Train and trac, relative motions...?

[rotchm]

In a recent thread about the usual train moving along a track scenario, I claimed that:

" ...there exist an inertial frame such that the train & track have the same speed. "

I wonder if this is true. Any inputs?


Someone (I dono who) claims otherwise, and I quote them:
" The train and the track are in relative motion, so there is no frame "such that the train & track have the same speed." "

[Mike Fontenot]

There is an inertial frame in which the train and track have the same
SPEED, but in OPPOSITE DIRECTIONS.

--
https://sites.google.com/site/cadoequation/cado-reference-frame

"Accelerated Observers in Special Relativity", PHYSICS ESSAYS,
December 1999, p629.

All you ever need to know about the twin "paradox".

[Dono,]

On Monday, August 14, 2017 at 10:41:22 AM UTC-7, rotchm wrote:
> In a recent thread about the usual train moving along a track scenario, I claimed that:
>
> " ...there exist an inertial frame such that the train & track have the same speed. "
>
> I wonder if this is true. Any inputs?
>
>

Yet another imbecility from Stephane Baune

[Tom Roberts]

On 8/14/17 12:41 PM, rotchm wrote:
> In a recent thread about the usual train moving along a track scenario, I claimed that:
> " ...there exist an inertial frame such that the train & track have the same speed."
> I wonder if this is true. Any inputs?

In the context of SR, for any pair of timelike objects moving inertially, it is
always possible to find an inertial frame relative to which they have
3-velocities with equal magnitudes and opposite directions. So relative to this
frame they have equal speeds.

Proof: Sum their 4-velocities; the result will be timelike.
The inertial frame in question is the rest frame of the sum.
This uses two well-known theorems: a) the sum of two timelike
4-vectors is timelike, and b) every timelike 4-vector has a
rest frame in which its spatial components are zero.

That frame is unique up to spatial rotations, unless the two objects are both at
rest in some inertial frame -- then they have equal speeds relative to any
inertial frame.

Tom Roberts

[Prudence Oppenheimer]

rotchm wrote:

> In a recent thread about the usual train moving along a track scenario,
> I claimed that:
>
> " ...there exist an inertial frame such that the train & track have the
> same speed. "

The axis of the wheels, what's wrong with you. Its "inertial" speed cannot
exceed ½c for obvious reasons.

[rotchm]

Correct Tom. I just wanted to "teach" or inform dono of the veracity
of my claim.

[Rafael Valls Hidalgo-Gato]

Do you consider that the time relative to an equivalent SR inertial system is
the same time defined by 1905 Einstein relative to a stationary system?

> Tom Roberts

RVHG (Rafael Valls Hidalgo-Gato)

[Dono,]

On Monday, August 14, 2017 at 1:01:58 PM UTC-7, rotchm wrote:
> Correct Tom. I just wanted to "teach" or inform dono of the veracity
> of my claim.

You are as imbecile as Tom on this one. Congratulations!

[Dono,]

You have reached a new low, Tom

Here is the counterproof:

In a given frame F the two objects have speeds u_1 \ne u_2
In any other frame F' moving with speed x wrt frame F the speeds are:

u'_1=\frac{u_1+x}{1+x*u_1/c^2}
u'_2=\frac{u_2+x}{1+x*u_2/c^2}

The only value of x that makes u'_1=u'_2 is x=c
You are just as stupid as Stephane Baune.

[Dono,]

On Monday, August 14, 2017 at 11:51:36 AM UTC-7, tjrob137 wrote:

About half of what you post is , in the words of W. Pauli, "Not even wrong" .

[rotchm]

On Monday, August 14, 2017 at 5:12:44 PM UTC-4, Dono, wrote:
> On Monday, August 14, 2017 at 11:51:36 AM UTC-7, tjrob137 wrote:

> > Tom Roberts
>
> You have reached a new low, Tom
>
> Here is the counterproof:

Good gawd idiot dono, you are so lost. I wont even look at your 'counterproof' since I'm sure thats its riddled with misconceptions and errors, as all ur proofs are .

[Dono,]

On Monday, August 14, 2017 at 2:37:30 PM UTC-7, rotchm wrote:
> On Monday, August 14, 2017 at 5:12:44 PM UTC-4, Dono, wrote:
> > On Monday, August 14, 2017 at 11:51:36 AM UTC-7, tjrob137 wrote:
>
> > > Tom Roberts
> >
> > You have reached a new low, Tom
> >
> > Here is the counterproof:
>

< I wont even look at your 'counterproof' since I'm sure thats its riddled with misconceptions and errors, as all ur proofs are .

You should look because it proves that you are an imbecile

[Dono,]

On Monday, August 14, 2017 at 10:41:22 AM UTC-7, rotchm wrote:

Stephane,

In the frame of the track the train has speed u. In any frame moving with speed x wrt the frame of the track the train has speed u'_1=\frac{u+x}{1+ux/c^2} and the track has speed u'_2=x

u'_1=u'_2 if and only if x=c

[rotchm]

Ok, I looked. As I claimed, your 'counterproof' is erroneous; you made a conceptual error thus goofed the eqs and thus the possible alternate solutions. A hint for you: Read carefully what Tom wrote; he describes it literally.


Can you now find your error?

[Dono,]

On Monday, August 14, 2017 at 4:44:33 PM UTC-7, rotchm wrote:
> On Monday, August 14, 2017 at 5:40:04 PM UTC-4, Dono, wrote:
> > On Monday, August 14, 2017 at 2:37:30 PM UTC-7, rotchm wrote:
>
> > < I wont even look at your 'counterproof' since I'm sure thats its riddled with misconceptions and errors, as all ur proofs are .
> >
> > You should look because it proves that you are an imbecile
>
> Ok, I looked. As I claimed, your 'counterproof' is erroneous;

The only error is in your claims. You are an imbecile, Stephane, live with it.

[rotchm]

On Monday, August 14, 2017 at 8:01:20 PM UTC-4, Dono, wrote:
> On Monday, August 14, 2017 at 4:44:33 PM UTC-7, rotchm wrote:

> > Ok, I looked. As I claimed, your 'counterproof' is erroneous;
>
>

> The only error is in your claims. You are an imbecile, mike, live with it.

Idiot dono, you cant even set up the correct equation, let alone solve it.
You are a failure, and you realize it. You just cant admit it and are too dumb to do anything about it. You really need psychiatric help ....

[danco...@gmail.com]

On Monday, August 14, 2017 at 3:39:58 PM UTC-7, Dono, wrote:
> In the frame of the track the train has speed u. In any
> frame moving with speed x wrt the frame of the track the
> train has speed u'_1=\frac{u+x}{1+ux/c^2} and the track

> has speed u'_2=x.

Right, assuming the positive sense of x is defined to be in the opposite direction of u.

> u'_1=u'_2 if and only if x=c.

True, but that's not the relevant case. The question is not to find x that gives the same velocity to track and train, but the same speed. So we set u'_1 = -u'_2, meaning the track and train are moving at equal speeds in opposite directions in the x frame. This gives the solution

x = -(c^2/u)[1 - sqrt(1 - (u/c)^2)]

For example, with u = 0.5c this gives x = -0.2679c, which is negative because you defined the positive sense of x in the opposite direction of u. The x frame is roughly about halfway between track frame and train frame, as we would expect.

More generally, if v and w are the speeds of track and train (same direction), the mid speed is given by the (negative of) the above formula, with u replaced by the relativistic composition of v and w.

[Dono,]

On Monday, August 14, 2017 at 5:15:32 PM UTC-7, rotchm wrote:
> On Monday, August 14, 2017 at 8:01:20 PM UTC-4, Dono, wrote:
> > On Monday, August 14, 2017 at 4:44:33 PM UTC-7, rotchm wrote:
>
> > > Ok, I looked. As I claimed, your 'counterproof' is erroneous;
> >
> >
> > The only error is in your claims. You are an imbecile, mike, live with it.
>

> you cant even set up the correct equation, let alone solve it.

The equations are correct, take another spoonful of shit , Stephane
.

[rotchm]

On Monday, August 14, 2017 at 8:17:53 PM UTC-4, danco...@gmail.com wrote:

> The question is not to find x that gives the same velocity to
> track and train, but the same speed.

And there you have it idiot dono. Someone explicitly had to give you the answer. Do you now see your mistake and did you learn from it. Are you now ready & man enough to excuse yourself to us all in this topic?

[rotchm]

No they are not. Didnt you read, or understand danco's reply?

[Dono,]

On Monday, August 14, 2017 at 5:17:53 PM UTC-7, danco...@gmail.com wrote:
> On Monday, August 14, 2017 at 3:39:58 PM UTC-7, Dono, wrote:
> > In the frame of the track the train has speed u. In any
> > frame moving with speed x wrt the frame of the track the
> > train has speed u'_1=\frac{u+x}{1+ux/c^2} and the track
> > has speed u'_2=x.
>
> Right, assuming the positive sense of x is defined to be in the opposite direction of u.
>
> > u'_1=u'_2 if and only if x=c.
>
> True, but that's not the relevant case. The question is not to find x that gives the same velocity to track and train, but the same speed.

u_1, u'_1, u_2, u'_2 are all SPEEDS. Nice try.

> So we set u'_1 = -u'_2, meaning the track and train are moving at equal speeds in opposite directions in the x frame.

This is not what the Stephane Baune cretin says in is problem statement, he say "same speed". Nice try, you fail.

[Dono,]

I read, he's wrong, you wrote an imbecility in your problem statement. Live with it, Stephane.

[Dono,]

On Monday, August 14, 2017 at 5:30:36 PM UTC-7, rotchm wrote:
> On Monday, August 14, 2017 at 8:17:53 PM UTC-4, danco...@gmail.com wrote:
>
> > The question is not to find x that gives the same velocity to
> > track and train, but the same speed.
>

> Someone explicitly had to give you the answer.

His answer is wrong, since he has not noticed the idiocy in your problem statement and changed it to a different one. Take another helping of shit, Stephane.

[rotchm]

On Monday, August 14, 2017 at 8:33:31 PM UTC-4, Dono, wrote:
> On Monday, August 14, 2017 at 5:17:53 PM UTC-7, danco...@gmail.com wrote:

> > True, but that's not the relevant case. The question is not to
> > find x that gives the same velocity to track and train, but the
> > same speed.
>
> u_1, u'_1, u_2, u'_2 are all SPEEDS. Nice try.

AMAZING how super convoluted you are. You just keep e'm coming dont you!
In the LT's, and the speed compo formulas, the u's etc are NOT speeds, but velocities. We wish to solve for x when the *speed* wrt your F' are the same.

> > So we set u'_1 = -u'_2, meaning the track and train are moving at
> > equal speeds in opposite directions in the x frame.
>

> This is not what the Mike cretin says in is problem statement,

> he say "same speed". Nice try, you fail.

Exactly, SAME SPEED we are seeking, not same velocity. Damn you are an infinite dunce!!! Your 'counterproof' is set up to find x such that the velocities are the same. But we want the *speeds* to be the same.
Understand the distinction?

[Dono,]

On Monday, August 14, 2017 at 5:42:18 PM UTC-7, rotchm wrote:
>
> In the LT's, and the speed compo formulas, the u's etc are NOT speeds,but velocities

LOL

Keep it up, Stephane, insert foot into mouth. Deeper.

[rotchm]

On Monday, August 14, 2017 at 8:36:39 PM UTC-4, Dono, wrote:
> On Monday, August 14, 2017 at 5:30:36 PM UTC-7, rotchm wrote:

> > Someone explicitly had to give you the answer.
>
> His answer is wrong

Of course you would say that...you are so clueless.

>since he has not noticed the idiocy in your problem statement
> and changed it to a different one.

A lie. He, nor I, changed the problem statement.
So you result to direct lies to try to support your claim.

> Take another helping of shit....

WoW! Spoken like a true scientist!

[rotchm]

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/veltran.html
"Velocities must transform..."

https://en.wikipedia.org/wiki/Velocity-addition_formula
"The addition law is also called a composition law for *velocities*. "

Did you ever read any books on physics or relativity?

[Dono,]

On Monday, August 14, 2017 at 5:46:50 PM UTC-7, rotchm wrote:
> On Monday, August 14, 2017 at 8:36:39 PM UTC-4, Dono, wrote:
>
> A lie. He, nor I, changed the problem statement.

danco did, without being aware of doings so. An embarrassing failure on his side.

> > Take another helping of shit....
>
> WoW! Spoken like a true scientist!

In your case, the shoe fits, Stephame. have another helping.

[Dono,]

On Monday, August 14, 2017 at 5:51:47 PM UTC-7, rotchm wrote:
> On Monday, August 14, 2017 at 8:45:17 PM UTC-4, Dono, wrote:
> > On Monday, August 14, 2017 at 5:42:18 PM UTC-7, rotchm wrote:
> > >
> > > In the LT's, and the speed compo formulas, the u's etc are NOT speeds,but velocities
> >
> >
> > LOL
> >
> > Keep it up, Stephane, insert foot into mouth. Deeper.
>

Keep inserting foot. Deeper, Stephane.

[rotchm]

On Monday, August 14, 2017 at 8:59:45 PM UTC-4, Dono, wrote:
> On Monday, August 14, 2017 at 5:46:50 PM UTC-7, rotchm wrote:
> > On Monday, August 14, 2017 at 8:36:39 PM UTC-4, Dono, wrote:
> >
> > A lie. He, nor I, changed the problem statement.
>
> danco did, without being aware of doings so.

No he didnt. You obviously failed to understand his simple reply. He even gave a numerical example that you seem to either not understand, or fear.

> Stephame.

Foaming? Cant control your rage?

[Dono,]

On Monday, August 14, 2017 at 7:54:21 PM UTC-7, rotchm wrote:
> On Monday, August 14, 2017 at 8:59:45 PM UTC-4, Dono, wrote:
> > On Monday, August 14, 2017 at 5:46:50 PM UTC-7, rotchm wrote:
> > > On Monday, August 14, 2017 at 8:36:39 PM UTC-4, Dono, wrote:
> > >
> > > A lie. He, nor I, changed the problem statement.
> >
> > danco did, without being aware of doings so.
>
> No he didnt.

Actually, he did. Most likely he realized his mistake , unlike you. You are too much of an idiot to understand it and too much of a poser to admit it.

>
> > Stephame.
>
> Foaming? Cant control your rage?

Calling you Stephane Baune is rage? You ARE Stephane Baune.

[Ross A. Finlayson]

On Monday, August 14, 2017 at 11:51:36 AM UTC-7, tjrob137 wrote:

> On 8/14/17 12:41 PM, rotchm wrote:
> > In a recent thread about the usual train moving along a track scenario, I claimed that:
> > " ...there exist an inertial frame such that the train & track have the same speed."
> > I wonder if this is true. Any inputs?
>

> In the context of SR, for any pair of timelike objects moving inertially, it is
> always possible to find an inertial frame relative to which they have
> 3-velocities with equal magnitudes and opposite directions. So relative to this
> frame they have equal speeds.
>
> Proof: Sum their 4-velocities; the result will be timelike.
> The inertial frame in question is the rest frame of the sum.
> This uses two well-known theorems: a) the sum of two timelike
> 4-vectors is timelike, and b) every timelike 4-vector has a
> rest frame in which its spatial components are zero.
>
> That frame is unique up to spatial rotations, unless the two objects are both at
> rest in some inertial frame -- then they have equal speeds relative to any
> inertial frame.
>
> Tom Roberts

This is educating me.

[Prudence Oppenheimer]

We don't even know what it is. This rotchm is not even quoting things
proving him an imbecile.

[Odd Bodkin]

On 8/14/17 12:41 PM, rotchm wrote:
> In a recent thread about the usual train moving along a track scenario, I claimed that:
>
> " ...there exist an inertial frame such that the train & track have the same speed."
>
> I wonder if this is true. Any inputs?
>
>

> Someone (I dono who) claims otherwise, and I quote them:

> " The train and the track are in relative motion, so there is no frame "such that the train & track have the same speed.""
>

The easiest way to catch a carp is to lower a hook that has a plastic
wiggly thing (speed) that might confuse the carp looking for a worm
(velocity.)

But the real question is, why would you want to catch a carp? They're
bottom-feeder crap fish.

--
Odd Bodkin -- maker of fine toys, tools, tables

[Tom Roberts]

On 8/14/17 4:12 PM, Dono, wrote:
> On Monday, August 14, 2017 at 11:51:36 AM UTC-7, tjrob137 wrote:

>> On 8/14/17 12:41 PM, rotchm wrote:
>>> In a recent thread about the usual train moving along a track scenario, I claimed that:
>>> " ...there exist an inertial frame such that the train & track have the same speed."
>>> I wonder if this is true. Any inputs?
>>

>> In the context of SR, for any pair of timelike objects moving inertially, it is
>> always possible to find an inertial frame relative to which they have
>> 3-velocities with equal magnitudes and opposite directions. So relative to this
>> frame they have equal speeds.
>>
>> Proof: Sum their 4-velocities; the result will be timelike.
>> The inertial frame in question is the rest frame of the sum.
>> This uses two well-known theorems: a) the sum of two timelike
>> 4-vectors is timelike, and b) every timelike 4-vector has a
>> rest frame in which its spatial components are zero.
>>
>> That frame is unique up to spatial rotations, unless the two objects are both at
>> rest in some inertial frame -- then they have equal speeds relative to any
>> inertial frame.
>

> Here is the counterproof:

It is not a counterproof, it is a mistake on YOUR part (actually two quite
different mistakes plus an omission).

> In a given frame F the two objects have speeds u_1 \ne u_2
> In any other frame F' moving with speed x wrt frame F the speeds are:
> u'_1=\frac{u_1+x}{1+x*u_1/c^2}
> u'_2=\frac{u_2+x}{1+x*u_2/c^2}
> The only value of x that makes u'_1=u'_2 is x=c

MISTAKE #1: You solved the wrong problem by confusing velocity with speed. The
formulas you give are for VELOCITIES along a single axis, not speed.

MISTAKE #2: You made a sign error in x, because we want velocities RELATIVE TO
F', given velocities relative to F. So we must subtract x from u_1 and u_2.

OMISSION: You didn't realize that x=-c is also a solution of your problem.

The statement was about SPEED, which along a single axis is the absolute value
of the velocity along that axis. Equating the SPEEDS RELATIVE TO F':

|(u_1-x)/(1-x*u_1/c^2)| = |(u_2-x)/(1-x*u_2/c^2)| (1)

The absolute values are difficult to handle algebraically, so it is simplest to
enumerate the possible signs in (1):

(u_1-x)/(1-x*u_1/c^2) = (u_2-x)/(1-x*u_2/c^2) (1a)
(u_1-x)/(1-x*u_1/c^2) = -(u_2-x)/(1-x*u_2/c^2) (1b)
-(u_1-x)/(1-x*u_1/c^2) = (u_2-x)/(1-x*u_2/c^2) (1c)
-(u_1-x)/(1-x*u_1/c^2) = -(u_2-x)/(1-x*u_2/c^2) (1d)

Clearly 1a and 1d yield the same values of x as your calculation: x=c and x=-c,
neither of which is valid. But 1b and 1c give a valid value of x:

x = (c^2+u_1*u_2-Sqrt[(c^2-u_1^2)*(c^2-u_2^2)]) / (u_1+u_2)

As it is not obvious that is a valid velocity, set u_1=0.8*c and u_2=0.9*c, then
x=0.857921*c, and the speeds relative to F' are 0.184661*c. This satisfies a
plausibility test, as x is between u_1 and u_2, and the relative speed is
comparatively small. Moreover, it satisfies a numerical test: composing
0.857921*c with +-0.184661*c yields 0.9*c and 0.8*c, as expected.

I used Mathematica to solve (1) for x. There are 4 solutions,
three of which are unphysical with |x| >= c. As I stated,
there is exactly one valid solution as long as u_1 != u_2; if
u_1 = u_2, all values -c < x < c are valid solutions to (1).

My original proof, of course, is valid in (3+1) dimensions, not just along a
single axis. The algebra along a single axis is hairy enough (even using
Mathematica), and I have no interest in doing it for the full (3+1) dimensions.


Later you said to me:

> About half of what you post is , in the words of W. Pauli, "Not even wrong" .

This is not true. It's just that YOU keep making errors and attributing them to
me (and to others around here), as above.

Unlike you, I admit my errors, but this is clearly not one of them.

Will you admit your errors here?

(In related posts in this thread you adamantly do not.)

Tom Roberts

[rotchm]

On Tuesday, August 15, 2017 at 1:27:40 PM UTC-4, Odd Bodkin wrote:

> The easiest way to catch a carp is to lower a hook that has a plastic
> wiggly thing (speed) that might confuse the carp looking for a worm
> (velocity.)
>
> But the real question is, why would you want to catch a carp? They're
> bottom-feeder crap fish.

:)

I wonder if idiot dono understands your above "analogy"!

[rotchm]

On Tuesday, August 15, 2017 at 1:53:15 PM UTC-4, tjrob137 wrote:

<snip>

I have nothing to add to all that since I agree with it all.

[Dono,]

On Tuesday, August 15, 2017 at 10:53:15 AM UTC-7, tjrob137 wrote:
>
> MISTAKE #1: You solved the wrong problem by confusing velocity with speed. The
> formulas you give are for VELOCITIES along a single axis, not speed.
>

Tom,

You are an imbecile, the formulas are for speed.

> MISTAKE #2: You made a sign error in x, because we want velocities RELATIVE TO
> F', given velocities relative to F.

Tom,

You are the same old cretin, both x and u are speeds, measured along the x axis. You are incapable of reading a simple exercise.

> So we must subtract x from u_1 and u_2.
>

This is the highest cretinism, F moves with SPEED x away from the origin of F'. The track and the train move with speeds u_1 and u_2 away from the origin of F. So, the speeds in F' are exactly what I posted.

> OMISSION: You didn't realize that x=-c is also a solution of your problem.
>

Yes, I agree to that. How about you admitting to your imbecilities above.

> The statement was about SPEED,

Yep, and you were incapable of setting the simple equations correctly. You are getting senile in your old age.

>
>
> Later you said to me:
> > About half of what you post is , in the words of W. Pauli, "Not even wrong" .
>
> This is not true.

Yep, it is true, look at the garbage you posted. Admit that you goofed and go away.

> Unlike you, I admit my errors,

Then admit to the above and go away. You goofed.

[Dono,]

You are both equally stupid <shrug>

[Dono,]

On Tuesday, August 15, 2017 at 10:53:15 AM UTC-7, tjrob137 wrote:

<a large number of goofs>

I made it easier for you Tom , by attaching frame F to the track

1. In F, the train moves with SPEED u
2. Frame F moves with speed x wrt frame F'

So, in frame F'

3. The track moves with speed x
4. The train moves with speed \frac{u+x}{1+ux/c^2}

The two speeds cannot be equal for any valid value of x.

[rotchm]

On Tuesday, August 15, 2017 at 2:19:36 PM UTC-4, Dono, wrote:
> On Tuesday, August 15, 2017 at 10:53:15 AM UTC-7, tjrob137 wrote:
> >
> > MISTAKE #1: You solved the wrong problem by confusing velocity
> > with speed. The formulas you give are for VELOCITIES along a
> single axis, not speed.
>
> Tom,
> You are an imbecile, the formulas are for speed.

No they are not. They are for velocities. This is a major confusion on your part. Note that throughout the years, you have shown to be totally confused on the meaning and the applicability of the concepts & formulas. And you are doing that same mistake for the "speed composition formula" which should be better called (as wiki points out) "velocity addition formula".

<rest of idiot donos confusions snipped & ignored>

[Dono,]

On Tuesday, August 15, 2017 at 11:47:35 AM UTC-7, rotchm wrote:
> On Tuesday, August 15, 2017 at 2:19:36 PM UTC-4, Dono, wrote:
> > On Tuesday, August 15, 2017 at 10:53:15 AM UTC-7, tjrob137 wrote:
> > >
> > > MISTAKE #1: You solved the wrong problem by confusing velocity
> > > with speed. The formulas you give are for VELOCITIES along a
> > single axis, not speed.
> >
> > Tom,
> > You are an imbecile, the formulas are for speed.
>
> No they are not.

Well, you made it quite clear that you don't know what speed is, Stephane.


> They are for velocities.

For the simple exercise I posted, the formula applies to speed, imbecile.

[Dono,]

Sucking up to "rotchm" (Stephane Baune) will only heap embarrassments upon yourself. The guy is an imbecile and you are sliding dangerously close.

[rotchm]

Perhaps, but irrelevant. YOU posted your wrong formula for MY setup.
And by your nomenclature, SR velocity formula is as you wrote v = (x+u)/(1+xu) [c=1 for simplicity]. where all vars are id oriented (positive).
Now if u is in the other direction (as in you other setup) then the velocity is negative: u < 0. But the formula v = (x+u)/(1+xu) remains the same.
OR, write is as v = (x - |u|)/(1-x|u|) which is what you meant (but didnt write). And the *speed* of all that, is |v| = |(x - |u|)/(1-x|u|)|.

I suggest you look at Tom's algebraic walkthrough & try to UNDERSTAND what he did. It is clear & thorough. The idea is to solve |v1| = |v2| as he did.

[Odd Bodkin]

I think you have mistaken sucking up to rotchm with my asking him why
his trying so persistently to hook a bottom-feeder crap fish, when that
is not normally something a good fisherman wastes time on. You have
focused entirely too much on what this means about you, perhaps.

[Dono,]

On Tuesday, August 15, 2017 at 12:08:37 PM UTC-7, rotchm wrote:
> On Tuesday, August 15, 2017 at 2:53:09 PM UTC-4, Dono, wrote:
> > On Tuesday, August 15, 2017 at 11:47:35 AM UTC-7, rotchm wrote:
>
> > > They are for velocities.
> >
> > For the simple exercise I posted, the formula applies to speed,
>
> Perhaps, but irrelevant. YOU posted your wrong formula for MY setup.

Hope, it is the correct formula but you are too much of an ignorant to get it.

> And by your nomenclature, SR velocity formula is as you wrote v = (x+u)/(1+xu) [c=1 for simplicity]. where all vars are id oriented (positive).
> Now if u is in the other direction (as in you other setup) then the velocity is negative: u < 0.

Not true, if x>|u| then x+u>0 and v>0
You are an imbecile.
Either way, your claim in the OP is invalid. One counterexample is sufficient to disprove your claim.

[rotchm]

On Tuesday, August 15, 2017 at 2:55:56 PM UTC-4, Dono, wrote:
> On Tuesday, August 15, 2017 at 10:27:40 AM UTC-7, Odd Bodkin wrote:

> > Odd Bodkin -- maker of fine toys, tools, tables
>

> Sucking up to "rotchm" (Mike) will only heap embarrassments

> upon yourself. The guy is an imbecile and you are sliding
> dangerously close.

idiot dono, aren't you able to have a respectful & sensible dialog?
With me, I dont mind, since I play it too. But Odd and especially Tom are very polite with you and yet you insult & potty mouth them too. Aren't you able to accept criticism? If you cant, then you should seek psychiatric help or something, because you are not well... You might not see it that way, but just in case, you should regularly consult a MD.

And another thing you probably didn't realize: if everyone disagrees with you, doesn't that ring a bell in your head and that maybe, maybe your mind is playing tricks on you!? How would you discover if that's the case? Consulting a shrink could help you discover that too btw. Just for preventive sake, do consult...

[Dono,]

On Monday, August 14, 2017 at 5:17:53 PM UTC-7, danco...@gmail.com wrote:
> So we set u'_1 = -u'_2,

Therein lies your (repeated) mistake. Speeds are positive entities, by definition, so, you cannot write the above, You keep comingling speeds and velocities.

[Dono,]

On Tuesday, August 15, 2017 at 12:16:01 PM UTC-7, rotchm wrote:
> On Tuesday, August 15, 2017 at 2:55:56 PM UTC-4, Dono, wrote:
> > On Tuesday, August 15, 2017 at 10:27:40 AM UTC-7, Odd Bodkin wrote:
>
> > > Odd Bodkin -- maker of fine toys, tools, tables
> >
> > Sucking up to "rotchm" (Mike) will only heap embarrassments
> > upon yourself. The guy is an imbecile and you are sliding
> > dangerously close.
>

> aren't you able to have a respectful & sensible dialog?

Why would I have a respectful dialog with the antirelativity crackpot named Stephane Baune? You are as stupid as Ken Shito, do you respect Ken Shito? I didn't think so.

[rotchm]

On Tuesday, August 15, 2017 at 3:14:49 PM UTC-4, Dono, wrote:

> Either way, your claim in the OP is invalid.

It is valid, as *I* and ALL others agree. You just totally misunderstand the setup, you misunderstand the meaning of the symbols used in SR and you just misunderstand the concepts of physics. But you can't accept that and you have no intention in learning physics, so there isn't much we can do for you at this point.

Yeah, perhaps I will let the crap fish go...it is after all the crappiest carp fish of them all.

[Dono,]

On Tuesday, August 15, 2017 at 12:23:42 PM UTC-7, rotchm wrote:
> On Tuesday, August 15, 2017 at 3:14:49 PM UTC-4, Dono, wrote:
>
> > Either way, your claim in the OP is invalid.
>
> It is valid, as *I* and ALL others agree.

Doesn't matter what YOU think (the others made YOUR mistakes even worse). You are a crank, Stephane. Live with it.

[rotchm]

On Tuesday, August 15, 2017 at 3:17:04 PM UTC-4, Dono, wrote:
> On Monday, August 14, 2017 at 5:17:53 PM UTC-7, danco...@gmail.com wrote:
> > So we set u'_1 = -u'_2,
>
> Therein lies your (repeated) mistake.

You mean yours...

> Speeds are positive entities, by definition,

Exactly what we have been telling you all along.

> You keep comingling speeds and velocities.

comingling? Foaming?

And as I & Tom have been telling you, YOU are confusing speeds with velocities. In SR's velocity composition formula, the vars are *velocities*.

In my OP, I essentially ask to find x such that the *speeds* are the same, that is, |v1| = |v2| and NOT v1 = v2 as you did. Tom explained this quite clearly to you already! Gees...

[Dono,]

On Monday, August 14, 2017 at 10:41:22 AM UTC-7, rotchm wrote:
> In a recent thread about the usual train moving along a track scenario, I claimed that:
>
> " ...there exist an inertial frame such that the train & track have the same speed. "
>
> I wonder if this is true. Any inputs?
>
>
> Someone (I dono who) claims otherwise, and I quote them:
> " The train and the track are in relative motion, so there is no frame "such that the train & track have the same speed." "

I made it easier for you, by attaching frame F to the track

1. In F, the train moves with SPEED u in the positive x direction
2. Frame F moves with speed x wrt frame F' in the positive x direction

So, in frame F':

3. The track moves with speed x in the positive x direction
4. The train moves with speed \frac{u+x}{1+ux/c^2} in the positive x direction

All above speeds are positive entities.

The two speeds cannot be equal for any valid value of x (unless you think x=\pm{c} is a legit frame speed).

[Dono,]

On Tuesday, August 15, 2017 at 12:27:52 PM UTC-7, rotchm wrote:
>
>
> And as I & Tom have been telling you, YOU are confusing speeds with velocities.

Nope, keep eating shit, Stephane.

> In SR's velocity composition formula, the vars are *velocities*.
>

For the simple counterexample that you are utterly unable to understand, the two
notions coincide. Have another helping.

> In my OP, I essentially ask to find x such that the *speeds* are the same, that is, |v1| = |v2| and NOT v1 = v2 as you did.

For v>0 |v| coincides with v. Idiot.

[Odd Bodkin]

rotchm <rot...@gmail.com> wrote:
>
> And another thing you probably didn't realize: if everyone disagrees with
> you, doesn't that ring a bell in your head and that maybe, maybe your
> mind is playing tricks on you!? How would you discover if that's the
> case? Consulting a shrink could help you discover that too btw. Just for
> preventive sake, do consult...
>

Just a thought here. For some people, a feud that has gone on as long as
the one you two have had begins to trump everything, and the content of the
conversation becomes secondary or even immaterial. After that point, the
symptom is that spectator participants are expected to polarize to one
"side" or the other. For at least one of you, possibly both, any comment is
viewed as, "Is this on my side or the other side?" without even
entertaining the prospect that the comment is not aligned along EITHER
side.

To be honest, rotchm, you have been shamelessly preying on both dono's ego
and escalating the feud, purely (it appears) for the fun of it. And
honestly, if there is a long back-and-forth chain between you two, I know
it's not really worth spending much time on, because the discussion is
usually 98% content-free. If you're interested in restoring more content to
your posts (and maybe you're having too much fun for whatever reason), then
perhaps taking a breather from poking Dino would help that.

I confess that I too am overly engaged with the nutjob Seto and the
anti-science troll Winn. In those cases, I try to point out why their ideas
are not scientific, no matter how much they pretend that they are. With
socially inept neurotics like Ken and Banerjee and Sefton, there is no hope
of rescue, but at least I can be some advocate for real science.


--

[rotchm]

On Tuesday, August 15, 2017 at 3:20:18 PM UTC-4, Dono, wrote:
> On Tuesday, August 15, 2017 at 12:16:01 PM UTC-7, rotchm wrote:
> > On Tuesday, August 15, 2017 at 2:55:56 PM UTC-4, Dono, wrote:
> > > On Tuesday, August 15, 2017 at 10:27:40 AM UTC-7, Odd Bodkin wrote:
> >
> > > > Odd Bodkin -- maker of fine toys, tools, tables
> > >
> > > Sucking up to "rotchm" (Mike) will only heap embarrassments
> > > upon yourself. The guy is an imbecile and you are sliding
> > > dangerously close.
> >
> > aren't you able to have a respectful & sensible dialog?
>
> Why would I have a respectful dialog with the antirelativity
> crackpot named Stephane Baune?

I dont care if you have dialogs with him... Not my concern.

But the question was, why do you foulmouth Tom & Bod? Just because they disagree with you, or agree with me, is not a reason, no?

> You are as stupid as Ken Shito,

Actually, as every one else here agrees, that it is YOU that is as dumb & confused as ken, even moreso.

> do you respect Ken Shito? I didn't think so.

Then you think wrong, because I do respect him. I never foul mouth him (unless he does first, since that's the game he wants to play).

[Dono,]

You are lying , Stephane. Again. As always.

[Dono,]

I think you got this wrong, "rotchm" (Stephane Baune) is a crank antirelativist who likes to masquerade as "mainstream" after being knocked upside the head many times. So, when he posts yet another one of his idiocies, he get his nose tweaked. That's all <shrug>

[Odd Bodkin]

Dono, please, you're only exemplifying what I was talking about. I'm not
part of your feud. Don't try to make me see things your way. I'm just
not going to engage that way.

[rotchm]

On Tuesday, August 15, 2017 at 3:35:03 PM UTC-4, Odd Bodkin wrote:


> To be honest, rotchm, you have been shamelessly preying on both dono's ego
> and escalating the feud, purely (it appears) for the fun of it.

:)

If the movie is good, I dont mind watching it a few times. If dono doesn't like the movie, why does he continue watching it over and over? Thats a weird behavior on his part.

> perhaps taking a breather from poking Dino would help that.

I do. If you noticed, I *never* engage nor poke him. It is he that always jumps on my legit posts and starts his stuff. Once he starts that, *then* I follow up to his game. Yes, this match has been going on for a week now!

> I confess that I too am overly engaged with the nutjob Seto and
> the anti-science troll Winn.

yes, I have noticed your plays too...

> In those cases, I try to point out why their ideas
> are not scientific,

And thats what I do for dono too in the first few posts. But he cant accept my constructive criticism and continues his foulmouth... So i then just "play the game" thenon. And whats interesting is that along the way he makes more and more blunders and I get to see better the root of his problems so as I might help him at a later time.

> there is no hope of rescue,

Indeed...it would seem that way.

[rotchm]

On Tuesday, August 15, 2017 at 3:29:22 PM UTC-4, Dono, wrote:

> I made it easier for you, by attaching frame F to the track
>
> 1. In F, the train moves with SPEED u in the positive x direction
> 2. Frame F moves with speed x wrt frame F' in the positive x direction
>
> So, in frame F':
>
> 3. The track moves with speed x in the positive x direction
> 4. The train moves with speed \frac{u+x}{1+ux/c^2} in the positive x direction
>
> All above speeds are positive entities.

Correct...

But what is the velocity of F' wrt F? Is it positive or negative?
What is then the *speed* of F' wrt F?

> The two speeds

STOP! The speeds of WHAT ?
Are you beginning to see where you goofed now!?

[Dono,]

On Tuesday, August 15, 2017 at 12:59:52 PM UTC-7, rotchm wrote:
>
> It is he that always jumps on my legit posts and starts his stuff.

But you posts are not "legit". They are imbecilities. Look at this thread.

[Dono,]

On Tuesday, August 15, 2017 at 1:18:56 PM UTC-7, rotchm wrote:
> On Tuesday, August 15, 2017 at 3:29:22 PM UTC-4, Dono, wrote:
>
> > I made it easier for you, by attaching frame F to the track
> >
> > 1. In F, the train moves with SPEED u in the positive x direction
> > 2. Frame F moves with speed x wrt frame F' in the positive x direction
> >
> > So, in frame F':
> >
> > 3. The track moves with speed x in the positive x direction
> > 4. The train moves with speed \frac{u+x}{1+ux/c^2} in the positive x direction
> >
> > All above speeds are positive entities.
>
> Correct...
>
> But what is the velocity of F' wrt F? Is it positive or negative?

-x. <0. What sort of imbecile are you?

> What is then the *speed* of F' wrt F?
>

x. Imbecile. Incurable.

[Prokaryotic Caspase Homolog]

Watching rotchm and Dono interact is rather like watching the "sport" of
bullbaiting. What's even worse, each of them considers himself to be the baiter
and the other to be the hapless bull. Same with so many of the participants on
these newsgroups. Knowing the truth doesn't make the show any less cruel to me.

I've justified my own participation in this debate in a manner similar to the
way that you have justified your interactions with Seto and Winn. But so much
of the time, I have to admit that it's really just an inhumane way of having fun
with clueless crackpots who believe that it is THEY who are having fun.

It's not an aspect of myself that I find very appealing.

[rotchm]

You're the imbecile since YOU are the one slandering.

Now, correct, the speed -x is negative. You dont see why I asked you this?
It means that not all speeds in the problem are positive; depends on which frame you are talking about. Now think carefully of what the question (the OP) is.

In your scenario ("all positive") you are *imposing* that F has speed > 0 wrt F'. But realize that this need not be the case. Your "imposition" is limiting (discarding) some possibilities.

Anw, if you havent understood with all our explanations, and Tom's succinct proof, then you will never get it :(

[Dono,]

On Tuesday, August 15, 2017 at 3:06:30 PM UTC-7, rotchm wrote:
>
> You're the imbecile since YOU are the one slandering.
>
> Now, correct, the speed -x is negative. You dont see why I asked you this?

Because you are a cretin who cannot follow a simple counterexample to your incorrect claim <shrug>

[Dono,]

For chuckles and grins (and for giving crackpot Stephane Baune and his cheerleaders yet another kick in the pants) let's allow for a change in the direction of motion for frame F:

> 1. In F, the train moves with SPEED u in the positive x direction

> 2. Frame F moves with speed x wrt frame F' in the NEGATIVE x direction
>
> So, in frame F':
>
> 3. The track moves with speed -x (note that -x>0)
> 4. The train moves with speed abs{\frac{u+x}{1+ux/c^2}} in the positive x direction

>
> All above speeds are positive entities.
>

Now, abs{\frac{u+x}{1+ux/c^2}}=\frac{\abs{u+x}}{1+ux/c^2}}

For u<-x abs{u+x}=-(u+x)

So, we have the equation:

-\frac{u+x}{1+ux/c^2}=-x

Which is exactly as the previous one:

\frac{u+x}{1+ux/c^2}=x

So, contrary to Stephane Baune idiotic claims, there is NO frame in "which the two speeds are equal".

Eat some more shit, Stephane.

[rotchm]

On Tuesday, August 15, 2017 at 6:49:00 PM UTC-4, Dono, wrote:

> Now, abs{\frac{u+x}{1+ux/c^2}}=\frac{\abs{u+x}}{1+ux/c^2}}
> For u<-x abs{u+x}=-(u+x)
> So, we have the equation:
> -\frac{u+x}{1+ux/c^2}=-x
>
> Which is exactly as the previous one:
>
> \frac{u+x}{1+ux/c^2}=x
>
> So, contrary to Stephane Baune idiotic claims, there is NO frame in "which the two speeds are equal".

Your comprehension of the variables and eqs are way off.
Lets start anew in a clean thread.... I will walk you through and perhaps we/you will see your error.

[Lofty Goat]

On Mon, 14 Aug 2017 10:41:20 -0700 (PDT), rotchm <rot...@gmail.com>

wrote:

>In a recent thread about the usual train moving along a track scenario, I claimed that:
>
>" ...there exist an inertial frame such that the train & track have the same speed. "

Sure, but not the same *velocity* unless the train is standing still on
the track.

I must have missed something: why this matters.

--
Goat

[Dono,]

On Tuesday, August 15, 2017 at 5:13:35 PM UTC-7, rotchm wrote:
> On Tuesday, August 15, 2017 at 6:49:00 PM UTC-4, Dono, wrote:
>
> > Now, abs{\frac{u+x}{1+ux/c^2}}=\frac{\abs{u+x}}{1+ux/c^2}}
> > For u<-x abs{u+x}=-(u+x)
> > So, we have the equation:
> > -\frac{u+x}{1+ux/c^2}=-x
> >
> > Which is exactly as the previous one:
> >
> > \frac{u+x}{1+ux/c^2}=x
> >
> > So, contrary to Stephane Baune idiotic claims, there is NO frame in "which the two speeds are equal".
>
> Your comprehension of the variables and eqs are way off.

Imbecile,

Every time you get stumped you start a new thread.

[Tom Roberts]

The subject is wrong, and this is really Dono compounding his errors.

On 8/15/17 8/15/17 1:19 PM, Dono, wrote:
> On Tuesday, August 15, 2017 at 10:53:15 AM UTC-7, tjrob137 wrote:
>> MISTAKE #1: You solved the wrong problem by confusing velocity with speed. The
>> formulas you give are for VELOCITIES along a single axis, not speed.
>
> the formulas are for speed.

Nope. LOOK IT UP, DON'T JUST RELY ON YOUR FAULTY MEMORY.

Hint: the basic formula is called "the Lorentz composition of
velocities".

Historically you are VERY BAD at remembering contexts for various formulas, and
this is just one more case of that. YOU ARE WRONG.

>> MISTAKE #2: You made a sign error in x, because we want velocities RELATIVE TO
>> F', given velocities relative to F.
>
> both x and u are speeds, measured along the x axis.

No, they are VELOCITIES. AGAIN, YOU ARE WRONG.

You claim that given a pair of timelike objects moving inertially along a single
axis with different velocities u_1 and u_2 relative to frame F, there is no
frame F' in which their speeds are equal.

That is an absurd claim. Have two friends drive along a
straight road at 20 MPH and 40 MPH. If you drive at 31 MPH
they will have speeds of 11 MPH and 9 MPH relative to you;
if you drive at 29 MPH they will have speeds of 9 MPH and
11 MPH relative to you. So CLEARLY there is some speed
between 29 and 31 MPH for which they will have equal speeds
relative to you. In Galilean relativity that speed is
30 MPH; in SR it is less than a part per billion higher.

So how do you explain the FACT that for these conditions in inertial frame F:
u_1 = 0.8*c
u_2 = 0.9*c
that the frame F' moving with velocity
x = 0.857921*c
sees those two objects moving with velocities
u'_1 = -0.184661*c
u'_2 = +0.184661*c
And thus they do indeed have the same speeds relative to F'.

The first step to wisdom is learning to recognize your own mistakes. You have
avoided that step for a long time.

[I'm done here, unless you come up with something sensible.]

Tom Roberts

[Dono,]

On Tuesday, August 15, 2017 at 7:38:50 PM UTC-7, tjrob137 wrote:
> The subject is wrong, and this is really Dono compounding his errors.
>
> On 8/15/17 8/15/17 1:19 PM, Dono, wrote:
> > On Tuesday, August 15, 2017 at 10:53:15 AM UTC-7, tjrob137 wrote:
> >> MISTAKE #1: You solved the wrong problem by confusing velocity with speed. The
> >> formulas you give are for VELOCITIES along a single axis, not speed.
> >
> > the formulas are for speed.
>
> Nope. LOOK IT UP, DON'T JUST RELY ON YOUR FAULTY MEMORY.
>
> Hint: the basic formula is called "the Lorentz composition of
> velocities".
>

For the PARTICULAR case in my counterexample, they apply to SPEED, narrow minded imbecile.

> Historically you are VERY BAD at remembering contexts for various formulas, and
> this is just one more case of that. YOU ARE WRONG.
>
> >> MISTAKE #2: You made a sign error in x, because we want velocities RELATIVE TO
> >> F', given velocities relative to F.
> >
> > both x and u are speeds, measured along the x axis.
>
> No, they are VELOCITIES. AGAIN, YOU ARE WRONG.
>

Repeating the same imbecility doesn't make it right , Tom.

> The first step to wisdom is learning to recognize your own mistakes. You have
> avoided that step for a long time.
>

Good advice, I suggest that you follow it.

> [I'm done here, unless you come up with something sensible.]
>

You'll come back, you can't take losing.

[rotchm]

On Tuesday, August 15, 2017 at 8:39:35 PM UTC-4, Lofty Goat wrote:
> On Mon, 14 Aug 2017 10:41:20 -0700 (PDT), rotchm

> >" ...there exist an inertial frame such that the train & track have the same speed. "
>
> Sure, but not the same *velocity* unless the train is standing still on
> the track.

Correct!!! For once the nym shifter says something smart & true! You even understand the physics better than idiot dono!

[Dono,]

On Tuesday, August 15, 2017 at 7:38:50 PM UTC-7, tjrob137 wrote:
>

> That is an absurd claim. Have two friends drive along a
> straight road at 20 MPH and 40 MPH. If you drive at 31 MPH
> they will have speeds of 11 MPH and 9 MPH relative to you;
> if you drive at 29 MPH they will have speeds of 9 MPH and
> 11 MPH relative to you. So CLEARLY there is some speed
> between 29 and 31 MPH for which they will have equal speeds
> relative to you. In Galilean relativity that speed is
> 30 MPH; in SR it is less than a part per billion higher.
>

Attempting to apply your Galilean intuition to SR is what took you into the ditch.


In SR the relative speed is a non-linear function of the relative speed between frames, x:

u_1=x, u_2=(u+x)/(1+ux)

v_{rel}=(u+x)/(1+ux)-x=u(1-x^2)/(1+ux)

In the above, u is the speed of the train in frame F and x is the speed of frame F wrt frame F'.

In Galilean kinematics the relative speed of the two cars is frame invariant:

v_{rel}=u_2-u_1=(u+x)-x=u

So, your Galiean intuition FAILED you.

[Dono,]

On Tuesday, August 15, 2017 at 8:13:23 PM UTC-7, rotchm wrote:
> On Tuesday, August 15, 2017 at 8:39:35 PM UTC-4, Lofty Goat wrote:
> > On Mon, 14 Aug 2017 10:41:20 -0700 (PDT), rotchm
>
> > >" ...there exist an inertial frame such that the train & track have the same speed. "
> >
> > Sure, but not the same *velocity* unless the train is standing still on
> > the track.
>
> Correct!!! For once the nym shifter says something smart & true!

He was just exposing your imbecility, Stephane.

[rotchm]

On Tuesday, August 15, 2017 at 11:14:04 PM UTC-4, Dono, wrote:
> On Tuesday, August 15, 2017 at 7:38:50 PM UTC-7, tjrob137 wrote:
> >
> > That is an absurd claim. Have two friends drive along a
> > straight road at 20 MPH and 40 MPH. If you drive at 31 MPH
> > they will have speeds of 11 MPH and 9 MPH relative to you;
> > if you drive at 29 MPH they will have speeds of 9 MPH and
> > 11 MPH relative to you. So CLEARLY there is some speed
> > between 29 and 31 MPH for which they will have equal speeds
> > relative to you. In Galilean relativity that speed is
> > 30 MPH; in SR it is less than a part per billion higher.
> >
>
>
> Attempting to apply your Galilean intuition to SR is what took
> you into the ditch.

Its an identical reasoning. Just different algebra.

> In SR the relative speed is a non-linear function of the relative
> speed between frames, x: u_1=x, u_2=(u+x)/(1+ux)

So what!!! Tom's point is that its a *continuous* function of x.
Two cars travel in the same direction, one with speed 20 MPH and the other 40 MPH. You (say that you are between them) drive at 21 MPH. So one is going 1 mph (backwards) and the other 19 mph wrt you. You now move at 29 mph. One is going at 9 mph and the other at 11 mph. You now go at 31 mph: the one that was at 9 mph is now at 11 mph wrt you and the other driver is now at +9 mph wrt you. So, being a continuous situation/functions/formulas, somewhere along the line their speeds are equal wrt you. Imagine drawing the curves (the plot of the functions, the speeds wrt you), the curves cross each other since they are continuous (in Gal.rel, and SR). gees... <sigh> ...

[Dono,]

On Tuesday, August 15, 2017 at 8:36:35 PM UTC-7, rotchm wrote:
> On Tuesday, August 15, 2017 at 11:14:04 PM UTC-4, Dono, wrote:
> > On Tuesday, August 15, 2017 at 7:38:50 PM UTC-7, tjrob137 wrote:
> > >
> > > That is an absurd claim. Have two friends drive along a
> > > straight road at 20 MPH and 40 MPH. If you drive at 31 MPH
> > > they will have speeds of 11 MPH and 9 MPH relative to you;
> > > if you drive at 29 MPH they will have speeds of 9 MPH and
> > > 11 MPH relative to you. So CLEARLY there is some speed
> > > between 29 and 31 MPH for which they will have equal speeds
> > > relative to you. In Galilean relativity that speed is
> > > 30 MPH; in SR it is less than a part per billion higher.
> > >
> >
> >
> > Attempting to apply your Galilean intuition to SR is what took
> > you into the ditch.
>
> Its an identical reasoning. Just different algebra.
>
> > In SR the relative speed is a non-linear function of the relative
> > speed between frames, x: u_1=x, u_2=(u+x)/(1+ux)
>
> So what!!!

It is another way of exposing your imbecility. <shrug>

[JanPB]

In other words, "I have no answer".

--
Jan

[rotchm]

I see you cowardly snipped the part that stumped you again. Here it is once more: So what!!! Tom's point is that its a *continuous* function of x.

Stumped?

[Dono,]

Not at all, imbecile. The fact that the function is continuous is totally irrelevant, what you need is to find out when do the two objects have the "same speed". That happens when their relative speed is 0. That happens when:

u (1-x^2)/(1+ux)=0

meaning x=\pm{1}

Cretin.

[rotchm]

On Wednesday, August 16, 2017 at 1:17:57 AM UTC-4, Dono, wrote:

> Not at all, imbecile. The fact that the function is continuous is
> totally irrelevant,

Nope. It proves that there exist a non trivial solution. Tom's explicit solution (& numerical values) is proof that there is a solution.
I shall continue this in the other cleaner thread if need be.

[Dono,]

On Wednesday, August 16, 2017 at 7:00:30 AM UTC-7, rotchm wrote:
> On Wednesday, August 16, 2017 at 1:17:57 AM UTC-4, Dono, wrote:
>
> > Not at all, imbecile. The fact that the function is continuous is
> > totally irrelevant,
>
> Nope. It proves that there exist a non trivial solution.

Not for u+x<0

Tom's explicit solution (& numerical values) is proof that there is a solution.
> I shall continue this in the other cleaner thread if need be.

Of course you will, you only opened 7 threads on this subject. Every time you get nailed in one thread , you open another one, in the hope of erasing the fact that you just got nailed.

[Prudence Oppenheimer]

Tom Roberts wrote:

> The subject is wrong, and this is really Dono compounding his errors.
>>

>> the formulas are for speed.
>
> Nope. LOOK IT UP, DON'T JUST RELY ON YOUR FAULTY MEMORY.
> Hint: the basic formula is called "the Lorentz composition of
> velocities".

I beg you to reconsider. A velocity would give a force/accl whereas speeds
are not giving it. Hence SR compositions (and Lorentz) must be around
speeds, NOT velocities. Dr. Dono has to be consistent. Or perhaps,
unfortunately, you forgot your theory.

[Prudence Oppenheimer]

JanPB wrote:

>> > [I'm done here, unless you come up with something sensible.]
>> >
>> You'll come back, you can't take losing.
>
> In other words, "I have no answer".

SR is not about velocities, but speeds. Velocities would evt give a force/
accl, which in turn, would satisfy SR concerns. You forgot your theory.

[Prudence Oppenheimer]

"would NOT satisfy the demands of SR"

[Dono,]

I posted the counterexample, imbecile. Why do you feel compelled to butt in?

[Larry Harson]

On Monday, August 14, 2017 at 10:12:44 PM UTC+1, Dono, wrote:
> On Monday, August 14, 2017 at 11:51:36 AM UTC-7, tjrob137 wrote:

> > On 8/14/17 12:41 PM, rotchm wrote:
> > > In a recent thread about the usual train moving along a track scenario, I claimed that:

> > > " ...there exist an inertial frame such that the train & track have the same speed."

> > > I wonder if this is true. Any inputs?
> >

> > In the context of SR, for any pair of timelike objects moving inertially, it is
> > always possible to find an inertial frame relative to which they have
> > 3-velocities with equal magnitudes and opposite directions. So relative to this
> > frame they have equal speeds.
> >
> > Proof: Sum their 4-velocities; the result will be timelike.
> > The inertial frame in question is the rest frame of the sum.
> > This uses two well-known theorems: a) the sum of two timelike
> > 4-vectors is timelike, and b) every timelike 4-vector has a
> > rest frame in which its spatial components are zero.
> >
> > That frame is unique up to spatial rotations, unless the two objects are both at
> > rest in some inertial frame -- then they have equal speeds relative to any
> > inertial frame.
> >
> > Tom Roberts
>
> You have reached a new low, Tom
>
> Here is the counterproof:
>
> In a given frame F the two objects have speeds u_1 \ne u_2
> In any other frame F' moving with speed x wrt frame F the speeds are:
>
> u'_1=\frac{u_1+x}{1+x*u_1/c^2}
> u'_2=\frac{u_2+x}{1+x*u_2/c^2}
>
> The only value of x that makes u'_1=u'_2 is x=c
> You are just as stupid as Stephane Baune.

That velocity composition formula restricts F' to moving in the negative x-axis direction. So you're right with your counter example where u_1 and u_2 are moving in the positive x-axis direction, giving speed x = c along -x-axis as the only solution. But Rotchm was considering any F'.

Try instead:

u'=\frac{u_1 - x}{1 - x*u_1/c^2}

where x is the *velocity* of F' in F along the x-axis.

You'll find there's always a solution for any u_1, u_2. Tom's argument is far more elegant and to the point, so you don't even have to bother with messy 3-vector manipulation.

Larry Harson

[Dono,]

You get the same problem:

|\frac{u-x}{1-ux/c^2}|=|x|

Consider x>0 and u-x<0, the above becomes:

\frac{x-u}{1-ux/c^2}=x

with the only solution x=\pm{c}

Tom's solutions suffer from the same exact issue, 4-vectors are not any different than 3-vectors in terms of UNDERLYING PHYSICS (they cannot produce solutions that disagree).

[Larry Harson]

On Tuesday, August 15, 2017 at 8:35:03 PM UTC+1, Odd Bodkin wrote:
> rotchm <rot...@gmail.com> wrote:
> >
> > And another thing you probably didn't realize: if everyone disagrees with
> > you, doesn't that ring a bell in your head and that maybe, maybe your
> > mind is playing tricks on you!? How would you discover if that's the
> > case? Consulting a shrink could help you discover that too btw. Just for
> > preventive sake, do consult...
> >
>
> Just a thought here. For some people, a feud that has gone on as long as
> the one you two have had begins to trump everything, and the content of the
> conversation becomes secondary or even immaterial. After that point, the
> symptom is that spectator participants are expected to polarize to one
> "side" or the other. For at least one of you, possibly both, any comment is
> viewed as, "Is this on my side or the other side?" without even
> entertaining the prospect that the comment is not aligned along EITHER
> side.
>
> To be honest, rotchm, you have been shamelessly preying on both dono's ego
> and escalating the feud, purely (it appears) for the fun of it. And
> honestly, if there is a long back-and-forth chain between you two, I know
> it's not really worth spending much time on, because the discussion is
> usually 98% content-free. If you're interested in restoring more content to
> your posts (and maybe you're having too much fun for whatever reason), then
> perhaps taking a breather from poking Dino would help that.
>
> I confess that I too am overly engaged with the nutjob Seto and the
> anti-science troll Winn. In those cases, I try to point out why their ideas
> are not scientific, no matter how much they pretend that they are. With
> socially inept neurotics like Ken and Banerjee and Sefton, there is no hope
> of rescue, but at least I can be some advocate for real science.
>
>
> --
> Odd Bodkin -- maker of fine toys, tools, tables

"I confess that I too am overly engaged with the nutjob Seto" which doesn't help things, does it?

I'm under the impression that everyone here is an adult with the power to decide who they want to engage with. We have a variety of personalities and characters that combine to contribute to the newsgroup in interesting ways if they decide to.

Rotchm and Dono might troll one another, but every now and again interesting points are raised in the threads that wouldn't be seen otherwise. For example, Tom and his elegant proof using 4-vectors on there always being a frame where the velocity of the train and track can sum to zero. As long as the conflict is contained within the newsgroup and has relevant content, then I don't see too much of a problem.

Larry Harson

[rotchm]

Dono, how can you claim that there are no non-trivial solutions, when you agree that my clear and step by step proof (in the other thread) is correct and its conclusion contradicts yours ?

Recall that I asked you to show where in my "Bolzano's " proof, that I erred. You never answered except for furious slanders! You provided no counterargument!

[Lofty Goat]

On Tue, 15 Aug 2017 20:13:21 -0700 (PDT), rotchm <rot...@gmail.com>
wrote:

What nym shifter? I posted using my real name for about fifteen years,
and I picked up a nym as a joke while discussing a science book for kids
on 'alt.pouting.sandwich', which I've used for the last decade or so.

(I thought it unique, then learned that it's the name of an English pub,
and is also the nym of a guy who posts on a rock-climbing board.)

I've been reading your posts for a long time. Seems like you're pretty
bright, but you get carried away too easily. Really, you've started
several threads and gone on foamily for several dozens of posts about
the difference between speed and velocity, the definition of functions
and so forth.

Take a pill. Go outside and play. Something. Flamewars, however
erudite, are a poor way of handling stress.

--
Goat

[rotchm]

On Wednesday, August 16, 2017 at 8:48:04 PM UTC-4, Lofty Goat wrote:


> I've been reading your posts for a long time. Seems like you're pretty
> bright,

Obviously!

> but you get carried away too easily.

Not at all. You must be mixing up threads.

> Really, you've started several threads

Of different topics.

> and gone on foamily for several dozens of posts about
> the difference between speed and velocity, the definition of functions
> and so forth.

I am patient with my patients.
Btw, I was discussing the definition of *equation* and not functions.
Just to see if you are worthy, does "f(x) = x²" satisfy the definition of *equation* ?

> Go outside and play.

Actually, I am outside, 60% of my replies are when I'm out.

> a poor way of handling stress.

You must be mistaken me for some else. I dont stress; the heated discussions I engage in are part of an experiment, a case study.

But try to remain on topic; you are invited in the discussions, but dont divert to unrelated subjects. The one at hand is in the OP:

"there exist an inertial frame such that the train & track have the same speed."

Agree or not? Why? proof?

[JanPB]

Because I feel like it.

--
Jan

[Dono,]

There are NO solutions for the 2D domain u+x<0

> Recall that I asked you to show where in my "Bolzano's " proof, that I erred.

I simply pointed out that your "Bolzano proof" is just hot air. This is a simple degree 2 equation. Stop puffing yourself up , Stephane, you are just the same old fraud.

[Prokaryotic Caspase Homolog]

On Wednesday, August 16, 2017 at 9:04:45 AM UTC-5, Dono, wrote:
> On Wednesday, August 16, 2017 at 7:00:30 AM UTC-7, rotchm wrote:

> > Nope. It proves that there exist a non trivial solution.
>
> Not for u+x<0

That is about as relevant as stating that there are no real solutions to
y = x^2 + 1
where x+y < 0

You have lost track of the problem setup and the meanings of the variables.

[Larry Harson]

You're considering the case where the LHS is positive so that
|lhs| = lhs which looks fine to me. You're then using the restriction x > 0 on the rhs so eliminating the other possibility x < 0:

\frac{x-u}{1-ux/c^2}|= -x

That is: F' moves at velocity x > 0 in F while u' moves at velocity -x in F'. But the speeds |x| and |-x|are still the same.

I agree with your solution for the case where u' moves with the same velocity as F'; but Rotchm was looking at speeds.

Larry Harson

[rotchm]

On Wednesday, August 16, 2017 at 10:05:22 PM UTC-4, Dono, wrote:

> > Recall that I asked you to show where in my "Bolzano's " proof,
> that I erred.
>
> I simply pointed out that your "Bolzano proof" is just hot air.

Thats not specifying which line or reasoning is wrong. It is just handwaving.

> This is a simple degree 2 equation.

Yes, and yet you failed miserably at solving it.
No matter the speed of the train, there is ALWAYS an inertial frame such that the train & track have the same speed, ALWAYS.

And using your nomenclature, the eqs |(u+x)/(1+ux)| = |x| ALWAYS has a non trivial solution for x, for any given |u| < 1.