Suppose a tall (100m high) rocket sits on the launch pad. It is equipped with launch boosters and a sustainer motor that can give the rocket a prolonged 1g length-wise acceleration in free space. Amongst others, it is also fitted with the following sensors: two identical, synchronized atomic clocks, one in the nose and one near the tail, plus three accurate identical accelerometers, one in the nose, one in the tail and one midway between the first two.
While on the launch pad (waiting for a long delayed launch) you monitor all the sensors and determine that the nose clock is marginally gaining time on the tail clock. You satisfy yourself that this is normal due to gravitational time dilation and amounts to 1 part in about 10^14 (coming from dt/t = gL/c^2, where g = 9.81 m/s^2 and L is the difference in height between the clocks). You also verify that the higher accelerometers read marginally lower accelerations than the lower ones, in agreement with the inverse square law of gravitational acceleration (a = -GM/r^2, where G is Newton’s gravitational constant M the mass of the Earth and r the distance from Earth’s center).
Eventually the system is launched into free space and all the boosters fall away. After verifying that everything operates as designed and synchronizing the nose- and tail clocks, you ignite the 1g-propulsion system at the back. After a fair time of monitoring exactly 1g of acceleration at the tail of the rocket, you read all your sensors again. Will the clocks and accelerometers be able to tell you that you are now being linearly accelerated at 1g in free space and no longer sitting stationary on Earth’s surface?
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Fred wrote: “This opens the door to another circular discussion about gravitational waves and energy. To all who have tired of that in previous threads, I apologize in advance.”
Not all of us participated in that debate, but I would like to hear Burt’s comments on gravitational waves. My impression is that apart from being predicted by Einstein, the existence of gravitational waves has been observed at the double pulsars, because without gravitational waves, there is no explanation for their in-spiraling nature.
Maybe I should open a new thread on that. (Edit: Have done that: https://scienceblog.com/cms/gravitational-waves-useful-or-wasteful-15144.html)
SL
Burt’s clarification is useful, and summarized as:
To me, the requirement of locality seems to be the core of Jim’s disagreement.
All agree that gravity can be distinguished from an accelerated frame of reference nonlocally. In that situation, Jim’s thesis is not novel.
Jim’s claim that gravity can be distinguished from an accelerated frame in a local measurement is novel. The problem is that he has not provided any evidence that supports his claim or proposed a test that could support it. So far, all experiments and observations indicate that the strong equivalence principle is a valid description of gravity.
This opens the door to another circular discussion about gravitational waves and energy. To all who have tired of that in previous threads, I apologize in advance.
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Fred wrote: “I just wanted to make sure that it was clear your agreement with Jim did not go as far as rejecting Einstein’s principle of equivalence.”
I’ve been very sloppy when referring to “Einstein’s equivalence principle” as the equivalence between gravity and acceleration. It actually reads:
1. “The trajectory of a falling test body depends only on its initial position and velocity, and is independent of its composition.“, i.e., the equivalence of gravitational and inertial mass.
and
2. “The outcome of any local non-gravitational experiment in a laboratory moving in an inertial frame of reference is independent of the velocity of the laboratory, or its location in spacetime.“, i.e., it excludes both acceleration and gravity.
The two parts of Einstein equivalence principles were later replaced by the so-called “strong equivalence principle“, reading:
1. The gravitational motion of a small test body depends only on its initial position in spacetime and velocity, and not on its constitution.
and
2. The outcome of any local experiment, whether gravitational or not, in a laboratory moving in an inertial frame of reference is independent of the velocity of the laboratory, or its location in spacetime.
This can be loosely translated to an equivalence between gravity and acceleration, but only in a local (infinitesimal) measurement where tidal gravity effects can be ignored.
Fred conluded with: “Hmmmm… Does that mean that your (Jim’s, I guess) discussion needs to include a cosmological constant to account for the fact that the universe itself may be expanding or contracting?”
Yep, Einstein obviously did consider this. Wikipedia says:
“Einstein’s theory of general relativity (including the cosmological constant) is thought to be the only theory of gravity that satisfies the strong equivalence principle.”
Burt Jordaan (www.Relativity-4-Engineers.com)
Hi Fred.
I thought we have thrashed it out in this thread that for extended frames, like a 100m tall rocket, Einstein’s equivalence principle does not quite hold. It was defined for infinitesimal size frames where you cannot, even in principle, detect a difference between being static in a gravitational field or being uniformly accelerated.
In an extended frame, there is a difference in the magnitude of the acceleration as a function of vertical distance (height) up the rocket, as well as Jim’s favorite, gravitational acceleration vectors pointing towards the center of mass in a gravitational field. Given a large lab and sensitive instruments, you can detect the non-equivalence by various means.
I also agree with Jim that we should not equate gravity with a force. The primary force we feel while sitting in front of our computers is that of the chair pushing us out of our present spacetime geodesic…
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
Burt, could you clarify what you mean by “I agree with you final statement about the non-equivalence.”
Jim’s claim, as I have been reading it, seems to dispute Einstein’s principle of equivalence. Yet all of the math you have been employing seems to assume that principle is true. Any single measurement can be interpreted as being due to a gravitational field or an equivalent accelerated reference frame, and there is no way to say that it is one, the other, or a combination of both.
So in what why are you agreeing about non-equivalence?
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Jim, you wrote: “I don’t see why the gravitational vertex isn’t (approximately) the same location in relation to both horizontal and vertical points on the rocket – the 4D geodesics of all are vectored in (approximately) the same direction, toward the same vertex.”
No, you can’t get out of this corner that easily! :-)
I have the feeling you are talking about a 3-d spatial vertex at the center of mass of the gravitating body. The gravitational spacetime vertex is not there, but far away on the opposite side of the mass. E.g. for Earth it lies about one light-year away, on the other side of Earth from the rocket. As I wrote before, in the low gravity limit, the spacetime vertex distance from each static point is D_v = (dc)^2/(GM). Double the distance (d) from the center of Earth and the gravitational spacetime vertex for that point sits about 4 light-years away. It is not only the vertex distance that changes, the vertices are not at the same spatial location for vertically separated test points.
I agree with you final statement about the non-equivalence.
Burt Jordaan (www.Relativity-4-Engineers.com)
Happy Arbitrary Orbital Fixation to Burt and everyone! And Oklahoma gonna beat West Virginia tomorrow, or gravitation isn’t non-equivalent to acceleration!
Burt wrote: “While the center of mass of the gravitating body can certainly be called a ‘vertex’, you are comparing apples and oranges here.â€
I don’t see why the gravitational vertex isn’t (approximately) the same location in relation to both horizontal and vertical points on the rocket – the 4D geodesics of all are vectored in (approximately) the same direction, toward the same vertex. Granted, the vertex-distance varies with the vertical, and the square law applies to the gravitational acceleration, and I’m definitely comparing (and distinguishing) apples and oranges – gravitational and non-gravitational acceleration.
“It is the gravitational accelerations of points on the rocket that have different spacetime vertices, while there is only one spacetime vertex for Born-rigid acceleration.â€
It seems we need to be more clear in distinguishing vertices and vertex-distances. I’m saying: gravitational acceleration has one vertex for all points, various vertex-distances; non-gravitational acceleration has multiple vertices, the same vertex-distance for each point.
In any case, my issue is the distinction of gravitation from force as is implied by the non-equivalence of gravitational and non-gravitational acceleration, and supported in either reading or interpretation of acceleration vertices.
Happy New Year to all participants and readers!
Burt wrote: “It is the gravitational accelerations of points on the rocket that have different spacetime vertices, while there is only one spacetime vertex for Born-rigid acceleration. Even if you do not agree that Born-rigid acceleration represents SL’s rocket, the gravitational case is indisputable. :)”
Hmmm…, I wondered how long it would take before someone gives Jim the tools to paint himself into a corner. Woof!
I also know Jim always finds a way out! :-)
SL
Burt asked me: “Where, in your view, is this common ‘vertex’ for a rocket sitting static in a gravitational field?â€
As I wrote earlier, each “point†of the rocket standing vertical at the surface of a gravitational field (in typical, representative situations where a rocket will be standing on the surface of a massive body) has, with insignificant variations, the same gravitational “vertex”, or center of mass, producing the orientation of the geodesics of the “points†of the rocket, and also the inertial resistance to the geodesics at the surface. This was to contrast the innumerable vertices that the “points” of the rocket would have in Mallinckrodt’s treatment of the rocket’s inertial acceleration in space, where the “vertex†is described as the location where the acceleration was initiated.
SL asked re: the question of the equivalence of grav & inertia: “Burt’s answer to my original question was an unequivocal ‘No’ [equivalence]…. So what is your point?†I won’t go into it again here(!) My point is that a thoroughgoing analysis of the non-equivalence reveals that gravitation is not a force, that gravitational waves don’t carry energy, that the common association of gravitation with force and energy is due to its common coincidence (in our everyday experience at the earth’s surface) with force.
I’m short on time, and anxious to get back to Burt, but I want to respond briefly to Fred, who wrote:
“Thanks, Burt. I just wanted to make sure that it was clear your agreement with Jim did not go as far as rejecting Einstein’s principle of equivalence…. The difference is that Jim presumes the non-equivalence is true in the limiting case, whereas you recognize that it is not.â€
I can’t speak for anyone else, but a “principle†that can only be invoked if we agree to limit the scope of our observations and the precision of our instruments just sufficiently to render an undesired aspect of our object undetectable – is no principle at all. It may be a license for experimental expedience, as when we might want to conduct an experiment factoring out the effects of gravitation. But there is no principle of chromatics that would hold all bow-ties are black in a dark enough room. If gravitation and force are non-equivalent, (if black ties and red ties are non-equivalent) the inability to detect a gravitational field (a color) in a limiting case doesn’t render them equivalent, or their non-equivalence relative.
Accidental duplicate. Can’t seem to delete
Thanks again, Burt and Scruffy.
I’m currently using a cranky old laptop with a small screen, so I won’t get back to this until my main machine gets a new power supply (I suspect). But you have addressed a number of things that can help anyone who wants to battle with the math and concepts of general relativity.
Jim’s question about the chase rockets also led to further clarification. If only he weren’t so adamant that he is seeing something that no one else ever has, then he, too, could contribute more useful ideas.
The best part of a blog like this is that people can ask questions that are new to them, even if they are old hat to others, and get valuable answers that teach readers who are silent in the discussion.
Science, after all, is about following questions. For the most part, I try to use this blog, including my own pages, to raise interesting questions and to guide readers to their own ways of following them.
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
At the risk of further distracting from Scruffy’s OP, a few comments to Jim.
Bell’s thought experiment does not sport differences in clock rates, because by definition, the two ships are accelerating identically and are always at the same speed in the reference frame, meaning identical clock rates. The only issue is whether they can do so without stretching the string.
BTW, Bell’s experiment is not about parallel acceleration, but rather tandem acceleration at the same rate. Your example on the surface of Jupiter is hence not quite applicable.
Burt Jordaan (www.Relativity-4-Engineers.com)
[Edit] PS: I forgot to ask you about this statement you made earlier: “But among other non-equivalences between inertial acceleration and gravitation, there is no common “vertex” in the acceleration of the body, whereas there is a common “vertex” in the gravitation of the body if it is “static” in a gravitational field.”
Where, in your view, is this common “vertex” for a rocket sitting static in a gravitational field?
Hi Fred, you asked: “Is my new phenomenological understanding serving me well or poorly here, Burt?”
Your phenomenological understanding works well for the clocks, but not so well for the accelerometer differences, I think.
Jim’s three chase-rockets can’t all accelerate at a proper 1g and stay static relative to the main rocket. The nose chase-rocket’s proper acceleration must be smaller. I think one can use Lorentz contraction in a phenomenological explanation of the acceleration differences, but I haven’t tried that.
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
Firstly, happy New Year Jim.
Your answer about the position of the vertex: “… each “point†of the rocket standing vertical at the surface of a gravitational field …[has] with insignificant variations, the same gravitational “vertex”, or center of mass, producing the orientation of the geodesics of the “points†of the rocket, ”
While the center of mass of the gravitating body can certainly be called a “vertex”, you are comparing apples and oranges here. This is a 3-d spatial vertex of the spacetime geodesics of points on the rocket separated horizontally. In Mallinckrodt’s treatment it was a 4-d spacetime “vertex” in 2-d form (one space and one time dimension) of points on the rocket separated “vertically”. To make a comparison, one must find the spacetime vertices for the gravity case for points separated vertically.
Interestingly, the spacetime vertices of points separated vertically on the static rocket will be in significantly different spatial locations, because the vertex distance follows a square law against spatial distance from the center of mass and not the linear law of non-gravitational acceleration. In the low gravity limit, the vertex distance of each static point (at distance d from the center of mass) is D_v = (dc)^2/(GM).
So, ironically, this is a very visualizable non-equivalence between gravitational and non-gravitational acceleration, but it operates in the opposite sense to what you seem to hold. It is the gravitational accelerations of points on the rocket that have different spacetime vertices, while there is only one spacetime vertex for Born-rigid acceleration. Even if you do not agree that Born-rigid acceleration represents SL’s rocket, the gravitational case is indisputable. :)
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
[I edited the post for clarity only]
I don’t want to distract from the original question of the equivalence of acceleration and gravitation, but I’d like to address an example of the self-certainty that seems to be ingrained and unrestrained in contemporary science. Take a look at the graph illustrating Bell’s thought experiment at http://en.wikipedia.org/wiki/Bell's_spaceship_paradox#_note-1. Aside from the uncritical application of tenets of special relativity to accelerating bodies, the graph (and the underlying mathematics) is claimed, at different times, to represent the perspective of a co-accelerating body and of one of the vectors, body A. What would a graph of either one actually look like? Parallel, vertical vectors. Do the math. You and I could sit (in principle) on the surface of Jupiter, accelerating crushingly against the surface, for years and Jupiter years, and never experience a divergence of simultaneity between us. Should we conclude that the parallel acceleration of two spacecraft would be different from the gravitational acceleration? On what basis? In any case, it seems clear to me that acceleration can only be objectively considered from a non-accelerating frame of reference, and even then, the principles and projections of Special Relativity should be handled with care and deliberation. Maybe there’s a problem of presumption in Bell’s paradox? Maybe there’s a problem of over-reliance on mathematics at the expense of conceptualization? And maybe there’s a problem of collective, mutually reinforcing self-certainty.
Well summarized, SL.
I apologize to you if my remark about Jim’s input sounded a bit “undignified”, spoiling a very good Blog thread. I’ll practice more restraint in the future…
Interesting how Jim wrongly credits me (“… that contrary to his assertion (of which no lurid characterization is necessary), neither the three chase-rockets nor the nose nor middle of the 100m rocket “share the same vertex†or vertex-distance at any point along the acceleration“) with the science behind the rigid rod acceleration issue. Neither me, nor Mallinckrodt dreamed up this stuff – it’s been standard science for a long time before us!
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
Jim,
Burt’s answer to my original question was an unequivocal “No”, because the accelerometer readings along the rocket’s length will differ between static gravity and acceleration. The same for the clock rate differences, given a strong enough gravity field. So what is your point?
Your quote: “a ‘clear consensus’ of the CERN theory division arrived at the answer that the string would not breakâ€, has long since been settled in favor of Bell. I have checked and could find no valid refutations of the work of Rindler, Bell, Born, Nikolic, Mallinckrodt on the effects we were discussing here.
I’ll let the other readers make up their own minds about your views.
SL
Given that “a ‘clear consensus’ of the CERN theory division arrived at the answer that the string would not breakâ€, Burt’s characterization of my assumption on a secondary point as “scientific nonsense†is both unjustified and undignified. My point remains, whether the illustration is accepted or not, that contrary to his assertion (of which no lurid characterization is necessary), neither the three chase-rockets nor the nose nor middle of the 100m rocket “share the same vertex†or vertex-distance at any point along the acceleration. Take away the 100m rocket; the three rockets accelerate the same as before; where is the “one†vertex? which is the “one†vertex from which vertex distance is measured? Mallinckrodt obviously extrapolated from a point-mass to a rod without making an adjustment; this needn’t invalidate his thesis, but it changes the specific results.
My issue goes to the original question: “[Are] Acceleration and Gravity Equivalent?†The non-equivalence of vertices for various points on an accelerating rod in contrast to the vertex of a rod “static†on a gravitational surface is one more reason the answer is “no.†Burt can get ad hominem ad nauseam, but the issue is no less dispassionate and rational when confronted rather than swept aside.
I think Burt has made the required points, backed with good references. Thanks.
To summarize what I’ve learned about the equivalence of gravity and acceleration.
1. Inside small frames, rockets, whatever, Einstein’s equivalence principle holds and one cannot say whether you are sitting statically in a gravity field or are being accelerated.
2. Inside a large frame, one can use primarily the accelerometer differences to identify in which scenario you are, because the differences are huge, meaning gravity gives a much larger horizontal difference in acceleration than what an accelerating frame would yield.
3. Clock rate differences between nose and tail of an accelerating rocket are practically the same as for low-g gravity, like on Earth. Hence, it is difficult to use clocks as a distinguishing test.
4. In high-g gravity, there are second order relativistic clock rate effects that differ from accelerating frames and a test is feasible.
5. The Minkowski diagram is well suited to very simply illustrate the acceleration and time measurements along a rigid rocket being accelerated.
Anything else that I could have mentioned from the discussions above?
SL
EDIT: Thanks to everyone (including Jim) that participated!