Scalar fields: why symmetric ener-mom. tensor?

I didn't see that. so the energy momentum tensor is symmetric because of the symmetry of the dot product. I understand now.In summary, the conversation discusses the properties of the energy momentum tensor for a scalar field in a theory involving only scalar fields. The statement is made that the energy momentum tensor is always symmetric, but there are doubts on how to demonstrate this. The conversation then goes through a particular example and uses the definition of the energy momentum tensor to show that it is indeed symmetric. The conversation also discusses the significance of the symmetric tensor and how it relates to the symmetry of the dot product. It is concluded that the energy momentum tensor is symmetric due to the symmetry of the dot product and the fact that it is a rank-2
  • #1
provolus
18
0
I'm studying the properties of the energy momentum tensor for a scalar field (linked to the electromagnetic field and corresponding energy-momentum tensor) and now I'm facing the statement:

"for a theory involving only scalar fields, the energy-momentum tensor is always symmetric". But I've some doubts on how to demonstrate it. So I started with a particular example.

Given a density of Lagrangian [tex]L[/tex] for a scalar field [tex]\phi[/tex] (our field is in Minkowsky space with metric tensor [tex] \eta^{\mu\nu}[/tex] +---) of the form:

[tex]L=\frac{1}{2}(\partial_{\mu}\phi\partial^{\mu}\phi-m^2\phi^2})-\frac{\lambda}{4!}\phi^4[/tex]

with real constant [tex]\lambda[/tex] and [tex]m[/tex], according to the definition of energy-momentum tensor [tex] T^{\mu\nu}[/tex], that is
[tex]T^{\mu\nu}=\frac{\partial L}{\partial(\partial_{\mu}\phi_i)}\partial^{\nu}\phi_i-\eta^{\mu\nu}L} [/tex]

(with latin index i running on space coordinates) for this [tex]L[/tex], substituing into the last formula, I should get (dropping the index i since [tex]\phi[/tex] is a scalar):

[tex]T^{\mu\nu}=\frac{1}{2}(\partial^{\mu}\phi\partial^{\nu}\phi-\partial^{\nu}\phi\partial^{\mu}\phi)-m^2\phi^2-\frac{\lambda}{4!}\phi^4[/tex]

Arrived at this point my first question is: is this [tex]T^{\mu\nu}[/tex] symmetric?

Trying to answer by myself but I'm new in tensor computing and I know that it would be a joke with a few of QFT and GR knowledge...

given that [tex]\partial^{\mu}\phi[/tex] is a vector, and then [tex]\partial^{\mu}\phi\partial^{\nu}\phi [/tex] is a dot product (but where are the contracted indices?), so a scalar, it should be

[tex]\frac{1}{2}(\partial^{\mu}\phi\partial^{\nu}\phi-\partial^{\nu}\phi\partial^{\mu}\phi)=0[/tex]

so with all scalars my tensor should be symmetric, isn't?

But, generally speaking, my second question is why a theory of only scalar field the energy-momentum tensor is always symmetric?

Trying to answer by myself I think that being the Langrangian Lorentz invariant, in certain way this is reflected to the tensor... but I'm not sure...

can someone be so kind to explain to help me in answering these two questions?
 
Physics news on Phys.org
  • #2
well. I'm a little unsure but i'll give it a go

[itex]\partial^\mu \phi[/itex] is a vector.
[itex]\partial^\mu \phi \partial^\nu \phi[/itex] isn't a dot product. if there are contracted indices it reduces to a scalar.
the indices are different so it corresponds to multiplying to matrices together.
a 4x1 and a 1x4 giving an overall 4x4 matrix

so the stress energy tensor is going to be a 4x4 matrix

so now you have to ask : is [itex]T^{\mu \nu}=T^{\nu \mu}[/itex]

but it appears not to be the case. hmmmmm, hopefully someone else can point out why!
 
  • #3
How did you use the definition of the energy momentum tensor to get that minus sign between the two derivative terms? That's not right.
 
  • #4
Considering the terms between the brackets, the first one is obtained deriving the Lagrangian respect to [tex]\partial\phi[/tex]; the second one, with the minus sign, is obtained raising the index of the first 4-gradient... with + sign the terms in bracket with the 1/2 factor would be the common decomposition of a symmetric tensor. so everything would be right, right?
 
  • #5
Think of it this way:
[tex]L = \frac{1}{2}\eta^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4[/tex]
In detail,
[tex]L = \frac{1}{2}[-(\partial_t\phi)^2 + (\partial_x\phi)^2 + (\partial_y\phi)^2 + (\partial_z\phi)^2] -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4[/tex]
so
[tex]\frac{\partial L}{\partial(\partial_t \phi)} = -\partial_t\phi[/tex]
[tex]\frac{\partial L}{\partial(\partial_x \phi)} = \partial_x\phi[/tex]
[tex]\frac{\partial L}{\partial(\partial_y \phi)} = \partial_y\phi[/tex]
[tex]\frac{\partial L}{\partial(\partial_z \phi)} = \partial_z\phi[/tex]
which is equivalent to
[tex]\frac{\partial L}{\partial(\partial_\mu \phi)} = \eta^{\mu\nu}\partial_\nu\phi = \partial^\mu\phi[/tex]
Try working through the rest of it from there. Also remember that your result is a tensor, so if you get something like [itex]m^2\phi^2[/itex], with no indices, you've done something wrong.
 
  • #6
ok, I've understood my error. Thanks! So, finally I get

[tex]
T^{\mu\nu}=\partial^{\mu}\phi\partial^ {\nu}\phi-\eta^{\mu\nu}(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-m^2\phi^2})-\frac{\lambda}{4!}\phi^4
[/tex]
and I'm sure that until this point the tensor is correct. But how can I see that
[itex]
T^{\mu \nu}=T^{\nu \mu}
[/itex]
?
 
Last edited:
  • #7
[tex]-\eta^{\mu \nu}[/tex] should multiply the the last term as well. Also you could change the contracted index to something else so there is no confusion. Symmetricity follows from the fact that [tex]\eta^{\mu \nu}[/tex] is symmetric and that derivatives of the field commute.
 
Last edited:
  • #8
Writing it as:
[tex]T^{\mu\nu}=\partial^{\mu}\phi\partial^ {\nu}\phi-\eta^{\mu\nu}L[/tex]
should make you see it directly.

Your metric is always symetric, and [tex]\partial^{\mu}\phi\partial^ {\nu}\phi=\partial^{\nu}\phi\partial^{\mu}\phi[/tex].
 
  • #9
your second term in post 6 is wrong-it violates the summation convetion. one of those indices should be a [itex]\nu[/itex] which means it will contract to an invariant.
 
  • #10
Ok thanks. I get everything, now. I was missing the point about the derivatives.

Can you help me, please, to understand why I get always a symmetric energy-momentum for a theory of scalar fields?
 
  • #11
sorry, can you be more explicit? I've compared that formula with several books, and the error is the last term out of the brackets...
 
  • #12
latentcorpse said:
your second term in post 6 is wrong-it violates the summation convetion. one of those indices should be a [itex]\nu[/itex] which means it will contract to an invariant.
What, like this?
[tex]\eta^{\mu\nu} \partial_\mu\phi \partial^\nu\phi[/tex]
I think you meant to say something else...

[tex]\eta^{\mu\nu} \partial_\alpha\phi \partial^\alpha\phi[/tex]
is what we're after. It doesn't contract to a scalar, and can't, because the answer has to be a rank-2 contravariant tensor.
 
  • #13
my bad.
 

Related to Scalar fields: why symmetric ener-mom. tensor?

1. What is a scalar field?

A scalar field is a mathematical function that assigns a scalar value to every point in space. It is used to describe physical quantities that only have magnitude, such as temperature or pressure.

2. What is a symmetric energy-momentum tensor?

A symmetric energy-momentum tensor is a mathematical object used in the theory of general relativity to describe the distribution of energy and momentum in space. It is a 4x4 matrix that contains information about the energy, momentum, and stress at each point in space.

3. Why are scalar fields important in physics?

Scalar fields are important in physics because they allow us to mathematically describe physical quantities that do not have a direction, but only a magnitude. This includes important quantities such as temperature, density, and pressure.

4. How are scalar fields related to symmetry?

Scalar fields are often used in physics because they possess certain symmetries, meaning that they remain unchanged when certain transformations are applied to them. This makes them useful for describing physical systems that exhibit symmetries.

5. What is the role of scalar fields in the theory of relativity?

In the theory of relativity, scalar fields are used to describe the curvature of spacetime, which is responsible for the effects of gravity. They are also used in the equations that describe the behavior of matter and energy in the universe.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
646
  • Advanced Physics Homework Help
Replies
5
Views
3K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
1K
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
30
Views
5K
  • Advanced Physics Homework Help
Replies
1
Views
633
  • Advanced Physics Homework Help
Replies
1
Views
794
  • Advanced Physics Homework Help
Replies
1
Views
784
Back
Top