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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?

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ I notice you've been asking some questions that ignore the fundamentals of physics lately - personally, I think these questions belong on Physics.SE and not WB as they're pure physics and have nothing to do with WB. Also, generally speaking, they're too broad, and if you want what you're describing to be possible then you're basically describing a portion of a magic system which usually isn't enough for any full answers. $\endgroup$ – Aify May 16 '16 at 15:29
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    $\begingroup$ These questions wouldn't fit on Physics.SE, as they aren't about real physics. They ask about a world where something is different. That's entirely on-topic here and not at all on a science site. $\endgroup$ – Brythan May 16 '16 at 22:19
  • $\begingroup$ @Brythan Again, them belonging on Physics is just my personal opinion on it - regardless of that, however, these questions are still way too broad; Like I said in my first comment, these questions basically describe an incomplete magic system which is definitely not enough for any good answers. An incomplete magic system, IMO, is typically grounds for closure. $\endgroup$ – Aify May 17 '16 at 1:32
  • $\begingroup$ Yes : see gyroscopes $\endgroup$ – Ewan May 17 '16 at 20:50
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    $\begingroup$ Sounds like you are describing angular momentum. The quantity is a vector which indeed is perpendicular to the actual motion. $\endgroup$ – JDługosz May 18 '16 at 2:40
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No.

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.

MOAR Proof

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.

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  • $\begingroup$ You argued brilliantly that a pure “side mass” can’t exist. But the Curl argument says nothing about some side-mass admixture to a positive-mass body. Ī̲ mean, the angle between momentum and velocity would be acute, not right. If you know complex numbers, think of the usual mass as of positive (real) numbers and of “side mass” as of imaginary ones. So, there is no purely “imaginary” mass, but what about the right complex half-plane? $\endgroup$ – Incnis Mrsi May 16 '16 at 18:35
  • $\begingroup$ By a "curl force", are you referring to a force determined by a cross product, as is the case with the magnetic component of the Lorentz force, $F=qv\times B$? $\endgroup$ – HDE 226868 May 16 '16 at 22:19
  • $\begingroup$ @HDE226868 No, I meant that the force field has a non-zero Curl ($\nabla \times$) $\endgroup$ – Aron May 17 '16 at 3:10
  • $\begingroup$ @IncnisMrsi The "imaginary" part and the real part can be treated separately, given that they are orthogonal. The "real" part will have the same treatment as normal mass, the "side" part will still have the same over-unity problems. $\endgroup$ – Aron May 17 '16 at 3:14
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    $\begingroup$ @JDługosz Massively Over-Abundant Reasoning. $\endgroup$ – timuzhti May 18 '16 at 11:57
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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 ($\hat a = \frac{\vec a}{|\vec a|}$)

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  • $\begingroup$ I remember a Scrooge McDuck story where he develops a genetically engineered seaweed that filters gold out of the sea water. I don't know who wrote Uncle Scrooge stories, but they had some good ideas. $\endgroup$ – Howard Miller May 18 '16 at 0:46
  • $\begingroup$ What if you add a preferred direction? Maybe it's like the apparent situation on the surface of a planet; not the true underlying physics but circumstantial. $\endgroup$ – JDługosz May 18 '16 at 10:21
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    $\begingroup$ @JDługosz: If I understand correctly, a preferred direction is a way to choose perpendicular in a non-smooth way. If you want this, then why not. $\hat x$ in $a + b \hat x$ already breaks the rotational symmetry. $\endgroup$ – BartekChom May 18 '16 at 11:17
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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.

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  • $\begingroup$ From the hard-science tag wiki excerpt: "All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Review the tag info before using this tag." $\endgroup$ – a CVn May 21 '16 at 15:07
  • $\begingroup$ Thanks for pointing that out. I added a citation and some formulas to help make my answer address the tag. $\endgroup$ – Jim May 21 '16 at 19:14
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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.

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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.

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  • $\begingroup$ -1 That is a very good approximation of under certain boundaries, such as t >> E and v << c. That approximation does not hold under QM nor SR/GR. $\endgroup$ – Aron May 16 '16 at 13:49
  • $\begingroup$ @Aron: Actually, depending on how you define $m$, that relation still holds under relativistic conditions. And even if you don’t, the directions of momentum and velocity still align. I can only guess why you think that this does not apply on the quantum scale. But there is no such thing as a momentum–velocity uncertainty (as the corresponding operators commute). Moreover, any uncertainty of movement does not yield a systematic effect such as asked for in the question. $\endgroup$ – Wrzlprmft May 16 '16 at 14:44
  • $\begingroup$ Momentum is the quantity conserved in the conservation of momentum. Or force times duration. The mass is a complicated notion not directly relevant to momentum. $\endgroup$ – Incnis Mrsi May 16 '16 at 17:39
  • $\begingroup$ @IncnisMrsi: The mass is a complicated notion not directly relevant to momentum. – A considerable amount of physics textbooks disagrees with you (and I know a few). There is such a thing as generalised momenta, I know, but I think it’s safe to say that this is not what the OP had in mind (not that the question would make more sense this way). — Or force times duration. – That hardly changes the fundamental problem of the question. If the movement in question is caused by a force, then this cannot be perpendicular to a force times duration. $\endgroup$ – Wrzlprmft May 16 '16 at 18:17
  • $\begingroup$ Aron presented some arguments on impossibility. You presented only some pettifogging on definitions, whereas the question implies clearly that OP’s “momentum” hasn’t necessarily to be collinear with velocity. $\endgroup$ – Incnis Mrsi May 16 '16 at 18:22
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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.

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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.

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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.

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  • $\begingroup$ No matter how you write a complex number, though, it still represents the same quantity, and you can go from one form to another without loss of accuracy or correctness (different forms are simply easier to work with for different problems). It's a little like writing $1+1$ or $2 \times 1$ or $2^1$ or even $\frac{10}{5}$ in that they are all just different representations of the same quantity. $\endgroup$ – a CVn May 21 '16 at 15:04
  • $\begingroup$ @MichaelKjörling It's not clear to me what you are trying to say. It reads like you are disagreeing ("though"), but you seem to be agreeing with "...it's not clear why a complex mass would have that effect." Could you clarify what you are saying? $\endgroup$ – Brythan May 21 '16 at 21:54
  • $\begingroup$ First, yes, I agree that it's not clear why, how, or even if, a complex mass would have the effect sought. Second, you wrote in your answer that "Even if you write a complex number in polar form (which gives a direction)", but the direction is simply a property of complex numbers, it's not a property of a specific form for writing complex numbers. Converting between, say, polar and rectangular forms is a matter of trigonometry, which doesn't introduce any new quantity or remove any existing quantity from the complex number, it just expresses it differently. Does this make more sense to you? $\endgroup$ – a CVn May 22 '16 at 10:21
  • $\begingroup$ @MichaelKjörling I added my response to the answer. $\endgroup$ – Brythan May 22 '16 at 11:07

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