- #141
Dale
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That is correct, it is the coordinate acceleration in the original inertial frame transformed to the rotating frame. It is not the proper acceleration.
DaleSpam said:That is correct, it is the coordinate acceleration in the original inertial frame transformed to the rotating frame.
It is not the proper acceleration.
That equation was not the line element, it was the worldline of the particle. A line element is a scalar, the worldline is a four-vector parameterized by some arbitrary scalar.starthaus said:May I suggest that you open a different thread that discusses how to obtain the proper acceleration from the correct line element (see Gron's book)? The line element "s" that you used in your derivation is incorrect.
DaleSpam said:That equation was not the line element, it was the worldline of the particle. A line element is a scalar, the worldline is a four-vector parameterized by some arbitrary scalar.
Do you or do you not agree that the worldline of a particle undergoing uniform circular motion in some inertial reference frame is given by:starthaus said:You volunteered to study the chapter on rotating frames in Gron's book. Why don't we reprise this discussion in a different thread, once you have studied the chapter?
DaleSpam said:Do you or do you not agree that the worldline of a particle undergoing uniform circular motion in some inertial reference frame is given by:
[tex](ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)[/tex]
DaleSpam said:You are really starting to irritate me with your repeated assertions that it is wrong followed by a completely evasive non-answer every time your assertion is challenged. You have stated in this thread that my derivation is wrong so defend your statement in this thread and stop trying to weasel your way out of it.
If that expression does not represent the worldline of a particle undergoing uniform circular motion in some inertial reference frame then what expression does?
DaleSpam said:So Gron agrees with me wrt the form of the worldline.
DaleSpam said:Why should I admit I am wrong when you have shown no evidence to support that assertion? So far the only "error" you have pointed out is that you disagreed with my expression for the worldline. Now we find that it is not, in fact, an error and that you agree with my expression for the worldline in a standard inertial frame.
So, given that we now agree on the expression for the worldline, do you agree or disagree with my expression for the four-velocity in the standard inertial frame:
[tex]\mathbf u=(\gamma c,\; -\gamma r \omega \; sin(\phi + \omega t),\; \gamma r \omega \; cos(\phi + \omega t),\; 0)[/tex]
where [tex]\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}[/tex]
And I already told you:starthaus said:I think that I have already told you several times that the error occurs at the last step of your derivation, when you need to use [tex]\gamma=1[/tex]
I think that you get hung up on symbols instead of paying attention to what they mean. For instance, you thought that my expression for the worldline of the particle was an expression for the line element simply because I used the symbol s to represent it, and you failed to notice that I had clearly stated that it was the worldline and you also failed to notice that it was a four-vector and not a scalar so it clearly was not the line element. Now, I suspect that you think that the expression [itex](1-\frac{r^2 \omega^2}{c^2})^{-1/2}[/itex] resulted from a Lorentz transform simply because I used the symbol [itex]\gamma[/itex] to represent it and you failed to notice that I did not do any Lorentz transforms. The substitution [itex]\gamma[/itex] is only there because I did not want to write a lot of nested fractions. I don't know why you have such a mental block and cannot realize that in my notation [itex]\gamma=1[/itex] would imply a particle at rest in the inertial frame, not a particle undergoing uniform circular motion.DaleSpam said:Since,[tex]\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}[/tex] the condition [itex]\gamma=1[/itex] would imply [itex]\omega=0[/itex] which is not true in general.
I deliberately didn't do any Lorentz transforms nor did I look at any other reference frame.
This tells me that you don't know the difference between proper acceleration and coordinate acceleration.starthaus said:Another way of looking at it: if you calculate the proper acceleration in RF [tex]\frac{d^2\Theta}{dT^2}=\frac{d^2\theta}{dt^2}[/tex] . Earlier in this thread (post 120), the coordinate acceleration in IF has been found to be [tex]R\omega^2[/tex]. What does this tell you?
[tex]\frac {d^2x}{dt'^2}[/tex] is not a physically meaningless entity. It is the PROPER centripetal acceleration which is what is measured by an accelerometer and therefore very physically meaningful. Yes, I do mean x without the prime and for the sake of this discussion we will take x to be the direction pointing to the centre of the circle.starthaus said:[tex]\frac {d^2x}{dt'^2}[/tex] is physically a meaningless entity, you are mixing frames. Can you write down the correct definition for [tex]a'[/tex]?
Well you need to learn to frame mix, because PROPER quantities (the quantities all observers agree on and coordinate independent are frame mixed entities.kev said:You effectively derive:
[tex]\frac{d^2x}{dt^2} = \gamma^{-2}\frac{d^2x}{dt'^2}[/tex]
in expression (6) of your attachment, although you probably don't realize that.
starthaus said:I would never write such frame-mixing nonsese.
kev said:Hi Starthaus,
I have looked back over your previous posts and I think I have now identified the root of all your misunderstandings and confusion.
all you have done in your blog document is transformed from one coordinate system to another,
but what you have NOT done is found the PROPER centripetal acceleration which the quantity everyone else in this thread is talking about.
These are the quotes that identify your confusion:[tex]\frac {d^2x}{dt'^2}[/tex] is not a physically meaningless entity.
It is the PROPER centripetal acceleration which is what is measured by an accelerometer and therefore very physically meaningful. Yes, I do mean x without the prime and for the sake of this discussion we will take x to be the direction pointing to the centre of the circle.
starthaus said::lol: You are jumping frames. Again.
DaleSpam said:This tells me that you don't know the difference between proper acceleration and coordinate acceleration.
kev said:I will quote Dalespam:
atyy said:Eqn 9.26 of http://books.google.com/books?id=MuuaG5HXOGEC&dq=Wolfgang+Rindler&source=gbs_navlinks_s gives the centrifugal force as gamma.gamma.r.w.w. , which seems to be in agreement with DaleSpam.
starthaus said:Correct. Rindler uses a different line element (9.26) than Gron. He also uses a completely different definition of the potential (9.13). The line element determines the force as I have shown in post 154, so you'll need to take your pick as to which one to choose.
starthaus said:Correct. Rindler uses a different line element (9.26) than Gron.
atyy said:Wouldn't one need to use the standard form for a stationary metric rather than a static one?
atyy said:Eqn 9.26 of http://books.google.com/books?id=MuuaG5HXOGEC&dq=Wolfgang+Rindler&source=gbs_navlinks_s gives the centrifugal force as gamma.gamma.r.w.w. , which seems to be in agreement with DaleSpam.
starthaus said:Correct. Rindler uses a different line element (9.26) than Gron. He also uses a completely different definition of the potential (9.13). The line element determines the force as I have shown in post 154, so you'll need to take your pick as to which one to choose.
You are repeating the same error as before: proper acceleration is equal to four-acceleration for [tex]\gamma[/tex]=1.DaleSpam said:The quantity starthaus derived is a frame-variant coordinate acceleration, not the frame-invariant proper acceleration.
The mistake is yours, the magnitude of the proper acceleration is equal to the norm of the four-acceleration in all reference frames and regardless of gamma. This should be obvious since the norm of the four-acceleration is a frame invariant scalar. You simply don't know what proper acceleration is. Also, the use of gamma in the metric is Gron's convention, not mine. You cannot seek to rely on Gron as an authority on the metric in the rotating system and then reject his metric in the rotating system.starthaus said:You are repeating the same error as before: proper acceleration is equal to four-acceleration for [tex]\gamma[/tex]=1.
Post #154 shows that your claim is not true.
Wrong, I calculated the force directly off the line element. This is standard procedure, you can see it in any book.DaleSpam said:In post 154 you once again calculated the coordinate acceleration and erroneously called it proper acceleration. All post 154 shows is that you don't know the difference between the two.
Yes, it is a standard procedure for calculating the coordinate acceleration, not the proper acceleration. That is the part that you just don't seem to understand.starthaus said:Wrong, I calculated the force directly off the line element. This is standard procedure, you can see it in any book.
DaleSpam said:So, from Gron (p. 89), using the convention that spacelike intervals squared are positive, in a rotating reference frame with cylindrical coordinates given by:
[tex](t,r,\theta,z)[/tex]
The line element is:
[tex]ds^2 = -\gamma^{-2} c^2 dt^2 + dr^2 + 2 r^2 \omega dt d\theta + r^2 d\theta^2 + dz^2[/tex]
where
[tex]\gamma = (1 - r^2 \omega^2/c^2)^{-1/2}[/tex]
And the metric tensor is:
[tex]\mathbf g =
\left(
\begin{array}{cccc}
-\gamma ^{-2} c^2 & 0 & r^2 \omega & 0 \\
0 & 1 & 0 & 0 \\
r^2 \omega & 0 & r^2 & 0 \\
0 & 0 & 0 & 1
\end{array}
\right)
[/tex]
NB [itex]\gamma[/itex] is given by Gron as part of the line element and metric for the rotating frame in equations 5.3-5.5, and is only equal to 1 for the special case of [itex]\omega=0[/itex].
Finally, some of the Christoffel symbols in the rotating reference frame are non-zero (Gron p. 149). Specifically:
[tex]\Gamma^{r}_{tt}=-\omega^2r[/tex]
[tex]\Gamma^{r}_{\theta \theta}=-r[/tex]
[tex]\Gamma^{r}_{\theta t}=\Gamma^{r}_{t \theta}=-\omega r[/tex]
[tex]\Gamma^{\theta}_{rt}=\Gamma^{\theta}_{tr}=\omega/r[/tex]
[tex]\Gamma^{\theta}_{\theta r}=\Gamma^{\theta}_{r \theta}=1/r[/tex]
Now, the worldline of a particle starting on the x-axis at t=0 and undergoing uniform circular motion at angular velocity [itex]\omega[/itex] in the x-y plane in an inertial frame is given by the following expression in the rotating frame:
[tex]\mathbf X = (t,r_0,0,0)[/tex]
From this we can derive the four-velocity in the rotating frame as follows:
[tex]\mathbf U = \frac{d \mathbf X}{d \tau} = i c \frac{d \mathbf X}{ds} = i c \frac{d \mathbf X}{dt} \frac{dt}{ds} = i c \; (1,0,0,0) \; \frac{1}{\sqrt{-\gamma^{-2} c^2}} = (\gamma,0,0,0)[/tex]
The norm of the four-velocity is given by:
[tex]||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = -c^2[/tex]
So this agrees with my previous results so far as expected since the norm is a frame invariant quantity.
Now we can derive the four-acceleration in the rotating frame as follows:
[tex]A^{\mu}=\frac{DU^{\mu}}{d\tau}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}[/tex]
[tex]\frac{d \mathbf U}{d\tau}= i c\frac{d \mathbf U}{ds}= i c\frac{d \mathbf U}{dt}\frac{dt}{ds}= i c \; (0,0,0,0) \; \frac{dt}{ds}=(0,0,0,0)[/tex]
There is only one non-zero component of:
[tex]\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma^{r}_{tt}U^{t}U^{t}=-\gamma^2 r \omega^2[/tex]
So, substituting back in we obtain the four-acceleration in the rotating frame:
[tex]\mathbf A = (0,-\gamma^2 r \omega^2,0,0)[/tex]
The norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by:
[tex]||\mathbf A||^2=A_{\mu} A^{\mu}= g_{\mu\nu} A^{\nu} A^{\mu} = \gamma^4 r^2 \omega^4[/tex]
So this also agrees with my previous results as expected since the norm is a frame invariant quantity.
In summary, if you use four-vectors it does not matter which frame you do the calculations in, they will all agree on the norms. The magnitude of the proper acceleration, which is equal to the norm of the four-acceleration, is a frame-invariant quantity, and it is given by the above expression. The quantity starthaus derived is a frame-variant coordinate acceleration, not the frame-invariant proper acceleration.
DrGreg said:When you compare different sources for a derivation of centripetal acceleration, you need to know what [itex]\omega[/itex] actually is.
In the Gron example above it is [itex]d\phi/dt[/itex]. In the Wikipedia article it is [itex]d\phi/d\tau[/itex]. The two are related by a factor of [itex]\gamma = dt/d\tau[/itex].
So in fact both derivations agree when you take that into account.
kev explicitly gave the definition of [itex]\omega=d\theta/dt[/itex] in the very first equation of the very first post. If you were going to use a different definition than everyone else was using then it would have been quite helpful for you to post your definition instead of assuming that everyone on the forum has mystical psychic powers and could read your mind.starthaus said:In the end, my derivation is correct and so is Dale's, we are differing on the definition of [itex]\omega[/itex].
DaleSpam said:kev explicitly gave the definition of [itex]\omega=d\theta/dt[/itex] in the very first equation of the very first post. If you were going to use a different definition than everyone else was using then it would have been quite helpful for you to post your definition instead of assuming that everyone on the forum has mystical psychic powers and could read your mind.
It isn't that hard to understand. The coordinate transformations are easy in terms of coordinate time, they would be much more difficult in terms of proper time. In fact, with your alternate definition of [itex]\omega = d\theta/d\tau[/itex], what exactly are the transformations between your coordinates and an inertial coordinate system? And, what is the metric in your coordinate system?starthaus said:It is hard to understand why one would define the [proper acceleration as a function of coordinate angular speed when proper angular speed is the natural choice (and produces a much more elegant expression).