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Implications of the statement Acceleration is not relative

by GregAshmore
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harrylin
#163
Feb20-13, 02:40 AM
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Quote Quote by GregAshmore View Post
True. But no one is claiming that the rocket is at rest in an inertial frame--not even the twin in the rocket.
The physical meaning of "in rest" is very well clarified in Einstein's 1918 paper.
The twin in the rocket feels the unbalanced force of the rocket engine, and he knows (or would know upon reunion) that the twin on earth feels no such force. Even without the formal definitions of inertial and non-inertial, the rocket twin would recognize that his situation is fundamentally different than that of his twin. Fully aware of that difference, he claims that he is at rest throughout the episode.
Once more, that is true for Einstein's example and completely wrong (even in two ways) for Langevin's example. "At rest" in the sense that you adopt here only makes sense in the way Einstein elaborates - and that isn't SR.
The rocket twin [..] can point to the spacetime diagram (which the earth twin accepts as valid) and show that he remains at rest in his own frame, even while not at rest in any one inertial frame. To prove the rocket twin wrong, [..]
For a last time, as we've been here twice before: everyone can always claim to be at rest in his own frame; such a statement cannot be disproved. You could just as well state that you're in your own world. That's physically meaningless.
harrylin
#164
Feb20-13, 02:48 AM
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Quote Quote by DaleSpam View Post
The two definitions are equivalent. The Smoot definition basically just pushes the use of accelerometers one step further. Instead of reading the proper acceleration directly off the accelerometer, you define an inertial frame by strapping accelerometers to your clocks and rods, ensuring that they read 0, and then reading the proper acceleration off the clocks and rods.
I prefer his definition as he doesn't confound a displacement with a force; and it does make a difference when using it in SR, due to the different definition of "inertial frame" in SR.
my_wan
#165
Feb20-13, 02:53 AM
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Quote Quote by GregAshmore View Post
It would be of consequence for the person who mistakenly believes that because either object can be the one that appears to turn around, the spacetime diagram can be drawn with either object having the bent worldline.
What has been repeated here is that either object CANNOT be the one that appears to turn around because all observers MUST agree on which one accelerated, just as the one that accelerated is also the only one to experience a g force as a result.

You are confusing Einstein's term, most probably with respect to the explanation given for the principle of equivalence. IF every agrees on what happened, even if not to what degree, we are by definition talking about an absolute, not relative event. The relative terms of that absolute involve only the quantitative value associated with it. That is the point we are trying to make with the distinction between coordinate acceleration and proper acceleration.
harrylin
#166
Feb20-13, 02:54 AM
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Quote Quote by DaleSpam View Post
Which he can do quite easily simply by stating the laws of physics in a covariant form and then using any coordinates where his state of motion does not change.
As I suspect that you also don't copy Einstein's physical explanation, I'm curious to know which physical explanation that you found in the literature you fancy for the moving and faster aging Earth with a stationary rocket (let's stay away from personal ideas). How can the firing of the rocket engine move the rest of the universe while keeping the rocket's state of motion unaffected?
DaleSpam
#167
Feb20-13, 07:28 AM
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Quote Quote by harrylin View Post
it does make a difference when using it in SR, due to the different definition of "inertial frame" in SR.
Any way you can determine if your frame is inertial or not is a way of determining your proper acceleration, i.e. it is an accelerometer. You cannot get away from using accelerometers.
harrylin
#168
Feb20-13, 07:34 AM
P: 3,187
Quote Quote by DaleSpam View Post
Any way you can determine if your frame is inertial or not is a way of determining your proper acceleration, i.e. it is an accelerometer. You cannot get away from using accelerometers.
We have discussed that before, and I found that your definition of "inertial frame" is at odds with that of classical mechanics wrt which SR is defined; as a matter of fact, it's already at odds with Einstein's 1905 paper and Langevin's 1911 paper. So, let's agree to disagree; but if you even disagree to agree to disagree, then I'll just add some links to earlier discussions later, as IMHO everything has been said already.

ADDENDUM: see
http://physicsforums.com/showthread.php?p=4117808
In post #190 I provided three operational ways with which such reference frames can be defined/determined.
See also post #200 there and a 4th defintion (by Einstein) in post #264:
http://physicsforums.com/showpost.ph...&postcount=264
DaleSpam
#169
Feb20-13, 07:35 AM
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Quote Quote by my_wan View Post
What has been repeated here is that either object CANNOT be the one that appears to turn around because all observers MUST agree on which one accelerated
The phrase "appears to turn around" (emphasis added) seems to refer to coordinate acceleration, in which case it would not be true that all observers must agree on it. If you intended the statement to refer to proper acceleration then it is a little confusing.

I think that GregAshmore understands the distinction between coordinate and proper acceleration, so I think that the rest is just miscommunication about which "flavor" of acceleration is being discussed at any one moment.
DaleSpam
#170
Feb20-13, 07:36 AM
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Quote Quote by harrylin View Post
We have discussed that before, and I found that your definition of "inertial frame" is at odds with that of classical mechanics; as a matter of fact, it's already at odds with Einstein's 1905 paper and Langevin's 1911 paper. So, let's agree to disagree; but if you even disagree to agree to disagree, then I'll just add some links to earlier discussions later.
Some links would help. I don't remember that discussion.
DaleSpam
#171
Feb20-13, 07:45 AM
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Quote Quote by harrylin View Post
As I suspect that you also don't copy Einstein's physical explanation,
Physical explanation of what? You never clarified exactly what you thought he meant by "gravitational field", and he wasn't explicit about it. Until you have defined your terms you are just giving physical explanations of flubnubitz.

Quote Quote by harrylin View Post
I'm curious to know which physical explanation that you found in the literature you fancy for the moving and faster aging Earth with a stationary rocket (let's stay away from personal ideas). How can the firing of the rocket engine move the rest of the universe while keeping the rocket's state of motion unaffected?
The firing of the rocket engine doesn't move the rest of the universe, the choice of coordinates does. The state of motion or rest is a coordinate-dependent quantity. Do you disagree that I can give any object any velocity profile I like simply by choosing the coordinates appropriately?
stevendaryl
#172
Feb20-13, 09:11 AM
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Quote Quote by harrylin View Post
We have discussed that before, and I found that your definition of "inertial frame" is at odds with that of classical mechanics wrt which SR is defined
What Einstein said by way of defining inertial frame was in his 1905 paper:
Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.
Presumably, since he is suggesting a modification to Newtonian mechanics, he means "approximately", in the low-velocity limit.

The way I interpreted Einstein's words are that an inertial coordinate system satisfies:
  1. Objects that are not acted upon by physical, external forces travel so that [itex]\dfrac{d^2 x}{dt^2} = \dfrac{d^2 y}{dt^2} = \dfrac{d^2 z}{dt^2} = 0[/itex]
  2. For objects moving slowly compared with the speed of light, the response of an object to a physical external force [itex]\vec{F}[/itex] is given (approximately, ignoring correction terms of order [itex]\dfrac{v^2}{c^2}[/itex]) by [itex]F^i = m \dfrac{d^2 x^i}{dt^2}[/itex]
  3. If an object exerts a force [itex]\vec{F}[/itex] on a second object, then the second object exerts a force [itex]-\vec{F}[/itex] on the first.

These conditions characterize an inertial Cartesian coordinate system. They imply that an accelerometer at rest in that coordinate system will show no acceleration. But the other way around may not be true. An accelerometer at rest showing no acceleration doesn't imply that your coordinate system is an inertial Cartesian coordinate system.
ghwellsjr
#173
Feb20-13, 09:20 AM
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Quote Quote by GregAshmore View Post
Quote Quote by ghwellsjr View Post
T&W never consider non-inertial frames or a frame in which the traveler is at rest, at least not when discussing the Twin Paradox. What they are dismissing is the explanation that I gave in post #23 of the thread you linked to in your OP. I considered three Inertial Reference Frames which included different rest states of the two twins. I attempted to show that any one of them was adequate to explain everything about the scenario, which is pretty much the classic way of explaining the Twin Paradox. But they think they have a better way which involves any observer calculating the spacetime interval between pairs of events. I don't think that helps at all but since they are writing the book, they get to decide when the objectors are happy. But real objectors, like you, remained unhappy.
I read that post in its entirety before opening this thread. I didn't catch on to what you were doing because in each of the diagrams the rocket twin is spoken of as moving for part of the trip.

What helped me was to realize that in the typical two-frame spacetime diagram, the world line of an inertial particle shows the particle both as moving and at rest. It is moving in one frame, and at rest in the other frame.
So because my spacetime diagrams only show one frame instead of the two that are more commonly shown in a Minkowski diagram, that prevented you from grasping what I was presenting, correct? But now that you realize the difference, does post #23 make perfect sense to you? Could you use it with further explanation to get your son to understand what I was presenting there?
Quote Quote by GregAshmore View Post
Thus, the one spacetime diagram actually shows the case I wanted to see-the case in which the rocket twin considers himself at rest.
Are you talking about this one spacetime diagram?



If so, wouldn't it have been just as confusing to you if you had not previously figured out that it was not a conventional Minkowski diagram with two frames in it?
Quote Quote by GregAshmore View Post
The symmetrical diagram (which is invalid) is not needed.
What are you calling a symmetrical diagram? A Minkowski diagram? And why would it be invalid? And why is it not needed? Now I'm confused.
ghwellsjr
#174
Feb20-13, 10:23 AM
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Quote Quote by GregAshmore View Post
Quote Quote by ghwellsjr View Post
Actually, Event B is of no consequence in the definition of your scenario. As long as we know the rocket's speed and where (or when) it turns around, that defines the end point.
It would be of consequence for the person who mistakenly believes that because either object can be the one that appears to turn around, the spacetime diagram can be drawn with either object having the bent worldline. Event B is where the Earth would turn around. I chose to show at this juncture that the Earth cannot change inertial frames because it experiences no unbalanced force; because it is inertial.
But when I drew the (non-inertial) spacetime diagram showing the Earth as the object that turns around, it doesn't happen at Event B, it happens at two other events. The Earth's bent worldline has two bends in it, not just one.
Quote Quote by GregAshmore View Post
Quote Quote by ghwellsjr View Post
However, I'd like you to consider an issue related to the one you just raised here and that is, how does the rocket know when to turn around? The rocket cannot know from any direct measurement when the Earth has traveled 10 units away in the rocket frame. That distance is the difference between the spacial components of Event B and C which are simultaneous in the Earth frame but which have a distance between them of 16.67 in the rocket frame. Even if the rocket had instant access to the remote information, it would still have to do some calculation if it's based on distance to determine when to fire its rockets.

To me, a much cleaner way to specify the Twin Paradox is to state the Proper Time on the traveler's clock when he should fire the rockets to turn around. This has the advantage that it doesn't require the specification of any reference frame. In fact, it doesn't even require fully specifying any events since we don't care about the spatial component. So if we know how long it takes for the traveler to get to the turnaround point and we know how fast he is traveling, those two parameters fully specify the complete Twin Paradox scenario (assuming of course that he is returning at the same speed he left at). I fully explained this in the thread you linked to in your opening post.
It is a cleaner solution.
Please note that I did not say it was a cleaner solution-just a cleaner specification. The cleaner specification does not imply any particular solution or explanation of the Twin Paradox.
Quote Quote by GregAshmore View Post
Yet even now I feel the [vestigial] reflexive urge to tune it out because it says the rocket twin "travels", "turn[s] around", "return[s]". The doubter has been told that in relativity the rocket has the same right to be at rest as the Earth has. Language of motion applied to the rocket speaks so loudly that it drowns out the perfectly valid point that is being made.
Well let's review your entire specification:
Quote Quote by GregAshmore View Post
Given:
G1. Earth and rocket are both at rest at same position. Earth clock and rocket clock are synchronised.
G2. At time 0.0, rocket fires a pulse.
G3. Earth and rocket separate at relative velocity 0.8c.
G4. At distance 10 units from Earth, as measured in Earth frame, rocket fires a pulse.
G5. Earth and rocket approach at relative velocity -0.8c.
G6. Upon reaching Earth, rocket fires a pulse, coming to rest on Earth.
G7. Gravitational effects of mass are to be ignored.
It's obvious that you are trying to specify the scenario in such a way that it does not imply which object is moving. However, G6 implies that it is the rocket that was moving because you say it comes to rest on Earth. Maybe you should say: At closest approach of Earth and rocket, rocket fires a pulse and once again, both are at rest at the same position.

But there are still problems. Saying that the rocket "fires a pulse" three times throughout the scenario implies that it is exactly the same pulse fired three times in the same direction which, of course, won't work. What could work is if you leave G2 alone and change G3 (using my suggestion) to say: "At time 7.5 according to the rocket's clock, the rocket turns around and fires two pulses" and then at G6 you could say: "At closest approach of Earth and rocket, rocket turns around again and fires a pulse and once again, both are at rest at the same position." If you don't like the term "turns around" then you will have to provide the rocket with thrusters at both ends and state which thruster is being used at each point.
Quote Quote by GregAshmore View Post
For perspective, I have read two or three explanations of the twin paradox to my 30+ son. He has some technical training, has a job that requires him to evaluate contractual language. He had exactly my reaction, without me making any comment.
I'm all for providing better explanations that actually communicate and maybe we could enlist your son in making that happen. Did he offer any suggestions such as the point you made that my explanation in post #23 was confusing because you expected two frames in one diagram? I can use that suggestion to improve my explanations in the future. More suggestions would be welcome.
ghwellsjr
#175
Feb20-13, 11:11 AM
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As a side note with regard to my diagrams showing only one frame compared to the more traditional Minkowski diagram showing two frames, let me show you what happened when I tried to make the point that my type of diagram was not a Minkowski diagram:
Quote Quote by ghwellsjr View Post
By the way, your graphs are not Minkowski diagrams, they are simply conventional position versus time graphs. And I'm not saying that simply because you are interchanging the time versus distance axes that is more common for a Minkowski diagram.
Read the posts following that one and you'll see why I no longer made a distinction between my one-frame spacetime diagrams and the more common two-frame Minkowski diagrams.
harrylin
#176
Feb20-13, 02:42 PM
P: 3,187
Quote Quote by DaleSpam View Post
Some links would help. I don't remember that discussion.
Done.
Quote Quote by stevendaryl View Post
What Einstein said by way of defining inertial frame was in his 1905 paper: [..] Presumably, since he is suggesting a modification to Newtonian mechanics, he means "approximately", in the low-velocity limit. [..]
Indeed, SR is defined wrt the same reference systems as Newton's mechanics.
The way I interpreted Einstein's words are that an inertial coordinate system satisfies:[*] Objects that are not acted upon by physical, external forces travel so that [itex]\dfrac{d^2 x}{dt^2} = \dfrac{d^2 y}{dt^2} = \dfrac{d^2 z}{dt^2} = 0[/itex][..]
Not sure if you mean that correctly; a coordinate system of reference is not "objects". For example the surface of the Earth is approximately a valid "Galilean" reference frame (neglecting its rotation and orbit), in which a cannon ball is acted upon by the force of gravity in accordance with Newton's second law. That is a classical textbook example of motion wrt to a valid reference system for classical mechanics, as your second point also stresses:
[*]For objects moving slowly compared with the speed of light, the response of an object to a physical external force [itex]\vec{F}[/itex] is given (approximately, ignoring correction terms of order [itex]\dfrac{v^2}{c^2}[/itex]) by [itex]F^i = m \dfrac{d^2 x^i}{dt^2}[/itex]
[..]
These conditions characterize an inertial Cartesian coordinate system. They imply that an accelerometer at rest in that coordinate system will show no acceleration.
It may however show gravitational force.
But the other way around may not be true. An accelerometer at rest showing no acceleration doesn't imply that your coordinate system is an inertial Cartesian coordinate system.
Right - as illustrated by Langevin. Note that neither he or Einstein would call that "at rest" (without qualifier) in the context of SR.
stevendaryl
#177
Feb20-13, 03:05 PM
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Quote Quote by harrylin View Post
Not sure if you mean that correctly; a coordinate system of reference is not "objects".
I didn't say otherwise. But the path of an object that is not acted on by any force can be described using a coordinate system, and if that coordinate system happens to be an inertial Cartesian coordinate system, then the path [itex]x^i(t)[/itex] will satisfy

[itex]\dfrac{d^2 x^i}{dt^2} = 0[/itex]

For example the surface of the Earth is approximately a valid "Galilean" reference frame (neglecting its rotation and orbit), in which a cannon ball is acted upon by the force of gravity in accordance with Newton's second law.
I don't see how that contradicts what I said. Of course, if you think of gravity as a force, then you get one inertial coordinate system, and if you don't think of it as a force, then you get a different inertial coordinate system.

That is a classical textbook example of motion wrt to a valid reference system for classical mechanics, as your second point also stresses:

It may however show gravitational force.
So are you saying that whether or not an accelerometer correctly identifies an inertial frame depends on whether you consider gravity to be a force or not?
harrylin
#178
Feb20-13, 03:05 PM
P: 3,187
Quote Quote by DaleSpam View Post
Physical explanation of what? You never clarified exactly what you thought he meant by "gravitational field", and he wasn't explicit about it. Until you have defined your terms you are just giving physical explanations of flubnubitz.
It strikes me now that, incredible as this is to me, you actually have no idea what Einstein or his critics were discussing... Basically the whole clock paradox is flubnubitz for you.

I feel like a policeman at a crossroad who shouts at a motorist to stay put - the motorist speeds away and after a chase he's arrested. To his defence the guy states that he stayed put on his bike. and it turns out that he's an English teacher. What to think? Can such a lack of understanding really exist, or is the biker just trying to talk his way out of it by stripping the term from its obviously intended contextual meaning?
The firing of the rocket engine doesn't move the rest of the universe, the choice of coordinates does. The state of motion or rest is a coordinate-dependent quantity. Do you disagree that I can give any object any velocity profile I like simply by choosing the coordinates appropriately?
While I certainly agree with the second sentence, the first one is plain nonsense to me: your choice of coordinates cannot physically move (affect) the universe. "In rest" has in this context the physical meaning that the laws of nature are valid wrt the object; its state of motion is null. Only a force can change its state of motion. We can ascribe a different state of motion to the object by a different choice of reference system; however you can of course not affect my state of motion by changing your choice of coordinates - imaginations of our brains cannot physically affect objects. That is what Einstein clarified as I mentioned to you earlier already in this thread:

To be sure, the accelerated coordinate systems cannot be called upon as real causes for the field, an opinion that a jocular critic saw fit to attribute to me on one occasion.
harrylin
#179
Feb20-13, 03:17 PM
P: 3,187
Quote Quote by stevendaryl View Post
I didn't say otherwise. [..] Of course, if you think of gravity as a force, then you get one inertial coordinate system, and if you don't think of it as a force, then you get a different inertial coordinate system. [..]
Yes indeed, that's what I meant.
So are you saying that whether or not an accelerometer correctly identifies an inertial frame depends on whether you consider gravity to be a force or not?
I gave 4 ways of identifying an inertial frame in SR, incl. how to interpret an accelerometer in that context. Isn't that all very basic, elementary physics??

As this is probably deviating from GregAshmore topic, I'll limit myself to this.
DaleSpam
#180
Feb20-13, 03:18 PM
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Quote Quote by harrylin View Post
ADDENDUM: see
http://physicsforums.com/showthread.php?p=4117808
In post #190 I provided three operational ways with which such reference frames can be defined/determined.
See also post #200 there and a 4th defintion (by Einstein) in post #264:
http://physicsforums.com/showpost.ph...&postcount=264
In post #190 the first definition was "Newton defined it as in uniform straight line motion wrt the 'fixed stars'" which of course is a bad definition since the stars aren't fixed (which Newton couldn't have known at the time).

The second definition was "the definition of inertial motion at places far away from massive bodies". Simply being away from massive bodies doesn't give you an inertial frame, so this definition is incomplete and would require something like an accelerometer to complete it.

I didn't see a third definition.

The fourth definition was "coordinate systems relative to which sufficiently isolated, material points move in straight lines and uniformly". That is an equivalent definition to the accelerometer definition. Any way that you can think of determining whether or not a material point is "sufficiently isolated" winds up being an experiment to whether or not its proper acceleration is 0, hence an accelerometer.


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