Neutral simultaneity for two frames.

[Genady]

1 minute ago, DimaMazin said:

use the same module of reverse velocities in two frames.

What is it and where was it done? Give the reference.

[DimaMazin]

1 hour ago, Genady said:

What is it and where was it done? Give the reference.

I have confused. We should create law of addition of non-smultaneities. We can image any velocity(and bigger than c)  as non-simultaneity. Then we can check any law of addition of velocities.

[Genady]

1 minute ago, DimaMazin said:

law of addition of non-smultaneities

I have no idea what it means.

[iNow]

14 minutes ago, DimaMazin said:

We can image any velocity(and bigger than c)  as non-simultaneity.

Velocities "bigger than c" are not possible based on current understanding and evidence. In fact, it's not even possible for a velocity to reach c if the object under discussion has non-zero mass (has any mass at all).

If you must invent fictions regarding how the universe behaves in order to support your assertion, then perhaps it's time to acknowledge that your assertion is itself also fictional. 

[DimaMazin]

37 minutes ago, Genady said:

I have no idea what it means.

On distance 0 time is 0. On distance 1 million km time is 1 second in S frame. What is velocity of non-simultaneity? Non-simultaneity is not velocity but it can have imaging velocity. In frame S' on distanse 0 time is 0 but on what distance and what time was or will be in co-located point relative to second point of frame S?

[swansont]

On 8/22/2023 at 11:01 AM, DimaMazin said:

A very simple  refutation of Einstein's law of addition of velocities.

Since the velocity addition formula can be derived from the Lorentz transforms, this would mean relativity is wrong.

Do you have any experimental evidence of this?

[Genady]

3 minutes ago, DimaMazin said:

On distance 0 time is 0. On distance 1 million km time is 1 second in S frame. What is velocity of non-simultaneity? Non-simultaneity is not velocity but it can have imaging velocity. In frame S' on distanse 0 time is 0 but on what distance and what time was or will be in co-located point relative to second point of frame S?

If I understand what you say, you are talking about two events, let's call them, event A and event B. 

In the frame S, the coordinates of the events are respectively:

xA=0, tA=0 and xB=10^6 km, tB=1 s.

I still have no idea what the question, "What is velocity of non-simultaneity?" means.

[Genady]

1 hour ago, DimaMazin said:

On distance 0 time is 0. On distance 1 million km time is 1 second in S frame. What is velocity of non-simultaneity? Non-simultaneity is not velocity but it can have imaging velocity. In frame S' on distanse 0 time is 0 but on what distance and what time was or will be in co-located point relative to second point of frame S?

Ignoring the references to "non-simultaneity", which I don't understand, my answer is:

In the frame S, 

xA=0, tA=0, xB=10^6 km, tB=1 s.

If the frame S' moves with velocity 100000 km/s relative to S, then in S',

xA'=0, tA'=0, xB'=954594 km, tB'=-.118 s.

[md65536]

10 hours ago, DimaMazin said:

We should create law of addition of non-smultaneities. We can image any velocity(and bigger than c)  as non-simultaneity.

9 hours ago, Genady said:

I still have no idea what the question, "What is velocity of non-simultaneity?" means.

Earlier in the thread, it was used like this:

Given a set of events, the velocity of non-simultaneity is the distance between pairs of subsequent events divided by the time between those events.

So for example if you have a train at rest with a bunch of synchronized clocks along its length, the events "noon" at each clock are all simultaneous and the "velocity of non-simultaneity" is undefined. In other frames, the events would propagate along the length of the train at a rate greater than c (the pairs of events are space-like).

 

SR can handle particle velocities greater than c just fine, you just can't accelerate something from slower than c to c or faster, or vice versa. I guess a frame of reference moving relative to another at v>c doesn't make sense? So a composition of velocity formula for v>c doesn't make sense? Anyway, DimaMazin didn't you already calculate the rate at which events would propagate along the x-axis for a given reference frame velocity? You would just use the existing composition formula for that velocity, which is less than c, and find the "velocity of non-simultaneity" using that.

[Genady]

42 minutes ago, md65536 said:

Earlier in the thread, it was used like this:

Given a set of events, the velocity of non-simultaneity is the distance between pairs of subsequent events divided by the time between those events.

So for example if you have a train at rest with a bunch of synchronized clocks along its length, the events "noon" at each clock are all simultaneous and the "velocity of non-simultaneity" is undefined. In other frames, the events would propagate along the length of the train at a rate greater than c (the pairs of events are space-like).

 

SR can handle particle velocities greater than c just fine, you just can't accelerate something from slower than c to c or faster, or vice versa. I guess a frame of reference moving relative to another at v>c doesn't make sense? So a composition of velocity formula for v>c doesn't make sense? Anyway, DimaMazin didn't you already calculate the rate at which events would propagate along the x-axis for a given reference frame velocity? You would just use the existing composition formula for that velocity, which is less than c, and find the "velocity of non-simultaneity" using that.

Thank you for the clarification. Finally, a clear statement. What a relief!

But what does it have to do with the relativistic formula for addition of velocities, which is the topic of discussion now?

[DimaMazin]

On 8/24/2023 at 6:21 PM, swansont said:

Since the velocity addition formula can be derived from the Lorentz transforms, this would mean relativity is wrong.

Do you have any experimental evidence of this?

It can be checked experimentaly. 3 travelers is traveling relative to each other. They send light with the same fraquency to each other and resieve the light with other fraquency. Then they can define their velocities and check the laws.

Energy factor has square power of velocity c, therefore delta energy can be variable in different frames. But velocity and gamma factor have unit power in formula of momentum. Because square root annihilate square power. Therefore delta momentum should be constant in any  frame.

[swansont]

5 hours ago, DimaMazin said:

It can be checked experimentaly. 3 travelers is traveling relative to each other. They send light with the same fraquency to each other and resieve the light with other fraquency. Then they can define their velocities and check the laws.

Why 3? The velocity addition formula is for two objects. But if it works for two (and it does), it will work for three. That’s a math issue, and math works that way.

And you have not provided evidence that there’s a problem.

 

[DimaMazin]

10 hours ago, swansont said:

Why 3? The velocity addition formula is for two objects. But if it works for two (and it does), it will work for three. That’s a math issue, and math works that way.

And you have not provided evidence that there’s a problem.

 

The equation has three velocities. How do you experimentally define them?

[swansont]

5 minutes ago, DimaMazin said:

The equation has three velocities. How do you experimentally define them?

Three velocities. Two objects. The third is the result of the addition of the two objects' velocities.

[DimaMazin]

On 9/15/2023 at 2:28 AM, swansont said:

Three velocities. Two objects. The third is the result of the addition of the two objects' velocities.

My experiment does experimentally define the three velocities. You don't understand my experiment because you don't understand relativity of observers and travelers. Wy the experiment can be bad for Relativity? When the experiment can prove Relativity,if it correct.

[DimaMazin]

Seems I have got absolutely neutral velocity 

V = 2×5^1/2×c/5

gamma=5^1/2

Then non-simultaneity

t =t₀ + (5^1/2 - 1)x/(2c)

is absolutely neutral simultaneity.

[DimaMazin]

15 hours ago, DimaMazin said:

Seems I have got absolutely neutral velocity 

V = 2×5^1/2×c/5

gamma=5^1/2

Then non-simultaneity

t =t₀ + (5^1/2 - 1)x/(2c)

is absolutely neutral simultaneity.

I cannot  correctly make complex exploring therefore that is incorrect.

Let's consider what is quantum velocity. Quantum velosity is velocity which is created by constant and instant acceleration from frame S in frame S' and back. For example velocity between S and S' is v then quantum velocity is v/2. Time of such observer is t/2+t/(2gamma) because the observer has equal time of travel in both frames.

[Bufofrog]

20 hours ago, DimaMazin said:

Seems I have got absolutely neutral velocity 

V = 2×5^1/2×c/5

gamma=5^1/2

It seems that this is meaningless.

4 hours ago, DimaMazin said:

I cannot  correctly make complex exploring therefore that is incorrect.

So after writing the equation you now agree it is meaningless?

4 hours ago, DimaMazin said:

Let's consider what is quantum velocity. Quantum velosity is velocity which is created by constant and instant acceleration from frame S in frame S' and back. For example velocity between S and S' is v then quantum velocity is v/2. Time of such observer is t/2+t/(2gamma) because the observer has equal time of travel in both frames.

This also seems like meaningless word salad.  Velocity is constant and instant acceleration?!?

[md65536]

On 12/30/2023 at 11:00 AM, DimaMazin said:

Then non-simultaneity

t =t₀ + (5^1/2 - 1)x/(2c)

is absolutely neutral simultaneity.

I'm struggling to figure out your meaning relating to previous posts. This represents the timing of events along the length of say a train, in a particular inertial reference frame? But the rate is slower than c, which would mean those events cannot be simultaneous in any frame.

On 12/31/2023 at 3:05 AM, DimaMazin said:

For example velocity between S and S' is v then quantum velocity is v/2.

Are you using the composition of velocity calculation to figure out the velocities or at least check that they make sense? It seems like you should be doing that here.

[DimaMazin]

Let's consider Minkowski diagram with general coordinates of two frames

[Bufofrog]

12 minutes ago, DimaMazin said:

Let's consider Minkowski diagram with general coordinates of two frames

I'm not sure what that graph is supposed to tell us, but it sure isn't a Minkowski diagram.

The only way that graph makes sense is if t'= -t and x' = -x

[DimaMazin]

22 minutes ago, Bufofrog said:

 

The only way that graph makes sense is if t'= -t and x' = -x

Why it cannot be so?

[Bufofrog]

2 hours ago, DimaMazin said:

Why it cannot be so?

Because that is how graphs are made.

The graph you supplied makes no sense at all.

[DimaMazin]

On 12/11/2022 at 9:57 AM, md65536 said:

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

The "Class A" section contains an entire class of applicable motions.

Instantaneous (not generally simultaneous) acceleration using the solution I proposed (different from OP's I think), does not satisfy Born rigidity, because it requires different parts of the train to accelerate at different times, in the 2 rest frames (before and after acceleration) of the train. Therefore the train's length changes in those frames, and its proper length doesn't even seem defined while accelerating. However, it should be the only solution that involves a single instantaneous acceleration at each point of the train, with which the train has the same proper length before and after the acceleration.

It would seem not even Born rigidity can be satisfied with an instantaneous acceleration. However, OP's proposal never mentioned anything about a requirement of rigidity. For that matter, they neither required that the proper length of the train is the same before and after, I assumed that.

 

The way I figure it, if you have the back of the train accelerate at time 0, the rest of the train accelerates over time until the front of the train finally accelerates when the length of the train has become L/gamma in the initial frame, where L is its original proper length. By that time, the back of the train has traveled a distance of (L - L/gamma) at velocity v. Therefore the time when the front of the train accelerates would be t=d/v = (L - L/gamma)/v.

Then the velocity of the "wave" would have to be d/t = Lv/(L - L/gamma) = v/(1 - 1/gamma)

For gamma=2, it seems intuitive that the wave would have to travel at 2v, to reach the front of the train at the same time that the back of the train traveling at v gets halfway there. With v = 3^0.5 c/2, I get a velocity of the wave equal to 3^0.5 c.

Like you said this is faster than c, so it's non-causal and needs local forces to accelerate the parts of the train.

Then t=t'=1second on distance 2v(1-1/gamma) between point of S and point of S'. Why so?

[DimaMazin]

t'=gamma(t-vx/c²)

Let's use velocity u as distance

t'=gamma(1-vu/c²)

u'=u/(gamma×t')=u/(gamma²×(1-vu/c²))

I have got equation for addition of any velocity, even when it is as nonsimultaneity bigger than c .

Edited November 2 by DimaMazin