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General relativity predicts that normal mass (positive mass) results in the curvature of spacetime which in return leads to gravitation. Since space and time are bonded together, any change on the fabric of space may inevitably lead to a change in time, as postulated by Einstein's theory of relativity. So the effect of positive mass on time is that it slows it down through the formulation of an attractive gravitational field, but what happens to time in the presence of negative mass?

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    $\begingroup$ From wiki: negative mass is a hypothetical concept of matter whose mass is of opposite sign to the mass of normal matter, e.g. −2 kg. Of couse, you being God n'all, I suppose anything's possible :) $\endgroup$ – user81619 Jun 3 '15 at 15:44
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    $\begingroup$ My gut feeling tells me that time becomes imaginary, which would mean a transition from a classical to a quantum mechanical regime. That may actually be perfectly physical, but it wouldn't mean that time would run backward, it doesn't do that in the quantum mechanical treatment of other fields, either. $\endgroup$ – CuriousOne Jun 3 '15 at 17:53
  • $\begingroup$ Have a look at this question about what tachyons actually do. $\endgroup$ – ACuriousMind Jun 3 '15 at 18:29
  • $\begingroup$ Imaginary Time is a new one to me. I'm going to have to read up on that a bit. $\endgroup$ – userLTK Jun 3 '15 at 19:23
  • $\begingroup$ General relativity is time symmetric physicsforums.com/threads/…, so there is no "arrow of time" in it. Whatever effect negative mass might have it can not reverse what isn't there in the first place. But the question of what effects negatively massive objects would have on proper times is interesting, some effects of negative mass are discussed here physics.stackexchange.com/questions/44934/… $\endgroup$ – Conifold Jun 4 '15 at 2:19
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The time dilation factor with respect to an observer at infinity is

$$\sqrt{1-\frac{\text{2 G M}}{\text{c}^2\text{ r}}}$$

so if we plug in G=1, c=1, r=10 and M=+1 we get the clocks running slower by a factor of 0.8944 if they are in a distance of 10GM/c² from the center of the positive mass.

If we change the sign of M to M=-1 we get a time dilation factor of 1.095 so mathematically the clocks should run faster near negative masses than they would for a field free observer.

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The arrow of time is believed to be related to the fact that the universe started in a state of low entropy and is evolving towards a state of larger entropy. The effect of negative mass will not change this. The reason is that any model of negative mass will leave the initial state of the universe as as state of low entropy. A rather uniform distribution of mass in the presence of gravity, be it either attractive or negative, will still constitute a state of low entropy, whereas states of high entropy will be created as mass clumps start to form either as an effect of positive or negative mass. The second law of thermodynamics is a statistical theorem that can be derived from very general principles, without the details of the force fields, thus there nothing a priori that would make a negative mass to reverse the arrow of time.

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  • $\begingroup$ Entropy always increases with time but there is no reason to assume that entropy controls the direction of time. Hawking acknowledged that he was wrong when he assumed time would reverse if the Universe contracted and entropy began to increase. $\endgroup$ – Peter R Jun 19 '16 at 0:31
  • $\begingroup$ @PeterR do you mean instead that if the universe contracts the entropy begins to decrease? otherwise why would the arrow reverse? $\endgroup$ – Wolphram jonny Jun 19 '16 at 0:33
  • $\begingroup$ @PeterR first, I will agree with you that using the second law to account for the arrow of time is not universally accepted, however I believe that the majority of physicists believe on that explanation. $\endgroup$ – Wolphram jonny Jun 19 '16 at 1:00
  • $\begingroup$ second, I do not believe that the arrow of time or the entropy decreases just because the universe is bouncing back. I never heard a coherent reason as to why this should be so. So, I disagree. $\endgroup$ – Wolphram jonny Jun 19 '16 at 1:01

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