Energy for moving mass up a centrifugal gravity barrier

In summary, if you want to move a point mass from a radial position to the center of a rotating frame, you need to account for centrifugal forces and the work done against them.
  • #1
Sunfire
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4
Hi,

I have been pondering on this -

what is the energy required to move a point mass m0 from a radial position R to the center of a rotating frame, if the linear velocity v of the point mass at position R is relativistic (close to c).

For an observer in the rotating frame, it should seem as if m is climbing against the centrifugal gravity. In the classical limit, the answer is [itex]\frac{mv^2}{2}[/itex]

m0 is the rest massThanks!
 
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  • #2
It would be the equivalent of stopping the point mass.
If you know the apparent mass of the particle as it is moving (ma), then the energy to decelerate it would be (m0-ma)[itex]c^2[/itex].
That would be a negative value, indicating that you are removing energy from the system.
 
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  • #3
Just use the relativistic centrifugal force that appears in the rotating frame to calculate the work done relative to this frame in order to move a point mass from some radius to the origin of the frame along a desired path e.g. a radial line.
 
  • #4
WannabeNewton said:
Just use the relativistic centrifugal force that appears in the rotating frame to calculate the work done relative to this frame in order to move a point mass from some radius to the origin of the frame along a desired path e.g. a radial line.

Don't you have to also include work done against the Coriolis force to keep the mass on the radial line?
 
  • #5
PeterDonis said:
Don't you have to also include work done against the Coriolis force to keep the mass on the radial line?

I don't think so because the Coriolis force is always perpendicular to the 3-velocity vector ##\vec{v}## of the path taken between two end points in the rotating frame so ##\vec{F}_{\text{Coriolis}}\cdot \vec{dr} = 0## i.e. it does no work.

But now that you mention it, there is a relativistic term that appears in non-inertial frames to second order in ##\vec{v}## of the form ##\vec{F}_{\text{rel}} = 2m \vec{v} (\vec{a}\cdot \vec{v})## which becomes negligible for ##||\vec{v}|| << c## but will have to be taken into account for ##||\vec{v}|| \sim c##; here ##\vec{a}## corresponds to the acceleration of the frame itself so ##\vec{F}_{\text{rel}}## is a correction to the inertial acceleration. It won't matter in this case because the rotating frame is non-accelerating so ##\vec{F}_{\text{rel}} = 0## but for more general settings it would be relevant.

See MTW exercise 6.8.
 
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  • #6
WannabeNewton said:
because the rotating frame is non-accelerating

I gather you meant the rotating frame moves with constant angular velocity?
 
  • #7
Sunfire said:
I gather you meant the rotating frame moves with constant angular velocity?

I'm assuming you're talking about starting with an inertial frame and rotating it about a fixed axis with some constant angular velocity ##\vec{\omega}## or equivalently performing the transformation ##\phi \rightarrow \phi - \omega t## of the cylindrical coordinates in the inertial frame.

In such a case the rotating frame so achieved is non-accelerating in the sense that an observer fixed at its origin has vanishing acceleration. This can be made more mathematically precise but it's not really important at the moment.

The main point is you still have a centrifugal potential in the rotating frame that gives rise to a centrifugal acceleration ##a_c = \gamma^2 \omega^2 r## where ##\gamma^2 = (1 - \omega^2 r^2)^{-1}##.
 
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  • #8
WannabeNewton said:
I'm assuming you're talking about starting with an inertial frame and rotating it about a fixed axis with some constant angular velocity ##\vec{\omega}## or equivalently performing the transformation ##\phi \rightarrow \phi - \omega t## of the cylindrical coordinates in the inertial frame.

Yes, the rotation is uniform (makes things easier :biggrin:).

WannabeNewton said:
In such a case the rotating frame so achieved is non-accelerating in the sense that an observer fixed at its origin has vanishing acceleration.

This is interesting, that you can actually think of the observer as rotating only, but not experiencing any acceleration because he is at the center. But he can still measure the centrifugal gravity, correct?
If I understand you correctly, additional relativistic effects are avoided by having the rotating observer at the center of the frame.

WannabeNewton said:
The main point is you still have a centrifugal potential in the rotating frame that gives rise to a centrifugal acceleration ##a_c = \gamma^2 \omega^2 r## where ##\gamma^2 = (1 - \omega^2 r^2)^{-1}##.

Then I guess the work to bring ##m## from ##R## to the center will be

##\int_R^0 m(r) a_c(r) dr= \int_R^0 \frac{m_0}{\sqrt(1-\frac{\omega^2r^2}{c^2})}\frac{\omega^2 r}{(1-\frac{\omega^2r^2}{c^2})}dr=\int_R^0 \frac{m_0 \omega^2 r dr}{(1-\frac{\omega^2r^2}{c^2})^{3/2}}=\frac{m_0c^2(\sqrt{c^2 - \omega^2 R^2}-c)}{\sqrt{c^2 - \omega^2 R^2}}##

which diverges, because ##\omega R=c##. This disagrees with post #2. Either ##a_c## is wrong or I did something wrong in the derivation.
 
  • #9
Sunfire said:
This is interesting, that you can actually think of the observer as rotating only, but not experiencing any acceleration because he is at the center.

If I'm floating in free space with my arms extended out and a pellet traveling at a constant speed hits one of my palms in a completely elastic collision then from conservation of momentum and angular momentum I will gain some constant speed and I will also start rotating. My acceleration is still zero but my rest frame is now rotating.

Sunfire said:
But he can still measure the centrifugal gravity, correct?

Yes.

Sunfire said:
If I understand you correctly, additional relativistic effects are avoided by having the rotating observer at the center of the frame.

Yes.

Sunfire said:
which diverges, because ##\omega R=c##.

I don't see why you're using ##\omega R = c##. You can't have the particle start out at the light barrier that's physically impossible.

Sunfire said:
This disagrees with post #2. Either ##a_c## is wrong or I did something wrong in the derivation.

The expression can be rewritten as ##m_0 c^2 (1 - \gamma)## which is exactly what Scott had. Note that the expression is negative, as we would expect.
 
  • #10
WannabeNewton said:
I don't see why you're using ##\omega R = c##. You can't have the particle start out at the light barrier that's physically impossible.

I meant ##\omega R \rightarrow c##
 
  • #11
Sunfire said:
I meant ##\omega R \rightarrow c##

Well that clearly diverges because it takes an infinite amount of energy to bring a particle to the light barrier.
 
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  • #12
WannabeNewton said:
Well that clearly diverges because it takes an infinite amount of energy to bring a particle to the light barrier.

Makes sense... Ah, and in the classical limit ##\omega R = v << c##, then

##\gamma = \frac{1}{\sqrt{1-\frac{\omega^2 r^2}{c^2}}} \approx 1+\frac{1}{2}\frac{\omega^2 r^2}{c^2}## then

##m_0c^2(\gamma-1) \approx m_0 c^2 (1+\frac{1}{2}\frac{\omega^2 r^2}{c^2} -1) = \frac{m_0 \omega^2 r^2}{2}##

This settles it then :smile:
Thank you to all who contributed!
 

Related to Energy for moving mass up a centrifugal gravity barrier

1. What is a centrifugal gravity barrier?

A centrifugal gravity barrier is a hypothetical concept in physics that suggests the existence of a barrier of centrifugal force that prevents objects from moving beyond a certain point in space.

2. How does energy affect the movement of mass up a centrifugal gravity barrier?

Energy plays a crucial role in moving mass up a centrifugal gravity barrier. The amount of energy required will depend on the barrier's strength and the mass of the object being moved. More energy is needed to overcome a stronger barrier or to move a larger mass.

3. Can energy be converted into mass to move objects up a centrifugal gravity barrier?

According to the theory of relativity, energy and mass are equivalent, so it is possible to convert energy into mass. However, this would require an immense amount of energy, and it is currently beyond our technological capabilities.

4. Is it possible to create a centrifugal gravity barrier artificially?

At this point in time, it is not possible to create a centrifugal gravity barrier artificially. The concept is still theoretical and has not been proven to exist in reality. However, some scientists are researching ways to manipulate gravity, which could potentially lead to the creation of a centrifugal gravity barrier.

5. How would the existence of a centrifugal gravity barrier impact space travel?

If a centrifugal gravity barrier does exist, it could greatly impact space travel. It could act as a natural barrier, making it more difficult for spacecraft to travel beyond a certain point in space. However, it could also potentially be used to our advantage, allowing for more efficient and faster space travel within our solar system.

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