Does the Lorentz Condition Apply to the Given Vector Field Lagrangian?

In summary, by considering a particular derivative of the field equations, it can be shown that the field \phi^\alpha(x) satisfies the Lorentz condition \partial_\alpha\phi^\alpha(x)=0.
  • #1
orentago
27
0

Homework Statement



Given the Lagrangian density

[tex]L=-{1 \over 2}[\partial_\alpha\phi_\beta(x)][\partial^\alpha\phi^\beta(x)]+{1\over 2}[\partial_\alpha\phi^\alpha(x)][\partial_\beta\phi^\beta(x)]+{\mu^2\over 2}\phi_\alpha(x)\phi^\alpha(x)[/tex]

for the real vector field [tex]\phi^\alpha(x)[/tex] with field equations:
[tex][g_{\alpha\beta}(\square+\mu^2)-\partial_\alpha\partial_\beta]\phi^\beta(x)=0[/tex]

Show that the field [tex]\phi^\alpha(x)[/tex] satisfies the Lorentz condition:
[tex]\partial_\alpha\phi^\alpha(x)=0[/tex]

Homework Equations



See above.

The Attempt at a Solution



[tex][g_{\alpha\beta}(\square+\mu^2)-\partial_\alpha\partial_\beta]\phi^\beta(x)=0[/tex]
[tex]\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=g_{\alpha\beta}(\square+\mu^2)\phi^\beta(x)[/tex]
[tex]\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=g_{\alpha\beta}(\partial^\beta\partial_\beta+\mu^2)\phi^\beta(x)[/tex]
[tex]\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=\partial_\alpha\partial_\beta\phi^\beta(x)+\mu^2g_{\alpha\beta}\phi^\beta(x)[/tex]
[tex]\Rightarrow\mu^2g_{\alpha\beta}\phi^\beta(x)=0[/tex]
[tex]\Rightarrow\mu^2\phi^\beta(x)=0[/tex]
[tex]\Rightarrow\mu^2\partial_\alpha\phi^\alpha(x)=0[/tex]

I think I've done it, but I don't know if my method is correct. Would anyone be able to validate or refute this?
 
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  • #2
orentago said:
[tex]\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=g_{\alpha\beta}(\square+\mu^2)\phi^\beta(x)[/tex]
[tex]\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=g_{\alpha\beta}(\partial^\beta\partial_\beta+\mu^2)\phi^\beta(x)[/tex]
[tex]
\Rightarrow\partial_\alpha\partial_\beta\phi^\beta (x)=\partial_\alpha\partial_\beta\phi^\beta(x)+\mu ^2g_{\alpha\beta}\phi^\beta(x)
[/tex]

You made a mistake when you used the index [tex]\beta[/tex] for [tex]\square = \partial^\gamma\partial_\gamma[/tex]. You have to use a dummy index and it's not the same as the one on [tex]\phi^\beta[/tex].

You can obtain the result by considering a particular derivative of the field equations.
 
  • #3
Ok how about this:

[tex]\Rightarrow\partial_\alpha\partial_\beta\phi^\beta (x)=g_{\alpha\beta}(\partial^\gamma\partial_\gamma+\mu^2)\phi^\beta(x)[/tex]
[tex]\Rightarrow\partial^\alpha\partial_\alpha\partial_\beta\phi^\beta (x)=g_{\alpha\beta}\partial^\alpha(\partial^\gamma\partial_\gamma+\mu^2)\phi^\beta(x)[/tex]
[tex]\Rightarrow\square\partial_\beta\phi^\beta (x)=\partial_\beta(\square+\mu^2)\phi^\beta(x)[/tex]
[tex]\Rightarrow\square\partial_\beta\phi^\beta (x)=\square\partial_\beta\phi^\beta(x)+\mu^2\partial_\beta\phi^\beta(x)[/tex]
[tex]\Rightarrow\mu^2\partial_\beta\phi^\beta(x)=0[/tex]
[tex]\Rightarrow\partial_\alpha\phi^\alpha(x)=0[/tex]
 

Related to Does the Lorentz Condition Apply to the Given Vector Field Lagrangian?

1. What is the Lorentz condition in QFT?

The Lorentz condition in QFT is a gauge fixing condition that is used to eliminate redundant degrees of freedom in the electromagnetic field. It ensures that the equations of motion for the field are consistent with special relativity.

2. Why is the Lorentz condition important in QFT?

The Lorentz condition is important in QFT because it allows for the consistent application of special relativity in the theory. It also simplifies the equations of motion for the electromagnetic field and makes them easier to solve.

3. How is the Lorentz condition derived in QFT?

The Lorentz condition is derived in QFT by imposing the requirement that the equations of motion for the electromagnetic field are invariant under Lorentz transformations. This leads to the condition that the divergence of the vector potential must be equal to zero.

4. Can the Lorentz condition be generalized to other fields in QFT?

Yes, the Lorentz condition can be generalized to other fields in QFT. In general, gauge fixing conditions are used to eliminate redundant degrees of freedom in various fields, and the Lorentz condition is just one example of such a condition.

5. What are the implications of not satisfying the Lorentz condition in QFT?

If the Lorentz condition is not satisfied in QFT, it means that the equations of motion for the electromagnetic field are not consistent with special relativity. This could lead to inconsistencies and contradictions in the theory, making it less reliable and accurate in describing physical phenomena.

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