Energy-Momentum Tensor for the Klein-Gordon Lagrangian

Click For Summary

Homework Help Overview

The discussion revolves around the energy-momentum tensor for the Klein-Gordon Lagrangian, specifically the expression for the tensor and the requirement to show that its divergence vanishes. Participants are exploring the implications of the Klein-Gordon equation in this context.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to compute the divergence of the energy-momentum tensor and are discussing the necessary steps and terms involved in the calculation. Questions arise about the placement of terms and the use of the Klein-Gordon equation to simplify the expression.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's attempts. Some participants have suggested relabeling variables for clarity and are verifying the correctness of the steps taken in the calculations. There is a recognition of the need to focus on the divergence of the tensor.

Contextual Notes

Participants are working under the constraints of the problem statement and are questioning the assumptions made in their calculations. The discussion reflects a collaborative effort to clarify the mathematical reasoning involved in deriving the result.

spaghetti3451
Messages
1,311
Reaction score
31

Homework Statement



The energy-momentum tensor ##T^{\mu\nu}## of the Klein-Gordon Lagrangian ##\mathcal{L}_{KG} = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}## is given by

$$T^{\mu\nu}~=~\partial^{\mu}\phi\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}_{KG}.$$
Show that ##\partial_{\mu}T^{\mu\nu}=0##.

Homework Equations



The Attempt at a Solution



$$\partial_{\mu}T^{\mu\nu} \\
=\partial_{\mu}[\partial^{\mu}\phi\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}_{KG}]\\
=\partial_{\mu}(\partial^{\mu}\phi\partial^{\nu}\phi-\partial^{\nu}\mathcal{L}_{KG})\\
=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)-\partial^{\nu}(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2})\\
=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)-\partial^{\nu}(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi)+m^{2}\phi(\partial^{\nu}\phi)$$

Where do I go from here? I know I need to use the Klein-Gordon equation, but using the KG equation cancels the first and third terms and leaves the second term.
 
  • Like
Likes   Reactions: JD_PM
Physics news on Phys.org
failexam said:
$$\partial_{\mu}T^{\mu\nu} \\
=\partial_{\mu}[\partial^{\mu}\phi\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}_{KG}]\\
=\partial_{\mu}(\partial^{\mu}\phi\partial^{\nu}\phi-\partial^{\nu}\mathcal{L}_{KG})$$
The right parenthesis is misplaced in last line above.

$$=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)-\partial^{\nu}(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2})$$
Did you miss a term when writing out ##\partial_{\mu}(\partial^{\mu}\phi\partial^{\nu}\phi)##?
 
  • Like
Likes   Reactions: spaghetti3451
μνμνφ + m2φ = 0
 
Ok. Let me start again.

##\partial_{\rho}T^{\mu\nu} = \partial_{\rho}(\partial^{\mu}\phi\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}_{KG})##

##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\eta^{\mu\nu}\partial_{\rho}\mathcal{L}_{KG}##

##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\eta^{\mu\nu}\partial_{\rho}(\frac{1}{2}\partial_{\sigma}\phi\partial^{\sigma}\phi-\frac{1}{2}m^{2}\phi^{2})##

##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\rho}\partial_{\sigma}\phi)(\partial^{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

Where do I go from here?
 
failexam said:
##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\rho}\partial_{\sigma}\phi)(\partial^{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

Where do I go from here?

Are the 3rd and 4th terms related?

You want to investigate the divergence ##\partial_{\mu}T^{\mu \nu}## rather than ##\partial_{\rho}T^{\mu \nu}##
 
  • Like
Likes   Reactions: JD_PM and spaghetti3451
TSny said:
Are the 3rd and 4th terms related?

Yes. The third and fourth terms are identical as follows:

##(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\rho}\partial_{\sigma}\phi)(\partial^{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\rho}\partial^{\sigma}\phi)(\partial_{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

TSny said:
You want to investigate the divergence ##\partial_{\mu}T^{\mu \nu}## rather than ##\partial_{\rho}T^{\mu \nu}##

Okay. Let me relabel ##\rho## to ##\mu##. Then, I have

##=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\mu}\partial^{\nu}\phi)-\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\mu}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\mu}\phi)##

##=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial_{\mu}\phi)(\partial^{\mu}\partial^{\nu}\phi)-(\partial_{\sigma}\phi)(\partial^{\nu}\partial^{\sigma}\phi)+m^{2}\phi(\partial^{\nu}\phi)##

##=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial_{\sigma}\phi)(\partial^{\sigma}\partial^{\nu}\phi)-(\partial_{\sigma}\phi)(\partial^{\nu}\partial^{\sigma}\phi)+m^{2}\phi(\partial^{\nu}\phi)##, where the second and third terms will now cancel

##=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)+m^{2}\phi(\partial^{\nu}\phi)##

##=(\partial_{\mu}\partial^{\mu}\phi+m^{2}\phi)(\partial^{\nu}\phi)##, where we will now use the Klein-Gordon equation

##=0##.

I think it's all correct now, isn't it?
 
  • Like
Likes   Reactions: JD_PM
Looks nice.
 
  • Like
Likes   Reactions: JD_PM
Thanks!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
8K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K