Relativistic centripetal force

[Dale]

That is correct, it is the coordinate acceleration in the original inertial frame transformed to the rotating frame. It is not the proper acceleration.

 

[starthaus]

That is correct, it is the coordinate acceleration in the original inertial frame transformed to the rotating frame.

Thank you.

It is not the proper acceleration.

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.

 

[Dale]

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.

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]

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.

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?

 

[Dale]

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?

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:
$$(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)$$

 

[starthaus]

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:
$$(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)$$

Why don't you spend some time understanding the Gron chapter on the subject? Once you do that, we can talk in a separate thread about how to calculate the proper acceleration. The value you calculated is wrong, there is no point in muddling this thread.

 

[Dale]

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?

 

[starthaus]

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.

Since you don't want to read the Gron chapter, I'll help you out:

The transformation between IF and RF is:

$$t=T$$
$$\theta=\Theta-\omega T$$

If you calculate the proper acceleration $$\frac{d^2\Theta}{dT^2}$$ you find out that it is equal to the coordinate acceleration in IF $$\frac{d^2\theta}{dt^2}$$ . Earlier in this thread (post 120), the coordinate acceleration in IF has been found to be $$R\omega^2$$. You can read the same exact value straight of Gron's line element. What does this tell you?

If that expression does not represent the worldline of a particle undergoing uniform circular motion in some inertial reference frame then what expression does?

The answer is found in the Nikolic paper. I provided a link to it several times.

 

[Dale]

OK, so according to Gron the worldline of a particle undergoing uniform circular motion:
In the rotating frame is given by:
$$(ct,r_0,\theta_0,0)$$

And in an inertial frame using cylindrical coordinates is given by:
$$(cT,r_0,\theta_0+\omega T,0)$$

Transforming this into the inertial frame using Cartesian coordinates we obtain:
$$(ct,\; r_0 \; cos(\theta_0 + \omega t),\; r_0 \; sin(\theta_0 + \omega t),\; 0)$$

Which is the same form as:
$$(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)$$

So Gron agrees with me wrt the form of the worldline.

 

[starthaus]

So Gron agrees with me wrt the form of the worldline.

...but not on your derivation for the acceleration. This is very simple stuff, why is so difficult for you to admit that you are wrong?

 

[Dale]

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:
$$\mathbf u=(\gamma c,\; -\gamma r \omega \; sin(\phi + \omega t),\; \gamma r \omega \; cos(\phi + \omega t),\; 0)$$
where $$\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}$$

 

[starthaus]

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:
$$\mathbf u=(\gamma c,\; -\gamma r \omega \; sin(\phi + \omega t),\; \gamma r \omega \; cos(\phi + \omega t),\; 0)$$
where $$\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}$$

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 $$\gamma=1$$

Another way of looking at it: if you calculate the proper acceleration in RF $$\frac{d^2\Theta}{dT^2}=\frac{d^2\theta}{dt^2}$$ . Earlier in this thread (post 120), the coordinate acceleration in IF has been found to be $$R\omega^2$$. What does this tell you?

 

[Dale]

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 $$\gamma=1$$

And I already told you:

Since,$$\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}$$ the condition $\gamma=1$ would imply $\omega=0$ which is not true in general.

I deliberately didn't do any Lorentz transforms nor did I look at any other reference frame.

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 $(1-\frac{r^2 \omega^2}{c^2})^{-1/2}$ resulted from a Lorentz transform simply because I used the symbol $\gamma$ to represent it and you failed to notice that I did not do any Lorentz transforms. The substitution $\gamma$ 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 $\gamma=1$ would imply a particle at rest in the inertial frame, not a particle undergoing uniform circular motion.

Another way of looking at it: if you calculate the proper acceleration in RF $$\frac{d^2\Theta}{dT^2}=\frac{d^2\theta}{dt^2}$$ . Earlier in this thread (post 120), the coordinate acceleration in IF has been found to be $$R\omega^2$$. What does this tell you?

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

 

[starthaus]

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:
$$\vec{F}=grad(\Phi)=-m\vec{r}\omega^2$$

 

[yuiop]

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:

$$\frac {d^2x}{dt'^2}$$ 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.

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 realize that.

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 learned 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.

 

[starthaus]

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

 

[yuiop]

:lol: You are jumping frames. Again.

I will quote Dalespam:

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

 

[starthaus]

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.

 

[atyy]

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]

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.

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.

 

[atyy]

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?

 

[George Jones]

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.

 

[starthaus]

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).

 

[yuiop]

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.

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.

 

[Dale]

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.

 

[starthaus]

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.

 

[Dale]

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.

 

[starthaus]

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.

 

[Dale]

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.

 

[starthaus]

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 $$a_0=r\omega^2$$.

 

[DrGreg]

When you compare different sources for a derivation of centripetal acceleration, you need to know what $\omega$ actually is.

In the Gron example above it is $d\phi/dt$. In the Wikipedia article it is $d\phi/d\tau$. The two are related by a factor of $\gamma = dt/d\tau$.

So in fact both derivations agree when you take that into account.

 

[starthaus]

When you compare different sources for a derivation of centripetal acceleration, you need to know what $\omega$ actually is.

In the Gron example above it is $d\phi/dt$. In the Wikipedia article it is $d\phi/d\tau$. The two are related by a factor of $\gamma = dt/d\tau$.

So in fact both derivations agree when you take that into account.

Gron (5.20) shows the potential to be :

$$\Phi=1/2r^2\omega^2$$

This gives an acceleration of :

$$r\omega^2$$

which is contradictory to his definition of $\omega=\frac{d\theta}{dt}$.

In the end, my derivation is correct and so is Dale's, we are differing on the definition of $\omega$. My two derivations reproduce the result from wiki, yet they use different approaches. Dale's claim that I cannot tell the proper from the coordinate acceleration is false.

 

[Dale]

In the end, my derivation is correct and so is Dale's, we are differing on the definition of $\omega$.

kev explicitly gave the definition of $\omega=d\theta/dt$ 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]

kev explicitly gave the definition of $\omega=d\theta/dt$ 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.

If it weren't for DrGreg, none of us would have figured out the difference.
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).

 

[Dale]

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).

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 $\omega = d\theta/d\tau$, what exactly are the transformations between your coordinates and an inertial coordinate system? And, what is the metric in your coordinate system?

Btw, the advantage of my approach is that it applies for any arbitrary worldline in any arbitrary coordinate system and will always give the correct proper acceleration. A derivation based on potentials only works for static spacetimes where potentials can be defined, and I am not sure that it works in any coordinates where the particle is not stationary. I don't see any advantage to it when the general approach is so straightforward.