# Is this 4-velocity tangential to the worldline

1. Mar 18, 2013

### Sunfire

Is this "4-velocity" tangential to the worldline

Hello,

In the Minkowski spacetime (x,y,ict) the spacetime velocity (the "4" velocity, which in this particular instance happens to have 2 spatial and 1 temporal component) is given by

V = γ(vx, vy,ic),

where v = (vx, vy) is the spatial velocity vector.

Will V be tangential to the worldline of the particle and why?

2. Mar 18, 2013

### Staff: Mentor

Re: Is this "4-velocity" tangential to the worldline

Yes, the 4-velocity is the tangent vector to the worldline. To see why, note that the 4-velocity is the derivative of position with respect to proper time.

3. Mar 18, 2013

### Fredrik

Staff Emeritus
That "ict" stuff is very rarely used these days. The idea behind it was that the i allows you to define the metric tensor using the formula for the Euclidean inner product, i.e. if $x=(x_1,x_2,x_3,ict)$ and $y=(y_1,y_2,y_3,ics)$, we have
$$g(x,y) =(x_1,x_2,x_3,ict)\cdot (y_1,y_2,y_3,ics) =x_1y_1+x_2y_2+x_3y_3-c^2ts.$$ But we can do without the i if we use matrix multiplication instead. You can e.g. write the above as $$g(x,y) = \begin{pmatrix}x_1 & x_2 & x_3 & ct\end{pmatrix} \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1\end{pmatrix} \begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ cs\end{pmatrix} = x_1y_1+x_2y_2+x_3y_3-c^2ts,$$ or as
$$g(x,y) = \begin{pmatrix}x_1 & x_2 & x_3 & t\end{pmatrix} \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -c^2\end{pmatrix} \begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ s\end{pmatrix} = x_1y_1+x_2y_2+x_3y_3-c^2ts.$$ Most people also prefer to use units such that c=1, and its also very common to put the time coordinate first, rather than last. People who do that number the components from 0 to 3 instead of from 1 to 4.

I'm one of those people. I would write the above as
$$g(x,y)=x^T\eta y =-x_0y_0+x_1y_1+x_2y_2+x_3y_3,$$ where
$$\eta = \begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$$ This is how I think about 4-velocity in SR: I will consider 1+1-dimensional Minkowski spacetime here for simplicity. We define the four-velocity as the normalized tangent vector to the world line. This means that in the comoving inertial coordinate system, its components are
$$\begin{pmatrix}1\\ 0\end{pmatrix}.$$ To find its components in an inertial coordinate in which the particle's velocity is v, we apply a Lorentz transformation with velocity -v:
$$\gamma\begin{pmatrix}1 & v\\ v & 1\end{pmatrix} \begin{pmatrix}1 \\ 0\end{pmatrix} =\gamma\begin{pmatrix}1\\ v\end{pmatrix}.$$

Last edited: Mar 18, 2013
4. Mar 18, 2013

### Sunfire

@PeterDonis

now I have to post something clever, else I will be gaining red points

how does one go about proving the above; the text I am reading defines proper time interval τ2=t2-(1/c2)(x2+y2+z2) and then defines the spacetime velocity V given above.

Worldline has only been mentioned but not defined yet; kind of hard to find out if V is tangential to the worldline without its definition.

The reason I am asking this is I am trying to make sense of the fact V$\bullet$ F = 0,

V is the spacetime velocity
F is the "4"- force

Last edited: Mar 18, 2013
5. Mar 18, 2013

### Sunfire

Fredrik, thank-you for your reply and for the detailed, easy-to-follow explanation. The text I am reading is an intro and a bit on the older side (Lawden), but very good price on Amazon . He uses the "ict" idea I guess for an easier introduction to SR.

Have you considered the relation V $\cdot$ F = 0

V is the spacetime velocity
F is the "4"-force given by

F = (f, i$\dot{m}$c), f is the 3-force

In 3-space, this means "the force does no work" and the 2 vectors are orthogonal; but in Minkowski spaces the geometry of the above two vectors is... well, non-intuitive. I am trying to make sense what does V $\cdot$ F = 0 mean.

6. Mar 18, 2013

### WannabeNewton

Hi Sunfire. You don't exactly need to know that the 4 - velocity is tangential to the worldline to see that. All you need is what Fredrik noted earlier, that the 4 - velocity is normalized so that $u^{a}u_{a} = -1$ (I'm using natural units). Then, $\frac{d}{d\lambda }(u^{a}u_{a}) = u_{a}\frac{\mathrm{d} u^{a}}{\mathrm{d} \lambda} +u^{a}\frac{\mathrm{d} u_{a}}{\mathrm{d} \lambda} = 2u_{a}\frac{\mathrm{d} u^{a}}{\mathrm{d} \lambda} = 0$ thus $u\cdot a= u_{a}\frac{\mathrm{d} u^{a}}{\mathrm{d} \lambda} = 0$ (this of course then implies the 4 - force is orthogonal to the 4 - velocity since the 4 - force is proportional to the 4 - acceleration).

I can show you what the tangent vector actually is and why it in fact is tangent to the wordline, using the coordinate basis viewpoint of the tangent space, but it would be a bit abstract and might use notation unfamiliar to you and I don't want to show you anything that might overly confuse you. If you are ok with it then I can try to explain.

7. Mar 18, 2013

### Staff: Mentor

Does the worldline get defined later on in your text? The simplest way to define it, at least if you want to see how 4-velocity is tangent to it, is to define it as four functions $t(\tau)$, $x(\tau)$, $y(\tau)$, $z(\tau)$, giving the coordinates of the object traveling on the worldline as a function of the object's proper time. Then the derivatives of the four functions are (I'm using units in which c = 1):

$$\frac{dt}{d\tau} = \gamma$$

$$\frac{dx}{d\tau} = \frac{dx}{dt} \frac{dt}{d\tau} = \gamma v_x$$

$$\frac{dy}{d\tau} = \frac{dy}{dt} \frac{dt}{d\tau} = \gamma v_y$$

$$\frac{dz}{d\tau} = \frac{dz}{dt} \frac{dt}{d\tau} = \gamma v_z$$

So if we consider the 4 coordinate functions as a 4-vector $x^{\mu}$, then the four derivatives give another 4-vector $u^{\mu} = d x^{\mu} / d\tau$, which is the 4-velocity. Obviously the 4-velocity must then be tangent to the worldline, since it's just the derivative of the coordinates of the worldline with respect to proper time, i.e., with respect to the parameter we're using to define the coordinate functions.

WannabeNewton's post addresses this so I won't go into it further; but what he said is consistent with what I said above. (He writes $\lambda$ where I write $\tau$, but it amounts to the same thing.)

8. Mar 18, 2013

### pervect

Staff Emeritus
ict is fairly common in introductory SR books. It will eventually get replaced if one gets to tensors, but it's perfectly adaquate unitl that point.

9. Mar 18, 2013

### nitsuj

I don't think I've ever seen math explained so clearly, feels like I can almost follow it

Very clear summary of the math ^

10. Mar 18, 2013

### Sunfire

This makes perfect sense The coordinates, being a function of τ, define the curve of the worldline, while their derivatives with respect to τ define the tangent to the curve.

11. Mar 18, 2013

### Sunfire

12. Mar 18, 2013

### Fredrik

Staff Emeritus
I suspect that WannabeNewton is referring to how tangent vectors are defined in differential geometry. That is indeed pretty abstract, and takes a lot of time to learn thoroughly. But it is what you have to learn if you want to understand these things in the context of general relativity.

However, things can be made much easier in special relativity. We don't have to define the spacetime of special relativity as a Lorentzian smooth manifold. It can also be defined as a vector space, or as an affine space. This choice is of no practical importance. I like to define it as a vector space, because this makes the mathematics easy.

If we define spacetime as the set ℝ4 with the standard vector space structure, and the bilinear form g defined by
$$g(x,y)=x^T\eta y$$ for all $x,y\in\mathbb R^4$, then the tangent vector of a curve $C:(a,b)\to\mathbb R^4$ at the point C(t) is just the derivative C'(t).

Last edited: Mar 18, 2013
13. Mar 18, 2013

### robphy

This may help:

In 3-space, that the 3-force is perpendicular to the 3-velocity also means that the magnitude of the 3-velocity (i.e. the spatial speed) is constant [assuming that the mass is constant].
In spacetime, that the 4-force is perpendicular to the 4-velocity means that magnitude of the 4-velocity is constant [assuming that the rest-mass is constant].)

14. Mar 18, 2013

### Staff: Mentor

Actually, the magnitude of the 4-velocity is constant, period; it's always 1 (or c in conventional units). The magnitude of the 4-momentum changes if the rest mass changes, since the rest mass *is* the magnitude of the 4-momentum. But that doesn't change the 4-velocity.

15. Mar 18, 2013

### robphy

Yes, I agree. The 4-velocity is normalized. So, it (edit: its magnitude) is automatically constant.

I was appealing to the typical intro-physics argument that zero-net-work means that [via the work-energy theorem] the speed is unchanged, assuming the mass is constant.
Then, use the analogue for the 4-dimensional case.

A fancier interpretation [from Tom Moore's Six Ideas] of the 3D-case is that the momentum-vector's magnitude (which has no special name) doesn't change when the force is perpendicular to the velocity. If that were more common, then I would have used your 4-momentum-interpretation for the spacetime analogue.

Last edited: Mar 19, 2013
16. Mar 19, 2013

### Sunfire

I have developed an intuition for the orthogonality between f and v in 3-space; If the projection of f along v is zero, then there is no action in the direction of v and consequently, v remains unchanged.

But this intuition no longer works for the 4-vectors F and V, because "orthogonality", a geometrical notion, makes no sense anymore, because the temporal axis takes imaginary values ict, which leads to a subtraction in the dot-product summation. One can no longer imagine F having zero projection onto V.

This is why I started this thread with lower dimensional spacetime (x1, x2, ict), because in it one can visualize F and V, and they are no longer having the familiar geometry, e.g. they do not form a 90-degree angle.

Last edited: Mar 19, 2013
17. Mar 19, 2013

### Fredrik

Staff Emeritus
The result that 4-acceleration is orthogonal to 4-velocity means only that the 0 component (i.e. the "temporal" component) of the 4-acceleration in the comoving inertial coordinate system is 0.

18. Mar 19, 2013

### robphy

(Technically, we should work in the tangent space.)

Think of a vector as the radius vector to a curve of equidistant points from the origin (the analogue of a "circle/sphere", which is a hyperbola/oid in Minkowski spacetime).
Where the tip of that radius vector meets the "circle", regard the vectors tangent to that circle at that point as orthogonal to that radius vector. (If the radial vector is timelike, then these tangent-vectors are spacelike, and specifically orthogonal to that particular radius vector.)

So, consider a future-timelike-momentum vector p [along the radial direction].
(Its tip lies on a hyperboloid [called the mass-shell of the particle].)
Additionally, consider a spacelike force-vector F orthogonal to p (i.e. F (dot) p=0 [along a tangent direction]). Since F has no radial component and F=(d/ds)p , the momentum-vector p can only change in direction (with its tip sliding along the same hyperboloid), but not change in magnitude (which would imply that tip moves to a different concentric hyperboloid).

[This elaborates on the reply of Fredrik.]

19. Mar 19, 2013

### Fredrik

Staff Emeritus
If we define the spacetime of SR as a vector space instead of as a manifold, we don't have to define a tangent space. For example, the derivative of a differentiable curve in spacetime is automatically a vector in spacetime. Now if you want to draw that vector as an arrow, you may want to draw it at the point where the derivative was computed. But we can think of this as nothing more than a choice of how to illustrate what we're doing.

I don't think I have thought of this way to find the direction of the orthogonal vector before. The image I visualize is simpler. Here's one I found using Google:

If the 4-velocity is a point in the almost vertical line in the picture, then the 4-acceleration is a point in the almost horizontal plane. In the comoving inertial coordinate system, that line is just the t axis and that plane is just the xy-plane. Every point in that line is orthogonal to every point in that plane. (Note that the two angles indicated in the picture are the same).

This is the web page where I found the image. I haven't read the text, but it has lots of nice pictures.

20. Mar 19, 2013

### robphy

Yes, for the specific case of Minkowski spacetime, that image is consistent... and possibly simpler (after you have accepted that formulation of orthogonality). (You can use the Lorentz transformations or a radar-experiment to justify that formulation.)

My method works for the Euclidean, Galilean, and Minkowskian cases.
It builds on the tangent-is-perpendicular-to-radius idea hopefully-familiar from Euclidean geometry, which beautifully generalizes to these other geometries.