Getting the Klein Gordon energy momentum tensor

[LCSphysicist]

Homework Statement

.

Relevant Equations

.

I want to get the stress energy tensor of a scalar field using the Hilbert method (namely, $T^{\mu v} = \frac{2}{\sqrt{-g}} \frac{\delta S}{\delta g_{\mu v}}$)

$$S = \int \frac{1}{2}(\partial_\mu \phi \partial^{\mu} \phi - m^2 \phi ^2)\sqrt{-g}d^4x$$
$$= \int \frac{1}{2}(\partial^{v} \phi \partial^{\mu} \phi g_{v \mu} - m^2 \phi ^2)\sqrt{-g}d^4x$$
$$\delta S / \delta g_{a b} =$$
$$ \int \frac{1}{2}(\partial^{a} \phi \partial^{b} \phi )\sqrt{-g} d^4x + \int \frac{1}{2}(\partial^{v} \phi \partial^{\mu} \phi g_{v \mu} - m^2 \phi ^2)\frac{\sqrt{-g} g^{a b}}{2}d^4x$$

Where i have used $\delta \sqrt{-g} = \sqrt{-g} g^{x y} \delta g_{x y} / 2$

$$T^{a b} = \frac{1}{2}(\partial^{a} \phi \partial^{b} \phi) + \frac{1}{2}(\partial^{v} \phi g_{v \mu} \partial^{\mu} \phi- m^2 \phi ^2) g^{a b}$$

This is not what i was expecting...

 

[Orodruin]

Homework Statement:: .
Relevant Equations:: .

Where i have used δ−g=−ggxyδgxy/2

This is missing a minus sign. You also forgot to multiply the first term by 2.

 

[LCSphysicist]

This is missing a minus sign. You also forgot to multiply the first term by 2.

But second this video
(see time 7:21)
There are no minus sign?

I know the minus sign can arise if, instead of $g^{xy}\delta g_{xy}$, we use $g_{xy}\delta g^{xy}$

 

[Orodruin]

Sorry, I was a bit quick because the stress-energy tensor is typically defined by variation wrt the inverse metric components so this threw me off a bit. If you want to use a definition wrt the metric components, you need to take into account that the derivatives of the field that are metric independent are $\partial_\mu \phi$, not $\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$. Therefore, the variation of the kinetic term wrt the metric is
$$
\delta((\partial_\mu \phi)(\partial_\nu\phi)g^{\mu\nu}) =(\partial_\mu \phi)(\partial_\nu\phi)\delta g^{\mu\nu}.
$$
Now, $g_{\rho\lambda}\delta g^{\mu\lambda} = - g^{\mu\lambda}\delta g_{\rho\lambda}$ so $\delta g^{\mu\nu} = - g^{\rho\nu} g^{\mu\lambda} \delta g_{\rho\lambda}$. The variation therefore becomes
$$
\delta((\partial_\mu \phi)(\partial_\nu\phi)g^{\mu\nu}) = - (\partial_\mu \phi)(\partial_\nu\phi)g^{\rho\nu} g^{\mu\lambda} \delta g_{\rho\lambda} = - (\partial^\mu \phi)(\partial^\nu \phi) \delta g_{\mu\nu}.
$$
This has a minus sign relative to your result.

Not having to rewrite the variation of the inverse metric components is the reason you will typically see the stress energy tensor defined in terms of the variation wrt the inverse metric components.

 

[LCSphysicist]

Sorry, I was a bit quick because the stress-energy tensor is typically defined by variation wrt the inverse metric components so this threw me off a bit. If you want to use a definition wrt the metric components, you need to take into account that the derivatives of the field that are metric independent are $\partial_\mu \phi$, not $\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$. Therefore, the variation of the kinetic term wrt the metric is
$$
\delta((\partial_\mu \phi)(\partial_\nu\phi)g^{\mu\nu}) =(\partial_\mu \phi)(\partial_\nu\phi)\delta g^{\mu\nu}.
$$
Now, $g_{\rho\lambda}\delta g^{\mu\lambda} = - g^{\mu\lambda}\delta g_{\rho\lambda}$ so $\delta g^{\mu\nu} = - g^{\rho\nu} g^{\mu\lambda} \delta g_{\rho\lambda}$. The variation therefore becomes
$$
\delta((\partial_\mu \phi)(\partial_\nu\phi)g^{\mu\nu}) = - (\partial_\mu \phi)(\partial_\nu\phi)g^{\rho\nu} g^{\mu\lambda} \delta g_{\rho\lambda} = - (\partial^\mu \phi)(\partial^\nu \phi) \delta g_{\mu\nu}.
$$
This has a minus sign relative to your result.

Not having to rewrite the variation of the inverse metric components is the reason you will typically see the stress energy tensor defined in terms of the variation wrt the inverse metric components.

Thank you. I have been able to follow the math reasoning you have showed. The thing i am not getting is the statement "If you want to use a definition wrt the metric components, you need to take into account that the derivatives of the field that are metric independent are $\partial_\mu \phi$, not $\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$
Therefore, the variation of the kinetic term wrt the metric is
$$
\delta((\partial_\mu \phi)(\partial_\nu\phi)g^{\mu\nu}) =(\partial_\mu \phi)(\partial_\nu\phi)\delta g^{\mu\nu}.
$$
"

My point is: I have a function f(x,y) and g(x,y) = 0 equation. If i want to know how $f(x,y)$ varies with $\delta x$, i would use $g(x,y) = 0$ to express $y = y(x)$, and so $f(x,y)$ goes to $f(x,y(x))$. What i am saying is, i would express my function as a function of the variables $(x)$ i am varying infinitesimally.

Why is here different? That is, if i want to vary with respect to $g_{\mu v}$, i need to express the terms as a function of $g^{\mu v}$ and not, as $f(x,y)$ above, as a function of $g_{\mu v}$?

 

[Orodruin]

Thank you. I have been able to follow the math reasoning you have showed. The thing i am not getting is the statement "If you want to use a definition wrt the metric components, you need to take into account that the derivatives of the field that are metric independent are $\partial_\mu \phi$, not $\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$"

My point is: I have a function f(x,y) and g(x,y) = 0 equation. If i want to know how $f(x,y)$ varies with $\delta x$, i would use $g(x,y) = 0$ to express $y = y(x)$, and so $f(x,y)$ goes to $f(x,y(x))$. What i am saying is, i would express my function as a function of the variables i am varying infinitesimally.

Why is here different? That is, if i want to vary with respect to $g_{\mu v}$, i need to express the terms as a function of $g^{\mu v}$ and not, as $f(x,y)$ above, as a function of $g_{\mu v}$?

It is not different, but you have ignored that $y$ depends on $x$ when asking the question how $f(x,y(x))$ depends on $x$.

 

[LCSphysicist]

It is not different, but you have ignored that $y$ depends on $x$ when asking the question how $f(x,y(x))$ depends on $x$.

Sorry to be late. I have been thinking about it, and i think i still didn't get it. Sorry
So, considering what you have said, "but you have ignored that $y$ depends on $x$ when asking the question how $f(x,y(x))$ depends on $x$.", we need to consider $g_{\mu v}$ e $g^{\mu v}$ both independent objects?
My doubt still remains on why $\partial_{\mu} \phi$ is the one which is metric independent. Why? Have i assumed that when i decided to vary with respect to $\delta g_{\mu v}$? How did i assumed it?

 

[Orodruin]

The coordinates are $x^\mu$ and the partial derivatives $\partial_\mu = \partial/\partial x^\mu$. You need no metric to define it. On the other hand, $\partial^\mu = g^{\mu\nu}\partial_\nu$ does depend on the metricby definition.

we need to consider gμv e gμv both independent objects?

No. This is exactly the point. What you have done is effectively to ignore that $g^{\mu\nu}$ in $\partial^\mu$ depends on $g_{\mu\nu}$.

 

[LCSphysicist]

The coordinates are $x^\mu$ and the partial derivatives $\partial_\mu = \partial/\partial x^\mu$. You need no metric to define it. On the other hand, $\partial^\mu = g^{\mu\nu}\partial_\nu$ does depend on the metricby definition.No. This is exactly the point. What you have done is effectively to ignore that $g^{\mu\nu}$ in $\partial^\mu$ depends on $g_{\mu\nu}$.

@Orodruin
Hello again... I was revising the problem, and i got confused with one thing :v

So i would like to confirm with you this:

For metric $$(+,-,-,-)$$
$$T_{\mu v} = \frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu v}}$$
$$T^{\mu v} = -\frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu v}}$$

Is that right? I am asking because the different conventions for metric is confusing me a lot...

 

[Orodruin]

There are at least three different sign conventions going around in GR (giving you a total of 8 possible choices as to where minuses pop up although I am not sure all can be found used in textbooks). Metric signature, definition of curvature, and definition of the stress energy tensor. The different conventions and where the signs pop up are discussed in MTW.

Edit: … but if one of your equations have a minus the other comes with a plus so at least that is internally consistent.