Brans-Dicke Action: Exploring the Metric Components

[Apashanka]

The brans-dicke action is

We have R which contains second order derivative in space and time of the metric components and product of first order derivatives in space and time of the metric components .
The second term in the figure also contains product of first order derivatives in space and time of the scaler field Φ(x,t).
To make it dimensionally consistent if we consider that the metric components g_ij vary over space, but how can be it is a function of time??
It is only if the space is curved for which locally the metric components g_ij doesn't vary locally but on a large scale (globally)on the manifold it varies (e.g g_ij(x) and if the space is expanding e.g g_ij(t) .
Is it so??

 

[PeterDonis]

how can be it is a function of time??

What, the scalar field? Why do you think it can't be a function of time?

 

[Apashanka]

What, the scalar field? Why do you think it can't be a function of time?

No the metric elements g_ij

 

[PeterDonis]

No the metric elements gij

Ok, why do you think those can't be functions of time?

 

[Apashanka]

Ok, why do you think those can't be functions of time?

Actually I am thinking if the space is curved then locally they (g_ij)doesnt vary with space (since locally we can assume the space to be flat) but globally they will vary with space(e.g g_ij=g_ij(x))and also if the space expands then only g_ij(t) hence as a whole for curved and expanding space g_ij(x,t) .
Is it so??
If so then only , as the second term in the figure contains product of first order derivative in space and time of Φ(x,t) and R contains second order derivative and product of first order derivative in space and time of g_ij(x,t) they will be dimensionally consistent .
Am I right?!

 

[PeterDonis]

if the space is curved then locally they (gij)doesnt vary with space (since locally we can assume the space to be flat) but globally they will vary with space(e.g gij=gij(x))

Yes, but you should be saying spacetime is curved, not just space. Globally $g_{\mu \nu}$ can vary with both space and time (and the split between "space" and "time" is not invariant anyway, so pure variation in "space" in one frame will be variation in both "space" and "time" in another frame).

if the space expands then only gij(t)

If you are thinking of a particular spacetime, such as the FRW spacetimes, which can be described as having expanding space, yes, in those particular spacetimes, you can find coordinates in which the metric only varies with time (not space). But only in those particular spacetimes. Brans-Dicke theory, if it were correct, would have to apply to all spacetimes, just like standard GR does.

Also, even in those particular coordinates in those particular spacetimes, the metric is a function of time, but it's not a function of space. But you were arguing that the metric should be a function of space but not time. So I don't understand what argument you are trying to make.

If so then only , as the second term in the figure contains product of first order derivative in space and time of Φ(x,t) and R contains second order derivative and product of first order derivative in space and time of gij(x,t) they will be dimensionally consistent

I don't understand. Why do you think this would rule out the metric being a function of time? Or of space, if that's what you're trying to argue?

 

[Apashanka]

Yes, but you should be saying spacetime is curved, not just space. Globally $g_{\mu \nu}$ can vary with both space and time (and the split between "space" and "time" is not invariant anyway, so pure variation in "space" in one frame will be variation in both "space" and "time" in another frame).
If you are thinking of a particular spacetime, such as the FRW spacetimes, which can be described as having expanding space, yes, in those particular spacetimes, you can find coordinates in which the metric only varies with time (not space). But only in those particular spacetimes. Brans-Dicke theory, if it were correct, would have to apply to all spacetimes, just like standard GR does.

Also, even in those particular coordinates in those particular spacetimes, the metric is a function of time, but it's not a function of space. But you were arguing that the metric should be a function of space but not time. So I don't understand what argument you are trying to make.
I don't understand. Why do you think this would rule out the metric being a function of time? Or of space, if that's what you're trying to argue?

No I am saying that g_ij has to be g_ij(x,t) for which the term containing R and the 2nd term in the Brans-Dicke action would be dimensionally consistent.
Am I right?

 

[haushofer]

What is "dimensionally consistent"?

 

[Apashanka]

What is "dimensionally consistent"?

What I am trying to say that R contains second order derivative and product of first order derivative in space and time of g_ij which has to be function of (x,t)
Since the second term in the figure also contains product of first order derivative in space and time of Φ(x,t)

 

[PeterDonis]

I am saying that gij has to be gij(x,t)

Ok, so you're saying the metric is a function of space and time? Of course it is. Why is this even a question?

R contains second order derivative and product of first order derivative in space and time of gij which has to be function of (x,t)

Yes.

the second term in the figure also contains product of first order derivative in space and time of Φ(x,t)

Yes, it does. So what exactly is your question?

 

[Apashanka]

Ok, so you're saying the metric is a function of space and time? Of course it is. Why is this even a question?
Yes.
Yes, it does. So what exactly is your question?

So there is a Φ related to R the first term and ω(Φ)/Φ related to the 2nd term ,now how these two terms are matched dimensionally since R and ∂^μΦ∂_μΦ(in figure) are matched dimensionally

 

[PeterDonis]

how these two terms are matched dimensionally since R and ∂μΦ∂μΦ(in figure) are matched dimensionally

$R$ and $\partial_\mu \varphi \partial_\mu \varphi$ aren't matched dimensionally; they don't have the same units. The things that have the same units are

$$
\varphi R
$$

and

$$
\frac{\omega(\varphi)}{\varphi} g^{\mu \nu} \partial_\mu \varphi \partial_\nu \varphi
$$

Each term must have units of length to the inverse fourth power (because the integrand as a whole must be dimensionless, and it multiplies each term by $d^4 x$). We know $R$ has units of inverse length squared. That means $\varphi$ must also have units of inverse length squared. (Note that these are not the usual units for a scalar field.) Each derivative has units of inverse length, so $\partial_\mu \varphi \partial_\nu \varphi$ has units of length to the inverse sixth power, which is not the same as the units of $R$. This also tells us that $\omega(\varphi)$ must be dimensionless (since $g^{\mu \nu}$ is dimensionless and the factor of $\varphi$ in the denominator of the factor in front of the second term makes the units for that term as a whole work out).

(Note: all this assumes that $G$ as it appears in what you wrote has no units. But $G$ is supposed to be Newton's gravitational constant, in appropriate units, which is not dimensionless. Most presentations that I've seen of Brans-Dicke theory leave out the $G$ altogether for this reason; basically they are taking the field $\varphi$ to substitute for $G$.)

 

[Apashanka]

Each term must have units of length to the inverse fourth power (because the integrand as a whole must be dimensionless, and it multiplies each term by d⁴x). We know R has units of inverse length squared

If so then how is the dimension in Einstein -hilbert action matched
∫R√|g|d⁴x??
Implies R has units of inverse length to power 4

 

[PeterDonis]

how is the dimension in Einstein -hilbert action matched

Because you left out the factor of $1 / 16 \pi G$, which has dimensions of inverse length squared in the Einstein-Hilbert action. (Note that I specifically commented about how $G$ must have different units than it usually does in the version of the Brans-Dicke action that you wrote down.)

 

[PeterDonis]

Because you left out the factor of $1 / 16 \pi G$, which has dimensions of inverse length squared in the Einstein-Hilbert action. (Note that I specifically commented about how GG must have different units than it usually does in the version of the Brans-Dicke action that you wrote down.)

See, for example, how the two actions are compared here:

https://en.wikipedia.org/wiki/Brans–Dicke_theory#The_action_principle