- #1
quZz
- 125
- 1
Consider a 4-coordinate system [tex]x=(x^0,x^1,x^2,x^3)[/tex], [tex]x^0[/tex] plays a role of time, [tex]x^1, x^2, x^3[/tex] are some kind of space coordinates. In what follows greek letters will be 1,2 or 3, latin will be 0,1,2,3, [tex]c = 1[/tex], proper time squared [tex] ds^2 = g_{ik}(x) dx^{i} dx^{k}[/tex].
Now, if the body is "at rest" we have
[tex]x^{\alpha}={\rm const}[/tex],
[tex] dx^{\alpha} = 0 [/tex],
so space components of the 4-velocity are zeroes
[tex] u^k = (dx^0/ds, 0, 0, 0)[/tex]
and space components of the contravariant 4-momentum [tex]p^k = m u^k[/tex] are zeroes as well.
But [tex]p^{\alpha}=0[/tex] is not equivalent to [tex]p_{\alpha}=0[/tex]:
[tex]p_{\alpha}=g_{\alpha k}p^k = g_{\alpha 0}p^0=g_{\alpha 0} g^{00} p_0 + g_{\alpha 0} g^{0 \beta} p_{\beta}[/tex],
so we have a system of the form (with non-trivial solutions in general)
[tex] (\delta^{\beta}_{\alpha} - g_{\alpha 0} g^{0 \beta}) p_{\beta} = g_{\alpha 0} g^{00} p_0[/tex].
On the other hand, momentum may be considered as a gradient of the action [tex]S(x)[/tex]:
[tex]\vec{p}=\nabla S(x)[/tex],
energy as minus time derivative of [tex] S(x) [/tex]:
[tex] E = -\partial S/\partial x^0 [/tex],
in other words [tex]p_k = -\partial S/\partial x^k = (E, -\vec{p})[/tex].
So we are having a very interesting situation when body is at rest ([tex] dx^{\alpha} = 0[/tex]), but has a non-zero momentum (that is not constant!).
Are my thoughts correct?
Now, if the body is "at rest" we have
[tex]x^{\alpha}={\rm const}[/tex],
[tex] dx^{\alpha} = 0 [/tex],
so space components of the 4-velocity are zeroes
[tex] u^k = (dx^0/ds, 0, 0, 0)[/tex]
and space components of the contravariant 4-momentum [tex]p^k = m u^k[/tex] are zeroes as well.
But [tex]p^{\alpha}=0[/tex] is not equivalent to [tex]p_{\alpha}=0[/tex]:
[tex]p_{\alpha}=g_{\alpha k}p^k = g_{\alpha 0}p^0=g_{\alpha 0} g^{00} p_0 + g_{\alpha 0} g^{0 \beta} p_{\beta}[/tex],
so we have a system of the form (with non-trivial solutions in general)
[tex] (\delta^{\beta}_{\alpha} - g_{\alpha 0} g^{0 \beta}) p_{\beta} = g_{\alpha 0} g^{00} p_0[/tex].
On the other hand, momentum may be considered as a gradient of the action [tex]S(x)[/tex]:
[tex]\vec{p}=\nabla S(x)[/tex],
energy as minus time derivative of [tex] S(x) [/tex]:
[tex] E = -\partial S/\partial x^0 [/tex],
in other words [tex]p_k = -\partial S/\partial x^k = (E, -\vec{p})[/tex].
So we are having a very interesting situation when body is at rest ([tex] dx^{\alpha} = 0[/tex]), but has a non-zero momentum (that is not constant!).
Are my thoughts correct?