A particle of mass $m = 1$ is launched from the origin in the $xy$ plane with speed $v_0 = 0.8$ at an angle $\theta = \pi/4$ with respect to the $+x$ axis. There is a constant force $\vec{F} = (F_x, F_y) = (-0.05, -0.1)$ on the particle. The speed of light $c = 1$ .

When the particle reaches $y = 0$ again at time $t > 0$ , what is the value of $x$ ?

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Bonus:
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How does this compare to the results one would get from non-relativistic equations?

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Problem Solving Notes:
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There are (at least) two equivalent ways of formulating this problem. In the expressions below, $\vec{v}$ is the vector velocity and $v$ is the scalar speed. Note that the "kinetic energy" term in the Lagrangian is somewhat counter-intuitive, but its derivation is beyond the scope of this problem.

The relativistic "Newtonian" formulation is:

$\vec{F} = \frac{d}{dt} \Big( \frac{m c \, \vec{v}}{\sqrt{c^2 - v^2}} \Big)$

The relativistic "Lagrangian" formulation is:

$\mathcal{L} = - m c^2 \sqrt{1 - \frac{v^2}{c^2}} + x F_x + y F_y \,\,\,\,\,\,\,\,\,\,\,\,\,\, \frac{d}{dt} \frac{\partial{\mathcal{L}}}{\partial{v_x}} = \frac{\partial{\mathcal{L}}}{\partial{x}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \frac{d}{dt} \frac{\partial{\mathcal{L}}}{\partial{v_y}} = \frac{\partial{\mathcal{L}}}{\partial{y}}$

The answer is 6.92.

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Let:

$\vec{V} = v_x \ \hat{i} + v_y \ \hat{j}$

$\vec{p} = p_x \ \hat{i} + p_y \ \hat{j}$

$p_x = \frac{v_x}{\sqrt{1-v_x^2-v_y^2}}$ $p_y = \frac{v_y}{\sqrt{1-v_x^2-v_y^2}}$

Applying Newton's second law gives: $F_x = \frac{dp_x}{dt} = \frac{\partial p_x}{\partial v_x} \dot{v}_x + \frac{\partial p_x}{\partial v_y} \dot{v}_y = m_{11}\dot{v}_x + m_{12}\dot{v}_y$ $F_y = \frac{dp_y}{dt} = \frac{\partial p_y}{\partial v_x} \dot{v}_x + \frac{\partial p_y}{\partial v_y} \dot{v}_y= m_{21}\dot{v}_x + m_{22}\dot{v}_y$

The partial derivative computations have not been shown here. This implies:

$\left[\begin{matrix} F_x\\F_y\end{matrix}\right] = \left[\begin{matrix} m_{11}&m_{12}\\m_{21} &m_{22} \end{matrix}\right] \left[\begin{matrix} \dot{v}_x\\\dot{v}_y\end{matrix}\right]$

This implies:

$\boxed{\left[\begin{matrix} \dot{v}_x\\\dot{v}_y\end{matrix}\right] = \left[\begin{matrix} m_{11}&m_{12}\\m_{21} &m_{22} \end{matrix}\right] ^{-1}\left[\begin{matrix} F_x\\F_y\end{matrix}\right]}$

Beyond this point, numerical integration does the rest. The Lagrangian formulation gives the same. The motion of the system can be summarised in the following plots:

One can see that in the non-relativistic case, the speed of the particle eventually exceeds the speed of light, which is an impossibility.

Just as general information, the 'kinetic energy' term can be derived by realising that in classical mechanics, the expression for generalised momentum (for 1D motion) is:

$\frac{\partial \mathcal{T}}{\partial \dot{x}} = p$ For the relativistic case:

$\frac{\partial \mathcal{T}}{\partial \dot{x}} = p = \frac{mc\dot{x}}{\sqrt{c^2-\dot{x}^2}}$

Integrating the above gives the required expression.