Dark energy question

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1. Feb 12, 2015

binderberger

not sure where to post this question... kind of a thought experiment... if a mass is given an initial force in a frictionless environment, like space. it would continue to move forward away from the applied force at a constant rate until acted upon by another force. if you were to remove some of the mass without affecting the remaining mass in any other way, the velocity of the mass should increase in order to keep the F=MA equation balanced. well, if the initial force is the big bang. and the stars/galaxies are the mass moving away from the big bang at a "constant" rate. as the stars slowly turn mass in to energy through fusion, shouldn't the stars slowly accelerate as they lose mass? or rather, wouldn't this be a better explanation as to why stars appear to be accelerating than the vague concept of dark energy? or, at least, could this explain away some of the acceleration?

Last edited: Feb 12, 2015
2. Feb 12, 2015

phinds

You are assuming a center to the universe. There is none. The big bang singularity did not happen at a point in space. I recommend the page linked to in my signature.

3. Feb 12, 2015

binderberger

even within the confines of the balloon theory my question is still valid. if two stars are moving away from each other because of an applied force that is no longer being applied, their velocity should still be constant. but if the mass of the stars are slowly reducing due to a conversion of mass to energy the velocity should still increase...???

4. Feb 12, 2015

phinds

There is, as far as I am aware, no such thing as a "balloon theory". There is a "balloon analogy" to help understand some things about cosmology.

I think your question is addressed by the fact that conservation of energy is a local phenomenon and has no meaning on cosmological scales.

5. Feb 12, 2015

binderberger

ok, thanks for that explanation. I guess the question i still have is why would it not have any effects on a cosmological scale? if a star is moving away from us at a certain velocity and that star is losing mass, why would it not appear to be moving faster away from us from our point of view? relative to us...

6. Feb 12, 2015

phinds

That would seem to imply a relationship between us and the star that does not exist. I'm beginning on one hand to share your confusion as to why it wouldn't but then I ask my self "why would it" and I see no answer that that either. I think you are trying to conserve momentum but that is a type of conservation of energy and we're back to my previous answer.

7. Feb 12, 2015

binderberger

well, there is no relationship between the two stars except that they are both moving away from each other. or, away from some imaginary arbitrary point between the two stars. but the motion of those two stars relative to each other or relative to any arbitrary fixed point should slightly increase if the mass of the star were to slightly decrease. i guess i dont understand why f=ma wouldn't apply on a cosmological scale.

8. Feb 12, 2015

phinds

F=ma uses acceleration. What acceleration are you talking about?

9. Feb 12, 2015

binderberger

the star's motion away from each other. or, from our point of view, the star's motion away from us. if that momentum is constant but the mass decreases, the velocity should increase in order to balance the equation. because the mass that is turned in to energy is dropped from the equation the remaining mass has that same amount of force divided among less mass...

10. Feb 12, 2015

Blackberg

When there is no force, there is no acceleration : F = MA becomes 0 = 0. This is Newton's first law, and it is independant of mass. In other words, reducing mass alone does not make anything accelerate. You need a force that is present. If it is different than 0, then there is acceleration, and reducing mass will increase acceleration.

11. Feb 12, 2015

binderberger

i guess i mean momentum. momentum is mass x velocity. if a star has a certain momentum away from an arbitrary point. if you reduce the mass, the velocity has to increase to keep the momentum constant. because you cant gain or lose momentum without being acted upon by an outside force...??? excuse my lack of knowledge of the terms...

12. Feb 13, 2015

DelcrossA

Isn't the radiation from stars isotopic? So why would the center of mass of the star get an impulse in any specific direction?
As an analogy, a bomb will only become a rocket if some mechanism for directing the mass ejection is engineered.

13. Feb 13, 2015

Staff: Mentor

That is not correct. The F=MA equation describes a relationship between an applied force, acceleration, and mass. Specifically that the magnitude of the acceleration is equal the applied force divided by the mass of the object. It does not say that removing mass will cause an object's velocity to increase. For that you need a force. Note that an equation is not an explanation, theory, or definition. It's just a mathematical tool. It's up to us to use it correctly.

14. Feb 13, 2015

Staff: Mentor

The momentum doesn't remain constant. A star is constantly losing mass and momentum by radiating it out into space in the form of EM radiation and solar wind.

15. Feb 13, 2015

Bandersnatch

$F=ma$ and $F=dp/dt$ are equivalent. Whether you mean acceleration or change in momentum, a force must act on the centre of mass of the system to change its state of motion.

There are many facets to this confusion but I think it mostly stems from the mistaken assumption that the conservation of momentum holds for magically 'disappearing' mass. It doesn't, because there is no such thing as simply disappearing mass. Mass (=energy) always leaves via some process, and carries away any momentum that the reminder of the system losses.

When you shoot a gun, the momentum of the bullet changes because there are forces acting on it, and similarly the momentum of the gun changes because there is an opposite force acting on the gun. Taken together the momentum is conserved, but the changes in the momenta of the components of the system are the result of concrete forces acting on each one of them.

If the net momentum of the lost mass is 0, as is the case with radiation (it has momentum too ) leaving the star isotropically, then the reminder of the system doesn't gain or lose any momentum. All the forces resulting from the isotropically leaving radiation cancel each other out.

Furthermore, as seen from the outside of the system, i.e., taken together the star plus the radiation it has so far emitted, its total mass remains unchanged (where we calculate the mass-energy contribution of the radiation).

But if that is not convincing enough, maybe this will disabuse you of the idea: consider the mass lost in the lifetime of a star due to radiation amounts to less than 0.7% of its total mass (the mass defect in hydrogen fusion). The expansion of the universe is observed to carry both stars and inert (non-fusing) gas in the galaxies at recession velocities many times the speed of light.
This tells you that even if what you envisaged was a real process it could not account for the velocities observed, and that it lies beyond the domain of applicability of simple Newtonian mechanics due to the velocities involved.

@Drakkith how is a star losing momentum (in some arbitrary reference frame)? See DelcrossA's post #12.

16. Feb 13, 2015

binderberger

Bandersnatch: the mass of a star lost is very small, but the "acceleration" due to dark energy is very small too. i am not trying to explain the total velocity of the star. i am trying to explain the increase in velocity which is now attributed to dark energy.
if a star has a given momentum at the start of its life with 100% of mass and constant velocity in relation to an arbitrary point.
at the end of its life it has that same momentum but only 99.3% solar mass so its velocity should be higher...???
picture a 1000kg mass moving at 100 m/s its momentum is 100*1000 so 100,000 Kg-m/sec
now imagine the 1000kg mass lost .7% of its mass so 993 kg is left. but its momentum was unchanged since the radiation is isotropically lost.
so 100,000 Kg-m/sec momentum = 993kg * its new velocity of 100.7049
this increase in velocity would be spread out over the course of the star's billions year life cycle so it would be a very slight increase in velocity but, to my understanding, that lines up perfectly with dark energy since dark energy only increases velocity by a slight amount also...

17. Feb 13, 2015

Bandersnatch

You focused on the least important bit. The most important one is the one mentioned in post #12: radiation is emitted equally in all directions, so the net impulse is 0. No force, no change of momentum.

This is not correct for the reasons stated earlier. No mass is ever 'lost' - you need to include the radiation that was emitted in the balance if you want to use the shortcut of conservation of momentum.
It's like saying that me and my bike are moving at 30km/h, so if I make the bike disappear I should start moving faster.
You can't just disregard parts of the system.

If you go the way of analysing forces you'll find out there was no net force acting on the centre of mass of the system, so it can't change momentum.

By the way, dark energy is not responsible for velocities of the recession increasing - they would do so even in a universe devoid of dark energy. It is responsible for the accelerated rate of change of those velocities.
This accelerated rate is a recent (cosmologically speaking) development - for the first 10 billion years or so of the history of the universe the net effect was that of deceleration. But still, even then the relative recession velocities of any two objects were increasing.

18. Feb 13, 2015

binderberger

The light is emitting in all directions so it doesn’t apply any force to the system. The point I am making is that the system is losing mass and so it has to increase in velocity to conserve momentum.
And, when I say losing mass I mean that the mass is being converted in to light energy which is emitted in all directions equally. Therefore it wouldn’t add or subtract anything from the momentum of the system. The momentum is constant throughout the life of the star. but a slight loss in mass to energy emitted in all directions equally would result in a slight increase in velocity. In order to conserve momentum.
And the fact that for the first 10 billion years ago the universe was slowing down actually agrees with my idea too.
For the first 10 billion years everything was closer together so gravity was stronger and trying to pull everything together. after 10 billion years gravity weakened due to distance so that the ever so slight increase in velocity that resulted from the loss of mass began to have a greater effect. That results in a slow increase in velocity that is currently blamed on dark energy.
In your analogy if your bike were to be transformed in to light and that light is emitted in all directions instantly it would not exert a net force on your system. The force would be exerted in all directions simultaneously and equally and would therefore cancel each other out. That would leave you alone with the same momentum as you had when you were with your bike. Assuming the bike weighed the same as you, the loss (conversion of your bike to light) of this bike would cut the mass of your system in half but your momentum would be the same so your speed would have to double…
100kg person plus bike moving at 30 m/s has a total momentum of 3000
50kg person without bike would have to move at 60 m/s to keep the momentum a constant 3000.
My idea is that the concept of dark energy is something that can be explained away using regular Newtonian physics. Conservation of momentum etc…

19. Feb 13, 2015

phinds

So you figure all those thousands of physicists who have been trying for 15+ years to figure out what dark energy is have not thought of this and ruled it out?

I realize that's not an answer to the question of WHY you are wrong, but you can maybe see how there's really good chance that you ARE wrong, yes?

EDIT: and by the way, if you were right and were the first to figure this out there would be a Nobel Prize in your future.

20. Feb 13, 2015

Bandersnatch

No.
Enough of imprecise words.

The system before the conversion and after has the same total momentum. It is composed of the person and the bike. The bike is a part of the system both before and after conversion (afterwards it's made of light, but it's still there).
Write down the conservation of momentum equation.

The same argument applies for the stars, so write it out for a star if you like.

21. Feb 13, 2015

Staff: Mentor

It's losing mass, so its momentum would necessarily be reduced. I'm not including the emitted radiation as part of the star's system.

What about it? The center of mass of the star doesn't change of course, nor does the velocity of the star.

Momentum is conserved. It is carried away by radiation. It does not simply disappear. There is no force acting on the star to accelerate it so its velocity does not change. (Ignoring expansion) Conservation of momentum tells us that momentum cannot disappear. If an object loses momentum, then it must be transferred somewhere else, as is the case here where the light carries away momentum. If I launch a hot container of soup into space at a certain velocity, it will still have that same velocity after cooling off, even though the mass has changed.

Your understanding of the law of conservation of momentum is incorrect, so your idea doesn't hold. Besides, cosmologists have been working on the issue of what the universe is like on a large scale for nearly a century. If we were able to use newtonian physics to explain our observations we would. We moved away from newtonian physics with the development of Relativity and Quantum Physics. The universe simply doesn't obey newtonian laws.

22. Feb 14, 2015

Bandersnatch

Ah. Sorry, got me some reading comprehension failure.
Never mind. Carry on.