The relativistic mechanics of spinning discs

[jartsa]

These are kind of my own thoughts. Does the theory of relativity disagree with any of these?

When a spinning disc is heated, the spinning rate decreases, and spinning energy is being converted into additional heat.

When a spinning disc is accelerated linearly, the spinning rate decreases, and spinning energy is being converted into additional linear kinetic energy.

 

[pervect]

If you add heat to a spinning disk so as not to disturb its angular momentum, I'd agree that you should increase its moment of inertia , and I'd expect it to reduce it's angular velocity.

The Newtonian formulas L=Iw and E=(1/2)I w^2 translate to E = L^2 / 2I, so if you keep L constant and increase I, I'd expect an energy loss. So there might be some justice to this argument.I'm not convinced that the second scenario will work as you describe it though - certainly the disk won't change it's angular velocity in its won frame!

...

I've thought about it a little, and what I think should happen is that, assuming you acclerate a disk along its axis of rotation, that the disk won't change it's rotation speed in its own frame, and that after you cease acceleration (making it an isolated system) , the angular momentum before and after the acceleration should be the same in the inertial frame as well.

 

[jartsa]

If you add heat to a spinning disk so as not to disturb its angular momentum, I'd agree that you should increase its moment of inertia , and I'd expect it to reduce it's angular velocity.

The Newtonian formulas L=Iw and E=(1/2)I w^2 translate to E = L^2 / 2I, so if you keep L constant and increase I, I'd expect an energy loss. So there might be some justice to this argument.I'm not convinced that the second scenario will work as you describe it though - certainly the disk won't change it's angular velocity in its won frame!

...

I've thought about it a little, and what I think should happen is that, assuming you acclerate a disk along its axis of rotation, that the disk won't change it's rotation speed in its own frame, and that after you cease acceleration (making it an isolated system) , the angular momentum before and after the acceleration should be the same in the inertial frame as well.

Hooray, the first case makes some sense.

But there's no fundamental difference between heating and accelerating:

Heating:
If all molecules in a spinning disc are forced to vibrate back and forth very fast, then the molecules cannot move around in a circle very fast, which means the disc cannot spin very fast.

Accelerating:
If all molecules in a spinning disc are forced to vibrate back and forth very fast and in sync, then the molecules cannot move around in a circle very fast, which means the disc cannot spin very fast.

hot disc = disc with vibrating molecules
vibrating disc = disc with molecules that vibrate in sync
moving disc = vibrating disc, where frequency of vibration is low

 

[pervect]

If you view things from the point of the view of the disk, though, there won't be any "slowing down" of the disk in the second case. Slowing down occurs from the POV of the lab frame, surely, but not in the disk frame.

Thomas precession comes to mind, but I don't think that should happen either...

So I don't see how there can be any "dissipation of energy" in the second case. I suspect the problem needs a more careful analysis, but I don't have one to offer at the moment.

 

[jartsa]

If we add heat to a spinning disc, whose molecules consist of radioactive atoms, we would say that the slowing down of the radioactive decay is caused by time dilation, which is caused by the thermal motion of molecules.

We would not be so ready to say that the decrease of the rate at which the molecules complete rounds around the center of the disc is caused by time dilation, which is caused by the thermal motion of the molecules.

But I suggest that we say the latter thing, because there's no rational reason not to say it, or is there?

 

[jartsa]

Let's consider a stack of three discs, first there is a disc made of some strong material, then there is a disc made of some explosive material, then there is a disc identical to the first disc.

This stack is spinning. What happens when the explosive explodes?

The rest mass of the stack does not change. The rotational inertia of the stack does not change. The angular velocity of the stack does not change. And the rotational energy of the stack does not change.

There was no change of the amount of mass-energy, or the distribution of mass-energy, except in the direction of the spinning axis.

Here's my maybe somewhat odd sounding conclusion:

From the fact that the angular velocity did not decrease, it follows that the linear velocity of a disc flying off can not exceed some value, which is not c, but some smaller value. (in the stack's center of mass frame) When a spinning disc is accelerated so that the spinning energy does not decrease, then the dynamics seems quite abnormal.

My point is that normally the spinning energy of an accelerated spinning disc does decrease, and things don't look abnormal.

So does relativity have a problem with the odd conclusion I arrived at there?

 

[Nugatory]

From the fact that the angular velocity did not decrease, it follows that the linear velocity of a disc flying off can not exceed some value, which is not c, but some smaller value. (in the stack's center of mass frame)

Why does this follow? I don't see how you come to this conclusion..

 

[jartsa]

Why does this follow? I don't see how you come to this conclusion..

An observer at the center of mass frame of the stack of discs sees some point on a disc to have a linear velocity x, and a velocity perpendicular to that: y.

x is the velocity caused by the linear motion of the disc, linear motion was caused by the explosion.
y is the velocity caused by the spinning of the disc.

The velocity of the point is calculated using the Pythagoran formula: v=sqrt(x²+y²)

v<c
so:
sqrt(x²+y²)<c
now we solve x:
x²+y²<c²
x<sqrt(c²-y²)

if y is close to c, then x must be close to zero

 

[Nugatory]

The velocity of the point is calculated using the Pythagorean formula: v=sqrt(x²+y²)

You might give it a try using the formula for adding relativistic velocities.
Also be sure that you've correctly transformed the position of the various points on the disk.

 

[jartsa]

You might give it a try using the formula for adding relativistic velocities.

That would wrong. I almost made that mistake. If an object moves a 1 light years distance, and at the same time a 1 light years distance in the perpendicular direction, then the object moves a distance of 1.41 light years, which takes time t, so the velocity of the object is 1.41 lys / t

 

[Nugatory]

That would wrong. I almost made that mistake. If an object moves a 1 light years distance, and at the same time a 1 light years distance in the perpendicular direction, then the object moves a distance of 1.41 light years, which takes time t, so the velocity of the object is 1.41 lys / t

You've spent enough time here to know that whenever you find yourself saying "at the same time" that's a danger signal...

Here you have the instantaneous velocity vector of a point on the perimeter of the disk as measured in a frame in which the center of the disk is at rest. You also have the velocity vector of the center of the disk, as measured in a frame in which the pre-explosion stuff was at rest. Because they're measured in different frames you can't just vector-add them the way you're trying to do above; you have to use the relativistic vector addition (not just the simple (u+v)/(1+uv) one-dimensional addition rule) which you'll find in wikipedia.

You might also consider that when you say the disk is flat, you're really saying that all points on the surface of the disk lie in a the same plane at the same time. That's not going to be true in all frames.

 

[jartsa]

You've spent enough time here to know that whenever you find yourself saying "at the same time" that's a danger signal...

Here you have the instantaneous velocity vector of a point on the perimeter of the disk as measured in a frame in which the center of the disk is at rest. You also have the velocity vector of the center of the disk, as measured in a frame in which the pre-explosion stuff was at rest. Because they're measured in different frames you can't just vector-add them the way you're trying to do above; you have to use the relativistic vector addition (not just the simple (u+v)/(1+uv) one-dimensional addition rule) which you'll find in wikipedia.

You might also consider that when you say the disk is flat, you're really saying that all points on the surface of the disk lie in a the same plane at the same time. That's not going to be true in all frames.

I have explained two times what velocities I'm adding. A velocity in x dimension and a velocity in y dimension, measured by one inertial observer. The observer was "standing in the center of mass frame of the stack of discs"

 

[Dale]

But there's no fundamental difference between heating and accelerating:

There is a fundamental difference: heating does not change the momentum of the system, accelerating does. Also heating can be done internally, acceleration requires an external force.

 

[Dale]

Let's consider a stack of three discs, first there is a disc made of some strong material, then there is a disc made of some explosive material, then there is a disc identical to the first disc.

This stack is spinning. What happens when the explosive explodes?

This is very poorly described, so I am guessing that I understand your intention.

The rest mass of the stack does not change

True.

The rotational inertia of the stack does not change

If you mean angular momentum then this is also true.

The angular velocity of the stack does not change

The stack no longer has a single angular velocity. The moment of inertia of the explosive disk has certainly changed so the angular velocity will have changed also. I am not convinced that the angular velocity of the strong disks does not change.

And the rotational energy of the stack does not change.

I am not convinced of this either. The explosion certainly added a lot of energy, I don't see why some of it couldn't go to rotational energy.

There was no change of the amount of mass-energy, or the distribution of mass-energy, except in the direction of the spinning axis.

There certainly is a change in the distribution of mass energy. Why would you think there wasn't?

So does relativity have a problem with the odd conclusion I arrived at there?

Your conclusion may or may not be right, but I don't think your reasoning makes sense. You really need to actually work the math and not just assume the conclusion.

 

[jartsa]

This is very poorly described, so I am guessing that I understand your intention.

The explosive is a source of complications, so let's remove it:

A helical spring is held compressed by some lock mechanism, scotch tape for example, the spring is spinning around its longitudinal axis.

What happens when the lock breaks?

I think I know what happens: the spring vibrates, no change of angular velocity.

 

[jartsa]

I feel the mechanics of the spinning discs is pretty clear to me now, thanks to this thought experiment:A loaded cannon is moving to the right. When the cannon fires, the cannon and the cannonball continue the motion to the right at unchanged speed. This is possible with suitable orientation of the cannon. (straight up or straight down)

If we attach many of these kind of cannons on the flat surface of a spinning disc, and fire the cannons, then immediately after the firing each cannonball has the same sideways velocity as it had just before the firing.

So if the cannonballs are attached to each other with chains, then the group of cannonballs will spin around its center axis at the same rate as the disc was spinning.

 

[Dale]

What happens when the lock breaks?

I think I know what happens: the spring vibrates, no change of angular velocity.

I don't think I know what happens. I agree that there is no change in angular momentum, but I am not convinced at all that there is no change in angular velocity.