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Relativistic centripetal force

by yuiop
Tags: centripetal, force, relativistic
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DaleSpam
#181
May12-10, 08:54 AM
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Those metrics are incomplete, please provide the remaining terms. Especially all of the terms involving [itex]\omega[/itex]. I don't know why you are being so evasive about this.
starthaus
#182
May12-10, 08:59 AM
P: 1,568
Quote Quote by DaleSpam View Post
Those metrics are incomplete, please provide the remaining terms.
Here.
You DON'T NEED the remaining terms.
starthaus
#183
May12-10, 09:06 AM
P: 1,568
Quote Quote by kev View Post
I think Starthaus is talking about the derivation in his blog attachment titled "acceleration in rotating frames". It is incomplete, but I completed it for him in #100 and he said my final solution is correct.

I am not 100% sure it is.
Yes, this is the method that uses coordinate transformations in rotating frames. You did complete the calculations after a lot of prodding and prompting and correcting your calculus errors.
DaleSpam
#184
May12-10, 11:05 AM
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Quote Quote by starthaus View Post
Here.
You DON'T NEED the remaining terms.
The post you linked to has only Gron's metric which you are not using and the "standard" metric which is incomplete.

I find your evasiveness very disturbing. It is not as though it is unreasonable to ask for the metric.
starthaus
#185
May12-10, 11:17 AM
P: 1,568
Quote Quote by DaleSpam View Post
The post you linked to has only Gron's metric which you are not using and the "standard" metric which is incomplete.
Of course I am using the Gron metric. It is the same as the metric used in Rindler (9.26). If you read this thread rather than jumping in, you would have seen that.
You only need the first term (the coefficient for [tex]dt^2[/tex]) of the "standard" metric, you get the potential through a simple identification.

[tex]-ds^2=\left(1+\frac{2\Phi}{c^2 }\right)^{-1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)-c^2 \left(1+\frac{2\Phi}{c^2}\right)dt^2[/tex]

I find your evasiveness very disturbing. It is not as though it is unreasonable to ask for the metric.
This is your problem, the same exact approach is used in Rindler chapter 11. It can also be used very successfully in deriving the gravitational acceleration of a radial field (much more elegant than the covariant derivative solution)
DaleSpam
#186
May12-10, 01:10 PM
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Thanks for posting the "standard" metric.

Quote Quote by starthaus View Post
Of course I am using the Gron metric.
If you are using the Gron metric then your result is wrong. The [itex]\omega[/itex] he uses is [itex]\omega=d\theta/dt[/itex] (see eq 5.2). The only way your result can be right is if you are using a different metric where [itex]\omega=d\theta/d\tau[/itex].
starthaus
#187
May12-10, 01:43 PM
P: 1,568
Quote Quote by DaleSpam View Post
Thanks for posting the "standard" metric.
You are welcome.


If you are using the Gron metric then your result is wrong. The [itex]\omega[/itex] he uses is [itex]\omega=d\theta/dt[/itex] (see eq 5.2). The only way your result can be right is if you are using a different metric where [itex]\omega=d\theta/d\tau[/itex].
I am using the metric present in the Gron book, also present in the Rindler book.You need to look at Rindler, chapter 11.
If you use the weak field approximation, you get the result I showed you.
If you use the strong field approximation:

[tex]ds^2=e^{2\Phi/c^2} dt^2-....[/tex]

you get :

[tex]\Phi/c^2=\frac{1}{2}ln(1-\frac{r^2\omega^2}{c^2})[/tex]

[tex]\vec{F}=-grad(\Phi)=\frac{r\omega^2}{1-r^2\omega^2}[/tex]

Now:

[tex]\omega=\frac{d\theta}{dt}=\frac{d\theta}{d\tau}\frac{d\tau}{dt}=\omega_ {proper}\sqrt{1-r^2\omega^2}[/tex]

So:

[tex]\vec{F}=r\omega_{proper}^2[/tex]

Same exact result as in chapter 92, expression (97) page 247 in Moller ("The General Theory of Relativity")
DaleSpam
#188
May12-10, 05:09 PM
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Quote Quote by starthaus View Post
If you use the strong field approximation:

[tex]ds^2=e^{2\Phi/c^2} dt^2-....[/tex]

you get :

[tex]\Phi/c^2=\frac{1}{2}ln(1-\frac{r^2\omega^2}{c^2})[/tex]

[tex]\vec{F}=-grad(\Phi)=\frac{r\omega^2}{1-r^2\omega^2}[/tex]

Now:

[tex]\omega=\frac{d\theta}{dt}=\frac{d\theta}{d\tau}\frac{d\tau}{dt}=\omega_ {proper}\sqrt{1-r^2\omega^2}[/tex]

So:

[tex]\vec{F}=r\omega_{proper}^2[/tex]

Same exact result as in chapter 92, expression (97) page 247 in Moller ("The General Theory of Relativity")
OK, your results using the strong field approximation are correct and agree with the covariant derivative approach.

Is this the correct full expression for the strong-field approximation metric:
[tex]-ds^2=\left(e^{\frac{2\Phi}{c^2 }}\right)^{-1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)-c^2 \left(e^{\frac{2\Phi}{c^2}}\right)dt^2[/tex]
starthaus
#189
May12-10, 06:08 PM
P: 1,568
Quote Quote by DaleSpam View Post
OK, your results using the strong field approximation are correct and agree with the covariant derivative approach.

Is this the correct full expression for the strong-field approximation metric:
[tex]-ds^2=\left(e^{\frac{2\Phi}{c^2 }}\right)^{-1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)-c^2 \left(e^{\frac{2\Phi}{c^2}}\right)dt^2[/tex]
yes,it is
yuiop
#190
May13-10, 02:46 PM
P: 3,967
Quote Quote by starthaus View Post
The result is incorrect, a correct application of covariant derivatives (as shown here) gives the result [tex]a_0=r\omega^2[/tex].
Quote Quote by starthaus View Post
....
Now:

[tex]\omega=\frac{d\theta}{dt}=\frac{d\theta}{d\tau}\frac{d\tau}{dt}=\omega_ {proper}\sqrt{1-r^2\omega^2}[/tex]

So:

[tex]\vec{F}=r\omega_{proper}^2[/tex]
OK, you have effectively defined proper centripetal acceleration as:

[tex]a_0=r\omega_{proper}^2[/tex]

using your definition:

[tex]\omega=\frac{d\theta}{dt}=\omega_{proper}\sqrt{1-r^2\omega^2}[/tex]

This means in your coordinates, the coordinate acceleration is:

[tex]a=r\omega_{proper}^2(1-r^2\omega^2) = r\omega^2 = a_0 \gamma^{-2} [/tex]

and the fundemental relationship

[tex]a_0 = a\gamma^2[/tex]

between proper and coordinate centripetal acceleration, given by myself in #1 and later by Dalespam and others is correct.
starthaus
#191
May13-10, 03:57 PM
P: 1,568
Quote Quote by kev View Post
OK, you have effectively defined proper centripetal acceleration as:

[tex]a_0=r\omega_{proper}^2[/tex]

using your definition:
It is not "my" definition, it is the standard definition.


[tex]\omega=\frac{d\theta}{dt}=\omega_{proper}\sqrt{1-r^2\omega^2}[/tex]

This means in your coordinates, the coordinate acceleration is:

[tex]a=r\omega_{proper}^2(1-r^2\omega^2) = r\omega^2 = a_0 \gamma^{-2} [/tex]

and the fundemental relationship

[tex]a_0 = a\gamma^2[/tex]

between proper and coordinate centripetal acceleration, given by myself in #1 and later by Dalespam and others is correct.
The difference is that you did not derive anything (unless we factor in the stuff that I guided you to derive from the rotating frames transforms). Putting in results by hand doesn't count as "derivation".
Besides, post #3 shows that the naive transformation of force you attempted is wrong.
yuiop
#192
May14-10, 02:59 PM
P: 3,967
Quote Quote by starthaus View Post
The difference is that you did not derive anything (unless we factor in the stuff that I guided you to derive from the rotating frames transforms). Putting in results by hand doesn't count as "derivation".
Besides, post #3 shows that the naive transformation of force you attempted is wrong.
I never claimed that I derived the equations in #1. I merely stated them and pointed out some relationships between the equations.

The transformation of force that you object to in #3 is the perfectly standard Lorentz transformation of force and unless you are claiming the Lorentz transformations are wrong, there is no need for me to derive them.

You seem to think that angular acceleration might somehow make the transformation of centripetal force different from the transformation of linear transverse force, but if you understood the ramifications of the clock postulate you would know that acceleration has no effect on time dilation or length contraction and we only need to consider the instantaneous tangential velocity of a particle to work out the transformations of an accelerating particle. So all I did was apply the Lorentz transformation (which does not need deriving because it is an accepted standard result) and the clock postulate (which does not need deriving because it is a postulate supported by experimental evidence).

Your main objection seems to be that, even though I get the correct result, the method I used is not complicated enough.
starthaus
#193
May14-10, 04:59 PM
P: 1,568
Quote Quote by kev View Post

The transformation of force that you object to in #3 is the perfectly standard Lorentz transformation of force and unless you are claiming the Lorentz transformations are wrong, there is no need for me to derive them.
What has been explained to you is that the respective transformation is derived from the Lorentz transforms for translation. As such, it does NOT apply to rotation. Physics is not the process of mindless application of formulas cobbled from the internet.


So all I did was apply the Lorentz transformation (which does not need deriving because it is an accepted standard result) and the clock postulate (which does not need deriving because it is a postulate supported by experimental evidence).
You applied the inappropriate Lorentz transform. I do not expect you to understand that.


Your main objection seems to be that, even though I get the correct result, the method I used is not complicated enough.
My main objection is that you don't seem to know the domains of application of different formulas. I taught you how to derive the force transformation starting from the correct formulas: the transforms for rotating frames. You can't plug in willy-nilly the Lorentz transforms for translation.
yuiop
#194
May17-10, 04:19 AM
P: 3,967
Quote Quote by starthaus View Post
What has been explained to you is that the respective transformation is derived from the Lorentz transforms for translation. As such, it does NOT apply to rotation. Physics is not the process of mindless application of formulas cobbled from the internet.
I showed you how the clock hypothesis means that the Lorentz transforms for translation can be applied to rotation. You basically have the same misconception as many beginners to relativity, that think Special Relativity can not be applied to cases involving acceleration. My process of application is not ""mindless". I used logic applied to information from reliable sources and then got a second opinion about the results I obtained from the more knowledgeable members of this forum like Dalespam.


Quote Quote by starthaus View Post
You applied the inappropriate Lorentz transform. I do not expect you to understand that.
I did not. If I had I would have got the wrong result, but I did not. Everyone but you in this thread says the results I got in #1 are correct. All you have done is changed the definitions to make your results look different. This is like saying the speed of light is 299792.458 km/s and not 299792458 m/s, when in fact both answers are correct but are using different units. You fail to understand that our results are in agreement and they only differ in that my results are obtained much quicker and more directly.
starthaus
#195
May17-10, 08:36 AM
P: 1,568
Quote Quote by kev View Post
I showed you how the clock hypothesis means that the Lorentz transforms for translation can be applied to rotation.
The point is that you shouldn't try to apply the transforms derived for translation to a rotation exercise. This is precisely why special transforms have been derived for the case of rotation.
The type of hacks you keep attempting don't count as correct derivations , even if they produce the correct results by accident.



You basically have the same misconception as many beginners to relativity, that think Special Relativity can not be applied to cases involving acceleration.

LOL. You "forgot" the files that I wrote about accelerated motion in SR. You "forgot" the relativistic transforms for rotation that I tried to teach you.
I simply tried to teach you how to use the correct SR equations as they apply to rotation.
starthaus
#196
May17-10, 09:03 AM
P: 1,568
Quote Quote by kev View Post

there is something wrong with rhs of your two eqautions:

if

[tex]\frac{d}{dt}( m_0 v)=q v x b[/tex]

is true, then by the properties of simultaneous equations, it must follow that:

[tex]\frac{d}{dt}(\gamma m_0 v)= (q v x b) \gamma [/tex]
lol.


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