# Wouldn't a negative mass be going faster than $c$ according to our current models of relativity?

I heard that there are some physicists trying to figure out, out least hypothetically, how things with positive and negative mass may interact with each other.

I'm really confused about how this can even be a possibility at all. I don't know much about relativity, so please bear with me if I'm completely wrong (I'm not one of those guys). But I thought mass and time were fundamentally linked. So if positive mass is going forward through time, wouldn't the hypothetical negative mass be moving backward? So how could such an interaction even take place in the first place?

Edit:

There has been some confusion about what exactly I'm talking about, and I think rightly so. I'm not talking about Newtonian mechanics or quantum mechanics, or even practicality. I am talking about our current understanding of relativity.

The moving backwards through time thing would require some sort of multidimensional graph thing, which I can't even imagine, so I'll try to explain what I was thinking another way.

So, from one observers p.o.v., as an object moves closer to $$c$$, like an electron in a particle accelerator, I thought it's mass moves toward infinity. Then the asymptote, where $$c$$ is, has 0 mass so it looks like it pops from +infinity mass to 0 mass right at $$c$$. Then, continuing with that, wouldn't it then do the inverse and pop from 0 mass right at $$c$$, to -infinity mass. Then, as it's negative mass moves toward 0, it's velocity increases toward infinity. That's the graph I was imagining.

There aren't things zipping around the universe blowing everything up, so the greater picture I'm imagining isn't correct. I just don't understand how this is not our current theory.

I read something about inertial mass versus?, so maybe that's key but I'm not really sure.

The part about time I was referring to was the clocks on the airplanes thing, where time moved slower and slower. It's a different type of idea and kind of only exists in my intuition, so I really don't know how to explain it or even if I'm just imagining things. It is still related to the idea above, though. But the crux is, if the negative mass somehow started on our side then it would move toward 0 time passed when moved close to $$c$$ and back again. But it would be on the opposite end of the multidimensional graph I'm somewhat imagining, so it would be moving backwards as it moved back to our 0 relative velocity.

• Have you read the Wikipedia article on negative mass? Commented Oct 13, 2020 at 2:55
• @Allure It doesn't talk about how negative mass would move through time. Am I just wrong on that concept? Commented Oct 13, 2020 at 3:12
• It does mention it - "In 1970, Jean-Marie Souriau demonstrated, using Kirillov's orbit method and the coadjoint representation of the full dynamical Poincaré group, i.e. the group action on the dual space of its Lie algebra (or Lie coalgebra), that reversing the arrow of time is equal to reversing the energy of a particle (hence its mass, if the particle has one)." I'm not an expert though, so I won't comment further. Commented Oct 13, 2020 at 3:18
• Your one statement is strange that positive mass moves forwards in time. Who says that it moves forwards in time. If you see with respect to time ,it's just there, it's not moving forward or backward. So if you record a video of a positive mass and you inverse that video, you'd still be able to see that the positive mass will follow all laws of physics correctly. Commented Oct 13, 2020 at 4:22
• @Ritanshu you are moving in forward direction in time and so is every other normal matter (with forward direction of time being that sequence of occurrence of phenomenas where entropy increases).
– user249968
Commented Oct 13, 2020 at 4:51

In relativistic quantum mechanics it can be shown that time reversal operator commutes the same way as parity inversion operator :

$${\text{T}}H{\text{T}}^{-1} \equiv {\text{P}}H{\text{P}}^{-1}$$

Where $$H$$ is energy operator, $$T$$ - time reversal operator and $$P$$ is parity transformation operator. What this means ? Consider this picture of a cannon shooting a projectile :

In A pic a time is reversed and in B pic - $$x$$ coordinate is reversed. It appears that both these cases produce same transformation to physical laws at hand. Or in layman terms, if you watch a projectile flying backwards in a movie, you can't distinguish if it is due to a movie being played backwards or simply due to a movie frame image flipping along X axis. Both transformations produce same effect.

In the quantum mechanics there can't be negative energy, because lowest possible energy is that of vacuum. It is greater than zero and can't be lower than that.

But if we look at it from classical Physics eyes, just out of curiosity, then according to second Newton law change in a particle speed can be expressed as :

$$\Delta v = m^{-1}F~(t_2-t_1)$$

So if you want to reverse a particle speed, keeping force fixed, you have to either reverse a time flow ($$t_2 < t_1$$) OR put a negative mass $$-m$$ into equation.

However it should be noted that negative mass is highly speculative thing. Because it is shown that two $$+$$ and $$-$$ masses would produce a "Runaway motion". Positive mass would be repelled from a negative one, but a negative one would be attracted to positive one at the same time ! This would put system in constant "runaway", where negative mass indefinitely tries to catch a positive one. Thus producing system self-acceleration with no external force or field applied. Put these opposite sign masses on a wheel, and you will get a perpetual motion machine. Which will break many laws, such as first or second law of thermodynamics or even a general relativity, because rotating device will get more massive for no apparent reason. (Albeit particle system total momentum and total kinetic energy remains zero). Gas composed from $$+-$$ mass particle mixture would also act very strange,- gas part composed of positive mass would increase in temperature without a bounds as well as second part composed of negative mass particles would gain a negative temperature also without a bound, balancing each other. There are some attempts to solve this runaway motion paradox, however this is still "a work in progress".

• I'm trying to look at this from the relativity model perspective. I think I edited my question since you made this answer, so it might might what I'm thinking a little clearer. Also, I'm not trained in physics so, while I can understand the math part of the equations, I have no idea what the letters mean. Though, I really wish I did :( . Commented Oct 13, 2020 at 14:38
• I've added description of quantum mechanical operators. Btw, first equation is from relativistic quantum mechanics, so it has some sort of connection to relativity. Commented Oct 13, 2020 at 18:11
• I've added a more visual and better explanation about time and parity symmetry transformation, check this out ;-) Commented Oct 13, 2020 at 20:49
• Thanks for the awesome picture! Commented Oct 14, 2020 at 13:13

It is difficult to know exactly what “moving backwards in time” means so I am going to interpret your question as “does replacing mass with negative mass give the same result as replacing $$t$$ with $$-t$$ ?”.

Newton’s second law $$F=ma$$ does not change if you replace $$t$$ with $$-t$$. In other words, if a force accelerates a mass from $$v_1$$ to $$v_2$$ then if we reverse time the same force accelerates the same mass from $$-v_2$$ to $$-v_1$$ - the acceleration is the same, and is still in the same direction as the direction of the applied force.

But if we replace $$m$$ with $$-m$$ then Newton’s second law becomes $$F=-ma$$. In other words a negative mass will accelerate in the opposite direction to the applied force. This is a different outcome than just reversing time.

• Why is it $-v_2$ to $-v_1$ instead of $v_2$ to $v_1$ Commented Oct 13, 2020 at 6:03
• @RyderRude Velocity is the first derivative of position w.r.t. time, so reversing time reverses all velocity vectors. If you play a film backwards all motions are reversed. But acceleration vectors are unchanged because acceleration is the second derivative w.r.t. time. Commented Oct 13, 2020 at 7:53
• I was thinking like a 1/x type graph. I thought that, as something approaches c, it's mass increases and, as a result, moves slower through time. Then 'c' has no mass, so it would be essentially 'timeless'? Then something on the other side of the asymptote would have negative mass, but be a 1/x type reflection so time would be negative as well. Commented Oct 13, 2020 at 12:48
• I feel like I am missing a key point (which is why I'm here, ty!). Commented Oct 13, 2020 at 13:05
• Or, maybe from the observers view, a negative mass would (hypothetically) be on the other side of c, and would only be going faster as it's -m moved closer to 0? So, again, how could they even meet? Commented Oct 13, 2020 at 13:27