What happens if a super fast rotating ball accelerates near speed of light?

Question

Assume we have a ball with diameter 1 meter and mass 1 kg rotating at 99,5% speed of light (Suppose I'm an observer placed above the ball, I see under me a disk wich external speed is 99,5% the speed of light)

Then we accelerate such ball to 99.5% of speed of light, to make clear from wich position we look at the ball when it's moving so fast I did a picture (there's a black ring around the ball, as you se the rotational axis is vertical to our observer)

I'm interested in a full explanation of what we observe (measured rotation speed, color etc.) and why that happens, as far as I know we should not measure something moving faster than light but I'm not sure how relativity constraints would alter the "measured state" of the ball as long as we accelerate it.

What we see change while the ball accelerates?

EDIT:

A precisation, I assume the ball is spherical when I start rotating it before even moving, this is because commenters made me to note that shape may change.

Answer 1

From an SR point of view, I suspect that two things will happen. If the ball is moving across the observer's field of view, rather than directly towards or away, it will appear to be a prolate ellipsoid of revolution due to apparent length contraction. Additionally, the rotation will appear to gain a phase shift caused by differential time of flight of light emitted from different parts of the surface.

Rather more interesting are the GR implications, which I'm not qualified to specify. But I will point out that, at some scale the equator of the sphere will begin to look rather like a Tipler cylinder, aka a time machine. That assumes, of course, that Tipler is right and Hawking is wrong, and I'm generally reluctant to bet against Hawking. Since a "practical" Tipler cylinder only required neutronium and a tangential velocity of something like 0.5c, a sphere made of unobtanium ought to be serviceable.

Answer 2

Intuitively, if you paint longitude/latitude lines on the ball, the linear movement will, if I remember the remnants of my special relativity class, result in a rotation of the ball. So far that's pretty tame. What can the fast rotation do? It will distort the coordinate system on the ball. What will happen to the colors? One side of the ball will be blue, the other side red-shifted. If I go by the analogy of wavelength to coordinates system distances, then the blue end of the ball will have the longitude lines closer together, the red end farther apart. Does that make any sense? I hope so.

There is, by the way, absolutely no need to chose impossible parameters for this problem. We have very precise atomic clocks that can measure these effects with high precision at velocities of a few km/s. Boosting things to near the speed of light does not change a iota about the theory. I am looking forward to somebody doing the actual calculation and discussion, so we can see if my intuition is even borderline correct.

Answer 3

The real problem here is what is meant by a rigid body in relativity. In Newtonian mechanics a rigid body is described as a system of mass points that maintain their relative distances to each other. You can count the degrees of freedom by adding the mass points one by one. For the first mass point you have no constraints, so you get 3 degrees of freedom (for the three spacial dimensions). For the second mass point you have one constraint which is the distance to the first point. Thus we get only 2 additional degrees of freedom (think the latitude and longitude on the sphere of a set radius around the first mass point). For the third mass point you have two constraints, i.e. the distances to the other points, so you only have one degree of freedom. All the remaining mass points have strictly defined coordinates by the relative distances to the other mass points, so you get no additional degrees of freedom. So we have 6 degrees of freedom for a rigid body, which can be interpreted as the center-of-mass position (3 DoF) and the three Euler angles (3 DoF). This is why we can talk about the meaning of things like "rotating the object" or "accelerating the object". From this definition you can among others derive the equations of motion for the object to be \begin{align} M\ddot{\mathbf{X}} &= \mathbf F \\ \dot{\mathbf{L}} &= \mathbf N \end{align} where \begin{align} M&: \text{Total mass} \\ \mathbf X&: \text{Center of mass position} \\ \mathbf F&: \text{Vector sum of all the forces applied to the system} \\ \mathbf L&: \text{(Orbital) angular momentum} \\ \mathbf N&: \text{Moment of force} \\ \end{align} from which you can see that angular momentum is conserved for an isolated system ($\mathbf F = \mathbf N = 0$).

Now when you go to relativity this definition is problematic, as exerting a slight push on one side of the rigid body would immediately move all the other points, thus giving us an information-transfer faster than the speed of light (infinite speed of sound).

There are a few attempts to define a relativistic notion of a rigid body. One of the most naive ones is probably Born-rigidity, which can been shown to be broken as soon as the object is put into rotation, and leads to the Ehrenfest-paradox.

Now for (total) angular momentum we know that this is conserved due to rotational invariance and Noether's theorem. However we don't get simple equations of motions as in the Newtonian case.

Thus I think your question cannot be answered unless we make certain assumptions about the forces that hold the object together, because unlike in Newtonian mechanics, it is not clear what is meant by a rotating sphere. A simple approximation would be to assume that the mass points are being held together by springs using a Hook force law.

I don't think this is what you were asking for, but I thought it is worth pointing out because it's rather interesting. People like Fritz Noether have worked on this problem: "Zur Kinematik des starren Körpers in der Relativtheorie".

A completely different approach rather than making the ball spin would be to spin the coordinate frame using general relativity and then perform a Lorentz transformation on the result. Then we can assume that each point emits light pulses of a certain frequency in the co-moving time frame (which is similar to the definition of Born-rigidity) and from that calculate how the ball "looks" like in such a helically or spirally (depending on the boost axis) moving coordinate frame. If anyone has the time to do that :)

Answer 4

What happens if a super fast rotating ball accelerates near speed of light?

The ball breaks up even before you accelerate it linearly. This is a bit of a problem with flywheels. In practice you just can't spin the ball at 99.5% of the speed of light. But never fear, we can still make progress.

Assume we have a ball with diameter 1 meter and mass 1 kg rotating at 99.5% speed of light (Suppose I'm an observer placed above the ball, I see under me a disk which external speed is 99.5% the speed of light)

Because it's so dangerous being near this 1 meter ball (which would typically have a mass of a tonne) let's replace it with light going round and round. To simplify matters let's make it a circle rather than a sphere. Then let's you and me look at it side on. So it looks like a vertical line, like this: |

Then we accelerate such ball to 99.5% of speed of light...

Instead of that, let's accelerate our gedanken ring of light so that it's moving bodily from left to right at 50% of the speed of light. When we look at it from the side, it will look something like this: /\/\/\/\/\ . It's a bit like stretching a spring. The light was going round a circular path, but now its path is helical: _{image courtesy of indiamart}

The light is moving round the helix at the speed of light, and the faster you move the ring, the more stretched the spring looks. Try this with a real spring. When the ring is moving at the speed of light the spring is fully stretched out into a straight horizontal wire. It isn't a ring any more. Now think about what our wire represents: light. Hold that thought for a minute.

I'm interested in a full explanation of what we observe (measured rotation speed, color etc.) and why that happens, as far as I know we should not measure something moving faster than light but I'm not sure how relativity constraints would alter the "measured state" of the ball as long as we accelerate it. What we see change while the ball accelerates?

Relativity is all about relative motion, wherein the simple inference of time dilation employs Pythagoras's theorem on right angle triangles like this /l. The hypotentuse is the speed of light, the base is your relative speed as a fraction of c, and the height is the Lorentz factor. And that speed is relative, so try to imagine what the rotating thing would look like if you moved past it fast. The ball is a bit too complicated, so think of that ring. When you're motionless with respect to it, it looks like a ring, like this | from the side. But when you go past it fast, it looks like a helix, like this /\/\/\/\/\ from the side. The faster you go by, the more stretched-out it looks, like the stretched out spring. And if you're going close to the speed of light the spring looks pretty much totally stretched out. It looks like a straight wire. And in our analogy that's what we used to represent light. So your very fast moving spinning body would look like it wasn't spinning. And it would look like light. Your spinning ball looks like a metre-wide streak of light, like this:

Answer 5

Being a good physics question answer we'll ignore a few things, what the ball is made of, why it isn't tearing it's self apart and where we get the energy to spin it up seem like good choices since we're interested in what it looks like not how we build one.

The trick to understanding how this looks is to realise that there are two different frames of reference here.

Firstly the rotational measure, if you're stationary relative to the ball (to it's centre) it's spinning at a speed such that the outer edges are 0.995c, we want to be side on to the spin so the first visual effect of this is that the side spinning towards us is blue shifted, and the side spinning away is red shifted.

As 0.995c is pretty darn fast by any measurement I imagine that the shifts will be pretty extreme, well outside the visible spectrum, I don't have the maths to hand to work this out exactly, but the colour shift should be smooth across the surface most shifted at the side and less shifted towards the closest point to you (which wouldn't be shifted at all)

Now we have the ball spinning what happens if is thrown past me at 0.995c as well, and how come nothing ever moves faster than 1c?

The effect that comes to the rescue is time dilation, and the easiest way I've heard of to think about this is to measure time in terms of the speed of light, using a light clock.

A light clock is basically to mirrors a fixed distance apart with a single photon of light bouncing endlessly between them, time is measured as the time it takes the light to move from one to another and back again.

If you place a light clock on the top of the ball and measure it's rotational speed as the time it takes for one rotation as 1 when you set the ball moving imagine that the light clock is now moving perpendicular to the direction that the light is bouncing inside it.

From the perspective of the ball nothing changes, bounces at the same rate (the speed of light is constant regardless of your frame of reference)

From your point of view watching the ball move past you at 0.995c however things look different. Because light moves at a fixed speed and the mirrors are moving sideways the light now has further to travel. The calculations for this are fun to derive, but rather than reproducing them I'll link to a nice article Google turned up on the subject.

https://sciencebasedlife.wordpress.com/2012/08/10/derive-time-dilation-yourself-feel-like-a-genius/

from which we learn:

$$t=t_0*\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

which tells us that the time for the ball looks 9.9874922% the usual rate, so the ball is now looks to be spinning more slowly, the outside is now needs to move at around 0.099375547c rather than our original 0.995c

That's all special relativity tells us, but it still leaves us with a problem. $$0.995c + 0.0993c = 1.0943c$$ so the edges still seem to be moving faster than light, this is as far as I've got thinking it through with the time I've got, there are other things to consider, the fact that there's spacial dilation caused by the movement which will reduce the speed of the edges, as well as the fact that the edges are technically an accelerated frame not a static one (they're moving at 0.995c, but they're accelerating towards the centre to prevent them flying off) which you can dilation based on general relativity by pretending that they're being held in place by a gravitational field.

So how does this all look (since that's part of the question!)

Well it apparently still looks like a sphere:

http://th.physik.uni-frankfurt.de/~scherer/qmd/mpegs/lampa_terrell_penrose_info.html

so I'm going to guess it looks like a sphere with a blue shifted and a red shifted side, anything more specific than that and I think you'd have to be considering how quickly eyes can respond, what colours and lights are in play, etc.

Hope someone can work out the last mile and get all the velocities under c, it's a little beyond me at the moment.

Answer 6

There are three main concepts to get clear to fully understand the issues raised in this question.

First, the part of the ball that has its tangential velocity in the same direction as the translational velocity of the ball will not exceed the speed of light. This is due to relativistic velocity addition. The formula is $$u = \frac{v+u'}{1+\frac{v u'}{c^2}}$$ where u is the velocity you measure as it goes past, u' is the velocity measured by someone comoving with the centre of the ball and v is the speed the ball goes past you. For your example the result is: $$u = \frac{2 \times 0.95}{1+0.95^2} \approx 0.998685939553 $$

Adding together two velocities in relativity will never result in a velocity greater than c. You can think of it like this. As you watch the ball go past you, it appears time dilated and appears to be rotating slower, so to you the tangential velocity of the ball is lot less than 0.95c and when you add this reduced tangential velocity to the relative linear velocity, the total is than c.

Secondly there is optical effect, whereby a a fast moving object appears to rotate and you can effectively see the back of an object.

There is a video demonstrating the effect here.

In the image above you can see the face of the dice with 4 spots on it even though its real orientation is with the faces with three spots and only spot facing you. This is due to light travel times and is purely optical. Measurement will reveal the object is length contracted. For a sphere, a sphere still looks like sphere in shape, but if there are surface patterns you will see them distorted and rotated. If one side of the ball is black and the other white and you can only see black when are above the ball and comoving with it, you will be able to see part of the white side when it is moving linearly relative to you. As others have already mentioned, you will see red and blue shifting of colours on the face of the ball so I wont elaborate any more on that.

Lastly, the person comoving with the ball sees a perfectly round ball. Since reality is the same for all observers, no other observer will see the ball as black hole.

Answer 7

It seems that none of the answers address the most interesting part of your question, that is: what happens to the side of the ball which is spinning in the "forward" direction? Obviously it cannot reach 99.5+99.5=199% of the speed of light.

While I am not an expert on relativity, we know that from a stationary observer's point of view, time on objects travelling close the speed of light appears to slow down. This has actually been confirmed in particle accelerator experiments: the observed lifetimes of particles travelling at the speed of light are much longer than the observed lifetimes of identical particles at rest.

Therefore the stationary observer will (from his point of view) see the ball spinning at a much slower rate than an observer riding on the ball, and he will therefore not see any part of the ball exceed the speed of light. Additionally, the ball will appear to be flattened in its direction of travel.

Finally, it's important to note that "stationary" relates to a completely arbitrary frame of reference. The point of view of the observer who sees the ball moving near the speed of light is no more or less valid than that of the observer riding on the ball.

Relevant links (the doppler effect is quite interesting.)

https://en.wikipedia.org/wiki/Length_contraction

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

Answer 8

You can't get the speed of the individual particles above speed of light. Once they get close, the amount of energy required to accelerate just a little is increasing very fast.

So I guess that, if the ball surface is rotating already very close to the speed of light, pushing sideways the ball will hardly accelerate in that direction, as if it is very heavy.

Answer 9

I remember wondering this exact same thing in high school. I did the math on it and figured that an infinitely thin bubble would take the shape of a peanut. The Lorentz Factor for 99,5% of C would shrink the equator to be 10cm in diameter, but if you look at the ball half way between the pole and the equator it will be traveling at 70% of the speed of light. The Lorentz factor at ,25m perpendicular to the pole will be ,71.

Different properties of your sphere, depending on things such as elasticity, density, or maybe even if it's hollow inside would change how close of a peanut shape it can be. The more resistant your sphere is to being squeezed, the closer it will resemble a prolate ellipsoid, like in WhatRoughBeast's answer.

Answer 10

How ever you will accelerate something, you will end up with photons as the mediator. This is the real reason why it is not possible to accelerate a body to the light velocity. What happens to a rotating ball? Suppose, you are looking in the direction of the light beam and the ball is rotating in a way, the left side is moving in your direction and the right side is moving from you away. Hitting the ball the photons give their momentum to the ball. This momentum will be higher for the left side of the ball and smaller for the right side. As mentioned DarioOO in his comment, the ball will describe a curve trajectory. Then more the ball get accelerated then more the ball get deflected to the right. At some time there will be balance between deflection and achievable speed.

Answer 11

as far as i could imagine the ball would be vibrating, it would have frequency, because it has a very large momentum or better to say that it has energy which is very huge with respect to its mass,

you can't see that ball because it is too fast to be observed by our necked eyes, you can only see it with a super super slow camera.

You see what the funny thing is? it would be present at many places at a single time instant! it is a way to say that there is an uncertainty in the position of the ball so you can not say that it is there, because you cant see it with your necked eyes so you can only say that it is somewhere in between this area! and you can only be certain about its position only when you use some super slow camera to see where it is. (well i don't know how the ball would look like in that camera)

it would be like a flash of light going in front of you and you will become deaf or may ever die if the ball in accelerated in air. see how:

swing your hand in air could you hear anything? well you won't, now if you have played badminton, then, when you make a smash you could hear sound of the swinging of the racket. Now if you increase the energy of swinging of your racket while making a smash you will hear more intense sound. That is you are giving more energy to the air particles so every time when you increase the energy you are increasing the amplitude of the sound wave. So same thing is applied to this ball. It is with such an energy that it will increase the amplitude of the sound wave to such a level which is dangerous to us it would be a sonic bomb.

You should read the about Schrodinger's thought experiment about its cat and its consequences

Answer 12

The geometry for a spinning ball changes. If you consider the circumference as small line segments, the fact that they are spinning means that the line segments are moving in the direction of motion and exhibit Lorentz contraction while the radii are not foreshortened since they are moving perpendicular to the spin. You are now dealing with non Euclidean geometry for the sphere since the circumference is no longer equal to pi times the diameter.