Gravitational Equivalent of Rindler Spaceby Chestermiller Tags: equivalent, gravitational, rindler, space 

#1
Mar2712, 08:29 AM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,435

In classical physics, the force on an object experiencing constant acceleration is equivalent to the "gravitational force" exerted on an object by a slab of material of uniform thickness and infinite lateral extent. The "gravitational field" is uniform above the slab, and the object's "acceleration" in free fall is constant.
Does this same type of equivalence apply in general relativity, in which the Rindler metric describes the geometry of an object's accelerated frame of reference (for constant acceleration)? IOW, for an observer at rest wrt a flat slab of infinite lateral extent, is the space above the slab described by the Rindler metric? I'm guessing that it is, since, when I evaluated the components of the RicciEinstein tensor for Rindler space, all the components came out to zero (if I did it right). Also, in Rindler space, the acceleration of the coordinate frame gradually decreases linearly in the direction of motion. Does the same type of effect happen in the region above a slab? I guess this would be interpreted as a tidal phenomenon that is not present in the classical analogy. Correct? Help!!!! 



#2
Mar2712, 01:47 PM

PF Gold
P: 4,081

Rindler coordinates are transformations of Minkowski coordinates and so they will give flatspace numbers just like Minkowski. It's an interesting question, although the GR case will not represent the same physics, giving a constant coordinate acceleration as opposed to the proper acceleration of the Rindler frame.
I think the slab spacetime is the Taub metric. It is a vacuum solution but the Weyl tensor is not zero. I'm going to look at the EOMs later, if I don't fall asleep first. 



#3
Mar2712, 01:55 PM

Sci Advisor
PF Gold
P: 4,862





#4
Mar2712, 02:24 PM

Sci Advisor
Thanks
P: 3,853

Gravitational Equivalent of Rindler Space
Chestermiller, If you're interested in the General Relativistic analog of the Rindler metric, take a look at the socalled C metric, which is the field of a uniformly accelerating Schwarzschild mass. Varieties of the Cmetric also exist for accelerating Kerr and ReissnerNordstrom.
The acceleration must be supplied by something external, and in the C metric this is represented by a singular strut which is attached to the particle and extends along the axis to infinity. 



#5
Mar2712, 03:16 PM

PF Gold
P: 4,081

The acceleration of a static observer in this spacetime
[tex] {ds}^{2}={dy}^{2}\,{\left( 13\,g\,z\right) }^{\frac{4}{3}}+{dx}^{2}\,{\left( 13\,g\,z\right) }^{\frac{4}{3}}\frac{{dt}^{2}}{{\left( 13\,g\,z\right) }^{\frac{2}{3}}}+{dz}^{2} [/tex] is [tex] \frac{g}{3\,g\,z1}\partial_z [/tex] I think g<0 so that is in the +ve zdirection. The metric is a vacuum solution of the region above ( z > 0 ) an infinite slab of constant density in the region z <= 0. I got the acceleration from [itex]\nabla_\nu U_\mu V^\nu[/itex] where U=(1,0,0,0) and V=(1,0,0,0). This is the proper acceleration required to remain at a constant zvalue. The acceleration isn't constant as I was led to believe by the article (cited above by PAllen). [correction: the article states the acceleration is constant for small z, which is true] 



#6
Mar2812, 06:39 AM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,435

[QUOTE=Mentz114;3836727]Rindler coordinates are transformations of Minkowski coordinates and so they will give flatspace numbers just like Minkowski.
Thank you and the other responders so much for your replies. Some of the replies have been a little over my head (for now), but I will follow up. The above quote was extremely enlightening, and has cleared up a source of confusion that I have been grappling with for months. As a retired engineer, I have become fascinated with relativity, and have been studying it on my own. One stumbling block I have encountered has been in trying to get an understanding of the relationship between accelerated reference frames in free space, and reference frames at rest in the vicinity of gravitationally massive bodies, within the framework of the equivalence principle. Somehow I got it in my head that, from the perspective of observers in an accelerated reference frame in free space, their space is curved. I studied accelerated reference frames in Chapter 6 of MTW, and found the development fascinating, but failed to recognize from the transformation equations and the development of the metric for the accelerated frame that flat space was also implied for the accelerated frame. In retrospect, I really should have, since, by applying the inverse transformation to the accelerated frame metric, you automatically obtain the Minkowski metric. However, MTW never articulated this point, which I regard as very important to my understanding. Your quote cleared all this up. From this I conclude that the pseudogravity associated with an accelerated frame of reference in free space is not the result of curvature, but the true gravity in the vicinity of massive objects is the result of curvature. In this respect, the two situations are fundamentally different. Now I have to go back and think a little more about free fall in which, somehow, the curvature vanishes (at least locally). 



#7
Mar2812, 08:22 AM

PF Gold
P: 4,081

http://en.wikipedia.org/wiki/Born_coordinates Speaking of curvature, the tidal forces seen by a static observer in the Taub spacetime above are given by [tex] T_{xx}=T_{yy}=\frac{2\,{g}^{2}}{{\left( 3\,g\,z1\right) }^{2}},\ \ T_{zz}=\frac{4\,{g}^{2}}{{\left( 3\,g\,z1\right) }^{2}} [/tex] So we have squishing in the zdirection and stretching in x and ydirections. The terms sum to zero so volume is preserved. 



#8
Mar2812, 08:36 AM

Sci Advisor
PF Gold
P: 4,862





#9
Mar2812, 08:37 AM

PF Gold
P: 4,081





#10
Mar2812, 08:47 AM

Sci Advisor
PF Gold
P: 4,862





#11
Mar2812, 09:22 AM

PF Gold
P: 4,081





#12
Mar2812, 09:10 PM

Sci Advisor
PF Gold
P: 1,806





#13
Mar2912, 07:28 AM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,435

Thanks very much DrGreg. You summarized the situation very well. Except for my conceptual "glitch" of failing to recognize that Rindler space is flat, I had already envisioned the remainder of what you so concisely and simply presented.
As someone studying relativity on my own, it has been hard to proceed without having someone to answer my questions. Discovering Physics Forums has been a real "find." I appreciate everyone's help. Now for one additional question. Since accelerated frames of reference are basically flat, and, in this respect, differ from reference frames at rest with respect to massive bodies, would it be at least conceptually possible for residents of an accelerated reference frame to experimentally measure the metrical properties of their space, and by doing so, ascertain that they are not experiencing gravitational curvature of spacetime? 



#14
Mar2912, 07:31 AM

Sci Advisor
PF Gold
P: 4,862





#15
Mar2912, 07:49 AM

P: 3,542

This might be relevant:
http://www.mathpages.com/home/kmath530/kmath530.htm 



#16
Mar2912, 11:49 AM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,435

One question: Is it correct to say that, statically, there is a unique solution for the metric in the region above the massive sheet? Chet 


Register to reply 
Related Discussions  
Question about event horizon and Rindler space  Special & General Relativity  5  
Is there a gravitational equivalent to magnetism?  Special & General Relativity  12  
Charge equivalent to a gravitational black hole???  General Physics  1 