Gravitational time dilation in Weinberg's "Cosmology"

[hiro0825]

I have a question about the gravitational time dilation explained in Appendix B of the book "Cosmology" written by S. Weinberg.
Why can the author say "In the negative gravitational potential at the surface of a star clocks therefore tick more slowly than in interstellar space, or in the much weaker gravitational potential at the surface of the earth." from his equation (B.22)?

Let me summarize his argument.

In a weak static gravitaional field,
$$ g_{00} \sim -1 - 2 \phi. \tag{B.21} $$
Consider a clock at rest in such a field. Its time interval between ticks in the absence of a gravitational filed is $dT$, but the time interval becomes $dt$ when the clock is put in the gravitational field. Then, due to the invariance of the space-time interval, we have
$$ -dT^2 = g_{00} dt^2 \sim (-1 - 2\phi) dt^2 .$$
Therefore we have
$$ dt \sim ( 1 - \phi ) dT . \tag{B.22} $$
Then the author says "In the negative gravitational potential at the surface of a star clocks therefore tick more slowly than in interstellar space... "

Why? As the author says, $\phi$ is negative. Therefore $dt > dT$, and it means that the clock in the gravitational field ticks faster than in the absence of the gravitational field. What am I misunderstanding?

I'm familiar with the usual view of the general relativity, the geometrical view of space-time. But I'm interested in the nongeometrical viewpoint of the general relativity, and so I'm reading this book and "Gravitation and Cosmology" of the same author. But I don't understand his argument of the gravitational time dilation... Please let me know what I'm misunderstanding.

Regards

 

[vanhees71]

What the (ideal) clock shows, is its proper time, i.e., for a clock at rest in a weak gravitational potential $\phi$ what you call $\mathrm{d} T$. A clock at rest far from the source of the gravitational field, where $\phi \rightarrow 0$ is $\mathrm{d} t$. Since $\phi<0$ you have $\mathrm{d} T <\mathrm{d} t$, i.e., the clock in the gravitational potential ticks slower than one far away, where $\phi=0$.

 

[hiro0825]

What the (ideal) clock shows, is its proper time, i.e., for a clock at rest in a weak gravitational potential $\phi$ what you call $\mathrm{d} T$. A clock at rest far from the source of the gravitational field, where $\phi \rightarrow 0$ is $\mathrm{d} t$. Since $\phi<0$ you have $\mathrm{d} T <\mathrm{d} t$, i.e., the clock in the gravitational potential ticks slower than one far away, where $\phi=0$.

Thanks for your comment.

As I summarized, Weinberg clearly says that $dt$ is the time interval of the clock at rest in the weak gravitational field, while $dT$ is that of the clock at in an inertial frame without gravity. Please look at the book if you have... So I'm not satisfied with your explanation. Sorry...

Actually, I can understand your explanation if the standard approach of the general relativity -- the geometrical approach --- is adopted.
But now I want to understand the gravitational dilation from the nongeometrical viewpoint of the general relativity that Weinberg adopts..

Thanks

 

[vanhees71]

Hm, maybe I'm biased, and English is not my mother tongue, but when I read Appendix B in Weinberg's books it reads just as what I said above. This must be so, because coordinates are something completely arbitrary, while the physical quantities are scalars (wrt. Lorentz transformations in special and wrt. arbitrary coordinate transformations in general relativity), and what an ideal clock shows are its proper time, $\mathrm{d} \tau$, i.e.,
$$\mathrm{d} \tau^2=-g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu},$$
using Weinberg's east-coast convention of the signature, $\mathrm{d} x^{\mu}$ denotes increments of the (necessarily timelike) worldline of the clock.

For a clock at rest you can use $x^0=t$ as a world-line parameter, while $x^{j}(t)=\text{const}$, and the proper time reduces trivially to
$$\mathrm{d} \tau^2 = -g_{00} \mathrm{d} t^2.$$

 

[hiro0825]

Yes, I understand what you mean from the standard viewpoint of the general relativity.

But the definition of Weinberg is different. In Weinberg's definition, the interval of two events, $ds = g_{\mu \nu} dx^\mu dx^\nu$, equals to the "wristwatch time" of the clock only when the clock is at rest in an inertial frame ($ds = \eta_{00} dt = - dt$, where $dt$ is the wristwatch time of the clock, not the "faraway time" or the "coordinate time" in usual textbooks.).
If the clock is at rest in a gravitational field, $ds = \sqrt{g_{00}} dt$, and therefore $ds \neq dt$.
That is the Weinberg's "nongeometrical" point of view, as far as I understand.
What do I misunderstand?

 

[vanhees71]

Again I read the complete opposite in Weinberg's book than what you seem to read :-)). In a local inertial frame you have $g_{\mu \nu}=\eta_{\mu \nu}$. Then the strong equivalence principle tells you that you can always find, at any space-time point, such a local inertial frame. Then GR is generally covariant, i.e., in any reference frame the infinitesimal interval $\mathrm{d} s^2 = g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}$ is (minus) the "writewatch time squared" (provided the vector given by the components $\mathrm{d} x^{\mu}$ is time-like. This argument was only brought by Weinberg to derive that this is the right definition of "proper time" by using the local inertial frame to argue within special relativity and then invoking the equivalence principle to generalize it to arbitrary reference frames. The value of $\mathrm{d} s^2$ of course stays the same, no matter whether you calculate it in the local inertial or any general frame.

 

[hiro0825]

Thanks.

Then how do you understand Weinberg's definitions of $dt$ and $dT$ in the book "Cosmology"? He says in page 517 "According to Eq. (B.2), if the time between ticks in the absence of a gravitational field is $dT$, then in the presence of the field it is $dt$, where $(-1 - 2\phi) dt^2 = - dT^2$".

 

[vanhees71]

Hm, that seems to me even wrong (although I'm a bit puzzled since I don't believe that Weinberg has errors in his textbooks). Checking another book (Fliessbach, Allgemeine Relativitätstheorie), however says almost the same I've stated above.

 

[PeterDonis]

Therefore $dt > dT$, and it means that the clock in the gravitational field ticks faster than in the absence of the gravitational field.

No, you have this backwards. $dt$ is the coordinate time interval; $dT$ is the proper time interval. The proper time interval $dT$ is what the clock in the gravitational field actually ticks. The coordinate time interval $dt$ is (heuristically) what the clock would have ticked in the absence of the gravitational field. So since $dT < dt$, the clock ticks slower in the gravitational field than it would have in the absence of the gravitational field.

 

[hiro0825]

No, you have this backwards. $dt$ is the coordinate time interval; $dT$ is the proper time interval. The proper time interval $dT$ is what the clock in the gravitational field actually ticks. The coordinate time interval $dt$ is (heuristically) what the clock would have ticked in the absence of the gravitational field. So since $dT < dt$, the clock ticks slower in the gravitational field than it would have in the absence of the gravitational field.

Thanks.
No, as I already said, Weinberg clearly defines $dT$ and $dt$ as "if the time between ticks in the absence of a gravitational field is $dT$, then in the presence of the field it is $dt$". Please look at page. 517 of "Cosmology" by S. Weinberg, if you have.
The author does not interpret $dt$ as the coordinate time as usual books on GR. I think this is because the author explains the GR in terms of "non-geometrical" view point.

 

[PeterDonis]

as I already said, Weinberg clearly defines $dT$ and $dt$ as "if the time between ticks in the absence of a gravitational field is $dT$, then in the presence of the field it is $dt$".

Then this equation that you wrote in the OP is wrong:

Then, due to the invariance of the space-time interval, we have
$$
-dT^2 = g_{00} dt^2 \sim (-1 - 2\phi) dt^2 .
$$

This equation is only correct if $dT$ is proper time and $dt$ is coordinate time, not the other way around.

Unfortunately I don't have Weinberg's book so I can't check it directly.

The author does not interpret $dt$ as the coordinate time as usual books on GR

"Coordinate time" as I'm using that term here is just shorthand for "the time the clock would have ticked in the absence of the gravitational field". There is no requirement that it be the actual coordinate time interval in any chart (although it's easy to see that in the particular case under discussion, the time the clock would have ticked in the absence of the field is in fact equal to the coordinate time interval in the standard Schwarzschild chart).

 

[hiro0825]

Then this equation that you wrote in the OP is wrong:
This equation is only correct if $dT$ is proper time and $dt$ is coordinate time, not the other way around.

Unfortunately I don't have Weinberg's book so I can't check it directly.
"Coordinate time" as I'm using that term here is just shorthand for "the time the clock would have ticked in the absence of the gravitational field". There is no requirement that it be the actual coordinate time interval in any chart (although it's easy to see that in the particular case under discussion, the time the clock would have ticked in the absence of the field is in fact equal to the coordinate time interval in the standard Schwarzschild chart).

Thanks.

The definition of $dt$ and $dT$ and the equation relating them are both written clearly in the book.
(If you have chance, please check it yourself... If you look for the PDF copy of the book through the web, you might be able to find it).
So, according to your comments, Weinberg seems to be wrong.

Yeah, all books include some mistakes. Even the book of Weinberg might be the case.
But... I'm not confident to say that this is the mistake of the author.
I feel that I don't understand something the author wants to claim, especially the "non-geometrical" view point of GR...

 

[PeterDonis]

all books include some mistakes. Even the book of Weinberg might be the case

It's possible that there's a misprint. Certainly it's easy to check in other textbooks, which will give you much the same math even if they are using a different interpretation, that the correct equation, in the weak field approximation you are using, will tell you that proper time in a gravitational field ticks slower than coordinate time (aka time in the absence of the gravitational field).

I feel that I don't understand something the author wants to claim, especially the "non-geometrical" view point of GR

The non-geometrical point of view is well known, although Weinberg's is the only textbook I'm aware of that adopts it as primary. It basically amounts to viewing gravity as a tensor field on flat spacetime; in other words, "gravity" is the difference between the actual metric coefficients $g_{\mu \nu}$ and the flat spacetime metric coefficients $\eta_{\mu \nu}$. Usually this is called $h_{\mu \nu}$, and is treated much the same as any other field, once you take its tensor character into account (other known fields are scalar or vector).

 

[vanhees71]

No, you have this backwards. $dt$ is the coordinate time interval; $dT$ is the proper time interval. The proper time interval $dT$ is what the clock in the gravitational field actually ticks. The coordinate time interval $dt$ is (heuristically) what the clock would have ticked in the absence of the gravitational field. So since $dT < dt$, the clock ticks slower in the gravitational field than it would have in the absence of the gravitational field.

Yes, that's what I also answered, but I've checked that indeed Weinberg has it indeed the other way around. I still think, that the standard point of view is right (i.e., what you wrote in the quotation above), but I'm a bit puzzled that Weinberg should get such a simple thing wrong in his book.

 

[vanhees71]

The non-geometrical point of view is well known, although Weinberg's is the only textbook I'm aware of that adopts it as primary. It basically amounts to viewing gravity as a tensor field on flat spacetime; in other words, "gravity" is the difference between the actual metric coefficients $g_{\mu \nu}$ and the flat spacetime metric coefficients $\eta_{\mu \nu}$. Usually this is called $h_{\mu \nu}$, and is treated much the same as any other field, once you take its tensor character into account (other known fields are scalar or vector).

I think this has nothing to do with a geometric vs. physics approach. I also prefer the latter, and precisely this approach clearly says that the standard point of view is correct, i.e., an ideal clock on a given worldline measures the proper time (that's already so within SR and thus it's also valid in GR (locally) due to the strong equivalence principle).

In this example of a clock at rest in the standard Scharzschild frame, you have
$$\mathrm{d} \tau = \sqrt{-g_{00}} \mathrm{d} t=\sqrt{1-r_S/r} \mathrm{d} t.$$
In this case, the coordinate time is thus the time of a clock at rest in this frame very far away, $r \rightarrow \infty$, from the star/black hole, i.e., where no gravity is present, while at any finite $r>r_S$ a clock at rest ticks slower. This is also in accordance with the very well measured redshift. So I come to the conclusion that Weinberg indeed has it wrong in his Appendix B.

 

[hiro0825]

Yes, that's what I also answered, but I've checked that indeed Weinberg has it indeed the other way around. I still think, that the standard point of view is right (i.e., what you wrote in the quotation above), but I'm a bit puzzled that Weinberg should get such a simple thing wrong in his book.

Yes, that is what I really thought. Throughout the famous two books, Weinberg took his own (?) viewpoint and was able to explain almost everything about GR and cosmology. Could his point of view be wrong? I don't think so... I feel that I misunderstand something in his explanation for the gravitational time dilation. That's why I asked the question in this forum.

 

[vanhees71]

Hm, then I've also misunderstood general relativity, and also the standard point of view has been veryfied by experiments like

https://en.wikipedia.org/wiki/Pound–Rebka_experiment

I'm puzzled, but on the other hand, why shouldn't even Weinberg making mistakes in his book?

 

[PeterDonis]

I've checked that indeed Weinberg has it indeed the other way around

If he has the symbols the other way around, I would think that must be a typo. I can't imagine Weinberg seriously claiming that gravitational time dilation works backwards from the way the standard view (which, as you say, has been experimentally verified) says it works.

 

[romsofia]

Dr Weinberg is still an active professor at UT Austin, and if you think youve found an error don't hesistate to contact him!

Here is a list of corrected errors in his cosmology book: http://zippy.ph.utexas.edu/~weinberg/swcorrections5.pdf

 

[hiro0825]

Dr Weinberg is still an active professor at UT Austin, and if you think youve found an error don't hesistate to contact him!

Here is a list of corrected errors in his cosmology book: http://zippy.ph.utexas.edu/~weinberg/swcorrections5.pdf

Thanks for your suggestion.
But still I feel I misunderstand what the author wants to claim...

 

[vanhees71]

If he has the symbols the other way around, I would think that must be a typo. I can't imagine Weinberg seriously claiming that gravitational time dilation works backwards from the way the standard view (which, as you say, has been experimentally verified) says it works.

The point is, he has it in the standard way on page 512 but in the other opposite way on page 517 in the context of a weak gravitational field. What's safe to say is that $\mathrm{d} T=\sqrt{-g_{\mu \nu} \mathrm{d}x^{\mu} \mathrm{d} x^{\nu}}$
is the time a clock is showing moving along an arbitrary timlike worldline.

If you take the Schwarzschild solution as an example
$$\mathrm{d} T^2=-(1-r_S/r) \mathrm{d} t^2 + \frac{\mathrm{d} r^2}{(1-r_S/r)} \mathrm{d} r^2 + r^2 \mathrm{d} \vec{\Omega}^2,$$
then for clocks at rest $\mathrm{d} T$ is the time shown by a clock somewhere in finite distance from the mass (gravity) and $\mathrm{d} t$ is shown by a clock at rest at infinity (no gravity). The clock near the mass is thus slower ticking than the clock asymptotically far away from the mass.

Note that here I assumed the mass/black hole as living in a spacetime which becomes asymptotically Minkowski spacetime at large distances from the mass.

 

[PeterDonis]

he has it in the standard way on page 512 but in the other opposite way on page 517 in the context of a weak gravitational field

Then I would say page 517 must have a typo. The gravitational field being weak doesn't change how gravitational time dilation works.

 

[PeterDonis]

still I feel I misunderstand what the author wants to claim

He is using the "gravity as a field" viewpoint, which says that gravitational time dilation happens because the gravitational field slows down the ticking of clocks. So a clock deep in a gravitational field ticks slower than a clock far away.

By contrast, the "gravity as spacetime curvature" viewpoint says that gravitational time dilation happens because of the geometry of spacetime--the length of the worldline of a clock deep in a gravitational field is shorter than the length of the worldline of a clock far away, because spacetime is curved.