Is Center of Mass a vector or scalar quantity?

[PHYSICS5502]

I am a little bit confused on Center of Mass and Center of Gravity about what are they vectors or scalars . As I may think they are neither because they are simply two points . Am I saying right or wrong .

 

[Gan_HOPE326]

I am a little bit confused on Center of Mass and Center of Gravity about what are they vectors or scalars . As I may think they are neither because they are simply two points . Am I saying right or wrong .

They're points, hence they're vectors. Points are defined by vectors, but of course since vectors represent more 'differences' between points, you need to define a reference frame first. If for example you have a cube of side $l$ and uniform density, and you place the origin in its bottom left corner the center of mass is going to be at $[l/2,l/2,l/2]$ which is a vector.

 

[PHYSICS5502]

I did not quite get it can you give more explanation please .

 

[PHYSICS5502]

I am talking about are they vector quantities.

 

[Gan_HOPE326]

I am talking about are they vector quantities.

See:

Points are defined by vectors

Namely, they are vector quantities.

 

[PHYSICS5502]

How are they vector quantities take any frame of your choice and please prove it to me .

 

[Gan_HOPE326]

How are they vector quantities take any frame of your choice and please prove it to me .

I gave you an example above, with the cube. There's nothing to prove really. Points in three-dimensional space require three numbers to be specified, therefore are vectors. The center of mass is a point. Therefore, the center of mass is specified by a vector.

 

[Dopplershift]

Yes, the center of mass is indeed a vector quantity

The center of mass is itself a position, thus it requires an exact position which is given by Coordinates x,y,z (if working in Cartesian coordinates).

The position vector R of the center of mass is given by:

\begin{equation}
R =\frac{1}{M} \sum_{\alpha = 1} ^N m_\alpha r_\alpha
\end{equation}

where m$_\alpha$ is the mass of an individual particle, and r$_\alpha$ as the position of the individual particle.

M is the total mass of the system.

Note that the each r$_\alpha$ is a position of each particle thus r = r(x,y,z)

If r is a function of the coordinates of (x,y,z), then the final position vector, R must be a function of x,y,z as well thus R = R(x,y,z).

The mass is a scalar function and r is a vector, a scalar times a vector is a vector.

Hope this helps!

 

[PHYSICS5502]

A point in 3 dimensional space is a vector ??

 

[Dopplershift]

The individual coordinates of the position vector, R, of the center of mass can be broken down into its 3d components (X,Y,Z) if each particle has a position (x,y,z)

Thus:
$$
X = \frac{1}{M} \sum_{\alpha = 1}^N m_\alpha x_\alpha \\
Y = \frac{1}{M} \sum_{\alpha = 1}^N m_\alpha y_\alpha \\
Z = \frac{1}{M} \sum_{\alpha = 1}^N m_\alpha x_\alpha
$$
<Moderator's note: LaTeX fixed>

 

[A.T.]

A point in 3 dimensional space is a vector ??

https://en.wikipedia.org/wiki/Three-dimensional_space#In_linear_algebra

 

[Dopplershift]

A point is any arbitrary position (0,4,6), a vector is the point's displacement relative to another point. a point would just be the displacement from the origin. If you move the origin and the point of your choice equally, the magnitude of the vector does not change. The direction is simply the components (x,y,z) of the point in space.

 

[Dopplershift]

Remember in 2D Kinematics (projectile motion) where you broke the velocity of the object to both the x and y components. Both the components had a magnitude and direction (either the x-direction or the y-direction). The position of the object at any given time had a x component and y component. (It had height above the ground and a distance from where it started).

The center of mass is a position and position (like velocity and acceleration) have a magnitude and direction thus it is a vector.

 

[PHYSICS5502]

These things are a little bit difficult to understand but as far as i get, the conclusion for this is that center of mass is position in 3D space so it will have both magnitude and direction .

 

[Gan_HOPE326]

These things are a little bit difficult to understand but as far as i get, the conclusion for this is that center of mass is position in 3D space so it will have both magnitude and direction .

Pretty much. Depending on what your origin of the reference system is, "magnitude" will be the distance from that origin, and "direction", well, the direction in which you have to go to reach it.

It's perhaps simpler to visualise in 2D space. If you have a treasure map and it's got an X marking the spot, and you landed in Dead Man's Bay, I can tell you "well, starting from Dead Man's Bay, to reach the treasure you have to walk 3 km to the North and 4 km to the East". And that's a vector with magnitude (5 km total) and direction (something like North-East-East).

 

[Dadface]

As I understand it the original question can be rephrased as:

Is a point in three dimensional space a vector or scalar?

As such I don't think the question makes much sense. If the question was more specific eg asking about the location of the point then it could be defined in terms of vectors.

 

[Dale]

I am a little bit confused on Center of Mass and Center of Gravity about what are they vectors or scalars . As I may think they are neither because they are simply two points . Am I saying right or wrong .

I would say they are neither. I would say that they are members of an affine space

https://en.m.wikipedia.org/wiki/Affine_space

Affine spaces are definitely closer to vector spaces, but with some subtle differences.

 

[Cutter Ketch]

I think there is some confusion in this thread. In order to specify position we have to compare to something, so we pick an arbitrary coordinate system with an arbitrary origin. Position is then a vector from the origin to the object. However this vector is not an intrinsic physical property of the object. The origin isn't anything real. The position vector is not an intrinsic property of the physical system. This is apparent when you pick a new origin. All the position vectors change even though the physical situation has not. This is distinguished from vector quantities like displacement, force, velocity, or electric field which intrinsically have direction. What's more the associated vector direction and magnitude does not depend on the choice of coordinate system. Those are intrinsically vector quantities. They can't be contemplated without direction. So when someone asks if center of mass is a vector quantity, I think the answer has to be no. It has no intrinsic directional physical property that is invariant under coordinate transform.

 

[Gan_HOPE326]

I think there is some confusion in this thread. In order to specify position we have to compare to something, so we pick an arbitrary coordinate system with an arbitrary origin. Position is then a vector from the origin to the object. However this vector is not an intrinsic physical property of the object. The origin isn't anything real. The position vector is not an intrinsic property of the physical system. This is apparent when you pick a new origin. All the position vectors change even though the physical situation has not. This is distinguished from vector quantities like displacement, force, velocity, or electric field which intrinsically have direction. What's more the associated vector direction and magnitude does not depend on the choice of coordinate system. Those are intrinsically vector quantities. They can't be contemplated without direction. So when someone asks if center of mass is a vector quantity, I think the answer has to be no. It has no intrinsic directional physical property that is invariant under coordinate transform.

True that. I had the impression the question was more at a beginner level and referring to practical matters such as "how should I treat the center of mass in calculations", to which the answer is "like a vector" because anyway any calculation is usually performed in a given system of reference and therefore the position vector of the center of mass will be all that features. But yeah, it's more of a habit to consider them as if they were one and the same while they actually are two slightly different things.

 

[ZapperZ]

A point in 3 dimensional space is a vector ??

Sometime I think the most elementary issue is taken way too far.

Let's to back to the foundational mathematics that is relevant here. Let's do this one step at a time, shall we?

@PHYSICS5502 : Look at this diagram below, which should not be anything new to you.

Do you understand that the vector r points to a location in space with respect to the coordinate axis defined in the diagram?

Zz.

 

[Dopplershift]

A point in 3D space itself is not a vector, but it's displacement from the origin is.

The displacement is the magnitude and the component (x,y,z) is the direction. Vectors are displacement from point a to point b.

In physics, we often are more concern with the displacement rather than the path of motion it took to get from point a to point b, If I started at position x = 0 walked to x = 5 and then walked back to x = 0 (my original position), then my displacement vector is zero. Likewise if I start at x = 0 and end at x = 3, my vector is 3$/hat{x}$

You will see in a lot of physics classes, that a lot of physical changes are only concern about the starting point vs the ending point. If you start at a bottom of the hill, and walk up, you are doing work to convert kinetic energy into potential energy, but as you go back down to your starting point, work is being done on you to change potential energy back into kinetic energy. Thus you end in the same amount of kinetic energy and by the work-energy theorem, no work was ever done. The point is although it may seem like you did work, you started at the same point and ended at the same point, so your displacement is zero, thus no work is done. Does this make sense? It doesn't matter if you walked straight up the hill or at an angle around the side of the hill, your starting height and ending height is all that matters, physics doesn't really care how you got up the hill or how you got down. (in terms of measuring kinetic and potential energy).

 

[pixel]

I agree with Zapper Z that this discussion has gone too far afield given the level of the question. From the Wikipedia article on center of mass, we have for a collection of n particles of mass m_i, total mass M, and position vectors r_i:

The center of mass, R, is a vector.

 

[beamie564]

This is distinguished from vector quantities like displacement, force, velocity, or electric field which intrinsically have direction.

How can you define the direction of displacement, velocity etc. without an external coordinate system? These quantities don't have an intrinsic direction.

Those are intrinsically vector quantities. They can't be contemplated without direction.

But then how can you describe position without direction; only by using a scalar quantity?
.
I'm sorry, but what you said doesn't make sense to me. In my view, position ( and therefore centre of mass) is a vector.

 

[Cutter Ketch]

How can you define the direction of displacement, velocity etc. without an external coordinate system? These quantities don't have an intrinsic direction.

I am surprised you don't see the difference. They HAVE direction whether you define a coordinate system or not. The electric field points from this charge toward that charge. A test charge placed in the field will move in that direction. The car is moving in that direction. In a time interval it will move from here to there. No choice of coordinates will change which direction the force points. Turning your head doesn't prevent the ball from falling down.

 

[Dopplershift]

I agree with Zapper Z that this discussion has gone too far afield given the level of the question. From the Wikipedia article on center of mass, we have for a collection of n particles of mass m_i, total mass M, and position vectors r_i:

The center of mass, R, is a vector.

I agree! But that's what makes math and physics so exciting is the discussions that come out of simple questions. Some of the greatest scientific breakthroughs came from simple questions. It's in our nature to try to go in depth as much as possible and to explore the validity of what we believe is true.

 

[beamie564]

They HAVE direction whether you define a coordinate system or not.

Maybe I wasn't clear enough. Obviously they have direction, but how would you define/specify that direction without an external frame of reference?

No choice of coordinates will change which direction the force points. Turning your head doesn't prevent the ball from falling down

But what is down, without a reference frame?

 

[Cutter Ketch]

Maybe I wasn't clear enough. Obviously they have direction, but how would you define/specify that direction without an external frame of reference?

But what is down, without a reference frame?

That's the whole point. The direction exists whether or not I specify it. That is not the case with position. A point in space has no physical quantity indicating any sort of directionality
outside the construct of an arbitrary coordinate system.

Let's start again. I'll make a set of statements and let's see if you disagree with any of them. I am considering some physical system with masses springs charges forces or whatever kind of physical entities you care to mention.

1) there is no physical entity associated with the origin of a coordinate system. It is a construct. It may be located arbitrarily in space without changing the physics of the described system.
2) there is no physical entity associated with the axes of a coordinate system. They are constructs. They may be oriented arbitrarily without changing the physics of the described system
3) a position vector is the vector from the origin of the coordinate system to a point in space
4) there is no physical entity associated with a position vector. It is a construct. Since the origin is arbitrary the position vector from the origin is arbitrary. Any position vector of any magnitude and direction can be associated with a point in space simply by moving the origin and orienting the coordinate axis. This has no impact on the physics of the described system.
5) in the system there is a ball falling under gravity toward the center of the earth. It experiences a force vector, an acceleration vector, it attains a velocity vector, and if you locate it at two points in time it has a displacement vector
6) None of the described vectors are arbitrary. They are defined by the physics. Under any coordinate transform they will maintain the same magnitude and they will always point from the ball toward the earth. You cannot change the magnitudes or make them point anywhere other than toward the Earth without changing the physics of the described situation.
7) those are real physical vector quantities in a way that position or center of mass are not. They have direction which is an immutable part of the physics and not an arbitrary and selectable construct.
8) there is no real physical direction associated with position or center of mass. They are points in space and do not point anywhere other than within the arbitrary construct of our coordinate system.

 

[robphy]

I am a little bit confused on Center of Mass and Center of Gravity about what are they vectors or scalars . As I may think they are neither because they are simply two points . Am I saying right or wrong .

I would say they are neither. I would say that they are members of an affine space

https://en.m.wikipedia.org/wiki/Affine_space

Affine spaces are definitely closer to vector spaces, but with some subtle differences.

I agree... neither.
Although to a beginner, one might answer "yes, a vector"...
I think the OP has picked up on a subtlety that is often glossed over... and should be validated (as @Dale has done).

An object having three components
to which one can add a vector to obtain another object of the same type
doesn't make that object a vector.

An affine space is often thought of as "a vector space that has forgotten its origin".
Generally speaking, it doesn't make sense to add two positions "as vectors" if one doesn't know where the origin is...
since the "sum" would depend on the choice of that origin. So, generally adding positions doesn't make any sense.
However, there is an exception, having to do with the center-of-mass (seen ahead).

Note that although one doesn't generally have the sum of two positions,
one can define the difference of two positions... that is a vector (the displacement vector from one position to the other)...
and that doesn't depend on any choice of origin.
If one now chooses an origin, then one now also has displacement-vectors from that specific origin to those positions.

Back to center of mass...

from https://en.m.wikipedia.org/wiki/Affine_space#Informal_description
referring to Alice and Bob using different origins to calculate a sum of positions

Similarly, Alice and Bob may evaluate any linear combination of a and b, or of any finite set of vectors, and will generally get different answers. However, if the sum of the coefficients in a linear combination is 1, then Alice and Bob will arrive at the same answer.

Similarly,
from these notes on "Basics of Affine Geometry" http://www.cis.upenn.edu/~cis610/geombchap2.pdf (the page labelled "10")

Thus, we have discovered a major difference between vectors and points:
The notion of linear combination of vectors is basis independent, but the
notion of linear combination of points is frame dependent. In order to salvage
the notion of linear combination of points, some restriction is needed:
The scalar coefficients must add up to 1.

This is called https://en.wikipedia.org/wiki/Affine_combination
and its reference
http://graphics.cs.ucdavis.edu/educ.../Affine-Combinations/Affine-Combinations.html
This is what occurs in the definition of the "center of mass" where the coefficients add up to 1.

 

[Dale]

This is what occurs in the definition of the "center of mass" where the coefficients add up to 1

I had not noticed this subtlety before, thanks!

 

[beamie564]

@Cutter Ketch I agree with your statements. In fact I was going to write something similar in my previous post. I get that they have a difference; that the direction for position doesn't even mean anything without a coordinate system, while for the other quantities it does. But is that a good enough reason to say that position is not a vector? I wasn't sure.
Seeing @robphy's post I understand the difference ( it's so subtle).

 

[Cutter Ketch]

@Cutter Ketch I agree with your statements. In fact I was going to write something similar in my previous post. I get that they have a difference; that the direction for position doesn't even mean anything without a coordinate system, while for the other quantities it does. But is that a good enough reason to say that position is not a vector? I wasn't sure.
Seeing @robphy's post I understand the difference ( it's so subtle).

Well, since we make vectors to represent position and add those vectors to find center of mass and many other manipulations, I have been careful not to say that position isn't a vector. I can certainly understand the point of view that things we represent with vectors are vector quantities. However if you ask me if it is a vector quantity that sounds like a different question to me. I hear does that physical property have a direction and a magnitude? In any case I'm pretty sure failing to make the distinction was the reason for the OPs confusion and incredulous responses before I chimed in, so I wanted to draw the line between math and physics.

 

[robphy]

Another way to distinguish the 3-component-objects that describe "position" from those that describe "displacement"
is to say that
a "displacement" (from one position to another) is a 3-d vector (whose magnitude is independent of choice of origin and orientation of axes).
but a "position" is merely a labeling by 3 numbers (whose values (and sum-of-squares) depend on the choice of origin and choice of orientation of axes).
Again, it wouldn't mean anything (independent of frame) to generally add two positions... but one can get a weighted average of positions.
Thus, everyone will agree on where the center of mass is located (its position in space)... but not necessarily agree on how to label it.

Similarly, an "elapsed time" (from one clock reading to another) is a 1-d vector (in some abstract space) whose magnitude is independent of choice of origin (of time).
However, a "clock reading" is merely a labeling by one number (whose value depends on the choice of origin of time).
Generally, it wouldn't mean anything (independent of origin) to generally add two clock readings ... but one can get an average of clock readings.
Everyone will agree on when the halfway-time occurs... but not necessarily agree on how to label it.