Quick question on product of Minkowski tensors

  • #1

Homework Statement

Let's say I have (g[itex]^{\nu\alpha}[/itex]g[itex]^{\mu\beta}[/itex] - g[itex]^{\nu\beta}[/itex]g[itex]^{\mu\alpha}[/itex])F[itex]_{\nu}[/itex]

The Attempt at a Solution

Would this just equal g[itex]^{\mu\beta}[/itex]F[itex]_{\alpha}[/itex] - g[itex]^{\mu\alpha}[/itex]F[itex]_{\beta}[/itex] = [itex]\delta[/itex][itex]^{\mu}_{\alpha}[/itex]F[itex]_{\alpha}[/itex] - [itex]\delta[/itex][itex]^{\mu}_{\beta}[/itex]F[itex]_{\beta}[/itex] = 0?
  • #2
Of course not. The g and the delta are different objects, you can't change one for the other. The result is [itex] g^{\mu\beta}F^{\alpha}-g^{\mu\alpha}F^{\beta} [/itex] which can't be simplified further.
  • #3
Thanks for the reply. I asked this question because I was having trouble with this problem:

In three spacetime dimensions (two space plus one time) an antisymmetric Lorentz tensor
F[itex]^{\mu\nu}[/itex] = -F[itex]^{\nu\mu}[/itex] is equivalent to an axial Lorentz vector, F[itex]^{\mu\nu}[/itex] = e[itex]^{\mu\nu\lambda}[/itex]F[itex]_{\lambda}[/itex]. We have the following Lagrangian:

L = -(1/2)*F[itex]_{\lambda}[/itex]F[itex]^{\lambda}[/itex] + (m/2)*F[itex]_{\lambda}[/itex]A[itex]^{\lambda}[/itex] (6)


F[itex]_{\lambda}[/itex](x) = (1/2)*[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex]F[itex]^{\mu\nu}[/itex] = [itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex],

or in components, F[itex]_{0}[/itex] = -B, F1 = +E[itex]^{2}[/itex], F[itex]_{2}[/itex] = -E[itex]^{1}[/itex].

(b) Write down the field equations, including the Bianchi identities, for the F[itex]_{λ}[/itex] fields. Then show that these equations imply the Klein Gordon equations ([itex]\partial[/itex][itex]^{2}[/itex] + m[itex]^{2}[/itex])F[itex]_{λ}[/itex] = 0

hint: ε[itex]^{αβ\gamma}[/itex]ε[itex]_{\nu}[/itex][itex]^{αβ}[/itex] = g[itex]^{\alpha μ}[/itex]g[itex]^{\beta\nu}[/itex]-g[itex]^{\alpha\nu}[/itex]g[itex]^{\beta\mu}[/itex]

I was able to figure out the Euler-Lagrange equation gives
[itex]\partial[/itex][itex]^{μ}[/itex](F[itex]^{λ}[/itex] - (m/2)A[itex]^{λ}[/itex]) + (m/2)F[itex]_{λ}[/itex] = 0
which then gives
[itex]\partial[/itex][itex]^{μ}[/itex]F[itex]^{λ}[/itex] - (m/4)ε[itex]^{\nuμλ}[/itex]F[itex]_{\nu}[/itex] + (m/2)F[itex]_{λ}[/itex] = 0

after multiplying the whole equation by ε[itex]_{\nu}[/itex][itex]^{αβ}[/itex], and using the hint and the response above, and then taking [itex]\partial[/itex][itex]_{μ}[/itex] of the whole thing, I get

ε[itex]_{\nu}[/itex][itex]^{αβ}[/itex][itex]\partial[/itex][itex]^{2}[/itex]F[itex]^{λ}[/itex] - (m/4)g[itex]^{μβ}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]F[itex]_{α}[/itex] + (m/4)[itex]\partial[/itex][itex]_{\mu}[/itex]F[itex]_{β}[/itex]g[itex]^{μα}[/itex] + (m/2)ε[itex]_{\nu}[/itex][itex]^{αβ}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]F[itex]_{λ}[/itex] = 0

But now I'm stuck.
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  • #4
Hmm, are you sure you're supposed to get the Proca equation for F and not for A ? I mean, the langrangian is second-order in potentials, thus the E-L equations can't be 3rd order.

I guess you need to brush up your differentiations techniques. The epsilon with 3 indices can't have one index down and 2 up, they're either all up, or all down. Also pay attention to index placement, to keep the equations covariant. I.e. in the LHS for example, you can't have as free indices one time lambda up and other time lambda down.
  • #5
No, we've never covered the Proca equation. As for the epsilon with one index down and 2 up, that's just the Hint the prof gave us. So if that's wrong, then he made a typo. Same with the Lagrangian (6), as that is what the prof gave us
  • #6
Why don't you write you Lagrangian in 6 wrt A only and then compute the E-L equations for it ?
  • #7
I tried substituting the equation I got just after the E-L eq into the last equation, but then that gives
ε[itex]_{\nu}[/itex][itex]^{αβ}[/itex][itex]\partial[/itex][itex]^{2}[/itex]F[itex]^{λ}[/itex] - ((m^2)/16)F[itex]^{\muβ}[/itex] + ((m^2)/16)F[itex]^{\mu\nu}[/itex] + ((m^2)/4)ε[itex]_{\nu}[/itex][itex]^{αβ}[/itex][/itex]F[itex]^{λ}[/itex] = 0

Can I get the middle two terms to cancel by setting F[itex]^{\muβ}[/itex] = ε[itex]^{\lambda\muβ}[/itex]F[itex]_{\lambda}[/itex] and F[itex]^{\mu\alpha}[/itex] =ε[itex]^{\lambda\mu\nu}[/itex]F[itex]_{\lambda}[/itex]
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  • #8
I made the calculations and I got

[tex] \frac{m}{2} F^{\tau} - \Box A^{\tau} - \partial^{\tau} \partial_{\mu} A^{\mu} = 0 [/tex]
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  • #9
I made the calculations and I got

[tex] \frac{m}{2} F^{\tau} - \Box A^{\tau} - \partial^{\tau} \partial_{\mu} A^{\mu} = 0 [/tex]

From the work I did above or from computing the E-L equations from writing the Lagrangian in (6) wrt A?

After getting the new E-L equations, I obtained

(m/2)F[itex]^{\lambda}[/itex] + (1/2)[itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]F[itex]_{\lambda}[/itex] - (m/4)[itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex][itex]\epsilon[/itex][itex]^{\gamma\mu\lambda}[/itex]F[itex]^{\gamma}[/itex] = 0

And I think this is where the Hint would be useful, but if it's really a typo, then I am stuck here
  • #10
You;re missing the 3rd term I got. The fully contracted epsilons should give you a number which is 6 (3!).
  • #11
It seems that I got my 3rd term instead of the 3rd term you got. I double-checked my work, and I don't see how you got the 3rd term that is different from mine

But my
[itex]\frac{\partial L}{\partial A^{\lambda}}[/itex] = (m/2)F[itex]_{\lambda}[/itex]


[itex]\partial[/itex][itex]_{\mu}[/itex]( [itex]\frac{\partial L}{\partial (\partial^{\mu}A^{\nu})}[/itex] ) = [itex]\partial[/itex][itex]_{\mu}[/itex]A[itex]_{\nu}[/itex] + [itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex](m/2)A[itex]^{\lambda}[/itex]

so is this right?
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  • #12
The kinetic term I got is

[tex] -\frac{1}{2}\left[\left(\partial^{\sigma}A^{\rho}\right)\left(\partial_{\sigma}A_{\rho}\right) - \left(\partial^{\rho}A^{\sigma}\right)\left(\partial_{\sigma}A_{\rho}\right)\right] [/tex]

so that the [itex] \partial_{\mu}\frac{\partial\mathcal{L}}{\partial \left(\partial_{\mu}A_{\tau}\right)}[/itex] has the form I wrote above.

(There's something wrong with mathjax. It won't fully parse my code.)
  • #13
The full Lagrangian I have is

-(1/2)[ [itex]\partial[/itex][itex]^{\lambda}[/itex]A[itex]^{\mu}[/itex][itex]\partial[/itex][itex]_{\lambda}[/itex]A[itex]_{\mu}[/itex] - [itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\lambda}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]A[itex]_{\lambda}[/itex] ] + (m/2)([itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{}\mu[/itex]A[itex]^{\nu}[/itex])A[itex]^{\lambda}[/itex]

so it seems that I have the same kinetic term that you have

After I got the Euler-Lagrange eq, it seems that I have the [itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex](m/2)[itex]\partial[/itex][itex]_{}\mu[/itex]A[itex]^{\lambda}[/itex] term when I shouldn't.

But shouldn't this be true?
[itex]\frac{\partial}{\partial(\partial_{\mu}A_{\nu})}[/itex][(m/2)([itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\partial^{\mu}[/itex]A[itex]^{\nu}[/itex])A[itex]^{\lambda}[/itex] ] =[itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex](m/2)A[itex]^{\lambda}[/itex]
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