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 $a_z = 0$.

 

[ravi#]

Interesting
You are right. Thiscan be solution if $u_y$ & $a_y$ are in opposite direction but condition in post 1 is
Fx/Fy is always equal to $\frac { \frac {v}{c^2} . u_y}{(1- \frac {v. u_x}{c^2})}$
here, $F_y$ , $a_y$ & $u_y$ are in same direction.
So, this can not be solution .
Point 1:-When I was working on this problem. I interact with one interesting situation.
It is wrong if I will not share it to this important forum. If I am wrong please correct me.
$F'_x$ = Fx -$\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$ is the transformation equation
Means if Fx = 0 then also there is -ve force in non prime frame.
as $F'_x$ = 0 - $\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$
so, $F'_x$ = - $\frac{ \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$
Now, just consider that $u_x =v$ or object velocity is same as frame velocity or for object whose velocity in X- direction is zero for non-prime framethen also there is -ve force in non prime frame because
$F'_x$ = -$\frac{ \frac {v}{c^2} .F_y u_y}{(1- \frac {v^2}{c^2})}$
Mean's for observer in train, for object in his cabin where $u_x =v$ for platform frame if there is forced acceleration in Y-direction then
$F'_x$ = - $\frac{ \frac {v}{c^2} .F_y u_y}{(1- \frac {v^2}{c^2})}$
This -ve force will automatically act on it.
Mean's for every forced acceleration of object in train cabin in Y-direction, there is -ve force formation in X-direction as per above formula due to motion with prime frame.

is this possible ?

 

[A.T.]

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 $a_z = 0$.

This can be solution if $u_y$ & $a_y$ are in opposite direction

Or if $u_x$ & $a_x$ are in opposite direction

 

[Ibix]

Your original problem specification is in a frame S. In this frame there is a body moving with 3-velocity $\vec{u}=(u_x,u_y,0)$. It is subject to a force $\vec{F}=(vu_yF/(c^2-vu_x),F,0)$, where F is some constant. This produces a 3-acceleration $\vec{a}=(a_x,a_y,0)$. In general, Orodruin's expression from #30 is not satisfied in this frame because the x-component of the force is not zero.

You then transform to a frame S', which moves with velocity v in the x direction relative to S. In this frame, the body is moving with 3-velocity $\vec{u'}=(u'_x,u'_y,0)$. It is subject to a force $\vec{F'}=(0,F'_y,0)$ - note that the x-component is zero because you have carefully chosen the direction of your force in S so that this happens. This produces a 3-acceleration $\vec{a'}=(a'_x,a'_y,0)$. In this frame I do expect Orodruin's expression (using primed quantities and noting that $\gamma$ is the $\gamma$ associated with u') from #30 to be satisfied.

You appear to have noted that Orodruin's expression is not satisfied in S, but do not appear to have checked it in S'. It is the latter that is of interest to you.

As for the rest of your post #31, if I understand correctly, you seem to be worried that a force in the y-direction is not parallel to y'. Isn't that an obvious consequence of the force transformation equations you quoted?

 

[ravi#]

(1)
Yes, Mr A.T. you are right I also think about this solution
but problem remains un-resolve when $u_x$ becomes more than $V$ in prime frame.
See transformation equation for F’x, there is –ve force created due to action of Fy or Fz perpendicular to X-direction. It is not form due to any acceleration & velocity has opposite direction. Following problem will make it very clear. I think that these are problem happen due to force can have without acceleration in that direction.

(2)
My second problem in post 31:- This is very important problem which I put in my previous post 31.
This problem can easily be understood by following paradox.
IMPORTANT NEW PARADOX:- On friction-less platform object is moving with constant velocity $u_x$ in X-direction & only magnetic force $F_y$ is acting in Y-direction & there is acceleration in Y-direction only.
For non-prime observer moving with velocity V in X-direction :-
For this observer by transformation of forces equation in x-direction
here, F'x =0 - $\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$
F'x = - $\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$
This force will act in –ve direction on object in non-prime frame. (Even there is no Fx in prime frame).
Mean’s in this non-prime frame, this standing force will act in x-direction.
Let, consider that $u_x=V$
Then also this force will not get cancel &
F'x = - $\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v^2}{c^2})}$ in X'-direction--------(1)
--------Now, this create problem because as $u_x=V$
Here, object velocity matches with observer velocity
Or There is no relative motion of object in this frame in X’-direction
but there is force F’x present in this non-prime frame by equation (1) of transformation.

--------But we know that in any frame if there is no motion in x-direction then
$F_x = 0$ because differentiation of $\gamma m_0 u_x$ gives
$F_x=\gamma m_0 a_x+\gamma^3 m_0 \frac { u_x }{c^2} (u o a)$ in any frame
& if acceleration & velocity is zero in X-direction
Then $F_x = 0$ -----------(2)
Now case(1) & (2) are completely opposite.
Transformation equation says that there can be force even there is no relative motion in X-direction but above equation says that this is impossible.
Interesting but opposite situation.
Now, somebody will say this situation is not possible in reality but I can give example...

 

[Ibix]

Your problem statement appears rather confused to me. Let me restate it to be clear that I understand.

You have a frame S in which there is a body moving with velocity $(u_x,u_y,0)$. It is subject to a force $(0,F_y,0)$.

You then transform to a frame S' where the object moves with velocity $(0,u'_y,0)$ and is subject to a force $(F'_x,F'_y,0)$.

Your claim is that $F'_x \neq 0$ when derived from the force transformation, yet $dP'_x/dt'=0$. The solution is simple - the second statement is false. In the case that $u'_x=0$, $dP'_x/dt \propto a'_x \propto a_x$, and $a_x$ is not zero, as you can see easily from $0=F_x=dP_x/dt$.