<p><em>NOTE: We ignore the energy it takes to break the binding energy between the electrons in the object required to break it into two pieces.</em></p>
<p>A clue to the answer is that the difference in question between the stationary object and the two moving objects is <em>very very small.</em> That usually indicates that something is either quantum or relativistic. Surprisingly, no matter how slowly the two masses move apart, the explanation is relativistic.</p>
<p>Einstein himself wondered about this when he considered an object that fell from a shelf. The discrepancy comes from the fact that the 2 moving objects are experiencing time passing at a slower rate.</p>
<p>Without going into the math of Lorentz groups, the 4-dimensional momentum (which includes the mass) is unchanged. The two halves merely rotated in 4 dimensions.</p>
<p>interestingly(?), ALL acceleration in any number of dimensions is simple rotation in N+1 dimensions. You can see that in the motion of a candle on the edge of a rotating record player. Turn out the light.</p>
<p>Seen from the side, the candle is accelerating back and forth endlessly in 1 dimension. You'd think all that continual acceleration consolidation and deceleration would take energy!</p>
<p>But seen from the top, by adding a 2nd dimension, it is seen to be merely rotating. And it will rotate forever without using any energy.</p>
<p>Momentum through space involves motion in space, but momentum through time is equivalent to mass. Feynman told me that in his Australia lecture and in the Easy Pieces book. Actually, he said, &quot;Persistence in time is the same thing as mass.&quot; Unquote.</p>
<p>He should have won the Nobel Prize just for saying that. For example, quantum particles that have no mass do not experience the passage of time. That's why.</p>

CC BY-SA 4.0
2024-06-21T16:54:29.940Z
Sorry for my way of asking, but are you really answering the above question?
KeShAw
CC BY-SA 4.0
2024-06-21T17:10:03.660Z
I did leave out the tensor equations, so I suppose that yes, this is incomplete. But if you want the ultimate truth, you have  to understand the language that god speaks - math.
Miss Understands
CC BY-SA 4.0
2024-06-05T12:03:33.833Z
2024-06-05T12:28:13.510Z
2024-06-05T12:28:13.510Z
Miss Understands
<p>Consider a Universe which has only one object (bigger particle). The object exploded (somehow) into two smaller particles (with some release of energy). Consider the two particles to have masses M and m, and their speed be V and v respectively. Momentum Conservation demands MV = mv.</p>
<p>From the frame of an observer (not included in Universe), the total energy of the Universe is 1/2MV^2 + 1/2mv^2. But from the frame of the particle with mass M, total energy of the Universe is 1/2m(v + V)^2, while it is still different w.r.t particle of mass m. Why this discrepancy? Cannot we consider the total energy of the Universe to be the same w.r.t every observer, or in other words, cannot we consider the total energy of the Universe to be a physical constant? (Have we to include the radiation energy of explosion as well?)</p>

CC BY-SA 4.0
2024-06-04T00:51:23.033Z
The sole reason for including this question here is because it consists of the term 'Universe'. Spare me on that part!
KeShAw
CC BY-SA 4.0
2024-06-04T03:52:32.157Z
Kinetic energy isn't frame-invariant. The answers here (& on the linked questions) may be helpful: https://physics.stackexchange.com/q/51220/123208
PM 2Ring
CC BY-SA 4.0
2024-06-04T12:17:37.440Z
@PM2Ring I am talking about total energy of the Universe. Including all forms of energy, shouldn't it be a constant?
KeShAw
CC BY-SA 4.0
2024-06-04T13:04:52.720Z
It *is* constant, in any given frame. But the constants are different in different frames. That's explained in some of the linked answers.
PM 2Ring
CC BY-SA 4.0
2024-06-04T13:13:14.467Z
Generally, we can't define an absolute energy of a system. But that doesn't actually matter, since it's the energy *differences* that tell us what's going on. In your simple closed system, we only have kinetic energy, and it never changes. In a more elaborate system, we'd also have potential energy, rest-mass energy, etc. Those various energies would change, but the total would be invariant, within a given frame.
PM 2Ring
<universe>
Is the total energy of the Universe not invariant?
KeShAw
# Is the total energy of the Universe not invariant?

(Asked by: username_0 on 2024-06-04)

Consider a Universe which has only one object (bigger particle). The object exploded (somehow) into two smaller particles (with some release of energy). Consider the two particles to have masses M and m, and their speed be V and v respectively. Momentum Conservation demands MV = mv.

From the frame of an observer (not included in Universe), the total energy of the Universe is 1/2MV^2 + 1/2mv^2. But from the frame of the particle with mass M, total energy of the Universe is 1/2m(v + V)^2, while it is still different w.r.t particle of mass m. Why this discrepancy? Cannot we consider the total energy of the Universe to be the same w.r.t every observer, or in other words, cannot we consider the total energy of the Universe to be a physical constant? (Have we to include the radiation energy of explosion as well?)

**Tags:** universe
> **Comments:**
> **username_0**: The sole reason for including this question here is because it consists of the term 'Universe'. Spare me on that part!  
> **username_1**: Kinetic energy isn't frame-invariant. The answers here (& on the linked questions) may be helpful: https://physics.stackexchange.com/q/51220/123208  
> **username_0**: @PM2Ring I am talking about total energy of the Universe. Including all forms of energy, shouldn't it be a constant?  
> **username_1**: It *is* constant, in any given frame. But the constants are different in different frames. That's explained in some of the linked answers.  
> **username_1**: Generally, we can't define an absolute energy of a system. But that doesn't actually matter, since it's the energy *differences* that tell us what's going on. In your simple closed system, we only have kinetic energy, and it never changes. In a more elaborate system, we'd also have potential energy, rest-mass energy, etc. Those various energies would change, but the total would be invariant, within a given frame.  

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## 1 answers

### Answer by username_2 on 2024-06-05. Score: 0

*NOTE: We ignore the energy it takes to break the binding energy between the electrons in the object required to break it into two pieces.*

A clue to the answer is that the difference in question between the stationary object and the two moving objects is *very very small.* That usually indicates that something is either quantum or relativistic. Surprisingly, no matter how slowly the two masses move apart, the explanation is relativistic.

Einstein himself wondered about this when he considered an object that fell from a shelf. The discrepancy comes from the fact that the 2 moving objects are experiencing time passing at a slower rate.

Without going into the math of Lorentz groups, the 4-dimensional momentum (which includes the mass) is unchanged. The two halves merely rotated in 4 dimensions.

interestingly(?), ALL acceleration in any number of dimensions is simple rotation in N+1 dimensions. You can see that in the motion of a candle on the edge of a rotating record player. Turn out the light.

Seen from the side, the candle is accelerating back and forth endlessly in 1 dimension. You'd think all that continual acceleration consolidation and deceleration would take energy!

But seen from the top, by adding a 2nd dimension, it is seen to be merely rotating. And it will rotate forever without using any energy.

Momentum through space involves motion in space, but momentum through time is equivalent to mass. Feynman told me that in his Australia lecture and in the Easy Pieces book. Actually, he said, "Persistence in time is the same thing as mass." Unquote.

He should have won the Nobel Prize just for saying that. For example, quantum particles that have no mass do not experience the passage of time. That's why.

> **Comments:**
> **username_0**: Sorry for my way of asking, but are you really answering the above question?  
> **username_2**: I did leave out the tensor equations, so I suppose that yes, this is incomplete. But if you want the ultimate truth, you have  to understand the language that god speaks - math.  

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