Acceleration of A Classical Particle without A Force Acting on The Particle

**trfrm** · April 14th, 2013, 10:07 PM

Excuse me ... .

In sight, it is impossible that a classical particle can be accelerated without a force acting on it ... . In other word ... , the velocity of the particle cannot change if the force acting on it is zero, sightly ... .

But ... , in fact, this opinion is not always true generally ... .

Really, the Newton’s second law of motion is a differential equation

where the

and the

, respectively, are the force and the momentum of the particle as function of the time

... .

If the velocity and mass of the particle, respectively, are

and

as function of time

, then

... . Thus, in general,

where

is rate of the mass, and

is its acceleration ... . The last expression doesn’t only obey in non-relativistic cases, but also in relativistic cases ... .

If we set

, then

... . In this case,

alias

... . Thus (by subtituting or by differentiating), this yields

... . [Here, the

and the

denote the initial mass and the initial velocity, respectively ... .]

In another case, if

, then

... .

Therefore, really, the force merely changes the momentum rather than the velocity ... .

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**Howard Roark** · April 14th, 2013, 11:08 PM

Originally Posted by trfrm

Excuse me ... .

In sight, it is impossible that a classical particle can be accelerated without a force acting on it ... . In other word ... , the velocity of the particle cannot change if the force acting on it is zero, sightly ... .

But ... , in fact, this opinion is not always true generally ... .

Really, the Newton’s second law of motion is a differential equation

where the

and the

, respectively, are the force and the momentum of the particle as function of the time

... .

If the velocity and mass of the particle, respectively, are

and

as function of time

, then

... . Thus, in general,

where

is rate of the mass, and

is its acceleration ... . The last expression doesn’t only obey in non-relativistic cases, but also in relativistic cases ... .

If we set

, then

... . In this case,

alias

... . Thus (by subtituting or by differentiating), this yields

... . [Here, the

and the

denote the initial mass and the initial velocity, respectively ... .]

In another case, if

, then

... .

Therefore, really, the force merely changes the momentum rather than the velocity ... .

This stuff is really bad. Care to try again, given the information that

. May I suggest that you forget about the nonsense

. This doesn't exist.
If you want to do "classical domain", just set

**trfrm** · April 14th, 2013, 11:52 PM

Originally Posted by xyzt

This stuff is really bad. Care to try again, given the information that

. May I suggest that you forget about the nonsense

. This doesn't exist.
If you want to do "classical domain", just set

O.K. ... . Thanks ... .

We have known that the relativistic momentum is

, where

and

... . [Here, the

is not the initial mass, but is the rest mass ... .] Then,

where

and

... , so

... .

Thus ... (by subtituting), this yields

which be known as the relativistic force ... .

[Note that the special relativity (without quantum aspect) is sometimes classified into the classical mechanics subject ... . I saw it in the Classical Mechanics textbook by Hebert Goldstein, 1980 ... .]

**Howard Roark** · April 14th, 2013, 11:58 PM

Originally Posted by trfrm

Originally Posted by xyzt

This stuff is really bad. Care to try again, given the information that

. May I suggest that you forget about the nonsense

. This doesn't exist.
If you want to do "classical domain", just set

O.K. ... . Thanks ... .

We have known that the relativistic momentum is

, where

and

... . [Here, the

is not the initial mass, but is the rest mass ... .] Then,

where

and

... , so

... .

Thus ... (by subtituting), this yields

which be known as the relativistic force ... .

[Note that the special relativity (without quantum aspect) is sometimes classified into the classical mechanics subject ... . I saw it in the Classical Mechanics textbook by Hebert Goldstein, 1980 ... .]

Better but still wrong, you should get:

**trfrm** · April 15th, 2013, 12:48 AM

Hence, if the force was zero, then the acceleration should not be zero ... . In other word (in the relativistic case without presence of force),

, so

alias

... .

N.B. : The

and the

, respectively, are the initial gamma and the initial velocity (at

) ... .

Because the final velocity has same direction with the initial velocity, then

Thus, in this case,

if

... .

**Howard Roark** · April 15th, 2013, 01:03 AM

Originally Posted by trfrm

Hence, if the force was zero, then the acceleration should not be zero ... . In other word (in the relativistic case without presence of force),

, so

alias

... .

N.B. : The

and the

, respectively, are the initial gamma and the initial velocity (at

) ... .

Because the final velocity has same direction with the initial velocity, then

Thus, in this case,

if

... .

I don't know where you are trying to get, you simply have:

meaning that you get

A little further algebraic manipulation shows that the above is impossible unless

.

**trfrm** · April 15th, 2013, 02:06 AM

Oh O.K. ... . From the identity of vectors

, then

alias

... .

Thus,

alias

... .

Since

, then

... .

In the last equation, we can see that there are transversal and longitudinal components of the force ... .

**Howard Roark** · April 15th, 2013, 02:11 AM

Originally Posted by trfrm

Oh O.K. ... . From the identity of vectors

, then

alias

... .

Thus,

alias

... .

Since

, then

... .

In the last equation, we can see that there are transversal and longitudinal components of the force ... .

It's getting worse, you should really arrive to :

**trfrm** · April 15th, 2013, 02:26 AM

Originally Posted by xyzt

I don't know where you are trying to get, you simply have:

meaning that you get

A little further algebraic manipulation shows that the above is impossible.

Here ... , incidentally,

... .

Originally Posted by xyzt

It's getting worse, you should really arrive to :

It turns out that the equation

is not different from the equation

... .

**Howard Roark** · April 15th, 2013, 09:01 AM

Originally Posted by trfrm

Originally Posted by xyzt

I don't know where you are trying to get, you simply have:

meaning that you get

A little further algebraic manipulation shows that the above is impossible.

It turns out that the equation

is not different from the equation

... .

If

this means

const, i.e.

(i.e

) so

. This is the correct chain of inference, what you posted so far, isn't.

**Markus Hanke** · April 15th, 2013, 09:34 AM

I think it is important to distinguish pure mathematics from physics in this context. Mathematically it is certainly possible to come up with all sorts of relations affecting speed and acceleration even in the absence of a force, but that does not necessarily mean those relations are physically meaningful. In the context of physics, changing the state of motion of an object requires changing its momentum, and, at least classically, that will require a force.

**trfrm** · April 15th, 2013, 11:52 AM

Originally Posted by xyzt

If

this means

const, i.e.

(i.e

) so

. This is the correct chain of inference, what you posted so far, isn't.

In this context (this relativistic case) ... , it is right ... .

But, in general case, if

, then

(where the index

denotes the initial state) ... , then ... ,

alias

... .

Because in general, the mass is not constant, then

which must not to be zero ... .

But, however, in the relativistic case,

, so the last equation becomes

... .

The possible solution of the last equation is

... .

Originally Posted by Markus Hanke

I think it is important to distinguish pure mathematics from physics in this context. Mathematically it is certainly possible to come up with all sorts of relations affecting speed and acceleration even in the absence of a force, but that does not necessarily mean those relations are physically meaningful. In the context of physics, changing the state of motion of an object requires changing its momentum, and, at least classically, that will require a force.

This seemly comes from the origin definition of the force as derivative of the linear momentum respect to time, that is the Newton’s second law of motion as a differential equation [Goldstein, Classical Mechanics, 1980, Chapter 1] ... .

**Howard Roark** · April 15th, 2013, 12:02 PM

Originally Posted by trfrm

Originally Posted by xyzt

If

this means

const, i.e.

(i.e

) so

. This is the correct chain of inference, what you posted so far, isn't.

In this context (this relativistic case) ... , it is right ... .

But, in general case, if

, then

(where the index

denotes the initial state) ... , then ... ,

alias

... .

Because in general, the mass is not constant, then

which must not to be zero ... .

This is nonsense.

But, however, in the relativistic case,

, so the last equation becomes

... .

The possible solution of the last equation is

... .

Originally Posted by Markus Hanke

I think it is important to distinguish pure mathematics from physics in this context. Mathematically it is certainly possible to come up with all sorts of relations affecting speed and acceleration even in the absence of a force, but that does not necessarily mean those relations are physically meaningful. In the context of physics, changing the state of motion of an object requires changing its momentum, and, at least classically, that will require a force.

This seemly comes from the origin definition of the force as derivative of the linear momentum, that is the Newton’s second law of motion as a differential equation [Goldstein, Classical Mechanics, 1980, Chapter 1] ... .

I don't know why you insist on using the antiquate notion of relativistic mass, this is the source of confusion. Using the modern notion of invariant mass makes things a lot more straightforward. And yes,

is the only acceptable solution for the equation:

**trfrm** · April 15th, 2013, 12:13 PM

Originally Posted by xyzt

I don't know why you insist on using the antiquate notion of relativistic mass, this is the source of confusion. Using the modern notion of invariant mass makes things a lot more straightforward. And yes,

is the only acceptable solution for the equation:

I’m sorry ... . There,

and

are the initial and final relativistic mass, respectively ... .

**Howard Roark** · April 15th, 2013, 12:15 PM

Originally Posted by trfrm

Originally Posted by xyzt

I don't know why you insist on using the antiquate notion of relativistic mass, this is the source of confusion. Using the modern notion of invariant mass makes things a lot more straightforward. And yes,

is the only acceptable solution for the equation:

I’m sorry ... . Hence,

and

... .

I know, no one is using "relativistic mass" anymore, it is messy and a never ending source of errors and misunderstandings.