- #1
npc
- 11
- 1
Have looked on the forum but can't find this query so hope I am not repeating something that might be here already. Also hope this is the right forum for the post.
I am working through a paper on particle motion in GR (Cohen PhysRevD 19,8,p2273) but am running into a few hurdles.
We need to work out radial motion for the general metric:
[tex]
ds^2=g_{tt}dt^2+g_{rr}dr^2
[/tex]
starting from the Lorentz force equation (for the r-component):
[tex]
\frac{dp_r}{d\tau}+\Gamma^r_{\alpha\beta}p_\alpha\frac{dx^\beta}{d\tau}\\=-qF^{r\nu}g_{\nu\beta}\frac{dx^\beta}{d\tau}
[/tex]
where
[tex]
F_{\mu\lambda}=\partial_\lambda A_\mu-\partial_\mu A_\lambda
[/tex]
and
[tex]
A_{\mu}=(\psi,0,0,0)
[/tex]
The first problem I have is that the paper gives a result:
[tex]
\frac{d^2r}{d\tau^2}+\left(\frac{\partial_rg_{tt}}{2g_{tt}}+\frac{\partial_rg_{rr}}{2g_{tt}}\right)\left(\frac{dr}{d\tau}\right)-\frac{\partial_rg_{tt}}{2g_{rr}g_{tt}}
=-\frac{q}{m}\frac{1}{g_{rr}}\frac{d\psi}{dr}\frac{dt}{d\tau}
[/tex]
whereas I get:
[tex]
\frac{d^2r}{d\tau^2}+\left(\frac{\partial_rg_{tt}}{2g_{tt}}+\frac{\partial_rg_{rr}}{2g_{tt}}\right)\left(\frac{dr}{d\tau}\right)^2-\frac{\partial_rg_{tt}}{2g_{rr}g_{tt}}
=-\frac{q}{m}\frac{1}{g_{rr}}\frac{d\psi}{dr}\frac{dt}{d\tau}
[/tex]
Workings in attachment.
just wondering if anyone can see where I am wrong.
Also, The next step in the paper is to say that the above equation has first integral:
[tex]
\left(\frac{dr}{d\tau}\right)^2=\frac{g_{tt}+(K-\frac{q}{m}\psi)^2}{-g_{tt}g_{rr}}
[/tex]
Where K is a constant of integration.
I can't see where this solution is from, any suggestions on getting it? (probably something obivious)
There is a second integral solution given in the paper:
[tex]
\frac{dt}{d\tau}=\frac{(K-\frac{q}{m}\psi)}{-g_{tt}}
[/tex]
Which I am able to get correctly from the [tex]\mu=t[/tex] component of the lorentz equation.
See attachment
Homework Statement
I am working through a paper on particle motion in GR (Cohen PhysRevD 19,8,p2273) but am running into a few hurdles.
Homework Equations
We need to work out radial motion for the general metric:
[tex]
ds^2=g_{tt}dt^2+g_{rr}dr^2
[/tex]
starting from the Lorentz force equation (for the r-component):
[tex]
\frac{dp_r}{d\tau}+\Gamma^r_{\alpha\beta}p_\alpha\frac{dx^\beta}{d\tau}\\=-qF^{r\nu}g_{\nu\beta}\frac{dx^\beta}{d\tau}
[/tex]
where
[tex]
F_{\mu\lambda}=\partial_\lambda A_\mu-\partial_\mu A_\lambda
[/tex]
and
[tex]
A_{\mu}=(\psi,0,0,0)
[/tex]
The first problem I have is that the paper gives a result:
[tex]
\frac{d^2r}{d\tau^2}+\left(\frac{\partial_rg_{tt}}{2g_{tt}}+\frac{\partial_rg_{rr}}{2g_{tt}}\right)\left(\frac{dr}{d\tau}\right)-\frac{\partial_rg_{tt}}{2g_{rr}g_{tt}}
=-\frac{q}{m}\frac{1}{g_{rr}}\frac{d\psi}{dr}\frac{dt}{d\tau}
[/tex]
whereas I get:
[tex]
\frac{d^2r}{d\tau^2}+\left(\frac{\partial_rg_{tt}}{2g_{tt}}+\frac{\partial_rg_{rr}}{2g_{tt}}\right)\left(\frac{dr}{d\tau}\right)^2-\frac{\partial_rg_{tt}}{2g_{rr}g_{tt}}
=-\frac{q}{m}\frac{1}{g_{rr}}\frac{d\psi}{dr}\frac{dt}{d\tau}
[/tex]
Workings in attachment.
just wondering if anyone can see where I am wrong.
Also, The next step in the paper is to say that the above equation has first integral:
[tex]
\left(\frac{dr}{d\tau}\right)^2=\frac{g_{tt}+(K-\frac{q}{m}\psi)^2}{-g_{tt}g_{rr}}
[/tex]
Where K is a constant of integration.
I can't see where this solution is from, any suggestions on getting it? (probably something obivious)
There is a second integral solution given in the paper:
[tex]
\frac{dt}{d\tau}=\frac{(K-\frac{q}{m}\psi)}{-g_{tt}}
[/tex]
Which I am able to get correctly from the [tex]\mu=t[/tex] component of the lorentz equation.
The Attempt at a Solution
See attachment