Simplified crash problem of Special Relativity - contradiction?

[dsep...@austin.rr.com]

In this gedanken experiment, as measured in one inertial reference frame two particles never crash, while in another inertial reference frame these same two particles crash into each other. Please explain the error.

In inertial reference frame, F0, a long device is built that can accelerate particles at a constant acceleration rate of 3 meters / second**2 as long as the particle has a speed as measured in F0 of less than 90 percent of the speed of light. The acceleration can start at any point along the x-axis, and the acceleration of the particle will be a constant rate of 3 meters per second squared.

Inertial reference frame, F1, moves along the x-axis of F0 with a relative velocity of V = 0.8 c. Frame F1 also observes that the acceleration of the particle is not dependent on the x' coordinate where the acceleration starts.

As measured in frame F1, there are two particles traveling at a constant velocity in the device of F0, separated by a distance of 0.75 light-seconds as measured in F1, or using 3 * 10**8 meters/second as the speed of light, the two particles are separated by 2.25 * 10**8 meters. The two particles are not accelerating, they are traveling with a constant and equal velocity.

At time t' equal to zero, observers in frame F1 simultaneously start the acceleration of both of these particles in the accleration device of F0. Since the acceleration is measured to be independent of the x' coordinate where the acceleration starts, and both particles started their acceleration simultaneously as measured in F1, observers in frame F1 measure that the distance between the two particles is always equal to 0.75 light-seconds, or 2.25 * 10**8 meters. In frame F1 the two particles never crash.

Using the Lorentz transform, if the particles started their acceleration simultaneously in F1 and are separated by a distance of 0.75 light-seconds as measured in F1, then frame F0 will measured that the two accelerations did not start accelerating simultaneously. Instead, F0 measures that one particle started accelerating one second before the other particle started accelerating.
delta t = (0 - (0.8 c / c**2) * 0.75 light-seconds) / 0.6 = 1 second.
The distance between the starts of the two accelerations as measured in F0 is:
0.75 light-seconds / 0.6 = 1.25 light-seconds or 3.75 * 10**8 meters.

Now how long does it take a crash to occur when two particles with the same initial velocity which are separated by 3.75*10**8 meters accelerating at a constant rate of 3 meters / second**2 when their accelerations start one second apart?

A particle accelerating at a constant rate travels a distance of 0.5 * a * t**2 due to the acceleration. So the difference in distance traveled between these two particles if they are moving in the same direction and the particle that started its acceleration first is accelerating in the direction towards the other particle and the accelerations start one second apart is:
0.5 * a * (t+1)**2 - 0.5 * a * t**2
or
distance = 0.5 * a * (2*t + 1) = a * t + 0.5 * a

So in frame F0, with a = 3 meters /second**2, the two particles will crash into each other in
(3.75 *10**8) / 3 = 1.25 * 10 ** 8 seconds.

The change in the particles speed during that time interval is a*t or 3.75 * 10**8 meters per second. Now at this point some readers of this problem might conclude that since the speed of light is 3 * 10**8 meters per second, that the speed of light was exceeded. However, in any given inertial reference frame, an object can change its speed by just under 2 * c and not exceed the speed of light. If a particle for example is moving along the x-axis in the negative x direction at a speed close to c, and then starts accelerating in the positive x direction, the change in speed will almost reach 2 * c before Einstein's limit of c is met. The initial speeds of the two particles with equal initial speeds have zero effect on how fast the crash occurs once the acceleration starts, so if each particle has a very high speed in the negative x direction and then accelerates in the positive x direction, the crash of the two particles will occur before a particle reaches Einstein's limit.

So where is the error?

Thanks,
David Seppala
Bastrop TX

[Dirk Van de moortel]

Op 22-apr-2017 om 22:58 schreef dsep...@austin.rr.com:

> In this gedanken experiment, as measured in one inertial reference
> frame two particles never crash, while in another inertial reference
> frame these same two particles crash into each other. Please explain
> the error.

[snip attempt #745]

>
> So where is the error?

In failing to address the problem with events and coordinates.

>
> Thanks, David Seppala Bastrop TX
>

You are welcome, even after two decades of failed attempts.

Dirk Vdm

[dsep...@austin.rr.com]

> > So where is the error?

Dirk responded:

> In failing to address the problem with events and coordinates.

In my posting:
I specified two events, the starting of acceleration of two particles.
I specified the distance as measured in F1 that the two particles were separated when the acceleration of each particle starts.
I specified the that the times of the two events in starting the acceleration of each of the two particles as measured in F1 was simultaneous.
I specified that in F1 the acceleration is independent of the x' coordinate where the acceleration starts.

I specified that F0 and F1 have a relative velocity of 0.8c along the x-axis.
I used the Lorentz transform to determine the separation of the two events as measured in F0.
I used the Lorentz transform to determine the time difference of the starts of the two accelerations as measured in F0.
I stated the acceleration rate of each particle was a constant rate of 3 meters per second squared as measured in F0, as long the particles had a velocity of less than 0.9c as measured in F0.

I used the equations for constant rate of acceleration to show that the particles as measured in F0 crash before there maximum velocity exceeded 0.9c.

So what events and coordinates did I fail to address?

David Seppala
Bastrop TX

[JanPB]

On Saturday, April 22, 2017 at 1:58:25 PM UTC-7, dsep...@austin.rr.com wrote:
> In this gedanken experiment,

David, you've been doing EXACT SAME THING for more than 20 years now. Your
questions change slightly but they are always fundamentally the same. What
are you trying to accomplish?

--
Jan

[dsep...@austin.rr.com]

When one inertial reference frame observes that two particles never crash and another inertial reference frame observes that the same two particles crash as in this posting, I am trying to find out either where the error in my calculations is or why people like you and Dirk don't think its contradictory. Why not just point on which line in my calculation is incorrect? Dirk posted that I didn't address the parameters of the problem, but he hasn't posted which parameters I didn't address. Why don't you?

David Seppala
Bastrop TX

[Dirk Van de moortel]

Op 24-apr-2017 om 04:40 schreef dsep...@austin.rr.com:

Show the events with their coordinates in the two frames.
Show the equations of motion of your particles in the
two frames. Show your calculation of the coordinates of
an intersection event, if there is one.
Don't use numbers. Use variables. The only numbers that
should appear in your text, are 0 and 1. You may use 2 to
denote the square of some variable. Numbers are for kids
and Usenet crackpots. I stopped reading after your second
number.

Dirk Vdm

[JanPB]

On Monday, April 24, 2017 at 9:29:25 AM UTC-7, dsep...@austin.rr.com wrote:
> On Monday, April 24, 2017 at 11:08:14 AM UTC-5, JanPB wrote:
> > On Saturday, April 22, 2017 at 1:58:25 PM UTC-7, dsep...@austin.rr.com wrote:
> > > In this gedanken experiment,
> >
> > David, you've been doing EXACT SAME THING for more than 20 years now. Your
> > questions change slightly but they are always fundamentally the same. What
> > are you trying to accomplish?
> >> --
> > Jan
> When one inertial reference frame observes that two particles never crash and another inertial reference frame observes that the same two particles crash as in this posting, I am trying to find out either where the error in my calculations is or why people like you and Dirk don't think its contradictory.

That's fine, but you've been asking this for 20 years now. This sort of thing
has a final resolution _in principle_, just like Euclidean geometry does.

> Why not just point on which line in my calculation is incorrect?

Because it's a waste of my time, just like examining some complicated claim
of perpetual motion machine or a complicated angle trisection construction.
You know there is an error hidden somewhere but it would take a lot of time
to find it, AND the end result is of no value as such resolution practically
never yields any interesting insight.

Your problem is you endlessly focus on minutiae which by general theorems
can never yield a self-contradiction. Learn more about those general
theorems and forget the doodling.

--
Jan

[dsep...@austin.rr.com]

On Monday, April 24, 2017 at 2:08:53 PM UTC-5, JanPB wrote:
> On Monday, April 24, 2017 at 9:29:25 AM UTC-7, dsep...@austin.rr.com wrote:
> > On Monday, April 24, 2017 at 11:08:14 AM UTC-5, JanPB wrote:
> > > On Saturday, April 22, 2017 at 1:58:25 PM UTC-7, dsep...@austin.rr.com wrote:
> > > > In this gedanken experiment,
> > >
> > > David, you've been doing EXACT SAME THING for more than 20 years now. Your
> > > questions change slightly but they are always fundamentally the same. What
> > > are you trying to accomplish?
> > >> --
> > > Jan
> > When one inertial reference frame observes that two particles never crash and another inertial reference frame observes that the same two particles crash as in this posting, I am trying to find out either where the error in my calculations is or why people like you and Dirk don't think its contradictory.
>
> That's fine, but you've been asking this for 20 years now. This sort of thing
> has a final resolution _in principle_, just like Euclidean geometry does.
>
> > Why not just point on which line in my calculation is incorrect?
>

Jan replied:

"You know there is an error hidden somewhere but it would take a lot of time to find it"

I do not know of any hidden error somewhere in my calculations. Apparently since you think its a "hidden" error, you can't find it either.

I do know that one inertial reference frame cannot say two particles never crash while another inertial reference frame observes the crash of these same two particles.

David Seppala
Bastrop TX

[Odd Bodkin]

On 4/24/2017 2:56 PM, dsep...@austin.rr.com wrote:
> Jan replied:
> "You know there is an error hidden somewhere but it would take a lot of time to find it"
> I do not know of any hidden error somewhere in my calculations. Apparently since you think its a "hidden" error, you can't find it either.

Not able, or not willing to dig through the morass for you.

David, the point that has been made repeatedly to you is that you are
using rules of thumb like length contraction and time dilation formulas
that are specific applications derived from Lorentz transformations that
work in SPECIFIC CONDITIONS. Then you studiously find puzzles where you
those specific conditions DO NOT APPLY and then wonder why the length
contraction and time dilation formulas do not work.

The proper way to address this is to start with labeling specific events
and use the full Lorentz coordinate transforms. Then you will find that
there is no paradox AND you may also learn in that process why the
specific conditions do not apply that would allow you to analyze it
using length contraction and time dilation formulas. You seem highly
resistant to this, and insist instead that people explain to you why
those specific conditions do not apply here.

I'll give you an analog. In simple 2D parabolic trajectories, there is a
very nice formula called the range formula.
It looks like this:
x = (v^2/g)(sin(2T)), where v is the launch speed and T is the launch
angle above horizontal.
This is much easier than the usual set of simultaneous equations
x = x0 + v0*cos(T)*t
y = y0 + v0*sin(T)*t - gt^2/2.
The problem is, the range formula only works when y=y0, whereas the
simultaneous equation set works all the time. If you try to apply the
range formula in a case where the specific condition is not met, then
you are going to come up with the wrong answer and possibly
contradictory conclusions.

>
> I do know that one inertial reference frame cannot say two particles never crash while another inertial reference frame observes the
> crash of these same two particles.
>
> David Seppala
> Bastrop TX

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

[dsep...@austin.rr.com]

In my posting, in inertial reference frame F1, where accelerations are independent of their position along the x axis, F1 simultaneously starts the acceleration of two particles that are separated by some distance L that have identical initial velocities. Therefore as measured in F1, the distance between the two particles is always constant and equal to L.

Using the Lorentz transform with Einstein's concepts, inertial reference frame observers in F0 that has a relative velocity V along the x-axis with respect to inertial reference frame F1, observe that the accelerations do not start simultaneously because of the separation between the two particles. Therefore, the distance as measured in F0 between the two particles is always changing, if the acceleration is independent of the starting position along the x-axis.

With the numbers I used, I show that observers in F0 observe that the two particles can crash before the particles exceed the speed of light as measured in F0.

Why not show my error in this problem? No one has done that yet, instead they say the error is "hidden", or I used actual numbers, or as you did state here's an example of a different problem where the transforms were used incorrectly.

David Seppala
Bastrop TX

[Odd Bodkin]

On 4/24/2017 5:20 PM, dsep...@austin.rr.com wrote:
> In my posting, in inertial reference frame F1, where accelerations are independent of their position along
> the x axis, F1 simultaneously starts the acceleration of two particles that are separated by some distance
> L that have identical initial velocities. Therefore as measured in F1, the distance between the two particles
> is always constant and equal to L.
>
> Using the Lorentz transform with Einstein's concepts, inertial reference frame observers in F0 that has a
> relative velocity V along the x-axis with respect to inertial reference frame F1, observe that the accelerations
> do not start simultaneously because of the separation between the two particles. Therefore, the distance as
> measured in F0 between the two particles is always changing, if the acceleration is independent of the starting
> position along the x-axis.
>
> With the numbers I used, I show that observers in F0 observe that the two particles can crash before the particles
> exceed the speed of light as measured in F0.
>
> Why not show my error in this problem? No one has done that yet, instead they say the error is "hidden", or I used
> actual numbers, or as you did state here's an example of a different problem where the transforms were used incorrectly.

I've already told you the mistake. You have not specified the events
whose coordinates you wish to transform, and you did not use the full
Lorentz coordinate transformations. Instead, you used time dilation and
length contraction formula, which do not apply here. The reasons they do
not apply will become apparent once you use the full Lorentz coordinate
transformations on the event coordinates you wish to consider. When you
do it correctly, and compare it with your initial analysis using time
dilation and length contraction formula, you will see the difference.

[dsep...@austin.rr.com]

The first event (start of particle 1's acceleration) occurs as measured in F1 at x = 0, t = 0.
The second event (start of particle 2's acceleration) occurs as measured in F1 at x = 0.75 light-seconds, t = 0

The transforms I used to find the coordinates of those events in F0 were:

t' = gamma * (t - V*x/c**2)
and
x' = gamma * (x - V*t)

What transformations should I have used to find the coordinates in the F0 frame?

David Seppala
Bastrop TX

[Paul B. Andersen]

Den 22.04.2017 22.58, skrev dsep...@austin.rr.com:
>
> In inertial reference frame, F0, a long device is built that
> can accelerate particles at a constant acceleration rate of
> 3 meters / second**2
>
>

> So in frame F0, with a = 3 meters /second**2, . . .

>
> So where is the error?
>

Right above!

As always you confuse yourself by making a seemingly very complicated
scenario to hide your very elementary error.

I suspect you knew what your error was, though.


--
Paul

https://paulba.no/

[Himère Bezuinig]

Paul B. Andersen wrote:

> Den 22.04.2017 22.58, skrev dsep...@austin.rr.com:
>>
>> In inertial reference frame, F0, a long device is built that can
>> accelerate particles at a constant acceleration rate of 3 meters /
>> second**2 So in frame F0, with a = 3 meters /second**2, . . .
>> So where is the error?
>
> Right above! As always you confuse yourself by making a seemingly very
> complicated scenario to hide your very elementary error.

So true indeed. People often forget complicated scenarios are not allowed
in Relativity.

[Poutnik]

As the most people have problems even with simple scenarios,
it make no sense to bring for them anything complicated.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.

[rotchm]

On Monday, April 24, 2017 at 3:56:57 PM UTC-4, dsep...@austin.rr.com wrote:
> On Monday, April 24, 2017 at 2:08:53 PM UTC-5, JanPB wrote:

> Jan replied:
> "You know there is an error hidden somewhere but it would take a
> lot of time to find it"
> I do not know of any hidden error somewhere in my calculations.

Yes you do, so stop lying. You are just too "dumb" to find your error, so you want others to find it for you. Well, NO! We helped you for 20 years, and you havent learned anything. We thus consider you a lost cause and brush you off. We have more interesting people to respond to.

[rotchm]

On Monday, April 24, 2017 at 6:20:12 PM UTC-4, dsep...@austin.rr.com wrote:

> Why not show my error in this problem? No one has done that yet,

Tell you what. Paypal me 100$ and I will show you where your mistakes are.
We have been "giving" to you for the past 20 yrs and you havent been giving back. So its time for you to give out if you still want to receive. Its only fair!

[rotchm]

On Monday, April 24, 2017 at 7:22:25 PM UTC-4, dsep...@austin.rr.com wrote:

> The first event (start of particle 1's acceleration) occurs as
> measured in F1 at x = 0, t = 0.
> The second event (start of particle 2's acceleration) occurs as
> measured in F1 at x = 0.75 light-seconds, t = 0
>
> The transforms I used to find the coordinates of those events in F0 were:
>
> t' = gamma * (t - V*x/c**2)
> and
> x' = gamma * (x - V*t)

Ok (and also meaning that your F0 frame is moving to the right wrt F1, since you have "-V" where V > 0).

So, you specified two events. You havent specified the set of events generated by each particle: you haven specified their motion in F1. These motions have spacetime coordinates too, like x=x(t) & t = t. Once you specify their motions in F1, then transform them into F0 via the LT's.

[dsep...@austin.rr.com]

Paul

I stated in frame F0, the particles in the F0 device accelerate at a = 3 meters / second squared. Why is that an error?

David Seppala
Bastrop TX

[dsep...@austin.rr.com]

I stated in that in F0 the F0 device accelerates a particle at 3 meters /second squared independent of the position of the particle in the device, once the acceleration of a particle starts.

The acceleration motion as measured in F1 is irrelevant to the problem since each particle has an identical acceleration independent of its position along the x-axis and the acceleration of each particle as measured in F1 starts simultaneously. Therefore the acceleration as measured in F1 has zero effect on the distance between the two particles.

David Seppala
Bastrop TX

[rotchm]

On Wednesday, April 26, 2017 at 8:42:12 AM UTC-4, dsep...@austin.rr.com

> > You havent specified the set of events generated by each particle:

> I stated in that in F0 the F0 device accelerates a particle at 3
> meters /second squared independent of the position of the particle
> in the device, once the acceleration of a particle starts.

Ok. I say to you gain: You havent specified the set of events generated by each particles... Restate your above declaration by using the "events language".

And, are you going to paypal me 100$?

> The acceleration motion as measured in F1 is irrelevant to the problem

According to you, it is relevant. So, why are you contradicting yourself now? YOU declared that the particles crash (or not) in F1. If you can make such a declaration, it is because you did the analysis, the calculations, right?

[rotchm]

On Wednesday, April 26, 2017 at 8:37:01 AM UTC-4, dsep...@austin.rr.com

> I stated in frame F0, the particles in the F0 device accelerate at a = 3
> meters / second squared. Why is that an error?

Well, not an error a priori. But incomplete. *when* did they start accelerating? What is their initial position in F0 etc... ?

As I asked you, SPECIFY

[dsep...@austin.rr.com]

If you read the original posting I stated:
The relative velocity between F0 and F1 is 0.8c

As measured in inertial reference frame F1
The two particles are separated by a distance of 0.75 light-seconds
The acceleration of each particles starts simultaneously
The acceleration is identical, independent of the x coordinate where the acceleration starts.

As measured in reference frame F0
The acceleration of one particle starts one second before the acceleration of the second particle.
The separation of these two events is 1.25 light-seconds
The acceleration rate of each particle once it starts accelerating as measured in F0 is a constant 3 meters / second squared.

David Seppala
Bastrop TX

[rotchm]

On Wednesday, April 26, 2017 at 10:03:24 AM UTC-4, dsep...@austin.rr.com wrote:
> On Wednesday, April 26, 2017 at 8:12:58 AM UTC-5, rotchm wrote:


> If you read the original posting I stated:

<snip>

before we continue,, why are you not answering my questions?
Specifically, are you going to paypal me 100$ ?
Its just fair: I help you, you help me.

[Himère Bezuinig]

Poutnik wrote:

>>> Right above! As always you confuse yourself by making a seemingly very
>>> complicated scenario to hide your very elementary error.
>>
>> So true indeed. People often forget complicated scenarios are not
>> allowed in Relativity.
>
> As the most people have problems even with simple scenarios,
> it make no sense to bring for them anything complicated.

Why are you bitching me, por favor, you are not my father.

[dsep...@austin.rr.com]

So far, you have ignored the info in the original posting, so there is no way that I can tell if you are even going to help. So no paypal payment.

David Seppala
Bastrop TX

[rotchm]

On Wednesday, April 26, 2017 at 10:31:22 AM UTC-4, dsep...@austin.rr.com

> So far, you have ignored the info in the original posting,

A lie. For the past 20 yrs, *I* have been answering & helping you in many of your questions. Google kept a record.

> so there is no way that I can tell if you are even going to help.
> So no paypal payment.

But its your turn to pay up *first*, since I did help you in the past.

You cant just take from everyone and give out nothing.
So, be honest and pay up what you owe, then I will answer the problem of this thread.

[Poutnik]

"The goose, that has been shot,
makes the loudest honking."

[Paul B. Andersen]

Since you specify a constant coordinate acceleration in F0,
the proper acceleration must increase with time.

Since you assume that the distance between the particles will stay
the same in F1, you assume that the coordinate acceleration
of both particles are equal and constant in F1 as well.
That is wrong.


--
Paul

https://paulba.no/

[dsep...@austin.rr.com]

I stated that the acceleration along the x-axis as measured in frame F1 (and frame F0) was independent of the x coordinate where the acceleration starts. I did not say that the acceleration of the particles as measured in F1 is constant. I simply assert that the acceleration is the same for both particles as measured in F1, and therefore the distance between the two particles as measured in F1 never changes. As long as the acceleration pattern is identical for the two particles and the acceleration of each particle starts simultaneously, the distance between the two particles remains constant.

David Seppala
Bastrop TX

[rotchm]

On Wednesday, April 26, 2017 at 8:03:28 PM UTC-4, dsep...@austin.rr.com w

>
> I stated that the acceleration along the x-axis as measured in
> frame F1 (and frame F0) was independent of the x coordinate where

> the acceleration starts....

Since you cant describe your setup via the more comprehensive approach of spacetime coordinates & equations, I propose this to you: I will translate your initial conditions via the equations. For this, then you can paypal me 25$. Once I receive it, I will then start to go through the solution and show where you erred. Then you can give me the remaining 75$. Want to do it this way?

[dsep...@austin.rr.com]

No. Don't think you can identify the error. I exchange emails with various physics professors so when they find the error, they will explain the mistake.

David Seppala
Bastrop TX

[rotchm]

On Wednesday, April 26, 2017 at 10:24:08 PM UTC-4, dsep...@austin.rr.com

> No. Don't think you can identify the error.

Seriously... many of us here found *many* errors already.

> I exchange emails with various physics professors so when they
> find the error, they will explain the mistake.

But you will have to wait, if they do answer you. I, I offer you now, here, to walk you through, for a small fee, a few you owe already.

[JanPB]

Nobody will bother reading your prose because it's obvious there is a mistake in there
somewhere. It's just like with trisecting angles or perpetual motion machines - there is
no reson to waste time analysing such things.

--
Jan

[Paul B. Andersen]

Den 27.04.2017 02.03, skrev dsep...@austin.rr.com:
> On Wednesday, April 26, 2017 at 4:08:01 PM UTC-5, Paul B. Andersen wrote:
>>
>> Since you specify a constant coordinate acceleration in F0,
>> the proper acceleration must increase with time.
>>
>> Since you assume that the distance between the particles will stay
>> the same in F1, you assume that the coordinate acceleration
>> of both particles are equal and constant in F1 as well.
>> That is wrong.
>>
>>
>> --
>> Paul
>>
>> https://paulba.no/
>
> I stated that the acceleration along the x-axis as measured in
> frame F1 (and frame F0) was independent of the x coordinate where
> the acceleration starts. I did not say that the acceleration of
> the particles as measured in F1 is constant.
> I simply assert that the acceleration is the same for
> both particles as measured in F1, and therefore the distance
> between the two particles as measured in F1 never changes.

This assertion is wrong.

> As long as the acceleration pattern is identical for the two particles

It isn't . .

> and the acceleration of each particle starts simultaneously,
> the distance between the two particles remains constant.

.. so it doesn't


--
Paul

https://paulba.no/

[dsep...@austin.rr.com]

So you are saying that if inertial reference frame starts the identical acceleration of two particles simultaneously the distance between the two particles does not remain constant.

So if one particle is at x1 and the other identical particle is at x2 some distance L away as measured in F1, and their accelerations start simultaneously, which particle accelerates faster as measured in F1?

David Seppala
Bastrop TX

[dsep...@austin.rr.com]

One of the Princeton physics professors I exchange email with has already responded, but hasn't found the error yet. And unlike you hasn't asked for money. He's more interested in knowledge and education.

David Seppala
Bastrop TX

[rotchm]

On Thursday, April 27, 2017 at 8:30:27 AM UTC-4, dsep...@austin.rr.com

> So you are saying that if inertial reference frame starts the
> identical acceleration of two particles simultaneously the
> distance between the two particles does not remain constant.

No, that is not what he is saying... Seriously, you cant even understand what you read. No wonder you are so confused. Are you able to accept that you have a severe mental impairment ? If you can admit to the possibility, then what line of action would you take for yourself if it were the case? Then, just to be on the safe side, go get checked out by a professional(s). If they do indeed confirm that you are impaired, initiate your line of action to resolve your problem(s).

> So if one particle is at x1 and the other identical particle is at
> x2 some distance L away as measured in F1, and their accelerations
> start simultaneously, which particle accelerates faster as measured in F1?

Why dont you say all that with the appropriate equations? It would be much more clear & concise. Like, your above declaration is translated to:

x1''(t) = x2''(t), x1(0)=0, x1'(0)=0, x2(0)=L>0, x2'(0)=0.

One line & clear! With that clear line, you (can you?) can show that
x2(t) - x1(t) = L.

So, I suggest to you to start your post anew in a new thread, where you will use the much clearer language of equations. Then the math leading to the answer will directly show where you made your mistakes.

[Python]

Are you harassing him with the very same kind of question based on the
very same kind of errors for twenty years?

[dsep...@austin.rr.com]

My conclusion is the same as yours that x2(t) - x1(t) = L for frame F1 since the accelerations as measured in F1 start simultaneously.

Why not respond to Paul's posting that that "assertion is wrong"

David Seppala
Bastrop TX

[danco...@gmail.com]

On Saturday, April 22, 2017 at 1:58:25 PM UTC-7, dsep...@austin.rr.com wrote:
> Now at this point some readers of this problem might conclude
> that … the speed of light was exceeded. However… if each particle
> has a very high speed in the negative x direction and then
> accelerates in the positive x direction, the crash of the two
> particles will occur before a particle reaches Einstein's limit.

Use units with c=1. In terms of F0, let V denote the initial speed of the two particles, v the speed of F1, and let the particles be at (0,0) and (X,T) when their accelerations begin. Then in F0 the crash “occurs” at time (1 – vV + (a/2)Xv^2)/(av), and at this event the velocity of the two particles would be 1/v + (a/2)vX and 1/v – (a/2)vX, independent of V. Thus at least one of the particle velocities exceeds 1.

[dsep...@austin.rr.com]

Your equations show that the time of the crash depends on the initial speed V of the two particles, and that the velocity of each of the particles when the crash occurs is independent of their initial speed. Please explain why you think the time of the crash is dependent on the particles initial speed when both particles have the identical speed prior to the acceleration.

Also, please explain why you think the particle's speed when the crash occurs is independent of their initial velocity.

I thought things are the other way around. The time of the crash is not dependent on the initial speed of the particles since both particles have the same initial speed. I thought during an acceleration the particle's speed depends on the acceleration AND its initial speed. Please clarify.

David Seppala
Bastrop TX

[JanPB]

This comes with the territory, it's like a tax you have to pay if you're
a faculty member at the physics department: suffer the cranks.

--
Jan

[danco...@gmail.com]

On Thursday, April 27, 2017 Dave "Barnpole" Seppala wrote:
> > Use units with c=1. In terms of F0, let V denote the initial
> > speed of the two particles, v the speed of F1, and let the
> > particles be at (0,0) and (X,T) when their accelerations
> > begin. Then in F0 the crash “occurs” at time
> > (1 – vV + (a/2)Xv^2)/(av), and at this event the velocity
> > of the two particles would be 1/v + (a/2)vX and 1/v – (a/2)vX,
> > independent of V. Thus at least one of the particle velocities
> > exceeds 1.
>
> Your equations show that the time of the crash depends on the
> initial speed V of the two particles, and that the velocity of
> each of the particles when the crash occurs is independent of
> their initial speed.

Correct.

> Please explain why you think the time of the crash is dependent
> on the particles initial speed when both particles have the
> identical speed prior to the acceleration.

The accelerations start at different times for the particles in different locations, so their initial speed affects the time to collision. To see by how much, just write down the simple equations for the positions of the particles as a function of time, set the positions equal, and solve for the time. That's the time of collision, as quoted in the previous message.

> Also, please explain why you think the particle's speed
> when the crash occurs is independent of their initial
> velocity.

Given the simple equations for the positions of the particles as a function of time, differentiate to get the velocities as a function of time, and then set the time to the time of collision computed above. The result is the speeds quoted in the previous message, which are independent of the initial velocity.

This shows that the scenario you described (the particles colliding) is impossible, because it would involve at least one of the particles exceeding the speed of light.

[dsep...@austin.rr.com]

Let the acceleration rate of each particle be 2 meters per second squared.
Let the initial velocity of each particle be zero.
Let one particle be at x=0 and let the other particle be at x = 3 meters.
Let the particle at x= 0 start accelerating at time t=0.
Let the particle at x = 3 start accelerating at time t = 1 second.

At time t = 2 seconds, the first particle has traveled 4 meters.
At time t = 2 seconds, the second particle has only accelerated for 1 second, so it has traveled a distance of 1 meter. Since the particles were separated by 3 meters before the acceleration started, the crash occurs at x = 4 meters.
The speed of the first particle is 4 meters / second, the speed of the second particle when the crash occurs is 2 meters / second.

Now use the same coordinates but let the initial speed of each particle be 1000 meters / second. Again the acceleration of the first particle and second particle start one second apart.
After 1 second the first particle has traveled a distance of 1001 meters while the second particle has traveled a distance of 1000 meters.

After two seconds the first particle has traveled a distance of 2004 meters and the second particle has traveled a distance of 2001 meters. The two particles started 3 meters apart, so the crash occurs at t = 2 seconds. The speed of the first particle when the crash occurs is 1004 meters per second and the speed of the second particle when the crash occurs is 1001 meters per second.

As shown in these two simple examples the speed of the particles when the crash occurs depends on their initial speeds and the acceleration rate.

The time of the crash is independent of the particles initial velocity.
The difference in distance traveled of the two particles until the crash occurs is independent of the particles initial speeds since the speeds are equal.

You say just the opposite.

David Seppala
Bastrop TX

[danco...@gmail.com]

On Thursday, April 27, 2017 Dave "Barnpole" Seppala wrote:
> Let the particle at x= 0 start accelerating at time t=0.
> Let the particle at x = 3 start accelerating at time
> t = 1 second.

No, those conditions imply that F1 is moving at 10^8 times the speed of light relative to F0. Remember, you specified that the particles start accelerating at the same time in frame F1, which is moving at speed v relative to F0. In units with c=1, if the particles are at (0,0) and (X,T) in terms of F0 when they start accelerating, then T = vX. Your conditions are T=1 sec and X=10^-8 sec, which implies v = 10^8 (dimensionless fraction of the speed of light).

[dsep...@austin.rr.com]

I did not post those numbers as the numbers of the problem I am trying to solve. Those numbers were just ordinary everyday numbers to show you that when two particles accelerate at a constant rate but start at different times that the initial speeds of the particles if they are identical do not affect the time before a crash occurs, nor do they affect the distance traveled during that time. These ordinary everyday numbers also show that the speed the particles have when a crash occurs depends on the initial velocity of the particles.

You stated that the of the particles when the crash occurs does not depend on the initial velocity of the particles. You also stated that the time of the crash once the acceleration starts depends on the initial speed of the particles even though the speeds are identical. The simple everyday numbers of my previous post simply show that what you stated isn't true.

I did not intend to imply that those simple everyday numbers are related to the problem of my original posting other than to show that the time from the start of the acceleration to the time of the crash does not depend on the initial speeds of the particles if those speeds are identical, and the speed of the particles when a crash occurs does indeed depend on the initial speed of the particles.

Your posting and equations show the opposite which is false.

David Seppala
Bastrop TX

[mlwo...@wp.pl]

Your bunch of idiots has been harassing the world with the
same idiocies for more than 100 years already.

[danco...@gmail.com]

On Thursday, April 27, 2017 dsep...@austin.rr.com wrote:
> I did not post those numbers as the numbers of the problem I am trying to solve. Those numbers were just ordinary everyday numbers to show you that when two particles accelerate at a constant rate but start at different times that the initial speeds of the particles if they are identical do not affect the time before a crash occurs, nor do they affect the distance traveled during that time. These ordinary everyday numbers also show that the speed the particles have when a crash occurs depends on the initial velocity of the particles.

Your calculations for the case of initial speed of 1000 m/sec are wrong. Since you're now talking about arbitrary starting points, this doesn't have anything to do with relativity at all.

In general, if V is the initial speed, and the particles begin their constant acceleration "a" at (0,0) and (X,T) respectively, their positions (after they start to accelerate) are given by x = Vt + (a/2)t^2 and (x-X) = V(t-T) + (a/2)(t-T)^2, and these intersect at the time (X-VT+(a/2)T^2)/(aT), which depends on V. The speeds of the particles at this collision event are found by taking the derivatives of the positions and evaluating at the time of collision. This gives the speeds X/T + (a/2)T and X/T - (a/2)T, which do not depend on V.

With these equations you can see that your calculations for V=0 are correct, but your calculations for V=1000 m/sec are not. Notice that, with the initial speed V=1000 m/sec, the 1st particle is already ahead of the second when it starts to accelerate, and there is no collision in this case. (The parabolic paths intersect in the extrapolated region prior to the accelerations.)

[Paul B. Andersen]

Den 27.04.2017 14.30, skrev dsep...@austin.rr.com:
> On Thursday, April 27, 2017 at 7:20:31 AM UTC-5, Paul B. Andersen wrote:
>> Den 27.04.2017 02.03, skrev dsep...@austin.rr.com:
>>> On Wednesday, April 26, 2017 at 4:08:01 PM UTC-5, Paul B. Andersen wrote:
>>>>
>>>> Since you specify a constant coordinate acceleration in F0,
>>>> the proper acceleration must increase with time.
>>>>
>>>> Since you assume that the distance between the particles will stay
>>>> the same in F1, you assume that the coordinate acceleration
>>>> of both particles are equal and constant in F1 as well.
>>>> That is wrong.

1. time

>>>>
>>>>
>>>> --
>>>> Paul
>>>>
>>>> https://paulba.no/
>>>
>>> I stated that the acceleration along the x-axis as measured in
>>> frame F1 (and frame F0) was independent of the x coordinate where
>>> the acceleration starts. I did not say that the acceleration of
>>> the particles as measured in F1 is constant.
>>> I simply assert that the acceleration is the same for
>>> both particles as measured in F1, and therefore the distance
>>> between the two particles as measured in F1 never changes.
>>
>> This assertion is wrong.
>>
>>> As long as the acceleration pattern is identical for the two particles
>>
>> It isn't . .

2. time

>>
>>> and the acceleration of each particle starts simultaneously,
>>> the distance between the two particles remains constant.
>>
>> .. so it doesn't
>>
>>
>> --
>> Paul
>>
>> https://paulba.no/
>
> So you are saying that if inertial reference frame starts
> the identical acceleration of two particles simultaneously
> the distance between the two particles does not remain constant.

No, I don't say so.

>
> So if one particle is at x1 and the other identical particle
> is at x2 some distance L away as measured in F1, and their
> accelerations start simultaneously, which particle accelerates
> faster as measured in F1?

The one with the highest acceleration.
Obvious, no?

Listen:
In F0, the particles have the same constant coordinate
acceleration a, but they do not start the acceleration
simultaneously. The consequence is that the distance
between the particles measured in F0 changes with time,
one particle may even overtake the other.

At any time, you can use the LT to transform the position
of the particles to F1, and of course will the distance between
the particles change with time in F1 as well.
For example, if the particles at any time are co-located in
in F0, so they are in F1 as well, and of course there is no
way you can make the LT say anything else.
(How can this not be obvious to you?)

But since the particles starts their acceleration simultaneously
in F1, the only way this can happen is if the accelerations of
the particles are different in F1.

So for the third time:
The acceleration patterns for the two particles are not
identical in F1.

It is indeed a weird idea that the different coordinate
acceleration patterns in F0 should transform to identical
coordinate acceleration patterns in F1. :-D

Coordinate acceleration is frame dependent!

Case closed.

--
Paul

https://paulba.no/

[dsep...@austin.rr.com]

Paul,
Let's use your arguments, but modified from F1 point of view. I am modifying your statements, so that you can tell me which
concept is incorrect. Here goes

Listen:
In F1, each particle has the same coordinate
acceleration a = f(t), and they start the acceleration

simultaneously. The consequence is that the distance

between the particles measured in F1 never changes with time,
the distance between the particles is always constant.

At any time, you can use the LT to transform the position

of the particles to F0, and of course the distance between
the particles will never change with time, in F1 as well.
The only way that can happen is if the identical acceleration of each particle as measured in F0 start simultaneously. Of course there is no

way you can make the LT say anything else.
(How can this not be obvious to you?

David Seppala
Bastrop TX

[danco...@gmail.com]

On Friday, April 28, 2017 Paul B. Andersen wrote:
> In F0, the particles have the same constant coordinate
> acceleration a, but they do not start the acceleration
> simultaneously. The consequence is that the distance
> between the particles measured in F0 changes with time,

Yes.

> one particle may even overtake the other.

No, in the scenario described, the particles will not collide, because that would require at least one of them to exceed the speed of light.

> At any time, you can use the LT to transform the position
> of the particles to F1,

Yes.

and of course will the distance between
> the particles change with time in F1 as well.

No, the distance between the particles in terms of F1 doesn't change (as long as they have constant acceleration in F0, which can't go on indefinitely).

> For example, if the particles at any time are co-located in
> in F0, so they are in F1 as well, and of course there is no
> way you can make the LT say anything else.

Right, but in this scenario they are never co-located, because that would require at least one of the particles to exceed the speed of light.

> But since the particles starts their acceleration simultaneously
> in F1, the only way this can happen is if the accelerations of
> the particles are different in F1.

The histories of accelerations are the same in F1, for as long as the accelerations remain constant in F0. But that cannot continue indefinitely, because eventually the particles would exceed the speed of light.

> It is indeed a weird idea that the different coordinate
> acceleration patterns in F0 should transform to identical
> coordinate acceleration patterns in F1. :-D

It isn't so weird. From the standpoint of F1, consider a particle at rest on the x axis, and then suppose at some instant the particle begins to accelerate at a rate that is constant in terms of a coordinate system moving with constant speed -v relative to F1. This will produce a certain pattern of acceleration. But notice that I didn't specify where along the x axis the particle began. It doesn't matter. We will get the same pattern of acceleration, where ever we start. The same applies with any initial velocity of the particle.

So, the reason the scenario breaks down before a contradiction is reached is that the particles cannot collide, because that would imply at least one of them exceeds the speed of light. In other words, the acceleration can't remain constant in F0 long enough for the particles to collide. The equations proving this were given in the previous post.

[bigal...@gmail.com]

On Saturday, April 22, 2017 at 1:58:25 PM UTC-7, dsep...@austin.rr.com wrote:
>

"in inertial reference frame, F0, a long device is built that can accelerate particles at a constant acceleration rate of 3 meters / second**2 as long as the particle has a speed as measured in F0 of less than 90 percent of the speed of light. The acceleration can start at any point along the x-axis, and the acceleration of the particle will be a constant rate of 3 meters per second squared.

Inertial reference frame, F1, moves along the x-axis of F0 with a relative velocity of V = 0.8 c. Frame F1 also observes that the acceleration of the particle is not dependent on the x' coordinate where the acceleration starts."

If the particle had constant acceleration in OUR non accelerating reference frame, the acceleration would have to be constantly INCREASING in the F0 frame to maintain that 3 meters per second per second in OUR non accelerating frame.
From F1 point of view, non accelerating like OUR non accelerating F0, the particle acceleration IS dependent on the x0 or x1 coordinates, constantly increasing.

[Paul B. Andersen]

Den 28.04.2017 15.15, skrev dsep...@austin.rr.com:
> On Friday, April 28, 2017 at 7:25:42 AM UTC-5, Paul B. Andersen wrote:
>>

#################################################################

>> The acceleration patterns for the two particles are not
>> identical in F1.

###############################################################
>>
>>
##################
>> Case closed.
##################

> In F1, each particle has the same coordinate
> acceleration a = f(t), and they start the acceleration
> simultaneously.

--
Paul

https://paulba.no/

[danco...@gmail.com]

On Saturday, April 29, 2017 at 6:03:50 AM UTC-7, Paul B. Andersen wrote:
> The acceleration patterns for the two particles are not
> identical in F1.

Not true. In F1 (x',t'), each particle has the same initial velocity, and at the same instant t' each particle is subjected to an acceleration which in frame F0 (x,t) moving at -v relative to F1 has the constant value a. As a result, each particle has the same pattern of acceleration as a function of t' in F1. For example, if the initial velocity in F1 of the particles is -v (meaning they are at rest in F0 prior to the acceleration), then the acceleration of each particle in F1 is

a' = a[(1-v^2)/(1 - 2avt'/g)]^3/2

where g = 1/sqrt(1-v^2). If you disagree, can you tell me what you think the accelerations of the two particles in F1 (as a function of t') are in this case?

[Odd Bodkin]

On 4/26/17 7:42 AM, dsep...@austin.rr.com wrote:

> I stated in that in F0 the F0 device accelerates a particle at 3 meters /second squared independent of the position of the particle in the device, once the acceleration of a particle starts.
>
> The acceleration motion as measured in F1 is irrelevant to the problem since each particle has an identical acceleration independent of its position along the x-axis and the acceleration of each particle as measured in F1 starts simultaneously. Therefore the acceleration as measured in F1 has zero effect on the distance between the two particles.
>
> David Seppala
> Bastrop TX
>

David, surely you know that if the acceleration is independent of x in
F0, then the acceleration cannot possibly be independent of x' in F1, right?

This is heart and soul of the Lorentz transform.


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

[Odd Bodkin]

On 4/27/17 7:30 AM, dsep...@austin.rr.com wrote:
> So if one particle is at x1 and the other identical particle is at x2 some distance L away as measured in F1, and their accelerations start simultaneously, which particle accelerates faster as measured in F1

Apply the Lorentz transform and you will find out.

[danco...@gmail.com]

On Monday, May 1, 2017 at 11:07:19 AM UTC-7, Odd Bodkin wrote:
> David, surely you know that if the acceleration is independent of x in
> F0, then the acceleration cannot possibly be independent of x' in F1, right?

He isn't saying the acceleration is independent of x', he's saying the pattern of acceleration, a'(t'), of each particle is independent of the starting x' position of the particle, provided the particles have the same initial speeds and begin to accelerate at the same t' and the acceleration has a constant value in F0. This is perfectly true. The error in Dave's analysis was explained last week. It isn't what you think.

[Odd Bodkin]

Yes, I see the error has been explained.