## Applied force less than acting force in Special Relativity

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### Applied force less than acting force in Special Relativity

Step 1
This problem can easily be understood by the following paradox.

Before stating this paradox, I want to derive a formula given in standard books on relativity. In Special Relativity, the four-force in the $x$-direction for any inertial frame is:
${F_x}=\frac{\text{d}}{\text{d}t}\left({\gamma}{m_0} {u_x}\right)$ where ${\gamma}{\equiv}{\left(1-{\frac{u^2}{c^2}}\right)}^{-0.5}$ is the Lorentz transformation.
So, after differentiation:
${F_x}={\gamma}{m_0}{\frac{\text{d}{u_x}}{\text{d}t}}+{{\gamma}^{3}}{m_0}{\frac{u_x}{c^2}}{u}{\frac{\text{d}u}{\text{d}t}$
${F_x}={\gamma}{m_0}{a_x}+{{\gamma}^{3}}{m_0}{\frac{u_x}{c^2}}{ua}$ -----(A)
We know,
${u^2}={u_x^2}+{u_y^2}+{u_z^2}$
So, after differentiation:
$2u{\frac{\text{d}u}{\text{d}t}}=2{u_x}{\frac{\text{d}{u_x}}{\text{d}t}+2{u_y}{\frac{\text{d}{u_y}}{\text{d}t}+2{u_z}{\frac{\text{d}{u_z}}{\text{d}t}$
$2ua=2{u_x}{a_x}+2{u_y}{a_y}+2{u_z}{a_z}$
$ua={u_x}{a_x}+{u_y}{a_y}+{u_z}{a_z}$ --------(B)
Combining (A) and (B):
${F_x}={\gamma}{m_0}{a_x}+{{\gamma}^{3}}{m_0}{\frac{u_x}{c^2}}{\left({u_x}{a_x}+{u_y}{a_y}+{u_z}{a_z}\right)}$ ------(1)
Mod note: Eq. (1) was rederived and confirmed in this response.

An object is moving with constant velocity ${u_x}$ in the $x$-direction on a frictionless platform. The only non-zero magnetic force is acting in $y$-direction, and there is acceleration in the $y$-direction only with velocity $u_y=0$ & $F_z=0$.

If we apply Eq. (1) to this case then result will be
${F_x}={{\gamma}^3}{m_0}{\frac{u_x}{c^2}}{u_y}{a_y}$ ---------- as $a_x=0$,
or
${F_x}={F}{a_y}$
as this force is form due to $a_y$ only.

This means that, if there is no magnetic force acting on object from outside in $x$-direction and no $a_x$, then the force will act on object in $+{v_e}$ direction of $x$-axis due to $a_y$.

Important point (1):
This means that the applied magnetic force on object in the $x$-direction is $0$ and that the acting force in the $x$-direction is
${F_x}={{\gamma}^{3}}{m_0}{\frac{u_x}{c^2}}{u_y}{a_y}+0$,
or
${F}{a_y}+0={F}{a_y}$.

Step 2
Now, the force acting in the $x$-direction is
${F_x}={{\gamma}^{3}}{m_0}{\frac{u_x}{c^2}}{u_y}{a_y}$ or ${F}{a_y}$.
Now, after this happens, a very small magnetic force of the same intensity,
$-{f_x}=-{{\gamma}^{3}}{m_0}{\frac{u_x}{c^2}}{u_y}{a_y}$,
or $-{F}{a_y}$, starts acting on the object in the direction opposite to above force (but velocity is still positive ${u}_{x}$) and cancels that above force.

So, Eq. (1) becomes
$0={\gamma}{m_0}{a_x}+{{\gamma}^{3}}{m_0}{\frac{u_x}{c^2}}{\left({u_x}{a_x}+{u_y}{a_y}\right)}$
or
$0={\gamma}{m_0}{a_x}\left(1+{{\gamma}^{2}}{\frac{u_x^2}{c^2}}\right)+{F}{a_y}$
Here,
${F}_{a_y}={\gamma}^{3}{m_0}{\frac{u_x}{c^2}}{u_y}{a_y}$
means
${F}_{a_y}={\gamma}{m_0}-{a_x}{\left(1+{\gamma}^{2}{\frac{u_x^2}{c^2}}\right)}$
This means that there must be acceleration in ${-}{v}_{e}$ $x$-direction to fulfill above equation of Special Relativity.

Now, see above equation carefully, it is of nature
$0=-{f}_{x}+{F}{a}_{y}$
Important point (2):
This means that the applied magnetic force on the object in $x$-direction is $-f_x$ and the acting force in the $x$-direction is $-{f}_{x}+{F}{a_y}=0$, or $0$.

Here, the resultant force in the $x$-direction is zero but there is acceleration.

Step 3
Same things happen for $+v_e$ force in the $x$-direction (for less than ${F}{a_y}$ or more).

Now, I am generalizing above result.

Steps 1&2 clearly show that when we apply any magnetic force $\left({F_{\text{m}}}_{x}\right)$ in the $x$-direction on the object, actual force acting on object is more & that quantity is $\left({{F}_{\text{m}}}_{x}+{{F}_{\text{a}}}_{y}\right)$.

Similarly, if we apply any magnetic force $\left({F_{\text{m}}}_{y}\right)$ in the $y$-direction on the object then actual force acting on object is more and that quantity is $\left({{F}_{\text{m}}}_{y}+{F_{\text{a}}}_{x}\right)$.

This is completely contradicts the results which say that the applied force and acting forces are different (larger) in Special Relativity.

Step 4
Force does work, consumes energy, gain energy, and we know that energy cannot be created. Energy can only be transferred.

It is clear that energy is transferred from the magnet to the object. But, if the applied force is less than the acting force, then the energy gain by the object will be more than the energy lose by the magnet. This means that more work is done by a larger force for the same displacement, so more energy is generated.

This excess energy generation is the problem. Where does this additional energy (or force) come from?
Last edited by Natural ChemE on April 11th, 2016, 6:38 am, edited 47 times in total.
Reason: Formatting.
mahesh
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### Re: Applied force less than acting force in Special Relativi

mahesh,

Welcome to the forums! I removed the caps lock from the thread title, which is to be avoided.

Could you type this up in an easier-to-read way? For example, I can't really understand
; I'm only able to recognize it as the Lorentz transformation (or four-force) based on the context. It would be much easier to read if written in $\TeX$, e.g.
$F_x={\frac{\text{d}}{\text{d}t}}{\left({{\gamma}m{u}_{x}}\right)}$ where ${\gamma}{\equiv}{\left(1-\frac{u^2}{c^2}\right)}^{-\frac{1}{2}}$.
Natural ChemE
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### Re: Applied force less than acting force in Special Relativi

In S.R., force is not related to change in the state of motion or acceleration but with change in moment.
So, even I move towards falling ball,
$F_x=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y$ -------- this force will act on the ball.
--------------------------------------------------------------------------------------------------
If mathematics of post 1 is true & applied force is always less than acting force then if old man pulls the cart on horizontal platform with force f and fx, fy are their components in X & Y direction respectively
Then above calculation says that actual forces acting on the cart are not fx, fy but
Fx=fx+ $\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y$ = fx +Fmay
& Fy=fy+ $\gamma^3 m_0 \frac { u_y }{c^2} u_x a_x$ = fy +Fmax

This will create problem to special theory of relativity as given in post 1 because force do work done, consume energy.
Where this additional energy comes from?
this is the problem
mahesh
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### Re: Applied force less than acting force in Special Relativi

Welcome mahesh :)

mahesh » 28 Mar 2016, 10:30 wrote:--------------------------------------------------------------------------------------------------
If mathematics of post 1 is true & applied force is always less than acting force then if old man pulls the cart on horizontal platform with force f and fx, fy are their components in X & Y direction respectively

If you want serious consideration of your calculations, please rewrite the equations in tex form, as Natural ChemE requested. Otherwise you may get no useful responses. Too much effort to decipher the math...
--
Regards
Jorrie

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### Re: Applied force less than acting force in Special Relativi

Thanks, I will try to re-write complete post 1 in tex form.......
Force without acceleration in S.R.
& acceleration without force in S.R.
& applied force is less than acting force in S.R.

STEP 1:-This problem can easily be understood by following paradox.
{Before starting this paradox, I want to put some relativity formula's
In any frame
Fx = $\frac { d }{d_t} ( \gamma m_0 u_x)$
After differentiation, we get
So, $F_x=\gamma m_0 a_x+\gamma^3 m_0 \frac { u_x }{c^2} (u_x a_x+u_y a_y+u_z a_z)$ ------(1)}
On frictionless platform, object is moving with constant velocity $u_x$ in X-direction & only magnetic force is acting in Y-direction & there is acceleration in Y-direction only with velocity $u_y$ & $F_z=0$
If we apply eq(1) to this case then result will be
$F_x=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y$
$F_x=F_a_y$ as this force is form due to $a_y$ only
Mean’s even there is no magnetic force acting on object from outside in x-direction & no $a_x$ then also above force will act on object in +ve direction of x-axis due to $a_y$
Important point (1):-
Mean’s applied magnetic force on object in X-direction is 0 & acting force in X-direction is
$F_x=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y +0$or $F_a_y+0=F_a_y$
-----------------------------------------------------------------------------------------------------------------------------------------
STEP 2:-Now, Force acting in X-direction is
$F_x=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y or F_a_y$
Now, after this happen, very small magnetic force of same intensity
-fx =- $\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y$ or $-F_a_y$ start acting on object in direction opposite to above force (but velocity is still positive $u_x$) & cancel that above force.
Mean’s equation (1) becomes
$0 =\gamma m_0 a_x+\gamma^3 m_0 \frac { u_x }{c^2} (u_x a_x+u_y a_y)$
Or $0 =\gamma m_0 a_x(1+\gamma^2 \frac { u_x^2 }{c^2}) +F_a_y$
(Here as $F_a_y=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y$)

Mean’s $F_a_y =\gamma m_0 (-a_x) (1+\gamma^2 \frac { u_x^2 }{c^2})$
Mean’s there must be acceleration in –ve X-direction to fulfill above equation of S.R.
Now, see above equation carefully, it is of nature
$0= -fx + F_a_y$
Important point (2):- Mean’s applied magnetic force on object in X-direction is -fx & acting force in X-direction is $-fx + F_a_y=0$ or 0.
Here, resultant force in X-direction is zero but there is acceleration.
STEP3:- same things happen for +ve force in X-direction (for less than $F_a_y$ or more)
Now, I am generalizing above result.
This clearly shows that when we apply any magnetic force (Fmx) in X-direction on the object, actual force acting on object is more & that quantity is (Fmx+ $F_a_y$)
Similarly,
If we apply any magnetic force (Fmy) in Y-direction on the object then actual force acting on object is more & that quantity is (Fmy+ $F_a_x$)
This is completely complicated results, which says that applied force & acting forces on objects are different in S.R.
STEP4:- Force does work, consume energy, gain energy & we must know that energy cannot be created. It can be transferred only:-
From above setup it must be clear that energy get transfer from magnet to object but if applied force is less than acting force then energy gain by object will be more than energy loose by the magnet. Means due to more work done by more force for same displacement, more energy get generated.
HERE, more energy(& force) is the problem.
Where does this additional energy (& force) comes from?
mahesh
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### Re: Applied force less than acting force in Special Relativi

Is there any significance in your applying of magnetic forces? Why not just force?
-J

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### Re: Applied force less than acting force in Special Relativi

mahesh » 30 Mar 2016, 11:18 wrote:
{Before starting this paradox, I want to put some relativity formula's
In any frame
$\large Fx = \frac { d }{d_t} ( \gamma m_0 u_x)$ ----()
After differentiation, we get
So, $\large F_x=\gamma m_0 a_x+\gamma^3 m_0 \frac { u_x }{c^2} (u_x a_x+u_y a_y+u_z a_z)$ ------(1)}

Some more clarifications required: How do you get from () to (1) and what is $a_x, a_y, a_z$ ?

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### Re: Applied force less than acting force in Special Relativi

BurtJordaan » March 30th, 2016, 8:44 am wrote:
mahesh » 30 Mar 2016, 11:18 wrote:
{Before starting this paradox, I want to put some relativity formula's
In any frame
$\large Fx = \frac { d }{d_t} ( \gamma m_0 u_x)$ ----()
After differentiation, we get
So, $\large F_x=\gamma m_0 a_x+\gamma^3 m_0 \frac { u_x }{c^2} (u_x a_x+u_y a_y+u_z a_z)$ ------(1)}

Some more clarifications required: How do you get from () to (1) and what is $a_x, a_y, a_z$ ?

I'm guessing that ${a}_{i}{\equiv}\frac{\text{d}{u}_{i}}{\text{d}t}$ is acceleration in direction $x_i$, so (1) is meant to follow from () through the chain rule.

I'm almost literally asleep right now, but I'm not positive about the second term. It should be equal to ${u}_{x}{m}_{0}{\frac{\text{d}{\gamma}}{\text{d}t}}$, whatever that is after simplification, so (1)'s correct iff
${\frac{\text{d}{\gamma}}{\text{d}t}}=\frac{{\gamma}^{3}}{{c}^{2}}{\left({u_x}{a_x}+{u_y}{a_y}+{u_z}{a_z}\right)}$.
Anyone wanna check the above equation for ${\gamma}{\equiv}{\left(1-\frac{u^2}{c^2}\right)}^{-\frac{1}{2}}={\left(1-\frac{{u_x^2}+{u_y^2}+{u_z^2}}{c^2}\right)}^{-\frac{1}{2}}$?

EDIT: Performed this check in (ii) of this response below. This relationship turns out to be correct.
Natural ChemE
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### Re: Applied force less than acting force in Special Relativi

Natural ChemE » 30 Mar 2016, 18:02 wrote:Anyone wanna check the above equation for ${\gamma}{\equiv}{\left(1-\frac{u^2}{c^2}\right)}^{-\frac{1}{2}}={\left(1-\frac{{u_x^2}+{u_y^2}+{u_z^2}}{c^2}\right)}^{-\frac{1}{2}}$?

If I calculate for a force and motion only in x-direction, I get simply

$F_x=\gamma^2 m_0 a_x$

So the OP's differentiation eq. (1) seems to be wrong.

PS: I think my calculation is wrong and it is in any case not suitable for the OP scenario. I will follow NCE's derivation see if the OPs conclusions stand up.
Last edited by BurtJordaan on April 1st, 2016, 1:54 am, edited 1 time in total.
Reason: Correction&retraction

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### Re: Applied force less than acting force in Special Relativi

I am giving detail calculation for differentiation
In any frame, for force in X-direction by S.R.
$F_x = \frac { d }{d_t} ( \gamma m_0 u_x) where \gamma= \(1-\frac{u^2}{c^2})^-^1^/2$

So, after differentiation
$F_x= \gamma m_o \frac{du_x}{dt} + \gamma ^3. mo. \frac{u_x}{c^2}. (u . \frac{du}{dt})$
$Fx= \gamma mo. a_x + \gamma ^3. mo. \frac{u_x}{c^2}. (u . a) -----(A)$

We know, $u^2=u_x^2+u_y^2+u_z^2$

So, after differentiation
$2 u. (du/dt) = 2.u_x (du_x/dt) +2.u_y (du_y/dt) + 2.u_z (du_z/dt)$
$2 u. a = 2.u_x a_x +2.u_y a_y + 2.u_z a_z$
$u. a = u_x a_x + u_y a_y + u_z a_z --------(B)$
from (A) & (B)

So,
$F_x=\gamma m_o a_x+\gamma ^3 m_o. \frac{u_x}{c^2} (u_x a_x+u_y a_y+u_z a_z) ------(1)$
we can differentiate directly without (A) & (B) step also
mahesh
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### Re: Applied force less than acting force in Special Relativi

mahesh,

Yup, you're right! I think my eyes tripped up on the ${\gamma}^{3}$ part, since I was thinking that that exponent looked a tad high. I missed that we pulled ${\gamma}^{-2}$ out of $\frac{{\text{d}}{\gamma}}{{\text{d}}t}$ rather than just $\gamma$ in the beginning of (ii) below.

With that cleared up, I'll try to read on later!

1. $F_x={\frac{\text{d}}{\text{d}t}}{\left[{{m_0}{\gamma}{u_x}}\right]}$
$={m_0}{\frac{\text{d}}{\text{d}t}}{\left[{{\gamma}{u_x}}\right]}$
$={m_0}\left({\gamma}{\frac{\text{d}}{\text{d}t}}{\left[{{u_x}}\right]}+{u_x}{\frac{\text{d}}{\text{d}t}}{\left[{{\gamma}}\right]}\right)$
$={m_0}\left({\gamma}{a_x}+{u_x}{\frac{\text{d}{\gamma}}{\text{d}t}}\right)$
$={m_0}{\gamma}{a_x}+{m_0}{u_x}{\frac{\text{d}{\gamma}}{\text{d}t}}$
2. ${\frac{\text{d}{\gamma}}{\text{d}t}}={\frac{\text{d}}{\text{d}t}}\left[{\left(1-\frac{{u_x^2}+{u_y^2}+{u_z^2}}{c^2}\right)}^{-\frac{1}{2}}\right]$
$=-{\frac{1}{2} } { { \left( 1-\frac{{u_x^2}+{u_y^2}+{u_z^2}}{c^2} \right) }^{-\frac{3}{2}}}{\frac{\text{d}}{\text{d}t}}\left[1-\frac{{u_x^2}+{u_y^2}+{u_z^2}}{c^2}\right]$
$=-{\frac{1}{2} } { { \left( {\gamma}^{-2} \right) }^{-\frac{3}{2}}}{\frac{\text{d}}{\text{d}t}}\left[1-\frac{{u_x^2}+{u_y^2}+{u_z^2}}{c^2}\right]$
$=-{\frac{1}{2} } { {\gamma} }^{3}{\frac{\text{d}}{\text{d}t}}\left[1-\frac{{u_x^2}+{u_y^2}+{u_z^2}}{c^2}\right]$
$={\frac{1}{2} } { {\gamma} }^{3}{\frac{\text{d}}{\text{d}t}}\left[\frac{{u_x^2}+{u_y^2}+{u_z^2}}{c^2}\right]$
$={\frac{{\gamma}^{3}}{2{c^2}} }{\frac{\text{d}}{\text{d}t}}\left[{u_x^2}+{u_y^2}+{u_z^2}\right]$
$={\frac{{\gamma}^{3}}{2{c^2}} }\left[2{u_x}{\frac{\text{d}{u_x}}{\text{d}t}}+{2{u_y}{\frac{\text{d}{u_y}}{\text{d}t}}}+{2{u_z}{\frac{\text{d}{u_z}}{\text{d}t}}\right]$
$={\frac{{\gamma}^{3}}{{c^2}} }\left[{u_x}{\frac{\text{d}{u_x}}{\text{d}t}}+{{u_y}{\frac{\text{d}{u_y}}{\text{d}t}}}+{{u_z}{\frac{\text{d}{u_z}}{\text{d}t}}\right]$
$={\frac{{\gamma}^{3}}{{c^2}} }\left[{u_x}{a_x}+{{u_y}{a_y}+{{u_z}{a_z}\right]$
3. $F_x={m_0}{\gamma}{a_x}+{m_0}{u_x}{\frac{\text{d}{\gamma}}{\text{d}t}}$
$={m_0}{\gamma}{a_x}+{m_0}{u_x}{\frac{{\gamma}^{3}}{{c^2}} }\left[{u_x}{a_x}+{{u_y}{a_y}+{{u_z}{a_z}\right]$
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### Re: Applied force less than acting force in Special Relativi

mahesh,

I'll start reformatting your original post as I have time.

I've saved the original text, so if you don't like the revisions - which you're completely, 100% allowed to disagree with - then I'll be happy to change them back. I can also make other edits if you'd like to change any of your wording, add/remove content, etc..

For now I'm just formatting stuff, rather than checking for correctness. It'll be easier to talk about clarifications and corrections once it's easier to read.

For reference, your post's original formatting:
STEP 1:-This problem can easily be understood by following paradox.
{Before starting this paradox, I want to put one relativity formula’s given in standard book of relativity
In any frame, for force in X-direction by S.R.
Fx = d/dt( y. mo. ux) where y=(1-u^2/c^2)^-0.5

So, after differentiation
Fx= y. mo. (dux/dt) + y^3. mo. {ux/c^2}. (u . du/dt)
Fx= y. mo. ax + y^3. mo. {ux/c^2}. (u . a) -----(A)
We know, u^2=ux^2+uy^2+uz^2

So, after differentiation
2 u. (du/dt) = 2.ux (dux/dt) +2.uy (duy/dt) + 2.uz (duz/dt)
2 u. a = 2.ux ax +2.uy ay + 2.uz az
u. a = ux ax + uy ay + uz az --------(B)
from (A) & (B)

So, Fx=y. mo. ax+y^3 mo. (ux/c^2} (ux ax+uy ay+uz az) ------(1)}

On frictionless platform, object is moving with constant velocity ux in X-direction & only magnetic force is acting in Y-direction & there is acceleration in Y-direction only with velocity uy & Fz=0
If we apply eq(1) to this case then result will be

Fx= y^3 mo. (ux/c^2) uy ay ---------- as ax=0

Or Fx=Fay as this force is form due to ‘ay’ only
Mean’s even there is no magnetic force acting on object from outside in x-direction & no ‘ax’ then also above force will act on object in +ve direction of x-axis due to ‘ay’
Important point (1):-
Mean’s applied magnetic force on object in X-direction is 0 & acting force in X-direction is
Fx= y^3 mo. (ux/c^2} uy ay+0 or Fay+0=Fay
---------------------------------------------------------------------------------------------------------------------

STEP 2:-Now, Force acting in X-direction is Fx= y^3 mo. (ux/c^2} uy ay or Fay
Now, after this happen, very small magnetic force of same intensity
-fx = -y^3 mo. (ux/c^2} uy ay or -Fay start acting on object in direction opposite to above force (but velocity is still positive ux) & cancel that above force.

Mean’s equation (1) becomes
0=y. mo. ax+y^3 mo. (ux/c^2} (ux ax+uy ay)
Or 0 =y. mo ax. (1+ y^2 {ux^2/c^2} ) +Fay
(Here as Fay= y^3 mo. (ux/c^2} uy ay)
Mean’s Fay = y. mo. -ax. (1+ y^2. {ux^2/c^2} )
Mean’s there must be acceleration in –ve X-direction to fulfill above equation of S.R.
Now, see above equation carefully, it is of nature
0= -fx + Fay
Important point (2):- Mean’s applied magnetic force on object in X-direction is -fx & acting force in X-direction is -fx + Fay = 0 or 0.
Here, resultant force in X-direction is zero but there is acceleration.

STEP3:- same things happen for +ve force in X-direction (for less than Fay or more)
Now, I am generalizing above result.
Step 1 & 2 clearly shows that when we apply any magnetic force (Fmx) in X-direction on the object, actual force acting on object is more & that quantity is (Fmx+Fay)
Similarly,
If we apply any magnetic force (Fmy) in Y-direction on the object then actual force acting on object is more & that quantity is (Fmy+Fax)
This is completely complicated results, which says that applied force & acting forces on objects are different & more in S.R.

STEP4:- Force does work, consume energy, gain energy & we must know that energy cannot be created. It can be transferred only:-
From above setup it must be clear that energy get transfer from magnet to object but if applied force is less than acting force then energy gain by object will be more than energy loose by the magnet. Means due to more work done by more force for same displacement, more energy get generated.

HERE, more energy (& force) is the problem.
Where this additional energy (or force) does comes from?
Natural ChemE
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### Re: Applied force less than acting force in Special Relativi

Sir, I am very thankful to you.
mahesh
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### Re: Applied force less than acting force in Special Relativi

I have not worked through the whole example, because I'm still concerned that there is a confusion in terms of reference frames in the calculations. Looking at the relativistic force equations in https://en.wikipedia.org/wiki/Relativistic_mechanics#Force:
The force described in this section is the classical 3-D force which is not a four-vector. This 3-D force is the appropriate concept of force since it is the force which obeys Newton's third law of motion. It should not be confused with the so-called four-force which is merely the 3-D force in the comoving frame of the object transformed as if it were a four-vector.

The equations derived differ from the OP's (which apply in the 'moving' frame). The force given for the inertial frame is

where the first term uses acceleration parallel to the movement and the second term acceleration normal to the movement. I think one should carefully define what we are doing in which frame in the rest of the OP's deductions.

-J

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### Re: Applied force less than acting force in Special Relativi

$F_x = \frac { d }{d_t} ( \gamma m_0 u_x)$ ------(1)
This is the general equation for force in X-direction in S.R. (mentioned in all standard S.R. book)
where direction of a & u will be in any direction.
-----------------------------------------------------------------------------------------
& equation given in above post is conditional equation. These equations are not different & can be easily derived from above equation (1) . For example, 1st term (equation) can be easily derived considering force, acceleration & velocity in same direction.
I can give that derivation also.
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### Re: Applied force less than acting force in Special Relativi

mahesh » 01 Apr 2016, 10:07 wrote:$F_x = \frac { d }{d_t} ( \gamma m_0 u_x)$ ------(1)
This is the general equation for force in X-direction in S.R. (mentioned in all standard S.R. book)
where direction of a & u will be in any direction.

Yes, as NCE has pointed out above, it is the "four force", which is valid in the 'moving' system's reference frame. However, your "paradox" implies that you are are using the "frictionless platform" as a reference system, in which case the equation that you use does not apply. The time for the platform and for the 'moving' system will not be the same. Such mixing of reference system is the source of 99.9% of all "paradoxes" in SR.

I also specifically asked the question on the significance of using a magnetic force rather than just a 'force in the y-direction', because it will make a difference in the specifics of force and acceleration calculations.

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Jorrie

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### Re: Applied force less than acting force in Special Relativi

This is not for magnetic force only but for any force.
For example, if observer is moving towards any vertically falling ball with horizontal velocity $u_x$ then he will observe that
$F_x=\gamma ^3 m_o. \frac{u_x}{c^2} (u_y a_y)$

This horizontal force is automatically acting on that ball.
Even there is no horizontal force or 0 force applied from outside in horizontal direction in observer frame.
This is because in SR, force is not due to change in state or acceleration but due to change in momenta.
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Now, for transfer of frame, we require two observers.
In all above case, there is one observer & all events happen in that single frame. It is not necessary & require that inertial observer will accelerate & move with object on which force is acting.
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### Re: Applied force less than acting force in Special Relativi

mahesh » 02 Apr 2016, 12:21 wrote:This is not for magnetic force only but for any force.
For example, if observer is moving towards any vertically falling ball with horizontal velocity $u_x$ then he will observe that
$F_x=\gamma ^3 m_o. \frac{u_x}{c^2} (u_y a_y)$

If you talk about a " vertically falling ball", it now sounds like you are bringing gravity in as well?

That aside, let's rather discuss your opening post's scenario. You wrote:
On frictionless platform, object is moving with constant velocity ${u_x}$ in the $x$-direction & only magnetic force is acting in $y$-direction & there is acceleration in $y$-direction only with velocity $u_y=0$ & $F_z=0$.

The equation that you gave is for the four-force, which is as measured in the inertial frame that is momentarily at rest relative to your object and then transformed back to the platform inertial frame. If $u_y=0$ as you specified, then $F_x = 0$ anyway. But as soon as your object gains any $u_y$ it requires a new inertial frame in which your object is at rest (otherwise you cannot use your eq. 1.)

Hence you cannot use the four force to calculate accelerations or forces relative to the platform frame. If you use the correct equations, as in https://en.wikipedia.org/wiki/Relativistic_mechanics#Force, you will find that for your scenario, $F_x =a_x=0$, irrespective of the value of $u_y$ or $a_y$.
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Last edited by BurtJordaan on April 3rd, 2016, 1:50 am, edited 1 time in total.
Reason: Clarification

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### Re: Applied force less than acting force in Special Relativi

$F_x=\gamma m_o a_x+\gamma ^3 m_o. \frac{u_x}{c^2} (u_x a_x+u_y a_y+u_z a_z) ------(1)$
This is very general equation of any force in X-direction in that frame. Here, object may be at rest or moving in that frame.
This equation is derived from definition of force in S.R.
Definition of force:- force acting in any direction is equal to rate of change of momentum in that direction.
You are right. Force (magnetic) is not remaining same due to change of time in that frame. If you see equation (1), it is clear that value of $\gamma$ will change accordingly in formula (1).
So, even object is not at rest with frame then also equation (1) will be applicable to it.
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Now, Transfer of frame:- it is interesting that if we apply eq (1) to inertial frame S & then transfer it properly to any frame S’ then also, you will find that eq (1) is true in frame S’
i.e. $F'_x=\gamma m_o a'_x+\gamma ^3 m_o. \frac{u'_x}{c^2} (u'_x a'_x+u'_y a'_y+u'_z a'_z) ------(1)$
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### Re: Applied force less than acting force in Special Relativi

What the equation does tell you is that if an object has a speed component transverse to the direction of an applied force $F_x$, then for a required acceleration $a_x$, you will have to apply a larger $F_x$ than when there is no transverse speed component.

It says nothing about forces being generated in the x-direction when a force is applied in the transverse direction, as you apparently try to demonstrate. The mere fact that you get an nonphysical result should warn you that your application of the equation is not correct.

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### Re: Applied force less than acting force in Special Relativi

BurtJordaan » 04 Apr 2016, 14:36 wrote:The mere fact that you get an nonphysical result should warn you that your application of the equation is not correct.

When I looked at Sparky's expert post (http://www.sciencechatforum.com/viewtopic.php?f=84&t=8584) again, I realized what the real source of the OP's "paradox" is. It is the fact that in inertial frames that are not stationary to the reference frame, the force and the acceleration may not be parallel to each other. So you can get a coordinate dependent 'fictitious force', as the OP claims.

However, one can always transform away the uniform motion and then the 'fictitious force' disappears. Now both $u_x = u_y = 0$ and so $F_x = 0$ when $a_x=0$.

My apologies for taking so long to come to the correct point. I must run, will return to it later.

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### Re: Applied force less than acting force in Special Relativi

Point 1:-This problem happen because in S.R., in any inertial frame, direction of force & acceleration is not require to be same & force is not dependent on change in state of motion (acceleration) in that direction but change of momentum in that direction.
point 2:-I differentiate forces as applied force & acting force. You differentiate as real & fictitious force. This indicate there is some problem in SR because physics did not allow such differentiation. Force is a force.
Point 3:-One observer is observing one event where 2 or 3 object are moving & accelerating. You can not attach frame of reference to all object where ux=0 & uy=0 but you require one inertial observer with related to that you have to take all observation & analysis the result.
Point 4:- This is very serious problem because force do work, consume energy, gain energy. You can not say that fictitious force do fictitious work & consume fictitious energy.
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### Re: Applied force less than acting force in Special Relativi

mahesh » 07 Apr 2016, 10:30 wrote:I differentiate forces as applied force & acting force. You differentiate as real & fictitious force. This indicate there is some problem in SR because physics did not allow such differentiation. Force is a force.

Sorry, I should have been more careful in my choice of words - "fictitious force" is fine in some cases, but in this case it is really coordinate dependent acceleration that should be distinguished from proper acceleration. Proper acceleration is what is measured by an accelerometer. Coordinate acceleration is that acceleration transformed to some relatively moving frame. As you describe your scenario, all the accelerations are coordinate and each will differ as observed in different frames, not only in magnitude, but also in direction.

BTW, force also transforms differently to each frame, but not exactly in the same way as acceleration. The reason for the difference is that a proper acceleration is a 4-vector, while the so-called 4-force is not a 4-vector. Therefore there is no "proper force" in physics and force is seldom used in modern relativistic physics. The same with work, which is no more than a change in kinetic energy. Now kinetic energy is also frame dependent, so work is frame dependent.

99.99% of all "paradoxes" in SR are caused by talking about frame dependent quantities as if they are 'proper'.

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### Re: Applied force less than acting force in Special Relativi

I am sorry. I can not visit your post because I was busy in some week end job work.

To prove that the force and work done are much important in relative physics, I am referring books on Relativity: Page #118 of Elements of Special Relativity by Dr. T.M. Karade-----

To prove mass energy equivalence $E=mc^2$, Einstein considered
$dw=F . dr$
where $F$ is considered as
$F =\frac { d }{dt} ( \gamma m_0 u)$ where ${\gamma}=\frac{1}{{\left(1-\frac{u^2}{c^2}\right)}^{0.5}}$
where $F$ is force and $dw$ is work done for displacement $dr$.

This shows the important of force and work done. Force is primary important for any action, work done & change in energy in any physic.

If Einstein uses force & displacement to find work done in that frame then these are important quantities.
& very important point is force & work are frame dependent & not depend on acceleration only.
Last edited by Natural ChemE on April 11th, 2016, 6:31 am, edited 2 times in total.
Reason: Formatting.
mahesh
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### Re: Applied force less than acting force in Special Relativi

mahesh,

I've applied more formatting and revisions to the OP. I think that I'm starting to understand it now, though I'm also fairly sure that I've made at least a few mistakes in translating it. If you see any errors, I'd be happy to correct them to better reflect your intent.

I skimmed your blog and PDF on this topic. The gist seems to be that the theory of Special Relativity is invalid due to being mathematically inconsistent with itself. In other words, you're not objecting to just its predictions, but rather its entire formulation.

If I'm understanding your argument correctly, this strikes me as particularly unlikely because self-consistency is one of the first things physicists and mathematicians check a theory for.
Special relativity, Wikipedia wrote:Special relativity is mathematically self-consistent, and it is an organic part of all modern physical theories, most notably quantum field theory, string theory, and general relativity (in the limiting case of negligible gravitational fields).
While it could be an oversight, the absence of a citation on such a strong claim suggests that it's well-established.

It's not uncommon for people to believe that they've found an inconsistent result of a complex theory, though it seems that pretty much all of these alleged findings are actually just mathematical mistakes or/and conceptual mistakes on the part of the reporter. I suspect that you've fallen into the same trap - there's some subtle mistake that's causing you to arrive at an inconsistent result, causing you to conclude that the theory itself is mathematically inconsistent.

In short, while I'm still not able to read the entire problem statement, I agree with BurtJordaan's perception that the apparent paradox is more likely a misunderstanding.
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### Re: Applied force less than acting force in Special Relativi

Mahesh, now that NCE has kindly edited your whole original post so that it is more readable, I think it is time to conclude this discussion. I think I have shown you where your misapplication lies - the fact that force and acceleration are not parallel if you consider orthogonal forces relative to a moving frame. Your 'paradox' stems from not treating these concepts correctly and hence leads you to false conclusions.

Add to this that your agenda is apparently to find a flaw in special relativity and have a personal theory as a substitute, I have closed this thread. Your are welcome to start a thread in Personal Theories, or you can request me to move some of your posts there.

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