Suvat vector versus the scalar form

[heroslayer99]

Homework Statement

This isn't really a set homework so I don't have anything to write here sorry.

Relevant Equations

Suvat formulae

Hi
I was just wondering about the suvat formulae and a question popped into my head, which I'd like someone to try and explain the reason as to why please.

So I know that when we have a formula such as F=ma or v = u + at, you can evaluate the magnitude of both sides and arrive at a scalar form of said equation, which is what I think that we use most of the time. However say if we had the scalar form of the first suvat equation, namely mod(v) = mod(u) + mod(a)t, I know that this is a correct mathamatical statement however we usually like to know the direction of our vector in physical situations, so we put a - sign Infront of each quantity which is going in our deemed negative direction.

However we also know that a scalar quantity (watch how I said quantity and not just a scalar) cannot be negative, so when we use a suvat formula and accompany a quantity with a negative sign, we must be using the vector form right? However it makes no sense to say that because we usually use the magnitude of each vector (which is a scalar) and not the vector in it's i/j or column form. So I'd just like to ask, why can we use negative signs in suvat formula if we aren't using the vector form, as a scalar quantity cannot be negative.

 

[PeroK]

The introductory SUVAT equations are for motion in one-dimension. In general, they apply to each component of the vector quantities:
$$x(t) = x(0) + v_x(0)t + \frac 1 2 a_x t^2$$$$y(t) = y(0) + v_y(0)t + \frac 1 2 a_y t^2$$$$z(t) = z(0) + v_z(0)t + \frac 1 2 a_z t^2$$And, of course, these can be combined into a single vector equation:
$$\vec r(t) = \vec r(0) + \vec v(0)t + \frac 1 2 \vec a t^2$$Where $\vec r(t) = (x(t), y(t), z(t))$.

 

[PeroK]

PS note that there is an important difference between components of a vector and scalar quantities.

 

[Orodruin]

I know that this is a correct mathamatical statement

You shouldn’t know that because it is false.

Edit: It is also false that scalars cannot be negative.

 

[BvU]

Hello hero,

$\qquad$ !​

Homework Statement:: This isn't really a set homework so I don't have anything to write here sorry.
Relevant Equations:: Suvat formulae

Yeah, well ...

we must be using the vector form right?

All this stuff is vector stuff. In cases (especially homework exercises ) we can simplify to one dimension. Hence the SUVAT business. With components of vectors as variables. And components of vectors are indeed scalars: a force (a force component) can be in one of two directions.

But a scalar is a numerical value, not necessarily non-negative. I.e. not just a magnitude. Check e.g. here.

I see others have already jumped in -- I'm a slow typist

$\$

 

[heroslayer99]

You shouldn’t know that because it is false.

Edit: It is also false that scalars cannot be negative.

I said a scalar quantity cannot be negative.
For example -2 is a negative scalar but it's not a measurable quantity, whereas speed is defined to be +ve only.

 

[vcsharp2003]

I said a scalar quantity cannot be negative.
For example -2 is a negative scalar but it's not a measurable quantity, whereas speed is defined to be +ve only.

A scalar quantity takes on a value that is always scalar i.e. a real number. And therefore, a scalar can be 0 or +ve or -ve.

However, the magnitude of a vector is always 0 or +ve, and of course the magnitude of a vector is a scalar. Saying that magnitude of a vector is negative makes no sense.

Temperature is a scalar quantity in Physics that can take - ve values if the temperature is below freezing point of $0^o C$.

 

[Orodruin]

I said a scalar quantity cannot be negative.
For example -2 is a negative scalar but it's not a measurable quantity, whereas speed is defined to be +ve only.

It depends on the scalar. Speed is a magnitude and as such cannot be negative, but that is not the only type of scalar.

For example, the velocity component in a particular direction is also a scalar and can certainly be negative. Do not confuse scalar with magnitude. Magnitude is a scalar but a scalar is not necessarily a magnitude.

 

[BvU]

Kudos for defending your point of view. Forces old hands to do some thinking instead of using autopilot.

I know that this is a correct mathematical statement

if with mod(v) you mean $|\vec v|$, I strongly disagree. Simply not true as Odo says.

By now I'm curious to see some examples of scalars that can have a negative value. (first thought: voltage ... )
And NASA helps out: temperature, energy, work, power are definitely not vectors, but they can be negative. Well...

I'll try to find some even higher authority... NIST, SI, ???

$\$

 

[PeroK]

I said a scalar quantity cannot be negative.
For example -2 is a negative scalar but it's not a measurable quantity, whereas speed is defined to be +ve only.

I prefer to think of motion in one dimension being described by one-dimensional vectors, rather than "scalars". Even in one-dimension, speed, $|v|$, is clearly different from velocity $v$. You must respect that $v$ can be positive or negative.

 

[heroslayer99]

Hello hero,

$\qquad$ !​

Yeah, well ...

All this stuff is vector stuff. In cases (especially homework exercises ) we can simplify to one dimension. Hence the SUVAT business. With components of vectors as variables. And components of vectors are indeed scalars: a force (a force component) can be in one of two directions.

But a scalar is a numerical value, not necessarily non-negative. I.e. not just a magnitude. Check e.g. here.

I see others have already jumped in -- I'm a slow typist

$\$

So when we have suvat in one direction the variables which come from a vector represent the component of the vector in that axis, so the suvat equations for one direction are made up of scalars which come from the components of the vectors?

 

[Orodruin]

By now I'm curious to see some examples of scalars that can have a negative value.

What was wrong with the x component of velocity for an object moving antiparallel to the x axis?

 

[Orodruin]

So when we have suvat in one direction the variables which come from a vector represent the component of the vector in that axis, so the suvat equations for one direction are made up of scalars which come from the components of the vectors?

Yes. You have also not addressed the big elephant in the room:

mod(v) = mod(u) + mod(a)t, I know that this is a correct mathamatical statement

To me this suggests that you do not understand how the modulus works for general vectors.

Edit: For example, your conclusion would mean that speed can only increase with time.

It is generally not true that mod(a+b) = mod(a) + mod(b). Not even in one dimension.

 

[heroslayer99]

Yes. You have also not addressed the big elephant in the room:To me this suggests that you do not understand how the modulus works for general vectors.

Well I'm sorry for getting it incorrect. However I will go back to the F=ma example. In this case we can say that mod(F) = m (mod(a)), so could you explain how this translates to the v=u+at equation?

 

[Orodruin]

so could you explain how this translates to the v=u+at equation?

It doesn’t.

 

[heroslayer99]

So this equation has no "scalar form"?

 

[Orodruin]

Which one of them? Each component of the suvat equation is a scalar equation.

 

[PeroK]

So this equation has no "scalar form"?

If you take any equation $X = Y$, then it must follow that $|X| = |Y|$. That's clear. And, if you have an equation $X = Y + Z$, then $|X| = |Y + Z|$. But, that's as far as you can go without looking at the cases where $X, Y$ and $Z$ take positive or negative values. That's a general mathematical point.

PS although you have the triangle inequality: $|Y + Z| \le |Y| + |Z|$.

 

[heroslayer99]

If you take any equation $X = Y$, then it must follow that $|X| = |Y|$. That's clear. And, if you have an equation $X = Y + Z$, then $|X| = |Y + Z|$. But, that's as far as you can go without looking at the cases where $X, Y$ and $Z$ take positive or negative values. That's a general mathematical point.

PS although you have the triangle inequality: $|Y + Z| \le |Y| + |Z|$.

Alright I see where my mistake was. It makes much more sense to me now to think of this whole suvat nonsense as just unidirectional vectors, or equivalently the component of the vector going in that direction. (I know the component is a scalar but it seems to be the same idea)

 

[BvU]

Alright I see where my mistake was. It makes much more sense to me now to think of this whole suvat nonsense as just unidirectional vectors, or equivalently the component of the vector going in that direction. (I know the component is a scalar but it seems to be the same idea)

It's not nonsense .

Alright I see where my mistake was.

I sense a (somewhat reluctant) aceptance that a component of a vector is a scalar. Good !

And I grant you that the frst link I gave was not very helpful

Check e.g. here.

fortunately there was entry d. in the test

$\$

 

[Steve4Physics]

Hi @heroslayer99. I’d like to add my tuppence worth...

The quantities in the suvat equations (apart from time) represent vectors in one dimension - with the sign of each quantity giving its direction.

Take vertical motion with the usual convention: up=positive, down=negative. Then, for example:
u= 4m/s tells us that initial velocity is 4m/s upwards.
a = -10m/s² tells us that acceleration is 10m/s² downwards.

After 3s we calculate s = ut+½at² = 4·3+½(-10)·3² = -33m, i.e. the object is 33m below its start position.

If you ignored the signs you would get s = 4·3+½(10)·3² = 57m, which is wrong.

Ignoring the signs (treating all vector quantities as positive) will only work when all the vectors point in the same direction.

‘F=ma’ contains no addition or subtraction operations. This guarantees that the vectors ‘F’ and ‘a’ always have the same direction.

 

[vcsharp2003]

I know the component is a scalar but it seems to be the same idea

Components of a vector are always vectors, but the magnitude of each component is always a scalar.

 

[heroslayer99]

Components of a vector are always vectors, but the magnitude of each component is always a scalar.

Is this correct, I've been told earlier that a vector component is a scalar. In the case of 2i m/s, 2 is the scalar and 2i is the vector, maybe this is the confision?

 

[Steve4Physics]

Is this correct, I've been told earlier that a vector component is a scalar. In the case of 2i m/s, 2 is the scalar and 2i is the vector, maybe this is the confision?

There is some ambiguity in the use of the term ‘component’. E.g. see discussion here: https://math.stackexchange.com/questions/3757723/why-are-components-of-vectors-scalars

 

[jbriggs444]

Temperature is a scalar quantity in Physics

Temperature in Physics is neither a vector nor a scalar.

First year physics can leave one with the impression that the quantities in physics are split into two categories: vectors (a list of numbers) and scalars (a single number).

Eventually one can take a course in linear algebra and learn that "vectors" and "scalars" are two pieces that go into defining a vector space. That "vectors" need not be lists of real numbers and that "scalars" are both more restrictive and more inclusive than just being real numbers.

To be fair, a finite dimensional vector space together with an inner product and an orthonormal basis can allow one to construct a coordinate space that matches the vector space. The result matches pretty well the simple picture that is presented to first year students. [One can graduate college without ever discarding the naive picture. I did just that. I was a math major who never took linear algebra. Everything I know about vector spaces, I've learned on the Internet]

A "scalar" is the set of values by which one can multiply the "vector" elements of a "vector space". Scalars need not be real numbers. They can be elements of any algebraic field. In the case of the field of the real numbers, scalars can be negative.

However...

Since temperatures do not form a vector space, temperatures are not vectors.
Since temperatures do not form a set of multipliers for a vector space, temperatures are not scalars either.

 

[Steve4Physics]

Since temperatures do not form a vector space, temperatures are not vectors.
Since temperatures do not form a set of multipliers for a vector space, temperatures are not scalars either.

I suspect the meaning of ‘scalar’ is context-dependent.

E.g. a scalar can be an element of an algebraic field and used as a multiplier for a vector.

E.g. a scalar can be a rank 0 tensor allowing, for example, a scalar temperature field to be defined.

 

[kuruman]

In the case of the field of the real numbers, scalars can be negative.

There is no question in my mind that scalars can be negative. A clear example of this is the scalar product of two vectors. It is positive or negative depending on the specifics. Furthermore, it does not depend on the choice of coordinate system in which the two vectors are described. For example, the work done by constant force $\mathbf{F}$ when it displaces an object by displacement $\mathbf{d}$ is $W=\mathbf{F}\cdot \mathbf{d}$ and independent of the coordinate system in which one writes the two vectors.

I always thought that the criterion for scalars is that they are independent of the choice of coordinate axes. The SUVAT equation $x=x_0+v_{0x}t+\frac{1}{2}a_xt^2$ is the x-component of the position vector of a mass moving at constant acceleration. One can argue that it fits the criterion above because it is the scalar product $\mathbf{r}\cdot \mathbf{\hat x}$. One can also argue that it doesn't fit the criterion because if I change the coordinate axes so that the constant acceleration is perpendicular to $\mathbf{\hat x},~$ then I have $x=x'_0+v'_{0x}t$ which is not same. However, what is invariant is the work per unit mass $\mathbf{a}\cdot \mathbf{r}$ done by the external unbalanced force on the object.

This thread confused me because I was under the impression that invariance under coordinate transformations was the criterion for identifying scalars. If that is too narrow a criterion, please set me straight and provide a more comprehensive criterion because I don't want to be stuck with a naive picture.

 

[vcsharp2003]

Is this correct, I've been told earlier that a vector component is a scalar. In the case of 2i m/s, 2 is the scalar and 2i is the vector, maybe this is the confision?

Component of a vector is always a vector.

In the example you mentioned, the vector is $\vec v = 2 \hat i$ and it's magnitude is 2. When looking at a vector you should know it's magnitude which is 2 in your example and it's direction which is in +ve x direction.
Please don't associate a scalar with a vector, since it will cause confusion and the vector concept will not be clear. A vector has magnitude and direction and that's all you need to know about the vector.

 

[PeroK]

This thread confused me because I was under the impression that invariance under coordinate transformations was the criterion for identifying scalars. If that is too narrow a criterion, please set me straight and provide a more comprehensive criterion because I don't want to be stuck with a naive picture.

I think we are using a more relaxed basic definition appropriate to a first exposure to physics that a scalar is a quantity represented by a single number. This surprised me too, as I'd already posted post #3.

 

[vela]

Is this correct, I've been told earlier that a vector component is a scalar. In the case of 2i m/s, 2 is the scalar and 2i is the vector, maybe this is the confision?

If you have a vector $\vec A = A_x \,\hat i + A_y \,\hat j$, I'd call $A_x\,\hat i$ the vector component of $\vec A$ in the x-direction while $A_x$ is the scalar component of $\vec A$ in the x-direction. I think typical usage is that one is talking about the scalar component if one just says "component."

 

[heroslayer99]

If you have a vector $\vec A = A_x \,\hat i + A_y \,\hat j$, I'd call $A_x\,\hat i$ the vector component of $\vec A$ in the x-direction while $A_x$ is the scalar component of $\vec A$ in the x-direction. I think typical usage is that one is talking about the scalar component if one just says "component."

Thank you.
And finally, if I use one of the suvat equations to compute a quantity, say s, and I get an answer of -3. Does this mean I've computed the displacement to be a one dimensional vector of -3 i, or does it mean that my displacement vector has a component of -3 in the i direction? Or does this mean the same thing...

 

[PeroK]

Thank you.
And finally, if I use one of the suvat equations to compute a quantity, say s, and I get an answer of -3. Does this mean I've computed the displacement to be a one dimensional vector of -3 i, or does it mean that my displacement vector has a component of -3 in the i direction? Or does this mean the same thing...

Those are the same thing.

 

[heroslayer99]

Those are the same thing.

Ok and just because I am pedantic, suvat in 1D is with the components of the vectors in that direction (which is a scalar and hence +ve or -ve)

 

[PeroK]

Ok and just because I am pedantic, suvat in 1D is with the components of the vectors in that direction (which is a scalar and hence +ve or -ve)

I don't know what you mean. Having a unit vector in one dimension is redundant. Vectors only really get interesting in two or more dimensions.

 

[kuruman]

If you have a vector $\vec A = A_x \,\hat i + A_y \,\hat j$, I'd call $A_x\,\hat i$ the vector component of $\vec A$ in the x-direction while $A_x$ is the scalar component of $\vec A$ in the x-direction. I think typical usage is that one is talking about the scalar component if one just says "component."

Suppose one has $\vec A = 3 \,\hat i - 4 \,\hat j$. I am using specific numbers instead of $A_x$ and $A_y$ to remove the sign ambiguity that is inherent in algebraic symbols.

I see two interpretations for the term $- 4 \,\hat j$ representing the vector component of $\vec A$ in the y-direction.

1. A vector of negative y-component pointing in +y-direction, i.e. $- 4 \,\hat j=(-4)(+\hat y).$
2. A vector of positive magnitude 4 pointing in the negative y direction, i.e. $- 4 \,\hat j=(+4)(-\hat y).$

We teach students to say that vector $\vec A = 3 \,\hat i - 4 \,\hat j$ has a negative y-component, which conforms with interpretation 1, yet in everyday life we conform with interpretation 2. When someone asks us, "which way did he go?", we extend our arm to form a unit vector with its tail at our shoulder and its tip at the end of our index finger and say, "he went that-a-way" whichever way that is. If he went South, we say and point "South" (interpretation 1), we don't say "North moving backwards" (interpretation 2.) Small wonder there is confusion.

Just a few thoughts $\dots$