Conceptual vector question from Halliday and Resnick

[loom91]

Why isn't time a vector?

Hi,

I found the following question in Halliday and Resnick's Physics Part I chapter on Vectors and the Laws of Motion.

5.We can order events in time. For example, event b may precede event c but follow event a, giving us a time order of events a, b, c. Hence there is a sense of time, distinguishing past, present and future. Is time a vector therefore? If not, why not?

I'm stumped. Can you help me? Thanks.

 

[daveb]

Just think about the criteria for a vector. Does time have both a direction that can be plotted on an x, y and/or z axis, and does it have magnitude?

 

[loom91]

Time CAN be plotted on a single axis, and it does have a magnitude once you select an origin, so why can't it be considered an one-dimensional vector?

 

[daveb]

Yes, but you can also plot the number of apples versus price, but neither is a vector. It's a question of does it have direction and magnitude?
In upper division physics, you'll find out a better definition of a vector is how it transforms under rotations, translations, etc.

 

[marlon]

Time CAN be plotted on a single axis, and it does have a magnitude once you select an origin, so why can't it be considered an one-dimensional vector?

To be able to speak about a vector, you need two things :

1) components : these are numbers denoted by x, y and z
2) unitvectors : these are vectors with magnitude 1 and they denote a spatial direction like the X, Y and Z axis.

The magnitude of a vector is defined as the squareroot of the sum of the squared components.

So, since time does not have a spatial direction, one does not express time as a vector.

This is just a very intuitive explanationand as you study on you shall learn more abstract definitions like vectors being a certain tensor or objects that transform in a specific way under certain transformations...

regards
marlon

 

[loom91]

I'm not getting these 'intuitive' explanations. I think I'll fare better with the more technical ones. What property must a vector possesses that is not possessed by time? I don't see why a vector must have spatial directions. It feels like saying time isn't a vector because it's time and not space. Surely not all vectors must indicate directions in space?

Molu

 

[marlon]

Surely not all vectors must indicate directions in space?

Molu

Nope, but one does need an appropriate base to describe vectors. This base contains the unit-vectors. The point is that time does NOT have a unit vector.

regards
marlon

 

[loom91]

Why not? Let us take th birth of Christ as our origin (0), and one gregorian year to be the unit of time. Now the 1D vwctor pointing from the origin to the point of time 1AD can be considered an unit vector?

 

[RadiationX]

Why not? Let us take th birth of Christ as our origin (0), and one gregorian year to be the unit of time. Now the 1D vwctor pointing from the origin to the point of time 1AD can be considered an unit vector?

You are wasting your time sir. I see where you are coming from but the logic is not correct

 

[loom91]

That is exactly what I want to know, why my logic is not correct? That was what my question was about. Why can't time be considered a vector?

 

[marlon]

Why not? Let us take th birth of Christ as our origin (0), and one gregorian year to be the unit of time. Now the 1D vwctor pointing from the origin to the point of time 1AD can be considered an unit vector?

Well, you can always define a vector like this, that is indeed true. But as i will explain you, this vector cannot be an Euclidean vector like the space vector or the acceleration vector.

Within the context of your original question coming from Halliday and Resnick, the vectors that you use are Euclidian vectors. Such vectors have unitvectors that are defined by their magnitude being equal to ONE and the fact that they are perpendicular to each other. Knowing this, do you still believe that time is a vector in Euclidian space ? What would be the physical relevance ?

So, in classical physics, time does not exhibit the same behaviour as a space vector because you cannot define a unitvector for T that has magnitude 1 AND that is perpendicular to all spatial directions (x, y and z)

Also, let me ask you this : if i were to ask you to perform a rotation onto your socalled time vector, what operation would you execute ? What would be the physical relevance ? Keep in mind that in classical mechanics, time and space are NOT connected what so ever !

You see, it's questions like this that define wether an object can be classified as a vector or not.

marlon

 

[loom91]

But I still do not see exactly which clauses and subclauses of the definition of vector are not obeyed by time. A straight-line is an one dimensional Euclidean Space, therefore it is not valid to ask what would happen if you rotated time. There is no extra dimension to rotate it onto. Also, as time and space are not connected in Classical Mechanics why would they all need to be perpendicular? Why would time need to be perpendicular to the spatial directions.

To put it a different way, is it not possible to construct a one-dimensional vector space over the reals and consider its elements time vectors?

 

[loom91]

Looks like I'm never going to understand this.

 

[daveb]

What you're trying to distinguish is the concept of vectors from a mathematical perspective versus a physics perspective. In mathematics, the reals are a vector space, and an individual real number is a vector (regardless of whether it represents time). But if those real numbers represent time, then in physics they are not considered vectors. The thing mathematics considers a vector is much more broad than what physics considers a vector.

 

[dx]

Yes, you can define things in such a way that time-intervals are vectors. choose an origin, and a unit time interval in the positive(or negative) direction of the real line. Now you can treat time intervals and space-intervals exactly alike, and yes, you can consider them as vectors.(provided that you define things such that they combine(like addition) and transform in the same way as other vectors, like displacement)

 

[loom91]

It looks to me that a real answer to the question doesn't exist. In that case, what should I write if I'm given this question in my exam?

 

[maverick280857]

I am not sure at what level is this question to be answered. I am presuming it is But Marlon's and daveb's explanations are the general ones. If you have trouble understanding this think of it from a physics point of view forgetting the operations (dot, cross, addition, scaling, dilation, etc) that you are familiar with in vectors.

If you denote two events A($t_{A}$) and B($t_{B}$) on the time axis with reference to an arbitrarily assigned $t=0$ datum and try to define a vector from A to B, it would indicate time duration between the two events. If you translate this "time duration vector" along the time axis then it can also represent the time duration vector for some two other events on the time axis which could well have occurred before A and B (and even before the datum). With normal length vectors this is not a problem since we are bothered with the direction the vector points in rather than its actual initial and terminal points. A vector from $(0,0,0)$ to $(1,1,1)$ is equivalent to a vector from $(1,1,1)$ to $(2,2,2)$ in length space for example. There is no way you can distinguish between these two vectors though you know that they are joining two different pairs of points.

Further, a time duration vector could also be a vector from B to A. As we're concerned with the absolute value $|t_{B}-t_{A}|$ it doesn't really matter what its direction is...you could assign it an arbitrary direction too, but then it means its not really a vector--its not zero vector after all. (This is not an explanation for your question but just a perspective.)

I don't think that at a school level such a question can be answered without understanding time and space notions coupled with Special Relativity. So don't worry too much if you are uncomfortable with some of the things written here. You will soon understand them.

 

[loom91]

Maverick,

Vectors don't necessarily have to be free vectors, though most vectors are. Position vector is a bound vector that is tied to a magnitude, a direction AND an endpoint. It is meaningless to move a position vector around because that destroys the vector.

 

[Curious3141]

Yes, you can draw directed arrows to represent time intervals if you like. Does that make them vectors? As Marlon said, not really.

Just because something has direction and a magnitude does not make it a vector. A current in a circuit takes one of two directions (clockwise/counterclockwise or forward/backward, according to your definition) and has a magnitude. Is it a vector? Why does every Physics text out there define a current as a scalar?

 

[maverick280857]

Maverick,

Vectors don't necessarily have to be free vectors, though most vectors are. Position vector is a bound vector that is tied to a magnitude, a direction AND an endpoint. It is meaningless to move a position vector around because that destroys the vector.

Precisely that's why there is no uniqueness...but I'm not moving the position vector. The "vector" I defined there just contains information about the time duration between two events. So far as the information is concerned, it is not "directed" so I really don't care what direction I attach to the vector. Thats all I meant.

 

[maverick280857]

Yes, you can draw directed arrows to represent time intervals if you like. Does that make them vectors? As Marlon said, not really.

Just because something has direction and a magnitude does not make it a vector. A current in a circuit takes one of two directions (clockwise/counterclockwise or forward/backward, according to your definition) and has a magnitude. Is it a vector? Why does every Physics text out there define a current as a scalar?

I said "time duration vector" and I did say that at the end it doesn't make sense to make this quantity a vector. It wasn't an explanation for time not to be a vector.

(As for current, we're not saying its a vector...charge flow is constrained by the conductor. I don't think there was ever any confusion regarding this.)

I don't see how one can perform mathematical vector operations (as defined in a vector space) on hypothetical "time vectors" and attach some physical signifiance to them. Thats what I was trying to explain without resorting to mathematics. The explanation is more subtle as Marlon as provided it. But at the "school" level, Special Relativity isn't taught so you have to give some weird kind of argument as an answer.

 

[Curious3141]

I said "time duration vector" and I did say that at the end it doesn't make sense to make this quantity a vector. It wasn't an explanation for time not to be a vector.

(As for current, we're not saying its a vector...charge flow is constrained by the conductor. I don't think there was ever any confusion regarding this.)

I don't see how one can perform mathematical vector operations (as defined in a vector space) on hypothetical "time vectors" and attach some physical signifiance to them. Thats what I was trying to explain without resorting to mathematics. The explanation is more subtle as Marlon as provided it. But at the "school" level, Special Relativity isn't taught so you have to give some weird kind of argument as an answer.

No, all that's fair enough, I agree with you. I was simply making the OP comfortable in realising and accepting that there are other directed quantities in physics that are treated as scalars. Hence, in principle, there is no reason why time cannot be treated as a scalar despite the apparent presence of a "direction".

 

[maverick280857]

No, all that's fair enough, I agree with you.

Thank you

Best wishes,

Cheers
Vivek

 

[loom91]

What I've gotten from this discussion is that it is possible to define time as a vector after making certain assumptions, but it is not usually defined as such because in classcial physics dealing with time does not require us to use the extended toolset provided by the vector formalism, scalars being able to do the job.

 

[siddharth]

I thought the answer to this is simple enough, and marlon has already given it. Time doesn't transform properly when we change the coordinates. So, it isn't a vector.

Also,

A current in a circuit takes one of two directions (clockwise/counterclockwise or forward/backward, according to your definition) and has a magnitude. Is it a vector? Why does every Physics text out there define a current as a scalar?

Griffiths does define current as a vector. That is,

$$\vec{I} = \lambda \vec{v}$$

where,
$\lambda$ is the line charge density
$\vec{v}$ is the velocity of the charge

Since, the path of the flow is dictated by the shape of the wire, we normally don't display the vector character of I.

 

[loom91]

I thought the answer to this is simple enough, and marlon has already given it. Time doesn't transform properly when we change the coordinates. So, it isn't a vector.

Also,


Griffiths does define current as a vector. That is,

$$\vec{I} = \lambda \vec{v}$$

where,
$\lambda$ is the line charge density
$\vec{v}$ is the velocity of the charge

Since, the path of the flow is dictated by the shape of the wire, we normally don't display the vector character of I.

That's strange, a staple question our teachers use when teaching vectors is Why isn't current a vector? The answer given is that addition of currents follows the law of scalar addition rather the triangle law of vector addition.

 

[maverick280857]

Griffiths does define current as a vector. That is,

$$\vec{I} = \lambda \vec{v}$$

This is Vector Current. Please read Feynman's Lectures for more details.

 

[maverick280857]

That's strange, a staple question our teachers use when teaching vectors is Why isn't current a vector? The answer given is that addition of currents follows the law of scalar addition rather the triangle law of vector addition.

Thats not a convincing physical explanation at all! In fact its like saying a vector is a vector because its not a scalar! Once you know that current cannot be treated as a vector quantity you can say that current addition follows scalar additon. Its like proving a logical statement.

 

[siddharth]

This is Vector Current. Please read Feynman's Lectures for more details.

I don't understand. What do you mean?

 

[loom91]

Thats not a convincing physical explanation at all! In fact its like saying a vector is a vector because its not a scalar! Once you know that current cannot be treated as a vector quantity you can say that current addition follows scalar additon. Its like proving a logical statement.

They draw two wires inclined at an angle and then say that the currents flowing inside them add up at the meeting point to give a scalar sum (independent of the angle) instead of varying with the cosine of the angle as would be required for vector addition.

 

[maverick280857]

They draw two wires inclined at an angle and then say that the currents flowing inside them add up at the meeting point to give a scalar sum (independent of the angle) instead of varying with the cosine of the angle as would be required for vector addition.

It has to be charge conservation How can the charge coming out of the second wire differ from the one entering in as the cosine of the angle that the first wire makes with it...this is funny (and I know many books I've seen have this example too) But charge conservation happens to be the critical reason. If you insert it after the phrase "add up at the meeting point" then you have the correct explanation mate.