Confused by different definitions of position

AI Thread Summary
The discussion centers around confusion regarding the definition of position in mechanics, particularly the claim that "position is scalar" and the concept of "absolute position." Participants clarify that position is typically defined as a vector, which includes both magnitude and direction, and that all reference frames are equally valid. The notion of "absolute position" is challenged, as it implies a fixed reference that contradicts the relativity of position in physics. The original poster seeks clarity on their textbook's definition, which aligns with the vector concept, while the online source is deemed likely incorrect. The conversation emphasizes the importance of context and reference points in understanding position in mechanics.
Sabra
Messages
2
Reaction score
0
Hello,

I am self-studying an introductory mechanics textbook and while I feel I understand the material there pretty well, I came across across an online definition of position which seems at odds with the explanations and definitions in my book.

The definition that confused me is:

""Position is scalar; it has neither magnitude nor direction.

EDIT: A position vector refers to a location relative to an arbitrary reference origin. This is not the same as position. Being arbitrary, the reference origin can be moved, without moving or changing the position (location) of the object, but this will change the position vector. Think of it this way: put an object on a glass table. Place a grid, with an arbitrary zero point under the table with the object centered on the zero point. The postion vector is zero relative to the origin on the grid. Now, move the grid such that the zero point is no longer under the object. The object's absolute position on the table is unchanged, but its vector relative to the grid is now different. ""

"Absolute position" seems to contradict the principle that no reference frame is inherently more correct than another. Also, if this "absolute position" is not defined relative to any reference point, how would you identify it?

I think this definition is wrong (and appropriately, position should be a vector quantity, not a scalar one). Am I right?

I'd very much appreciate it if someone could shed some light on this for me. Thanks!
 
Physics news on Phys.org
Sabra said:
The object's absolute position on the table is unchanged
This isn't absolute position either - it's defined with respect to the table. Imagine moving the table under the object. The object's 'absolute' position with respect to the walls of the room is unchanged, but the vector relative to the table is now different.

I'm a little unsure of what exactly your textbook is saying, but I suppose it is colloquially possible to specify a kind of absolute position without magnitude and direction ("the object is here", "the object is there") and have people know what you mean (although you could still argue that 'here' and 'there' are relative to the person talking). However, you are right that all reference frames are equally valid, so to describe a position without reference to some chosen origin would be meaningless in a mathematical context.
 
I'll take a stab at it, but I am guessing at what the author had in mind and may be giving him more credit than he deserves.

I would say that an object has a position/location even if we are not there to give it position coordinates. That may seem like a silly statement to make, but it leads to some consequences that are not immediately obvious. One motivation of tensors is to define something that has the same physical meaning and rules regardless of the coordinate system used. Once one coordinate system is used to measure a tensor object, there are specific rules that must be followed if the coordinate system is changed. Those rules allow us to say that that tensor has a physical meaning independent of coordinate system.
 
Sabra said:
an introductory mechanics textbook

Which textbook?
 
Just to clarify, the quote about "absolute position" came from an online source - my book defines position as a vector and does not mention "absolute position". The quote just made me think I had misunderstood the book, so I wanted to check here.

sk1105 - That's what I thought. Thanks very much for your reply!

FactChecker - I think it's likely the online source is mistaken. I didn't know that about tensors - I'll hopefully study them in the future. Thanks!

Edit: jtbell, didn't see your post there. The textbook is Newtonian Mechanics by A.P. French. But like I said, the quote that says position is a scalar and talks about absolute position is not from the book.
 
Last edited:
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top