The answer for the total mass of the Universe?

[WeirdUniverse]

TL;DR Summary

The principle of relativity posits that the fundamental laws governing physical phenomena remain invariant across all inertial frames of reference. However, a pertinent quandary about mass arises.

it requires less energy to impart acceleration to an object than to the entirety of the universe, This would break the principle of relativity since the amount of energy to move an object should be the same as the amount of energy to move the rest of the universe. This observation implies a diminutive total mass for the universe.

Denoting the mass of the universe as 'M', the mass of an object as 'm', and the acceleration of that object as 'a' (vector), we can express the resultant force acting on the object and the rest of the universe as:
F = ma = (M - m)(-a).

Thus, by equating the opposing forces, we get:
-aM = 0, which means M equals zero.

Consequently, if this proposition is true then the total mass of a relative universe is zero and the negative masses are real.
(information about the nothingness universe: https://en.wikipedia.org/wiki/Zero-energy_universe)

 

[Hill]

F = ma = (M - m)(-a)

In the equation above, you assume that "the rest of the universe" has acceleration (-a). What is a justification for this assumption?

 

[Dale]

TL;DR Summary: The principle of relativity posits that the fundamental laws governing physical phenomena remain invariant across all inertial frames of reference. However, a pertinent quandary about mass arises.

This would break the principle of relativity since the amount of energy to move an object should be the same as the amount of energy to move the rest of the universe

This analysis is incorrect. To show that something breaks the principle of relativity you have to work it out in two different reference frames and show that the measurable outcome is different. In the case of energy, what you find is that all measurable outcomes are the same, the energy is different in different frames, but the conservation of energy holds in each frame.

 

[WeirdUniverse]

In the equation above, you assume that "the rest of the universe" has acceleration (-a). What is a justification for this assumption?

I'm still wondering, But If you replace it with (a) instead, M will equal to 2m which is wrong because the total mass does not depend on a single object. (-a) might make sense if we dive a little closer to negative masses.

Another answer is because observers moving at constant velocities relative to one another will observe the same physical phenomena (same for all inertial frames of reference of an object) we can conceptually accelerate the entire universe by any desired amount and the laws of physics will remain consistent. So aM = 0M and M=0.

This analysis is incorrect. To show that something breaks the principle of relativity you have to work it out in two different reference frames and show that the measurable outcome is different. In the case of energy, what you find is that all measurable outcomes are the same, the energy is different in different frames, but the conservation of energy holds in each frame.

I don't understand, because that's what im trying to express.

 

[jbriggs444]

Denoting the mass of the universe as 'M', the mass of an object as 'm', and the acceleration of that object as 'a' (vector), we can express the resultant force acting on the object and the rest of the universe as:
F = ma = (M - m)(-a).
Thus, by equating the opposing forces, we get:
-aM = 0, which means M equals zero.

@Hill (not @Dale) has already asked you to justify your assumption that the acceleration ($a$) of the object is equal and opposite to the acceleration ($-a$) of the rest of the universe. I want to set that fundamental misunderstanding to one side and address the algebra instead.

There is a sign error in the algebra.

You equated the two forces. But the two forces are equal and opposite.

What is the force on the object? $F = ma$
Assume (erroneously) that the object's acceleration $a$ is equal and opposite to the universe $-a$.
By Newton's third law, the force on the rest of the universe is equal and opposite: $-F=(M-m)(-a)$
Simplify: $F=(M-m)a$
Substitute the first equation into this: $ma = (M-m)a$
Simplify. $2ma = Ma$

Done correctly, the $ma$ does not cancel.

One can check the result in the case of two balls, each of mass $m$ alone in a universe that totals mass $M = 2m$. Here, the accelerations are equal and opposite. And indeed, $2ma = Ma$.

With any other mass ratio, things go awry since the acceleration $a$ of the one will not be equal and opposite to the acceleration $-a\frac{m}{M-m}$ of the other.

Edit: One ought always to be suspicious of mathematical reasoning that leads to a physical result without requiring experiment.

 

[Hill]

I'm still wondering, But If you replace it with (a) instead, M will equal to 2m which is wrong because the total mass does not depend on a single object. (-a) might make sense if we dive a little closer to negative masses.
Another answer is because observers moving at constant velocities relative to one another will observe the same physical phenomena (same for all inertial frames of reference of an object) we can conceptually accelerate the entire universe by any desired amount and the laws of physics will remain consistent. So aM = 0M and M=0.

The body and "the rest of the universe" are two different objects and there is no reason for them to have the same acceleration. The body has acceleration a, and "the rest of the universe" has acceleration 0. Thus, its mass is infinitely large.

 

[Hill]

@Dale has already asked you to justify your assumption that the acceleration (a) of the object is equal and opposite to the acceleration (−a) of the rest of the universe.

No, he did not.

 

[jbriggs444]

No, he did not.

Sorry, you are right. That was you, not @Dale. I apologize for the mistake.

Edit: I am reminded of the inscription on Margaret Houlihan's wedding ring.

 

[Hill]

Sorry, you are right. That was you, not @Dale. I apologize for the mistake.

No problem.

 

[Dale]

I don't understand, because that's what im trying to express

Where are your calculations for two different frames? Where are your calculations for the conservation of energy in the two different frames? You did neither of those. So you have no basis to claim "This would break the principle of relativity".

Instead of making grandiose claims, you need to work on the very basics. This is not even special or general relativity, it is just standard Newtonian relativity that you have wrong here.

 

[WeirdUniverse]

The body and "the rest of the universe" are two different objects and there is no reason for them to have the same acceleration. The body has acceleration a, and "the rest of the universe" has acceleration 0. Thus, its mass is infinitely large.

Why acceleration 0?

With any other mass ratio, things go awry since the acceleration a of the one will not be equal and opposite to the acceleration −amM−m of the other.

That's not true because from the object's viewpoint, the universe must have an acceleration of -a equal to its acceleration.

 

[PeterDonis]

F = ma = (M - m)(-a).

This is the relativity forum. This equation is Newtonian physics, which is certainly not a good approximation for the universe as a whole.

by equating the opposing forces

Gravity is not a force in GR.

Your whole analysis is invalid because of these two errors.

 

[PeterDonis]

it is just standard Newtonian relativity

There is no valid Newtonian model for the whole universe, and the OP is trying to analyze the whole universe. So I don't think it's even "Newtonian relativity" (or Galilean relativity) that the OP has wrong here; the whole basis for the OP's analysis is wrong.

 

[WeirdUniverse]

There is no valid Newtonian model for the whole universe, and the OP is trying to analyze the whole universe. So I don't think it's even "Newtonian relativity" (or Galilean relativity) that the OP has wrong here; the whole basis for the OP's analysis is wrong.

Then how do I approach solving this question "it requires less energy to impart acceleration to an object than to the entirety of the universe"?

 

[PeterDonis]

how do I approach solving this question "it requires less energy to impart acceleration to an object than to the entirety of the universe"?

The question itself doesn't make sense. You can't "impart acceleration" to the entire universe.

Why do you care about this question? What are you actually trying to figure out?

 

[WeirdUniverse]

Because that's what Galilean relativity about, How can we discern whether we are in motion or if the world moves beneath our feet?

 

[PeterDonis]

Because that's what Galilean relativity about, How can we discern whether we are in motion or if the world moves beneath our feet?

If you are, as I suspect, sitting at your computer as you post here, you are not moving relative to the Earth, so the world is not moving beneath your feet. The world and you are certainly moving relative to other things, such as the Sun, or the Andromeda galaxy. All motion is relative, yes, that is what relativity says.

What I still don't understand is how the scenario you were trying to pose in your OP of this thread has anything to do with relativity.

 

[Dale]

You can't "impart acceleration" to the entire universe.

Per Newton’s 3rd law, you can impart acceleration to the rest of the entire universe (excluding yourself).

 

[WeirdUniverse]

What I still don't understand is how the scenario you were trying to pose in your OP of this thread has anything to do with relativity.

Consider an object 'O' with mass 'm' initially at rest in space, observed from our point of view. When 'O' is accelerated to a vector 'a', two scenarios arise: either 'O' moves relative to us, leading to a force of F = ma acting upon it, or the rest of the universe (with a mass of M-m) appears to accelerate in the opposite direction, resulting in a force of F = (M-m)(-a). According to the principle of relativity, which asserts that the laws of physics are consistent across all inertial frames of reference, both scenarios yield identical forces which means ma = (M-m)(-a) and then conclude M = 0

 

[jbriggs444]

appears to accelerate in the opposite direction,

Relative to the non-inertial frame where you are always at rest, despite your acceleration. But Newton's laws require that you do the analysis in an inertial frame.

@PeterDonis might well note that there is no such thing as a global inertial frame spanning our entire universe. You will need to restrict your attention to a more limited region where inertial frames exist as a locally valid approximation.

 

[Dale]

Consider an object 'O' with mass 'm' initially at rest in space, observed from our point of view. When 'O' is accelerated to a vector 'a', two scenarios arise: either 'O' moves relative to us, leading to a force of F = ma acting upon it, or the rest of the universe (with a mass of M-m) appears to accelerate in the opposite direction, resulting in a force of F = (M-m)(-a). According to the principle of relativity, which asserts that the laws of physics are consistent across all inertial frames of reference, both scenarios yield identical forces which means ma = (M-m)(-a) and then conclude M = 0

The $a$’s are not the same. As you have already been told. And when you invoke the principle of relativity you need to look at different reference frames. As you have also already been told.

 

[WeirdUniverse]

The $a$’s are not the same. As you have already been told. And when you invoke the principle of relativity you need to look at different reference frames. As you have also already been told.

Why are the a's different? They must be the same. if A moves with velocity 'v' relative to B, then from A's perspective, B also appears to move with velocity v.

 

[Dale]

This analysis is ridiculous in the extreme because it treats the universe as a rigid body with a mass of $M$.
The results should be understood not as showing anything about the universe, but rather just how to set up a problem in standard Newtonian physics regarding the principle of relativity.

Consider two objects, mass $M$ and mass $m$ respectively, that interact with a force. By Newton's 3rd law $$\vec F_m =- \vec F_M$$$$m\vec a=-M \vec A$$giving an acceleration of$$\vec a=\frac{\vec F_m}{m}=\frac{-\vec F_M}{m}=\frac{-M\vec A}{m}$$$$\vec A=-\frac{m}{M}\vec a$$

If that force acts for a time $t$ then in a reference frame where $M$ and $m$ were initially stationary they would accelerate to a final velocity $$\vec v_f = t \vec a$$$$\vec V_f = t \vec A=-t\frac{m}{M}\vec a$$so the measurable difference in their velocity would be $$\vec v_f-\vec V_f=t \vec a + t \frac{m}{M}\vec a=t\frac{M+m}{M}\vec a$$

Similarly, in a reference frame where $M$ and $m$ were initially travelling at velocity $\vec v_i$ they would accelerate to final velocities $$\vec v_f = t \vec a+\vec v_i$$$$\vec V_f = t \vec A+\vec v_i=-t\frac{m}{M}\vec a+\vec v_i$$so the measurable difference in their velocity would be $$\vec v_f-\vec V_f=t \vec a +\vec v_i + t \frac{m}{M}\vec a - \vec v_i=t\frac{M+m}{M}\vec a$$ So the measurable final result is independent of the reference frame. Again, this is a ridiculous analysis because we are treating the universe as a rigid body of mass $M$, but it shows:

1) The accelerations are not equal
2) The principle of relativity holds just fine
3) There is no implication that $M=0$ (nor that $M+m=0$)

Why are the a's different? They must be the same. if A moves with velocity 'v' relative to B, then from A's perspective, B also appears to move with velocity v.

The fact that they agree on their relative velocity has absolutely nothing to do with their respective accelerations. Your reasoning here is a non-sequitur.

 

[WeirdUniverse]

So the measurable final result is independent of the reference frame.

Yes, it does because both objects were added with the same velocity so the difference in their velocities don't change.
But I'm not talking about their collisions, If we push a box, either the box gained the acceleration of a or the rest masses gained the acceleration of -a as we have experienced, but the total energy must be the same for that two inertial frames so there must be something to do with the masses.
There is another proof for this concept as I've said before:

Another answer is because observers moving at constant velocities relative to one another will observe the same physical phenomena (same for all inertial frames of reference of an object) we can conceptually accelerate the entire universe by any desired amount and the laws of physics will remain consistent. So aM = 0M and M=0.

 

[jbriggs444]

But I'm not talking about their collisions, If we push a box, either the box gained the acceleration of a or the rest masses gained the acceleration of -a as we have experienced

Neither is correct. Suppose that the box has mass $m$ and we (the pusher) have mass $M$.

Then in any inertial frame, the box will have acceleration $a_\text{b} = \frac{M}{M+m}a$ and we will have acceleration $a_\text{us} = -\frac{m}{M+m}a$.

In any inertial frame, the accelerations be the same. It does not matter what inertial frame you choose. In all of them, unless $M=m$ the two accelerations will not be equal and opposite.

The force on the box will be given by $F=ma_\text{b}=m\frac{M}{M+m}a$
The force on us will be given by $F=Ma_\text{us}=-M\frac{m}{M+m}a$

The forces are equal and opposite as predicted by Newton.

The difference between the two accelerations will be $\frac{M}{M+m}a - (-\frac{m}{M+m}a) = \frac{M+m}{M+m}a = a$

The relative acceleration is $a$ as prescribed by your scenario.

This is all using the rules of Newtonian mechanics.

Under special relativity, there is a distinction between relative velocity (the velocity of one thing in the rest frame of another) and closing velocity (the velocity difference between two things as measured in an inertial frame). For every day non-relativistic velocities, this is a minor correction and the Newtonian prediction is a very good approximation.

 

[Nugatory]

@WeirdUniverse Before you take on the problem you have posed (loosely, accelerating an object relative to the rest of the universe in a general relativistivc context) you might try analyzing a more intuitive problem: Two unequal masses being pushed apart by a repulsive force between them - for definiteness you can assume that the force is electrostatic - both masses are positively charged and repel one another. This will be purely classical physics, just Newton's laws and Galilean relativity.

Try your $F=ma$ calculations using an inertial frame in which both are initially at rest, the again using an inertial frame in which either one of them is momentarily at rest (note that there is no inertial frame in which either one remains at rest).

Once you have satisfied yourself that Galilean relativity works here, you will be better placed to think about your original problem. (And chances are that you will see the mistake in your initial analysis without further help from us).

 

[PeterDonis]

Consider an object 'O' with mass 'm' initially at rest in space, observed from our point of view. When 'O' is accelerated to a vector 'a'

Note that you are conflating two different types of acceleration here: coordinate acceleration, meaning the rate of change of O's coordinate position with respect to coordinate time in our reference frame; and proper acceleration, meaning the acceleration that O actually feels and which is measured by an accelerometer attached to O. They're not the same thing. Fortunately, however, we don't need to distinguish between them to see where you are making a mistake. See below.

two scenarios arise

No. There is only one scenario here: O's motion relative to us changes. The rest of the universe is irrelevant.

either 'O' moves relative to us, leading to a force of F = ma acting upon it

Yes--but note that F = ma assumes that a is proper acceleration, not coordinate acceleration.

or the rest of the universe (with a mass of M-m) appears to accelerate in the opposite direction

Nonsense. Applying a force to O can't change the motion of the rest of the universe relative to us. It can only change the motion of O relative to us.

 

[PeterDonis]

if A moves with velocity 'v' relative to B, then from A's perspective, B also appears to move with velocity v.

A and B here are O (= A) and us (= B), not (O and us = A) and (the rest of the universe = B).

 

[PeterDonis]

both objects were added with the same velocity

Nonsense. Only one object has a force exerted on it in your scenario: O. Nothing else has "velocity added".

 

[WeirdUniverse]

I'm confused, I think I'll analyze it mathematically later on. However, what could potentially refute the conclusions drawn from my second proof?

Another answer is because observers moving at constant velocities relative to one another will observe the same physical phenomena (same for all inertial frames of reference of an object) we can conceptually accelerate the entire universe by any desired amount and the laws of physics will remain consistent. So aM = 0M and M=0.

 

[Ibix]

However, what could potentially refute the conclusions drawn from my second proof?

In Newtonian physics, acceleration is frame invariant, so all frames measure the same acceleration $a$.

 

[PeterDonis]

what could potentially refute the conclusions drawn from my second proof?

Um, the fact that it's wrong?

observers moving at constant velocities relative to one another will observe the same physical phenomena

This is not correct. What is correct is that the laws of physics will be the same in all inertial frames. But that does not mean that all physical phenomena will be the same. For example, if you and I are moving relative to each other, we will measure the same light beam to have different energy and frequency--different physical phenomena. But the relationship between the light beam's energy and momentum will be the same for both of us--same laws of physics.

So your so-called "proof" is based on a false premise and is invalid.

 

[WeirdUniverse]

But If they have different physics phenomena in different inertial frames, doesn't mean that we would know if we are moving or not?

 

[PeterDonis]

If they have different physics phenomena in different inertial frames, doesn't mean that we would know if we are moving or not?

You would know you were moving or not moving relative to something else--such as the light source in my example. "Moving" is relative. You would not know you were moving or not moving in any absolute sense.

 

[WeirdUniverse]

You mean the universe is actually absolute, but it appears relative to us?