- #1

PhysicsRock

- 117

- 18

- Homework Statement
- Consider a relativistic particle of mass ##m_0## in ##(1+1)##-spacetime dimensions. In an inertial frame of reference with spacetime coordinates ##(t,x)##, we define the potential as ##V(x) = \alpha x## with ##\alpha > 0##. The particle is at rest at time ##t=0## and it's position is ##x(0) = 0##. Determine the trajectory ##x(t)## of the particle.

Hint: Use the conservation of total energy.

- Relevant Equations
- ##E_\text{tot} = \sqrt{ c^2 p^2 + m_0^2 c^4 } + \alpha x##

Since energy is conserved and the particle is initially at rest, we can determine that ##E(0) = m_0 c^2##, so

$$

m_0 c^2 = \sqrt{ c^2 p^2 + m_0^2 c^4 } + \alpha x.

$$

Squaring this eqation gives

$$

m_0^2 c^4 = \alpha^2 x^2 + c^2 p^2 + m_0^2 c^4 + 2 \alpha x \sqrt{ c^2 p^2 + m_0^2 c^4 }

\Rightarrow 0 = \alpha^2 x^2 + c^2 p^2 + 2 \alpha x ( E - \alpha x ).

$$

Using ##p = \gamma m_0 \dot{x}##, I was able to simplify this equation to

$$

0 = -\alpha^2 x^2 + \frac{E^2 \dot{x}^2}{c^2 - \dot{x}^2} + 2 \alpha E x

$$

This is the point where I'm stuck. I have doubled checked and I'm pretty sure that this final expression is correct, however, I cannot guarantee that it actually is. If it is, I have no clue how to solve this equation.

$$

m_0 c^2 = \sqrt{ c^2 p^2 + m_0^2 c^4 } + \alpha x.

$$

Squaring this eqation gives

$$

m_0^2 c^4 = \alpha^2 x^2 + c^2 p^2 + m_0^2 c^4 + 2 \alpha x \sqrt{ c^2 p^2 + m_0^2 c^4 }

\Rightarrow 0 = \alpha^2 x^2 + c^2 p^2 + 2 \alpha x ( E - \alpha x ).

$$

Using ##p = \gamma m_0 \dot{x}##, I was able to simplify this equation to

$$

0 = -\alpha^2 x^2 + \frac{E^2 \dot{x}^2}{c^2 - \dot{x}^2} + 2 \alpha E x

$$

This is the point where I'm stuck. I have doubled checked and I'm pretty sure that this final expression is correct, however, I cannot guarantee that it actually is. If it is, I have no clue how to solve this equation.