four dimensional notation of lagrangian

[haqnomi]

Hi, all

 

I would like to have understanding for the above notation please. The Lagrangian density is expressed for acoustic waves in four d notation. Further how to differentiate this Lagrangian for Lagrangian equation. I am very confused with indices mix up.

[elfmotat]

I'm afraid I don't see any equations.

[haqnomi]

L= 1/2 (μ_(o ) γ_(i,0 ) γ_(i,0)+2P_o γ_(i,i)-αP_o γ_(i,i) γ_(j,j) ),

 

Lagrangian density for acoustic waves in four dimensional notation. The form in vector notation can be found on the following article please:

 

http://physics.stackexchange.com/questions/36249/speed-of-sound-and-the-potential-energy-of-an-ideal-gas-goldstein-derivation

[Enthalpy]

Already a few elements - without having put the necessary time into the linked page, sorry for that...

 

It's the enthalpy that matters in the propagation equation, not the internal energy. The enthalpy variation tells the work in an adiabatic transformation, and this woks converts to and from kinetic energy in a wave.

 

Neither the internal energy would be entirely kinetic.For a monoatomic gas yes, but already a diatomic one like air contains more contributions. Rotations can be understood as kinetic energy, vibrations only in part, as the other part is a change in atomic distances or angles. Vapour and carbon dioxide, for instance in air, do contain vibration energy even at room temperature.

 

I don't grasp the difficulty with "entirely kinetic" and "microscopic versus macroscopic". If you mean 1/2µV² by "macroscopic", this collective oriented movement is not included in the thermodynamic internal energy (nor enthalpy), which accounts for individual unoriented movements (and more) versus the collective one. Did you mean something else?

[haqnomi]

@ enthalpy, what are you talking about? My post was not about the topic you started. Are you replying for my topic? If yes then you may not understand this and if no then you are on the wrong place.

[elfmotat]

L= 1/2 (μ_(o ) γ_(i,0 ) γ_(i,0)+2P_o γ_(i,i)-αP_o γ_(i,i) γ_(j,j) ),

 

Lagrangian density for acoustic waves in four dimensional notation. The form in vector notation can be found on the following article please:

 

http://physics.stackexchange.com/questions/36249/speed-of-sound-and-the-potential-energy-of-an-ideal-gas-goldstein-derivation

 

That Lagrangian is not Lorentz invariant (which should be obvious given the formula for kinetic energy they used), so I'm not sure why you're trying to use four-vector notation. If an equation doesn't have Lorentz symmetry then putting it in four-vector form isn't going to look pretty! It's also probably not going to help very much. The next thing I should point out is that the second term in the Lagrangian you posted will not contribute to the action, because it's a total divergence (remember the divergence theorem + boundary conditions).

 

I'll use the notation that the guy on stackexchange used. With signature (-+++) we have:

 

[math]\mathcal{L}=\frac{1}{2} \left ( \mu_0 \, \dot{\vec{\eta}} \cdot \dot{\vec{\eta}} - \gamma P_0 \, (\nabla \cdot \vec{\eta})^2 \right ) = \frac{1}{2} \left ( \mu_0 \, g_{ij} \, \partial_0 \eta^i \, \partial_0 \eta^j - \gamma P_0 \, (\partial_i \eta^i)^2 \right )[/math]

 

I'm going to drop the index notation now, because it's a hassle to use when the Lagrangian isn't relativistic. I'm sure you can work it out yourself by analogy if you really want it in that notation. So for the action we have:

 

[math]S= \frac{1}{2} \int d^4x \left ( \mu_0 \, \dot{\vec{\eta}} \cdot \dot{\vec{\eta}} - \gamma P_0 \, (\nabla \cdot \vec{\eta})^2 \right )[/math]

 

To get the equations of motion, we vary the action and set it to zero:

 

[math]\delta S= \frac{1}{2} \int d^4x \left (2 \mu_0 \, \dot{\vec{\eta}} \cdot \delta \dot{\vec{\eta}} - 2 \gamma P_0 \, \nabla \cdot \vec{\eta} ~ \delta \left ( \nabla \cdot \vec{\eta} \right ) \right ) = \int d^4x \left (\mu_0 \, \dot{\vec{\eta}} \cdot \left [ \frac{d}{dt} \delta \vec{\eta} \right ] - \gamma P_0 \, \nabla \cdot \vec{\eta} ~ \left [ \nabla \cdot \delta \vec{\eta} \right ] \right )[/math]

 

When we do the integral over time, we can integrate the first term by parts and set the boundary terms to zero. Similarly, when we do the integral over space we can integrate the second term by parts. What we get is:

 

[math]\delta S = \int d^4x \left (-\mu_0 \, \ddot{\vec{\eta}} \cdot \delta \vec{\eta} + \gamma P_0 \, \nabla^2 \vec{\eta} \cdot \delta \vec{\eta} \right ) = \int d^4x \left (-\mu_0 \, \ddot{\vec{\eta}} + \gamma P_0 \, \nabla^2 \vec{\eta} \right ) \cdot \delta \vec{\eta}=0[/math]

 

The above must hold for all [math]\delta \vec{\eta}[/math], so we find the following equations of motion:

 

[math]\mu_0 \, \ddot{\vec{\eta}} = \gamma P_0 \, \nabla^2 \vec{\eta}[/math]

[Enthalpy]

@ enthalpy, what are you talking about? My post was not about the topic you started. Are you replying for my topic? If yes then you may not understand this and if no then you are on the wrong place.

 

No, I was responding to some points in the webpage you linked, and didn't realize you were not necessarily the one who wrote the linked webpage. So indeed, it wasn't an answer to the topic you started here, sorry.

Edited October 30, 2014 by Enthalpy