan Fultoni]

On Thursday, September 29, 2022 at 1:25:23 AM UTC-7, Richard Hachel wrote:
> I said that the theory of relativity was a true theory, but the proper time
> calculated in the Tau Ceti problem is 4.776 years and not 3.139 years...

You contradict yourself, twice. First, you say special relativity is true, and then in the same sentence you say it is false. Second, your belief in 4.776 implies 1=0, which is self-contradictory. Hence you are completely irrational. Agreed?

> For example, the general relativistic velocity addition equation that
> physicists use is mathematically correct.

You contradict yourself again: Special relativity is a subset of general relativity in regions of constant gravitational potential, so if the general relativity composition (not addition) is correct, then so is the special relativity composition in its domain of applicability.

> I prefer mine, more aesthetic.

One of the standard elementary relations of special relativity is u'^2 = 1 - (1-u^2)(1-v^2)/[1-uv cos(q)]^2. Is "your" equation preferable to this? Why do you refuse to type your equation? You claim that you want to promote "your" equation, but when invited to present it, you run away. So you are totally irrational, right?

[Dono.]

On Wednesday, September 28, 2022 at 11:06:46 PM UTC-7, Stan Fultoni wrote:
> On Wednesday, September 28, 2022 at 10:06:54 PM UTC-7, Dono. wrote:
> > > Again, physical effects are not derived from definitions.
> > I never said it was..
>
> Yes you did. You said if v1-v2 were called the closing speed, there would be no Sagnac effect. (Scroll back up to your message if you've forgotten.)

You are outright lying.

> > I showed you how it is done using the correct definition of closing speed.
> The derivation you recited (obviously without comprehension) makes explicit use of quantities of the form v1 - v2. It's amazing that you don't even realize this.

You must be a total idiot, it makes use of the closing speed equations, two entities covering a common expanse of space (2Pi*R)

> > What you are explaining is the useless "relative speed".
> Again, the term "relative speed" of two objects is the speed of one object in terms of the inertial coordinates in which the other is at rest. This was explained to you before. Remember?
>

I fully understand "relative speed". You cheated in trying to use this concept in the explanation of the Sagnac effect and you got caught. Remember?

> > You couldn't use it for the derivation of the phase difference...
>
> Again, you pleaded with me for a derivation of the Sagnac effect and I provided it to you.

I did not "plead". I challenged you. You squirmed for a long time and you ended up cheating. Remember?

>The bizarre notion that this (or any other) derivation somehow implies that quantities of the form v1-v2 can or cannot be called closing speeds is entirely yours... and entirely insane.

The insane person is you. The claim was that the quantity v1-v2 is useless in deriving the Sagnac effect. You made a lame attempt by cheating and you got caught. Remember?

[Dono.]

On Thursday, September 29, 2022 at 5:30:33 AM UTC-7, rotchm wrote:
> On Thursday, September 29, 2022 at 1:18:02 AM UTC-4, Dono. wrote:
> > On Wednesday, September 28, 2022 at 9:55:09 PM UTC-7, Stan Fultoni wrote:
> > > The definition of the term "closing speed" is v1 - v2 as explained to you before. This is simply a fact, and it
> >> explains why your response to the OP was erroneous.
> >...I told the OP that the most straightforward way is to calculate the closing speed u=v1+v2 in the rocket
> > frame and to transform it into the observer frame.
> Yes you said to transform u=v1+v2 to the obs frame.
> But there were no v1 nor v2 in the op.

that is because you are an idiot, Stephane
v1,v2 are the speeds of the two bullets in the rocket frame.

> Can you state here the definition (or formulas) of closing speed?
> Perhaps this will clear things up.

I did, early on in the thread. Go find it.

[Richard Hachel]

Le 29/09/2022 à 15:28, Stan Fultoni a écrit :
> On Thursday, September 29, 2022 at 1:25:23 AM UTC-7, Richard Hachel wrote:
>> I said that the theory of relativity was a true theory, but the proper time
>> calculated in the Tau Ceti problem is 4.776 years and not 3.139 years...
>
> You contradict yourself, twice. First, you say special relativity is true, and
> then in the same sentence you say it is false. Second, your belief in 4.776
> implies 1=0, which is self-contradictory. Hence you are completely irrational.
> Agreed?

Je n'ai pas dis ça.

J'ai dis que les bases sont vraies.

Mais qu'on les applique très mal.

Avec, in fine, des résultats clairement faux.

R.H.

[rotchm]

On Thursday, September 29, 2022 at 10:18:31 AM UTC-4, Dono. wrote:
> On Thursday, September 29, 2022 at 5:30:33 AM UTC-7, rotchm wrote:

> > Yes you said to transform u=v1+v2 to the obs frame.
> > But there were no v1 nor v2 in the op.

> v1,v2 are the speeds of the two bullets in the rocket frame.

Here is an excerpt of the scenario:

" A rocket is moving at speed V in the positive x direction of an inertial
coordinate system S, and inside the rocket there are two bullets
fired from rear and front, with velocities v1' and v2' respectively in
terms of the inertial coordinate system S' in which the rocket is at rest. "

See? In the frame of the rocket, frame S', the speeds of the bullets are v1' & v2', not v1 & v2.
The rocket (frame) is the primed frame, as per conventions, and as per the enunciation of the
scenario.

> > Can you state here the definition (or formulas) of closing speed?
> > Perhaps this will clear things up.

> I did, early on in the thread. Go find it.

Be honest and helpful; state it here.
Then we can take it from there.

[rotchm]

On Thursday, September 29, 2022 at 11:26:23 AM UTC-4, Richard Hachel wrote:
> Le 29/09/2022 à 15:28, Stan Fultoni a écrit :

> > You contradict yourself, twice. First, you say special relativity is true, and
> > then in the same sentence you say it is false. Second, your belief in 4.776
> > implies 1=0, which is self-contradictory. Hence you are completely irrational.
> > Agreed?
> Je n'ai pas dis ça.

Why do you change language as you go along?
Do you know that that is dishonest and disrespectful?

[Dono.]

On Thursday, September 29, 2022 at 8:54:49 AM UTC-7, rotchm wrote:
> On Thursday, September 29, 2022 at 10:18:31 AM UTC-4, Dono. wrote:
> > On Thursday, September 29, 2022 at 5:30:33 AM UTC-7, rotchm wrote:
>
> > > Yes you said to transform u=v1+v2 to the obs frame.
> > > But there were no v1 nor v2 in the op.
> > v1,v2 are the speeds of the two bullets in the rocket frame.
> Here is an excerpt of the scenario:
>
> " A rocket is moving at speed V in the positive x direction of an inertial
> coordinate system S, and inside the rocket there are two bullets
> fired from rear and front, with velocities v1' and v2' respectively in
> terms of the inertial coordinate system S' in which the rocket is at rest. "
>
> See? In the frame of the rocket, frame S', the speeds of the bullets are v1' & v2', not v1 & v2.

How dumb are you, Stephane?

Don't answer that, it was a rhetorical question. Does the extra tick make any difference, pedantic imbecile?

> The rocket (frame) is the primed frame, as per conventions, and as per the enunciation of the
> scenario.
> > > Can you state here the definition (or formulas) of closing speed?
> > > Perhaps this will clear things up.
>
> > I did, early on in the thread. Go find it.
> Be honest and helpful; state it here.
> Then we can take it from there.

I published it at least 3 times. Not my problem that you are too lazy and too stupid to find it.

[Dart Ruzzier]

Dono. wrote:

>> See? In the frame of the rocket, frame S', the speeds of the bullets
>> are v1' & v2', not v1 & v2.
>
> How dumb are you, Stephane?

extreme.

[rotchm]

On Thursday, September 29, 2022 at 12:07:19 PM UTC-4, Dono. wrote:
> On Thursday, September 29, 2022 at 8:54:49 AM UTC-7, rotchm wrote:


> > See? In the frame of the rocket, frame S', the speeds of the bullets are v1' & v2', not v1 & v2.

> Does the extra tick make any difference, pedantic imbecile?

You confuse the notations, you confuse the values from one frame to another.
No wonder you get the wrong answer and are totally confused about the scenario.

> > The rocket (frame) is the primed frame, as per conventions, and as per the enunciation of the
> > scenario.

No rebuttal?

> > > > Can you state here the definition (or formulas) of closing speed?
> > > > Perhaps this will clear things up.

> > Be honest and helpful; state it here.

So you still fail to give here explicitly the definition of closing speed.
I must then agree with Stan, that you have no clue on its meaning.

I give you my understanding of definition of 'closing speed' :

The definition of closing speed in terms of system S for two objects with speeds v1 and v2 in terms of S is v1-v2 with suitable sign conventions.

Does this definition correspond to yours?
Does this definition corresponds to the physics (SR) community's definition?

[Dono.]

On Thursday, September 29, 2022 at 10:02:53 AM UTC-7, rotchm wrote:


> The definition of closing speed in terms of system S for two objects with speeds v1 and v2 in terms of S is v1-v2 with suitable sign conventions.
>
> Does this definition correspond to yours?

No , it doesn't , Stephane.

BTW, the definition of "closing speed" has come up on this forum multiple times. Do a search, lazy imbecile.

> Does this definition corresponds to the physics (SR) community's definition?

The one I posted, yes. The one Stan posted, no.

[Mark-T]

On September 28, Dono. wrote:
> I told the OP that the most straightforward way is to calculate the
> closing speed u=v1+v2 in the rocket frame and to transform it into the observer frame.

But, due to the nonlinear relativity addition formula, the transform
of (v1+ v2) doesn't equal the transform of v1 plus the transform of v2.


Mark

[Dono.]

...there is no reason why it should.

[Mark-T]

On September 28, Richard Hachel wrote:

> HELP!

MAYDAY!

Mark
long ago francais student

[Richard Hachel]

My position is clear. The theory of relativity, (ie the fact that we do
not live in a Newtonian universe but in a universe where space and time
have relative properties) is true.

In this sense, I detach myself from people who think that everything is
wrong, and that there is no relativity of times, lengths, etc., depending
on the speed and position of the observers.

I am not an anti-relativist.

There are also things that are perfectly correct in SR, like the
relativity of chronotropy, like the Lorentz transformations, like the
relativity of impulses and energies.

But all is not true.

At times, there are conceptual errors, and things that are very poorly
explained by the teachers.

It is all these things that I re-explain, in a clearer and more true way.

R.H.

[Mark-T]

On September 27, rotchm wrote:
>> by a shrinking rod... length contraction, hmmm....
>> So, is relativistic length contraction germane to this question
>> of closing speed?
>
> Relativistic length contraction & closing speed are different concepts.
> But you can "germane" then in the sense that you are thinking:
> Let a rod shrink from the tips of the approaching bullets all wrt S'.
> Now just translate that (the LTs) to frame S.
> But the basic idea/math for all these problems is to view the scenarios via the relevant events.
> For instance, in your bullet scenario, wrt S' you state that the eqs of motion of the bullets are:
> x1'(t') = v1' t'
> x2'(t') = L' - v2' t'.
> Where L' is the proper length of the rocket (initial separation of the bullets).
> IOW, you have the set of events (t', v1' t') and (t', L' - v2' t') as described in S'.
> Just use the LT's to translate these two sets to frame S .
> Then you will be able to find their eqs of motion in S.
>
> Same idea for a rod, moving or changing shape or whatever.
> Note that length [of an object] means the distance between
> its endpoints, taken *simultaneously* in the chosen frame.

I think that's the right idea.
But there's a complication.

The rocket frame is S. Observer is S'. v1 = v2, for simplicity.
The bullets approach at speed 2L(v), as seen in S', where L is
the velocity transform.

Now let their separation be the rod's endpoints. The rod is
shrinking, not 'shrunk'; i.e. therefore its velocity increases,
it accelerates.

But in the original scenario, there is no acceleration of rocket
or bullets. So something is inconsistent here.

Mark

[rotchm]

On Thursday, September 29, 2022 at 1:37:00 PM UTC-4, Dono. wrote:
> On Thursday, September 29, 2022 at 10:02:53 AM UTC-7, rotchm wrote:
>
>
> > The definition of closing speed in terms of system S for two objects with speeds v1 and v2 in terms of S is v1-v2 with suitable sign conventions.
> >
> > Does this definition correspond to yours?
> No , it doesn't

Then, what is your definition/concept of closing speed?
State it here.

> BTW, the definition of "closing speed" has come up on this forum multiple times. Do a search,

Did. And on the net. Conclusion: its the definition/concept I (& stan) stated above.
Also, an equivalent one I adhere to is, the rate of change of the distance (separation)
of two things; this is d/dt of ( x1 - x2) , bearing sign & ordering conventions. But d/dt of ( x1 - x2) = dx1/dt - dx2/dt = v1-v2.

> > Does this definition corresponds to the physics (SR) community's definition?
> The one I posted, yes. The one Stan posted, no.

Evidence shows otherwise.
What is your definition of closing speed, and what evidence do you have to support it?
And, since its the op that invoked it, we can ask him what he meant by closing speed, right?

[Dono.]

On Thursday, September 29, 2022 at 12:46:23 PM UTC-7, rotchm wrote:
> On Thursday, September 29, 2022 at 1:37:00 PM UTC-4, Dono. wrote:
> > On Thursday, September 29, 2022 at 10:02:53 AM UTC-7, rotchm wrote:
> >
> >
> > > The definition of closing speed in terms of system S for two objects with speeds v1 and v2 in terms of S is v1-v2 with suitable sign conventions.
> > >
> > > Does this definition correspond to yours?
> > No , it doesn't
> Then, what is your definition/concept of closing speed?
> State it here.

Stephane

It is stated (3 times). In this thread. You are so inept you cannot find it.



> Evidence shows otherwise.

Evidence shows you are an imbecile, Stephane.

[rotchm]

On Thursday, September 29, 2022 at 2:10:20 PM UTC-4, Mark-T wrote:
> On September 27, rotchm wrote:

> > For instance, in your bullet scenario, wrt S' you state that the eqs of motion of the bullets are:
> > x1'(t') = v1' t'
> > x2'(t') = L' - v2' t'.
> > Where L' is the proper length of the rocket (initial separation of the bullets).
> > IOW, you have the set of events (t', v1' t') and (t', L' - v2' t') as described in S'.
> > Just use the LT's to translate these two sets to frame S .
> > Then you will be able to find their eqs of motion in S.
> >
> > Same idea for a rod, moving or changing shape

> I think that's the right idea.
> But there's a complication.

> The rocket frame is S. Observer is S'. v1 = v2, for simplicity.

Realize now that you are changing the notation from the previous discussions in this thread; S was the observer's
frame and S' was the rocket frame.

Adopting now your notation (S is the rocket...)

> The bullets approach at speed 2L(v), as seen in S', where L is
> the velocity transform.

?? L was the proper length of the rocket. Now its a transform?
Keeping L as the proper length of the rocket and not the transform (whatever you meant by that).

In the rocket frame, the eqs of motion of the bullets are:

x1(t) = v t
x2(t) = L- vt

Agreed?

Thus, the distance (separation) between them is x2-x1 = ... = L - 2vt
Agreed?

Thus their closing speed is d(L - 2vt)/dt = -2v ~ 2v [due to our conventions; speeds are magnitudes].
This is the v1 +- v2 formula for closing speeds. v + v = 2v [since they are approaching, 'closing in'].

> Now let their separation be the rod's endpoints.

OK, thus the eqs of motion of the endpoints are thus

x1(t) = v t
x2(t) = L- vt

Agreed?

> The rod is shrinking,

Ambiguous verbiage and unnecessary. Be it true or not, the eqs of motion of the
endpoints of the rod (in the frame of the rocket) as stated by you are

x1(t) = v t
x2(t) = L- vt

Agreed?

> i.e. therefore its velocity increases, it accelerates.

Does it now? What is "it" exactly .
Does the center of the rod accelerate? Do its endpoints? Other of its material points?
Are you "forcing" the rod to shrink in S? The consequence of this force in S' is thus?

And does all this change what you posited as

x1(t) = v t
x2(t) = L- vt

?

Think about all that.
This will lead you to the notions of rigidity/elasticity/materials and how those
are treated in relativity. Also of that topic: Born rigid motion.
https://en.wikipedia.org/wiki/Born_rigidity

> But in the original scenario, there is no acceleration of rocket
> or bullets. So something is inconsistent here.

Having considered what I just told you above, are things starting to clear up?