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Four-vector problem

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Wox
#1
Mar8-12, 10:24 AM
P: 71
The four-velocity 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 arc-length (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 arc-length (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 four-velocity but in fact [itex]\bar{v}(\tau)=\frac{U}{c}[/itex] where U the four-velocity as defined in textbooks. What am I missing?
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Matterwave
#2
Mar8-12, 11:26 AM
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If you are using units where c is not 1, then certainly you want the 4-velocity 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).
PAllen
#3
Mar8-12, 11:38 AM
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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 4-position 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.

PAllen
#4
Mar8-12, 05:25 PM
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Four-vector problem

Quote Quote by Matterwave View Post
If you are using units where c is not 1, then certainly you want the 4-velocity 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).
I'll add one more thing. The idea of norming U to c has led to Brian Greene's "speed through space-time equals c", which has led to numerous confusions and debates on these forums, as well as facilitating cranks. The convention of Einstein and Bergmann that U has norm 1, irrespective of the value of c sidesteps all of this lunacy.
Matterwave
#5
Mar8-12, 06:09 PM
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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?
PAllen
#6
Mar8-12, 06:18 PM
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Quote Quote by Matterwave View Post
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?
No. The conventions:

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.
Matterwave
#7
Mar8-12, 07:25 PM
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Are you talking about adding a factor of c into the metric? o.o
PAllen
#8
Mar8-12, 07:51 PM
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Quote Quote by Matterwave View Post
Are you talking about adding a factor of c into the metric? o.o
Don't know what you are asking. The two common conventions for the metric are:

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.
Wox
#9
Mar9-12, 07:11 AM
P: 71
Quote Quote by PAllen View Post
Given your units for v, the integral is c * tau, not tau.
Ok, so because [itex]\bar{w}(t)=(ct,\bar{x}(t))[/itex] is in space units, [itex]\left\| \bar{v}(t)\right\|[/itex] has units space/time and the integral is in space units. Then the arc-length (proper time) parameterization in time units is given by
[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 four-velocity in space/time units
[tex]\bar{v}(\tau)=\gamma(c,\bar{u}(t))[/tex]
Is this the correct explanation? As I understand, the four-velocity in the other convention (norm=1) is unitless, isn't it? So how is it used then?
PAllen
#10
Mar9-12, 07:27 AM
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Quote Quote by Wox View Post
Ok, so because [itex]\bar{w}(t)=(ct,\bar{x}(t))[/itex] is in space units, [itex]\left\| \bar{v}(t)\right\|[/itex] has units space/time and the integral is in space units. Then the arc-length (proper time) parameterization in time units is given by
[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 four-velocity in space/time units
[tex]\bar{v}(\tau)=\gamma(c,\bar{u}(t))[/tex]
Is this the correct explanation? As I understand, the four-velocity in the other convention (norm=1) is unitless, isn't it? So how is it used then?
Yes, this is correct.

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.
Wox
#11
Mar9-12, 07:38 AM
P: 71
Thanks, you've been a great help!

Quote Quote by PAllen View Post
The magnitude of U really has no real meaning in either convention.
And what about the magnitude of space-component [itex]\gamma\bar{u}(t)[/itex] of the four-velocity? I'm not quite sure how all this relates to some physical reality. Can I interpret the space-component of the four-velocity as the classical velocity (at least when not choosing the norm=1 convention)?
PAllen
#12
Mar9-12, 12:26 PM
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Quote Quote by Wox View Post
Thanks, you've been a great help!



And what about the magnitude of space-component [itex]\gamma\bar{u}(t)[/itex] of the four-velocity? I'm not quite sure how all this relates to some physical reality. Can I interpret the space-component of the four-velocity as the classical velocity (at least when not choosing the norm=1 convention)?
Let's say you have U as a tangent vector, considered a coordinate independent quantity. You want to know the spatial velocity measured in some frame defined by a 4 orthonormal unit vectors, one timelike the others spacelike. You take U dot <x unit vector>/ U dot <t unit vector>, same for y and z. Clearly, the norm of U drops out.

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.
PAllen
#13
Mar9-12, 03:18 PM
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Quote Quote by PAllen View Post
No. The conventions:

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.
Error here. In this convention you simply have position: (t,x,y,z)

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).
Chestermiller
#14
Mar9-12, 05:09 PM
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[QUOTE=Wox;3804757]The four-velocity 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.
PAllen
#15
Mar10-12, 09:52 AM
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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 non-canonic metric, you have only one natural form, with norm 1:

gamma * (1, u) // mixed dimensions, as is characteristic of non-canonic metric
// from d (t,x,y,z) / d tau

With non-canonic metric, a form with norm c is simply unnatural.

It happens that I learned SR with non-canonic metric, 4-velocity 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 4-space of the tangent vector plays any role in computing observables.]
Matterwave
#16
Mar10-12, 05:33 PM
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Quote Quote by PAllen View Post
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 non-canonic metric, you have only one natural form, with norm 1:

gamma * (1, u) // mixed dimensions, as is characteristic of non-canonic metric
// from d (t,x,y,z) / d tau

With non-canonic metric, a form with norm c is simply unnatural.

It happens that I learned SR with non-canonic metric, 4-velocity 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 4-space of the tangent vector plays any role in computing observables.]
Yes, this what I meant when I said "are you talking about adding a factor of c to the metric?" I always learned to us diag(-1,1,1,1) for my metric so any factors of c's I need are in the 4-vectors themselves.
Wox
#17
Mar12-12, 10:07 AM
P: 71
Quote Quote by PAllen View Post
You want to know the spatial velocity measured in some frame defined by a 4 orthonormal unit vectors, one timelike the others spacelike. You take U dot <x unit vector>/ U dot <t unit vector>, same for y and z. Clearly, the norm of U drops out.
Not sure what you mean: there is no dot product in Minkowskian space-time... Do you mean that [Ux/Ut,Uy/Ut,Uz/Ut] is the spatial 3-velocity? Why?
PAllen
#18
Mar12-12, 10:12 AM
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Quote Quote by Wox View Post
Not sure what you mean: there is no dot product in Minkowskian space-time... Do you mean that [Ux/Ut,Uy/Ut,Uz/Ut] is the spatial 3-velocity? Why?
Sure there is a dot product. It is defined by the metric. For example, if the metric is diag(1,-1/c^2, -1/c^2,-1/c^2), then the dot product of X and Y is:

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


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