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## Relativistic centripetal force

Sigh,

Let's try a different way. Earlier I mentioned that you can read all the necessary info straight off the line element. Now, if you look at Gron's line element (5.5) :

$$ds^2=(1-\frac{r^2\omega^2}{c^2})(cdt)^2+2r^2\omega dt d\theta+(r d \theta)^2+z^2$$

and you compare this against the standard metric:

$$ds^2=(1+\frac{2\Phi}{mc^2})(cdt)^2+....$$

You get the potential $$\Phi=-1/2mr^2\omega^2$$
This gives you immediately the force:
$$\vec{F}=grad(\Phi)=-m\vec{r}\omega^2$$

Blog Entries: 6
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. Dalespam is correctin that 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:

 Quote by starthaus $$\frac {d^2x}{dt'^2}$$ is physically a meaningless entity, you are mixing frames. Can you write down the correct definition for $$a'$$?
$$\frac {d^2x}{dt'^2}$$ 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.

Quote by kev
You effectively derive:

$$\frac{d^2x}{dt^2} = \gamma^{-2}\frac{d^2x}{dt'^2}$$

in expression (6) of your attachment, although you probably don't realise that.
 Quote by starthaus I would never write such frame-mixing nonsese.
Well you need to learn to frame mix, because PROPER quantities (the quantities all observers agree on and coordinate independent are frame mixed entities.

Relativity 101 especially for Starthaus:

In Minkowski spacetime take two inertial frames S and S' with relative linear velocity.

Let there be a rod in S such that the two ends of the rod x2 and x1 are both at rest in frame S. The quantity x2-x1 is the proper length of the rod. (dx)

A stationary clock in S' moves from x1 to x2 in time t2'-t1' as measured by the clock. The time interval t2'-t1' is the proper time of the clock. (dt')

The PROPER velocity of the clock is dx/dt'.

Proper velocity is a "frame mixed" quantity. If you never write "frame mixed" quantities then it is about time you learnt to use them, as they are very useful and at the heart of all four vectors.

dx/dt is the coordinate velocity of the clock in frame S.

dx'/dt' is the coordinate velocity of the clock in frame S' and equal to zero in this case.

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 Quote by kev 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.
There is no confusion and no misunderstanding. I have taken you step by step through all your errors in doing the coordinate transformation from IF to RF. This is what I have corrected in your derivation starting from post #3 : you can't apply the Lorentz transforms derived for translation to a rotation problem.

 all you have done in your blog document is transformed from one coordinate system to another,
....and this is precisely what I have been telling you all along I am doing. I am showing you how to use the appropiate Lorentz transforms. I have been telling you the same exact thing from post #3.

 but what you have NOT done is found the PROPER centripetal acceleration which the quantity everyone else in this thread is talking about.
This is a separate issue, to be discussed only after I corrected all your errors.
You are also incorrect, in the last posts I have provided several methods that produce the proper acceleration or the centripetal force. You only had to look up, at post #154.

 These are the quotes that identify your confusion: $$\frac {d^2x}{dt'^2}$$ is not a physically meaningless entity.
:lol: You are jumping frames. Again.

 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.
:lol:

Sorry, I had to snip your "lesson" , you are in no position to offer lessons. You wrote so many incorrect things that it prompted me to write a detailed followup of my file on accelerated motion in SR. I have just posted it under "Accelerated Motion in SR part II". You have a lot to learn

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 Quote by starthaus :lol: You are jumping frames. Again.
I will quote Dalespam:

 Quote by DaleSpam This tells me that you don't know the difference between proper acceleration and coordinate acceleration.

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 Quote by kev I will quote Dalespam:
Ad-hominems make very poor scientific arguments :lol:
It is especially bad form when I spent so much time teaching you the appropiate physical formalism and correcting your calculus errors.
 Recognitions: Science Advisor Eqn 9.26 of http://books.google.com/books?id=Muu...gbs_navlinks_s gives the centrifugal force as gamma.gamma.r.w.w. , which seems to be in agreement with DaleSpam.

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 Quote by atyy Eqn 9.26 of http://books.google.com/books?id=Muu...gbs_navlinks_s gives the centrifugal force as gamma.gamma.r.w.w. , which seems to be in agreement with DaleSpam.
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.

Recognitions:
 Quote by starthaus 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.
Wouldn't one need to use the standard form for a stationary metric rather than a static one?

Mentor
 Quote by starthaus Correct. Rindler uses a different line element (9.26) than Gron.
No, line elements (5.3) from Gron and Hervik and (9.26) from Rindler are exactly the same.

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 Quote by atyy Wouldn't one need to use the standard form for a stationary metric rather than a static one?
Yes, this is what both authors use. Rindler rearranges his metric in a strange way, in order to line up with his definition (9.13).

Blog Entries: 6
Quote by atyy
Eqn 9.26 of http://books.google.com/books?id=Muu...gbs_navlinks_s gives the centrifugal force as gamma.gamma.r.w.w. , which seems to be in agreement with DaleSpam.
 Quote by starthaus 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.
Do you understand proper acceleration is independent of choice of cooordinate system?

If Rindler states the proper centripetal acceleration is $\gamma^2R\omega^2$ then the proper centripetal acceleration is $\gamma^2R\omega^2$ in any coordinate system including the one used by Gron, in agreement with Dalespam, myself, Jorrie, Pervect etc.
 Mentor So, from Gron (p. 89), using the convention that spacelike intervals squared are positive, in a rotating reference frame with cylindrical coordinates given by: $$(t,r,\theta,z)$$ The line element is: $$ds^2 = -\gamma^{-2} c^2 dt^2 + dr^2 + 2 r^2 \omega dt d\theta + r^2 d\theta^2 + dz^2$$ where $$\gamma = (1 - r^2 \omega^2/c^2)^{-1/2}$$ And the metric tensor is: $$\mathbf g = \left( \begin{array}{cccc} -\gamma ^{-2} c^2 & 0 & r^2 \omega & 0 \\ 0 & 1 & 0 & 0 \\ r^2 \omega & 0 & r^2 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$ NB $\gamma$ 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 $\omega=0$. Finally, some of the Christoffel symbols in the rotating reference frame are non-zero (Gron p. 149). Specifically: $$\Gamma^{r}_{tt}=-\omega^2r$$ $$\Gamma^{r}_{\theta \theta}=-r$$ $$\Gamma^{r}_{\theta t}=\Gamma^{r}_{t \theta}=-\omega r$$ $$\Gamma^{\theta}_{rt}=\Gamma^{\theta}_{tr}=\omega/r$$ $$\Gamma^{\theta}_{\theta r}=\Gamma^{\theta}_{r \theta}=1/r$$ Now, the worldline of a particle starting on the x axis at t=0 and undergoing uniform circular motion at angular velocity $\omega$ in the x-y plane in an inertial frame is given by the following expression in the rotating frame: $$\mathbf X = (t,r_0,0,0)$$ From this we can derive the four-velocity in the rotating frame as follows: $$\mathbf U = \frac{d \mathbf X}{d \tau} = i c \frac{d \mathbf X}{ds} = i c \frac{d \mathbf X}{dt} \frac{dt}{ds} = i c \; (1,0,0,0) \; \frac{1}{\sqrt{-\gamma^{-2} c^2}} = (\gamma,0,0,0)$$ The norm of the four-velocity is given by: $$||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = -c^2$$ 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: $$A^{\mu}=\frac{DU^{\mu}}{d\tau}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu \lambda}U^{\nu}U^{\lambda}$$ $$\frac{d \mathbf U}{d\tau}= i c\frac{d \mathbf U}{ds}= i c\frac{d \mathbf U}{dt}\frac{dt}{ds}= i c \; (0,0,0,0) \; \frac{dt}{ds}=(0,0,0,0)$$ There is only one non-zero component of: $$\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma^{r}_{tt}U^{t}U^{t}=-\gamma^2 r \omega^2$$ So, substituting back in we obtain the four-acceleration in the rotating frame: $$\mathbf A = (0,-\gamma^2 r \omega^2,0,0)$$ The norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by: $$||\mathbf A||^2=A_{\mu} A^{\mu}= g_{\mu\nu} A^{\nu} A^{\mu} = \gamma^4 r^2 \omega^4$$ 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.

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 Quote by DaleSpam The quantity starthaus derived is a frame-variant coordinate acceleration, not the frame-invariant proper acceleration.
You are repeating the same error as before: proper acceleration is equal to four-acceleration for $$\gamma$$=1.
Post #154 shows that your claim is not true. One can read the potential straight off the line element and calculate the force through a simple derivative.
Anyway, this is going nowhere , so it is time for me to give up. I wrote another attachment that corrects all of kev misconceptions about proper acceleration.

Mentor
 Quote by starthaus You are repeating the same error as before: proper acceleration is equal to four-acceleration for $$\gamma$$=1. Post #154 shows that your claim is not true.
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.

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.

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 Quote by DaleSpam 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.
Wrong, I calculated the force directly off the line element. This is standard procedure, you can see it in any book.
Anyway, this is going nowhere, let's agree to disagree.

Mentor
 Quote by starthaus Wrong, I calculated the force directly off the line element. This is standard procedure, you can see it in any book.
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.

Note that the line element depends on the choice of coordinates as does the force you calculated. The proper acceleration does not. So the force you calculated cannot possibly be the proper acceleration.

Blog Entries: 9
 Quote by DaleSpam So, from Gron (p. 89), using the convention that spacelike intervals squared are positive, in a rotating reference frame with cylindrical coordinates given by: $$(t,r,\theta,z)$$ The line element is: $$ds^2 = -\gamma^{-2} c^2 dt^2 + dr^2 + 2 r^2 \omega dt d\theta + r^2 d\theta^2 + dz^2$$ where $$\gamma = (1 - r^2 \omega^2/c^2)^{-1/2}$$ And the metric tensor is: $$\mathbf g = \left( \begin{array}{cccc} -\gamma ^{-2} c^2 & 0 & r^2 \omega & 0 \\ 0 & 1 & 0 & 0 \\ r^2 \omega & 0 & r^2 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$ NB $\gamma$ 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 $\omega=0$. Finally, some of the Christoffel symbols in the rotating reference frame are non-zero (Gron p. 149). Specifically: $$\Gamma^{r}_{tt}=-\omega^2r$$ $$\Gamma^{r}_{\theta \theta}=-r$$ $$\Gamma^{r}_{\theta t}=\Gamma^{r}_{t \theta}=-\omega r$$ $$\Gamma^{\theta}_{rt}=\Gamma^{\theta}_{tr}=\omega/r$$ $$\Gamma^{\theta}_{\theta r}=\Gamma^{\theta}_{r \theta}=1/r$$ Now, the worldline of a particle starting on the x axis at t=0 and undergoing uniform circular motion at angular velocity $\omega$ in the x-y plane in an inertial frame is given by the following expression in the rotating frame: $$\mathbf X = (t,r_0,0,0)$$ From this we can derive the four-velocity in the rotating frame as follows: $$\mathbf U = \frac{d \mathbf X}{d \tau} = i c \frac{d \mathbf X}{ds} = i c \frac{d \mathbf X}{dt} \frac{dt}{ds} = i c \; (1,0,0,0) \; \frac{1}{\sqrt{-\gamma^{-2} c^2}} = (\gamma,0,0,0)$$ The norm of the four-velocity is given by: $$||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = -c^2$$ 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: $$A^{\mu}=\frac{DU^{\mu}}{d\tau}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu \lambda}U^{\nu}U^{\lambda}$$ $$\frac{d \mathbf U}{d\tau}= i c\frac{d \mathbf U}{ds}= i c\frac{d \mathbf U}{dt}\frac{dt}{ds}= i c \; (0,0,0,0) \; \frac{dt}{ds}=(0,0,0,0)$$ There is only one non-zero component of: $$\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma^{r}_{tt}U^{t}U^{t}=-\gamma^2 r \omega^2$$ So, substituting back in we obtain the four-acceleration in the rotating frame: $$\mathbf A = (0,-\gamma^2 r \omega^2,0,0)$$ The norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by: $$||\mathbf A||^2=A_{\mu} A^{\mu}= g_{\mu\nu} A^{\nu} A^{\mu} = \gamma^4 r^2 \omega^4$$ 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.
The result is incorrect, a correct application of covariant derivatives (as shown here) gives the result $$a_0=r\omega^2$$.