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For this problem, the solution is: $$y=\frac{m}{qE}\gamma[\text{cosh}(\frac{qEx}{mv\gamma})-1]$$ where $\gamma = 1/\sqrt(1-v^2)$. Here something seems to be wrong. I think that for the nonrelativistic limit where $v$ is very small in magnitude, the trajectory of $y$ must be a parabola. However, this kind of behavior happens when $v$ is close to $1$, when $1/(v\gamma)$ becomes small, so the taylor expansion works for some range of $x$. How should I account for this phenomenon? In Newtonian mechanics, it seems clear that the trajectory of $y$ is a parabola.

For this problem, the solution is: $$y=\frac{m}{qE}\gamma\left[\cosh\left(\frac{qEx}{mv\gamma}\right)-1\right]$$ where $\gamma = 1/\sqrt{1-v^2}$. Here something seems to be wrong. I think that for the non-relativistic limit where $v$ is very small in magnitude, the trajectory of $y$ must be a parabola. However, this kind of behavior happens when $v$ is close to $1$, when $1/(v\gamma)$ becomes small, so the taylor expansion works for some range of $x$. How should I account for this phenomenon? In Newtonian mechanics, it seems clear that the trajectory of $y$ is a parabola.

For this problem, the solution is: $$y=(m/qE)\gamma[\text{cosh}(qEx/(mv\gamma))-1]$$ where $\gamma = 1/\sqrt(1-v^2)$. Here something seems to be wrong. I think that for the nonrelativistic limit where $v$ is very small in magnitude, the trajectory of $y$ must be a parabola. However, this kind of behavior happens when $v$ is close to $1$, when $1/(v\gamma)$ becomes small, so the taylor expansion works for some range of $x$. How should I account for this phenomenon? In Newtonian mechanics, it seems clear that the trajectory of $y$ is a parabola.

For this problem, the solution is: $$y=\frac{m}{qE}\gamma[\text{cosh}(\frac{qEx}{mv\gamma})-1]$$ where $\gamma = 1/\sqrt(1-v^2)$. Here something seems to be wrong. I think that for the nonrelativistic limit where $v$ is very small in magnitude, the trajectory of $y$ must be a parabola. However, this kind of behavior happens when $v$ is close to $1$, when $1/(v\gamma)$ becomes small, so the taylor expansion works for some range of $x$. How should I account for this phenomenon? In Newtonian mechanics, it seems clear that the trajectory of $y$ is a parabola.

For this problem, the solution is $y=(m/qE)\gamma[\text{cosh}(qEx/(mv\gamma))-1]$ where $\gamma = 1/\sqrt(1-v^2)$. Here something seems to be wrong. I think that for the nonrelativistic limit where $v$ is very small in magnitude, the trajectory of $y$ must be a parabola. However, this kind of behavior happens when $v$ is close to $1$, when $1/(v\gamma)$ becomes small, so the taylor expansion works for some range of $x$. How should I account for this phenomenon? In Newtonian mechanics, it seems clear that the trajectory of $y$ is a parabola.

For this problem, the solution is: $$y=(m/qE)\gamma[\text{cosh}(qEx/(mv\gamma))-1]$$ where $\gamma = 1/\sqrt(1-v^2)$. Here something seems to be wrong. I think that for the nonrelativistic limit where $v$ is very small in magnitude, the trajectory of $y$ must be a parabola. However, this kind of behavior happens when $v$ is close to $1$, when $1/(v\gamma)$ becomes small, so the taylor expansion works for some range of $x$. How should I account for this phenomenon? In Newtonian mechanics, it seems clear that the trajectory of $y$ is a parabola.