<p>Gravitational wave sources do lose energy (and angular momentum) by emitting gravitational waves. That is what causes the orbits of binary systems that emit gravitational waves to spiral inwards.</p>
<p>An isolated mass <em>can</em> emit gravitational waves, but in order to do so it must have a distorted, non-axisymmetric geometry <em>and</em> be spinning. An example could be a spinning sphere with a lump on one side.</p>
<p>Such an object will emit gravitational waves that carry away both energy and angular momentum. The net result is that the spin of the object will slow, or the mass would relax into a more symmetric configuration; both of which would have less energy and angular momentum than the initial state. Note that this would happen even if there were no other possible ways of losing energy (e.g., through heating).</p>
<p>If the isolated asymmetric mass were not spinning then it would not emit gravitational waves.</p>

CC BY-SA 4.0
2024-12-09T21:28:49.000Z
An object with mass doesn’t decay if there is no external force the object would remain with the same mass hence the same force of gravity. The orbit mentioned is attraction or curvature of space time of another object if there is no such object in the vicinity then mass and gravity will remain the same over time.
Solo Man
CC BY-SA 4.0
2024-12-09T21:32:02.443Z
@SoloMan If there were  no other object there would be no gravitational waves.
ProfRob
CC BY-SA 4.0
2024-12-09T22:32:01.287Z
@SoloMan As Rob says, a body at rest emits no gravitational radiation. But even for typical moving systems, gravitational wave energy is pretty small. Eg, if we consider the Earth-Moon system in isolation, the average gravitational radiation power of the Earth & Moon orbiting their common centre of mass is only ~7.4 microwatts, which is totally negligible relative to their orbital kinetic energy.
PM 2Ring
CC BY-SA 4.0
2024-12-11T14:18:00.003Z
@ProfRob How does an object "know" that there's no other objects in order to not emit any gravitational waves?
ajd138
CC BY-SA 4.0
2024-12-11T14:25:19.863Z
@ajd138 because the other object would have a gravitational influence.
ProfRob
CC BY-SA 4.0
2024-12-11T17:57:35.370Z
@ProfRob I think that the concept of an "isolated mass" is questionable to begin with -- most nuclei are "isolated" from each other, aren't they. The key is asymmetry and rotation, and that's only possible with spatially distributed mass.
Peter - Reinstate Monica
CC BY-SA 4.0
2024-12-09T18:08:57.393Z
2024-12-10T10:35:25.927Z
2024-12-10T10:35:25.927Z
ProfRob
<p>A single motionless mass does not emit gravitational waves.  Such a mass has a gravitational field around it, but it emits no radiation. It will remain the same over time. It will not gain or lose energy or mass.</p>
<p>A pair of masses in orbit do emit gravitational radiation. This causes the system of the two masses to get closer together.  Energy is conserved. It is changed from potential energy to gravitational radiation. The total energy is unchanged.</p>

CC BY-SA 4.0
2024-12-10T13:28:23.867Z
"

A single mass does not emit gravitational waves": taken at face value, this is false. Suppose you have two bodies orbiting each other, emitting gravitational waves. Now according to your claim, if we tied a piece of string between them, that would make them a single mass, so they would stop emitting gravitational waves. But this is clearly absurd. It is true that a single spherically symmetric mass, or a non-rotating mass, doesn't emit gravitational waves; but that is not what you wrote.
TonyK
CC BY-SA 4.0
2024-12-10T22:15:02.550Z
edited. But this seems to go to the heart of the OP misconception.  They seem to suppose that the production of a gravitational field requires energy to maintain the field, in the form of gravitational waves.  I've seen this kind of misconception before: that a gravitational field is produced by the emission of gravitational waves that then impinge on other objects and in doing so attract them.  Needless to say, that's not what gravitational waves are.
James K
CC BY-SA 4.0
2024-12-11T19:34:51.993Z
If that's the issue, this seems no different from the equivalent question for electromagnetism: why doesn't a charged particle lose energy from radiating photons? (Answer: because although the electromagnetic field is mediated by photons in a certain sense, a stationary lone charge does not radiate real photons.)
Glenn Willen
CC BY-SA 4.0
2024-12-09T22:29:47.933Z
2024-12-10T22:15:34.167Z
2024-12-10T22:15:34.167Z
James K
<p>If gravitational waves exists shouldn’t the object keep losing energy? But since objects that aren’t radioactive do not lose energy, how is law of conservation energy upheld for objects at rest if they are emitting gravitational waves ?</p>

CC BY-SA 4.0
2024-12-10T21:38:44.667Z
"If" they exist? I think physicists are pretty confident that they do.
Barmar
<gravity>
Law of conservation of energy with gravitational waves
Solo Man
# Law of conservation of energy with gravitational waves

(Asked by: username_0 on 2024-12-09)

If gravitational waves exists shouldn’t the object keep losing energy? But since objects that aren’t radioactive do not lose energy, how is law of conservation energy upheld for objects at rest if they are emitting gravitational waves ?

**Tags:** gravity

> **Comments:**
> **username_1**: "If" they exist? I think physicists are pretty confident that they do.  

---

## 2 answers

### Answer by username_2 on 2024-12-09. Score: 18

Gravitational wave sources do lose energy (and angular momentum) by emitting gravitational waves. That is what causes the orbits of binary systems that emit gravitational waves to spiral inwards.

An isolated mass *can* emit gravitational waves, but in order to do so it must have a distorted, non-axisymmetric geometry *and* be spinning. An example could be a spinning sphere with a lump on one side.

Such an object will emit gravitational waves that carry away both energy and angular momentum. The net result is that the spin of the object will slow, or the mass would relax into a more symmetric configuration; both of which would have less energy and angular momentum than the initial state. Note that this would happen even if there were no other possible ways of losing energy (e.g., through heating).

If the isolated asymmetric mass were not spinning then it would not emit gravitational waves.

> **Comments:**
> **username_0**: An object with mass doesn’t decay if there is no external force the object would remain with the same mass hence the same force of gravity. The orbit mentioned is attraction or curvature of space time of another object if there is no such object in the vicinity then mass and gravity will remain the same over time.  
> **username_2**: @username_0 If there were  no other object there would be no gravitational waves.  
> **username_3**: @username_0 As Rob says, a body at rest emits no gravitational radiation. But even for typical moving systems, gravitational wave energy is pretty small. Eg, if we consider the Earth-Moon system in isolation, the average gravitational radiation power of the Earth & Moon orbiting their common centre of mass is only ~7.4 microwatts, which is totally negligible relative to their orbital kinetic energy.  
> **username_4**: @username_2 How does an object "know" that there's no other objects in order to not emit any gravitational waves?  
> **username_2**: @username_4 because the other object would have a gravitational influence.  
> **username_5**: @username_2 I think that the concept of an "isolated mass" is questionable to begin with -- most nuclei are "isolated" from each other, aren't they. The key is asymmetry and rotation, and that's only possible with spatially distributed mass.  

---

### Answer by username_6 on 2024-12-09. Score: 7

A single motionless mass does not emit gravitational waves. Such a mass has a gravitational field around it, but it emits no radiation. It will remain the same over time. It will not gain or lose energy or mass.

A pair of masses in orbit do emit gravitational radiation. This causes the system of the two masses to get closer together. Energy is conserved. It is changed from potential energy to gravitational radiation. The total energy is unchanged.

> **Comments:**
> **username_7**: "  A single mass does not emit gravitational waves": taken at face value, this is false. Suppose you have two bodies orbiting each other, emitting gravitational waves. Now according to your claim, if we tied a piece of string between them, that would make them a single mass, so they would stop emitting gravitational waves. But this is clearly absurd. It is true that a single spherically symmetric mass, or a non-rotating mass, doesn't emit gravitational waves; but that is not what you wrote.  
> **username_6**: edited. But this seems to go to the heart of the OP misconception.  They seem to suppose that the production of a gravitational field requires energy to maintain the field, in the form of gravitational waves.  I've seen this kind of misconception before: that a gravitational field is produced by the emission of gravitational waves that then impinge on other objects and in doing so attract them.  Needless to say, that's not what gravitational waves are.  
> **username_8**: If that's the issue, this seems no different from the equivalent question for electromagnetism: why doesn't a charged particle lose energy from radiating photons? (Answer: because although the electromagnetic field is mediated by photons in a certain sense, a stationary lone charge does not radiate real photons.)  

---

