Four-Vector Problem: What Am I Missing?

  • Thread starter Wox
  • Start date
In summary: Norm 1 makes it easy to work with spatial dimensions, for example, but norm 0 can be more general. Norm 1 also facilitates cranks by making it seem like velocity is constant.
  • #1
Wox
70
0
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,\frac{\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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  • #2
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).
 
  • #3
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.
 
Last edited:
  • #4
Matterwave said:
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.
 
  • #5
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
Matterwave said:
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.
 
  • #7
Are you talking about adding a factor of c into the metric? o.o
 
  • #8
Matterwave said:
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.
 
  • #9
PAllen said:
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?
 
  • #10
Wox said:
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.
 
  • #11
Thanks, you've been a great help!

PAllen said:
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)?
 
  • #12
Wox said:
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.
 
  • #13
PAllen said:
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).
 
Last edited:
  • #14
Wox said:
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.
 
  • #15
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.]
 
Last edited:
  • #16
PAllen said:
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.
 
  • #17
PAllen said:
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?
 
  • #18
Wox said:
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
 
  • #19
Depends on your definition of the dot product, but I see what you mean. But I don't see why [Ux/Ut,Uy/Ut,Uz/Ut] would correspond to a spatial velocity.
 
  • #20
Wox said:
Depends on your definition of the dot product, but I see what you mean. But I don't see why [Ux/Ut,Uy/Ut,Uz/Ut] would correspond to a spatial velocity.

In a metric space there is one definition of the dot product. The Euclidean one looks the way it does solely because the Euclidean metric is diag(1,1,1).

If U is some tangent vector, and x,y,z,t are unit vectors for some frame, then the dot product of U with such unit vectors expresses U in that frame basis. Then Ux/Ut gives the x speed (well, actually, x-speed/c , but that is just as good). Look at the tangent vector itself expressed in your starting coordinates (c-normed, canonic metric; works the same in any other convention):

U = gamma(c,u)

Ux = gamma * ux is not the x speed; but note Ux/Ut = ux/c. This feature will be true in any other basis. In particular, in an orthonormal basis with U itself taken as the time unit vector, you get spatial speed of zero - the particle has no spatial speed in its own basis.
 
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1. What is a Four-Vector problem?

A Four-Vector problem is a mathematical problem that involves the use of four-dimensional vectors, which are mathematical quantities that have both magnitude and direction in four-dimensional space. These problems are commonly encountered in the fields of physics and engineering.

2. What is the purpose of using Four-Vectors?

Four-Vectors are used to describe physical quantities in four-dimensional space, such as time and space coordinates, energy, and momentum. They allow for a more accurate and comprehensive understanding of physical phenomena, particularly in the realm of relativity and quantum mechanics.

3. How are Four-Vectors represented mathematically?

Four-Vectors are typically represented as four-component column or row matrices, with each component representing a different physical quantity. The components are usually labeled as (t, x, y, z) or (ct, x, y, z), where c is the speed of light and t represents time.

4. What are some common applications of Four-Vectors?

Four-Vectors have various applications in physics and engineering, including special and general relativity, quantum mechanics, electromagnetism, and particle physics. They are also used in fields such as fluid dynamics, astrophysics, and signal processing.

5. How can I solve a Four-Vector problem?

To solve a Four-Vector problem, you will need to use mathematical operations such as addition, subtraction, multiplication, and division on the four components of the vector. It is important to pay attention to the units of the components and use proper notation to ensure accurate calculations. Additionally, understanding the physical principles behind the problem can help in finding the correct approach to solving it.

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