Big Bang vs Antimatter: Exploring the Mystery

In summary: This article discusses how the angular correlation of the CMB in the universe can be explained without invoking cosmic variance. It uses the theory of the Rh = ct universe.
  • #36
Chalnoth said:
The gravitational matter density would be negative in that case, yes.
And therefore would behave like a cosmological constant, gravitationally repelling(i). But that's confusing, because as you said, an anti-matter universe shouldn't be distinguishable from a matter universe. So, perhaps (i) is wrong or anti-matter doesn't mean negative mass, or something else?
 
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  • #37
timmdeeg said:
And therefore would behave like a cosmological constant, gravitationally repelling(i). But that's confusing, because as you said, an anti-matter universe shouldn't be distinguishable from a matter universe. So, perhaps (i) is wrong or anti-matter doesn't mean negative mass, or something else?
A cosmological constant acts very differently. In Newtonian terms, it's effectively like adding an extra term to the force equation:

[tex]F = G m_1 \left(-{m_2 \over r^2} + {\Lambda r \over 3}\right)[/tex]

This is the force of mass 2 on mass 1 (so the cosmological constant adds a pure acceleration term of any two objects away from one another).
 
  • #38
Chalnoth said:
A cosmological constant acts very differently. In Newtonian terms, it's effectively like adding an extra term to the force equation:

[tex]F = G m_1 \left(-{m_2 \over r^2} + {\Lambda r \over 3}\right)[/tex]

This is the force of mass 2 on mass 1 (so the cosmological constant adds a pure acceleration term of any two objects away from one another).
Yes, I understand, but instead had the Friedmann term ##(\rho c^2 + 3p)## in my mind. As we agreed that the gravitational matter density ##\rho## is negative, it should act repelling, like negative pressure. On the other side a cloud consisting of particles with negative gravitational mass each, should gravitate attractive due to the acceleration law (because of ##m^2##). So, there must be a severe misconception. I suspect, I can't interpret the Friedmann term in that way and it would be great, if you could help me out.
 
  • #39
timmdeeg said:
Yes, I understand, but instead had the Friedmann term ##(\rho c^2 + 3p)## in my mind. As we agreed that the gravitational matter density ##\rho## is negative, it should act repelling, like negative pressure. On the other side a cloud consisting of particles with negative gravitational mass each, should gravitate attractive due to the acceleration law (because of ##m^2##). So, there must be a severe misconception. I suspect, I can't interpret the Friedmann term in that way and it would be great, if you could help me out.
I think you'd have to re-derive that equation with the new mass concept. I'm not entirely sure how it would turn out. I think it could be done in a consistent manner, but I'm not sure.
 
  • #40
Chalnoth said:
I think you'd have to re-derive that equation with the new mass concept. I'm not entirely sure how it would turn out. I think it could be done in a consistent manner, but I'm not sure.
Or, perhaps, there is a real contradiction, revealing that the conjecture to assign a negative gravitational mass for anti-matter is unphysical.
 
  • #41
How is mass defined at the current moment?
Is it the result of attractive forces or the result of how we weigh it or perhaps of how the matter interacts with gravitation?

In case it is the result of attractive forces it will remain positive even if it is perhaps anti-matter.
In case it is the result of interactions with gravitation is will change depending on whether or not the antimatter falls towards Earth or remain still(moves away perhaps)
if it is the result of how we weigh it... Gravitation, for all the fun people out there.

Sorry if this is an unwanted reply
 
  • #42
Can we stop talking about negative gravitational mass? It's been brought up at least twice that this is a non-starter because it disagrees with data.
 
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