# Magnitude of a 4-vector

1. Mar 27, 2004

### Severian596

My direct question is at the bottom of this post, but I thought I would set the scene. Skip the first part if you wish.

The magnitude of any 4-vector is scalar and therefore the same in all frames. For the velocity vector
$$u = (\gamma c, \gamma v)$$
where v is a 3-vector velocity term, we can therefore switch frames to a frame where the object in question is at rest. In this frame
$$u = (c, 0, 0, 0)$$
because $$v = (0, 0, 0)$$ and $$\gamma = 1$$.

So the magnitude of u is c.

My text justifies this strange derivation by saying that because magnitudes are scalar we need 4-vector magnitudes to be c, because c does not change with our frame of reference. Furthermore he says that "speeds (magnitudes of 3-vectors) are relative but the magnitude of a 4-vector must be invariant."

So my question: what is the meaning of a 4-vector's magnitude? It's always c, but what is 'it'?
(I believe this is related to a recent thread where an author claimed "all things are moving around at the speed of light...")

2. Mar 27, 2004

### pmb_phy

To me it means nothing more than what you said. It's a quanity which remains unchanged by a change of coordinates

The magnitude of a 4-vector defined through the scalar product of the 4-vector. I.e. Let A represent the magnitude of the 4-vector A. Then

$$A^2 = \mathbf{A}\bullet\mathbf{A} = g_{\alpha\beta}A^{\alpha}A^{\beta} = A_{\alpha}A^{\alpha}$$

where $$g_{\alpha\beta}$$ is the metric tensor. For a definition of this tensor see
http://www.geocities.com/physics_world/ma/intro_tensor.htm

In Lorentz coordinates (ct, x, y, z) this becomes

$$A^2 = (A^0)^2 - (A^1)^2 - (A^2)^2 - (A^3)^2$$

If this is 4-velocity then A = c. To me it has no real physical signifigance beyond the fact that it's the speed of light. I see some people say that this is the rate at which a particle moves through spacetime - But you won't catch me explaining it like this.

Pete

Last edited: Mar 27, 2004
3. Mar 27, 2004

### HallsofIvy

Staff Emeritus
First, there is no frame in which "all things are moving about at the speed of light"- not if you are using "moving about" in the standard sense.

The reason why the magnitude of the 4-vector velocity is the speed of light is because that is the factor relating the fourth (time) dimension with the other three. Given any object "moving about" with any given 3-vector velocity, there exist a frame in which IT is stationary- its 3-vector velocity is 0. That means the magnitude of the 4-vector velocity is entirely in the fourth (time) dimension- and everything "moves through" time at the same rate.

4. Mar 28, 2004

### turin

If I understand your question correctly, then you might want to read the (first?) chapter in Misner, Thorne, and Wheeler's "Gravitation." They describe it sort of as the number of surfaces pierced by an arrow, where the arrow is the vector, and the collection of surfaces is the contraction of the metric with the vector.

Though, the magnitude of an arbitrary 4-vector is not c. The magnitude of the 4-velocity is c.

5. Mar 28, 2004

### pmb_phy

This can be made more precise by using 4-vectors and including the observer. Let the observers 4-velocity be Uobs and let the 4-velocity of the particle by U. Define the 4-vector Vobs as

$$\mathbf{V}_{obs} = \frac { c^2\mathbf{U} - (\mathbf{U}\bullet\mathbf{U}_{obs})\mathbf{U} } { (\mathbf{U}\bullet\mathbf{U}_{obs})}$$

Then

$$|\mathbf{V}_{obs}|^2 = -v^2$$

where v is the speed of the particle as measured by the observer. So it is Vobs, and not U which you want to think of when you think of the speed of a particle in the normal sense of the word. Notice that Vobs is a 4-vector and defined for the observer whose 4-velocity is Uobs.

Last edited: Mar 28, 2004
6. Mar 28, 2004

### lethe

forgetting special relativity for a second, you can parametrize any curve with its arc length s. with this choice of parametrization, the magnitude of the tangent vector to the curve is always 1

let's see how that happens:

start with any parametrization you like for your curve $\alpha(t)[/tex]. the the arc length of the curve is [itex]s=\int|\dot{\alpha}(t)|dt$

then you reparametrize your curve by inverting to find t in terms of s, and you find that the velocity vector is $\dot{\alpha}/|\dot{\alpha}|$. the magnitude of this vector is always 1.

in special relativity, this is exactly what happens. you parametrize the worldline by proper time, and the arc length is given by the proper time times c. you could choose to parametrize your worldline by another parameter, but then your tangent vector would not transform like a vector under Lorentz rotations.

7. Mar 29, 2004

### Severian596

Very nice. Thank you all for the excellent explanations.

And thank you turin for reminding me that I mispoke and referred to the magnitude of 4-vectors when I should've specified velocity 4-vectors. That's very important and I assume the magnitude of 4-vectors is not always c.

For example what if we refer to the vector (u) that connects two events. In this case the magnitude of u is defined as

$$(|u|){^2} = (c \Delta t)^{2} - (\Delta x)^{2} - (\Delta y)^{2} - (\Delta z)^{2}$$

and this particular vector is referred to as the interval, right? Furthermore we know that the interval is invarient for any reference frame, and that this is appropriate for a magnitude because magnitudes are scalar, correct?

Last edited: Mar 29, 2004
8. Mar 29, 2004

### turin

I don't think this is a vector, though, strictly speaking. Lethe will no doubt revise/correct/improve my reply. This can only be considered as related to or described by a vector, and then only if you assume that the space is flat. I think the more appropriate/general/correct way to do it is to consider some parametrization of some curve, like Lethe did in his previous post. Then, instead of talking about things like Δx, you should really be talking about dx, or more correct still, dx/dλ, where λ is the parameter that you chose.

9. Mar 29, 2004

### Severian596

Okay thanks again. I'm only beginning my study of SR so I'm only accustomed to working with flat space, and that explains my ignorance with some of the more in depth definitions. And I admit I understood very litte of what lethe posted!

I understand that an arc's length is the sum of all differential lengths with respect to t $\dot{\alpha}(t)|dt$, but I don't remember parametrizing any curves or what that means, much less parametrizing it and then reparametrizing any way I like!

10. Mar 29, 2004

### turin

Take, as an example, some arbitrary curve in 3-space. You can parametrize it by some arbitrary parameter, λ, so that you can describe the curve in Cartesian coordinates as the triplet:

curve = x(λ) = (x(λ),y(λ),z(λ))

The tangent vector to this curve is:

(dx/dλ) / |dx/dλ| = (dx/λ,dy/dλ,dz/dλ) / √{(dx/λ)2 + (dy/λ)2 + (dz/λ)2}

To find the arclength for some Δλ:

Δs = integral of |dx/dλ|dλ

From this, you can infer that:

ds = |dx/dλ|dλ

and then you can use s as the parameter.

11. Mar 29, 2004

### pmb_phy

Severian596 - That is not a 4-vector. That is the spacetime interval. It represents the magnitude of a 4-vector call the spacetime displacement 4-vector. It's a qualified tensor. In this case it's called a Lorentz 4-vector. I.e.

$$\Delta \mathbf{X} = (c \Delta t, \Delta x, \Delta y, \Delta z)$$

In fact this is the prototype of a Lorentz 4-vector. It represents a spacetime displacement from between two events. One event can be a reference point for all other events. That event can be the origin of a Lorentz coordinate system. The tail of the vector is then at the origin. All other events are then represented by the tip of the vector. Then the vector can be expressed as

$$\mathbf{X} = (ct, x, y, z)$$

This is called the coordinate 4-vector. It is the prototype of the Lorentz 4-vector. i.e. any set of quantities Xu[/sup] which transforms in the same way as the xu is called a Lorentz 4-vector.

This applies to all homogeneous Lorentz transformations between Lorentz coordinate systems. A Lorentz coordinate system is an inertial coordinate system for which the spatial axes are Cartesian.

Regarding the magnitu