Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Proper time is identical with the spacetime interval of a timelike movement.

A spacetime interval is the dot product of two vectors and thus a scalar. Proper time however is always pointing exactly in time direction, all space dimensions are 0. Also it may be added or multiplied with scalars.

As outlined in the answer of Ross Millikan, proper time does not take part in Lorentz transforms. That means that it is not a vector within Minkowski vector space. But with its above mentioned characteristics, is there something missing for proper time being a vector?

Or, in case that the physical definition of proper time is too restricted, what else is this vector pointing in time direction with the magnitude of proper time?

Edit: The most confusing fact is the comparability of time (vector) and proper time (scalar?), example: A space traveler returns after 30 years, but he got only 20 years older, thus he "saved" 10 years (substraction vector / scalar??).

share|improve this question
    
A vector in Minkowski space needs four components, not just one. A scalar has just one component, a vector (rank 1 tensor) has four, a rank 2 tensor has 16, etc. Why does that fact that it can be multiplied by a scalar have to make it something like a vector? –  Ross Millikan Jun 4 '14 at 19:50
    
@Ross Millikan, regarding your first sentence: I agree that proper time is not a vector within Minkowski spacetime (see my question). I agree also that the spacetime interval is not a vector. However, it seems to me that proper time has a direction, and all its space coordinates are zero (thus a four vector). The corresponding vector space might be similar to Minkowski spacetime, but Lorentz transforms don't apply, and the observer's time + 3Dspace is replaced by proper time + 3D space. –  Moonraker Jun 4 '14 at 20:16
2  
No, (proper time,0,0,0) is not a four vector. A four vector has to transform properly under Lorentz transformations. Proper time does transform properly as a scalar (not at all). –  Ross Millikan Jun 4 '14 at 20:21
    
Sorry, but you are talking about Minkowski and I am not. We do not talk about the same thing. –  Moonraker Jun 4 '14 at 20:40
2  
@moonraker, in your 2nd comment, you're essentially saying, "I know proper time is not a four-vector however, it seems to me that it is a four-vector". Honestly, this is nonsense. If something is a four-vector, Lorentz transforms apply period. From Wiki: a four-vector is defined as a quantity which transforms according to the Lorentz transformation. Thus, if as you write, the Lorentz transforms don't apply to the object, it is analytically true that the object isn't a four-vector. –  Alfred Centauri Jun 4 '14 at 22:00

6 Answers 6

Proper time is the dot product of two four-vectors (actually one vector with itself). As such it is a scalar. You can see it is not a part of a vector by the fact that it is not changed by rotations or boosts.

share|improve this answer
    
Thank you for the great answer! I edited my question with regard to your observations. –  Moonraker Jun 4 '14 at 19:02

You say:

Proper time however is always pointing exactly in time direction

but this is not so. It is certainly true that in an observer's rest frame the proper time is numerically equal to the coordinate time, but this does not mean that the proper time and the coordinate time are the same. The proper time is still defined by:

$$ d\tau^2 = g_{ab}dx^a dx^b $$

so it is still a scalar. It's just that in the rest frame only $x^0$ (i.e. $dt$) is non-zero so we have:

$$ d\tau^2 = dt^2 $$

In other inertial frames $dx$, $dy$ and $dz$ may not be zero, but $d\tau$ will be the same (because it's invarient under Lorentz transformations) so in general $d\tau$ will not be equal to $dt$.

share|improve this answer
    
Read my question, Your first sentence does misunderstands what I said. My sentence you highlighted wants to say: proper time is just the absolute part (the quintessence) of observer's time for which all observers agree that it is pointing in time direction. Or: time is relative (submitted to Lorentz transforms), but proper time is absolute. –  Moonraker Jun 4 '14 at 19:37
    
@Moonraker: well yes, but my point is that the magnitude of a vector is not the same as the vector. For the stationary observer $d\tau$ is the magnitude of $\vec{dt}$ - the two are not the same object. –  John Rennie Jun 5 '14 at 10:07
    
OK, we get to the same point as below with Alpha Centauri, my decisive question is: what is the vectorlike observable behaving like propertime (see my double comment below, see also my example of a diagram of the age of the universe in the comment to NeuroFuzzi) –  Moonraker Jun 5 '14 at 11:10

Is proper time a vector?

Unequivocally, no. Proper time is a scalar, not a vector. From the Wikipedia article "Scalar (physics)":

Examples of scalar quantities in relativity include electric charge, spacetime interval (e.g., proper time and proper length), and invariant mass.


Proper time however is always pointing exactly in time direction

Proper time, as a scalar, is a number without direction; proper time does not point, period.

This is elementary and, evidently, at the root of your misunderstanding.

Again, from the Wikipedia article "Scalar (physics)":

In physics, a scalar is a one-dimensional physical quantity, i.e. one that can be described by a single real number (sometimes signed, often with units), unlike (or as a special case of) vectors, tensors, etc. which are described by several numbers which characterize magnitude and direction

So, your conception of proper time is flawed and you must give up this idea that proper time points in direction, time or otherwise. It does not.

share|improve this answer
    
Formally, you are right. But it seems that there must be an observable which is nearly identical with spacetime interval but which is a vector: If an observer is observing 3 objects moving near light speed in his Minkowski diagram, he will assign a time dimension to them and thus a spacetime vector. The time dimension is increasing each second by one second. Now this observer calculates the proper time of the objects, which is e.g. increasing each second by half a second and after one hour half an hour. And now take the point of view of one of the objects: –  Moonraker Jun 5 '14 at 8:56
    
In their Minkowski diagram half an hour has passed. --- What I want to say is that you cannot distinguish a difference between observer's and proper time, it is the same units, the same sense of evolution, and at each moment you can change from one to another with only a quantitative difference. So, you can call it "proper time" or refuse to call it "proper time", there must be something very similar to proper time that is a vector. –  Moonraker Jun 5 '14 at 8:57
    
@Moonraker, your use of certain terms is so unusual that it is often difficult tease the actual meaning out. For example: "he will assign a time dimension to them..." is quite odd as is "The time dimension is increasing each second...". Now, assuming uniform motion, the world line of an object is essentially the time axis of the object's frame of reference. And, of course, there is a four-vector that is parallel to that axis and, in the object's frame of reference, this four-vector has zero spatial components. Further, there is a unit four-vector in this direction. So what? –  Alfred Centauri Jun 5 '14 at 11:29
    
Further this vector (which is no four-vector) is invariable for all other observers. –  Moonraker Jun 5 '14 at 11:37
1  
@Moonraker, it is my opinion that your conception of "vector" is inchoate at best. If you think that your can group the proper time together with the spatial components of a four-vector and call it a vector (but not a four-vector) you don't understand what a vector is. A vector is a geometric object that exists independent of the coordinate system one wishes to express the components of the vector in. You can't simply group some numbers together and call it a vector. Please do some more reading and thinking about the nature of vectors. –  Alfred Centauri Jun 5 '14 at 12:03

A timelike vector of unit length can have $(ct)^2-x^2=1$, and this traces out a hyperbola of two sheets symmetric after reflections through the $x$ axis. There are an infinite number of vectors with the proper time of 1, but they point in vastly different directions.

Generally, "time coordinate" and "proper time" are treated as two completely separate things. In the same way, "x coordinate" and "length" are two completely separate things. The x coordinate can be negative or undefined while length is still defined, and length is defined so that it is always positive. It's a scalar, not a vector. Proper time is a scalar, not a vector. It does not "point in the time direction with all space coordinates zero".

If you're still in doubt, try to write down a mathematical definition of proper time, or try to find some way to apply it to a problem. If your definition is like the ones given by John Rennie or the second paragraph of this post, you will find that proper time is a scalar and not a vector. (By the way, when $\tau^2$ is negative, we call $l^2=x^2+y^2+z^2-(ct)^2$ a proper length)

share|improve this answer
    
Thank you! "Magnitude of a vector but not a vector": Lets take a Minkowski diagram, an object is going upwards from (0;0) to (0;2). The magnitude of proper time equals the magnitude of the object's coordinate movement. If you say that proper time is a scalar, there must at least be something similar to a proper time vector. –  Moonraker Jun 4 '14 at 20:41
    
You asked for a practical application, here is one: Draw a diagram of the age of the world. 13 billion years is OK for us, but not for the electrons who made their way from the beginning. Moving near light speed the age of these parts of the universe is not 13 billion years. A Minkowski diagram is not appropriate for such a representation. –  Moonraker Jun 4 '14 at 20:41
    
@NeuroFuzzy Every text on SR I've ever seen agrees that the term "spacetime interval" is a term reserved for the invariant squared length. It is not like an interval in the real line, it is just a scalar. People disagree in the units (meters, seconds, etc.) and in the sign (timelike as positive, timelike as negative, etc.), but all texts I've seen agree that it is a scalar. So I disagree with your first paragraph (except the last sentence). –  Timaeus Dec 30 '14 at 7:56
    
@Timaeus You're right. I'll remove the first paragraph. –  NeuroFuzzy Dec 30 '14 at 17:18

Moonraker: "Edit: [...] comparability" --

How to compare

  • the magnitude $s[ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} ]$ of a particular time-like interval between two particular events "$\mathcal{E}$" where the indicated participants met each other; i.e. at one $A$ and $J$ met each other (but not $Q$), and at the other event $A$ and $Q$ met each other (but not $J$) to

  • the duration $\tau A[ \circ_J, \circ_Q ]$ of some particular participant $A$ from $A$'s indication at one initial event (the meeting with participant $J$) until $A$'s indication at the other, subsequent final event (the meeting with participant $Q$)

?
That's an important question (which surely has been raised and addressed on this site, too).
And there's a cute, somewhat superficial and decidedly mathematical answer:

The value of the ratio

$$ \tau A[ \circ_J, \circ_Q ] ~ / ~ s[ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} ] $$

is (equal to) the limit of the sum of ratios

$$ \text{Limit}_{ \mathscr{\hat S} \rightarrow \mathscr{A}_{J~Q}; \text{and for successive event pairs } \in (\mathscr{\hat S} \cup \{ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} \}): s[ \mathcal{E}_{A\hat K}, \mathcal{E}_{A\hat P} ] ~ / ~ s[ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} ] \rightarrow 0 } \large{[} \sum_{ \text{successive event pairs } \in (\mathscr{\hat S} \cup \{ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} \} } s[ \mathcal{E}_{A\hat K}, \mathcal{E}_{A\hat P} ] ~ / ~ s[ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} ] \large{]}, $$

where

  • set $\mathscr{A}_{JQ}$ is the set of all events in which $A$ took part from (including) the initial event $\mathcal{E}_{AJ}$ of having met $J$ until (including) the final event $\mathcal{E}_{AQ}$ of having met $Q$,

  • set $\mathscr{\hat S}$ is a (any variable) subset of $\mathscr{A}_{JQ}$ consisting of discrete successive events (in which $A$ took part; such as $\mathcal{E}_{A\hat K}$ for any suitable (variable) participant $\hat K$ and $\mathcal{E}_{A\hat P}$ for any suitable (variable) participant $\hat P$),

  • and the limit (if it exists, for the particular participant $A$, the particular initial event $\mathcal{E}_{AJ}$ and the particular final event $\mathcal{E}_{AQ}$) is taken as ever more (discrete successive) events of $\mathscr{A}_{JQ}$ are included in $\mathscr{\hat S}$, and

  • all ratios $s[ \mathcal{E}_{A\hat K}, \mathcal{E}_{A\hat P} ] ~ / ~ s[ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} ]$ between the magnitude of an interval between any two consecutive events in set $\mathscr{\hat S} \cup \{ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} \}$ and the magnitude of the interval between initial and final event approaches $0$.

This limit (if it exists) constitutes a Riemann integral and may accordingly be written as

$$ \tau A[ \circ_J, \circ_Q ] ~ / ~ s[ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} ] := \int_{\mathscr{A}_{J~Q}} ds_{JQ}.$$

So far, so good. (Hopefully.) However, it remains the task to compare magnitudes of (time-like) intervals to each other in the first place; i.e. the question should be addressed how the real number values of ratios $$s[ \mathcal{E}_{A\hat K}, \mathcal{E}_{A\hat P} ] ~ / ~ s[ \mathcal{E}_{AJ}, \mathcal{E}_{AQ} ]$$ ought to be determined, by geometrical physical measurement.

This primary question (of physics) must be addressed without presuming or requiring results of comparisons discussed above, of course.
Not surprisingly, that's quite difficult in general; some basics approach to an answer is sketched for instance in (my answer to the question) "Deriving formula for time dilation".

(That question as well as my indicated answer presume and require the notion of certain participants having pairwise been and remained "at rest to each other"; which therefore must in turn be determined without presuming or requiring results of comparisons discussed above.)

share|improve this answer
    
As far as I see, my question does not imply any measurement issue - let's suppose roughly that after 30 years our astronaut looks only 20 years older - this rough measurement by visual check might be sufficient. What would be your answer to my question? –  Moonraker Jun 5 '14 at 20:25
    
Moonraker: "let's suppose roughly that after 30 years our astronaut looks only 20 years older [...]" -- Fair enough, as such: that's a (rough, or even thorough) comparison of individual appearances (Result, for instance: "At the departure the astronaut and his stay-at-home twin looked alike; but upon returning the astronaut looked just like the stay-at-home twin had looked 10 summers before.") But what I'm trying to get at is separate and distinct from comparing appearances; namely the measurement: How long did the astronaut's trip take in comparison to the stay-at-home duration? –  user12262 Jun 5 '14 at 21:03

Let's measure our vectors in meters, just to pick a convention (afterwards I'll do it in seconds so that you can see that it doesn't affect the basic ideas).

Given a point in spacetime $(ct_1,x_1,y_1,z_1)$ and another point $(ct_2,x_2,y_2,z_2)$, we can easily construct three things.

1) A vector $\vec{v}=(c(t_2-t_1),x_2-x_1,y_2-y_1,z_2-z_1)$ that points from one to the other.

2) The proper length $l=\sqrt{c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2}$ of that timelike vector $\vec{v}$.

3) And finally, the unit vector $\vec{u}=\vec{v}/l$, which equals $\frac{(c(t_2-t_1),x_2-x_1,y_2-y_1,z_2-z_1)}{\sqrt{c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2}}$.

As numbers (scalars) and vectors (geometric displacements), everyone will agree. The vectors might appear as different quadruples of numbers to different observers (imagine picking a different direction to call your $x$-axis, the geometric entity is the same displacement, but you get different numbers).

Now if you didn't want to measure in meters you can do it again, dividing everything by $c$, except it turns out $u$ will be the same.

1) A vector $\vec{V}=(t_2-t_1,(x_2-x_1)/c,(y_2-y_1)/c,(z_2-z_1)/c)$ that points from one to the other.

2) The proper length $L=\sqrt{(t_2-t_1)^2-(x_2-x_1)^2/c^2-(y_2-y_1)^2/c^2-(z_2-z_1)^2/c^2}$ of that timelike vector $\vec{V}$.

3) And finally, the unit vector $\vec{U}=\vec{V}/L=\vec{u}$, which again equals $\frac{(c(t_2-t_1),x_2-x_1,y_2-y_1,z_2-z_1)}{\sqrt{c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2}}$.

So you have a vector $\vec{v}$ between two spacetime points, as a vector it has a length $l$, and a corresponding unit vector $\vec{u}=\vec{v}/l$ that keeps track of the direction in spacetime, but ignores how much dispalcement (or even which units you use, just the direction in spacetime).

OK, so now lets get to your question. Given two points we have the spacetime displacement $\vec{v}$, the unit displacement $\vec{u}$ that tells us which direction in spacetime to go (but has no informative length or even units), and the length $l$ telling us how far in spacetime to go (and has units of length or time). When someone says proper time, they are referring to the length of the timelike vector, maybe in units of time instead of length.

A common reason people say proper length for spacelike entities and proper time for timelike entities in textbooks isn't because they care a great deal about the units of the answer (it's just a measure of length) but because the squared length $l^2=c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2$ is what is easy to compute and for a timelike spacetime displacement $c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2$ is positive, whereas for a spacelike displacement $-c^2(t_2-t_1)^2+(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$ is positive. We like to have square roots of positive numbers, so people like to have formulas like $\tau=c^-1\sqrt{c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2}$ and $l=\sqrt{-c^2(t_2-t_1)^2+(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. The only real reason to have two formulas is because then you are taking the square root of a positive number. The idea for either case is the same: the displacement has a magnitude (proper length or proper time) and a direction (unit vector in spacetime), their product gives you the displacement.

Just like in three-space, a vector $(\Delta x,\Delta y,\Delta z)$, has a magnitude, $r=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}$, and a unit vector (for direction) $((\Delta x)/r,(\Delta y)/r,(\Delta z)/r)$.

Edit: I guess I didn't actually answer the questions.

First question: Is there something missing for proper time being a vector?

Yes, proper time is a magnitude, not a vector, it is just a length in spacetime, it by itself has no direction, there are many directions you could travel all with the same length/magnitude.

Second question: What else is this vector pointing in time direction with the magnitude of proper time?

The vector pointing from one event to the other will look like $(\tau,0,0,0)$ in the frame that inertially moves from one event to the other. But the vector $(\tau,0,0,0)$ is different than the scalar $\tau$. It's like if you walked north a distance of five meters, the distance is 5, everybody agrees. And to you the space vector might be $(5,0,0)$ but what if someone else like to put east first and north second? Then they will say that the vector is $(0,5,0)$. A vector has to tell you the direction. If all you tell me is you walked 5 units I can't tell if you walked $(5,0,0)$, $(0,5,0)$, $(0,0,5)$, or even $(3,4,0)$. The vector tells you the direction and the magnitute, but only when you say what the 3 (or 4) numbers mean first.

share|improve this answer
2  
What is your answer? Do you want to say the space time interval is a scalar, but proper time is a vector? –  Moonraker Dec 29 '14 at 6:47
    
Your question never asked a question about a spacetime interval. Every text I've examined uses the term "spacetime interval" to mean plus or minus the (square of) the invariant proper length ( or invariant proper time). People don't agree on whether timelike is positive or negative, whether to measure in seconds or meters, but they agree on the rest, it's the (squared) length, a "spacetime interval" is not a vector, and it's not an interval like an interval of numbers. –  Timaeus Dec 30 '14 at 8:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.