Relativity of simultaneity in motionless frame and factor of motion.

[swansont]

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'.

Repeating the equation doesn't answer the question. You need to show where it came from.

[DimaMazin]

Repeating the equation doesn't answer the question. You need to show where it came from.

Length of string between the rockets in frameS is r

Length of the string in frameS' relative to frameS is r/gamma

Contracted distance is r-r/gamma or (gamma-1)r/gamma

t=contracted distance/v

t=(gamma-1)r/(gamma_*v)

 

Sorry. I have mistaken about simultaneity of stops in frameS'. Of course the forward rocket stops later than back rocket in S'.

This is single case when the accelerated rockets don't create force to break off or bend the string.

[swansont]

Length of string between the rockets in frameS is r

Length of the string in frameS' relative to frameS is r/gamma

Contracted distance is r-r/gamma or (gamma-1)r/gamma

t=contracted distance/v

t=(gamma-1)r/(gamma_*v)

 

Sorry. I have mistaken about simultaneity of stops in frameS'. Of course the forward rocket stops later than back rocket in S'.

This is single case when the accelerated rockets don't create force to break off or bend the string.

 

Between the rockets? You said there is 1 rocket, and we are looking at the difference between the tail moving and the nose moving.

 

Your equation (or the application of it) makes no sense. What frame is S'? That of the tail? If the front of the rocket and back of the rocket have different accelerations, there is no well-defined v for the situation. The tail may be moving at v, but the next bit of the rocket will be moving slightly slower, and so on, all the way to the nose. So the entire length can't be contracted as if it were a rigid rod moving at v. Just the infinitely thin tail, which is moving at v.

 

And v is very small, is it not?

[DimaMazin]

 

Between the rockets? You said there is 1 rocket, and we are looking at the difference between the tail moving and the nose moving.

 

Your equation (or the application of it) makes no sense. What frame is S'? That of the tail? If the front of the rocket and back of the rocket have different accelerations, there is no well-defined v for the situation. The tail may be moving at v, but the next bit of the rocket will be moving slightly slower, and so on, all the way to the nose. So the entire length can't be contracted as if it were a rigid rod moving at v. Just the infinitely thin tail, which is moving at v.

 

And v is very small, is it not?

Part of distance disappears with any speed and at the speed. In my math is no different between 1 rocket or 2 rockets, because theoretical laws correctly work independently and each with other.The different acceleration is only in time , moving tail is losing length and approaching to the nose, some instant without acting force to the nose(due to losing length) when we consider 1 rocket.

I think you don't understand that contracted length of traveler can be created only by two way: with unsimultaneous accelerations or with force of connection when the accelerations try to be simultaneous.

[swansont]

In my math is no different between 1 rocket or 2 rockets,

I find that confusing, partly because you've only described a scenario with 1 rocket. I have no idea how you would describe what's going on with 2 rockets.

 

I think you don't understand that contracted length of traveler can be created only by two way: with unsimultaneous accelerations or with force of connection when the accelerations try to be simultaneous.

There is much of what you write I don't understand. Yet when I ask questions, you either repeat what you've said or change the discussion. Neither of these tactics is helpful. This is a discussion board, not a blog.

[DimaMazin]

I find that confusing, partly because you've only described a scenario with 1 rocket. I have no idea how you would describe what's going on with 2 rockets.

 

 

There is much of what you write I don't understand. Yet when I ask questions, you either repeat what you've said or change the discussion. Neither of these tactics is helpful. This is a discussion board, not a blog.

With two simultaneously accelerated rockets there is no distance contraction between them when they have weak connection.

Your question was about derivation of the math, I showed. What is your next question?

Why do you think a tactic of mistakes correction isn't helpful?

Edited September 20, 2015 by DimaMazin

[swansont]

Your question was about derivation of the math, I showed. What is your next question?

Oh, please. You didn't derive anything, you wrote the same equations down and changed a few words.

 

Why is that the contraction when most of the rocket is not moving at v? What is the contraction of the middle of the rocket?

[DimaMazin]

Oh, please. You didn't derive anything, you wrote the same equations down and changed a few words.

 

Why is that the contraction when most of the rocket is not moving at v? What is the contraction of the middle of the rocket?

The equation is derived from law of length contraction, all data are there .

Only moving part of rocket is contracted by gamma factor to motionless part of the rocket. Still middle of the rocket isn't moving therefore it isn't contracted but it is moving to nose as mathematical middle.

Edited September 20, 2015 by DimaMazin

[swansont]

The equation is derived from law of length contraction, all data are there .

Only moving part of rocket is contracted by gamma factor to motionless part of the rocket. Still middle of the rocket isn't moving therefore it isn't contracted but it is moving to nose as mathematical middle.

How is it that the tail is moving but the middle isn't? And if the middle isn't moving, why is there length contraction for the whole length?

[DimaMazin]

How is it that the tail is moving but the middle isn't? And if the middle isn't moving, why is there length contraction for the whole length?

Tail, losing own length,approaches to middle.

whole length=tail / gamma + length of most of the rocket

[Mordred]

The above makes absolutely no sense.

[swansont]

Tail, losing own length,approaches to middle.

whole length=tail / gamma + length of most of the rocket

 

length contraction is "length of what's moving"/gamma

 

if only the tail is moving for some unexplained reason, then only the tail is contracted

[DimaMazin]

 

length contraction is "length of what's moving"/gamma

 

if only the tail is moving for some unexplained reason, then only the tail is contracted

Even when engines are on rocket nose their force with connection force reduce tail length and move the tail earlier than nose. Still I think I do make math of length contraction. You don't make.

Edited September 21, 2015 by DimaMazin

[swansont]

Even when engines are on rocket nose their force with connection force reduce tail length and move the tail earlier than nose. Still I think I do make math of length contraction. You don't make.

 

But the midpoint has to move earlier, too, since it takes time for the force to propagate. But at a reduced v as compared to the tail.

 

The math you used is wrong.

[DimaMazin]

 

But the midpoint has to move earlier, too, since it takes time for the force to propagate. But at a reduced v as compared to the tail.

 

The math you used is wrong.

Agen you make math of break, it is when connection force is weaker than force of acceleration. The forces initially pull tail.

 

If tail is very massive then the forces some pull nose to tail.

New math is always wrong for simple folk.

[Mordred]

Simple folk like I hadn't heard that one before. For one thing Swansort is a professional physicist. Secondly not a single statement you have made this thread has made any sense whatsoever.

 

Length contraction.

 

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

 

Those are the length contraction formulas according to GR. Your formulas are nonsense. To put it mildly. GR is extremely well tested. Your formulas are not.

Agen you make math of break, it is when connection force is weaker than force of acceleration. The forces initially pull tail.

 

If tail is very massive then the forces some pull nose to tail.

New math is always wrong for simple folk.

The amount of force is already applied in achieving relativistic velocity. Force applies to changes in velocity. Newtons three laws of inertia.

 

An object can already be at a relatistic velocity and have length contraction without additional forces exerted upon it at the time of measurement.

 

Secondly the above equations apply to whatever length of measurement you apply. This talk of tail end/ front end is nonsense. The ship is defined by front to back end. A to B. Even if you decide to devide the ship into individual portions. The same equations above apply to whatever measuring rod you choose to measure. The ratio of measurement/length contraction between whatever you set as L and [latex]L_o[/latex] does not change from the ratio of source to observer defined in the above equations. Deviding the ship into seperate measurement rods DO NOT affect the ratio above

Let's put this to math.

 

Define your ship length a to c with b being the centre.

 

Replace L and [latex]L_o[/latex] with a,b,c and [latex]\acute{a}\acute{b}\acute{c}[/latex] respectively.

 

Lets see [latex] a,b=\frac{\acute{a}\acute{b}}{\gamma (v)}[/latex] as the ship is devised in two the same ratio applies to the second measuring rod therefore..

 

[latex] b,c=\frac{\acute{b}\acute{c}}{\gamma (v)}[/latex]

 

this means that [latex](a,b)+(b,c)=(a,c)[/latex] and respectively [latex](\acute{a},\acute{b})+(\acute{b},\acute{c})=\acute{a},\acute{c}[/latex]

 

This means [latex] a,c=\frac{\acute{a}\acute{c}}{\gamma (v)}[/latex]

If you wish to check read.

 

https://en.m.wikipedia.org/wiki/Length_contraction

You can devide the ship into however many chunks as you want. The ratio of change on the measuring stick follows the same relations. As the velocity is the same for every portion

 

 

The problem you seemingly have and this is only a guess, is that you are confusing velocity and acceleration.

"A non-inertial reference frame is a frame of reference that is undergoing acceleration with respect to an inertial frame. An accelerometer at rest in a non-inertial frame will in general detect a non-zero acceleration."

 

The above formulas apply to inertial frames. Ie velocity not acceleration.

Edited September 22, 2015 by Mordred

[swansont]

Agen you make math of break, it is when connection force is weaker than force of acceleration. The forces initially pull tail.

 

If tail is very massive then the forces some pull nose to tail.

If only 1 cm of the tail is moving, then only 1 cm will be length contracted.

 

New math is always wrong for simple folk.

Yeah, whatever.

[DimaMazin]

The problem you seemingly have and this is only a guess, is that you are confusing velocity and acceleration.

"A non-inertial reference frame is a frame of reference that is undergoing acceleration with respect to an inertial frame. An accelerometer at rest in a non-inertial frame will in general detect a non-zero acceleration."

 

The above formulas apply to inertial frames. Ie velocity not acceleration.

I consider instant acceleration to v. And my equation is created for it. All parts of object are instantly accelerated to v , but they can't be simultaneously accelerated to have length contraction. Why do you think they are obligated be simultaneously accelerated? Events can be non-simultaneous.

If only 1 cm of the tail is moving, then only 1 cm will be length contracted.

 

 

Yeah, whatever.

You even don't know where is length contracting to.

[swansont]

I consider instant acceleration to v. And my equation is created for it. All parts of object are instantly accelerated to v , but they can't be simultaneously accelerated to have length contraction. Why do you think they are obligated be simultaneously accelerated? Events can be non-simultaneous.

That's not consistent with what you were saying before. If they are instantly accelerated to v (which, of course is not actually possible), why are they not simultaneously accelerated to v?

 

You even don't know where is length contracting to.

I sure as hell don't know what you're talking about, and everybody else in the thread has said similar things. Maybe because you do a crappy job of explaining what you think is going on. I was under the impression the nose and tail had different speeds because the force can't propagate faster than c, but now you are saying something inconsistent with that.

[DimaMazin]

That's not consistent with what you were saying before. If they are instantly accelerated to v (which, of course is not actually possible), why are they not simultaneously accelerated to v?

 

 

I sure as hell don't know what you're talking about, and everybody else in the thread has said similar things. Maybe because you do a crappy job of explaining what you think is going on. I was under the impression the nose and tail had different speeds because the force can't propagate faster than c, but now you are saying something inconsistent with that.

They are not simultaneously accelerated because forces don't instantly transfer momentum especially when losing length of parts creates delay of subsequent collisions. Therefore speed of force propagation is v.

Edited September 22, 2015 by DimaMazin

[swansont]

They are not simultaneously accelerated because forces don't instantly transfer momentum especially when losing length of parts creates delay of subsequent collisions. Therefore speed of force propagation is v.

How does the object instantly accelerate to v without there being a force changing its momentum?

[DimaMazin]

How does the object instantly accelerate to v without there being a force changing its momentum?

I didn't say that.

Sorry. Seems I have made mistake about propagation at v, I will reconsider it.

Do you think all railroad cars in train are simultaneously accelerated after a push?

[swansont]

I didn't say that.

Yes, you did.

 

"All parts of object are instantly accelerated to v" and "They are not simultaneously accelerated because forces don't instantly transfer momentum"

 

Sorry. Seems I have made mistake about propagation at v, I will reconsider it.

Do you think all railroad cars in train are simultaneously accelerated after a push?

No, but you eliminated that consideration when you claimed that "All parts of object are instantly accelerated to v" and when I asked you what was happening with the middle of the rocket, you said "middle of the rocket isn't moving"

[Mordred]

Your adding unnecessary confusion.

 

If different cars are moving at different velocities, they are in different inertial frames and must be calculated seperately for each car.

 

However all the cars can also be moving at the same velocity. Then they can be included in the same calculation.

 

The formula for length contraction rely on the velocity. Not the acceleration. If you have an accelerating object you can use the same formulas I posted by using the accelating objects instantaneous velocity.

Force isn't particularly useful in relativity the amount of force will vary depending on the reference frames

 

[latex]F=\frac{dp}{dt}=\frac{d}{dt}\gamma mv[/latex].

 

One reason why the force to accelerate an object will vary in different reference frames is due to changes in inertial mass.

Edited September 22, 2015 by Mordred

[DimaMazin]

I confused some things.

Speed of force propagation = r/t = r/[(gamma-1)r/(gamma_*v)] = gamma_*v/(gamma-1)