# Relativity of simultaneity in motionless frame and factor of motion.

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I thought observer makes math of moving object relative to own frame.And when observer can't make the math then he/she isn't a scientist.

The observer in this case is in the same frame as the object, according to your setup of the problem. There is no "relative to own frame"

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The observer in this case is in the same frame as the object, according to your setup of the problem. There is no "relative to own frame"

I exactly know that an observer doesn't see contracted gap between simultaneously accelerated rockets. Therefore you can't deceive me.

Once you have admitted with brag that you were breaking laws of USA. Therefore it's not surprising when you make crime in science.

Edited by DimaMazin
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!

Moderator Note

Dima Mazin

We realise that English is not your first language - but this argument is beginning to seem fruitless and I am no longer sure it is due to any language mis-comprehension.

It appears that you have decided to push your own ideas regardless of current theory, counter-arguments presented, and requests for maths/clarification.

If this does not change in the next few posts then the thread will be locked. Do not respond to this post - report it if you feel it is unfair.

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• 1 month later...

Length contraction is not a force; cooling the rod is not part of relativity nor is it a good analogy. The rod is in its own frame, and as such it does not contract.

My understanding of relativity is not the issue here. It's your inability to articulate your argument in a rigorous fashion.

Even if tail and head are simultaneously accelerated in frameS and from frameS to frameS' then they aren't simultaneously stoped in frame S'

T' tail=T'head - gamma * v * length/c2

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What I have been asking for is a justification, not just plopping down an equation with an assertion that it applies. Why does the location of a clock matter in the frame, regarding the amount of time that passes for it, if the accelerations are equal? Can you derive a justification for this?

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What I have been asking for is a justification, not just plopping down an equation with an assertion that it applies. Why does the location of a clock matter in the frame, regarding the amount of time that passes for it, if the accelerations are equal? Can you derive a justification for this?

T'tail - is time of tail at instant of stop of the tail.

T'head - is time of head at instant of stop of the head.

The events are not simultaneous in frameS'. But indications of tail clock and head clock are equal and simultaneous at instant of stop of the head.

Edited by DimaMazin
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T'tail - is time of tail at instant of stop of the tail.

T'head - is time of head at instant of stop of the head.

The events are not simultaneous in frameS'. But indications of tail clock and head clock are equal and simultaneous at instant of stop of the head.

That's an assertion. You have included no physics in your post. There is no justification for your claim.

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That's an assertion. You have included no physics in your post. There is no justification for your claim.

I just used relativity of simultaneity

https://en.wikipedia.org/wiki/Relativity_of_simultaneity The acceleration is instant therefore time of tail acceleration in frameS and time of tail stop in frameS' are equal. Events of tail and head accelerations are simultaneous in frameS therefore the tail stops earlier than head in frameS'.
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I just used relativity of simultaneity

https://en.wikipedia.org/wiki/Relativity_of_simultaneity

The acceleration is instant therefore time of tail acceleration in frameS and time of tail stop in frameS' are equal. Events of tail and head accelerations are simultaneous in frameS therefore the tail stops earlier than head in frameS'.

Why aren't the tail and head in the same frame? You appear to have simply assumed the answer, rather than showing it.

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Why aren't the tail and head in the same frame? You appear to have simply assumed the answer, rather than showing it.

Simply because there again paradox appears when I use relativity.

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Simply because there again paradox appears when I use relativity.

Presumably because you are using it wrong. It is impossible to create a paradox in relativity. It is proven to be mathematically consistent.

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Simply because there again paradox appears when I use relativity.

Not giving the answer you want is not a paradox.

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• 2 weeks later...

Not giving the answer you want is not a paradox.

I think there is such phenomenon when nonsimultaneity exists between two clocks but the clocks don't show it.

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• 4 months later...

Two clocks can show the same time at nonsimultaneity in different frames and in motionless frame on distance.

In motionless frame a simultaneity can be defined by speed, therefore it is different if it is defined by different speeds.

Edited by DimaMazin
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• 4 weeks later...

Traveling simultaneity reduces a length of accelerating object . Clocks simultaneity isn't real simultaneity. Tail and head are simultaneously accelerated at clocks simultaneity, but tail is accelerated earlier than head at traveling simultaneity. We can't consider a motion of contracted object in our frame without traveling simultaneity.

Edited by DimaMazin
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Traveling simultaneity reduces a length of accelerating object . Clocks simultaneity isn't real simultaneity. Tail and head are simultaneously accelerated at clocks simultaneity, but tail is accelerated earlier than head at traveling simultaneity. We can't consider a motion of contracted object in our frame without traveling simultaneity.

Ah, I have mistaken again. Object is contracted in clocks simultaneity therefore tail and head aren't simultaneously accelerated in clocks simultaneity. They are simultaneously accelerated in perfect simultaneity.

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Observer is on zero point therefore simultaneity of acceleration from frame S to frame S' =t+(gamma-1)x/|v|

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Observer is on zero point therefore simultaneity of acceleration from frame S to frame S' =t+(gamma-1)x/|v|

Perfect simultaneity of acceleration from S into S' =t+(gamma-1)x/(gamma|v|)

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• 4 weeks later...

Then relativity of simultaneity is:

t'=t - 2(gamma-1)x/(gamma * v)

Therefore head is earlier stoped in S' relative to tail, and head is later accelerated relative to tail in S

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• 2 weeks later...

If we apply the relativity of simultaneity :

t'=t - (gamma-1)x/(gamma*v)

Then we can explain different time in two frames and forget Einstein's relativity of simultaneity.

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• 1 month later...

In condition of simultaneity x=tv

gamma(t-vx/c2)=t-(gamma-1)x/(gamma*v)

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So, DM, seriously, which one is it? 4 posts, 4 different equations:

S into S' =t+(gamma-1)x/(gamma|v|)

t'=t - 2(gamma-1)x/(gamma * v)

t'=t - (gamma-1)x/(gamma*v)

gamma(t-vx/c2)=t-(gamma-1)x/(gamma*v)

and then a head smacking smiley at the end.

This is now how science goes.

1) equations are normally derived. start from first principles, and then make and justify some assumptions, and then see where it leads. Not "let's add and x here, a 2 there, take away a t there" as you seen to be doing

2) equations do change. Especially as assumptions change. But you know how we tell if those assumptions are any good? By testing the equation. Using the equation to make predictions and then comparing those predictions to measurements. Again, not just adding and removing terms and throwing the result up and seeing what reactions it gets... which really is none here because you aren't doing science.

Instead of going another 2 weeks and posting another random equation, take some time and see if the equation is any good yourself by seeing if it makes good predictions.

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• 3 weeks later...

So, DM, seriously, which one is it? 4 posts, 4 different equations:

and then a head smacking smiley at the end.

This is now how science goes.

1) equations are normally derived. start from first principles, and then make and justify some assumptions, and then see where it leads. Not "let's add and x here, a 2 there, take away a t there" as you seen to be doing

2) equations do change. Especially as assumptions change. But you know how we tell if those assumptions are any good? By testing the equation. Using the equation to make predictions and then comparing those predictions to measurements. Again, not just adding and removing terms and throwing the result up and seeing what reactions it gets... which really is none here because you aren't doing science.

Instead of going another 2 weeks and posting another random equation, take some time and see if the equation is any good yourself by seeing if it makes good predictions.

This is now how my simultaneity works:

Two rockets are simultaneously accelerated from S to S'.

First rocket is accelerated on x1 , second rocket is accelerated on x2 .

t=(gamma-1)(x2 - x1)/(2*gamma*v)

t is time of simultaneity of motion of first rocket from observer and motion of observer reference frame from second rocket. After this time of the simultaneity we have a contracted distance between the rockets and usual motion of contracted gap between them .

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t=(gamma-1)(x2 - x1)/(2*gamma*v)

Did you even read and comprehend my post? Because after another few weeks, yet another equation.

And no evidence posted that this one works any better or worse than the other random combinations of gamma and t you've thrown against the wall here.

Please re-read my previous post and understand what I am asking before just slapping together another permutation of symbols.

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• 2 weeks later...

Did you even read and comprehend my post? Because after another few weeks, yet another equation.

And no evidence posted that this one works any better or worse than the other random combinations of gamma and t you've thrown against the wall here.

Please re-read my previous post and understand what I am asking before just slapping together another permutation of symbols.

That is genius method.

I am not a genius, therefore I use random math. But random math has more chances than nothing.

We should use general simultaneity(it is when nothing is contracted relative each other)

tg=t-(gamma-1)x/(gamma|v|)

tg - is time of General Simultaneity

Rocket x1 and rocket x2 are simultaneously accelerated in S, then t >tg on x1 and t >>tg on x2. The same t on distance means later time relative to tg when x aims to plus infinity. I do math of length contraction.

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