Energy-Momentum Tensor for the Klein-Gordon Lagrangian

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SUMMARY

The energy-momentum tensor for the Klein-Gordon Lagrangian, defined as ##T^{\mu\nu} = \partial^{\mu}\phi\partial^{\nu}\phi - \eta^{\mu\nu}\mathcal{L}_{KG}##, satisfies the divergence-free condition ##\partial_{\mu}T^{\mu\nu} = 0##. The derivation involves applying the Klein-Gordon equation, which simplifies the expression to zero. Key steps include recognizing the cancellation of terms and correctly applying the product rule for derivatives. The final result confirms the conservation of energy-momentum in the context of the Klein-Gordon field.

PREREQUISITES
  • Klein-Gordon Lagrangian formulation
  • Energy-momentum tensor definition
  • Partial derivatives and product rule
  • Klein-Gordon equation
NEXT STEPS
  • Study the derivation of the energy-momentum tensor for other field theories, such as the Dirac field.
  • Explore the implications of the conservation law ##\partial_{\mu}T^{\mu\nu} = 0## in general relativity.
  • Learn about Noether's theorem and its relation to symmetries and conservation laws in physics.
  • Investigate the role of the energy-momentum tensor in cosmology and its applications in the Friedmann equations.
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The discussion is beneficial for theoretical physicists, graduate students studying quantum field theory, and researchers focusing on the mathematical foundations of field theories.

spaghetti3451
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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.
 
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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)##?
 
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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}##
 
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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?
 
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Looks nice.
 
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Thanks!
 

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