Special Relativity: Going from A to B on a helix

[Passionflower]

OK, it's not familiar terminology for me. Doing a google search for "pace" and "helix" doesn't turn up anyone else using "pace" in these sense in the first two pages of results, are you sure this is standard usage? I also didn't know why you put (2*pi*c) after the word "pace", although now that you quote the mathworld page I see it does give parametric equations for the helix in which z=ct, and where "r is the radius of the helix and 2*pi*c is a constant giving the vertical separation of the helix's loops" (note that the c here has nothing to do with the speed of light, so it was a bit confusing for you to write 2*pi*c without explaining this!)

Yes it is called pace.
See for instance http://books.google.com/books?id=s9...v=onepage&q=helix pace cos sin length&f=false

Anyway, if the length of a coil is 2 light-seconds in the frame where the speed of the sphere is 0.6c, then the time for a rotation in this frame must be 2/0.6 = 3.333... seconds. So, in the sphere's rest frame the rotation time must be 3.333...*0.8 = 2.666... seconds. That means the tangential speed in the sphere's rest frame is (2*pi*0.3)/2.666... = 0.70686c...where did you get a tangential speed of 0.57c?

Ah, I did not apply the gamma.

And if the height of a coil is 2 light-seconds, then the length along a single coil must be sqrt(2^2 + (2*pi*0.3)^2) = 2.7483 light-seconds, for a total length of 6*2.7483 = 16.49 light-seconds. Thus if the time is 20 seconds in the frame where the sphere moves at 0.6c, the speed of the clock in this frame must be 16.49/20 = 0.82c.

Where on that page do you see anything about the length of the helix? The page I quoted said the length of a single coil with height H and radius R is given by L = sqrt(H^2 + (2 pi R)^2), do you think this formula is incorrect?

Well shouldn't it be L = 2 pi * sqrt( R^2 + (H/(2 pi))^2)?

 

[JesseM]

Well shouldn't it be L = 2 pi * sqrt( R^2 + (H/(2 pi))^2)?

No, why do you think so? Did you follow what I said about a helix being drawn on a cylinder, and since a cylinder has no intrinsic curvature you can "unwrap" it into a rectangle with the same height H as the cylinder and a width equal to the circumference of the cylinder 2*pi*R?

 

[Passionflower]

I took the formula from: http://mathworld.wolfram.com/Helix.html

It reads:

for t in [0,2$\pi$), where r is the radius of the helix and $2\pi c$ is a constant giving the vertical separation of the helix's loops.
...
The arc length is given by

$$s=\sqrt{r^2+c^2} t$$

 

[JesseM]

I took the formula from: http://mathworld.wolfram.com/Helix.html

It reads:

for t in [0,2$\pi$), where r is the radius of the helix and $2\pi c$ is a constant giving the vertical separation of the helix's loops.
...
The arc length is given by

$$s=\sqrt{r^2+c^2} t$$

This is when you parametrize the helix using the equations x=r*cos(t), y=r*sin(t), and z=c*t, where 2pi*c is the height of a single coil. So if you start at t=0, you reach a single coil when t=2pi, meaning the arc length of that coil is $$2\pi \sqrt{r^2 + c^2}$$
= $$\sqrt{(2\pi)^2}*\sqrt{r^2 + c^2}$$ = $$\sqrt{(2\pi r)^2 + (2 \pi c)^2$$ = $$\sqrt{(2 \pi r)^2 + H^2 }$$ where H is the height of a coil, the same formula I quoted from the other page, and which you can see is correct by considering the argument I gave about unwrapping a cylinder with a helix drawn on.

 

[Passionflower]

So then the formula I wrote down is correct you just used a different form.

 

[JesseM]

So then the formula I wrote down is correct you just used a different form.

Ah yes, I didn't catch that the formula you wrote down in post #36 was equivalent to my formula, I just assumed that since you had previously gotten the wrong value for the length of the helix, and seemed to be disagreeing with the formula I wrote down, that you were using a non-equivalent formula.

 

[Passionflower]

Looks like I got it down (finally), so now I can start comparing proper times along straight and helical paths.

It was certainly a good exercise for me. It turned out to be much simpler than I thought it was but that did not prevent me from going through a bunch of mistakes.

Thanks everybody for your help.