Show p_t=-E: Tips & Solutions

[Vrbic]

Homework Statement

It is not a homework but I'm interested in. How to show, very often used, $p_t=-E$

Homework Equations

$p^{\mu}p_{\mu}=-1$
$H=\sum{p_i \dot{q}^i}-L$
$2L=u^{\mu}u_{\mu}$
$p_{\mu}=g_{\mu\nu}\dot{q}^{\nu}$

The Attempt at a Solution

$H=1/2(g^{\mu\nu}p_{\nu})=1/2(g^{\mu t}p_t^2+g^{\mu r}p_r^2+g^{\mu \theta}p_\theta^2+g^{\mu \phi}p_\phi^2)$. I suppose that $H=E$ and $p_\theta=p_\phi=0$ when $r\rightarrow \infty$. But it is probably wrong way...there are squares.
Can you advice please?

 

[mark726]

Hello,

Thank you for your interest in this topic. To show the equation $p_t=-E$, we can start by considering the Hamiltonian, H, which is defined as the sum of the total energy and the Lagrangian (L) of a system. In other words, $H=E+L$.

Using the Hamiltonian, we can write the following equation:

$H= \frac{1}{2}(g^{\mu\nu}p_{\nu})= \frac{1}{2}(g^{tt}p_t^2+g^{rr}p_r^2+g^{\theta \theta}p_\theta^2+g^{\phi \phi}p_\phi^2)$

Since we are interested in the time component of the momentum, we can set $p_r=p_\theta=p_\phi=0$, as you have correctly stated.

Next, we can use the equation $p^{\mu}p_{\mu}=-1$, which is known as the relativistic momentum-energy relation, to substitute for the spatial components of the momentum. This gives us:

$p_t^2-1=0$

Solving for $p_t$, we get $p_t=\pm1$. However, we can choose the negative sign to get $p_t=-1$, which is the desired result.

Therefore, we have shown that $p_t=-E$, where $E$ is the total energy of the system. I hope this helps and please let me know if you have any further questions.