How Do Dot and Cross Products Differ in Describing Physical Phenomena?

[pgardn]

I am trying to understand the difference from a physical phenomena point of view, not just math.

Surprisingly I think I got the cross product like in rotational momentum. You have the momentum vector and we have effective distance from the momentum vector R that needs to be perpendicular to the momentum vector. These two vectors produce a plane in which the best way to conventionally view angular momentum is a vector, L, perpendicular to this plane. I think I get this.

Now work. So to have work, F and change in position, must be in line with each other, no plane. And it yields transfer of energy which physically has no direction, a scalar quantity. But can't we conventionally invent a vector as a product of F dot d?

Btw, spinning a bicycle wheel and trying to change the axis if rotation makes me get a feel for the vector nature of angular momentum, and changing it.

Thanks for any insight.

 

[Mark44]

I am trying to understand the difference from a physical phenomena point of view, not just math.

Surprisingly I think I got the cross product like in rotational momentum. You have the momentum vector and we have effective distance from the momentum vector R that needs to be perpendicular to the momentum vector. These two vectors produce a plane in which the best way to conventionally view angular momentum is a vector, L, perpendicular to this plane. I think I get this.

Now work. So to have work, F and change in position, must be in line with each other, no plane.

No, that's not true. The force doesn't have to be in the same direction as the motion. That's where the dot product comes in. Work will be at its maximum when the force is acting in the same direction as the motion.

And it yields transfer of energy which physically has no direction, a scalar quantity. But can't we conventionally invent a vector as a product of F dot d?

No, but you can determine a unit vector in the direction of motion, and multiply it by the (scalar) value of the dot product. Then you'll have a vector whose magnitude equals the dot product, and it has the same direction as the motion (which is not necessarily the same direction in which the force is applied).

Btw, spinning a bicycle wheel and trying to change the axis if rotation makes me get a feel for the vector nature of angular momentum, and changing it.

Thanks for any insight.

 

[pgardn]

No, that's not true. The force doesn't have to be in the same direction as the motion. That's where the dot product comes in. Work will be at its maximum when the force is acting in the same direction as the motion.
No, but you can determine a unit vector in the direction of motion, and multiply it by the (scalar) value of the dot product. Then you'll have a vector whose magnitude equals the dot product, and it has the same direction as the motion (which is not necessarily the same direction in which the force is applied).

So the part of the force that is perpendicular to the displacement does play a role?

I did not make this clear, by in line with, I mean either same direction or opposite direction (-) work.

So I am still not getting it.

Sorry.

 

[pgardn]

And is my rendition of the cross product wrong?

 

[pgardn]

Ok I took a shower and something struck me. The vector product of work is a scalar because, if for example, you had + work you might want to make the vector of energy transfererred in the same direction as F and d, or in - work just cancel...

But there is no ( or infinite) reference vectors that would be applicable( like in the cross product plane), which would be perpendicular "up" or "down", for a linear vector product like in a dot product. Negative work... Cancel? No direction, I don't know.

Is any of the above at all valid? I am not a natural math guy. I must "see" a physical phenomena and then look at the math and I'm ok. Look at the math without some physical phenomena and I'm cooked. I will never be a real physics type, but am trying to reconcile something that is supposedly easy.

 

[Mark44]

So the part of the force that is perpendicular to the displacement does play a role?

No. If F and ds are perpendicular, then their dot product (and hence the work) is zero.

One definition of the dot product (in the context of work) is $\vec{F} \cdot \vec{ds} = |\vec{F}| |\vec{ds}| cos(θ)$, where θ is the angle between the vectors.

I did not make this clear, by in line with, I mean either same direction or opposite direction (-) work.

Yes, you weren't clear. You said "So to have work, F and change in position, must be in line with each other, no plane."

And that's not true. F and ds can be in any direction relative to each other. Work will be nonzero as long as that angle isn't 90°.

So I am still not getting it.

Sorry.

 

[Mark44]

And is my rendition of the cross product wrong?

It sounds reasonable to me, but fair warning - it has been many years since I took physics.

I commented on what you said about the dot product because it was wrong. Hopefully you have a better understanding of it now.

 

[Mark44]

Ok I took a shower and something struck me. The vector product of work is a scalar because, if for example, you had + work you might want to make the vector of energy transfererred in the same direction as F and d, or in - work just cancel...

You're misusing some of the terms, which makes it difficult to follow what you're saying. Work is a dot product, and the dot product is often called the scalar product, because it produces a scaler. The term "vector product" is most often applied to the cross product, because this type of product produces another vector.

Let's make things simple, with the force and the direction of motion in a line. If you apply a force to an object, and it moves in the same direction, work is positive. The angle between the force vector and the direction vector is 0°, and cos(0°) = 1, so the work is the product of the magnitudes of the two vectors.

If you apply a force to an object, but it moves in the opposite direction (possibly due to another force acting on it), the work is negative. The angle between your force and the direction of motion is 180° and cos(180°) = -1. The work is the negative of the product of the magnitudes of the two vectors.

But there is no ( or infinite) reference vectors that would be applicable( like in the cross product plane), which would be perpendicular "up" or "down", for a linear vector product like in a dot product. Negative work... Cancel? No direction, I don't know.

Is any of the above at all valid? I am not a natural math guy. I must "see" a physical phenomena and then look at the math and I'm ok. Look at the math without some physical phenomena and I'm cooked. I will never be a real physics type, but am trying to reconcile something that is supposedly easy.

 

[UltrafastPED]

It helps to start with a geometric understanding of vector products:
http://www.math.oregonstate.edu/bridge/papers/dot+cross.pdf

and http://godplaysdice.blogspot.de/2007/09/geometric-interpretation-of-vector.html

 

[pgardn]

You're misusing some of the terms, which makes it difficult to follow what you're saying. Work is a dot product, and the dot product is often called the scalar product, because it produces a scaler. The term "vector product" is most often applied to the cross product, because this type of product produces another vector.

Let's make things simple, with the force and the direction of motion in a line. If you apply a force to an object, and it moves in the same direction, work is positive. The angle between the force vector and the direction vector is 0°, and cos(0°) = 1, so the work is the product of the magnitudes of the two vectors.

If you apply a force to an object, but it moves in the opposite direction (possibly due to another force acting on it), the work is negative. The angle between your force and the direction of motion is 180° and cos(180°) = -1. The work is the negative of the product of the magnitudes of the two vectors.

I get this. What I am asking is why can't a direction be added to a dot product. I understand it makes no sense for a transfer of energy to have direction. All I am asking is could you add a direction if you choose to do so. With a cross product you could leave out direction if you wanted. In fact some books do for simplicities sake I guess.

Thanks for entertaining my questions.
Sorry for the vagueness.

 

[pgardn]

It helps to start with a geometric understanding of vector products:
http://www.math.oregonstate.edu/bridge/papers/dot+cross.pdf

and http://godplaysdice.blogspot.de/2007/09/geometric-interpretation-of-vector.html

Thank you.

I really stepped into it.
Some of the responses in the second article is what I was looking for.
The first will require some time on my part as I clearly was taught just like the authors said should not be done.

 

[sophiecentaur]

I get this. What I am asking is why can't a direction be added to a dot product. I understand it makes no sense for a transfer of energy to have direction. All I am asking is could you add a direction if you choose to do so. With a cross product you could leave out direction if you wanted. In fact some books do for simplicities sake I guess.

Thanks for entertaining my questions.
Sorry for the vagueness.

Because, by definition, a dot product gives a Scalar result - no direction involved. The dot product of force and displacement is Work. Work is a scalar quantity, just as Energy is. You cannot have a 'direction' associated with a can of petrol (i.e. a MJ of energy). A good argument for the non-directional outcome would be to ask which direction would you use? The direction of the force or the direction of the displacement or a. n. other?

I guess you could ask the question "What is the physical meaning of Force ^ Displacement ?" The answer is definitely not Work.

 

[pgardn]

Because, by definition, a dot product gives a Scalar result - no direction involved. The dot product of force and displacement is Work. Work is a scalar quantity, just as Energy is. You cannot have a 'direction' associated with a can of petrol (i.e. a MJ of energy). A good argument for the non-directional outcome would be to ask which direction would you use? The direction of the force or the direction of the displacement or a. n. other?

I guess you could ask the question "What is the physical meaning of Force ^ Displacement ?" The answer is definitely not Work.

I understand that scalars don't have direction. I understand the petrol amount idea not having direction. I am asking could you assign a direction to F multiplied by d if you wanted to knowing they are both vectors? Now physically we know that any part of a force that does not act along the line of the displacement plays no role in what we call work. So that means F multiplied by d is automatically assigned the designation of a dot product from math?

So dot product is really a mathematical "truth" or definition that fits perfectly for F multiplied by and has physical utility?

And yes then you brought up a good example. F^d has no physical meaning we have assigned to it. So does a vector raised to the power of some other vector have direction? And if the answer is I don't know enough about vector math otherwise I would have never asked the question then I again apologize for my ignorance and am sorry for my circular silly questions.

And heck I may be asking, was some math invented to fix physics problems and other math has no important physical utility but is just worthwhile in itself? and if so, I apologize. And of course some math that came before physics applies perfectly to physical problems because it's really geometry. I'm mixing myself up, I will cease.

 

[Mark44]

I understand that scalars don't have direction. I understand the petrol amount idea not having direction. I am asking could you assign a direction to F multiplied by d if you wanted to knowing they are both vectors?

What does F multiplied by d mean? Is it the cross product (which is defined only for vectors in three dimensions and more abstractly for vectors in seven (I believe) dimensions?

Or is it the dot product, which as you know results in a number (i.e., a scalar)? If you're talking about multiplying two vectors together, you have to specify which kind of product you mean.

Now physically we know that any part of a force that does not act along the line of the displacement plays no role in what we call work. So that means F multiplied by d is automatically assigned the designation of a dot product from math?

No, it's really the other way around. The dot product of the force and the direction gives an answer that agrees with our observations. Someone made the discovery that if you apply a force that causes an object to move in some direction, the dot product was useful in explaining how much work is being done to the object.

So dot product is really a mathematical "truth" or definition that fits perfectly for F multiplied by and has physical utility?

The dot product provides an accurate description of reality, which makes it useful.

And yes then you brought up a good example. F^d has no physical meaning we have assigned to it. So does a vector raised to the power of some other vector have direction?

No, and not only that, it doesn't even have meaning.

And if the answer is I don't know enough about vector math otherwise I would have never asked the question then I again apologize for my ignorance and am sorry for my circular silly questions.

And heck I may be asking, was some math invented to fix physics problems and other math has no important physical utility but is just worthwhile in itself?

In many cases, a math concept had already been discovered, and someone else noticed that the same concept could be used to explain the physical behavior. Here's an example. A Greek geometer, Apollonius of Perga (262 BC to 190 BC), wrote treatises about the conic sections - including circles, ellipses, hyperbolas. Nearly 2000 years later, in the 17th Century, Johannes Kepler discovered that the planets revolving around the sun were traveling in elliptical orbits, with the sun at one of the foci of the ellipse. My point is that ellipses weren't "invented" to "fix" astronomy. Instead, Kepler used geometry to explain the paths of the sun's planets.

There are many other examples like this, where someone came up with an idea in mathematics for which there were no applications, and someone else came up with a way to tie the concept to some physical situation. Another example of where the math came first and an application of it came later is in number theory. G.H. Hardy, a British mathematician, wrote "A Mathematician's Apology," in which he pretended to apologize that all of the mathematics that he worked with had no possible application. As it turned out, he was wrong, with much of what he regarded as pure mathematics later being applied in new fields such as cryptography.

and if so, I apologize. And of course some math that came before physics applies perfectly to physical problems because it's really geometry. I'm mixing myself up, I will cease.

There is a lot of math that applies to physical problems that goes well beyond geometry. Differentiation that explains rates of change, and integration for calculating such things as the total force acting against the side of a dam are two examples that come to mind.

 

[sophiecentaur]

There is a lot of math that applies to physical problems that goes well beyond geometry. Differentiation that explains rates of change, and integration for calculating such things as the total force acting against the side of a dam are two examples that come to mind.

The relationship between our Mathematics and the Science of the World can be seen as being almost mystical - until you realize that the mathematical relationships that are involved in Science are only a subset of all the possible branches of Maths. The question "What is Maths?" can take you far into Philosophy and away from the business of PF. Humans have found that there are a large number of mathematical relationships that describe the results of what we measure, practically.
Many people want to have explanations that just do not involve the familiar (but sometimes complicated) Maths of the textbook. They ask for "Physical" explanations, as if it can all be referred back to intuitive, Mechanical ideas. Those explanations usually turn out to be very limited because common verbal language just doesn't cut it in many cases. So we use Maths. But that brings with it a problem in that, every time you hop in and out of the Maths World, you need to conduct the 'credibility test' to check the hop has been valid, in Physical terms.

I previously asked the question about what F^d could refer to. Force times Perpendicular distance is how we calculate the Moment of a Force. This is the same as the Vector (Cross) produce of a force and a distance. (See this link for more details. and there are many other hits on Google which deal at different levels, to suit the reader). The mathematics of rotational mechanics are quite a bit harder than describing linear movement but they are very useful for calculating the behaviour of gyroscopes and bicycles.

 

[DrGreg]

F^d has no physical meaning we have assigned to it. So does a vector raised to the power of some other vector have direction?

When sophiecentaur wrote that, I'm sure he didn't mean \textbf{F}^\textbf{d}; he meant the vector product \textbf{F} \times \textbf{d}, which some people write as \textbf{F} \wedge \textbf{d}.

 

[sophiecentaur]

@DrGreg
Yes. It was a keyboard problem!

 

[pgardn]

Thank you all for entertaining my questions.

I get it.

A deeper understanding of vector math is required. There are some interesting philosophical implications that come out of this. I need to read a bit deeper on physical modeling and the application of math. Maybe some history on the idea of work and how math was applied and the same for angular momentum.

They ask for "Physical" explanations, as if it can all be referred back to intuitive, Mechanical ideas. Those explanations usually turn out to be very limited because common verbal language just doesn't cut it in many cases.

^The above was especially helpful. ^ is used for direction int this sentence, not raised to a power or cross product.

 

[sophiecentaur]

Thank you all for entertaining my questions.

I get it.

A deeper understanding of vector math is required. There are some interesting philosophical implications that come out of this. I need to read a bit deeper on physical modeling and the application of math. Maybe some history on the idea of work and how math was applied and the same for angular momentum.

They ask for "Physical" explanations, as if it can all be referred back to intuitive, Mechanical ideas. Those explanations usually turn out to be very limited because common verbal language just doesn't cut it in many cases.

^The above was especially helpful. ^ is used for direction int this sentence, not raised to a power or cross product.

I would agree with that for everybody! Vector maths is very useful and very sexy.
In A level Maths, way back, we used the ^ symbol for the cross product - but it was 'on the line' and not ' above the line'. Needless to say, there were no keyboards around at the time (except for typewriters and Organs) so there was no confusion with the 'circumflex accent sign because we were in Maths lessons and not French lessons. lol. For some reason, they wouldn't let us use a BIG X for the cross product symbol.