Difference between these 4-vector derivatives?

In summary, the two Lagrangian densities are equal when the action integral is taken over all of space.
  • #1
Dixanadu
254
2
Hey everyone,

So I've come across something in my notes where it says that these two Lagrangian densities are equal:

[itex]\mathcal{L}_{1}=(\partial_{mu}\phi)^{\dagger}(\partial^{\mu}\phi)-m^{2}\phi^{\dagger}\phi[/itex]

[itex]\mathcal{L}_{2}=-\phi^{\dagger}\Box\phi - m^{2}\phi^{\dagger}\phi[/itex]

where [itex]\Box = \partial^{\mu} \partial_{\mu}=\frac{1}{c^{2}}\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}[/itex]

How does this come about? I know that the second term in each density is equal of course, but the first term in each...how are they equal? can someone explain please?

Thank you!
 
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  • #2
The first one has the square of the first derivative of the field where the second one has has the second derivative with power one multiplied by another field variable. This suggests there have been an integration by parts and so by equal, the note means the action integrals for this two Lagrangian densities are equal. Of course provided that the fields vanish at infinity.
 
  • #3
Hey, erm I'm sorry I don't really understand lol...:S
 
  • #4
For getting the equation of motion from the Lagrangian, at first the action integral should be written.
[itex]
S=\int \mathcal L \ d^4 x
[/itex]
Now for your case, we have:
[itex]
S=\int \partial_\mu \phi^\dagger \partial^\mu \phi \ d^4 x - \int m^2 \phi^\dagger \phi \ d^4 x
[/itex]
Integrating the first integral by parts, gives:
[itex]
S=\phi^\dagger \partial^\mu \phi |_{(0,-\vec{\infty})}^{(\infty,\vec{\infty})}-\int \phi^\dagger \partial_\mu \partial^\mu \phi -\int m^2 \phi^\dagger \phi \ d^4 x
[/itex]
But the first term is zero, so we get your second Lagrangian density.

The actions coming from the two Lagrangians are equal and so give the same equations of motion. Your Lagrangian densities are equal in this sense!
 
  • #5
Omg thank you so much that's so simple! The only thing is, how do we know that the action integral is from -infty to +infty? and the fields are defined to vanish in these limits right?
 
  • #6
The action integral is [itex] S=\int_{t_1}^{t_2}L dt [/itex](The time limits in the integral of my previous post was wrong). Varying this w.r.t. whatever dynamical variables, gives us the equations governing the evolution of those variables from [itex]t_1 [/itex] to [itex] t_2 [/itex]. But for continuous systems, we define Lagrangian density([itex]\mathcal L[/itex]) by [itex] L=\int \mathcal L d^3 x [/itex]. To make the procedure manifestly relativistic, people abandon Lagrangian and work with Lagrangian density. Anyway, just like when you want to find total energy from energy density, you should integrate over all of space. The space integrations in the action integral and its limits come from here.
 
  • #7
Wow man, thanks sooo much honestly you're the boss :D !
 

1. What are 4-vector derivatives?

4-vector derivatives are mathematical operations used to describe the change in a 4-dimensional vector quantity with respect to another variable, such as time or space. They are commonly used in the fields of relativity and quantum mechanics.

2. What is the difference between a covariant and contravariant 4-vector derivative?

A covariant 4-vector derivative is defined as the change in a 4-vector component with respect to a change in coordinate position, while a contravariant 4-vector derivative is defined as the change in a 4-vector component with respect to a change in coordinate time. In other words, a covariant derivative measures the change in a vector as it moves through space, while a contravariant derivative measures the change in a vector as it moves through time.

3. How do 4-vector derivatives relate to special relativity?

4-vector derivatives are an essential tool in describing the effects of special relativity, which is the theory of how space and time are perceived differently by observers in different reference frames. They are used to calculate quantities such as velocity, acceleration, and momentum in relativity.

4. Can 4-vector derivatives be applied to classical mechanics?

Yes, 4-vector derivatives can be applied to classical mechanics, but they are most commonly used in the context of relativistic mechanics. In classical mechanics, 4-vector derivatives may be used to describe the motion of a particle in a curved spacetime, such as in general relativity.

5. Are there any real-world applications of 4-vector derivatives?

Yes, 4-vector derivatives have many practical applications in fields such as physics, engineering, and computer science. They are used in the design and analysis of systems that involve high speeds, strong gravitational fields, or quantum effects.

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