Relativity of simultaneity in motionless frame and factor of motion.

[DimaMazin]

T'=T - r/c

T' is simultaneous time of motionless remote object

T is time of observer

r is distance between observer and remote object

c is speed of light

 

Moving object to observer creates additional rate of time. Object motion from observer creates additional slowing of time by factor of motion.

 

t=gamma _* t'+ dr/(c_*gamma)

[Mordred]

[latex]t=\gamma*\acute{t}+\frac{dr}{c*\gamma}[/latex]

 

So far you defined the variables, and posted a metric.

 

Where is the discussion, is this your own form? This doesn't define any reference to coordinate change.

 

Moving object to observer creates additional rate of time. Object motion from observer creates additional slowing of time by factor of motion.

Define the observer, the above equation doesn't Edited July 23, 2015 by Mordred

[DimaMazin]

[latex]T=\gamma*\acute{t}+\frac{dr}{c*\gamma}[/latex]

 

So far you defined the variables, and posted a metric.

 

Where is the discussion, is this your own form? This doesn't define any reference to coordinate change.

 

Define the observer, the above equation doesn't

Don't confuse T with t .

T is clock indication or age

t is time or dT

Any motionless observer has relative top of age. The equation defines reason why brought clock from remote place has such defined indication.

Edited July 23, 2015 by DimaMazin

[swansont]

How can this be tested? Any experimental confirmation of this? Does it agree with any existing relativity experiment?

[Mordred]

Don't confuse T with t .

T is clock indication or age

t is time or dT

Any motionless observer has relative top of age. The equation defines reason why brought clock from remote place has such defined indication.

I simply placed your formula into latex. Spell check placed the capital.

 

You still haven't shown how you derived the equation. Nor shown how this relates to Lorentz transformation

Edited July 23, 2015 by Mordred

[DimaMazin]

I simply placed your formula into latex. Spell check placed the capital.

 

You still haven't shown how you derived the equation. Nor shown how this relates to Lorentz transformation

T'=T-r/c is equation of relativity of simultaneity in motionless frame, this doesn't relate to Lorentz transformation. This defines that seeing events are simultaneous. I think we need to create another non-simultaneities also for explanation of relativity in one frame.

Second equation explains reason of indication of moving or moved clock for the non-simultaneity.Here you should use gamma if it considerably more than one.

Sorry,I am confused in derivation.

[Mordred]

I posted this in another thread but it suits here. As this is an example of how Lorentz transformations is derived.

 

note you have coordinates. What you posted cannot describe relativity of simultaneaty as you haven't included coordinates. ( you've only included the time coordinates) Neither have you explained as to how you arrived at that equation. (How did you derive it). Did you just put letters together without substance?

 

Lorentz transformation.

 

First two postulates.

 

1) the results of movement in different frames must be identical

2) light travels by a constant speed c in a vacuum in all frames.

 

Consider 2 linear axes x (moving with constant velocity and [latex]\acute{x}[/latex] (at rest) with x moving in constant velocity v in the positive [latex]\acute{x}[/latex] direction.

 

Time increments measured as a coordinate as dt and [latex]d\acute{t}[/latex] using two identical clocks. Neither [latex]dt,d\acute{t}[/latex] or [latex]dx,d\acute{x}[/latex] are invariant. They do not obey postulate 1.

A linear transformation between primed and unprimed coordinates above

in space time ds between two events is

[latex]ds^2=c^2t^2=c^2dt-dx^2=c^2\acute{t}^2-d\acute{x}^2[/latex]

 

Invoking speed of light postulate 2.

 

[latex]d\acute{x}=\gamma(dx-vdt), cd\acute{t}=\gamma cdt-\frac{dx}{c}[/latex]

 

Where [latex]\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[/latex]

 

Time dilation

dt=proper time ds=line element

 

since [latex]d\acute{t}^2=dt^2[/latex] is invariant.

 

an observer at rest records consecutive clock ticks seperated by space time interval [latex]dt=d\acute{t}[/latex] she receives clock ticks from the x direction separated by the time interval dt and the space interval dx=vdt.

 

[latex]dt=d\acute{t}^2=\sqrt{dt^2-\frac{dx^2}{c^2}}=\sqrt{1-(\frac{v}{c})^2}dt[/latex]

 

so the two inertial coordinate systems are related by the lorentz transformation

 

[latex]dt=\frac{d\acute{t}}{\sqrt{1-(\frac{v}{c})^2}}=\gamma d\acute{t}[/latex]

 

So the time interval dt is longer than interval [latex]d\acute{t}[/latex]

 

 

The above is what I would expect to see when one presents his own equation. The above isn't a full derivitave.

 

Several missing steps. It was for another post. However it provides a better explanation of the Lorentz transformations than merely posting a formula.

If your not using Lorentz then you need to define the coordinate transformation rules.

Here is relativity of simultaneaty coordinate transformation in Lorentz.

 

[latex]\acute{t}=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}[/latex]

 

[latex]\acute{x}=\frac{x-vt}{\sqrt{1-v^2/c^2}}[/latex]

 

[latex]\acute{y}=y[/latex]

[latex]\acute{z}=z[/latex]

Edited July 24, 2015 by Mordred

[swansont]

T'=T-r/c is equation of relativity of simultaneity in motionless frame, this doesn't relate to Lorentz transformation.

 

Not with Einstein clock synchronization. If you want to re-work relativity with another version of synchronization, go for it, but show your work.

[DimaMazin]

 

Not with Einstein clock synchronization. If you want to re-work relativity with another version of synchronization, go for it, but show your work.

Length of spaceship before motion = r

Length of spaceship at v = r/gamma

Then time of motion of tail before nose motion=(gamma-1) r / (gamma _* v)

then T' = T + (gamma-1)r/(gamma_*v)

[swansont]

Why would that be the time between the tail moving and the nose moving?

 

And I thought we were talking about a motionless frame. With Einstein synchronization, it's the same time everywhere in a frame. If you have a different answer, then you have to have another clock synchronization protocol.

[DimaMazin]

Why would that be the time between the tail moving and the nose moving?

 

And I thought we were talking about a motionless frame. With Einstein synchronization, it's the same time everywhere in a frame. If you have a different answer, then you have to have another clock synchronization protocol.

It is when motionless observers trace an immovability and motion of objects in motionless coordinates of own structure.

[swansont]

It is when motionless observers trace an immovability and motion of objects in motionless coordinates of own structure.

 

Can't parse what that means.

[Strange]

It is in standard DimaMazinese.

[DimaMazin]

 

Can't parse what that means.

It means dualism of simultaneity in motionless frame.

[swansont]

It means dualism of simultaneity in motionless frame.

 

That doesn't help.

[DimaMazin]

 

That doesn't help.

Then all this means that spaceship tail can't be simultaneously accelerated with spaceship nose because it breaks off the spaceship.

It additionally works in gravitational fall.

Edited August 27, 2015 by DimaMazin

[Strange]

Then all this means that spaceship tail can't be simultaneously accelerated with spaceship nose because it breaks off the spaceship.

 

If it is accelerating, then it is not a motionless frame as described in the OP.

 

I really have no idea what you are trying to say or ask. (Are you asking a question? Or stating a fact?)

[DimaMazin]

 

If it is accelerating, then it is not a motionless frame as described in the OP.

 

I really have no idea what you are trying to say or ask. (Are you asking a question? Or stating a fact?)

Study this: https://en.wikipedia.org/wiki/Frame_of_reference

Moving frame is contracted in our frame and it is in our frame.

And at all: show math of contraction of moving object in our frame please. If you don't make the math then you can use my math.

[Strange]

It is very unlcear what you are trying to say.

 

Study this: https://en.wikipedia.org/wiki/Frame_of_reference

 

If you are talking about special relativity then you shoudl really be talking about intertial frames.

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

Moving frame is contracted in our frame and it is in our frame.

 

It is well know that length contraction occurs in a frame of reference that is moving relative to an obsever.

 

I don't understand what you mean by "it is in our frame". What is in our frame?

 

It is also confusing because you started talking about a stationary frame of reference and then introduce a moving frame and then an accelerating one.

 

And at all: show math of contraction of moving object in our frame please.

 

That would be the Lorentz transform.

 

However, when you start talking about relativity of simultaneity and acceleration, then things start getting more complex. And that is where you seem to getting confused (or not explaining yourself clearly).

 

You seem to be talking about a variant of the spaceship paradox: https://en.wikipedia.org/wiki/Bell's_spaceship_paradox

Edited August 27, 2015 by Strange

[DimaMazin]

It is very unlcear what you are trying to say.

 

 

If you are talking about special relativity then you shoudl really be talking about intertial frames.

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

 

 

It is well know that length contraction occurs in a frame of reference that is moving relative to an obsever.

 

I don't understand what you mean by "it is in our frame". What is in our frame?

 

It is also confusing because you started talking about a stationary frame of reference and then introduce a moving frame and then an accelerating one.

 

 

That would be the Lorentz transform.

 

However, when you start talking about relativity of simultaneity and acceleration, then things start getting more complex. And that is where you seem to getting confused (or not explaining yourself clearly).

 

You seem to be talking about a variant of the spaceship paradox: https://en.wikipedia.org/wiki/Bell's_spaceship_paradox

I don't see confused me. Our frame is frame S,agen you don't understand that contracted moving objects are in frame S.

They don't make necessary math.They didn't make even math of non-simultaneity of start of object motion.

T_{start of tail motion}=T_{start of head motion} -(gamma-1)_*length/(gamma_*v)

[Strange]

I don't see confused me.

 

You presumably know what you are trying to say. I don't think anyone else does.

 

Our frame is frame S,agen you don't understand that contracted moving objects are in frame S.

 

If they are moving, then I would say they are in different frames of reference. We are in frame S, the moving object is in frame S'.

 

They don't make necessary math.They didn't make even math of non-simultaneity of start of object motion.

 

Who don't?

[DimaMazin]

 

If they are moving, then I would say they are in different frames of reference. We are in frame S, the moving object is in frame S'.

 

Moving object is moving and contracted in frame S, it is motionless in own frame S'.

[swansont]

Our frame is frame S,agen you don't understand that contracted moving objects are in frame S.

 

Anything in frame S is not moving relative to us.

[DimaMazin]

You seem to be talking about a variant of the spaceship paradox: https://en.wikipedia.org/wiki/Bell's_spaceship_paradox

Events of starts of motions are not simultaneous relative to one of two frames. Why do you think non-simultaneity of starts of motions doesn't cause force between the rockets?

[DimaMazin]

Why would that be the time between the tail moving and the nose moving?

 

And I thought we were talking about a motionless frame. With Einstein synchronization, it's the same time everywhere in a frame. If you have a different answer, then you have to have another clock synchronization protocol.

Length of spaceship before motion = r

Length of spaceship at v = r/gamma

Then time of motion of tail before nose motion=(gamma-1) r / (gamma _* v)

 

I have understood this is math of non-simultaneity of instant starts of motions of rockets at v relative to frameS and this is math of simultaneity of stops of the rockets in frameS'.