Special Relativity: Going from A to B on a helix

[Dale]

So a helix in (a space diagram in say frame A) is the same as a helix in a (spacetime diagram in say frame B) is the same as uniform circular motion (in a space diagram in frame B).

If that is not what Dalespam meant, then my apologies in advance for paraphrasing. At least that is what I think he meant, but not 100% sure.

That is what I meant. Thanks for clarifying it!

 

[JesseM]

Ok, that makes sense.

However I am still not out of the woods:

Let's assume we have a rotating test ball of radius r moving from A to B (in the direction of the axis of rotation), there is one clock in the center and one on the edge of the ball so that it circles around the axis of rotation over the center of the ball.

Distance A-B: 12
Velocity of the ball: 0.6
Radius of the ball: 0.3
Pace (2c\pi) of the helix: 2

So we got:
Coordinate time A-B: 20
Proper time ball (clock at the center): 16
Number of full turns between A and B: 6
Length of the helix: 16.48
Tangential speed:0.57
Helix velocity: 0.82
Gamma: 1.77
Proper time: 11.32

If the pace of rotation is one rotation every 2 seconds (i.e. 0.5 rotations per second) in the frame where it's moving at 0.6c, then in its own rest frame the ball rotates at 0.5/0.8 = 0.625 rotations per second, so the tangential velocity in the rest frame of the ball is (2*pi*0.3)*0.625 = 1.178 light-seconds per second, so it's still faster than light (yuiop gave a speed of 1.18c in post #24 above so I think we did the calculations the same way). If you instead wanted the pace to be 0.5 rotations per second in the rest frame of the ball, in that case the tangential velocity would be 0.94c in this frame, but in that case the pace in the frame where the ball is moving at 0.6c would only be 0.5*0.8 = 0.4 rotations per second, or one rotation every 2.5 seconds. So if the coordinate time is 20 seconds to get from A to B, there will be 8 full rotations. So, the height of each coil would be 12/8 = 1.5 light-seconds, so according to this page the length of each coil must be sqrt(1.5^2 + (2*pi*0.3)^2) = 2.409, so the total length of the helix is 8*2.409 = 19.27. And if the time in the rest frame of A and B is 20 seconds, the speed of the clock in this frame must be 19.27/20 = 0.96c.

 

[Dale]

Passionflower, the worldline of the clock on the end of the ball in the "circle" frame is given by:
\left( ct, 0, r \; cos(t\omega), r \; sin(t\omega) \right)

Then Lorentz transforming into the "helix" frame gives:

\left( ct\gamma, vt\gamma, r \; cos(t\omega), r \; sin(t\omega) \right) = \left( ct', vt', r \; cos\left(\frac{t'\omega}{\gamma}\right), r \; sin\left(\frac{t'\omega}{\gamma}\right) \right)

 

[Passionflower]

If the pace of rotation is one rotation every 2 seconds (i.e. 0.5 rotations per second) in the frame where it's moving at 0.6c, then in its own rest frame the ball rotates at 0.5/0.8 = 0.625 rotations per second, so the tangential velocity in the rest frame of the ball is (2*pi*0.3)*0.625 = 1.178 light-seconds per second, so it's still faster than light (yuiop gave a speed of 1.18c in post #24 above so I think we did the calculations the same way).

The pace of a helix is defined as the length of a coil it is not the rate of rotation.

So, the height of each coil would be 12/8 = 1.5 light-seconds, so according to this page the length of each coil must be sqrt(1.5^2 + (2*pi*0.3)^2) = 2.409, so the total length of the helix is 8*2.409 = 19.27.

I get 3.48 for each coil and a total length of 27.85.

I took the radius and the height (pace) of a coil divided by 2pi. See for instance: http://mathworld.wolfram.com/Helix.html

 

[JesseM]

The pace of a helix is defined as the length of a coil it is not the rate of rotation.

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 in the context of a relativity discussion without explaining this!)

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?

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.

I took the radius and the height (pace) of a coil divided by 2pi. See for instance: http://mathworld.wolfram.com/Helix.html

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? It's not hard to see that this formula is correct, since you can imagine the helix as a curve drawn on a cylinder, and then if you slice the surface of cylinder on an axis parallel to the central axis and "unwrap" it (which is possible since a cylinder has zero intrinsic curvature, just like a flat plane--see here), you'll have a rectangle with width 2*pi*R, and the vertical height of a single coil will be H.

 

[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.