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You're right; in the frame moving with the charges there will be no magnetic attraction. There will however be an electrostatic repulsion, so you will see the two particles move away from each other.

Consider another frame (the rest frame) where both particles move side by side in the x direction (I assume this is what you are imagining). The particles still repel each other electrostatically, but now they attract one another magnetically as well!

This may seem like a problem, but it isn't because even though the force measured in the rest frame is weaker since the electrostatic repulsion is partially canceled by the magnetic attraction, the time experienced by moving observers is also different due to the witchcraft of special relativity. If the frame moving with the charges is primed, and the rest frame is unprimed, then

$ t' = \gamma t $

$ F = q(E + v\times B) = q(E - vB) $

$ F' = qE' $

What we seek to show is that since force is inversely proportional to the square of the time, F and F' are related by a factor of $\gamma^2$ in which case observers in either frame will calculate matching trajectories using Maxwell's equations.

Since for a point charge

$$ E = \frac{1}{4\pi\epsilon}\frac{q}{r^2} $$

$$ B = \frac{\mu}{4\pi}\frac{qv\times\hat{r}}{r^2} = \frac{\mu}{4\pi}\frac{qv}{r^2}$$

Now we have all we need. Plugging B in to F and using $ \mu = 1/(c^2\epsilon) $ and $ \gamma = 1/\sqrt{1-\frac{v^2}{c^2}} $

$$ F = q(E - vB) = q(E - \frac{v^2}{c^2}\frac{1}{4\pi\epsilon}\frac{q}{r^2}) = q\gamma^2E$$

$$ F' = qE' = qE = \frac{1}{\gamma^2}F $$

Which is what we wanted all along since

$$ F = \frac{d^2y'}{dt'^2} = \frac{d^2y}{(\gamma dt)^2} = \frac{1}{\gamma^2}F $$

(noting that dy is unaffected by a boost in the x direction)

To solve more complicated problems of this sort, you usually want to talk about faraday tensors, 4-currents, and einstein notation, but I hope this simple example gave you some insight into how electromagnetism and special relativity are related. You may have wondered what if the particles go so fast that the magnetic force exceeds the electrostatic repulsion, but indeed you can see that to do so you must exceed the speed of light!

If you want to understand why electromagnetism is manifestly robust against relativity paradoxes, you'll want to learn about tensor analysis in 4D spacetime, which is must-know knowledge for any physicist worth his salt and can be found in the back of most mechanics or electrodynamics textbooks or in the beginning of most general relativity books. If you want to gain some physical intuition for how EM works I recommend purcell simplified

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

You're right; in the frame moving with the charges there will be no magnetic attraction. There will however be an electrostatic repulsion, so you will see the two particles move away from each other.

Consider another frame (the rest frame) where both particles move side by side in the x direction (I assume this is what you are imagining). The particles still repel each other electrostatically, but now they attract one another magnetically as well!

This may seem like a problem, but it isn't because even though the force measured in the rest frame is weaker since the electrostatic repulsion is partially canceled by the magnetic attraction, the time experienced by moving observers is also different due to the witchcraft of special relativity. If the frame moving with the charges is primed, and the rest frame is unprimed, then

\begin{align} t^\prime &= \gamma t \\ F &= q(E + v\times B) = q(E - vB) \\ F^\prime &= qE^\prime \end{align}

What we seek to show is that since force is inversely proportional to the square of the time, F and F' are related by a factor of $\gamma^2$ in which case observers in either frame will calculate matching trajectories using Maxwell's equations.

Since for a point charge

$$ E = \frac{1}{4\pi\epsilon}\frac{q}{r^2} $$

$$ B = \frac{\mu}{4\pi}\frac{qv\times\hat{r}}{r^2} = \frac{\mu}{4\pi}\frac{qv}{r^2}$$

Now we have all we need. Plugging B in to F and using $ \mu = 1/(c^2\epsilon) $ and $ \gamma = 1/\sqrt{1-\frac{v^2}{c^2}} $

$$ F = q(E - vB) = q\left(E - \frac{v^2}{c^2}\frac{1}{4\pi\epsilon}\frac{q}{r^2}\right) = q\gamma^2E$$

$$ F^\prime = qE^\prime = qE = \frac{1}{\gamma^2}F $$

Which is what we wanted all along since

$$ F = \frac{\mathrm d^2y^\prime}{{\mathrm dt^\prime}^2} = \frac{\mathrm d^2y}{(\gamma ~\mathrm dt)^2} = \frac{1}{\gamma^2}F $$

(noting that dy is unaffected by a boost in the x direction)

To solve more complicated problems of this sort, you usually want to talk about faraday tensors, 4-currents, and einstein notation, but I hope this simple example gave you some insight into how electromagnetism and special relativity are related. You may have wondered what if the particles go so fast that the magnetic force exceeds the electrostatic repulsion, but indeed you can see that to do so you must exceed the speed of light!

If you want to understand why electromagnetism is manifestly robust against relativity paradoxes, you'll want to learn about tensor analysis in 4D spacetime, which is must-know knowledge for any physicist worth his salt and can be found in the back of most mechanics or electrodynamics textbooks or in the beginning of most general relativity books. If you want to gain some physical intuition for how EM works I recommend Purcell simplified.

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

You're right; in the frame moving with the charges there will be no magnetic attraction. There will however be an electrostatic repulsion, so you will see the two particles move away from each other.

Consider another frame (the rest frame) where both particles move side by side in the x direction (I assume this is what you are imagining). The particles still repel each other electrostatically, but now they attract one another magnetically as well!

This may seem like a problem, but it isn't because even though the force measured in the rest frame is weaker since the electrostatic repulsion is partially canceled by the magnetic attraction, the time experienced by moving observers is also different due to the witchcraft of special relativity. If the frame moving with the charges is primed, and the rest frame is unprimed, then

$ t' = \gamma t $

$ F = q(E + v\times B) = q(E - vB) $

$ F' = qE' $

What we seek to show is that since force is inversely proportional to the square of the time, F and F' are related by a factor of $\gamma^2$

Since for a point charge

$$ E = \frac{1}{4\pi\epsilon}\frac{q}{r^2} $$

$$ B = \frac{\mu}{4\pi}\frac{qv\times\hat{r}}{r^2} = \frac{\mu}{4\pi}\frac{qv}{r^2}$$

Now we have all we need. Plugging B in to F and using $ \mu = 1/(c^2\epsilon) $ and $ \gamma = 1/\sqrt{1-\frac{v^2}{c^2}} $

$$ F = q(E - vB) = q(E - \frac{v^2}{c^2}\frac{1}{4\pi\epsilon}\frac{q}{r^2}) = q\gamma^2E$$

$$ F' = qE' = qE = \frac{1}{\gamma^2}F $$

Which is what we wanted all along since

$$ F = \frac{d^2y'}{dt'^2} = \frac{d^2y}{(\gamma dt)^2} = \frac{1}{\gamma^2}F $$

(noting that dy is unaffected by a boost in the x direction)

To solve more complicated problems of this sort, you usually want to talk about faraday tensors, 4-currents, and einstein notation, but I hope this simple example gave you some insight into how electromagnetism and special relativity are related. You may have wondered what if the particles go so fast that the magnetic force exceeds the electrostatic repulsion, but indeed you can see that to do so you must exceed the speed of light!

If you want to understand why electromagnetism is manifestly robust against relativity paradoxes, you'll want to learn about tensor analysis in 4D spacetime, which is must-know knowledge for any physicist worth his salt and can be found in the back of most mechanics or electrodynamics textbooks or in the beginning of most general relativity books. If you want to gain some physical intuition for how EM works I recommend purcell simplified

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

You're right; in the frame moving with the charges there will be no magnetic attraction. There will however be an electrostatic repulsion, so you will see the two particles move away from each other.

Consider another frame (the rest frame) where both particles move side by side in the x direction (I assume this is what you are imagining). The particles still repel each other electrostatically, but now they attract one another magnetically as well!

This may seem like a problem, but it isn't because even though the force measured in the rest frame is weaker since the electrostatic repulsion is partially canceled by the magnetic attraction, the time experienced by moving observers is also different due to the witchcraft of special relativity. If the frame moving with the charges is primed, and the rest frame is unprimed, then

$ t' = \gamma t $

$ F = q(E + v\times B) = q(E - vB) $

$ F' = qE' $

What we seek to show is that since force is inversely proportional to the square of the time, F and F' are related by a factor of $\gamma^2$ in which case observers in either frame will calculate matching trajectories using Maxwell's equations.

Since for a point charge

$$ E = \frac{1}{4\pi\epsilon}\frac{q}{r^2} $$

$$ B = \frac{\mu}{4\pi}\frac{qv\times\hat{r}}{r^2} = \frac{\mu}{4\pi}\frac{qv}{r^2}$$

Now we have all we need. Plugging B in to F and using $ \mu = 1/(c^2\epsilon) $ and $ \gamma = 1/\sqrt{1-\frac{v^2}{c^2}} $

$$ F = q(E - vB) = q(E - \frac{v^2}{c^2}\frac{1}{4\pi\epsilon}\frac{q}{r^2}) = q\gamma^2E$$

$$ F' = qE' = qE = \frac{1}{\gamma^2}F $$

Which is what we wanted all along since

$$ F = \frac{d^2y'}{dt'^2} = \frac{d^2y}{(\gamma dt)^2} = \frac{1}{\gamma^2}F $$

(noting that dy is unaffected by a boost in the x direction)

To solve more complicated problems of this sort, you usually want to talk about faraday tensors, 4-currents, and einstein notation, but I hope this simple example gave you some insight into how electromagnetism and special relativity are related. You may have wondered what if the particles go so fast that the magnetic force exceeds the electrostatic repulsion, but indeed you can see that to do so you must exceed the speed of light!

If you want to understand why electromagnetism is manifestly robust against relativity paradoxes, you'll want to learn about tensor analysis in 4D spacetime, which is must-know knowledge for any physicist worth his salt and can be found in the back of most mechanics or electrodynamics textbooks or in the beginning of most general relativity books. If you want to gain some physical intuition for how EM works I recommend purcell simplified

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

You're right; in the frame moving with the charges there will be no magnetic attraction. There will however be an electrostatic repulsion, so you will see the two particles move away from each other.

Consider another frame (the rest frame) where both particles move side by side in the x direction (I assume this is what you are imagining). The particles still repel each other electrostatically, but now they attract one another magnetically!

This may seem like a problem, but it isn't because even though the force measured in the rest frame is weaker since the electrostatic repulsion is partially canceled by the magnetic attraction, the time experienced by moving observers is also different due to the witchcraft of special relativity. If the frame moving with the charges is primed, and the rest frame is unprimed, then

$ t' = \gamma t $

$ F = q(E + v\times B) = q(E - vB) $

$ F' = qE' $

What we seek to show is that since force is inversely proportional to the square of the time, F and F' are related by a factor of $\gamma^2$

Since for a point charge

$$ E = \frac{1}{4\pi\epsilon}\frac{q}{r^2} $$

$$ B = \frac{\mu}{4\pi}\frac{qv\times\hat{r}}{r^2} = \frac{\mu}{4\pi}\frac{qv}{r^2}$$

Now we have all we need. Plugging B in to F and using $ \mu = 1/(c^2\epsilon) $ and $ \gamma = 1/\sqrt{1-\frac{v^2}{c^2}} $

$$ F = q(E - vB) = q(E - \frac{v^2}{c^2}\frac{1}{4\pi\epsilon}\frac{q}{r^2}) = q\gamma^2E$$

$$ F' = qE' = qE = \frac{1}{\gamma^2}F $$

Which is what we wanted all along since

$$ F = \frac{d^2y'}{dt'^2} = \frac{d^2y}{(\gamma dt)^2} = \frac{1}{\gamma^2}F $$

(noting that dy is unaffected by a boost in the x direction)

To solve more complicated problems of this sort, you usually want to talk about faraday tensors, 4-currents, and einstein notation, but I hope this simple example gave you some insight into how electromagnetism and special relativity are related. You may have wondered what if the particles go so fast that the magnetic force exceeds the electrostatic repulsion, but indeed you can see that to do so you must exceed the speed of light!

If you want to understand why electromagnetism is manifestly robust against relativity paradoxes, you'll want to learn about tensor analysis in 4D spacetime, which is must-know knowledge for any physicist worth his salt and can be found in the back of most mechanics or electrodynamics textbooks or in the beginning of most general relativity books. If you want to gain some physical intuition for how EM works I recommend purcell simplified

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

You're right; in the frame moving with the charges there will be no magnetic attraction. There will however be an electrostatic repulsion, so you will see the two particles move away from each other.

Consider another frame (the rest frame) where both particles move side by side in the x direction (I assume this is what you are imagining). The particles still repel each other electrostatically, but now they attract one another magnetically as well!

This may seem like a problem, but it isn't because even though the force measured in the rest frame is weaker since the electrostatic repulsion is partially canceled by the magnetic attraction, the time experienced by moving observers is also different due to the witchcraft of special relativity. If the frame moving with the charges is primed, and the rest frame is unprimed, then

$ t' = \gamma t $

$ F = q(E + v\times B) = q(E - vB) $

$ F' = qE' $

What we seek to show is that since force is inversely proportional to the square of the time, F and F' are related by a factor of $\gamma^2$

Since for a point charge

$$ E = \frac{1}{4\pi\epsilon}\frac{q}{r^2} $$

$$ B = \frac{\mu}{4\pi}\frac{qv\times\hat{r}}{r^2} = \frac{\mu}{4\pi}\frac{qv}{r^2}$$

Now we have all we need. Plugging B in to F and using $ \mu = 1/(c^2\epsilon) $ and $ \gamma = 1/\sqrt{1-\frac{v^2}{c^2}} $

$$ F = q(E - vB) = q(E - \frac{v^2}{c^2}\frac{1}{4\pi\epsilon}\frac{q}{r^2}) = q\gamma^2E$$

$$ F' = qE' = qE = \frac{1}{\gamma^2}F $$

Which is what we wanted all along since

$$ F = \frac{d^2y'}{dt'^2} = \frac{d^2y}{(\gamma dt)^2} = \frac{1}{\gamma^2}F $$

(noting that dy is unaffected by a boost in the x direction)

To solve more complicated problems of this sort, you usually want to talk about faraday tensors, 4-currents, and einstein notation, but I hope this simple example gave you some insight into how electromagnetism and special relativity are related. You may have wondered what if the particles go so fast that the magnetic force exceeds the electrostatic repulsion, but indeed you can see that to do so you must exceed the speed of light!

If you want to understand why electromagnetism is manifestly robust against relativity paradoxes, you'll want to learn about tensor analysis in 4D spacetime, which is must-know knowledge for any physicist worth his salt and can be found in the back of most mechanics or electrodynamics textbooks or in the beginning of most general relativity books. If you want to gain some physical intuition for how EM works I recommend purcell simplified

http://physics.weber.edu/schroeder/mrr/MRRtalk.html