Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dot products - why does the vector become scalar

  1. Sep 8, 2006 #1
    As Gokul43281 correctly pointed out to me, the dot product of two vectors is a scalar.

    I looked up this page to read about this.

    Thing is, I think conceptually. I could simply accept that this is so, but I don't "see" why it so. As my screen name implies, I must know why something is so, or I am not satisfied.

    The subject that started this was KE. My thinking follows the lines that since KE is a result of motion, momentum, it must contain a vector quantity. Yet thanks to Gokul I've researched this "dot product" and discover the result is a scalar.

    This doesn't make sense to me. How can one ignore the "direction" of the motion which is responsible for KE?

    What am I missing?
  2. jcsd
  3. Sep 8, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Do you think a car moving at 100 [itex] \frac {km} {h} [/itex] North has a different KE then the same car moving at a 100 [itex] \frac {km} {h} [/itex] South or East?

    Look at the basic definition of a dot product (sum of the components) the resultant does not have a vector nature. If you understand how to compute the dot product of 2 vectors the scaler nature of the result should be obvious.
    Last edited: Sep 8, 2006
  4. Sep 8, 2006 #3


    User Avatar
    Homework Helper

    You can think of the scalar product as a mapping from V x V into the set of real numbers. So, a real number can tell you something about the vectors, for example [tex]\vec{a} \cdot \vec{b} = 0[/tex] => the vectors a and b are orthogonal.
    Think of the fact that in R^n you first define the scalar product, which then implies the norm, or the distance between two points, if you like.
  5. Sep 8, 2006 #4


    User Avatar

    Staff: Mentor

    We don't. There's a second dynamical quantity related to the motion of an object, namely the momentum, which is a vector. A complete description of how an object interacts with other objects must involve both energy and momentum.
  6. Sep 8, 2006 #5
    I follow this intellectually; your example is clear. The scalar value of their KE is the same.

    Yet, the very fact that these cars are moving in different directions still seems significant to me. To illustrate:

    Consider a situation in a vacuum, no gravity. I have an object A I want to move through some plane, in a direction I arbitraily define as North. There are two other objects, B & C, of equal mass moving at equal speed. B is moving West directly away from A, while C is moving North directly away from A. By this discussion, B & C have the same KE. I have a massless tether I can attach from A to either B or C. If I attach the tether to B, I will not get the result I want, but if I attach it to C, then I get what I was after; A is moved North.

    Considering KE to be scalar ignores this glaring difference in the usefulness of the Work B and C can do for me. Yes, I get it that the same Work would be done if I attach to B, but it is not the direction I desire. You can see that the direction KE is applied is not the same.

    Do you see where my concept is sitting?
  7. Sep 8, 2006 #6
    Is not PE also a scalar? If an object's PE is due to gravity, for example, then we know the potential work is "down". PE is a scalar, yet it has no meaning without the context of a direction. We can't use this example of PE to lift something without realizing we need a pulley or some such to reverse the direction the work will be applied in.

    As I said to Integral, I get the "math", it just seems meaningless without a vector. It is like someone just arbitrarily decided to divorce the scalar from the vector. I don't see the practical sense of it.
  8. Sep 8, 2006 #7
    Folks, I think I'm asking this question the wrong way. Let me take a similar example to perhaps better explain...

    In my very limited math, I understand the idea of taking the absolute value of a number, and it so happens that I understand why doing so is valid in certain instances. I can see WHY the sign of a number doesn't matter when applied to certain problems.

    What y'all are doing is explaining to me why abs(-2) equals 2. What I see is that by taking the absolute value, we have lost information about the number, and the equivalent question here would be "why doesn't that matter?"

    I follow the math that results in a scalar from a term that includes a vector. But there is information lost (or ignored) here. Why is that acceptable? In what circumstances does the scalar remain a useful value?
  9. Sep 8, 2006 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Don't confuse yourself with execss words. Look at the math and get your understanding there. That is where the rubber meets the road.
  10. Sep 8, 2006 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Do we?

    Grab a rubber band between your two hands. Pull your hands away from each other, stretching the rubber band, and increasing its PE. Now tell me the direction of this PE. Does it point to the left or to the right?

    If energy were defined as a vector quantity (which is what you want to do), it would no longer be conserved. Consider a ball tossed up in the air at two instants of time: first, at some height h, going up; and then at the same height h, falling down. By your rules, in both cases the PE would be a downward pointing vector, but the KE would switch from pointing up to pointing down. The total energy vector would no longer be the same at the two points in time! "Energy" would be quite a useless quantity if defined in that manner.
    Last edited: Sep 8, 2006
  11. Sep 8, 2006 #10
    Ok, I actually laughed out loud when I read that and realized how neatly you make your point. I'm not finished with that issue, but I think answering your second part is a more clear example of where yet I am confused.
    Be patient with me :confused: First, I erred badly when I named this thread. I am no longer having a problem with seeing the scalar come from
    a vector product. It is why we can get away with doing this that I am not yet able to see. Unfortunately, your question seems to me to argue for my point, so if anything my confusion has deepened.

    I see what you are saying. Let us suppose the total energy at the start of this experiment is 10 (of some unit). Let us also assume that at this point PE = 5, and KE = 5. At the top of the ball's arc, PE = 10 and KE =0. But by my idea, when the ball has fallen to the same height as when this experiment started, its E total is 0 because I'm putting a minus sign in front of KE to indicate the direction it applies to. Actually, I would do this the other way around. If at the start I'd arbitrarily choosen "down" as my positive vector, the PE would be +5, and KE would be -5, total 0 (they cancel each other out). On the way down however, both values would be positive, yielding a total of +10 for energy. Oops!

    You see, I get the math. I get that energy is not conserved when one applies a vector concept. Intellectually I don't have a problem.

    Conceptually, however, these two moments just don't seem to be the same. At the start, PE is due to gravity, which is pulling down, while KE is due to momentum, which is decidedly up. At the end, gravity is the same, but now momentum is down.

    I know energy must be conserved, but it doesn't feel right.

    I guess that, for now, I'll just have to learn to live with frustration.
  12. Sep 8, 2006 #11


    User Avatar

    Staff: Mentor

    Not necessarily. Consider a hollow spherical shell of uniform thickness. An object inside the shell has the same gravitational potential energy (GPE) regardless of its location. Related to this, the net gravitational force exerted on the object by the shell is zero, and there is no associated direction. (The different parts of the shell pull on the object in different directions, in such a way that the pulls all cancel out.)

    In order for there to be a net gravitational force on an object, its GPE has to vary from one location to another. The gravitational force is in the direction in which the object's GPE decreases the most rapidly, per unit of distance traveled. This is where the directional aspect of the gravitational force comes from.

    We can describe this using the gravitational potential [itex]V_g[/itex], which is an object's GPE divided by its mass. For a uniform vertical gravitational field, [itex]V_g = gh[/itex]. [itex]V_g[/itex] doesn't depend on the mass of the object, and can be thought of as a property of the space surrounding the source of the gravitational field. The gravitational force on an object is its mass times the gradient (a sort of three-dimensional derivative with respect to position) of [itex]V_g[/itex]:

    [tex]\vec{F} = - m \vec{\nabla} V_g[/tex]

    (The minus sign is because we want the direction in which [itex]V_g[/itex] decreases, and the gradient [itex]\vec{\nabla} V_g[/itex] gives the direction in which it increases.)

    To put it another way, although [itex]V_g[/itex] and GPE at every location are scalar quantities, the way in which they vary from one location to another gives rise to a vector quantity, the gravitational force.

    I don't know how much calculus you've had, but you'll encounter the gradient when you get to multivariable calculus, if you haven't seen it yet. The equation above really represents three equations, one for each component of [itex]\vec F[/itex]:

    [tex]F_x = - m \frac{dV_g}{dx}[/tex]

    and similarly for the y and z components.
    Last edited: Sep 8, 2006
  13. Sep 8, 2006 #12


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It might help to [re]visit the origin of Kinetic Energy, via the Work-[Kinetic]Energy Theorem. You probably want the more general vector-based derivation (involving dot products).
  14. Sep 8, 2006 #13


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I was hoping, with my previous argument, to convince you that there is no good reason for you to choose a direction for the PE. For the case of the ball in the air, the PE is strictly speaking, the potential energy of interaction of the ball and the earth, and is given by PE = -GmM/r. It's NOT the potential energy of the ball; it's the potential energy of the ball+earth system. The earth is pulling the ball down, while the ball is pulling the earth up.

    I can come up with tons of other situations where you can not assign a direction to the PE. Consider the following : a pair of twin starts orbiting about a common center of mass, a stretched spring, the electrostatic interactions between the free electrons in a metal, the energy of an ideal gas compressed inside a cylinder (in addition to the examples provided by jtbell).

    Besides these arguments, it is important to recognize that the point of physics is to define quantites that are useful in the quest of identifying symmetries or patterns in the way nature works. Sure, you may be able to go ahead and define quantities like KE and PE as vectors (so long as the vector is well-defined; i.e, its magnitude and direction need to be uniquely specified) but of what use are such quantites? Can you come up with any useful explanations of nature based on these quantites? You lose the all powerful conservation law that helps unmask the workings of a myriad phenomena. What do you gain?

    Your intuition based on a preformed bias is what is causing your frustration. There is no reason that all quantites that are related to vectors also be vector quantites. The study of physics often involves the vanquishing of an old intuition to replace it with a new one, based on a better understanding of the workings of things. This happens time and time again, as one delves deeper and deeper into the subject. You are not alone.
    Last edited: Sep 8, 2006
  15. Sep 8, 2006 #14
    KE has no direction, it depends solely on mass and velocity. Momentum has direction, you might be confusing it with that.

    Suppose you have a ramp. A car has a certain velocity but is shut off and assume no frictional forces. The car will go up a certain amount on the ramp until it moves no more, where its KE=0 and its PE=max. Since the height that the car is above the original surface (non-ramp) is the same for any direction the car is travelling when it starts up the ramp, where the ramp is in the correct position for the car, the KE has no direction.
  16. Sep 8, 2006 #15

    I'm afraid your math is over my head. I'm in my final year of a Software Engineering degree which, sadly, went no further than College Algebra. I did Calculus in High School, but that was more than 20 years ago and is buried in cobwebs.

    Somewhere, I need to find time to re-learn calculus; then I might be able to benefit from comments like yours.


    That sounds like good advice. Going back to basics may help. Strangely, though, it was just that which constributed to my present confusion. I used HyperPhysics a lot recently to attempt to gain some knowledge of several things, and they tend to jump directly from formulae denoting a value for Force (a vector, yes?) to a representation of the same situation for KE (or PE). It was a natural (mis)step for me to view this as confirmation that KE was a vector quantity. Seems some sites assume one has a lot of supporting knowledge and experience. Of course, I can't blame them exclusively. I'm just having a "bad day" with this one.


    I hope you will be relieved to know that your paragraph makes some sense to me, and the highlighted portion is what I would request you to expand upon.

    I recognise some of the examples you have given me speak to that sentence. I suspect the "gem" of information I need to resolve this in my mind is hiding in there somewhere. May ignorance on this matter seems profound, but I'm not without a modicum of intelligence.

    Pick one of your examples, if you would, and develop your point from the basics. Begin as you would with a child. I am not easily offended and besides, I just asked for it :) You won't insult me.

    Build it up from there. I'll tell you when I "get it".

    And, thank you in advance for your patience.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Dot products - why does the vector become scalar
  1. Dot Products (Replies: 3)