Energy-Momentum Tensor for the Klein-Gordon Lagrangian

So, in summary, we can show that the divergence of the energy-momentum tensor for the Klein-Gordon Lagrangian, ##T^{\mu \nu} = \partial^{\mu}\phi\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}_{KG}##, is equal to 0 by using the Klein-Gordon equation.
  • #1
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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  • #2
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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  • #3
μνμνφ + m2φ = 0
 
  • #4
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?
 
  • #5
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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  • #6
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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  • #7
Looks nice.
 
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  • #8
Thanks!
 

Related to Energy-Momentum Tensor for the Klein-Gordon Lagrangian

What is the Energy-Momentum Tensor for the Klein-Gordon Lagrangian?

The Energy-Momentum Tensor for the Klein-Gordon Lagrangian is a mathematical object that describes the energy and momentum of a field in space-time. It is derived from the Klein-Gordon Lagrangian, which is a mathematical expression that describes the dynamics of a scalar field.

Why is the Energy-Momentum Tensor important in physics?

The Energy-Momentum Tensor is important because it allows us to calculate the energy and momentum of a field, which are crucial quantities in understanding the behavior of physical systems. It also plays a key role in the theory of general relativity, where it is used to describe the gravitational field.

How is the Energy-Momentum Tensor calculated from the Klein-Gordon Lagrangian?

The Energy-Momentum Tensor is calculated by taking the derivative of the Klein-Gordon Lagrangian with respect to the metric tensor. This produces a set of equations that can be solved to obtain the components of the Energy-Momentum Tensor.

What are the physical interpretations of the components of the Energy-Momentum Tensor?

The components of the Energy-Momentum Tensor have different physical interpretations. The diagonal components represent the energy and momentum density of the field, while the off-diagonal components represent the flow of energy and momentum in different directions. The trace of the Energy-Momentum Tensor represents the total energy density, and the divergence of the Energy-Momentum Tensor represents the rate of change of energy and momentum in a given region of space.

How does the Energy-Momentum Tensor relate to the conservation of energy and momentum?

The Energy-Momentum Tensor is closely related to the conservation of energy and momentum. In fact, the conservation laws can be derived from the Energy-Momentum Tensor by taking the divergence of the tensor and setting it equal to zero. This means that any changes in energy and momentum in a system must be balanced by corresponding changes in the field described by the Klein-Gordon Lagrangian.

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