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## Implications of the statement "Acceleration is not relative"

 Quote by GregAshmore The problem is that he has not proposed any law at all. Or at least ... no other law of physics that I have learned looks like that.
I can make it look more like a standard law of physics quite easily:$$\frac{d p^{\mu}}{d\tau} = f^{\mu} - {\Gamma^{\mu}}_{\nu\lambda} u^{\nu} p^{\lambda}$$Where f is the sum of the real four-forces acting on the particle, p is the four-momentum, u is the four-velocity, τ is the proper time along the particle's worldline, and $\Gamma$ is the Christoffel symbols in the coordinate system in question.

 Quote by GregAshmore Furthermore, the statement borders on the delusional (I tried to find a neutral word; I could not; sorry) in that it denies the obvious causal connection between the firing of the rocket and the movement of the earth and stars (which I made bold in the quote of your post).
It may be an obvious connection, but it is not a causal connection, as I clearly demonstrated earlier. If you would like to actually address the points that I made instead of making a blatantly fallacious rebuttal then I would be glad to discuss it.

 Quote by GregAshmore The problem is not that the law of physics proposed by DaleSpam to explain the sudden movement of the Earth and stars at the firing of the rocket is not simple, or is not intuitive. The problem is that he has not proposed any law at all. Or at least, I do not recognize the statement "the movement of the Earth and stars was not caused by the firing of the rocket; it was caused by my choice of a certain set of coordinates" as a law of physics; certainly no other law of physics that I have learned looks like that.
That's because you probably have used inertial Cartesian coordinates in physics. With inertial Cartesian coordinates, the relationship between applied force and coordinate acceleration is, as Newton wrote:

$m \dfrac{dV^\mu}{d \tau} = F^\mu$

where $V^\mu$ is the 4-velocity.

When you use noninertial or curvilinear coordinates, the relationship between applied force and coordinate acceleration is more complicated:

$m \dfrac{dV^\mu}{d \tau} +$fictitious force terms $= F^\mu$

So even when the applied force $F^\mu$ is zero, the coordinate acceleration $\dfrac{dV^\mu}{d \tau}$ can be nonzero due to "fictitious force" terms. Examples of such fictitious forces are the "g forces" due to acceleration, the "centrifugal force" and the "coriolis force". These "forces" are not due to any kind of physical interaction, but are artifacts of your choice of coordinate systems.

 Quote by Nugatory That's done just to simplify the example. It's not a fundamental assumption of the explanation. With instantaneous turnarounds, the proper distance along each leg is just algebra: $\sqrt{\Delta t^2-\Delta x^2}$. If we don't assume instantaneous turnarounds, we have to evaluate some sort of line integral. It's fairly easy to prove that in the limit as the turnaround time approaches zero, the line integral reduces to the simple algebraic calculation, so we use the latter when the details of the turnaround aren't important to the problem at hand.
In order to get a point through, simplification is fundamental to explanations...

 Quote by DaleSpam As you make your $\delta \tau$ small the SR predicted accelerometer reading becomes large while the actual accelerometer reading remains 0. [..]
 Quote by DaleSpam [..]SR predicts a very large accelerometer reading during the turnaround, and real free falling accelerometers read 0.
SR doesn't predict that an accelerometer in free fall will indicate a large acceleration. If you insist in this thread instead of starting it as a topic, I'll start that topic for you.

 Quote by GregAshmore [..] But this much I believe to be undeniably true of a purely SR treatment of a scenario in which two bodies, one inertial and the other non-inertial, separate from each other and then approach to reunion: the non-inertial body must experience unbalanced force at the transition from separation to approach. [..]
SR uses the inertial frames of classical mechanics. If you know classical mechanics, then you certainly understand that if you accelerate freely in a gravitational field, your accelerometer will read approximately zero. If you don't know that, we can discuss this in the classical forum.

 Quote by GregAshmore [..] The problem is not that the law of physics proposed by DaleSpam to explain the sudden movement of the Earth and stars at the firing of the rocket is not simple, or is not intuitive. The problem is that he has not proposed any law at all. Or at least, I do not recognize the statement "the movement of the Earth and stars was not caused by the firing of the rocket; it was caused by my choice of a certain set of coordinates" as a law of physics; certainly no other law of physics that I have learned looks like that. Furthermore, the statement borders on the delusional (I tried to find a neutral word; I could not; sorry) in that it denies the obvious causal connection between the firing of the rocket and the movement of the earth and stars (which I made bold in the quote of your post). [..]
You evidently understand the question that Einstein attempted to address in 1918. Regretfully, few people who try to answer you understand the question. But in any case, nobody here gave support for the answer that Einstein gave, and neither does the physics FAQ.

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 Quote by harrylin SR doesn't predict that an accelerometer in free fall will indicate a large acceleration. If you insist in this thread instead of starting it as a topic, I'll start that topic for you.
The entire Langevin scenario is off-topic, but at this point it would take too much effort to split off and it doesn't make sense to do so, IMO.

Yes, SR does predict that. According to SR the proper acceleration is:$$a^{\mu}=\frac{d^2x^{\mu}}{d\tau^2}$$Where x is the worldline in an inertial frame and τ is the proper time along that worldline. That quantity is non-zero.

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 Quote by harrylin You evidently understand the question that Einstein attempted to address in 1918. Regretfully, few people who try to answer you understand the question. But in any case, nobody here gave support for the answer that Einstein gave, and neither does the physics FAQ.
It is hard to see how you can believe that there was any definite answer since you don't even know what he meant by the term "gravitational field".

 Quote by DaleSpam It is hard to see how you can believe that there was any definite answer since you don't even know what he meant by the term "gravitational field".
I even cited his answer several times.

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 Quote by harrylin I even cited his answer several times.
Yes, you did. But you never were able to identify what you thought he meant. Seems strange to claim that a quote is an answer when you don't claim to know what the quote is even referring to.

 Quote by DaleSpam [..] at this point it would take too much effort to split off [..]
A misunderstanding of something so basic and simple surely requires discussing - much more than the topic of this thread.

 Quote by DaleSpam Yes, you did. But you never were able to identify what you thought he meant. Seems strange to claim that a quote is an answer when you don't claim to know what the quote is even referring to.
Instead I claimed to know what he was referring to; however I don't try hard anymore to explain other people's explanations - that is usually futile.

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 Quote by harrylin I claimed to know what he was referring to
So, according to you, what exactly was he referring to with the term "gravitational field"? I believe it was the Christoffel symbols. You believe he was referring to _______.?

 Quote by Mentz114 See this topic http://www.physicsforums.com/showthread.php?t=490163
Thanks for that link. Very interesting. Lots of good stuff.

 Quote by DaleSpam So, according to you, what exactly was he referring to with the term "gravitational field"? I believe it was the Christoffel symbols. You believe he was referring to _______.?
Einstein definitely referred to a field of force that possesses the property of imparting the same acceleration to all bodies; according to his theory, the gravitation-field generates the accelerated motion.
- http://en.wikisource.org/wiki/The_Fo...ity-postulate.

 Quote by harrylin Einstein definitely referred to a field of force that possesses the property of imparting the same acceleration to all bodies; according to his theory, the gravitation-field generates the accelerated motion. - http://en.wikisource.org/wiki/The_Fo...ity-postulate.
Yes, in Einstein's original discussion of the twin paradox, with elevators and all that, the way he put it was something like this: (paraphrased)

If an elevator in outer space accelerates downward, the people in the elevator will feel an apparent upward force lifting them toward the ceiling. This is the inertial force due to being at rest in an accelerated frame.

If you have the same elevator falling in a gravitational field, it's accelerating downward, but the people feel no forces, because the upward inertial force is exactly canceled by the downward gravitational force.
I can't find an online reference to the original argument, but I remember reading it once, and it seemed that Einstein talked about freefall as not the absence of any forces, but as an exact balance between gravitational forces and inertial forces so that they cancel. From the standpoint of today, that seems like a convoluted way of describing it.

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