# Composition of linear and rotational motion

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Summary:
How do linear and circular motion compose in SR?
I have a pencil of Iron of length ##L## rotating about its center in a plane at constant angular velocity ##\omega##. The tip of the pencil in Newtonian mechanics has linear velocity ##\frac{\omega L}{2}##. It can exceed ##c##, of course.

Now let us complicate this. Assume the center of the pencil (thus the pencil as a whole) moves linearly along the Ox axis (while rotating at ##\omega##) at constant velocity ##c-\epsilon## in the Lab frame.

What is the linear speed of tip while in motion, calculated in the lab frame in terms of ## c, \epsilon, \omega, L##? Pretty sure it cannot exceed ##c##.

It is tempting to say that in the lab frame the radius of the disk generated by the rotating pencil shrinks to ##\gamma L/2##, but how would the linear velocity of the tip be? It cannot be simply by multiplication with ##\omega##, it would exceed ##c##.

PeterDonis
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What is the linear speed of tip while in motion
Whatever the relativistic velocity addition law says it is. (Note that if you are using SR, the product ##\omega L## is restricted since the tip speed must be less than ##c## in the rest frame of the other end of the pencil--assuming the other end is on the axis of rotation.)

vanhees71
PeterDonis
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I have a pencil of Iron of length ##L## rotating about its center in a plane at constant angular velocity ##\omega##.
Note that in SR, you have to carefully define what you mean by ##L## and ##\omega##. Are they length and angular velocity in the rest frame of the axis of rotation? In the rest frame of the tip (which is rotating about the axis)?

vanhees71
hutchphd
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Also your pencil could be the hand of a clockface......

PeterDonis
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It is tempting to say that in the lab frame the radius of the disk generated by the rotating pencil shrinks
One should never yield to such temptations. Once you have a description of the motion in one frame, transforming it to any other frame is just a matter of math. Intuitive guesses are (a) not the same as doing the math, and (b) quite likely to be wrong since our natural intuitions are Newtonian, not relativistic.

vanhees71
Orodruin
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Summary:: How do linear and circular motion compose in SR?

It is tempting to say that in the lab frame the radius of the disk generated by the rotating pencil shrinks to γL/2, but how would the linear velocity of the tip be? It cannot be simply by multiplication with ω, it would exceed c.
If the disk is orthogonal to the motion, the radius will remain the same. If the direction of motion is in the plane of the disk then the disk will contract in that directon only. It will not look like a rotating disk.

The relativistic wheel on the Wikipedia length contraction page should give you an idea. https://en.wikipedia.org/wiki/Length_contraction

vanhees71 and pervect
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Note that in SR, you have to carefully define what you mean by ##L## and ##\omega##. Are they length and angular velocity in the rest frame of the axis of rotation? In the rest frame of the tip (which is rotating about the axis)?
True. ##\omega## in the inertial frame attached to the center of rotation. L is in the same frame.

vanhees71
Ibix
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You can write down the instantaneous four velocity of the pencil tip in the frame where the center of mass is at rest and then apply the Lorentz transform.

Note that the velocity of the tip cannot exceed ##c##, so ##\omega L/2<c##. In practice, ##\omega L/2\ll c##, since an iron rod will disintegrate at much lower spin rates.

vanhees71 and dextercioby
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No restrictions on ##\omega##, so how does the tip of the pencil in the frame attached to the center of rotation (middle of the pencil) have a linear speed under ##c##?

PeterDonis
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No restrictions on ##\omega##
Yes, there are. It is physically impossible to achieve an ##\omega## that would make the tip move faster than ##c##. Or, to put it in more invariant terms, it is impossible to achieve an ##\omega## that would cause the tip's worldline to not be timelike. If you try to do it, the pencil will be torn apart by stresses.

vanhees71
Ibix
2020 Award
No restrictions on ##\omega##
Yes there are - ##\omega L/2<c##.

Homework Helper
So you're both telling me that, if the material is ideal, then the tip can exceed "c", and it's only the fact that, as the angular velocity increases, the stress in the material would cause it to disintegrate, thing which you cannot put as "Special Relativity explains tearing"? This is interesting.

Homework Helper
Isn't it like a paradox about "speed of propagation of effects", like this one: say I have a ruler of 1 light years length and made of the lighest possible material. I push it at one end, then the other end would start moving immediately, not after more than 1 year, right?

Ibix
2020 Award
So you're both telling me that, if the material is ideal, then the tip can exceed "c"
No - we're telling you that "ideal" in SR can't be the same as it is in Newtonian physics. You cannot have a perfectly rigid rod, for example, because that would require an infinite speed of sound, a concept which is not well defined in a relativistic universe.

One way to look at this is to consider a small part of the rod tip. Its momentum is ##\gamma mv##. If ##v## approaches ##c##, the centripetal force required to keep this little bit in orbit tends to infinity - so the rod must disintegrate before the tip reaches ##c##, ideal or not.

Ibix
2020 Award
I push it at one end, then the other end would start moving immediately, not after more than 1 year, right?
No - the other end doesn't move for at least a year because the mechanical shock from the push propagates along the rod at the speed of sound in the rod. Typically this is on the order of kilometres per second, but it cannot even in principle exceed ##c## because it depends on the electromagnetic field between the atoms, disturbances in which propagate at or below ##c##.

Ultimately, all of this is expressions of the fact that causal influences cannot propagate faster that ##c## in a relativistic universe.

dextercioby
PeterDonis
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So you're both telling me that, if the material is ideal, then the tip can exceed "c"\
No. Even an ideal material must still travel on timelike worldlines. The tip cannot exceed ##c##, period.

it's only the fact that, as the angular velocity increases, the stress in the material would cause it to disintegrate
Not "only". This is what will happen, yes.

, thing which you cannot put as "Special Relativity explains tearing"?
I don't know what you mean here. SR does explain why the stress in the material must cause it to disintegrate: because internal forces in the material can only propagate at speeds less than or equal to ##c##. Or, to put it another way, SR imposes a finite limit on the strength of materials; roughly, the limit is that the speed of sound in the material (which is the speed at which internal forces in the material propagate) must be less than or equal to ##c## (and "equal to" is really an edge case that no actual material will realize). This has been well known for decades.

PeterDonis
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say I have a ruler of 1 light years length and made of the lighest possible material. I push it at one end, then the other end would start moving immediately, not after more than 1 year, right?
Wrong. This a classic SR "paradox" which is treated in most SR textbooks (and has been the subject of multiple previous PF threads).

It looks like you need to become more familiar with SR.

PeterDonis
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The tip of the pencil in Newtonian mechanics has linear velocity ##\frac{\omega L}{2}##. It can exceed ##c##, of course.
This might be the original source of your problem. Newtonian mechanics is not SR, and you can't take a scenario that you constructed assuming Newtonian mechanics and expect it to always be possible in SR.

Homework Helper
Wrong. This a classic SR "paradox" which is treated in most SR textbooks (and has been the subject of multiple previous PF threads).

It looks like you need to become more familiar with SR.
Yes, hence the question. The „SR education” was not part of my university physics curriculum, astonishingly (neither were complex analysis and functional analysis), so even the elementary problems are unfamiliar to me. :)

pervect
Staff Emeritus
The math gets somwhat messy, but basically it's just an example of relativistic velocity addition. I'd suggest the wiki page https://en.wikipedia.org/wiki/Velocity-addition_formula.

There are many approaches, but the most basic observation may be breaking down the formula into how the parallel and perpendicular components of the velocities are affected by a boost, parallel components being parallel to the boost.

Let ##\beta = u/c## be the normalized velocity, and let the parallel and perpendicular components be ##\beta_{\parallel} = u_{\parallel}/c## and ##\beta_{\bot} = u_{\bot}/c##. Let ##\alpha## be the normalized boost, ##v/c##.

Then wiki gives the relationship between ##\beta## and ##\beta'##, the normalized velocities in the unboosted and boosted frames, as:

$$\beta_{\parallel} = \frac{\beta'_{\parallel} + \alpha}{1+ \alpha \beta} \quad \beta_{\bot} = \frac{\sqrt{1-\beta^2} \beta'_{\bot} } {1+\alpha\beta}$$

This gives the usual relativistic velocity addition equation in one spatial dimension where there are no perpendicular components to the velocity. The results are, of course, that boosting a velocity always results in a velocity lower than the speed of light, c.

PeterDonis
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The „SR education” was not part of my university physics curriculum, astonishingly (neither were complex analysis and functional analysis), so even the elementary problems are unfamiliar to me. :)
Taylor & Wheeler's Spacetime Physics is the classic SR textbook (it's the one I learned it from, way back when I had to walk to school uphill both ways ). I believe there is an online edition available now.

vanhees71
Gold Member
Yes, there are. It is physically impossible to achieve an ##\omega## that would make the tip move faster than ##c##. Or, to put it in more invariant terms, it is impossible to achieve an ##\omega## that would cause the tip's worldline to not be timelike. If you try to do it, the pencil will be torn apart by stresses.
The dynamical argument is interesting too. For such examples we always have our beloved (or hated ;-)) model of the rigid body from Newtonian mechanics in mind. Of course there cannot be a rigid body in this literal sense in relativity, because the speed of sound in such a material should be infinite, because if I push on one end of a literary rigid body it must move as a whole, i.e., the "signal" telling the other end of the body that I pushed on one end must be instantaneous. That's impossible within relativity. The best approximation you can have is (another fictitious) elastic body with a sound velocity of ##c##. In reality you just have elastic bodies with some finite speed of sound less than ##c##.

A somewhat artificial mathematical playground is provided by the famous Born rigid body. It has been completely described in the 1910th by Fritz Noether (Emmy Noether's brother) and Gustav Herglotz, i.e., all possible motions of such a body under Born's rigidity constraints have been identified. For your example of a rod rotating around its center of momentum you must have a constant angular velocity, i.e., to describe how the rod gets rotating from rest you have to consider it as an elastic body until the final constant angular velocity is reached. The body is necessarily deformed (as seen in the momentaneous rest frame) compared to it's form when it is at rest in an inertial frame. This also resolves the apparent rigid-disk paradox. If I remember right there's a nice discussion in the old textbooks by Synge and/or Eddington. If needed, I can check.

alantheastronomer
pervect
Staff Emeritus
If I'm reading the wiki article https://en.wikipedia.org/w/index.ph..._formula&oldid=1021873364#Hyperbolic_geometry correctly, one can regard the "velocity space" parameterized by the components of the velocity in cartesian coordinates ##\beta_x,\beta_y,\beta_z## as the interior of a 3-sphere. The magnitude of the velocity has a value of ##\beta^2 = \beta_x^2 + \beta_y^2 + \beta_z^2 < 1## relative to the origin of the space at "zero velocity".

If one introduces standard spherical coordinates ##\beta, \theta, \phi## to replace the cartesian coordinates, wiki gives the line element as

$$ds^2 =\frac{ d\beta^2}{\left( 1-\beta^2 \right)^2} + \frac{\beta^2}{1-\beta^2} \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$$

where ds is the relative velocity between two nearby points in the "velocity space".

Wiki also gives the metric in terms of rapidities, which I won't repeat here.

It passes at least one common-sense test. If we set ##d\theta=d\phi=0##, ##ds=d\beta/(1-\beta^2)## reduces to the usual velocity subtraction formmula ##\frac{\beta2-\beta1}{1-\beta1\,\beta2}##.

vanhees71
Demystifier