Positive mass moves in the same direction as its momentum and if negative energy exists then it would move in the opposite direction from its momentum.
I was thinking of a type of mass, known as "side mass," that would move in a direction perpendicular to its momentum.
Could this type of mass exist and if so what would determine which direction perpendicular to its momentum it moves in?
It is pretty simple to prove this point.
If the velocity is perpendicular to the momentum, as a scientist, my first question is "which perpendicular"?
In Physics, there are plenty of forces that are perpendicular to the vector that causes it. Only thing is, in each and every case, those forces are also perpendicular to another vector as well.
Another point. Imagine an object that is moving around a massive object.
Clearly it would accelerate towards the massive object (momentum increases in the direction of the massive object).
This causes it to move increasingly quickly around the massive object.
This means that for your imaginary particle with side mass, gravity is what we call a Curl force.
In general curl force fields cannot have a point source, which the massive object IS. The reason being is that we can then create energy out of nothing by moving the object in a circle around the point source.
It is an interesting question. I have thought about it since I read many years ago a comic in which Magica De Spell made Donald Duck and Scrooge McDuck fall sidewards.
You can always find some laws that work, but maybe they cannot be mathematically elegant. I tried today to derive some mathematically elegant laws and got some serious constrains.
Your own question
what would determine which direction perpendicular to its momentum it moves in?
is the first problem. You cannot find a smooth way of doing it according to the hairy ball theorem: "you can't comb a hairy ball flat without creating a cowlick" i.e. "Every smooth vector field on a sphere has a [point with zero value]." - assigning direction perpendicular to given force is a vector field on a sphere with no zero value, so it cannot be smooth. We could use vector multiplication that gives a vector perpendicular to the given one, but its result's length is not constant and the length is zero if we multiply a vector parallel (or antiparallel) to the one by which we multiply.
More generally, we could use tensor mass (this is something like having both normal and "perpendicular" mass, taking cross product is also equivalent to multiplying by a tensor): $$\vec p = \hat m \vec v$$ (see note) that is $$p_i = \sum_{j=1}^3 m_{i j} v_j$$ (Similar equation is used for the moment of inertia: $\vec L = \hat I \vec \omega$ (6.3), p. 162)
Gravitational acceleration of body 1 (caused by body 2 with scalar mass (i.e. normal mass, a number)) $$\vec g_1 = - \frac{G m_2 \vec r_{21}}{r^3}$$ where $\vec r_{21}$ is distance from 2 to 1, $\vec r_{12}=\vec r_{21}$ and $r = |\vec r_{12}| = |\vec r_{21}|$
($\frac{\vec r_{21}}{r^3}$ has magnitude of $\frac{1}{r^2}$)
Force from 2 at 1 (we are here assuming the equality of the gravitational and inertial mass) $$\vec F_{21} = \hat m_1 \vec g_1$$ so (from Newton's third law of motion ("When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.") $$\vec F_{12} = - \vec F_{21} = m_2 \vec g_2 = - \hat m_1 \frac{G m_2 \vec r_{12}}{r^3}$$ so $$\vec g_2 = - \frac{G \hat m_1 \vec r_{12}}{r^3}$$
For a tensor $\hat m_2$ $$\vec F_{12} = - \frac{G \hat m_2 \hat m_1 \vec r_{12}}{r^3}$$ but we get a contradiction: $$\vec F_{21} \stackrel{\text{from analogy}}{=} - \frac{G \hat m_1 \hat m_2 \vec r_{21}}{r^3} \neq - \frac{G \hat m_2 \hat m_1 \vec r_{21}}{r^3} = - \vec F_{12} = \vec F_{21}$$ because tensors are not commutative ($\hat m_1 \hat m_2 \neq \hat m_2 \hat m_1$), so the situation is hard — we cannot keep Newton's third law and equality of the gravitational and inertial mass, at least in the general case. (We could keep everything in the form $a + b \hat x$ for a given $\hat x$ — such tensors commute — but I am not sure that there are no other problems.) We have to abandon something and then we risk loosing conservation of energy. This is still not so bad — we can write simulations without conservation of energy and, in the worst case, add an effect that keeps everything both stable and moving — but there are very many possibilities.
Besides, note that I did not even try to keep Einstein relativity.
note: $\overset{\scriptscriptstyle\leftrightarrow}{a}$ (I saw $\vec {\vec a}$ in equation (6.3) for the first time) can be used instead of $\hat a$ for a tensor and $\hat a$ can be used for an unit vector ($\hat a = \frac{\vec a}{|\vec a|}$)
Could there be a type of mass that moves in a direction perpendicular to its momentum?
I think a similar phenomenon may be possible, but it would need to be modified: Momentum that is perpendicular to the direction of movement of the mass
I'm not sure it's actually possible - but of the answers I've read here so far, these things seem to overlook or ignore an answer based on real physics: electromagnetism
Consider that there exists a force in electromagnetism where the force is perpendicular to the direction of movement/momentum. When electrons move, magnet fields are generated perpendicular to their direction of motion. There is a significant body of research on this, it's well known and proven. Do all the research you like.
So, if energy and momentum or mass and momentum... some similar combination... were in some way related like electricity and magnetism, then it may be that "momentum" could be perpendicular to the movement of the mass, like magnetism is perpendicular to the movement of electrons.
This may not be what you were asking, but I just thought you would want an answer that provided the possibility of something similar to what you asked. I hope this is helpful.
*I will mention that I majored in physics, and this sounds ridiculous, but this is a site about "world building" and what might be possible, not what can be proven.
EDIT:
I assumed electromagnetism was sufficiently easy to research that I didn't need citations, but since a comment was made, I will provide a reference to a wiki: https://en.wikipedia.org/wiki/Magnetic_field#Magnetic_field_due_to_moving_charges_and_electric_currents
And note that in the wiki, it discusses the "Lorentz Force":
F = q v X B
This is fairly analogous to the force equation:
F = m a = m v / t
Note that the cross product in the Lorentz Force means that the force is perpendicular to the velocity vector. However, this is constant in time because it describes the interaction of a quantity of charge (q) with a given velocity (v).
For your question, the problem is determining what this "field" is that would interact with the "momentum of mass" perpendicular to it's direction of movement. I provide this as an analogy - it may exist, and we may be like scientist were over a hundred years ago when they were first learning about the relationship between electricity and magnetism. But.. probably not. It's just a curious parallel to me.
This question's answer is reminiscent of that for others you have asked: Noether's theorem says no.
What you describe would be a thing completely and utterly unrelated in every way shape or form from momentum. It would not operate in any way related to momentum.
If you want a simple puzzler for this, consider that we live in 3-space. Given a single direction of motion, there's an infinite number of possible vectors at right angles to the motion. If your particle is moving "up," east, west, north, and south are all at right angles, as well as all the variants inbetween. So what is the particle's momentum?
Alternatively, fix the direction of momentum, which way is the particle going?
In either case, the result of this will looks so very much unlike momentum that it should not be thought of as a momentum replacement at all. There are other concepts, such as angular momentum, for which a vector, or sometimes a dyad, perpendicular to a direction of interest is meaningful. They just aren't momentum.
I may misunderstand what you are after, but the direction of momentum and movement are the same by definition:
$$ \vec{p} := m \vec{v}$$
An object cannot move perpendicular to its momentum, because the new direction of movement would then also be the direction of its momentum.
t >> E and v << c. That approximation does not hold under QM nor SR/GR.
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Maybe as an element of science fantasy, you might have a transuranic element that is stable because it contains subatomic particles that extend into other spatial dimensions besides the three common spatial dimensions. This extension causes it to look like it's moving perpendicular to the impulse, but when the extra dimensions are taken into consideration, it's actually still moving straight.
I'm going with no.
This would break one of the the fundamental rules of mechanics which basically says; perpendicular forces are entirely independent of each other.
Or as Newton put it: "Every action has an equal and opposite reaction", but you're asking for a totally unrelated reaction.
This is also why you can't tell which direction it's going to go after energy is added to the equation.
Why does negative mass move in the opposite direction from its momentum? It's because momentum is the product of mass and velocity. If the mass is negative, then the velocity (the motion) has a different sign from the momentum. But positive/negative is unique in that respect. It's the only simple change that can change the direction.
In order for the velocity to be anything but along the line of the momentum, you have to not only add a direction but remove the existing one. If you can remove the existing direction, there seems no reason for the new direction to be perpendicular to the old direction. It could go in any direction at that point.
I was thinking that a complex mass might work, but I don't see how it could. A complex number defines one extra axis, but for this you'd need two. Even if you write a complex number in polar form (which gives a direction), you'd still need a default direction. Sort of a universal north. And it's not clear why a complex mass would have that effect. Particularly as the more normal way of thinking about a complex number would have the direction in the same plane as the velocity, where you want it to be in a plane perpendicular to the velocity.
The only reason that it makes any sense at all is that it turns the mass into a magnitude and a direction. The idea of negative mass also does this but only along the line of the velocity/momentum. Note that a side effect of this would be that in this universe, all momentum would be perpendicular to velocity (and force and acceleration). Because the direction would always be perpendicular. There's no complex number where the direction is not in that plane. Not even the trivial ones where $a$ or $b$ is $0$ in $a + bi$.
That establishes that this can't be an undiscovered property in our universe (whereas negative mass could be). We know that velocity and momentum are usually in the same direction in our universe. In this perpendicular universe, they never would be. Whether it is possible to have such a perpendicular universe is a different question. Some of the other answers argue that that isn't possible either.
Note that it is not even established that negative mass exists. That's a purely theoretic construct at this point.
