I Acceleration without force is possible in relativity

[ravi#]

In prime frame, if Fz =0 & ratio Fx/Fy is equal to ( v/c² . Uy)/(1-V .Ux/c²) then after transformation in S’ frame F’x becomes F’x = 0 because
F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²) ----transformation equation
Let, consider one situation
In frame S :- Now, just consider that on magnetic substance on frictionless platform magnetic forces are acting in X-direction & in Y-direction. Magnetic force Fx is so adjusted by software program that ratio Fx/Fy is always equal to ( v/c² . Uy)/(1-V .Ux/c²).

Then, Forces Fx (very small) & Fy in this frame will create acceleration ax & ay in direction x & y.

Observer frame S’ is moving with velocity V with relative to frame S then in frame S’ :-

There is acceleration in X' direction because ax’= ax/{r³. (1-ux. v/c²) ³ } where r =1/(1-v²/c²) ^0.5 but there is no force in X'- direction because

as F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²) & as Fx/Fy=( v/c² . Uy)/(1-V .Ux/c²)

So, F’x =0

Means, in this case in frame S’
there is acceleration in X’-direction but no force is present in X’-direction.
In relativity, is this possible that in some frame, there is acceleration but no force in X-direction.

 

[Orodruin]

In relativity, is this possible that in some frame, there is acceleration but no force in X-direction.

I did not have time to read your entire setup, but yes, it is possible to have acceleration but no force in the x-direction. The force is given by the change in momentum, not the change in velocity. The momentum in the x-direction is given by $p_x = m\gamma v_x$, where $\gamma = 1/\sqrt{1-v^2}$ generally depends on all the velocity components. It is therefore possible to change $v_x$ and keep the same momentum in the x-direction if you change the other velocity components so that $\gamma$ changes accordingly to compensate.

 

[Ibix]

Acceleration is not necessarily parallel to the force producing it in relativity, no. This follows from the definition of force as $$\vec{F}=\frac{d\vec{p}}{dt}$$where $\vec{p}=\gamma m\vec{v}$.

 

[ravi#]

Thanks, interesting

 

[Dale]

This is one reason to use four-vectors instead of three-vectors.

 

[Orodruin]

This is one reason to use four-vectors instead of three-vectors.

4-acceleration is also not necessarily proportional to 4-force. It only holds of the mass of the object is constant.

 

[PWiz]

4-acceleration is also not necessarily proportional to 4-force. It only holds of the mass of the object is constant.

Even at relativistic speeds four-acceleration is related to the four-force such that

where m is the invariant mass of a particle.

 

[Orodruin]

This is not the general definition, it is only true when the mass $m$ is constant. The general expression is $F = dP/d\tau = mA + \dot m V$.

 

[PWiz]

This is not the general definition, it is only true when the mass mmm is constant.

Oh, I didn't realize that you were talking about objects which are gaining/losing invariant mass. (You'd only mentioned mass in your post.) But even in that case, the equation is identical to its non-relativistic version with just the 3-vectors replaced by 4-vectors, and this corroborates Dale's point.

 

[CrackerMcGinger]

This is very interesting, but there are things that still are not clear to me. At which point would the acceleration no longer affect the object(or, more importantly, is the object gaining speed)? I've figured that momentum had to do with acceleration without force, but in what way?

 

[Ibix]

The object can always consider itself to be instantaneously at rest in some inertial frame, so it can always accelerate. The velocity will be asymptotic to c.

You'll never have an acceleration without a force. However, you may have an acceleration which has components perpendicular to the impressed force.

 

[CrackerMcGinger]

Is there any possible way you can apply momentum in a way that you essentially "fall" forward(or the direction equal to forward)?

 

[Ibix]

I have no idea what you are asking. What forces are you thinking about on what objects?

 

[CrackerMcGinger]

In the event that the force(propulsion) acting upon the object(any object that has descent mass and density) no longer acts on the object, is it possible that momentum can be used to allow the object to "fall" forward? The object is in space, therefor the only forces would be inertia and the propulsion.

 

[Ibix]

Inertia isn't a force, not in Newtonian nor relativistic physics. If there are no forces on an object (e.g. a rocket in space with the engines off far from any mass) then it will continue in a straight line at constant speed. This is true in Newtonian and relativistic physics.

 

[ravi#]

I am happy, many intelligent understand that
When we transform forces for Fx =0 , F'x is not zero but when acceleration is transformed then for ax = 0 , a'x =0 (By transformation equation)
this create above situation after transformation that there may be acceleration in X-direction but no force in X-direction.
When I read some initial post, I think problem is solved, but now I think problem is not solved.

In SR, in any frame:-
Fx = d/dt { r. mo . Ux } where r = (1- U²/C²)^-1/2
Now, I differentiate above
Fx = mo. r . dUx/dt + mo. Ux . dr/dt
Fx = mo. r. ax + mo. Ux . r³. (U/C²) . a
This equation clearly shows that
i.e. if 'ax' is not zero then 'Fx' can not be zero.
Mean' if there is acceleration in X-direction then there is Fx in X-direction.

Is there any alternative mathematics is available which proves that even ax is present, Fx may not present in SR.

 

[A.T.]

I am happy, many intelligent understand that
When we transform forces for Fx =0 , F'x is not zero but when acceleration is transformed then for ax = 0 , a'x =0 (By transformation equation)
this create above situation after transformation that there may be acceleration in X-direction but no force in X-direction.
When I read some initial post, I think problem is solved, but now I think problem is not solved.

In SR, in any frame:-
Fx = d/dt { r. mo . Ux } where r = (1- U²/C²)^-1/2
Now, I differentiate above
Fx = mo. r . dUx/dt + mo. Ux . dr/dt
Fx = mo. r. ax + mo. Ux . r³. (U/C²) . a
This equation clearly shows that
i.e. if 'ax' is not zero then 'Fx' can not be zero.
Mean' if there is acceleration in X-direction then there is Fx in X-direction.

Please use LaTeX, or nobody will read your math. What did you assume about the change of U when Ux changes?

 

[Ibix]

Hint: re-read Orodruin's reply.

There is a guide to how to use LaTeX linked immediately below the box where you type your reply. Please read it and use it. Then we will not have to struggle to understand straightforward maths. You can write your final expression (quote my post to see how I did it) as $$F_x=\gamma m_0 a_x+\gamma^3 m_0u_x\frac {u}{c^2}a$$which is a lot clearer, and may help you to spot your error.

 

[ravi#]

This is interesting. (for every acceleration in x-direction, force Fx exist or not.)
Point 1:- Orodruin is true. If we use transformation equation in relativity, then for every set up, there is one value of Fx for which F'x =0
(as given in post 1)
but for every acceleration in prime frame there is acceleration in non-prime frame.
This happen due to nature of transformation equations in relativity for force & acceleration.
This can create following situation in non-prime frame,
there is acceleration in X-direction but no force in X-direction --------(1)

Point 2:- Now, I do calculation in same frame only,
In any frame, for getting force in X-direction, I have to differentiate Px by time t.
( I have not assumed anything. I am just plainly differentiating)
p_x = m_0\gamma u_x
after differentiation. I find following equation
F_x=\gamma m_0 a_x+\gamma^3 m_0u_x\frac {u}{c^2}a
Means, in any situation
If there is acceleration a_x then there must be Fx because 1St portion \gamma m_0 a_x can not be zero.
At every case, if there is acceleration in X-direction then there is force in X-direction---------(2)

Now, (1) & (2) out puts are completely opposite. What is wrong.

 

[Ibix]

Try writing $u=\sqrt {u_x^2+u_y^2+u_z^2}$ and doing the differentiation again.

 

[yogi]

The subject poses a curious side question - whether accelerating expansion requires dark energy - if (d/dt)[mv] = 0 during accelerated expansion, an increase in the inertial value of existing mass, ... well ...it might be something to muse over after a few drinks, or maybe more than a few.

 

[Orodruin]

The subject poses a curious side question - whether accelerating expansion requires dark energy - if (d/dt)[mv] = 0 during accelerated expansion, an increase in the inertial value of existing mass, ... well ...it might be something to muse over after a few drinks, or maybe more than a few.

This has absolutely nothing to do with accelerated expansion, that is a different concept entirely.

 

[ravi#]

This differentiation is much simple, if $u^2 = u_x^2+u_y^2+u_z^2$ then $u a = u_x a_x +u_y a_y + u_z a_z$
So, just replace "u a" by above values then
$F_x=\gamma m_0 a_x+\gamma^3 m_0\frac {u_x}{c^2} (u_x a_x + u_y a_y + u_z a_z)$
Means, in any situation
If there is acceleration $a_x$ then there must be Fx because 1St portion $\gamma m_0 a_x$ can not be zero.
At every case, if there is acceleration in X-direction then there is force in X-direction---------(2)
Now, (1) [refer post 19] & (2) out puts are completely opposite. What is wrong.

 

[A.T.]

$F_x=\gamma m_0 a_x+\gamma^3 m_0\frac {u_x}{c^2} (u_x a_x + u_y a_y + u_z a_z)$
Means, in any situation
If there is acceleration $a_x$ then there must be Fx because 1St portion $\gamma m_0 a_x$ can not be zero.

Did you consider that the two summands can have opposite signs?

 

[Dale]

This differentiation is much simple, if $u^2 = u_x^2+u_y^2+u_z^2$ then $u a = u_x a_x +u_y a_y + u_z a_z$

No. Consider what happens if the vectors are perpendicular.

 

[Ibix]

No. Consider what happens if the vectors are perpendicular.

Agree that the bit you quoted is not an equality. It is ravi replacing an incorrect expression with a correct one. He's determining $d\gamma/dt$, which requires him to determine $d(u^2)/dt$. He had written that as $ua$, but I think that it should be what he now has - $\vec {u}.\vec {a}$. So, for clarity, I believe that ravi's final expression is correct, although the explanation of what he was doing contains the error you highlighted.

A.T. has pointed out the issue in his interpretation, I think.

Unless I missed something...

 

[CrackerMcGinger]

Try writing $u=\sqrt {u_x^2+u_y^2+u_z^2}$ and doing the differentiation again.

This is one of the reasons I love this forum. People help each other instead of criticizing their inaccuracies. Ibix, I would like to start a conversation about some things I've thought about for some time.

 

[ravi#]

Thanks Mr Dale, you are completely right.
but in differentiation when we differentiate vector & we get two vectors as out put in different direction then that is scalar product or dot product of vectors.
$u^2 = u_x^2+u_y^2+u_z^2$
after differentiation
$u o a = u_x o a_x + u_y o a_y + u_z o a_z$ scalar products ------eq (1)
Now, I prove that this is true.
just as in train cabin gravity g just act on ball in moving train with velocity ux.
means, acceleration & velocity are perpendicular (this situation is possible)
As you say if $u_x$ & $a_y$ are only present as per above situation in perpendicular direction & other quantities are zero .
Then eq (1) changes as
$u. a. cos 90 = u_x a_x + u_y a_y + u_z a_z$ ----eq(2)
This is write differentiation
$0 = u_x . 0 + 0 . a_y$ ... as cos 90 =0
0 = 0
This shows that result of differentiation =$u_x a_x + u_y a_y + u_z a_z$
is right
&
$F_x=\gamma m_0 a_x+\gamma^3 m_0\frac {u_x}{c^2} (u_x a_x + u_y a_y + u_z a_z)$
is completely right.
Means, in any situation If there is acceleration $a_x$ then there must be Fx because 1St portion $\gamma m_0 a_x$ can not be zero.
At every case, if there is acceleration in X-direction then there is force in X-direction---------(2)
Now, (1) [refer post 19] & (2) out puts are completely opposite. What is wrong.

 

[A.T.]

$F_x=\gamma m_0 a_x+\gamma^3 m_0\frac {u_x}{c^2} (u_x a_x + u_y a_y + u_z a_z)$
Means, in any situation If there is acceleration $a_x$ then there must be Fx because 1St portion $\gamma m_0 a_x$ can not be zero.

Did you read post #24?

 

[Orodruin]

Now, (1) [refer post 19] & (2) out puts are completely opposite. What is wrong.

Your math is correct. This conclusion is wrong:

Means, in any situation If there is acceleration $a_x$ then there must be Fx because 1St portion $\gamma m_0 a_x$ can not be zero.
At every case, if there is acceleration in X-direction then there is force in X-direction---------(2)

It is perfectly possible to have
$$
\gamma m_0 a_x = - \gamma^3 m_0 \frac{u_x}{c^2}\vec u \cdot \vec a.
$$
You only have to let
$$
a_y = - \frac{c^2 a_x}{u_x u_y}\left(\gamma^{-2} + \frac{u_x^2}{c^2}\right)
$$
and $a_z = 0$.