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Implications of the statement Acceleration is not relative 
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#289
Feb2713, 07:31 PM

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So what makes a claim to be at rest "legitimate"? If the only requirement is the development of one set of equations that can be used by all atrest observers to correctly predict the behavior of physical systems, then I don't think I have any quarrel with relativity. (I'm past worrying about whether time can vary with velocity.) But if the requirement is that the observed universe be "real" to every atrest observer, then I'm not so sure. As I read the posts in this thread regarding what the observer in the resting rocket observes, I saw the word "fictitious" many times. I'm fine with fictitious quantities if they allow me to correctly calculate what is going to happen. (I'm an engineer, not a theoretician.) But if the proceedings seen by the observer in the resting rocket are fictitious, I am not comfortable affirming the statement that all coordinate systems are equally valid. It seems to me that a coordinate system that has the stars violating a law of physicstraveling faster than light speedhas a lesser validity than a coordinate system that has the stars behaving within the laws of physics. The two coordinate systems may be equally useful, depending on circumstances, but they are not equally valid: one of them is telling a lie about the stars. The premise is that every observer may legitimately consider himself to be at rest. For the premise to have any meaning at all, "at rest" must mean "absolutely at rest". (Einstein says, "permanently at rest".) Every observer develops the laws of physics on the assumption that he is at rest in absolute space. If all such observers are able to agree on one set of physical laws ("of the same form"), then the premise that motion is not absolute is proven to be true. If all observers are absolutely at rest, none are. But if just one observer can show that in his circumstances the laws of physics are unique, then the premise is falsified. If all we need to do is explain the relative motion of the rocket and Earth, I'm good with the force applied to the rocket. There is no reason at all to deal with the headache of induced gravitational fields in flat spacetime. But if the rocket must be allowed to rest, then the Earth must be moving by itself, and the headache must be endured. 


#290
Feb2713, 08:20 PM

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Once again, you appear to want to have it both ways; you want the "laws of physics" to look simple, but you want to be able to choose any coordinates you like. You can't have both of those things. 


#291
Feb2713, 08:22 PM

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#292
Feb2713, 09:22 PM

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 The rocket cannot consider themselves inertial. This means that the simplest form of laws of physics cannot be used. A more complex way of expressing laws, that is also true (by natural vanishing of extra terms) for inertial motion, can be used. Thus, if you choose the more complex expression, laws are the same for all motion; however, this in no way changes that inertial and noninertial motions are inequivalent.  The rocket is clearly at rest relative relative to itself. There is no escaping this, so it is clearly a legitimate thing to recognize. I am not going to address the rest of your post because I am curious to your reaction to the above, first. 


#293
Feb2713, 09:48 PM

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For example, Newton's laws of motion, when described using inertial Cartesian coordinates x and y, look like this: [itex]\dfrac{d^2 x}{dt^2} = \dfrac{1}{m} F^x[/itex] [itex]\dfrac{d^2 y}{dt^2} = \dfrac{1}{m} F^y[/itex] If we change to a new coordinate system [itex]r = \sqrt{x^2 + y^2}[/itex] [itex]\theta = arctan(\dfrac{y}{x})[/itex] then the same equations of motion look like this: [itex]\dfrac{d^2 r}{dt^2}  r (\dfrac{d\theta}{dt})^2 = \dfrac{1}{m} F^{r}[/itex] [itex]\dfrac{d^2 \theta}{dt^2} + \dfrac{2}{r} \dfrac{dr}{dt} \dfrac{d \theta}{dt} = \dfrac{1}{m} F^{\theta}[/itex] They're the same laws of motion, except written in different coordinates. The form of the laws change in different coordinates, but the physical content does not. Similarly, the rule of lightspeed is, in differential form: If an object travels a distance [itex]\delta x[/itex] in time [itex]\delta t[/itex], then [itex](c \delta t)^2  (\delta x)^2 \geq 0[/itex] That's what the law looks like in Cartesian coordinates. In general coordinates, the same law looks like this: [itex]g_{\mu \nu}\ \delta x^\mu\ \delta x^\nu \geq 0[/itex] (summed over all indices [itex]\mu[/itex] and [itex]\nu[/itex]) where [itex]g_{\mu \nu}[/itex] are the components of the metric tensor in the new coordinate system. In an inertial Cartesian coordinate system, the metric tensor has the simple form [itex]g_{tt} = c^2[/itex] [itex]g_{xx} = g_{yy} = g_{zz} = 1[/itex] (with all other components zero). The laws of physics look different in noninertial or curvilinear coordinates, but they have the same physical content. 


#294
Feb2813, 12:55 AM

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#295
Feb2813, 06:43 AM

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#296
Feb2813, 07:35 AM

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#297
Feb2813, 05:43 PM

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After the resting rocket twin fires his engine and sees the Earth accelerate away, he eventually sees his target star approach. He can measure the distance to the star at intervals and verify that it is indeed getting closer. You have said repeatedly that the motion of the Earth is caused by the choice of coordinates, independent of the firing of the rocket. I give you the very same scenario, except without an engine in the rocket. Can you make the very same events happen? Can you cause the Earth and stars to move by selecting a certain coordinate system? 


#298
Feb2813, 07:44 PM

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What you can't do is get them moving at *different* velocities just by changing coordinate systems. But nobody was claiming that you could; that is, nobody was claiming that you could change their *relative* velocities just by changing coordinates. That's what takes the rocket engine; which is why relative motion is "physical" in a way that "motion" by itself is not. 


#299
Feb2813, 07:44 PM

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#300
Feb2813, 07:46 PM

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I'm going to skip replying to some of your responses. I need more time, and probably more study, to give a good answer. As to why the rocket twin insists on claiming that he is at rest, I guess you'ld have to ask Einstein. I certainly never thought to make an issue of it until I read his book. If no plausible explanation for the workings of the gravitational field can be given, the field must be considered a pure fiction; adhoc handwaving. In that case, the absoluteness of acceleration is not removed, at least with respect to SR. Personally, I have no stake in the argument. I don't care if there is an absoluteness to acceleration. But the issue was important to Einstein, and having come this far, I'd like to be able to evaluate whether he succeeded in eliminating the problem. 


#301
Feb2813, 07:59 PM

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#302
Feb2813, 08:09 PM

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#303
Feb2813, 08:11 PM

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#304
Feb2813, 08:48 PM

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#305
Mar113, 04:04 AM

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Mar113, 05:29 PM

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