# Definition of four-velocity

• I
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In a formal manner the 4-velocity defined as ##\vec{u} = \frac{d\vec{x}}{cd\tau}##. Now this is also equal to the unit tangent vector of the worldline. My confusion is actually more geometrical. ##d\vec{x}## is the infitesimal distance between two points in the worldline and ##cd\tau## is the infitesimal distance along the worldline. How can their ratio can give us the four-velocity ?

Note that in this definiton four-velocity its unitless.

Edit: For instance ##cd\tau = |d\vec{x}|_g = \sqrt{-g(d\vec{x}, d\vec{x})}## if ##d\vec{x}## is future directed.
This means that we are actually calculating $$\vec{u} = \frac{d\vec{x}}{|d\vec{x}|}$$ which seems strange. I guess 4-velocity in SR is just defined as the tangent vector to the worldline rather then the usual definition of the velocity in our world which is decribed as meter per second.

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Now this is also equal to the unit tangent vector of the worldline. My confusion is actually more geometrical. d→xdx→d\vec{x} is the infitesimal distance between two points in the worldline and cdτcdτcd\tau is the infitesimal distance along the worldline. How can their ratio can give us the four-velocity ?
Because it’s not a ratio. ##dx## is the infinitesimal separation vector between two points, not the distance. In the limit where ##dx## approaches the 0 vector, the length of that vector equals the length along the curve between those two approaching points. When any vector is scaled by the reciprocal of its length, you get a unit vector.

Arman777
PeroK
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A four-velocity is a four-vector. It's not the three-velocity, which is a three-vector. The notation ##\vec u## is generally used for three-vectors. A common notation for the four-velocity is:
$$\mathbf{u} \ \text{or} \ \mathbf{U} = \frac{d \mathbf x}{d\tau}$$
Where ##\mathbf{x} = (ct, x, y, z) = (x^0, x^1, x^2, x^3)##. Alternatively, in component form:
$$u^{\mu} = \frac{d x^{\mu}}{d\tau} \ \ (\mu = 0, 1, 2, 3)$$

vanhees71, strangerep and Arman777
Ibix
You can do a Euclidean equivalent as well. Make equally spaced marks on a piece of string and then drop it on a piece of graph paper. If the marks of length (say they mark distances of ##\delta L##) are very very short, you can draw a straight line on the graph paper between two adjacent marks - thus defining a ##\delta x## and a ##\delta y## and having a length ##\delta L##. This lets you define a normalised vector ##(\delta x/\delta L,\delta y/\delta L)##. Taking the limit of infinitesimally small steps (letting ##\delta\rightarrow d##), you've now defined a Euclidean "two-velocity".

I must say I dislike the "four velocity" name because a normalised tangent vector doesn't feel like a velocity to me. However, the last time I complained about it, someone said that "velocity" was a standard geometrical term for a tangent vector to a curve.

Arman777
PeterDonis
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I guess 4-velocity in SR is just defined as the tangent vector to the worldline
Yes, exactly.

rather then the usual definition of the velocity in our world which is decribed as meter per second
Yes, the 4-velocity is not the same as an ordinary velocity, and does not have the same units.

vanhees71 and Arman777
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Thanks

robphy
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The 4-velocity is the future UNIT timelike tangent-vector to a worldline. (Unfortunate exception: the text by Callahan uses (c,v) as “4-velocity” (which is not a 4-vector) and “proper 4-velocity” $\gamma(c,v)$ .)

While the 4-velocity as a unit vector
$(\gamma, \gamma v/c) =(\cosh\theta,\sinh\theta)$ is dimensionless (which is convenient for taking dot products and extracting components), some authors introduce the speed of light factor so that the spatial component [and thus its temporal component] has units of velocity:
$(\gamma c, \gamma v)$.

Arman777 and vanhees71
PAllen
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It’s not really different in Galilean physics. Considering, e.g. x,t, the speed is the tangent of an angle in the X,T plane. Scaling adds redundancy: 10 meters per 5 seconds is the same speed as 2 meters per second. Using a single angle in this plane to represent speed, with a speed vector of (sin,cos), you have unit magnitude and no loss of information. Going to 4 dimensions does not change things - if you represent a path as curve in space by time, rather than a curve in 3 space parameterized by time, you have a scaling redundancy which can be removed by normalization.

Arman777
robphy
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It’s not really different in Galilean physics. Considering, e.g. x,t, the speed is the tangent of an angle in the X,T plane. Scaling adds redundancy: 10 meters per 5 seconds is the same speed as 2 meters per second. Using a single angle in this plane to represent speed, with a speed vector of (sin,cos), you have unit magnitude and no loss of information.
You actually have to use “Galilean Spacetime trigonometry”, using functions of a Galilean Angle, to make this work (which uses Galilean line-elements) ... but once you do, a lot of ideas carry over.

The tips of the Galilean 4-velocity lie on a line (hyperplane) whereas the tips of the Minkowskian 4-velocity lie on a hyperbola (hyperboloid).

Arman777
vanhees71
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The 4-velocity is the future UNIT timelike tangent-vector to a worldline. (Unfortunate exception: the text by Callahan uses (c,v) as “4-velocity” (which is not a 4-vector) and “proper 4-velocity” $\gamma(c,v)$ .)

While the 4-velocity as a unit vector
$(\gamma, \gamma v/c) =(\cosh\theta,\sinh\theta)$ is dimensionless (which is convenient for taking dot products and extracting components), some authors introduce the speed of light factor so that the spatial component [and thus its temporal component] has units of velocity:
$(\gamma c, \gamma v)$.
Note that it's not unique in the literature. Some define ##U^{\mu} = \mathrm{d} x^{\mu}/\mathrm{d} \tau##. Then it has the dimension of a velocity, some define it as ##u^{\mu}=\mathrm{d} x^{\mu}/c \mathrm{d} \tau##, then it's a "unit vector" in the sense of the Minkowski product. You simply have ##U_{\mu} U^{\mu}=c^2## or ##u_{\mu} u^{\mu}=1## (in the (1,3) west-coast convention of the Minkowski product). I prefer the latter, because then it's simple to project to time-like and space-like parts of four-vectors relative to an observer moving with four-velocity ##u##.

The most simple solution of this quibble is of course to use natural units with ##c=1## and forget about the man-made arbitrary conversion factors of the new SI ;-)).

robphy and Arman777
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I think pedagogically its more reasonable to study by not setting ##c=1##. Otherwise some people, including me, can confuse the four-velocity something similar or measurable as 3-velocity. However its not like 3-velocity. After explaining 4-velocity not the same as the usual velocity which can be done by ##c \ne 1##, we can set ##c=1## to make calculations easier.

robphy
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I think pedagogically its more reasonable to study by not setting ##c=1##. Otherwise some people, including me, can confuse the four-velocity something similar or measurable as 3-velocity. However its not like 3-velocity. After explaining 4-velocity not the same as the usual velocity which can be done by ##c \ne 1##, we can set ##c=1## to make calculations easier.
Strictly speaking, “c=1” seems likely to cause confusion since c is not dimensionless. Keeping x as a length and t as a time, “c=1 ft/ns” is better. Alternatively, one can use “c = 1 light-sec/sec” or “c = 1 light-year/year”.

One could use x as a time (time for light to travel that length) ... then the slope x/t is dimensionless.

PeterDonis
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Strictly speaking, “c=1” seems likely to cause confusion since c is not dimensionless.
From the point of view of relativity, ##c## is dimensionless; speeds in general are dimensionless in the natural units of relativity because the units of lengths and times (or more precisely spacelike intervals and timelike intervals) are the same. These units are "natural" to relativity because they make manifest the symmetry involved.

Whether it is better pedagogically to introduce this concept at the outset, or only after presenting other material, is a separate question. Back when I first learned relativity, I actually found it much easier to grasp other aspects once I had internalized the idea that spatial and temporal lengths had the same units.

Orodruin
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Strictly speaking, “c=1” seems likely to cause confusion since c is not dimensionless.
This is not true, but depends on your system of units. It is true in SI units, in natural units speeds are dimensionless - length and time has the same physical dimension.

robphy
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From the point of view of relativity, ##c## is dimensionless; speeds in general are dimensionless in the natural units of relativity because the units of lengths and times (or more precisely spacelike intervals and timelike intervals) are the same. These units are "natural" to relativity because they make manifest the symmetry involved.

Whether it is better pedagogically to introduce this concept at the outset, or only after presenting other material, is a separate question. Back when I first learned relativity, I actually found it much easier to grasp other aspects once I had internalized the idea that spatial and temporal lengths had the same units.
Of course, in fact I prefer to express spatial coordinates in units of time. But my comment refers to the fact that pedagogically for today’s students (which is what I took the context of @Arman777 ’s comment to be), “c=1” is potentially confusing because c has already been assigned a value in SI units and we teachers continually suggest that they reason with units as a way to check for consistency (and maybe an interpretation). So, my suggestion maintains the dimensions of Length/Time but assigns the numerical (calculator) value of 1.

strangerep and Arman777
Ibix
I personally thought of ##c## as one light second per second until it sank in how self-referential the units were.

Arman777
Han Solo made the Kessel Run in 12 parsecs.

Ibix and strangerep
vanhees71