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Imagine two positive charges in a space ship moving with a velocity,v with respect to an observer on earth.

according to the person in the spaceship,the electrostatic force between the charges is $F'=(\frac{1}{4π\epsilon_0})\times \frac{q_1q_2}{r^2} $

But if you are the observer on earth,then the equation becomes $F=\gamma \times F' =\gamma\times(\frac{1}{4π\epsilon_0})\times \frac{q_1q_2}{r^2} $ enter image description here
So,the force is greater in the frame on earth. But actually shouldn't the force be lesser because with respect to earth the charges are moving and hence have an inward magnetic attractive force which partially cancels the outwards repulsive force?

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    $\begingroup$ You should take the magnetic force into account as well. $\endgroup$
    – my2cts
    Apr 25 at 12:55
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But if you are the observer on earth,then the equation becomes $F=\gamma \times F' $

The above is not correct. The actual formula in this example would be $F= F' / \gamma$

In the frame of the person in the spaceship the force will indeed be as you have written $F' =(\frac{1}{4π\epsilon_0})\times \frac{q_1q_2}{r^2}$

EDIT : I have looked back on this and the rest of the answer is not accurate and just plain wrong. Hence , i am deleting the rest of what i had written.

The following question has a good answer which seems to be correct and provides a good answer for this
Is Magnetic force really the relativistic correction for electrostatic force?

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  • $\begingroup$ This does not answer the question. It only restates my earlier comment without reference. $\endgroup$
    – my2cts
    Apr 25 at 15:21
  • $\begingroup$ Fair enough . i had read your comment to that answer, not your comment to the original question.. I will edit my answer to make it complete . $\endgroup$ Apr 25 at 15:23
  • $\begingroup$ But $F=\gamma ma $ Or $\frac{F}{a} = \gamma m $ since $\gamma$ is greater for the observer on earth,$\frac{F}{a}$ should also be greater.and as acceleration 'a' is same for both observers,this must mean that Force in his frame must be larger. How is this possible if $F=\frac{ F'}{\gamma}$ ? $\endgroup$ Apr 26 at 7:01
  • $\begingroup$ @GeneralPhysics " acceleration 'a' is same for both observers " This is not true. Acceleration is not a lorentz invariant . Acceleration of each charge would be higher in the spaceship frame, than in the earth frame $\endgroup$ Apr 26 at 7:10
  • $\begingroup$ oh...I thought acceleration was lorentz invariant..Thanks for the clarification $\endgroup$ Apr 26 at 7:31
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A common class of error, in these sorts of problems, is getting the transformation backwards and putting $\gamma$ where you should have $1/\gamma$. That's what's happened to you here.

To see this intuitively, consider a pair of opposite charges instead of same-sign charges. Released from rest, the opposite charges will accelerate towards each other and collide after some finite time. An observer in a different frame will see this collision happen more slowly, due to time dilation.

How the moving observer explains this delayed collision is sensitive to the geometry of the problem. If the line between the charges is perpendicular to the velocity, as in your diagram, the moving observer believes there is a magnetic interaction between the charges which counteracts their electrical attraction. (A classic problem from Griffiths: consider parallel line charges, and find the speed at which the magnetic repulsion exactly cancels the electrical attraction.) If the line between your point charges were parallel to the velocity vector, there would be zero magnetic force between them, but the electrical attraction would have been reduced thanks to the Lorentz boost. At other angles, the boosted electrical force and the magnetic interaction conspire together so that your opposite charges collide after exactly the right dilated amount of time.

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Since the vector F is perpendicular to v it is not affected by the Lorentz transformation and it is the same in both reference frames.

You can yourself a big favour by using the [covariant formulation of electromagnetism][1]. The force is given by $$f^\mu = F^{\mu\nu} j_\nu$$ where $F^{\mu\nu}$ is the Lorentz covariant, antisymmetric force tensor and $j^\nu$ the current. Form this it is clear that you don't have to worry about the detailed transformation of the fields. You only have to transform the force. The Lorentz transformation is $$ t' = \gamma \left( t - \frac{vx}{c^2} \right) \\ x' = \gamma \left( x - v t \right)\\ y' = y \\ z' = z \,. $$ This shows that if v is along x, then the perpendicular components of the force, unlike those of the antisymmetric tensor fields, are not affected. [1]: https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

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  • $\begingroup$ Oh..Thanks for the reply $\endgroup$ Apr 25 at 13:38
  • $\begingroup$ That does not really matter. He could just as easily draw the 2 charges placed along the axis of the spaceship, in such a way, that the line joining them is parallel to v $\endgroup$ Apr 25 at 13:54
  • $\begingroup$ But then,you need to take into account lorentz contraction.. $\endgroup$ Apr 25 at 14:22
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    $\begingroup$ @silverrahul That would be an entirely different question with a different answer. In my opinion it does matter what the actual question is. $\endgroup$
    – my2cts
    Apr 25 at 15:14
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    $\begingroup$ @silverrahul I am just sticking to the talking points raised by the OP. $\endgroup$
    – my2cts
    Apr 25 at 15:22

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