Orbits of the Lorentz group

  • Thread starter parton
  • Start date
  • #1
parton
83
1
I want to determine the orbits of the proper orthochronous Lorentz group [itex] SO^{+}(1,3) [/itex].

If I start with a time-like four-momentum [itex] p = (m, 0, 0, 0) [/itex]

with positive time-component [itex]p^{0} = m > 0 [/itex],

the orbit of [itex] SO^{+}(1,3) [/itex] in [itex] p [/itex] is given by:

[tex] \mathcal{O}(p) \equiv \lbrace \Lambda p \mid \Lambda \in SO^{+}(1,3) \rbrace [/tex]

Now the point is: how do you show that

[tex] \mathcal{O}(p) = \lbrace q \mid q^{2} = m^{2}, q^{0} > 0 \rbrace [/tex] ?

Essently, the question is: why does a Lorentz transformation [itex] \Lambda \in SO^{+}(1,3) [/itex] exist
such that two four-vectors [itex] p [/itex] and [itex] q [/itex] with [itex] p^{2} = q^{2} = m^{2} [/itex] and [itex] p^{0}, q^{0} > 0 [/itex] are related via [itex] q = \Lambda p [/itex] ?

In fact, it is possible to answer my question(s) by brute-force calculations. But I am searching for an "elegant way", e.g. with the help of group theory.

I already searched in the literature, but in most cases it seems to be "trivial" for the authors
and I see now explicit proof.

Does anyone know of anything like that?
 

Answers and Replies

  • #2
dauto
1,948
201
Isn't that the very definition of the group SO+(1,3)?
 
  • #3
WannabeNewton
Science Advisor
5,829
547
I agree with the above. It's trivial according to those authors because it's the definition.
 
  • #4
SophusLie
1
0
Isn't that the very definition of the group SO+(1,3)?

No, it is not the definition of [itex] \mathrm{SO}^{+}(1,3) [/itex]. It is:

[tex] \mathrm{SO}^{+}(1,3) = \lbrace \Lambda \in \mathrm{GL}(4, \mathbb{R}) \mid \Lambda^{t} \eta \Lambda = \eta, \mathrm{det} \, \Lambda = 1, \Lambda^{0} \, _{0} \geq 1 \rbrace [/tex]
The elements of this group will not change [itex] p^{2} [/itex] and the sign of [itex] p^{0} [/itex] by definition.

So we can say that the set
[tex] \lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace [/tex] remains invariant under Lorentz transformations.

But this does not necessarily mean that for any two fourvectors [itex]p, q [/itex] with [itex]p^{2} = q^{2} = m^{2}[/itex] and [itex] p^{0}, q^{0} [/itex] there exists a transformation [itex] \Lambda \in \mathrm{SO}^{+}(1,3) [/itex] with [itex] q = \Lambda p [/itex]. In other words this would mean that [itex] \mathrm{SO}^{+}(1,3) [/itex] acts transitivly on [itex] \lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace [/itex]. And you need to show this in order to proof that
[tex] \mathcal{O}_{p} = \lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace [/tex]

Actually I also don't see how to do this without using brute force. I think it is not that trivial.
 

Suggested for: Orbits of the Lorentz group

Replies
6
Views
469
Replies
33
Views
1K
Replies
120
Views
3K
  • Last Post
Replies
32
Views
1K
  • Last Post
Replies
5
Views
422
Replies
15
Views
316
Replies
4
Views
895
Replies
3
Views
351
Replies
6
Views
456
  • Last Post
Replies
9
Views
680
Top