Is the Spin Connection Problem Affecting My Derivation of the Dirac Equation?

In summary, the Dirac equation in space-time (1,-a^{2},-a^{2},-a^{2}) is given by (i\bar{\gamma}^{\mu}(\partial_{\mu}+\Gamma_{\mu})-m)\psi=0, where \Gamma_{\mu}=\frac{1}{2}\sum ^{\alpha \beta} \left( e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial x^{\mu}}e_{\beta \nu}) + {e_\alpha}_\nu {e_b}^\sigma {\Gamma^\nu}_{\sigma\mu}\right). The components of e_{\alpha
  • #1
pleasehelpmeno
157
0
In trying to derive the Dirac equation in space-time [itex] (1,-a^{2},-a^{2},-a^{2})[/itex], I have read that the Dirac equation is [itex](i\bar{\gamma}^{\mu}(\partial_{\mu}+\Gamma_{\mu})-m)\psi=0 [/itex] where,

[itex]\Gamma_{\mu}=\frac{1}{2}\sum ^{\alpha \beta}e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial x^{\mu}}e_{\beta \nu}) [/itex]

Is it correct that [itex]e_{\beta\nu}[/itex] is equal to [itex](1,a^{2},a^{2},a^{2}) [/itex], [itex]e_{\alpha}^{\mbox{ }\nu}[/itex] equal to [itex](1,1/a^{2},1/a^{2},1/a^{2})[/itex] and finally [itex]\sum^{\alpha\beta}=\frac{1}{4}(\gamma^{\alpha}\gamma^{\beta} -\gamma^{\beta} \gamma^{\alpha})[/itex]?

With my metric choice [itex]\gamma_{\mu}[/itex] should equal [itex]\frac{3}{2}\frac{\dot{a}}{a}[/itex].

I don't see how with this [itex]\sum[/itex] term that this is possible, have I made a mistake and can anyone help?
 
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  • #2
pleasehelpmeno said:
In trying to derive the Dirac equation in space-time [itex] (1,-a^{2},-a^{2},-a^{2})[/itex], I have read that the Dirac equation is [itex](i\bar{\gamma}^{\mu}(\partial_{\mu}+\Gamma_{\mu})-m)\psi=0 [/itex] where,

[itex]\Gamma_{\mu}=\frac{1}{2}\sum ^{\alpha \beta}e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial x^{\mu}}e_{\beta \nu}) [/itex]


Is it correct that [itex]e_{\beta\nu}[/itex] is equal to [itex](1,a^{2},a^{2},a^{2}) [/itex], [itex]e_{\alpha}^{\mbox{ }\nu}[/itex] equal to [itex](1,1/a^{2},1/a^{2},1/a^{2})[/itex] and finally [itex]\sum^{\alpha\beta}=\frac{1}{4}(\gamma^{\alpha}\gamma^{\beta} -\gamma^{\beta} \gamma^{\alpha})[/itex]?

These are both matrices, so it is more accurate to say that

$$e_{\beta\nu} = \mathrm{diag} (1,a^{2},a^{2},a^{2}),$$
$$e_{\alpha}^{\mbox{ }\nu}=\mathrm{diag} (1,1/a^{2},1/a^{2},1/a^{2}).$$

Note that if these were not diagonal, then we would have to be careful about which indices labeled the rows vs columns.

With my metric choice [itex]\gamma_{\mu}[/itex] should equal [itex]\frac{3}{2}\frac{\dot{a}}{a}[/itex].

I don't see how with this [itex]\sum[/itex] term that this is possible, have I made a mistake and can anyone help?

Well, [itex]\gamma_{\mu}[/itex] (or even [itex]\Gamma_{\mu}[/itex]) is a collection of 4x4 matrices, so it cannot be equal to a scalar. However, it would appear that

$$e_{\alpha}^{\mbox{ }\nu}\left(\frac{\partial}{\partial t}e_{\beta \nu}\right) = 6\frac{\dot{a}}{a} \mathrm{diag} (0,1,1,1),$$

where this is a matrix in the frame indices ##\alpha,\beta##. You can no doubt work out the other components and then multiply against ##\Sigma^{\alpha\beta}##.
 
  • #3
Sorry I think I may have made a mistake isn't: [itex]e_{\alpha}^{\mbox{ }\nu}[/itex] equal to [itex](1,1/a,1/a,1/a)[/itex] and so on. I understand these e terms but I can't see how to deal with [itex] \sum ^{\alpha\beta} [/itex] because surely it would give a gamma matrix but it shouldn't be there
 
  • #4
pleasehelpmeno said:
Sorry I think I may have made a mistake isn't: [itex]e_{\alpha}^{\mbox{ }\nu}[/itex] equal to [itex](1,1/a,1/a,1/a)[/itex] and so on. I understand these e terms but I can't see how to deal with [itex] \sum ^{\alpha\beta} [/itex] because surely it would give a gamma matrix but it shouldn't be there

You're right about the powers of ##a##, I missed that. So

$$e_{\alpha}^{\mbox{ }\nu}\left(\frac{\partial}{\partial t}e_{\beta \nu}\right) = 3\frac{\dot{a}}{a} \mathrm{diag} (0,1,1,1),$$

The components of [itex] \sum ^{\alpha\beta} [/itex] and ##\Gamma_\mu## are 4x4 matrices in spinor space, acting on the spinor ##\psi##. I'm not sure what it is that you're implying.
 
  • #5
I know that [itex] \Gamma_{\mu}=\frac{3\dot{a}}{2a} [/itex] so how can one get rid of the matrix?
 
  • #6
pleasehelpmeno said:
I know that [itex] \Gamma_{\mu}=\frac{3\dot{a}}{2a} [/itex] so how can one get rid of the matrix?

I'm not sure what you mean by "get rid of the matrix," ##\Gamma_\mu## is a matrix in the spinor space. What you've written down is possibly the value of the nonzero components, but we can't change the fact that it's a matrix.
 
  • #7
thx I am still struggling with this sum term.
Surely when [itex]\alpha = 1 [/itex] then [itex] \nu =1 [/itex] then [itex]\beta =1 [/itex].
and the same for [itex]\beta =[/itex] to 2 and 3.

So when [itex]\beta= 1 [/itex] wouldn't [itex] \sum ^{11} =0 [/itex] and the same for beta =2,3 thus all these sum terms would become zero?
 
  • #8
pleasehelpmeno said:
thx I am still struggling with this sum term.
Surely when [itex]\alpha = 1 [/itex] then [itex] \nu =1 [/itex] then [itex]\beta =1 [/itex].
and the same for [itex]\beta =[/itex] to 2 and 3.

So when [itex]\beta= 1 [/itex] wouldn't [itex] \sum ^{11} =0 [/itex] and the same for beta =2,3 thus all these sum terms would become zero?

I should have looked it up earlier, but you appear to be missing a term in your spin connection, namely

$$\Gamma_{\mu}=\frac{1}{2}\sum ^{\alpha \beta} \left( e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial x^{\mu}}e_{\beta \nu}) + {e_\alpha}_\nu {e_b}^\sigma {\Gamma^\nu}_{\sigma\mu}\right). $$

The first term might vanish by symmetry arguments (I could have also made a mistake), but presumably the 2nd term does not.
 
  • #9
So [itex]\frac{1}{8}( \gamma^{\alpha}\gamma^{\beta} - \gamma^{\beta}\gamma^{\alpha}) ( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma\mu})[/itex]
 

What is the spin connection problem?

The spin connection problem is a theoretical issue in physics that arises when trying to reconcile the principles of general relativity and quantum mechanics. It refers to the difficulty in integrating spin (a quantum mechanical property) into the theory of gravity (general relativity).

Why is the spin connection problem important?

The spin connection problem is important because it is a fundamental issue in the search for a unified theory of physics. The successful integration of spin into the theory of gravity could lead to a better understanding of the nature of spacetime and the behavior of particles at a quantum level.

How is the spin connection problem being addressed?

Scientists are currently exploring different approaches to address the spin connection problem, including string theory, loop quantum gravity, and spin foam models. These theories attempt to incorporate both general relativity and quantum mechanics, and provide a framework for including spin in the theory of gravity.

What are some proposed solutions to the spin connection problem?

One proposed solution to the spin connection problem is the use of supersymmetry, which introduces new particles with both spin and mass. Another approach is to consider that spacetime is a discrete lattice, rather than continuous, which could help to incorporate spin into the theory of gravity.

What are the potential implications of solving the spin connection problem?

If the spin connection problem is successfully resolved, it could lead to a better understanding of the fundamental forces and particles in the universe. It could also have practical applications, such as improving our understanding of black holes and the development of new technologies based on quantum mechanics.

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