
#1
Mar812, 10:24 AM

P: 71

The fourvelocity as defined for example here, is given by
[tex] U=\gamma(c,\bar{u}) [/tex] but I get [tex] U=\gamma(1,\frac{\bar{u}}{c}) [/tex] Consider the timelike curve [itex]\bar{w}(t)=(ct,\bar{x}(t))[/itex] with velocity [itex]\bar{v}(t)=(c,\bar{x}'(t))\equiv (c,\bar{u}(t))[/itex] and the arclength (proper time) [tex] \tau\colon I\subset \mathbb{R}\to \mathbb{R}\colon t\mapsto \int_{t_{0}}^{t}\left\ \bar{v}(k)\right\dk [/tex] for which (by First Fundamental Theorem of Calculus (1), the Minkowskian inner product (2) and the definition of the Lorentz factor (3) ) [tex] \Leftrightarrow \frac{d\tau}{dt}=\left\ \bar{v}(t)\right\=\sqrt{c^{2}\bar{u}^{2}(t)}\equiv \frac{c}{\gamma} [/tex] then the velocity of the curve after arclength (proper time) parameterization, is given by [tex] \bar{v}(\tau)=\frac{d\bar{w}}{d\tau}=\frac{d\bar{w}}{dt}\frac{dt}{d\tau }=\frac{\bar{v}(t)}{\left\ \bar{v}(t)\right\}=\frac{(c,\bar{u}(t))}{\frac{c}{\gamma}}=\gamma(1,\f rac{\bar{u}(t)}{c}) [/tex] I would think that my [itex]\bar{v}(\tau)[/itex] is the fourvelocity but in fact [itex]\bar{v}(\tau)=\frac{U}{c}[/itex] where U the fourvelocity as defined in textbooks. What am I missing? 



#2
Mar812, 11:26 AM

P: 2,043

If you are using units where c is not 1, then certainly you want the 4velocity to be normalized to c (or c depending on signature of the metric), and not 1 which is not in units of velocity (if, again, c is not set to 1).




#3
Mar812, 11:38 AM

Sci Advisor
PF Gold
P: 4,860

I think it is purely a convention. I learned that U is meant to be a unit vector, always, even when c is not taken to be 1. Some people like norm of U = c, some like 1. Norm of one amounts to units of time rather than distance for positions.
However, since you start with 4position in units of distance, you should get a U whose norm is c. The flaw is your d tau/ dt computation. It is 1/gamma not c/gamma. Then you get U with norm of c. Specifically, your integral formula is not right. Given your units for v, the integral is c * tau, not tau. This, then, is the initial (and only) error, from which all else follows. 



#4
Mar812, 05:25 PM

Sci Advisor
PF Gold
P: 4,860

Fourvector problem 



#5
Mar812, 06:09 PM

P: 2,043

If you parameterize your curve with the proper time in seconds, and your proper distances are measured in meters, don't you necessarily get the norm condition in U to be c?
What I mean is, if you use units of distance the same as your units of time, aren't you necessarily setting c=1? 



#6
Mar812, 06:18 PM

Sci Advisor
PF Gold
P: 4,860

position: (t, x/c, y./c, z/c) line element: d tau^2 = d t^2  (dx^2 + dy^2 + dz^2)/c^2 and the consequence that U = d(position) / d tau has norm 1 in no way have c=1. 



#7
Mar812, 07:25 PM

P: 2,043

Are you talking about adding a factor of c into the metric? o.o




#8
Mar812, 07:51 PM

Sci Advisor
PF Gold
P: 4,860

ds^2 = dx^2 + dy^2 + dz^2  c^2 t^2 and d tau^2 = dt^2  dx^2/c^2  dy^2/c^2  dz^2/c^2 I have always used the latter. 



#9
Mar912, 07:11 AM

P: 71

[tex] \tau\colon I\subset \mathbb{R}\to \mathbb{R}\colon t\mapsto \frac{1}{c}\int_{t_{0}}^{t}\left\ \bar{v}(k)\right\dk [/tex] and the fourvelocity in space/time units [tex]\bar{v}(\tau)=\gamma(c,\bar{u}(t))[/tex] Is this the correct explanation? As I understand, the fourvelocity in the other convention (norm=1) is unitless, isn't it? So how is it used then? 



#10
Mar912, 07:27 AM

Sci Advisor
PF Gold
P: 4,860

The magnitude of U really has no real meaning in either convention. All information about measured velocity in any basis (frame) is contained in the direction of U as a tangent vector. The norm 1 convention makes this explicit: it is literally a unit tangent vector to a world line. There are pros and cons to either convention. 



#11
Mar912, 07:38 AM

P: 71

Thanks, you've been a great help!




#12
Mar912, 12:26 PM

Sci Advisor
PF Gold
P: 4,860

In the coordinates you initially used to express U, the spatial velocity is just the u you started with (which you would get by executing the procedure above). The quantity gamma*u would be rather meaningless: the rate of change of distance in a given frame by a particle's proper time. This could exceed c by a large factor. 



#13
Mar912, 03:18 PM

Sci Advisor
PF Gold
P: 4,860

Then U, with norm 1, becomes: gamma * (1,u) This is not necessarily taking c=1, because the covariant metric diagonal is still (1, 1/c^2, 1/c^2,1/c^2). Contravariant diagonal obviously (1,c^2,c^2,c^2). 



#14
Mar912, 05:09 PM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,403

[QUOTE=Wox;3804757]The fourvelocity as defined for example here, is given by
[tex] U=\gamma(c,\bar{u}) [/tex] but I get [tex] U=\gamma(1,\frac{\bar{u}}{c}) I have a simpler answer for you. Some people use the first version (and your derivations of this version are correct), and some people use the second version, which is what I call the dimensionless four velocity. Both versions are OK to use, provided you tell which one you are using in advance. I think that most physicists prefer using the dimensionless version (see e.g., MTW), although I personally prefer the dimensional version. When you use the dimensionless version, the 4 velocity is equal to a unit vector in the time direction of the object's rest frame. When you use the dimensional version, the 4 velocity is c times a unit vector in the time direction of the object's rest frame. I hope this is helpful. 



#15
Mar1012, 09:52 AM

Sci Advisor
PF Gold
P: 4,860

Probably no one interested anymore, but I have clarified a few things in my own mind.
Given the desire to express a 4  tangent vector to a world line in terms of u = (dx/dt, dy/dt, dz/dt) with conventional meanings, two separate conventions affect the form it takes:  how you norm it  is your metric canonic (all +1,1) or not (you have c^2 or 1/c^2 in your metric). The signature of the metric is irrelevant for this situation. With canonic metric, you have a factor of c in your tangent vector components, so you have two natural choices: gamma * (c, u) // norm c; dimensions distance/time; from d (ct,x,y,z) / d tau gamma * (1, u/c) // norm 1; dimensionless; from d (t, x/c, y/c, z/c) / d tau With noncanonic metric, you have only one natural form, with norm 1: gamma * (1, u) // mixed dimensions, as is characteristic of noncanonic metric // from d (t,x,y,z) / d tau With noncanonic metric, a form with norm c is simply unnatural. It happens that I learned SR with noncanonic metric, 4velocity being a unit vector, and the concept of 'speed through spacetime' not remotely meaningful. It appears that almost all modern books use canonic metric. [EDIT: For emphasis, note something I derived in an earlier post: the norm of c or 1 plays no role at all in computing any observable. You could even normalize to 42 and it would make no difference. Only the direction in 4space of the tangent vector plays any role in computing observables.] 



#16
Mar1012, 05:33 PM

P: 2,043





#17
Mar1212, 10:07 AM

P: 71





#18
Mar1212, 10:12 AM

Sci Advisor
PF Gold
P: 4,860

x0*y0  x1*y1/c^2  x2*y2/c^2  x3*y3/c^2 


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