Why the Lorentz factor?

[MaximThibodeau]

Its clearly disrespectful, close this thread, sorry...

[Strange]

I don't understand people who say, "Hey, I have an idea . . . but I don't want to talk about it"

3 minutes ago, MaximThibodeau said:

Its clearly disrespectful, close this thread, sorry...

Why is it disrespectful to ask where your expression (m c² / i^x) came from?

[Strange]

Where did your expression (m c² / i^x) come from?

[MaximThibodeau]

2 hours ago, MaximThibodeau said:

I came from me

1 hour ago, Strange said:

You are your own father?

Should read: It came from me

 

[Strange]

1) what is x in i^x ?

2) it is impossible for v to be greater or equal to c

[MaximThibodeau]

1- x is a value to achieve the equation, i dont understand ?

2- so, what is happening to the acceleration, so the speed, when we are close to the speed of light, do we decelerate, with the gas pedal still pushed, in the rocket referential? 

[studiot]

14 hours ago, MaximThibodeau said:

Could someone explain me what is mean?

 

13 hours ago, MaximThibodeau said:

What is happening to the quantity that is acceleration when the object is close to light speed ? I mean, why when we push on the acceleration pedal we actually slow down, close to light speed ?

 

It is clear that your English is not perfect so here is the answer to the question I think you originally wanted to ask, which equivalent to

 

"What happens if you apply the Lorenz tranformation twice (or many times)?"

Mathematically that means "what happens if you compose Lorenz tranformations as in acceleration?"

 

Well if you make the substitutions and do the algebra you find that you have another Lorenz transformation.

That is the Lorenz transformation is an eigenfunction of the system.

Edited March 31, 2018 by studiot
spelling

[MaximThibodeau]

I went at "Université de Montreal" in mathematic, so i understand your point about eigenfunction, but my question is not answered: Do we lost kinetic energy while pushing more, when we are close to the light speed ?

[studiot]

By pushing I assume you mean acceleration?

Questions of energy depend upon the frame in which you are making the measurements, but to a general observer the KE increases as the mass appears to increase.

Don't forget that kinematics is mass independent it deals with accelerations and velocities; distances and times without regard to the masses and forces involved.
Such considerations lie in the province of dynamics.

[MaximThibodeau]

In the rocket referential, with constant acceleration or not?

[studiot]

How would the rocket master be able to tell if his acceleration is constant or not in his own frame?

 

Surely this is one of the fundamental principles of relativity?

[StringJunky]

6 minutes ago, studiot said:

How would the rocket master be able to tell if his acceleration is constant or not in his own frame?

 

Surely this is one of the fundamental principles of relativity?

Is that highlighted  not an oxymoron?

[studiot]

2 minutes ago, StringJunky said:

Is that highlighted  not an oxymoron?

No, not really.

The formulae v2 = u2 +2fs, and v = u +ft  is taught to high school pupils who study kinematics of constant acceleration.

When the constant refers to the velocity we have the simpler formulae v = u and s = ut.

Maxim, since you say you have some higher Mathematics you should be able to cope easily with this modern book on University Mechanics which covers Relativity in the second half with really clear explanations and working.

 

Dynamics and Relativity

W D McComb

Oxford University Press

 

Here is a sample page on relativistic acceleration

 

 

 

 

[MaximThibodeau]

It is in the rest frame, what about the rocket frame, again?

[StringJunky]

7 minutes ago, studiot said:

No, not really.

The formulae v2 = u2 +2fs, and v = u +ft  is taught to high school pupils who study kinematics of constant acceleration.

When the constant refers to the velocity we have the simpler formulae v = u and s = ut.

 

OK. Cheers.

[studiot]

4 minutes ago, MaximThibodeau said:

It is in the rest frame, what about the rocket frame, again?

Sorry communication difficulty I think.

The rocketmaster = pilot so is automatically in the rocket frame.

Edited March 31, 2018 by studiot

[MaximThibodeau]

11 minutes ago, studiot said:

 

proper acceleration = the magnitude of the acceleration in the rest frame

[studiot]

9 minutes ago, MaximThibodeau said:

proper acceleration = the magnitude of the acceleration in the rest frame

So?

[MaximThibodeau]

1 hour ago, MaximThibodeau said:

so, what is happening to the acceleration, so the speed, when we are close to the speed of light, do we decelerate, with the gas pedal still pushed, in the rocket referential?

 

[Strange]

10 hours ago, MaximThibodeau said:

1- x is a value to achieve the equation, i dont understand ?

That looks completely wrong, I assume what you mean is:

[math]\sqrt{1 - \frac {v^2}{c^2} } = i x[/math] (when v > c)

where [math]x = \sqrt{\frac {v^2}{c^2} - 1} [/math]

Quote

2- so, what is happening to the acceleration, so the speed, when we are close to the speed of light, do we decelerate, with the gas pedal still pushed, in the rocket referential? 

No, you do not decelerate but the rate of acceleration , as seen from another frame of reference, is not as great. In your own frame you will see no change (because in your own frame you are not moving at the speed of light - you are not moving at all).

As "acceleration" can mean different things, let's try and be more explicit about what happens.

Let's say your rocket has a "gas pedal" that gives you 1g of acceleration when you press it all the way down. In other words, you will feel yourself pressed against your seat with a force equivalent to 1g (ie the same force that pushes on you because of Earth's gravity.

When you are moving slowly, relative to observers on Earth, they will see your speed increase ("acceleration") as 9.8 ms^-2 But when you are moving faster (relative to the observers on Earth) then they will see your speed increase as less than 9.8 ms^-2

The speed increase they see will get less and less as you get closer to the speed of light (relative to the observers on Earth). Which means that (relative to them) you will never reach the speed of light.

Meanwhile, you will continue to feel exactly the same 1g force whenever you accelerate. 

You understand, I hope, that speed is relative. So you can only be moving at the speed of light as seen by a particular observer.

For example, after a few months you slowly pass another rocket that is moving away from the Earth at half the speed of light (relative to the Earth) then that rocket will see your speed increase as 9.8 ms^-2 (when you accelerate at 1g).

So the decreasing "acceleration" (speed increase) is what other observers see and depends on your speed relative to them.

Does that help?

9 hours ago, studiot said:

How would the rocket master be able to tell if his acceleration is constant or not in his own frame?

Because he has an accelerometer that measures the force created by the acceleration.

Quote

Surely this is one of the fundamental principles of relativity?

Acceleration is not relative (but the speed increase caused by that acceleration is).

Edited April 1, 2018 by Strange

[michel123456]

It is Lorentz, not Lorenz

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

[Strange]

12 hours ago, MaximThibodeau said:

proper acceleration = the magnitude of the acceleration in the rest frame

Exactly. So if you accelerate away from Earth at 1g, your proper acceleration will always be 1g.

But your coordinate acceleration, as seen from Earth, will continually decrease (but never become negative; i.e. you will not appear to decelerate).

[studiot]

4 hours ago, Strange said:

14 hours ago, studiot said:

How would the rocket master be able to tell if his acceleration is constant or not in his own frame?

Because he has an accelerometer that measures the force created by the acceleration.

Perhaps you would like to comment on the highlighted paragraph?

 

[Strange]

5 minutes ago, studiot said:

Perhaps you would like to comment on the highlighted paragraph?

Exactly.

(I was going to point out that constant acceleration is indistinguishable from gravity; that being in free fall you are not accelerating; that, in GR, standing on the ground is equivalent to constant acceleration; etc but thought it wasn't relevant and, given the OPs poor grasp of physics, would only confuse things.)

This is why I drew a distinction between an increase in speed (which is not necessarily acceleration; e.g. free fall) and experiencing a constant force (because of the effect of your rocket).

[studiot]

16 minutes ago, Strange said:

Exactly.

(I was going to point out that constant acceleration is indistinguishable from gravity; that being in free fall you are not accelerating; that, in GR, standing on the ground is equivalent to constant acceleration; etc but thought it wasn't relevant and, given the OPs poor grasp of physics, would only confuse things.)

This is why I drew a distinction between an increase in speed (which is not necessarily acceleration; e.g. free fall) and experiencing a constant force (because of the effect of your rocket).

 

I am wondering if Maxim's grasp is weaker in English than Physics.

He is trying to get a grip on the Physics and seems to grasp clear statements on the subject, but doesn't seem to me to be always expressing exactly what he means.

(A PM in French might help him)

So attacking poorly worded English as poor thinking is counterproductive.

 

The rocket was not mine.