# A Help with Weinberg p. 72 -- time dt for a photon to travel a distance d⃗x

1. Oct 22, 2016

### Kostik

Hi all, and thanks in advance. I am an old guy learning GR for fun. Reading Weinberg's "Gravitation and Cosmology". PhD in math 1998, so I read all books like I read math books: every character, every word, every line, every page extremely carefully.

I am stuck on the stupidest thing. On p.72, he writes out a quadratic equation for the time $dt$ for a photon to travel a distance $d\vec{x}$:
$$0 = g_{00}dt^2 + 2g_{i0}dx^i dt + g_{ij}dx^i dx^j$$
($i=1,2,3$.), which follows immediately from (3.2.9) or (3.2.6). He then gives the solution in (3.2.10) using the quadratic formula.

What stumps me is that he has used the minus (-) square root instead of the (+) square root. How does he know to do that? If you test it using the simplest coordinate transformation $x^α=ξ^α$ and hence the metric $g_{μν}=η_{μν}$, then of course he IS right, because $g_{00} = -1$. But the $g_{μν}$ could be anything, so how does he justify taking the negative square root?

2. Oct 22, 2016

### Orodruin

Staff Emeritus
Please write out the equations you are refering to. Everybody does not have a copy of Weinberg easily accessible.

3. Oct 22, 2016

### Kostik

Sure. (3.2.6) is simply the statement of proper time $dτ^2$:
$$dτ^2 = -g_{μν}dx^μdx^ν.$$
From this follows the equation for the time $dt$ for a photon to travel a distance $d\vec{x}$:
$$0 = g_{00}dt^2 + 2g_{i0}dx^i dt + g_{ij}dx^i dx^j$$
Weinberg says, "the solution is":
$$dt = \frac1 {g_{00}} \left[ -g_{i0}dx^i - \sqrt { (g_{i0}g_{j0} - g_{ij}g_{00}) dx^i dx^j } \right]$$
What stumps me is how he justifies taking the negative squate root, given
$$g_{μν} = \frac {∂ξ^α} {∂x^μ} \frac {∂ξ^β} {∂x^μ} η_{αβ}$$ can be anything. Note Weinberg specifically remarks on the previous page that the coordinate system $x^μ$ can be "a Cartesian coordinate system at rest in the laboratory, but also may be curvilinear, accelerated, rotating, or what we will."

Last edited: Oct 22, 2016
4. Oct 22, 2016

### strangerep

Solving the eqn in @Kostik's post #1 as a quadratic in $dt$, we get: $$dt ~=~ \frac{-2g_{i0} \pm \sqrt{(4g_{i0}g_{j0} - 4 g_{00} g_{ij}) dx^i dx^j}}{2 g_{00}} ~=~ \frac{-g_{i0} \pm \sqrt{(g_{i0}g_{j0} - g_{00} g_{ij}) dx^i dx^j}}{g_{00}} ~,~~~~~~ [3.2.10]$$ IIUC, the "$\pm$" merely expresses the 2 possibilities that light could travel forward in time or backward in time, and still cover a spatial displacement given by $dx^i$.

The context of this is that Weinberg is considering the "time required for light to travel along any path". So he's arbitrarily picking 1 of the 2 possible directions along the path.

5. Oct 22, 2016

### Orodruin

Staff Emeritus
So what is the point of this exercise? The coordinate time generally does not have the interpretation of a time difference. In particular not for a geodesic with changing spatial coordinates. What is coordinate time and spatial coordinates is an arbitrary convention.

6. Oct 22, 2016

### strangerep

The section is titled "Gravitational Forces". He's basically just working through an exercise to show that "all effects of gravitation are comprised in $\Gamma^\lambda_{~\mu\nu}$ and $g_{\mu\nu}$." (See para at end of that section, on p73.) In the next section, he goes on to develop the relation between metric and connection...

However, the para in question begins with "Incidentally..." so he could also be taking a somewhat self-indulgent Weinberg-esque detour.

7. Oct 22, 2016

### Kostik

strangerep: the two values of $dt$ will NOT in general be equal in magnitude and opposite in sign. In fact, they could both be positive. All I can see is that, if I choose the simplest metric
$$g=\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$
then it's true that you need to take the minus sign to get $dt = \sqrt {d\vec x^2}$.
I'm guessing there must be some constraints among the components of $g$ that require the minus sign. Not seeing it.

8. Oct 22, 2016

### strangerep

If the metric has $g_{i0}=0$, then they will be. But a nonzero $g_{i0}$ means there's some counter-intuitive interplay occurring between space and time. Nevertheless, there will still be 2 directions along any path.

BTW, I will say that I've never found any of Weiberg's textbooks to be good as introductions. They tend to be better suited as more advanced texts, where the reader already knows something about the subject.

You might consider switching to Wald's textbook on "General Relativity".

9. Oct 22, 2016

### Orodruin

Staff Emeritus
I still do not see the point of computing the differential of an arbitrarily chosen timelike coordinate. It simply has no physical meaning.

10. Oct 22, 2016

### strangerep

The section is showing (among other things) that metric and connection (in an arbitrary coord system) provide enough information to determine locally inertial coords in the neighborhood of a point.

But perhaps this should wait until you have a copy of Weinberg at hand? Otherwise, I'll end up re-typing the whole section.

11. Oct 22, 2016

### Kostik

Right. But $g_{i0}=0$ is not true in general. And even when it is, you would choose the minus sign if $g_{00} < 0$ and the plus sign if $g_{00} > 0$. I think I'm missing something.

I liked Weinberg's summary of S.R. in Chapter 2; but admittedly, I knew a little about it already. The derivation of the Lorentz transformations as the only ones that preserve proper time (hence the constancy of the speed of light) is very simple and elegant, and so much better than pictures of trains and flashbulbs. I would like to try to stick it out with this book if I can.

12. Oct 22, 2016

### strangerep

Your choice would correspond to a choice of which direction is "forward in time".

Maybe also try to think of it in terms of finding locally inertial coords near a point, in which the metric is (approximately) diagonal.

Tbh, I wouldn't get too hung up on this point. It's just saying that given a quadratic expression in $dt$, you can integrate along any given path in 2 possible directions.

13. Oct 22, 2016

### Orodruin

Staff Emeritus
It is not what the section says I am interested in. It is the motivation for calling this arbitrary coordinate differential "time" in the first place. The construction of a locally inertial coordinate system from the metric is rather trivial. Just take a local orthonormal basis and base the coordinate system on the geodesics.

14. Oct 22, 2016

### strangerep

In that case, it sounds like a thread fork is in order (since this thread was about something in that section of Weinberg).

An excellent subject for discussion in a different thread: "What is time?"

15. Oct 22, 2016

### Orodruin

Staff Emeritus
You still misunderstand me - I know the answer to this question already (and you have been around long enough to know that I do). I am questioning the validity of calling this arbitrary coordinate differential dt "time" without additional qualifiers. Now I can guess that the t is going to be the time coordinate of Weinberg's constructed local inertial frame but really I have no way of knowing for sure. If so, please just say so instead of going off on a tangent.

Edit: So looking at that section, t is just an arbitrary timelike coordinate in an arbitrary coordinate system. It does not necessarily have the interpretation of an actual physical time. This answers my question - it is just a coordinate time.

Last edited: Oct 22, 2016
16. Oct 22, 2016

### DrGreg

The sign of $g_{00}$ determines whether $x^0$ is timelike or spacelike. I'm assuming it has already been specified that $t$ is timelike, therefore $g_{00} < 0$.

So he's choosing the solution that gives the maximum value of $dt$, on the further assumption that coordinate $t$ increases with time.

17. Oct 22, 2016

### strangerep

Tbh, I perceived that you were the one drifting off-topic, and I was puzzled since I know that your GR knowledge exceeds mine by a considerable margin. I will leave it at that.

18. Oct 22, 2016

### PAllen

While I don't have a copy of Weinberg, from the quoted sections, I see nothing forcing the coordinate called t to be timelike. That would depend on the sign of g00. The derivation would not hold for a light like t coordinate (g00 = 0), but it would for t being space like. In which case you have solved for what distance along a funny spatial coordinate would be necessary for light given the other coordinate differentials. Note that these other coordinates could all be light like. The derivation only breaks down if the zeroth coordinate happens to be light like.

[edit: I see dr. Greg just made part of this point.]

19. Oct 22, 2016

### Orodruin

Staff Emeritus
This is true. I just assumed this since Weinberg calls it "time". If it is not timelike, calling it that makes even less sense.

Weinberg mentions the coordinates possibility to be "cartesian coordinates in the lab frame" but generally states that he accepts any coordinate system. He might be implicitly expecting the reader to take a system with a timelike coordinate. I find "Cartesian coordinates in the lab frame" quite imprecise anyway.

20. Oct 23, 2016

### Kostik

strangerep, thanks. I won't get hung up on this. Can I ask another question from the section that follows? This is quite interesting where he shows that in a neighborhood of a point $X$, you can find the correct locally inertial coordinates $\xi^\alpha$ (up to a Lorentz transformation) by expanding in a Taylor series - eqn (3.2.12). What I don't understand is why eqn (3.2.14) "determines the $b^\alpha~_\lambda$ up to a Lorentz transformation". Certainly eqn (3.2.14) remains correct if you transform the $b^\alpha~_\lambda$ by a Lorentz transformation. But (3.2.14) is 10 quadratic equations in 16 variables (10 because $g_{\mu\nu}$ is symmetric). How do we know that (3.2.14) COMPLETELY determines the $b^\alpha~_\lambda$ up to a Lorentz transformation?

21. Oct 23, 2016

### strangerep

Was that a typo? In my copy, he says that eq(3.2.13) determines the $b^\alpha_{~\lambda}$ up to a Lorentz transformation.

I'm not sure if this will help, but a Lorentz transformation has 6 degrees of freedom (being an exponentiation involving 3 rotation generators and 3 boost generators). Perhaps this resolves the apparent discrepancy of "6" that you've noticed?

Last edited: Oct 23, 2016
22. Oct 23, 2016

### Kostik

strangerep: yes the book says (3.2.13) but I believe THAT is a typo, it should be (3.2.14). I'm sure the missing six is due to the arbitrary "up to a Lorentz transformation". I just wish I could see how that follows algebraically from (3.2.14). (PS thanks I appreciate your help.)

23. Oct 23, 2016

### Orodruin

Staff Emeritus
It does follow directly from (3.2.14). Just use the fact that the Lorentz transformation is defined as the set of coordinate transformations that leave the Minkowski metric invariant. You will find that if $b^\alpha_{\ \mu}$ satisfies the equation, then so does $\Lambda^\alpha_{\ \beta} b^\beta_{\ \mu}$, telling you that the equation only determines $b$ up to an arbitrary six-parameter transformation.

24. Oct 23, 2016

### strangerep

After I posted my last message, I began to think the same thing.

Well, a Lorentz transformation leaves the LHS of (3.2.14) form-invariant, since, e.g., if we have a Lorentz transformation which turns $b^\alpha_{~\mu}$ into $\Lambda^\alpha_{~\beta} b^\beta_{~\mu}$, then the LHS of (3.2.14) is unchanged, since $$\eta_{\alpha\beta} b^\alpha_{~\mu} b^\beta_{~\nu} ~\to~ \eta_{\alpha\beta} \Lambda^\alpha_{~\rho} b^\rho_{~\mu} \Lambda^\beta_{~\sigma} b^\sigma_{~\nu} ~=~ \eta_{\rho\sigma} b^\rho_{~\mu} b^\sigma_{~\nu} ~,$$ which is the same as the original after a renaming of the dummy summation indices $\rho, \sigma$.

[Edit: I see Orodruin got in first...]

25. Oct 23, 2016

### Kostik

Gents: yes, I saw that, it was easy to verify that (3.2.14) remains unchanged if a Lorentz transformation is applied. I wasn't asking whether "up to a Lorentz solution" follows from (3.2.14)--clearly it does. What isn't clear is why (3.2.14) is enough to *determine* a solution $b^\beta_{~\mu}$. As I said, ten *quadratic* equations in 16 variables sounds right knowing that the solution is only good up to any Lorentz transformation, but why is it clear that (3.2.14) *is* sufficient to determine $b^\beta_{~\mu}$?

To put it differently, can you show a constructive algebraic solution that *produces* any $b^\beta_{~\mu}$ (which of course can be multiplied by any Lorentz transformation)?

Eqn. (3.2.14) is necessary, but is it sufficient to determine a single solution to all 16 $b^\beta_{~\mu}$?

Just for fun, pretend we're in only two dimensions, so $\{\mu, \nu\} \in \{0, 1\}$. Then writing out (3.2.14) in detail, using $\eta = diag(-1,1)$:
$$-b^0_{~\mu} b^0_{~\nu} + b^1_{~\mu} b^1_{~\nu} = g_{\mu\nu}, ~~~ \{\mu, \nu\} \in \{0, 1\}$$ and the three equations corresponding to $\mu, \nu = (0,0), (1,1), (0,1)$ are:
\begin{align} -{b^0_0}^2 + {b^1_0}^2 &= g_{00}\\ -{b^0_1}^2 + {b^1_1}^2 &= g_{11}\\ -b^0_0 b^0_1 + b^1_0 b^1_1 &= g_{01} \end{align}
It's not so obvious that there is a unique solution to these equations, defined up to a Lorentz transformation. (Although, again, it *is* clear that if $b^\alpha_{~\mu}$ is a solution, then so is the same under a Lorentz transformation.)

Weinberg starts this discussion in the middle of p. 72 by saying "The values of the metric tensor $g_{\mu\nu}$ and the affine connection $Γ^\lambda_{\mu\nu}$ at a point X in an arbitrary coordinate system $x^\mu$ PROVIDE ENOUGH INFORMATION TO DETERMINE THE LOCALLY INERTIAL COORDINATES $\xi^\alpha$ in a neighborhood of X." *This* is what I am not yet convinced of.

Thanks again for any help.

Last edited: Oct 23, 2016