2D Relativistic Dynamics

A particle of mass m = 1 m = 1 is launched from the origin in the x y xy plane with speed v 0 = 0.8 v_0 = 0.8 at an angle θ = π / 4 \theta = \pi/4 with respect to the + x +x axis. There is a constant force F = ( F x , F y ) = ( 0.05 , 0.1 ) \vec{F} = (F_x, F_y) = (-0.05, -0.1) on the particle. The speed of light c = 1 c = 1 .

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

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


Problem Solving Notes:

There are (at least) two equivalent ways of formulating this problem. In the expressions below, v \vec{v} is the vector velocity and v 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:

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

The relativistic "Lagrangian" formulation is:

L = m c 2 1 v 2 c 2 + x F x + y F y d d t L v x = L x d d t L v y = L y \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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1 solution

Karan Chatrath
Apr 25, 2020


V = v x i ^ + v y j ^ \vec{V} = v_x \ \hat{i} + v_y \ \hat{j}

p = p x i ^ + p y j ^ \vec{p} = p_x \ \hat{i} + p_y \ \hat{j}

p x = v x 1 v x 2 v y 2 p_x = \frac{v_x}{\sqrt{1-v_x^2-v_y^2}} p y = v y 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 = d p x d t = p x v x v ˙ x + p x v y v ˙ y = m 11 v ˙ x + m 12 v ˙ y 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 = d p y d t = p y v x v ˙ x + p y v y v ˙ y = m 21 v ˙ x + m 22 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:

[ F x F y ] = [ m 11 m 12 m 21 m 22 ] [ v ˙ x v ˙ y ] \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:

[ v ˙ x v ˙ y ] = [ m 11 m 12 m 21 m 22 ] 1 [ F x F y ] \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:

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

T x ˙ = p = m c x ˙ c 2 x ˙ 2 \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.

Just a correction: you wrote that in the relativistic case the particle eventually exceeds the speed of light whereas in the graph it shows that in the non relativistic case it exceeds the speed of light, not in the relativistic case.

Brilliant solution. This is how I did it, although it took me 4 hours to solve this problem! Even more. I had to really read about special relativity. Wow. This was a really hard problem. @Steven Chase keep making cool mechanics problems!

Man, you two are geniuses.

Krishna Karthik - 1 year, 1 month ago

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Thanks for your kind words and for pointing out the mistake. These are concepts that even I am grappling with.

Karan Chatrath - 1 year, 1 month ago

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I've been really getting into physics recently, by trying my best to solve both your problems. It's fun, and hard. I'm training for IB HL grade 11 physics. I think schools undervalue the mathematical side of physics and leave the fun calculus out of it!

Krishna Karthik - 1 year, 1 month ago

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@Krishna Karthik The other thing I try to emphasize is that computers take the mechanical difficulty out of things, so it's OK to try more ambitious exercises.

Steven Chase - 1 year, 1 month ago

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@Steven Chase Yes, I've been trying to do some numerical integration recently. Solving problems by numerically evolving the systems through time is really fun! I plan to run it later on the GPU.

Krishna Karthik - 1 year, 1 month ago

By the way, I posted a new solution to your problem "System of Differential Equations-4"

Krishna Karthik - 1 year, 1 month ago

Thanks, glad you liked the problem, and that you got the satisfaction of solving after devoting time to it.

Steven Chase - 1 year, 1 month ago

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