GR as a Graded Time Dilation Field in Euclidean Space?

[WannabeNewton]

There's actually a bit of a subtlety here, which I first encountered in the MTW exercise that asks you to compare the scalar, vector, and tensor theories...

That's really cool. Do you remember which exercise it is explicitly?

 

[PeterDonis]

Do you remember which exercise it is explicitly?

I just went and looked it up, it's Exercise 7.2. (7.1 deals with the scalar theory, 7.3 covers the tensor theory.)

 

[Dale]

To answer your other question, no it doesn't have to do with the degrees of freedom. Both the gravitational and electromagnetic fields have two degrees of freedom at each point. This is the case for massless fields.

How do you figure that? The EM potentials have four degrees of freedom. I could see using the gauge invariance to reduce one degree of freedom, but that still leaves three. How do you get rid of the last degree of freedom?

 

[WannabeNewton]

How do you figure that? The EM potentials have four degrees of freedom. I could see using the gauge invariance to reduce one degree of freedom, but that still leaves three. How do you get rid of the last degree of freedom?

http://www.damtp.cam.ac.uk/user/tong/qft/six.pdf

I just went and looked it up, it's Exercise 7.2. (7.1 deals with the scalar theory, 7.3 covers the tensor theory.)

Thanks!

 

[PAllen]

If it helps, consider the special cases of gravitational waves and electromagnetic waves. The two independent degrees of freedom of the associated fields at each point of space just manifest themselves as the two independent polarization states (classically) of the waves i.e. both gravitational and electromagnetic waves have two independent polarizations. This is a consequence of describing EM and gravity through 4-vector and 4-tensor classical field theories respectively (and the associated gauge symmetries).

By the way I think you might have just misstated it but GR is a classical field theory-it's just not a scalar or vector theory on a flat background.

Speaking of gravitational phenomena as a whole, the metric has 6 degrees of freedom. In what sense can gravity per GR be described as having fewer than this? Does your statement of two apply only to gravitational waves?

[edit: Note, the addendum to this: http://www.staff.science.uu.nl/~hooft101/gravitating_misconceptions.html
presents the view the EM has 3 degrees of freedom, two of which can propagate; and gravity has 6 degrees of freedom, two of which can propagate. ]

 

[WannabeNewton]

[edit: Note, the addendum to this: http://www.staff.science.uu.nl/~hooft101/gravitating_misconceptions.html
presents the view the EM has 3 degrees of freedom, two of which can propagate; and gravity has 6 degrees of freedom, two of which can propagate. ]

Yes. See also p.266 of Wald.

 

[PAllen]

Yes. See also p.266 of Wald.

Ok, so my point was if you are going to 'get going at all' with an attempt to reproduce all GR predictions with scalar functions, you could not possibly use less than 6, due to the 6 degrees of freedom. Whether it can be done meaningfully at all ultimately depends how loose you want to be on 'meanfingful'.

 

[H_A_Landman]

Let's use the Schwartzschild metric as a simple exact example solution of GR. If you take the low-speed (v << c) and weak-field (r >> R_s) limits, you end up with a metric that is just the flat-space metric plus a time term. In this metric, ALL the curvature is in the time dimension, so it looks exactly like flat Euclidean space plus curved time. You can recover Newtonian gravity from that metric, and it matches what we see in the vicinity of the Earth to better than 1 part per billion.

Approximately, therefore, if you are not moving at relativistic speeds and are not close to a black hole or neutron star, mass causes time dilation and time dilation causes gravity (i.e. curved geodesics). The time dilation gradient due to Earth's mass is what's pressing you into your seat right now. The idea that "gravity causes time dilation" is precisely backwards.

Anyway, in this limit GR gravity really is just a graded time dilation field in Euclidean space, which I think may be the best answer to the original question.

 

[pervect]

Let's use the Schwartzschild metric as a simple exact example solution of GR. If you take the low-speed (v << c) and weak-field (r >> R_s) limits, you end up with a metric that is just the flat-space metric plus a time term. In this metric, ALL the curvature is in the time dimension, so it looks exactly like flat Euclidean space plus curved time. You can recover Newtonian gravity from that metric, and it matches what we see in the vicinity of the Earth to better than 1 part per billion.

Approximately, therefore, if you are not moving at relativistic speeds and are not close to a black hole or neutron star, mass causes time dilation and time dilation causes gravity (i.e. curved geodesics). The time dilation gradient due to Earth's mass is what's pressing you into your seat right now. The idea that "gravity causes time dilation" is precisely backwards.

Anyway, in this limit GR gravity really is just a graded time dilation field in Euclidean space, which I think may be the best answer to the original question.

Necro posts are us. The last post in this thread was March, 2014.

Additionally, I don't think I agree that taking the low-speed, high R limit gives only the time componenents. For instance, take gravitational wave detection ala Ligo - the spatial components of the metric are very important for gravitational waves.

 

[rjbeery]

H_A_Landman: Thanks for the timely response ;) Anyway time dilation "causing" gravity is precisely the idea I was exploring, I just didn't word it in that way...

 

[PeterDonis]

I don't think I agree that taking the low-speed, high R limit gives only the time componenents.

For the large R limit, this is true; there is also an extra factor in front of the space components. This factor shows up in, for example, the bending of light by the Sun; it makes the GR prediction twice the Newtonian prediction.

However, in the slow motion limit (which excludes phenomena like light bending, since light is not "slow moving"--neither are gravitational waves), the extra factor in front of the space components becomes negligible, because we are restricting attention to spacetime intervals for which the spatial differentials $dx$, $dy$, $dz$ are much smaller than the time differential $dt$ (we are using "natural" units here in which $c = 1$). So the effective metric in the weak field, slow motion limit does have only an extra time-time component in addition to the flat spacetime metric.

Carroll discusses this briefly in Chapter 4 of his online lecture notes on GR:

https://arxiv.org/abs/gr-qc/9712019

 

[H_A_Landman]

... take gravitational wave detection ala Ligo - the spatial components of the metric are very important for gravitational waves.

Gravitational waves, like light, travel at the speed of light and hence violate the low-speed assumption. You wouldn't expect them to necessarily be handled correctly by the simplified metric.

 

[H_A_Landman]

Carroll discusses this briefly in Chapter 4 of his online lecture notes on GR:
https://arxiv.org/abs/gr-qc/9712019

Yes he does, specifically around equations 4.10 to 4.22 on pages 105-106. Thanks for the reference.