2
$\begingroup$

As illustrated in the following diagram (A, B, C, D are 4 specified space points, and C is close to a black hole), a small ball at distance of a black hole is stationary (suppose now it's mass is m0) and begins to move towards the black hole along the path ABC due to the gravity of the black hole, so the ball moves faster and faster (now the gravity of the black hole is the unique force exerted on the ball), and its mass increases according to the special relative theory.

A-----------------------------> B ---------------------> C Black Hole
A<----------------------------- D <--------------------- C

After a long time (I think we don't need to care how long the time is), it arrives nearby the black hole at a very high speed, say, 0.99c. But we stop it now in any way (thus its mass reduces to m0; how to stop it? I think we don't need to care this neither -- I just want to release its kinetic energy so that its mass restores to m0; we don't consume our energy for its stopping, on the contrary, we collect kinetic energy from it), and drag it back very slowly along the path CDA (here "very slowly" means we can consider its mass stays at m0).

When the ball was moving on the path ABC, it had a much higher speed and more mass than it was on the path CDA, so, according to the special relative theory again, the black hole did more positive work when the ball was moving along ABC than the negative work done by the black hole when the ball was dragged back along CDA. So, the net work done by the gravity potential field of the black hole is not zero, which contradicts with basic physical laws – the gravity potential field does zero work on an object moving on a closed cycle. What is wrong?

Note: in my question, the black hole could be replaced by a normal star if you like; but using a black hole can enlarge the effect of the special relative theory so that we can image that the mass of the ball experiences great change during its motion.

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ "But we stop it now" - do you want to stop it without doing any work? - there's the hole in your reasoning. $\endgroup$ – Mithoron Mar 15 '15 at 23:58
  • $\begingroup$ That in which way we stop it doesn't matter. In the revised version of my question, I answered Mithoron. We don't need do any work but can gain energy from the ball. $\endgroup$ – adamant07 Mar 16 '15 at 9:32
  • $\begingroup$ @Mithoron "the hole in your reasoning", A black hole. ;p $\endgroup$ – Mark N Jun 12 '15 at 19:32
  • $\begingroup$ Wouldn't the distance from the black hole just be potential energy for the ball. Which you then let convert to kinetic energy. Which you then negate/remove (THERE'S YOUR PROBLEM)? Similarly, if you hold a ball off the ground, it has potential. If you dropped it onto a table, the table absorbs the kinetic energy, and it still has some potential left from the perspective of the earth. These would sum to the total potential energy you started with. [Note that special relativity should have no effect on these scenario's, just more possible conversions] $\endgroup$ – Mark N Jun 12 '15 at 19:37
  • $\begingroup$ I think you're mistaken in considering the sole application of gravity here. External forces are also acting on the system (for example- the work done to stop the ball). On another note, you are using the concept of relativistic mass too casually (please read this). $\endgroup$ – Prish Chakraborty Sep 16 '15 at 15:54
1
$\begingroup$

As a comment has hinted, there are several essential holes that need to be filled in the reasoning:

The ball moves faster and faster, and its mass increases according to the relative theory.

The ball moves faster according to whom? The mass increases according to whom?

After a long time, it arrives at the black hole at a very high speed, say, 0.99c.

After a long time according to whom? Has a very high speed according to whom?

But we stop it now, and drag it back very slowly along the path CDA.

Who's "we"? How does "we" do this?

What is wrong?

In general, 'energy' is not conserved in GR.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I modified my original question description according to Alfred's comments. I hope now it is more clear. Alfred's said, "In general, 'energy' is not conserved in GR." -- in this question, I didn't doubt energy is conserved; I just asked why the gravity potential field of the black hole did non-zero net work on the ball which moving on a closed cycle. Do you think this doesn't hold in GR? $\endgroup$ – adamant07 Mar 16 '15 at 9:27
  • $\begingroup$ A few years ago, I asked a professor of USTC(a China university famous for its physics, especially its quantum physics and astrophysics; see en.ustc.edu.cn ) the same question. He gave me a brief answer, saying this is not a paradox, but to understand it needs the knowledge of GR. As his reputation, I think probably he was right. $\endgroup$ – adamant07 Mar 16 '15 at 9:43
0
$\begingroup$

If we drag up also the kinetic energy of the ball, then dragging will be very hard.

If we leave the kinetic energy of the ball in the hole, then dragging the ball up is not very hard. The kinetic energy will now reside in a very low gravitational potential. The kinetic energy fell into a hole seemingly gaining energy while falling.

Let's say we drop and lift the ball many times, we drop it into a pool of water. That will cause lot of energy to become stored in the water, water becomes boiling hot. It kind of seems like energy is being created in this process. If the energy in the water actually is a large amount of energy, then energy is created. But the energy of large pool of hot water is not a large amount of energy when the energy resides very low. Pool full of hot water deep in a gravity well is equivalent to mug full of hot water at higher altitude.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Your answer does not catch the point. "dragging is hard" or "not hard" is not essential for this question. $\endgroup$ – adamant07 Mar 16 '15 at 9:21
  • $\begingroup$ It's quite essential! Ok answer was not good. How about now? $\endgroup$ – stuffu Mar 16 '15 at 10:22
  • $\begingroup$ I feel that user7027 and I are considering different things. I just focus on the ball, or more specifically, how much work the black hole has done on the ball when the ball finished a cyclic travel along the closed path ABCDA. In this case, it seems that the work is not 0; but to my best of knowledge, the gravity potential field must do zero work on an object moving on a closed cycle. However, it seems that user7027 spent his most time on explaining the kinetic energy of the ball, thus deviated from our topic. $\endgroup$ – adamant07 Mar 16 '15 at 21:35
  • $\begingroup$ The kinetic energy you extract from the ball is negated by increasing the balls potential energy along the path cda, $\endgroup$ – Triatticus Jun 12 '15 at 22:29
0
$\begingroup$

Here is a partial analysis of what happens in the process described by the person who asked the question. To make it a complete analysis would require the analysis of the energy extraction and energy storage machinery down in the gravity well.

Process1: A system consisting of a ball and a black hole pulls itself together. If the system is doing any work, it is doing work on itself, so the energy of the system stays constant.

Process2: An external agent stretches a system consisting of a ball and a black hole. Work is being done on the ball-black hole system by the external agent.

Now if we take one ball and one black hole, and perform the following series of operations with those two objects:

prosess1, process2, process1, process2 ... and so on, then the energy of the ball-black hole system increases as long as we continue the process, and energy of the motor driving the process decreases.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I like the concept of the processes and the ball-black hole system, which makes the question more clear and easier to understand. But I think user7027's final conclusion is wrong: "and so on, then the energy of the ball-black hole system increases as long as we continue the process," -- should be "and so on, then the energy of the ball-black hole system DECREASE as long as we continue the process,"; recall that the ball release its kinetic energy at location C. In a word, the system releases energy if we iteratively perform these two processes. $\endgroup$ – adamant07 Mar 17 '15 at 9:49
  • $\begingroup$ The released energy from the ball at location C is more than what we need to drag the ball back to location A. $\endgroup$ – adamant07 Mar 17 '15 at 10:04
  • $\begingroup$ Is the released energy perhaps radiated to space? Because there's an interesting little fact: Radiation climbing uphill does work on the gravitating object. $\endgroup$ – stuffu Mar 17 '15 at 13:45
  • $\begingroup$ :You gave me a great inspiration. It seems that the released energy should be separated from the ball-black hole system, so we won't consider it. But in fact we should not separate them, so the released energy is still in the ball-black hole system. We have neglected the fact that the ball is not that ball -- it is always changing its mass when it is moving from A to C. The final effect is that (the ball + release energy) becomes heavier and heavier, the black hole becomes lighter and lighter, the former absorbs energy from the later, and the whole process is like an energy pump! $\endgroup$ – adamant07 Mar 18 '15 at 3:47
-1
$\begingroup$

I think Mithoron's answer is correct and the question should apply in any gravity, not just a black hole.

For any object in free fall or orbit, the potential energy plus kinetic energy will remain the same (ignoring air resistance and tidal/sheering forces). So the object a mile up with no motion has some potential energy (extra mass), while the object, that was dropped from a mile in the air, if you measure it 1 inch off the ground, it's full of kinetic energy but the net energy and mass is the same.

But once you stop the ball, the ball loses mass. A ball 1 inch off the ground has less mass than a ball 1 mile off the ground.

A fun question to think about is, how does the ball 1 mile up and the ball 1 inch up, know where the ground is, and know how much potential energy it should have?

:-)

Not sure that's really an answer, but it's a fun question. Gravity confused a lot of scientists for a long time. A few physicists even told Einstein not to work on it after he'd published his special, before he'd published his general. They warned him, Gravity is "too hard".

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.