Zero Momentum in Elastic Collisions

[rkmurtyp]

Is it possible to have a zero vector? The question arises in the context of conservation of momentum. In center of mass reference frame, the momentum of two masses involved in an elastic collision is zero before (as well as after) collision. It is because the two masses have equal and oppposite values of momenta. Together they have zero momentum. What is the direction of this zero momentum? Since there are two possible directions opposite to one another which one do we choose. If we choose one of those directions what valid reason can we have to reject the opposite direction?

 

[Nugatory]

Yes it is possible to have a zero vector, and it has no direction - for roughly the same reason that zero is neither a positive number nor a negative number.

 

[A.T.]

zero is neither a positive number nor a negative number.

In computer science it is both.

 

[DeIdeal]

Not only is it possible, it is necessary by definition that any vector space V (that is, even if you don't consider eg vectors in 3d-space, but more general vectors) has a neutral element of sum, denoted (here) by 0 and defined s.t. v+0=v for all v∈V, that is, a zero vector.

In fact, a vector space (such as R³) can be defined by considering something called an (Abelian) group equipped with additional structure, and it is required that every group G has an identity element e s.t. e.g=g.e=g, for all g∈G where . is something called the law of composition (it's the addition of vectors in the case of a vector space). If the group is equipped with the additional structure to make it a vector space, the e here is the zero vector of the vector space. (As a consequence of the definition of the identity, it turns out that it (and therefore the zero vector) is also unique, that is, there always exists exactly one zero vector in a specific vector space.)

 

[rkmurtyp]

Not only is it possible, it is necessary by definition that any vector space V (that is, even if you don't consider eg vectors in 3d-space, but more general vectors) has a neutral element of sum, denoted (here) by 0 and defined s.t. v+0=v for all v∈V, that is, a zero vector.

In fact, a vector space (such as R³) can be defined by considering something called an (Abelian) group equipped with additional structure, and it is required that every group G has an identity element e s.t. e.g=g.e=g, for all g∈G where . is something called the law of composition (it's the addition of vectors in the case of a vector space). If the group is equipped with the additional structure to make it a vector space, the e here is the zero vector of the vector space. (As a consequence of the definition of the identity, it turns out that it (and therefore the zero vector) is also unique, that is, there always exists exactly one zero vector in a specific vector space.)

Please tell me: What is the difference between zero vector and zero scalar? If zero vector has no specific direction it is no different from a scalar zero and then we cannot add that scalar to any vecotor because addition of a vector and a scalar is forbidden. On the other, hand if zero vector has a specific direction how do we know which direction it is.

If possible an explanation with reference to binary elasic collisions would be very helpful to me.

I want a simple answer as I am not a mathematician to follow mathematical language.

 

[rkmurtyp]

Yes it is possible to have a zero vector, and it has no direction - for roughly the same reason that zero is neither a positive number nor a negative number.

If that is the case, then How do we distinguish a scalar zero from a scalar vector?

 

[scoobmx]

If that is the case, then How do we distinguish a scalar zero from a scalar vector?

The vector has implied dimensionality. Zero scalar means 0 in 1 dimension. Zero vector means 0 in all components of the vector.

 

[DeIdeal]

Please tell me: What is the difference between zero vector and zero scalar? If zero vector has no specific direction it is no different from a scalar zero and then we cannot add that scalar to any vecotor because addition of a vector and a scalar is forbidden. On the other, hand if zero vector has a specific direction how do we know which direction it is.

If possible an explanation with reference to binary elasic collisions would be very helpful to me.

I want a simple answer as I am not a mathematician to follow mathematical language.

First of all, sorry if I was too technical with the answer, I just thought I'd provide some more details since the previous posters had already pretty much answered your original question. It's by no means essential, so just ignore it if you want.

The difference is that they are, well, completely different objects. Scalars are somehting you use to multiply vectors with, whereas vectors are typically [Unless you introduce new operations] added together. The zero scalar in R³ is 0, the zero vector is (0,0,0).

What do you mean by saying that "it is no different from a scalar zero"? The (lack of) set direction for a zero vector is irrelevant, you can already see that it is a different object from the zero scalar because you can add it to any other vector, but you can't add a vector and the zero scalar together.

This is probably the most direct definition of both: When you multiply any vector by the zero scalar, you get the zero vector, but when you add the zero vector to any vector, you get the original vector.

Unfortunately I can't think of any way to explain the difference between the zero vector and the zero scalar by relying on binary collisions.

(Btw, if you ever plan on studying quantum mechanics, for example, you're very likely to encounter more abstract vectors and the definition of a general vector space I mentioned earlier, they are not "just for mathematicians", but also used in physics a lot.)

 

[rkmurtyp]

First of all, sorry if I was too technical with the answer, I just thought I'd provide some more details since the previous posters had already pretty much answered your original question. It's by no means essential, so just ignore it if you want.

The difference is that they are, well, completely different objects. Scalars are somehting you use to multiply vectors with, whereas vectors are typically [Unless you introduce new operations] added together. The zero scalar in R³ is 0, the zero vector is (0,0,0).

What do you mean by saying that "it is no different from a scalar zero"? The (lack of) set direction for a zero vector is irrelevant, you can already see that it is a different object from the zero scalar because you can add it to any other vector, but you can't add a vector and the zero scalar together.

This is probably the most direct definition of both: When you multiply any vector by the zero scalar, you get the zero vector, but when you add the zero vector to any vector, you get the original vector.

Unfortunately I can't think of any way to explain the difference between the zero vector and the zero scalar by relying on binary collisions.

(Btw, if you ever plan on studying quantum mechanics, for example, you're very likely to encounter more abstract vectors and the definition of a general vector space I mentioned earlier, they are not "just for mathematicians", but also used in physics a lot.)

Thanks

Let us say, we have two equal and opposite vectos parallel to X-axis. Their sum is a zero vector.

Now we consider two equal and opposite vectors in an arbitrary direction. Their sum is a zero vector.

Are the above two zero vectors equal?

 

[DeIdeal]

Thanks

Let us say, we have two equal and opposite vectos parallel to X-axis. Their sum is a zero vector.

Now we consider two equal and opposite vectors in an arbitrary direction. Their sum is a zero vector.

Are the above two zero vectors equal?

As I said, there is only one zero vector in a vector space. So yes, for any vector the sum of it and its additive inverse (the 'equal and opposite vector') is the same zero vector. To perhaps make it more transparent, let $\mathbf{v},\mathbf{w}$ be two (different) vectors. Then $\mathbf{v}+(-\mathbf{v})=\mathbf{0}=\mathbf{w}+(-\mathbf{w})$ Because that is how you define what $-\mathbf{v}$ means.

[This is probably irrelevant for you, so feel free to ignore it, but for completeness, here's a proof why the zero vectors are unique. Let $\mathbf{0},\mathbf{0'}$ be zero vectors. Remember that zero vector fulfills $\mathbf{v}+\mathbf{0}=\mathbf{v}$ for every vector by definition. So $\mathbf{0}=\mathbf{0}+\mathbf{0'}$ because $\mathbf{0'}$ is a zero vector. But addition of vectors is commutative, so we may switch the order to say $\mathbf{0}=\mathbf{0'}+\mathbf{0}$. But the RHS is just $\mathbf{0'}$, since $\mathbf{0}$ is also a zero vector. So $\mathbf{0}=\mathbf{0}+\mathbf{0'}=\mathbf{0'}+\mathbf{0}=\mathbf{0'}$)

 

[rkmurtyp]

The vector has implied dimensionality. Zero scalar means 0 in 1 dimension. Zero vector means 0 in all components of the vector.

Can we resolve zero vector into two equal and opposite vectors of arbitrary magnitude in any direction of our choice?

Is it possible to compare two zero vectors? If yes, how do we do that?

 

[rkmurtyp]

As I said, there is only one zero vector in a vector space. So yes, for any vector the sum of it and its additive inverse (the 'equal and opposite vector') is the same zero vector. To perhaps make it more transparent, let $\mathbf{v},\mathbf{w}$ be two (different) vectors. Then $\mathbf{v}+(-\mathbf{v})=\mathbf{0}=\mathbf{w}+(-\mathbf{w})$ Because that is how you define what $-\mathbf{v}$ means.

[This is probably irrelevant for you, so feel free to ignore it, but for completeness, here's a proof why the zero vectors are unique. Let $\mathbf{0},\mathbf{0'}$ be zero vectors. Remember that zero vector fulfills $\mathbf{v}+\mathbf{0}=\mathbf{v}$ for every vector by definition. So $\mathbf{0}=\mathbf{0}+\mathbf{0'}$ because $\mathbf{0'}$ is a zero vector. But addition of vectors is commutative, so we may switch the order to say $\mathbf{0}=\mathbf{0'}+\mathbf{0}$. But the RHS is just $\mathbf{0'}$, since $\mathbf{0}$ is also a zero vector. So $\mathbf{0}=\mathbf{0}+\mathbf{0'}=\mathbf{0'}+\mathbf{0}=\mathbf{0'}$)

Let us say V and W have different directions. Then the magnitude of the sum of V and -V, and the magnitude of the sum of W and -W are equal. But how can we can we equate their directions since V and W have different directions? And, we know two vectors are said to be equal if and only if they both have the same direction and magnitude.

Please understnad these are my genuine problems and not for just for argument sake.

 

[DeIdeal]

Let us say V and W have different directions. Then the magnitude of the sum of V and -V, and the magnitude of the sum of W and -W are equal. But how can we can we equate their directions since V and W have different directions? And, we know two vectors are said to be equal if and only if they both have the same direction and magnitude.

Please understnad these are my genuine problems and not for just for argument sake.

Remember that the direction of the vector can change when you add it to another vector, so the direction of a sum of two vectors does not necessarily have to be the direction of either of the initial vectors.

Sure, V,W may have different directions, but because the sums V+(-V), W+(-W) are both the zero vector, they have no particular direction attached to them.

 

[rkmurtyp]

Remember that the direction of the vector can change when you add it to another vector, so the direction of a sum of two vectors does not necessarily have to be the direction of either of the initial vectors.

Sure, V,W may have different directions, but because the sums V+(-V), W+(-W) are both the zero vector, they have no particular direction attached to them.

When we add two vectors we get a unique resultant vector with a definite magnitude and direction. Such being the case, V +(-V) has a direction parallel to the direction of V. This direction is different from the direction of the resultant vector W +(-W. Hence the question: How can V + (-V) be equal to W +(-Wwhen V and W are non colinear?

 

[DeIdeal]

When we add two vectors we get a unique resultant vector with a definite magnitude and direction.

This seems to be your main problem here. The zero vector simply does not have a set direction. An alternative convention is that it points to every direction, if that helps you.

The "definition" of vectors as things "that have both magnitude and direction" can not only cause inconsistencies in things like this, it is also difficult (I'm tempted to say impossible) to generalize outside "regular" vectors of Rⁿ.

 

[jbriggs444]

And, we know two vectors are said to be equal if and only if they both have the same direction and magnitude.

Many of the folks posting on this thread have an understanding of vectors based on linear algebra and "vector spaces". When you first learn about vectors in high school and first year college physics, the presentation is entirely different.

Given a presentation in which vectors are considered to consist of a magnitude and a direction, the zero vector is special. It has no direction. In such a presentation, two vectors are equal if either

1: They have the same magnitude and direction.
or
2: They both have zero magnitude.

Note that the "polar coordinates" presentation that you are using is not incompatible with the linear algebra presentation. The linear algebra presentation is more general. Vectors considered as a magnitude and a direction with the usual addition and scalar multiplication operations fulfil the axioms of a vector space. [As do vectors considered as n-tuples of coordinate values].

http://en.wikipedia.org/wiki/Vector_space

 

[BruceW]

When we add two vectors we get a unique resultant vector with a definite magnitude and direction. Such being the case, V +(-V) has a direction parallel to the direction of V. This direction is different from the direction of the resultant vector W +(-W. Hence the question: How can V + (-V) be equal to W +(-Wwhen V and W are non colinear?

it's true that V +(-V) is a scalar multiple of V and is also a scalar multiple of W +(-W) which is also a scalar multiple of W. But this scalar multiple is the zero scalar. you should not be surprised that 0*V + 0*W = 0 even though V and W are linearly independent.

 

[rkmurtyp]

This seems to be your main problem here. The zero vector simply does not have a set direction. An alternative convention is that it points to every direction, if that helps you.

The "definition" of vectors as things "that have both magnitude and direction" can not only cause inconsistencies in things like this, it is also difficult (I'm tempted to say impossible) to generalize outside "regular" vectors of Rⁿ.

Yes, it is my main problem.

If zero vector does not have a set direction it becomes an ill defined vector. On the other hand, on the alternative convention, if it points in every direction, then it looses uniqueness.

Either way it gives rise to problems in conceptulising a zero vector. If we impose deliberately ceratin properties to zero vector, they could lead to inconsistencies in usage in various contexts.

You would perhaps better appreciate my problem in the context of elastic collision analysis.

In center of mass (CM) reference frame, a system of two masses has zero momentum before collision and it keeps on changing direction as the process of collision progresses in time. So momentum conservation which demands the momentum be zero at any instant of time is accounted for, with infinite number of zero vectors with different directions.

In my openion a paradox is escaping scrutiny under the pretext of 'all zero vectors are equal'.

 

[Dale]

If zero vector does not have a set direction it becomes an ill defined vector. On the other hand, on the alternative convention, if it points in every direction, then it looses uniqueness.

The zero vector is perfectly well defined and unique. It is defined under the axioms of a vector space. Heuristically vectors may be described as little arrows with a magnitude and a direction, but that is merely a descriptive analogy. The actual definition of a vector has to do with the axioms regarding vector addition and scalar multiplication. The directionality of a vector is not a defining feature.

You simply need to let go of the misconception you have about how vectors are defined. The zero vector is not only perfectly well defined, it is an essential part of a vector space. Please pay attention to DeIdeal's excellent posts in this thread.

 

[BruceW]

You would perhaps better appreciate my problem in the context of elastic collision analysis.

In center of mass (CM) reference frame, a system of two masses has zero momentum before collision and it keeps on changing direction as the process of collision progresses in time. So momentum conservation which demands the momentum be zero at any instant of time is accounted for, with infinite number of zero vectors with different directions.

In my openion a paradox is escaping scrutiny under the pretext of 'all zero vectors are equal'.

it looks like your problem is not with the zero vector. to me, it looks like your problem is that you are expecting to be able to know the individual momenta, if you are given the total momenta. But this is definitely not true. the individual momenta can be expressed as two vectors $(P_{1x},P_{1y},P_{1z})$ and $(P_{2x},P_{2y},P_{2z})$ And the total momentum can be expressed as just one vector $(P_{x},P_{y},P_{z})$. So if I tell you only the total momentum, then of course you can't guess what are the individual momenta.

 

[D H]

Yes, it is my main problem.

No, you have two main problems. One is your persistent claim that there are a multiplicity of zero vectors when you have been shown that there is only one. See DeIdeal's post #10.

If zero vector does not have a set direction it becomes an ill defined vector.

You other main problem is that you are using the rather naive definition of a vector as something that has magnitude and direction. This is not an essential characteristic of what it means for something to be a "vector". On the other hand, the existence of the zero vector is an essential characteristic. The zero vector is provably unique. Once again, see post #10.

 

[atyy]

If zero vector does not have a set direction it becomes an ill defined vector. On the other hand, on the alternative convention, if it points in every direction, then it looses uniqueness.

Either way it gives rise to problems in conceptulising a zero vector. If we impose deliberately ceratin properties to zero vector, they could lead to inconsistencies in usage in various contexts.

Vectors are arrows attached to an origin, and the zero vector is the origin. You can think of the zero vector as having a special rule to define it as a vector.

Then the only questions are whether such a definition is consistent, and whether it is useful. I don't know of a proof of consistency, so someone else will have to answer that, but it certainly has been useful.

 

[Redbelly98]

If zero vector does not have a set direction it becomes an ill defined vector. On the other hand, the alternative convention, if it points in every direction, then it looses uniqueness.

You could just as well be talking about the number zero, as was mentioned in the first reply to your original post. It is neither positive or negative, yet is a perfectly well-defined number.

Or (according to post #3 of this thread), it is defined by some to be both positive and negative, but is still a unique number.

It seems like you are letting semantics get in the way of understanding a conceptually simple idea.

 

[A.T.]

Or (according to post #3 of this thread), it is defined by some to be both positive and negative, but is still a unique number.

I was rather jokingly referring to:
http://en.wikipedia.org/wiki/Signed_zero

It's more of a engineering workaround, for limited accuracy in digital number representations. In math as such, zero has no sign. And the zero vector has no direction.

 

[Nugatory]

I was rather jokingly referring to:
http://en.wikipedia.org/wiki/Signed_zero

It's more of a engineering workaround, for limited accuracy in digital number representations. In math as such, zero has no sign. And the zero vector has no direction.

And as the author of post #2, I knew what you were doing .

We also have the CDC 6x00 series machines of 40 years ago, which used one's-complement integer arithmetic and so had positive and negative representations of integer(!) zero... Picked up an entire clock cycle through the ALU that way, which is a big deal at 10 MHz.

 

[rkmurtyp]

it's true that V +(-V) is a scalar multiple of V and is also a scalar multiple of W +(-W) which is also a scalar multiple of W. But this scalar multiple is the zero scalar. you should not be surprised that 0*V + 0*W = 0 even though V and W are linearly independent.

Please tell me: When are two vectors said to be linearly independent?

I tried to understand the answer to this question from the information available on the net. But it deals with three or more vectors and brings in matrices and determinants to explain, which I don't understand well. In a different context I learned that two quantites are said to be linearly independent if one contains no information contained in the other. This is very general and easy to appreciate and I did not find much difficulty in applying it in algebra. But when it came o vectors I face problems. Here any two intersecting vectors seem to be considered linearly independent. Then I find each vector will have a component in the direction of the other vector, implying each vector has some information contained in the other vector, unless the two vectors are perpendicular to one another.

A reply by way of geometry is more welcome.

 

[rkmurtyp]

No, you have two main problems. One is your persistent claim that there are a multiplicity of zero vectors when you have been shown that there is only one. See DeIdeal's post #10.


You other main problem is that you are using the rather naive definition of a vector as something that has magnitude and direction. This is not an essential characteristic of what it means for something to be a "vector". On the other hand, the existence of the zero vector is an essential characteristic. The zero vector is provably unique. Once again, see post #10.

Thanks.

You other main problem is that you are using the rather naive definition of a vector as something that has magnitude and direction. This is not an essential characteristic of what it means for something to be a "vector".

This is only answer that is rightly directed to my problem.

My concept of a vector as something that has magnitude and direction might be naive, but any advanced concpept of a vector, I think, can not violate it. But I am not in a position to contest it. Moreover, I face no problem with this concept for any non-zero vector.

 

[rkmurtyp]

Vectors are arrows attached to an origin, and the zero vector is the origin. You can think of the zero vector as having a special rule to define it as a vector.

Then the only questions are whether such a definition is consistent, and whether it is useful. I don't know of a proof of consistency, so someone else will have to answer that, but it certainly has been useful.

Thanks.

Yes, it looks as if a special rule is required to define a zero vector.

Since vectors are dimensional quantities (having units), when we superpose two vector diagrams (which I think is not forbidden) say, of forces and the accelerations or velocity changes they produce, the zero vectors create problems because of the so called uniquness of a zero vector.

 

[jbriggs444]

Since vectors are dimensional quantities (having units), when we superpose two vector diagrams (which I think is not forbidden) say, of forces and the accelerations or velocity changes they produce, the zero vectors create problems because of the so called uniquness of a zero vector.

Can you elaborate on exactly what problems this creates?

Yes, you can create a drawing that puts position, velocity and acceleration on the same graph. Given a system of units, you could set up a natural correspondence so that, for instance, 1 m eastward, 1 m/sec eastward and 1 m/sec² eastward are all depicted at the same point on the graph. Yes, with the obvious natural correspondence, all three zero vectors would be depicted at the same point.

Are you, perhaps, concerned that having the "zero vector" be unique stands at odds with the notion of there being three "zero vectors" in the first place? It should be understood that the zero vector is unique within a particular vector space.

 

[voko]

Since vectors are dimensional quantities (having units), when we superpose two vector diagrams (which I think is not forbidden) say, of forces and the accelerations or velocity changes they produce, the zero vectors create problems because of the so called uniquness of a zero vector.

Zero velocity vector, zero acceleration vector and zero force vector still have different units, so they are different vectors.

 

[D H]

rkmurtyp, I strongly recommend you read the rules of this site. When you argue against the uniqueness of the zero vector you are, perhaps unknowingly, violating the rules of this site. If you are doing this knowingly you will not last long here. Our fundamental goal at this site is to help students learn. That's a bit tough when the student is being obstinate. This obstinacy is your main problem. It is keeping you from learning.

Here are the rules for what it means for something to be a vector. An element of a set V is a vector v (and the set V is a vector space) if the set is endowed two operations, addition of elements of V and scalar multiplication by an element of a field F, and if the set has an additive identity 0, subject to the following:

1. Vector addition must be:
   • Closed: u+v is an element of V for all u, v in V,
   • Commutative: u+v = v+u for all u, v in V, and
   • Associative: (u+v)+w=u+(v+w) for all u, v, w in V.
2. 0 is an additive identity: v+0=v for all v in V.
3. Every vector v in V has an additive inverse: v+(-v)=0.
4. Scalar multiplication must be:
   • Closed: av is an element of V for all a in F and all v in V,
   • Associative: a(bv)=(ab)v for all a, b in F and all v in V,
   • Distributive: a(u+v)=au+av and (a+b)v=av+bv for all a, b in F and all u, v in V.
5. The field's multiplicative identity is a multiplicative identity for scalar multiplication: 1v=v.

Note well: The concept of magnitude (or length) and direction is not in the above list. Nor is the concept of an inner product (or dot product). The existence of these concepts are not necessary conditions for something to qualify as a "vector". On the other hand, the existence of a (unique) zero vector is essential.

The zero vector is provably unique. Assume there exists some vector 0'≠0 that also satisfies condition #2. Now consider the sums 0+0' and 0'+0. By condition #2, 0+0'=0 and 0'+0=0'. Since addition is commutative, 0+0'=0'+0. This means 0'=0, which violates the assumption that 0'≠0. This is a contradiction, and hence the assumption that there exists some some vector 0'≠0 that also satisfies condition #2 is false. In other words, 0 is unique.

The naive definition of a vector as something with length and direction is a specialization of the general concept of a vector. When you look at vectors from this narrower perspective, the zero vector is still there, and it is still unique. For vector spaces that have this concept of length and direction, the zero vector has a well defined length (0), but its direction is necessarily undefined.

 

[voko]

Here are the rules for what it means for something to be a vector

With all due respect, this is not the only definition possible.

Vector

geometric

A directed segment of a straight line in a Euclidean space, one end of which (the point A) is said to be the origin, while the other (the point B) is said to be the end of the vector.

http://www.encyclopediaofmath.org/index.php/Vector

A similar definition is given in Ito's Encyclopedic Dictionary Of Mathematics:

http://books.google.ru/books?id=azS2ktxrz3EC&pg=PA1678#v=onepage&q&f=false

The very first treatise on vectors, Gibbs's Elements of Vector Analysis, opens with the following passage:

1. Definition.—If anything has magnitude and direction, its magnitude and direction taken together constitute what is called a vector.

The numerical description of a vector requires three numbers, but nothing prevents us from using a single letter for its symbolical designation. An algebra or analytical method in which a single letter or other expression is used to specify a vector may be
called a vector algebra or vector analysis.

Def.—As distinguished from vectors the real (positive or negative) quantities of ordinary algebra are called scalars.

As it is convenient that the form of the letter should indicate whether a vector or a scalar is denoted, we shall use the small Greek letters to denote vectors, and the small English letters to denote scalars. (The three letters, i, j, k, will make an exceptin, to be mentioned more particularly hereafter. Moreover, $\pi$ will be used in its usual scalar sense, to denote the ratio of the circumference of a circle to its diameter.)

2. Def.—Vectors are said to be equal when they are the same both in direction and in magnitude. This equality is denoted by the ordinary sign, as $\alpha = \beta$. The reader will observe that this vector equation is the equivalent of three scalar equations.

A vector is said to be equal to zero, when its magnitude is zero. Such vectors may be set equal to one another, irrespectively of any considerations relating to direction.

3. Perhaps the most simple example of a vector is afforded by a directed straight line, as the line drawn from A to B. We may use the notation AB to denote this line as a vector, i.e., to denote its length and direction without regard to its position in other
respects. The points A and B may be distinguished as the origin and the terminus of the vector. Since any magnitude may be represented by a length, any vector may be represented by a directed line; end it will often be convenient to use language relating to vectors, which refers to them as thus represented.

You termed such definitions "naive", but, frankly, I see nothing naive in them, especially in the context of 3D physics. Would you care to explain why they are "naive"?

 

[rkmurtyp]

rkmurtyp, I strongly recommend you read the rules of this site. When you argue against the uniqueness of the zero vector you are, perhaps unknowingly, violating the rules of this site. If you are doing this knowingly you will not last long here. Our fundamental goal at this site is to help students learn. That's a bit tough when the student is being obstinate. This obstinacy is your main problem. It is keeping you from learning.

Here are the rules for what it means for something to be a vector. An element of a set V is a vector v (and the set V is a vector space) if the set is endowed two operations, addition of elements of V and scalar multiplication by an element of a field F, and if the set has an additive identity 0, subject to the following:

1. Vector addition must be:
   • Closed: u+v is an element of V for all u, v in V,
   • Commutative: u+v = v+u for all u, v in V, and
   • Associative: (u+v)+w=u+(v+w) for all u, v, w in V.
2. 0 is an additive identity: v+0=v for all v in V.
3. Every vector v in V has an additive inverse: v+(-v)=0.
4. Scalar multiplication must be:
   • Closed: av is an element of V for all a in F and all v in V,
   • Associative: a(bv)=(ab)v for all a, b in F and all v in V,
   • Distributive: a(u+v)=au+av and (a+b)v=av+bv for all a, b in F and all u, v in V.
5. The field's multiplicative identity is a multiplicative identity for scalar multiplication: 1v=v.

Note well: The concept of magnitude (or length) and direction is not in the above list. Nor is the concept of an inner product (or dot product). The existence of these concepts are not necessary conditions for something to qualify as a "vector". On the other hand, the existence of a (unique) zero vector is essential.

The zero vector is provably unique. Assume there exists some vector 0'≠0 that also satisfies condition #2. Now consider the sums 0+0' and 0'+0. By condition #2, 0+0'=0 and 0'+0=0'. Since addition is commutative, 0+0'=0'+0. This means 0'=0, which violates the assumption that 0'≠0. This is a contradiction, and hence the assumption that there exists some some vector 0'≠0 that also satisfies condition #2 is false. In other words, 0 is unique.

The naive definition of a vector as something with length and direction is a specialization of the general concept of a vector. When you look at vectors from this narrower perspective, the zero vector is still there, and it is still unique. For vector spaces that have this concept of length and direction, the zero vector has a well defined length (0), but its direction is necessarily undefined.

Thanks
I assure you, I have no intention of flouting any rules of the forum. It is such a useful forum to learn, to share and discuss what we know and so on. One should be happy to be a part of the forum.

I don't want to be obstinate either. My urge to understand things in an elementary fashion, perhaps, gave you that impression.

It is better, I think, I don't pursue this thread further.

 

[atyy]

Yes, it looks as if a special rule is required to define a zero vector.

Yes, in D H's post #31, this special rule is his second rule:
"0 is an additive identity: v+0=v for all v in V."

Since vectors are dimensional quantities (having units), when we superpose two vector diagrams (which I think is not forbidden) say, of forces and the accelerations or velocity changes they produce, the zero vectors create problems because of the so called uniquness of a zero vector.

The zero vector is not unique in that sense. While you can add and subtract two forces, you cannot add and subtract a force and a velocity. The vectors for force and velocity live in different vector spaces, with different zero vectors.

 

[BruceW]

Please tell me: When are two vectors said to be linearly independent?

For two vectors in Euclidean space, they are linearly independent when they do not lie on the same line. For example, if you have two linearly independent vectors v and w then the linear combination av+bw can be used (with scalars a and b) to give any vector in 2D Euclidean space. If v and w were not linearly independent, then you could only make some subset of 2D Euclidean space, using them.

Therefore, the zero vector 0 is not linearly independent to any other vector. Using some vector v and the zero vector, the possible combinations you can make using av+b0 are no different to the possible combinations you could make using only the vector v, without using the zero vector (i.e. av).