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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I have recently learned that all the effects inside an accelerating laboratory are exactly equivalent to the gravitational effects of an ‘infinitely long’ cylindrical mass (at least in two of the three spatial dimensions).
If the lab is kept statically at a radial distance r perpendicular to the cylinder’s long axis, the gravitational acceleration scales with 1/r, just as is the case in a tall accelerating lab.
In the transverse direction parallel to that axis, there is no tidal forces focusing to the ‘center’ of the cylinder (an infinitely long thing has no defined center), exactly as inside the accelerating lab.
So purely inside such a lab, there will be no way to distinguish between gravity and linear, uniform acceleration, unless you use the third dimension, perpendicular to the both the radial and the axis. In that direction, there will be the squeezing effect (of tidal gravity) to the centerline of the cylinder.
Interesting.
Burt Jordaan (www.Relativity-4-Engineers.com)
I agree that the Principle of Equivalence has most widely come to be applied to that of inertial and gravitational mass. But originally it was more ambitious, it was Einstein’s seminal concept that led to the General Theory. I don’t have it at-hand, but he stated explicitly that he wanted to demonstrate the relativity of all accelerating frames of reference, just as he had uniform frames – and accordingly, gravitation in one frame of reference was held to be equivalent to inertial acceleration from another. That was the original “general” in the General Theory.
The original principle, that all forms of acceleration are equivalent, has morphed into various more moderate interpretations. But even that of inertial and gravitational mass is open to criticism. Given that the weight we experience at the earth’s surface is the resistance to our geodesic motion, unless we are sinking in water or quicksand, it is exactly equal to the force needed to offset the geodesic motion of our inertial mass. Gravitational mass is equivalent to inertial mass because gravitational mass IS inertial mass.
The fact that Jim claims to be able to distinguish between gravity and an accelerated frame in local measurements means that he can distinguish between gravitational and inertial mass.
At least that’s my reading of his statements, especially as regards the (ir)relevance of Einstein’s thought experiment that leads to that equivalence.
Now onward to Scruffy’s new thread.
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
The fact that Jim claims to be able to distinguish between gravity and an accelerated frame in local measurements means that he can distinguish between gravitational and inertial mass.
At least that’s my reading of his statements, especially as regards the (ir)relevance of Einstein’s thought experiment that leads to that equivalence.
Now onward to Scruffy’s new thread.
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Hi Jim, also consider my last reply to Fred.
Not sure one should read too much into the “principle of equivalence” between gravity and acceleration. It’s rather about the “principle of equivalence” between gravitational- and inertial mass.
Burt Jordaan (www.Relativity-4-Engineers.com)
SL wrote: “Maybe I should open a new thread on that. (Edit: Have done that: https://scienceblog.com/cms/gravitational-waves-useful-or-wasteful-15…)”
Cool idea! Will take a peek…
Burt Jordaan (www.Relativity-4-Engineers.com)
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 donรขโฌโขt see why the gravitational vertex isnรขโฌโขt (approximately) the same location in relation to both horizontal and vertical points on the rocket รขโฌโ 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 isnรขโฌโขt non-equivalent to acceleration!
Burt wrote: รขโฌลWhile the center of mass of the gravitating body can certainly be called a รขโฌหvertexรขโฌโข, you are comparing apples and oranges here.รขโฌย
I donรขโฌโขt see why the gravitational vertex isnรขโฌโขt (approximately) the same location in relation to both horizontal and vertical points on the rocket รขโฌโ 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 Iรขโฌโขm definitely comparing (and distinguishing) apples and oranges รขโฌโ gravitational and non-gravitational acceleration.
รขโฌลIt 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.รขโฌย
It seems we need to be more clear in distinguishing vertices and vertex-distances. Iรขโฌโขm 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: รขโฌลWhere, in your view, is this common รขโฌหvertexรขโฌโข for a rocket sitting static in a gravitational field?รขโฌย
As I wrote earlier, each รขโฌลpointรขโฌย 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 รขโฌลpointsรขโฌย 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 Mallinckrodtรขโฌโขs treatment of the rocketรขโฌโขs inertial acceleration in space, where the รขโฌลvertexรขโฌย is described as the location where the acceleration was initiated.
SL asked re: the question of the equivalence of grav & inertia: รขโฌลBurt’s answer to my original question was an unequivocal รขโฌหNoรขโฌโข [equivalence]รขโฌยฆ. So what is your point?รขโฌย I wonรขโฌโขt 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 donรขโฌโขt 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.
Iรขโฌโขm short on time, and anxious to get back to Burt, but I want to respond briefly to Fred, who wrote:
รขโฌลThanks, 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 equivalenceรขโฌยฆ. The difference is that Jim presumes the non-equivalence is true in the limiting case, whereas you recognize that it is not.รขโฌย
I canรขโฌโขt speak for anyone else, but a รขโฌลprincipleรขโฌย 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 รขโฌโ 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 doesnรขโฌโขt 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 รขโฌลpointรขโฌย 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 รขโฌลpointsรขโฌย 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 Mallinckrodtรขโฌโขs 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 donรขโฌโขt want to distract from the original question of the equivalence of acceleration and gravitation, but Iรขโฌโขd 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 Bellรขโฌโขs 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 thereรขโฌโขs a problem of presumption in Bellรขโฌโขs paradox? Maybe thereรขโฌโขs a problem of over-reliance on mathematics at the expense of conceptualization? And maybe thereรขโฌโขs 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 รขโฌลshare the same vertexรขโฌย 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: รขโฌลa รขโฌหclear consensusรขโฌโข of the CERN theory division arrived at the answer that the string would not breakรขโฌย, 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 รขโฌลa รขโฌหclear consensusรขโฌโข of the CERN theory division arrived at the answer that the string would not breakรขโฌย, Burtรขโฌโขs characterization of my assumption on a secondary point as รขโฌลscientific nonsenseรขโฌย 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 รขโฌลshare the same vertexรขโฌย 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 รขโฌลoneรขโฌย vertex? which is the รขโฌลoneรขโฌย vertex from which vertex distance is measured? Mallinckrodt obviously extrapolated from a point-mass to a rod without making an adjustment; this neednรขโฌโขt invalidate his thesis, but it changes the specific results.
My issue goes to the original question: รขโฌล[Are] Acceleration and Gravity Equivalent?รขโฌย The non-equivalence of vertices for various points on an accelerating rod in contrast to the vertex of a rod รขโฌลstaticรขโฌย on a gravitational surface is one more reason the answer is รขโฌลno.รขโฌย 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!
For everyone concerned, Jim Arnold has posted seven points of mainly unscientific nonsense here, which I will comment on only in the interest of proper science, not as an argument with Jim (which is a useless exercise).
Jim’s three rockets will not stay at a constant proper distance from each other – they will pull apart. Only the tail chase-rocket will stay with SL’s rocket, the other two will pass and leave the rocket behind. In order for them to keep station, the center and nose chase-rockets have to accelerate at slightly less than 1g, depending on their distance from the vertex.
Readers don’t have tot believe me; see this on the authoritative MathPages: http://www.mathpages.com/home/kmath422/kmath422.htm
Quote: “Of course, nothing prevents us from accelerating every part of an arbitrarily long rod, all in the same direction, in unison, but this sort of acceleration does not maintain Born rigidity. The instantaneous rest length of the rod increases, i.e., the rod is stretched.”
Another applicable article is about “Bell’s Spaceship Paradox” on Wikipedia: http://en.wikipedia.org/wiki/Bell's_spaceship_paradox, showing why Jim’s rockets will not all stay on station.
Quote:“An equivalent problem is more commonly mentioned in textbooks. This is the problem of Born rigid motion. Rather than ask about the separation of spaceships with the same acceleration, the problem of Born rigid motion asks “what acceleration profile is required by the second spaceship so that the distance between the spaceships remains constant in their proper frame”. The accelerations of the two spaceships must in general be different. In order for the two spaceships, initially at rest in an inertial frame, to maintain a constant proper distance, the lead spaceship must have a lower proper acceleration.”
Jim will undoubtedly have the last word here, but readers of Scruffy’s Blog are advised to weigh up his statements against scientific papers and writings.
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
I’m sure Burt will have a more mathematical reply, but it seems to me that the chase rocket clocks are not synchronized for the same reasons the three clocks in the larger rocket are are not synchronized.
My earlier discussion of red/blue shifts when light eams are sent parallel to the direction of acceleration seems to apply equally well to the three chase rockets as to the larger one.
The only clocks that remain in sync would be the three sets of paired clocks, the one in the large rocket paired with the appropriate chase rocket’s closk.
Another way to think of it is that the chase rockets, since they are not moving with respect to the main rocket, can be considered part of the same rigid body or to be in the same noninertial frame of reference.
Jim is making what seems to be natural assumption of synchrony of the chase rocket clocks, but it seems to be the same error I was making early on in this discussion.
Is my new phenomenological understanding serving me well or poorly here, Burt?
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Burt wrote: “[the clocks of the chase-rockets] will correspond to the clocks on the rocket and they will all share the same vertex; the one that of all the parts of the ‘rigid’ rocket share. How else?”
Hereรขโฌโขs how else. Mallinckrodt uses รขโฌลvertex distanceรขโฌย to calculate the relativistic effects of acceleration on a rigid body.
1. Three small chase-rockets simultaneously begin accelerating alongside a 100m rocket, all with identical accelerations of 1g.
2. One chase-rocket is alongside the nose of the 100m rocket, one is at the middle, one is at the tail.
3. The acceleration effects on the chase-rockets are calculated according to the รขโฌลvertex distanceรขโฌย of each. Because each chase-rocket begins accelerating at a different location, each has a different vertex. (How else? The clocks of the chase-rockets are independent of the 100m rocket, and of each other; they are independent of the vertex of the 100m rocket and of the vertices of the others.)
4. The clocks of the three chase-rockets are synchronous at any moment in their accelerations, because the accelerations are equivalent, and each is equidistant from its own vertex.
6. The clocks on the 100m rocket รขโฌโ at the nose, the middle, and the tail รขโฌโ will be synchronous with the clock of the accompanying chase-rocket, as each pair has an identical รขโฌลvertex distanceรขโฌย รขโฌโ the distance from the beginning of their accelerations.
7. Therefore, the effects described by Mallinckrodt are improperly derived from the assignment of one vertex and one vertex distance to the three clocks.
What about Special Relativity? The mass of the rocket increases and the length is shorter. The time on the rocket is slower than the time for an outside observer, etc.
Scruffy, when you say Jim is part right, that’s always the case. But he is almost always not quite right or unclear about part of his discussion.
That’s when the old merry-go-round starts spinning.
Here, as usual, Jim seems to be saying that at any instant along the motion and at any point in space, he can distinguish between whether an observer is seeing the effects of a gravitational field or being in an accelerated frame of reference.
Specifically, he seems to be saying that the clock rate is different in the two cases of gravity or an accelerated frame.
If you want to suffer through the past discussion, you will see that David Halliday eventually concluded that Jim had proposed nothing novel.
i still haven’t figured out, and I no longer care, whether Jim accepts or challenges the principle of equivlance. If he challenges it, he rejects Einstein’s thought experiment as unrealistic rather than accepting it the limiting case.
Since we’ve been through that discussion before, and since Jim always seems to want the last word, I’m glad to let him have it.
Reply at your peril!
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Looking at Jim’s reply, he agrees with almost everything with: “Of course.”
Then he poses what looks like the “ultimate question”: “Imagine three small chase-rockets accelerating alongside the 100m rocket, one at the nose, one at the middle, one at the tail. Don’t they each have a “vertex” according to Mallinckrodt? How would their clocks correspond to the clocks on the rocket that each is riding alongside?”
The short answer is: their clocks will correspond to the clocks on the rocket and they will all share the same vertex; the one that of all the parts of the “rigid” rocket share. How else?
The longer answer (we don’t want the long answer, do we?) is that once a rocket has compressed under acceleration and the inevitable oscillations have dampened out, the rocket is behaving in a mode named “Born rigid acceleration”. This means that every lengthwise point on the rocket accelerates slightly differently. Likewise, the three “chase-rockets accelerating alongside the 100m rocket” will each need a different acceleration to keep station relative to the rocket. However, since they are at different positions relative to the rocket’s length, they will all share the same vertex.
If there is anyone else that would like to get more information on this effect, I’ll be more than happy to continue discussing it. To keep on explaining it to Jim seems to be fairly unproductive… Maybe Jim should ask NASA to explain it – they have plenty of experience with rockets and clocks!
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
“It is not only Mallinckrodt that presumes the single spacetime vertex – the rest of the physics world does so too!”
Why do you find that so compelling that it relieves you of the need for a direct response?
“There’s actually an event horizon that forms behind an accelerating rocket.”
Of course. The rocket is accelerating, whether there’s one vertex or many – the latter corresponding to the initial position of each “point” on the rocket.
“At 1g it’s about one light-year (d = a/c^2 m) behind the rocket, exactly the radius of spacetime curvature we experience here on Terra-firma, i.e., exactly the distance to our spacetime vertex. Cool!”
Of course. They’re both 1g.
Imagine three small chase-rockets accelerating alongside the 100m rocket, one at the nose, one at the middle, one at the tail. Don’t they each have a “vertex” according to Mallinckrodt? How would their clocks correspond to the clocks on the rocket that each is riding alongside?
Scruffy, when you say Jim is part right, that’s always the case. But he is almost always not quite right or unclear about part of his discussion.
That’s when the old merry-go-round starts spinning.
Here, as usual, Jim seems to be saying that at any instant along the motion and at any point in space, he can distinguish between whether an observer is seeing the effects of a gravitational field or being in an accelerated frame of reference.
Specifically, he seems to be saying that the clock rate is different in the two cases of gravity or an accelerated frame.
If you want to suffer through the past discussion, you will see that David Halliday eventually concluded that Jim had proposed nothing novel.
i still haven’t figured out, and I no longer care, whether Jim accepts or challenges the principle of equivlance. If he challenges it, he rejects Einstein’s thought experiment as unrealistic rather thn as accepting it the limiting case.
Since we’ve been through that discussion before, and since Jim always seems to want the last word, I’m glad to let him have it.
Reply at your peril!
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Hi SL, tough questions you ask! I’ll do the 2nd one first, because it’s the easiest.
“…Incidentally, how does it [your diagram] tie up with your objections against Jim’s diagram?”
Mine is a “space-propertime” diagram, slightly different from Jim’s. The two orthogonal axes are labeled XJ and tau, indicating that they do not belong to the same observer. It is the accelerating observer’s propertime plotted against the reference observer’s space. See figure below. Jim’s diagram keeps the two axes for each observer orthogonal and hence cannot be used quite like this.
The X-double-dot is the constant proper acceleration as measured at the rocket’s propulsion system [edit]by a local accelerometer.[/edit]
Back to your first question: “…it appears that it gives a parabolic curve for a uniformly accelerated object. How does this tie up with the hyperbolic curves in Mallinckrodt’s presentation?”
You are right. Instead of tending towards the 45 degree light-cone (hyperbolic), the curve tend towards the horizontal (parabolic). This is the essence of a space-propertime diagram. In the end, it gives the same results as a Minkowski diagram.
When your rocket is plotted like this, you will find that the nose and tail parabolas also get closer to each other and have different curvatures (hence different proper accelerations). The nose clock will get ahead of the tail clock (in time) for the same coordinate time (which is the length of the individual curves and cannot be read directly off the graph).
“…can I consider the Lorentz contraction in the reference frame and hence the slower movement of the nose of my rocket in the references frame, giving less velocity time dilation, as the motivation for faster running clocks in the nose?”
Yes you can, but I’m hesitant to propagate that, because it is not as simple as it looks on the surface. It is not easy to transform the times and accelerations from the reference frame to the “Rindler observers” on board the accelerating rocket.
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
Burt, I see you could not resist replying to Jim yourself! :)
In your eBook’s chapter 3 (linear acceleration), you use a diagram much like Jim’s (discredited) diagram and it appears that it gives a parabolic curve for a uniformly accelerated object. How does this tie up with the hyperbolic curves in Mallinckrodt’s presentation? Incidentally, how does it tie up with your objections against Jim’s diagram?
Another question: can I consider the Lorentz contraction in the reference frame and hence the slower movement of the nose of my rocket in the references frame, giving less velocity time dilation, as the motivation for faster running clocks in the nose? (Sorry, badly worded question, but I must run, so I hope you find it intelligible!)
SL
PS: I wonder if Jim actually read the Nikolic paper that you referenced above: http://aps.arxiv.org/pdf/physics/9810017
Jim wrote: “But Burt doesn’t want to touch my point that the accelerating body doesn’t have a single “vertex”, as Mallinckrodt presumes.”
But Burt knows how circular arguments with Jim can get:-(
It is not only Mallinckrodt that presumes the single spacetime vertex – the rest of the physics world does so too! There’s actually an event horizon that forms behind an accelerating rocket. At 1g it’s about one light-year (d = a/c^2 m) behind the rocket, exactly the radius of spacetime curvature we experience here on Terra-firma, i.e., exactly the distance to our spacetime vertex. Cool!
Burt Jordaan (www.Relativity-4-Engineers.com)
“BTW, don’t pay too much attention to Jim Arnold’s ‘rebuttal'”
Interesting how everyone is always advising everyone else to pay no attention.
Yes, of course, “a radio signal sent from the tail of the accelerating ship and received at the nose will be red-shifted, similar to gravitational redshift.”
But Burt doesn’t want to touch my point that the accelerating body doesn’t have a single “vertex”, as Mallinckrodt presumes. It’s a simple point, really. Disturbing, apparently. But wrong? I’m (un)afraid not.
Yes, I think you are right that second order effects on vertically separated clock rates will differ between gravity and acceleration. The acceleration-related clock rate differences correspond to the low field limit of the gravitational redshift equation: dtau/dt = 1-GM/(rc^2). The strong field gravitational redshift equation is dtau/dt = [1-2GM/(rc^2)]^(0.5), which approximates to the former when 2GM << rc^2.
BTW, don't pay too much attention to Jim Arnold's "rebuttal". I used to know about two people who dispute the validity of John Mallinckrodt's presentation. Now I know three. The former two are well recognized crackpots.
If you can't resist arguing with Jim, ask him if he agrees that a radio signal sent from the tail of the accelerating ship and received at the nose will be red-shifted, similar to gravitational redshift. But then, I recall that you have already "barked out" of a previous discussion with Jim…
Burt Jordaan (www.Relativity-4-Engineers.com)
For anyone interested, Mallinckrodtรขโฌโขs presentation (http://www.csupomona.edu/~ajm/professional/talks/relacc.ppt ) has a fundamental flaw: An accelerating (rigid) body doesnรขโฌโขt have one รขโฌลvertex.รขโฌย Each point on the accelerating body is equidistant from its own รขโฌลvertex,รขโฌย i.e., its location at the initiation of the acceleration. There is a finite time differential in the communication of the acceleration from one end of the body to the other, resulting in a compression in the direction of acceleration due to tensile stresses, but only until the rigidity of the body has accommodated the acceleration. There is therefore no relativistic contraction of the body in the reference frame of an observer sharing the acceleration, at least so long as the body remains rigid, and the clocks at the nose and the tail remain synchronized except for the infinitesimal difference due to the bodyรขโฌโขs initial non-relativistic contraction.
In contrast, a body standing vertical at the surface of a gravitational field undergoes more time dilation at its base than its top, as predicted by the General Theory, because the intensity of the field produces actual time dilation according to the distance from the gravitational “vertex”, and with insignificant variations, each point in the rigid body has the same gravitational “vertex”, or center of mass. (There is also time dilation in this case due to the acceleration of the body from its geodesic as it presses against the surface, so there is actually a combination of effects, but that’s a secondary point here.)
Time dilation in both cases is real. Return the clocks to the origin of the acceleration, or bring the clock of the vertical body to the surface, and they will be asynchronous with clocks that remained at the origin in the one case or on the surface in the other. 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.
Fred, I think Jim is at least partially right, because uniform acceleration and gravity are not completely (generally) equivalent, as Burt has pointed out, e.g. the accelerometer readings. As he said, Einstein’s equivalence principle does not hold for extended systems of reference.
On Jim’s “non-equivalence of the behavior of clocks accelerating through space and ‘at rest’ in a gravitational field.”, I reckon clock rate changes along the length of the rocket are the same (g times Delta d/c^2) for Earth-like gravity and uniform acceleration, to first order at least. I’m not sure what will happen in a very highly curved space, e.g. near a neutron star – there may be second order effects that differ. Perhaps Burt will enlighten us here.
SL
You’re right about the impossibility of separating the two. It’s just that I had not previously considered that clocks might run at different rates at different places in a uniform gravitational field.
It makes no sense to compare the effects of an erroneous calculation to a correct one, so I withdraw my question.
Jim Arnold wants to add an additional consideration, since he rejects the equivalence of an accelerated frame of reference and a gravitational field.
If you care to open this up to the old merry-go-round from which everyone else has exited at Jim”s blog, perhaps you or Burt can address the supposed “non-equivalence of the behavior of clocks accelerating through space and ‘at rest’ in a gravitational field.”
From what I can tell from the math you have presented and from Einstein’s classic work, the two cases are equivalent, and there is no need to introduce non-equivalence to satisfactorily describe the universe.
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Tx Fred, it seems like an interesting place to frequent…
On your curiosity: “…but I’m curious about which effect is larger. I suspect it’s the non-uniformity on a relatively small rock like Earth. It might not be so if you could perform the same experiment on the surface of the Sun.”
I don’t quite understand how (and why) you want to divide the gravitational time dilation into two parts. For the Earth and the Sun, the low field approximation is the same: dtau/dt = 1 – GM/r. I’m actually not sure what you mean by “uniform gravity” – is it when the slope of dtau/dt (i.e. d^2tau/dt^2) is constant?
SL
Nice of you to say. And pay no attention to the non-equivalence of the behavior of clocks accelerating through space and “at rest” in a gravitational field.
After slogging along with jarnold’s muddled definitions and refusal to adopt standard terminology, it has been a delight to share this discussion with Burt and SL (Scruffy).
It’s also interesting to see how physics has both a mathematical side and a phenomenological side. Some people work better in one realm, and some in the other, but real insight comes from being able to combine the two.
One thing I have learned is that when the rocket is on the ground, there are two effects to consider in difference in clock speed: non-uniformity in the gravitational field and the effect that would be present even if the gravity were uniform.
I haven’t done the math, but I’m curious about which effect is larger. I suspect it’s the non-uniformity on a relatively small rock like Earth. It might not be so if you could perform the same experiment on the surface of the Sun.
Thanks for signing up here, Scruffy. I look forward to more dogged pursuits with you. Unlike jarnold, you don’t bark up the wrong tree. :)
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
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.
You also write: “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.” That contact force, and its “reaction” partner, the force our butt exerts on the chair, would not exist in the absence of either a gravittional field, a nongravitational force on us or our chairs, or our being in an accelerated frame of reference.
Leaving out the nongravitational force, the other two situations as well as any combunation of those two are indistinguishable in the limit of an infinitesimal observer.
For phenomenological thinkers like me, the way I would go with this is that Einstein started with a limiting case and then was faced with the very difficult problem of finding a general solution that matched the boundary conditions set by that limiting case. Jim, in contrast, takes the approach of rejecting the limiting case as irrelevant.
Hmmmm… Does that mean that your discussion needs to include a cosmological constant to account for the fact that the universe itself may be expanding or contracting?
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)