# I Brans-Dicke action

1. Feb 14, 2019 at 1:45 AM

### 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 gij 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 gij doesn't vary locally but on a large scale (globally)on the manifold it varies (e.g gij(x) and if the space is expanding e.g gij(t) .
Is it so??

Last edited: Feb 14, 2019 at 2:14 AM
2. Feb 14, 2019 at 12:05 PM

### Staff: Mentor

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

3. Feb 14, 2019 at 12:05 PM

### Apashanka

No the metric elements gij

4. Feb 14, 2019 at 12:08 PM

### Staff: Mentor

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

5. Feb 14, 2019 at 12:21 PM

### Apashanka

Actually I am thinking 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))and also if the space expands then only gij(t) hence as a whole for curved and expanding space gij(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 gij(x,t) they will be dimensionally consistent .
Am I right?!

6. Feb 14, 2019 at 12:28 PM

### Staff: Mentor

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?

7. Feb 14, 2019 at 12:45 PM

### Apashanka

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

8. Feb 14, 2019 at 1:01 PM

### haushofer

What is "dimensionally consistent"?

9. Feb 14, 2019 at 1:06 PM

### Apashanka

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

10. Feb 14, 2019 at 1:33 PM

### Staff: Mentor

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?

11. Feb 14, 2019 at 1:39 PM

### Apashanka

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

12. Feb 14, 2019 at 2:02 PM

### Staff: Mentor

$R$ 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 $G$.)

13. Feb 14, 2019 at 2:18 PM

### Apashanka

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

14. Feb 14, 2019 at 2:31 PM

### Staff: Mentor

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.)

15. Feb 14, 2019 at 2:32 PM

### Staff: Mentor

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

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