2

Consider the Lagrangian density

$$L=\frac{1}{12}A^{\alpha \beta \gamma}A_{\alpha \beta \gamma}$$

and $B_{\alpha \beta}$, an antisymmetric two-indices, 4 dimensional, free field; moreover $A_{\alpha \beta \gamma}= \partial_\alpha B_{\beta \gamma} + \partial_\beta B_{\gamma \alpha} + \partial_\gamma B_{\alpha \beta}$

$A_{\alpha \beta \gamma}$ is also invariant under a gauge transformation $B_{\alpha \beta} \to B_{\alpha \beta} + \partial_\alpha f_\beta - \partial_\beta f_\alpha $

Find the equations of motion, spin and mass of the particle described by this field.

In order to find the eom one is to solve the Euler-Lagrange equations, that is:

$$\partial_{\mu} \left[\frac{\partial L}{\partial({\partial_{\mu}B_{\sigma \rho}})}\right]=0$$

(already omitted the second term of the equation since its contribution is null in this case)

  1. What's left to do is to use $\frac{\partial({\partial_\alpha B_{\beta \gamma} })}{\partial({\partial_{\mu}B_{\sigma \rho}})}= \delta^\mu _\alpha \delta^\sigma _\beta \delta^\rho _\gamma$ to solve it, although this doesn't look like the smartest way to evaluate these eom. I get I'm supposed to use the antisymmetry of $B_{\mu \sigma}$ but I'm not sure how. What is the best way to deal with this kind of problems?

  2. Regarding the mass is it correct and enough to say that since there's no mass term in the Lagrangian density the particle will be massless, or do I have to obtain the eom and show that they imply $p^2=0$, with $p$ being the 4-momentum of the particle?

  3. I have quite a few doubts regarding the spin as well

Edit for point 3: in order to verify the spin I would get the eom and study the polarization tensors to obtain their dof, as suggested (if I understood it correctly) by Andrew's post.

Moreover, according to the solution given in the other thread (How many degrees of freedom in a massless $2$-form field?) this field is equivalent to a massless scalar field.

My doubt is the following: up to this point I was used to know the right spin beforehand, so if I were to study spin 1 massive particles I would start with 4 dof (that is, the components of a 4-vector $A_μ$) and lower it to 3 (with $∂^μA_μ=0$ for instance) because I know that's the right dimension for spin 1 representations. Then, the photon being massless brings the dof to 2 with another condition. But I knew the correct spin right from the start in that case, while now I would evaluate the dof and obtain the spin without knowing it a priori, so how do I deduce I'm dealing with a spin 0 field?

J. Doee
  • 21
  • 1
    What have you already tried? Basically the answers to your questions is that you are on the right track and you should keep pushing. 1 is a perfectly valid way to proceed. 2, either way works, if you don't have confidence that the lack of a mass term in the (quadratic, relativistic, 2-derivative-per-field) Lagrangian will end up meaning that the EOMs imply $p^2=0$, it doesn't hurt to work through the math to see it. You gave no details for 3 so I don't know what to say. – Andrew Feb 19 '21 at 15:03
  • I was evaluating each term in the EL equation when I realized that not exploiting the symmetry of the Lagrangian is probably not so efficient, especially when the Lagrangian contains a lot of terms, but I can't see how to simplify the calculus by myself
  • That's what I deduced from the Lagrangian density I know, but I was wondering whether I was perhaps assuming some conditions that are not always met
  • It's that up to this point the spin corresponding to a certain representation was always given a priori, so I found myself a bit lost now that I'm supposed to deduce it myself
    • – J. Doee Feb 19 '21 at 15:28
  • 1
    For 1, basically you can antisymmetrize the indices on the derivative if you want, so the right hand side can be taken to be $\delta^{[\mu}{[\alpha}\delta^{\sigma}{\beta}\delta^{\rho]}_{\gamma]}$. But it's better to be correct than efficient. For 2, you will end up with massless particles, there's no real subtlety. The tricky thing is really part 3, which is what degrees of freedom you end up with, and also 4, which is how you couple this thing to other matter. This question is relevant: https://physics.stackexchange.com/q/457269/ – Andrew Feb 19 '21 at 15:47
  • 1
    I'm assuming $A_{\mu\nu\rho}$ is totally antisymmetric. – Andrew Feb 19 '21 at 15:49
  • You are right that as long as I can solve it the method ultimately does not matter, but I'd like to become more relaxed with this kind of calculations rather than solving everything by using brute force, as that would lower the chances to make mistakes and make me more experienced in general.

    • Mind explain what you meant with those delta? I did not get it quite yet. 3. I'm not too familiar with the math used in the first answer; anyway, basically I want to study the polarization tensors and verify their dof, am I correct? In that case I obtain 1 dof which is the same as a spin 0 particle?

    – J. Doee Feb 19 '21 at 16:37
  • Regarding the dof and spin, for example if I were to study spin 1 massive particles I would start with 4 dof (that is, the components of a 4-vector $ A^\mu $) and lower it to 3 (with $\partial_{\mu}A^{\mu}=0$ for instance) because I know that's the right dimension for spin 1 representations. Then, the photon being massless brings the dof to 2 with another condition. But I knew the correct spin right from the start in that case, while now I would evaluating the dof and obtain a value without knowing anything else, so how do I deduce I'm dealing with a spin 0 field? – J. Doee Feb 19 '21 at 16:51
  • 1
    OK I think your comments have clarified for me enough what you want to know that I can try to write a full answer later. However it is simple enough for me to say that the square brackets $[]$ mean to antisymmetrize (with unit weight, ie dividing by $n!$) the indices. So for example, $\delta^{[\mu}{\alpha}\delta^{\nu]}{\beta}=\frac{1}{2!}(\delta^{\mu}\alpha\delta^\nu\beta-\delta^\nu_\alpha\delta^\mu_\beta)$. In my comment above I only should have put square brackets on the top and not the bottom, sorry. – Andrew Feb 19 '21 at 17:22
  • I think I understood the antisymmetrize part, and that also makes the $\frac{1}{12}$ coefficient's role clear (when you couple this with matter, that is) Unfortunately can't say the same for the spin/dof relationship – J. Doee Feb 20 '21 at 23:29