Does relativity change the rules for time being a vector or a scalar?

[joenitwit]

In the equation x = vt it is generally accepted that x and v are vectors and that they have a common eigenvector. Each vector is the product of a scalar and a unitary eigenvector. Dividing both sides by v works because in x/v = t the x and v vectors have identical and canceling eigenvectors. This would indicate then that t is the quotient of two scalar values, yielding a scalar value. If t is a vector then the multiplication of v and t doesn't work and the division of x/v doesn't work.

Flat Minkowski space has a matrix of:

| x 0 0 0 | - - - | 1 0 0 0 |
| 0 y 0 0 | or - | 0 1 0 0 |
| 0 0 z 0 | - - - | 0 0 1 0 |
| 0 0 0 -ct | - - | 0 0 0 -c |

This standard matrix is composed of vectors. Since x, y, z are vectors then -ct must also be a vector. The speed of light (c) is a scalar therefore t must be a vector for this matrix. Yet, the previous discussion indicates that t must be a scalar?

Is Time (t) a vector or a scalar? If t is a scalar then there can be no time travel?

 

[PeterDonis]

In the equation x = vt it is generally accepted that x and v are vectors and that they have a common eigenvector.

Huh? Vectors don't have eigenvectors. Matrices have eigenvectors.

Each vector is the product of a scalar and a unitary eigenvector. Dividing both sides by v works because in x/v = t the x and v vectors have identical and canceling eigenvectors. This would indicate then that t is the quotient of two scalar values, yielding a scalar value.

This is all incorrect. Where are you getting this from?

If t is a vector then the multiplication of v and t doesn't work

This is true, but it doesn't have anything to do with the rest of what you wrote above.

and the division of x/v doesn't work.

I'm not sure that viewing t as the quotient of x and v is the best way to look at it, although it's not exactly wrong; the equation x = vt does say that the vector x is the product of the vector v and the scalar t, so mathematically t does work out as the quotient of x and v. But that way of looking at it leaves out the fact that t can vary; it's different for different events. See further comments below.

Flat Minkowski space has a matrix of:

| x 0 0 0 | | 1 0 0 0 |
| 0 y 0 0 | or | 0 1 0 0 |
| 0 0 z 0 | | 0 0 1 0 |
| 0 0 0 -ct | | 0 0 0 -c |

The second "matrix" (the correct term is "metric tensor") is correct (though the -c is usually written as -1 since we usually use units in which c = 1 in relativity). The first one makes no sense at all; where are you getting it from?

This standard matrix is composed of vectors.

No, it isn't; a matrix is not "composed" of vectors, it's a matrix that you can multiply by a vector to get another vector (with some technicalities that I don't think we need to go into at this point).

Is Time (t) a vector or a scalar?

It's a coordinate; x, y, z, and t are all coordinates, at least as you're using those symbols now. When you wrote the equation x = vt above, you were implicitly using "x" to mean the coordinate triple (x, y, z), which can be thougt of as labeling a vector--but it's a 3-vector, not a 4-vector. In spacetime we use 4-vectors, so the 4-tuple (x, y, z, t) labels a 4-vector; it's the 4-vector that goes from the origin (0, 0, 0, 0) to the point labeled by the coordinates (x, y, z, t). t is therefore one component of this 4-vector, i.e., it's a coordinate.

If t is a positive scalar then there can be no time travel?

None of the above has anything to do with whether or not there can be time travel.

 

[Dale]

Is Time (t) a vector or a scalar?

Hi joenitwit, welcome to PF!

t is neither a vector nor a scalar. It is a coordinate. Coordinates map positions to points in R4. Positions are not a vector space (what is the vector sum of the positions of New York and Paris). Coordinates are also not generally a vector space although they can be in specific cases, like Cartesian coordinates on an infinite flat manifold with simple topology.

 

[Abhilash H N]

Consider a duration of time, say I took 2 minutes to type this. Then the duration is a scalar right?

 

[Dale]

Consider a duration of time, say I took 2 minutes to type this. Then the duration is a scalar right?

Proper time is a scalar. I believe that you are referring to proper time here, i.e. the time measured by your wristwatch as you typed.

 

[Meir Achuz]

Time is one component of the four-vector X=(ct,x,y,z).

 

[joenitwit]

Coordinates are always vectors. Coordinates refer to a space-time location in reference to an origin. Without an origin and an axis system there can be no coordinates. Drawing a line from the origin to the coordinate always produces a vector. If it were not a vector then the sign, plus or minus, would not have any meaning since the coordinate would then just be a distance.

My question concerned the flat-space Minkowski matrix since the first row is a multiplier for the x-vector, the second for the y-vector, the third for the z-vector and the fourth for the t-vector... yet t, time, is not and cannot be a vector? There seems to be a fundamental flaw here concerning the nature of time? Yes, time is definitely a coordinate, but time cannot be a vector.

No need to worry about a tuple (it just confuses things but you can do it if you want). For a single coordinate you can always rotate the axis so the location is on the x-axis.

A vector is always the multiplication of an eigenvector (vector of length one - unitary) and a scalar which expresses the length or magnitude of the vector. This is the only way to divide vectors. Mathematically, the only permissible quotient of two vectors is if they have identical eigenvectors. Eigenvectors are usually expressed as i-hat, or j-hat or k-hat, but I don't know how to make the symbols on this forum.

My ultimate point is that there can never be negative time. Since x and v must always be in the same direction (common eigenvectors), t must always be a positive scalar.

 

[joenitwit]

Hi joenitwit, welcome to PF!

t is neither a vector nor a scalar. It is a coordinate. Coordinates map positions to points in R4. Positions are not a vector space (what is the vector sum of the positions of New York and Paris). Coordinates are also not generally a vector space although they can be in specific cases, like Cartesian coordinates on an infinite flat manifold with simple topology.

You could add these vectors. I don't know why you would want to but it is mathematically permissible.

t is a scalar multiplier. If time can speed up or slow down then the multiplier might change in relation to a location in space (the depth of a g-field) or to the particular motion of the traveler. In any case, time is a multiplier, not a vector. Time has no direction, rather like entropy, it just expands.

 

[joenitwit]

Right! This is where I want to go...

Hi joenitwit, welcome to PF!

t is neither a vector nor a scalar. It is a coordinate. Coordinates map positions to points in R4. Positions are not a vector space (what is the vector sum of the positions of New York and Paris). Coordinates are also not generally a vector space although they can be in specific cases, like Cartesian coordinates on an infinite flat manifold with simple topology.

Proper time is a scalar. I believe that you are referring to proper time here, i.e. the time measured by your wristwatch as you typed.

What other time is there, other than proper time? Maybe this is the answer to my question?

 

[Nugatory]

What other time is there, other than proper time?

Coordinate time, which is the value of the t coordinate of a given point after we've made a particular choice of coordinate system.

We usually choose our coordinate systems so that the coordinate time corresponds to the proper time on a particular worldline, but that correspondence doesn't make them the same thing.

My wristwatch measures proper time along my worldline. When I say that the sweep hand went around the dial once so one minute has passed, that's one minute of proper time as experienced by me.

When I use my wristwatch to attach times to events off my worldline ("The neighbor's dog barked at the same time that my wristwatch read 1:00 AM"), those times are coordinate times. I'm choosing to use a coordinate system in which the time coordinate is equal to the reading of my wristwatch (and I'm making another arbitrary choice as well, namely how I define "at the same time").

Proper time is unambiguously and clearly a scalar; it's the distance along a worldline.

Coordinate time is treated on an equal footing with the spatial coordinates; I disagree with your statement that "coordinates are vectors", but if it were correct, then internal consistency would require that it apply to the time coordinate as well as the spatial coordinates.

 

[D H]

Coordinates are always vectors.

No, they are not. The coordinates (of a point) is *not* a vector. Affine spaces are not vector spaces. It doesn't make sense to add points.

My ultimate point is that there can never be negative time. Since x and v must always be in the same direction (common eigenvectors), t must always be a positive scalar.

Nonsense. The point in time one chooses to represent t=0 is arbitrary.

This mathematical discussion has absolutely nothing to do with whether time travel is possible.

 

[joenitwit]

No, they are not. The coordinates (of a point) is *not* a vector. Affine spaces are not vector spaces. It doesn't make sense to add points.

My comment was that coordinates mean nothing without an origin. The relation of coordinates with the associated origin does define a vector. But, that's not my question.

Nonsense. The point in time one chooses to represent t=0 is arbitrary.

t=0 is a reference, not a time. I'm trying to think of time in any context which is not actually Δt. Is there any equation where t is not actually Δt? When Δt is multiplied by the speed of light the result is a scalar distance, once again not associated with an axis since there is no direction information.

My question concerned the fourth row of the Minkowski matrix. This row corresponds to a time vector/axis or perhaps it is better to talk about a vector multiplier (sometimes -ct is put in the first row). Each of the other "multipliers" correspond to an axis, but everyone here has agreed that time is a scalar. A scalar can't correspond to an axis. My point was that the row corresponding to time (-ct) is unlike the other three (x,y,z). Why?

 

[Dale]

You could add these vectors. I don't know why you would want to but it is mathematically permissible.

No, you cannot, at least not without imposing some additional structure.

If you disagree then please answer the question: "What is the vector sum of the positions of New York and Paris?"

Unless you define some additional structure beyond that inherent in positions there is no well-defined vector sum.

 

[PeterDonis]

My question concerned the flat-space Minkowski matrix since the first row is a multiplier for the x-vector, the second for the y-vector, the third for the z-vector and the fourth for the t-vector...

That's not how multiplying a matrix by a vector works. If I multiply a matrix M by a vector V to get a product P, then P is also a vector; and the first component of P (the x component) is obtained by multiplying the first row of M, component-wise, by the components of V, and adding those four numbers together. That is, $P^x = M^x_x V^x + M^x_y V^y + M^x_z V^z + M^x_t V^t$, where the upper index on the matrix is the row and the lower index on the matrix is the column. Similarly for the other components of P.

 

[Dale]

What other time is there, other than proper time? Maybe this is the answer to my question?

The other time is coordinate time, which I believe is what you are discussing here.

 

[Dale]

Coordinates are always vectors.

Ahh! No! Coordinates are most defnitely NOT always vectors. For example, polar coordinates in a plane do not form a vector space. For polar coordinates 0<r and -π<θ<π, so any polar coordinate multiplied by -1 gives an r which is outside the space of polar coordinates, so scalar multiplication does not follow the axioms of vectors for polar coordinates. Similarly, for two vectors each with θ<π/2 the sum is outside the space, so addition of two polar coordinates also does not follow the axioms of vectors.

 

[joenitwit]

That's not how multiplying a matrix by a vector works. If I multiply a matrix M by a vector V to get a product P, then P is also a vector; and the first component of P (the x component) is obtained by multiplying the first row of M, component-wise, by the components of V, and adding those four numbers together. That is, $P^x = M^x_x V^x + M^x_y V^y + M^x_z V^z + M^x_t V^t$, where the upper index on the matrix is the row and the lower index on the matrix is the column. Similarly for the other components of P.

This is exactly my point! You just imposed a Vector on Time, $V^t$, which cannot be done! Time is not, cannot be, a vector!

 

[joenitwit]

Ahh! No! Coordinates are most defnitely NOT always vectors. For example, polar coordinates in a plane do not form a vector space. For polar coordinates 0<r and -π<θ<π, so any polar coordinate multiplied by -1 gives an r which is outside the space of polar coordinates, so scalar multiplication does not follow the axioms of vectors for polar coordinates. Similarly, for two vectors each with θ<π/2 the sum is outside the space, so addition of two polar coordinates also does not follow the axioms of vectors.

? Why must r be greater than zero ?

Never mind... this is getting off on a tangent. My question was about Time being viewed as a vector in the Minkowski flat space matrix (flat space is when it is a diagonal matrix). So far everyone has agreed that Time must be a scalar, not a vector. Something seems amiss with the given matrix and thus with General Relativity.

 

[Nugatory]

t=0 is a reference, not a time. I'm trying to think of time in any context which is not actually Δt. Is there any equation where t is not actually Δt? When Δt is multiplied by the speed of light the result is a scalar distance, once again not associated with an axis since there is no direction information.

My question concerned the fourth row of the Minkowski matrix. This row corresponds to a time vector/axis or perhaps it is better to talk about a vector multiplier (sometimes -ct is put in the first row). Each of the other "multipliers" correspond to an axis, but everyone here has agreed that time is a scalar. A scalar can't correspond to an axis. My point was that the row corresponding to time (-ct) is unlike the other three (x,y,z). Why?

You are still confusing the scalar proper time with the coordinate time and with the t component of a four-vector. That "Minkowski matrix" (which is really the representation of the metric tensor for flat spacetime, using Cartesian coordinates) acts on the four-vector, and it treats all four components of the four-vector equivalently.

 

[PeterDonis]

This is exactly my point! You just imposed a Vector on Time, $V^t$, which cannot be done! Time is not, cannot be, a vector!

I don't understand. $V^t$ is just the 4th component of a 4-vector $V$ that is being multiplied by the matrix $M$. Are you saying that 4-vectors can't have a 4th component? That makes no sense.

Also, I only labeled the 4th component $V^t$ because when you wrote down the components of a 4-vector, you wrote them as (x, y, z, t); you'll note that I labeled the other three components of $V$ as $V^x$, $V^y$, and $V^z$. In other words, my labels are taken from your labels. Are you saying your labels are wrong?

 

[Dale]

So far everyone has agreed that Time must be a scalar

No, everyone has agreed that time is a coordinate. Nobody has agreed that time must be a scalar.

 

[WannabeNewton]

A time coordinate is a smooth map $t:U \rightarrow \mathbb{R}$, where $U\subseteq M$ is an open subset of a space-time $M$ (so $t$ is a scalar field), such that $\frac{\partial}{\partial{t}}$ is a time-like vector field.

 

[joenitwit]

In v=x/t, t must be a scalar quantity like speed, not a vector like velocity. Time cannot be a coordinate, but it could be the difference between two coordinates, delta-t.

When (in what equation) would time ever be a coordinate?

In the Minkowski matrix, what is time (-ct)?

 

[Dale]

When (in what equation) would time ever be a coordinate?

As was mentioned earlier, in relativity there are two notions of time. The first is coordinate time which is a coordinate. The second is proper time which is a scalar.

It is a very strong convention in formulas in relativity that the variable $t$ (or $dt$) denotes coordinate time and the variable $\tau$ (or $d\tau$) denotes proper time.

In the Minkowski matrix, what is time (-ct)?

In keeping with the standard convention, that is coordinate time.

 

[pervect]

In pre-relativistic physics, time is a separate entity and doesn't mix with space, so it's never a vector. However, in post relativity physics, with the concept of space-time, time and space are combined in a unified entity called space time.

See for instance the "Parable of the Surveyor", in its original textbook source in "Space Time Physics" or some of the various versions that may be on the web. (If you can't find it, and are motivated enough to read it, please ask.) for a more detailed description of how this unification works in practice.

Because of the unification of space and time into space-time, the concept of vectors that was previously used to represent displacements and velocities in space is now used to represent space-time intervals (in SR the details of this part are more complex in GR), and 4-velocities in relativistic theory, which are 4-dimensional analogs to the 3-dimesional classical concepts of displacements and velocities.

 

[pervect]

I've thought of another approach that may clarify things. One way of describing vectors in pre-relativity physics is this. We create a framework of spatial vectors, called variously basis vectors, or frame vectors, that we represent via special symbols with arrows over their heads, thus: $\vec{x}, \vec{y}, \vec{z}$. A general spatial vector can be represented as a multiple of the basis vectors, i.e. $x \vec{x} + y \vec{y} + z \vec{z}$.

This isn't the only way to talk about vectors, but it is a good one.

Now, in this framework of pre-relativistic physics, there is no basis vector for time, there is no $\vec{t}$. At least - not yet.

When we move to relativistic physics, we unify space + time into space-time and we DO create the above basis vector for $\vec{t}$. As a consequence, to specify an event, we use a format like $x \vec{x} + y \vec{y} + z \vec{z} + t \vec{t}$.

So it is correct to notice that we've changed the rules to introduce basis vectors for time we go to relativity, but it seems counterproductive to me to complain about it a lot. It's a difference, but one can get used to it. And looking at the big picture, we might say the reason we introduce basis vectors for time is a consequence of the unification of space and time into space-time.

It may also be worthwhile to point out that in pre-relativistic time was the same for everyone, and was absolute. In relativistic physics, time is not absolute, so a basis vector for time in some system S, $\vec{t}$, is NOT the same as the basis vector for time in some system S' moving with respect to S. If a physical system S' is moving in the z direction, using the formula for the Lorentz boost, we can write $\vec{t'} = \alpha \vec{t} + \beta \vec{z}$, where $\alpha$ and $\beta$ are some constants that I'm too lazy to compute at the moment (they can be derived from the Lorentz transform, but the basis vectors transform a bit differently than the coordinates).