# Gravity and Energy

1. Apr 29, 2010

### ZirkMan

I would like to explore the relation of the effect of gravity (I deliberately do not say force) and energy. There is a simple situation I would like to understand in detail what happens.

The situation: I'm staying on a solid ground and then I jump up.

Now there are two sample enviroments I could do it.

A. a very small asteroid in deep space. For this situation of relatively small gravity and low energies (alternatively high energies in little stronger gravity) the law of conservation of energy will prevail and energy from my jump will be converted into kinetic energy. Therefore I will continue to move away from the asteroid until some other form of matter/energy will not interfere with me and change my relative speed.

B. I'm on the Moon. The situation is different here. After I jump up my new kinetic energy starts to diminish quickly until I stop relative to the surface of the Moon. Then without me anything doing and without interaction with any kind of energy I can detect I start to receive kinetic energy with opposite direction to my jump and keep receiving it more and more until I hit the surface again where the new kinetic energy is released as the energy of the impact.

Is this the correct depiction of the situation B.? Do I really start to loose my new kinetic energy (if yes, then the question is where does it go?) until it is transferred away entirely and then receive the new "gravity energy" from an undetectable source (if yes, then where does this energy come from)?

Or is it more like the gravity changes only the vector of my kinetic energy and the energy from my jump is the same energy as the energy of the impact?

Or something totally different happens?

2. Apr 29, 2010

### Matterwave

You need to also consider the gravitational potential energy. If you include that, you will find that KE+PE=constant (except for energy lost to heat or w/e other dissipative processes are present).

As you jump up, your KE, seeded from the mechanical energy produced by your legs, will start to decline as you increase your PE. After you hit maximum, your PE will star to decline as your KE increases. Once you hit the ground the energy is released into energy of impact.

3. Apr 29, 2010

### ZirkMan

I was expecting this answer. The problem is that I do not see how the potential energy (PE) can be really the energy I have produced in my jump. Does this PE contribute to my higher mass at the point where all my KE is converted to PE? Can this PE be converted to other forms of energy or mass other then KE?

And what about a scenario when I am the astronaut from the situation A. and happen to jump directly towards the Moon from situation B. I move towards the Moon first with a constant relative KE but as I move closer my KE rises and rises due to gravity until it is much more higher then what it was after I jumped from the asteorid. Where did this new energy come from? If I had this tremendous surplus energy as PE already on the asteorid does that mean that I was heavier on the surface of the asteroid then I was on the surface of the Moon after my impact? The loss of energy/mass would represent the difference in PE from state A. to state B.

4. Apr 29, 2010

### tiny-tim

All forms of energy are interchangeable.

uhh? your mass does not increase.

(even at relativistic speeds, your rest-mass stays the same)
All forms of energy are interchangeable.
You obviously think that energy and mass somehow change together. They don't.

Tell us what you think the relation between them is, so that we can check it.

5. Apr 29, 2010

### ZirkMan

You cannot measure rest-mass of a system at relativistic speed (or any relative speed for that matter), can you? You can only measure its relative speed and calculate its total energy from the rest mass you measured when the system was at relative rest to you.

If you measure mass of a system in relative motion the mass you get is not same as its rest-mass (I suppose the rest-mass is the mass you measure when the system is at rest i.e. with zero relative KE). The measured mass of a system with positive KE will be always higher because the KE has its mass as given by m=E/c2. This is how I understand it.
Ok, then we have the gravitational PE as a completely legitimate form of energy that is of the same nature as the energy all around us. As stated above that also must mean that this kind of energy must have relativistic mass as stated by m=E/c2. But you don't seem to like it. Tell me please what is wrong with the assumption that the PE could also be expressed in terms of mass like all other forms of energy.

6. Apr 29, 2010

### tiny-tim

The relativistic formula for energy of motion is mc2/√(1 - v2/c2).

(which incidentally = mc2 + 1/2 mv2 + smaller terms in v4/c2

the first term is a constant, and the second term is the Newtonian KE)

It is this energy which appears in the equation total energy = constant, and from which energy "lost" reappears as PE.

There is no exchange of energy and mass when something moves faster.

Your example of an astronaut pushing off, even at relativistic speeds, does not involve any exchange of energy for mass (unless, of course, his propulsion system involves nuclear fusion or fission, in which case obviously it does! ).
Energy can be converted into mass by creating new mass (or by annihilating mass), not by changing the speed of existing mass.

To include the equivalent-energy of the mass of the astronaut (which is the same before as after) is as pointless as including the chemical energy of the bonds in his molecules.

Yes, PE (and any other form of energy) can be measured in units of mass.

But if you do that in the astronaut's equation of motion, KE + PE = constant, you're simply changing everything in the equation by a factor of c2. ​

7. Apr 29, 2010

### ZirkMan

Oh now it makes sense! For some (strange) reason you substitute mass with matter where matter is a form of energy with properties like rest-mass, inertia etc. If you make this substitution then all you say above is true, of course. But for me mass is not matter. When I refer to mass I mean the relativistic mass "m" from E=mc^2. And this mass changes with change in total energy. Since KE as well as PE are all forms of the same energy E, PE should also be manifested as change in mass. For any observer this change of mass only due to change of energy is as real as it gets. Particles in accelerators get heavier the higher speed they have.

So the question is again: Is the astronaut on the asteroid heavier than the same astronaut after the crash on the moon because he has all that potential energy "on him"?

Or the same situation with an astronaut shot from the moon who just reached the point where all his KE has been transformed to PE. Is his mass heavier now as when he was on the moon before he was shot up because of all that PE he now carries?

8. Apr 29, 2010

### tiny-tim

PE has nothing to do with it …

the "heaviness" of the astronaut ("m" in the special relativity formula E = mc2) depends on his rest-mass, and on his speed as measured by the observer.

If the asteroid is moving faster than the Moon, relative to the observer, then yes the astronaut on the asteroid is "heavier" than the same astronaut on the Moon; but if the asteroid is moving slower, then he is less "heavy".

9. Apr 29, 2010

### ZirkMan

You see, one time you say that "All forms of energy are interchangeable" and "PE could also be expressed in terms of mass like all other forms of energy" and now you say that this is not the case for the gravitational PE and that only KE can change into mass.

Either one of these is not true or gravitational PE is not transformed KE.

10. Apr 29, 2010

### tiny-tim

Gravitational PE, like any other PE, can be exchanged with KE according to the formula PE + KE = constant
"heaviness" is not a different form of energy, it is simply a different name for KE/c2 (plus the rest-mass, which for a given body is constant) …

E/c2 = KE/c2 + mrest

it is not exchanged with KE, because it is KE (divided by a constant, and plus a constant).

As I said before, PE has nothing to do with "heaviness" of the astronaut.

11. Apr 29, 2010

### ZirkMan

Yes in this "mathematical universe" it works like this so you have no reason to doubt it. But I'm interested if it works like this in the real universe supported by experiment. Sometimes (very rarely) "map is not the territory". I'm not saying this is the case now.

Ok, there is no PE in the formula. Can you tell me why it is not there? Why should't the PE be part of E which equals mc^2 when PE + KE = constant?

12. Apr 30, 2010

### ZirkMan

Nobody is interested to explain why gravitational potential energy does not seem to have any relativistic mass, even when it is just transformed kinetic energy which does have relativistic mass? And I thought it was such an interesting topic!

13. Apr 30, 2010

### Ich

In static coordinates in a static spacetime, the total energy of a freely falling body is conserved.
For example in Schwarzschild coordinates, a body with energy E at rest at infinity will keep the same energy when it's fallig in. Part of the energy would then be attributed to its motion.
If it collides with the gravitating mass, it releases a large amount of energy (its kinetic energy). After this energy has been radiated away, the gravitating mass is indeed not increased by the rest mass of the body, but by its rest mass minus said energy. So in this sense, its "coordinate rest mass" (http://en.wikipedia.org/wiki/Komar_mass" [Broken]) is in fact reduced.

The issue is tricky, however, as a measurement of the body's rest mass will locally always yield the same result. Also, the released energy is much larger when measured at the impact site than when measured from far away, because there it arrives redshifted.

Last edited by a moderator: May 4, 2017
14. Apr 30, 2010

### ZirkMan

Ok, thank you for replies. In the meantime I was doing research on this topic and found a satisfying answer to my original question elsewhere. For the record here is a short summary in case somebody would find it helpful:

Situation B.

The loss and gain of energy of the jumping astronaut on the surface of the Moon is caused by gravitational time dilation which causes all energy in astronaut's body on the surface to move slower as when he is further from from the center of mass than on surface. That means his total energy on the surface is lower than when he is further away from it. His KE of the jump is consumed because he needs to raise his overall energy to match the faster rate of movement of energy associated with further distance from the center of mass. The KE is consumed as he gets higher and higher until it is completely used up and his motion upwards stops.

Because there is no force left to keep him on that level he starts falling down again. As his energy now slows down due to time dilation the redundant energy is released in form of the KE which is finally released away at the impact. Only that much energy is released how much was added from beginning and consumed during the jump.

Note that even though the total energy is conserved no PE is necessary to explain it! That means that gravitational PE is NOT a real form of energy. The PE is just a mathematical trick to maintain the law of energy conservation. And that is good so, because if the gravitational PE would indeed exist and be on par with other forms of energy which have relativistic mass such as KE we would be in big trouble. Imagine carrying on your back the PE of all the black holes in the Universe

15. May 1, 2010

### tiny-tim

Yes, potential energy is a mathematical trick …

it is simply another name for (minus) the work done by a conservative force.

So long as gravitation is conservative …

ie effectively, it only affects recoverable energy, and does not produce heat etc …

eg it is not conservative for the Moon's orbit, since that is gradually receding from the Earth because of energy "loss" through tidal forces

… it is mathematically convenient to supplement mechanical energy by this "potential energy" …

whether you wish to regard them as the same thing ("energy") is really only a matter of opinion and preference.

But this has nothing to do with relativistic mass, or any other aspect of relativity

it is valid for any conservative force.​
Since gravitation extends infinitely far, you are affected by their gravitation, and your momements do require work to be done against it, and so their PE must be taken into account.

(And your "Situation B" has nothing to do with time dilation. with "energy moving slower (?)" or with any other aspect of relativity)

16. May 1, 2010

### ZirkMan

Please, can you explain in more detail why this mechanism is not correct? It makes perfect sense for me, though I know it's just a simplification for static spacetime and there are more details involved such as length contractions that go hand in hand with the gravitational time dilation.

17. May 1, 2010

### tiny-tim

Gravitational potential energy is defined as minus the work done by a gravitational field.

Change in kinetic energy = work done (the work-energy theorem).

So change in kinetic energy = minus change in potential energy.

That's all it is!

(of course, we have to be able to define gravitational potential energy … it depends only on position, so it can only be defined in "static" general relativities)

18. May 1, 2010

### ZirkMan

The mechanism where gravity is caused by curved spacetime and its different total energy levels based on this curvature says the same. The main difference is that in this model the PE doesn't have to have relativistic mass because it simply doesn't exist. If it existed and had relativistic mass (as all forms of energy do) you would indeed had to carry the PE mass of all the matter in the Universe on your back which would make no sense.

19. May 1, 2010

### sheaf

OK, does this sound right ?

At the start of my jump, I have a velocity, so at this point my relativistic mass includes a contribution from my rest mass and my momentum. I continue upwards to my highest point, where my momentum goes to zero. At that point, my relativistic mass only includes my rest mass, the momentum contribution has gone away. The energy must have gone somewhere - it has been transferred to an increase in potential energy of the system consisting of me and the gravitational field. Energy conservation applies to the whole system - (me & field), not to me alone or to the field alone.

20. May 1, 2010

### tiny-tim

Hi sheaf!
Yes, that's fine

but relativistic mass = rest-mass, plus kinetic energy divided by c2 = total energy divided by c2 (m = m0 + KE/c2 = E/c2),

so change in relativistic mass = change in kinetic energy divided by c2 = change in total energy divided by c2 (∆m = ∆KE/c2 = ∆E/c2),

so you can substitute "energy" for "relativistic mass" throughout.
I don't really understand this …

when you say "the PE … simply doesn't exist", what have this and all your other posts, eg …

21. May 1, 2010

### ZirkMan

The problem with your and sheaf's explanation of the situation is that it makes perfect sense from mathematical point of view and for some limited situations (like static spacetime) it will give predictions that are in match with observations.

But because it uses purely mathematical entities like PE and "gravitational field" you can easily fall into a false feeling that these mathematical entities are real and that they fully explain what really happens. They are the reason everybody was fine with Newton's depiction of gravity for so long.

Problems with this model start when you dig in details and start to ask question like "what happens to the mass of the KE when the astronaut is in the highest point?", you answer: the energy is now in the gravitational field. Then I ask: And what is this gravitational field, how did the energy got there? Does it contain relativistic mass of transformed KE of all objects that are in the field? What is the energy and mass distribution of this gravitational field? etc. You would not be able to answer these questions from the framework of your mathematical theory, would you? That's why a new framework must come to scene.

To say that:
is like saying that Einstein has nothing to say about gravity, Newton explains everything.

22. May 1, 2010

### tiny-tim

No, I answer "whatever do you mean by 'the mass of the KE?' "
No, I said that Einstein (and relativistic mass, or any other aspect of relativity) has nothing to say about potential energy being simply another name for (minus) the work done by a conservative force.

23. May 1, 2010

### ZirkMan

That's correct. By this definition of PE Einstein would certainly not be necessary to explain anything.

But we have also used another definition that states that PE+KE=Constant. Because KE has relativistic mass it is natural to expect that PE(mass)+KE(mass)=Constant. Therefore it is also logical to ask what happens to PE(mass) when KE=0. And Einstein has certainly something to say from this point on.

24. May 1, 2010

### tiny-tim

That's not a definition, that's the work-energy theorem.
I have no idea what either of these means.

25. May 1, 2010

### ZirkMan

Ok but it is an important theorem that basically states that PE and KE are the same thing and that one changes proportionally into another. What is important is that this transformation should also include and conserve also the relativistic mass. This practically defines that the sum of relativistic mass of PE should equal relativistic mass of KE. That's what I meant when I wrote: PE(relativistic mass)+KE(relativistic mass)=Constant

Sorry I'm not able to state it more clearly than this.