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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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?”
Here’s how else. Mallinckrodt uses “vertex distance†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 “vertex distance†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 – at the nose, the middle, and the tail – will be synchronous with the clock of the accompanying chase-rocket, as each pair has an identical “vertex distance†– 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, Mallinckrodt’s presentation (http://www.csupomona.edu/~ajm/professional/talks/relacc.ppt ) has a fundamental flaw: An accelerating (rigid) body doesn’t have one “vertex.†Each point on the accelerating body is equidistant from its own “vertex,†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 body’s 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)