- #1
Matter_Matters
- 36
- 2
Question Background:
I'm considering the Eddington-Robertson-Schiff line element which is given by
[tex] (ds)^2 = \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) dt^2 - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (dr^2 + r^2 d\theta^2 + r^2 \sin^2{\theta} \;d\phi^2 ),[/tex]
where [itex]\mu = GM = \text{const.}[/itex] and [itex] r=|\mathbf{r}|.[/itex]
I'm interested in determining the equations of motion for such a line element which can be obtained from the least action principle. The classical action [itex]S[/itex] is the integral along the particle trajectory
[tex] S = \int ds, [/tex]
which can be equivalently expressed as
[tex] S = \int \left( \frac{ds}{dt} \right) dt \equiv \int L \; dt. [/tex]
We can see from the above that
[tex] L = \left[ \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}}) \right]^{1/2}, [/tex]
where [itex]L[/itex] is the associated Lagrangian over time.
Problem and question
The associated equations of motion are given by (Eq. 20)
[tex]\frac{d^2\mathbf{r}}{dt^2} = \frac{\mu}{r^3} \left[ \left(4 \frac{\mu}{r} - v^2 \right) \mathbf{r} + 4 (\mathbf{r}\cdot \mathbf{\dot{r}} ) \mathbf{\dot{r}}\right]. [/tex]
I cannot for the life of me obtain this using the Euler-Lagrange equations.
Attempt at a solution:
The Euler-Lagrange equations are given by
[tex] \frac{d}{dt} \left( \frac{\partial L}{\partial \mathbf{\dot{r}}} \right) - \frac{\partial L}{ \partial \mathbf{r}} =0. [/tex]
I note that the equations of motion should be equivalent for either
[tex] L = \sqrt{g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu}, [/tex]
or
[tex] L = g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu. [/tex]
Bearing this in mind and working through the process using
[tex] L = \left[ \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}}) \right], [/tex]
I find
[tex] \frac{d}{dt} \left( \frac{\partial L}{ \partial \mathbf{\dot{r}}} \right) = -2 \left[ \left( 1 + 2 \frac{\mu}{r} \mathbf{\ddot{r}} \right) - 2 \frac{\mu}{r^3} (\mathbf{r}\cdot \mathbf{\dot{r}}) \mathbf{\dot{r}} \right], [/tex]
and
[tex] \left( \frac{\partial L }{\partial \mathbf{r}} \right) = 2\frac{\mu}{r^3} \mathbf{r} - 4 \frac{\mu^2}{r^4} \mathbf{r} + 2 \frac{\mu}{r^3} \mathbf{r} (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}} ). [/tex]
Clearly, adding these together does not give the desired result. Any suggestions?
I'm considering the Eddington-Robertson-Schiff line element which is given by
[tex] (ds)^2 = \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) dt^2 - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (dr^2 + r^2 d\theta^2 + r^2 \sin^2{\theta} \;d\phi^2 ),[/tex]
where [itex]\mu = GM = \text{const.}[/itex] and [itex] r=|\mathbf{r}|.[/itex]
I'm interested in determining the equations of motion for such a line element which can be obtained from the least action principle. The classical action [itex]S[/itex] is the integral along the particle trajectory
[tex] S = \int ds, [/tex]
which can be equivalently expressed as
[tex] S = \int \left( \frac{ds}{dt} \right) dt \equiv \int L \; dt. [/tex]
We can see from the above that
[tex] L = \left[ \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}}) \right]^{1/2}, [/tex]
where [itex]L[/itex] is the associated Lagrangian over time.
Problem and question
The associated equations of motion are given by (Eq. 20)
[tex]\frac{d^2\mathbf{r}}{dt^2} = \frac{\mu}{r^3} \left[ \left(4 \frac{\mu}{r} - v^2 \right) \mathbf{r} + 4 (\mathbf{r}\cdot \mathbf{\dot{r}} ) \mathbf{\dot{r}}\right]. [/tex]
I cannot for the life of me obtain this using the Euler-Lagrange equations.
Attempt at a solution:
The Euler-Lagrange equations are given by
[tex] \frac{d}{dt} \left( \frac{\partial L}{\partial \mathbf{\dot{r}}} \right) - \frac{\partial L}{ \partial \mathbf{r}} =0. [/tex]
I note that the equations of motion should be equivalent for either
[tex] L = \sqrt{g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu}, [/tex]
or
[tex] L = g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu. [/tex]
Bearing this in mind and working through the process using
[tex] L = \left[ \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}}) \right], [/tex]
I find
[tex] \frac{d}{dt} \left( \frac{\partial L}{ \partial \mathbf{\dot{r}}} \right) = -2 \left[ \left( 1 + 2 \frac{\mu}{r} \mathbf{\ddot{r}} \right) - 2 \frac{\mu}{r^3} (\mathbf{r}\cdot \mathbf{\dot{r}}) \mathbf{\dot{r}} \right], [/tex]
and
[tex] \left( \frac{\partial L }{\partial \mathbf{r}} \right) = 2\frac{\mu}{r^3} \mathbf{r} - 4 \frac{\mu^2}{r^4} \mathbf{r} + 2 \frac{\mu}{r^3} \mathbf{r} (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}} ). [/tex]
Clearly, adding these together does not give the desired result. Any suggestions?