1
$\begingroup$

I'm a student working through a special relativity book. In a chapter of the book, they are deriving each of the components of a 4-velocity vector in terms of ordinary 3-velocity.

(Here $\vec{x}$ represents all the space dimensions of an event, so for example in 3 space dimensions it could actually represent $x^2+y^2+z^2$.)

What I currently understand is that:

Each component of the 4-velocity is the derivative of its respective component in a 4-dimensional time/space vector with respect to proper time, so for instance if the 4-velocity vector is $(U^0,U^1,U^2,U^3)$ then

$$U^0=\frac{dt}{d \tau}=\frac{1}{\frac{d \tau}{dt}}$$

Proper time is invariant and defined as $$\tau = \sqrt{t^2-\vec{x}^2}$$

However, after this, the book states that

$$d \tau = \sqrt{dt^2-d\vec{x}^2}$$

and hence

$$\frac{d \tau}{dt}=\frac{\sqrt{dt^2-d\vec{x}^2}}{dt}=\sqrt{1-\vec{v}^2}$$

directly follows from the definition of proper time.

I don't understand how you can take that definition of proper time and get $$d \tau = \sqrt{dt^2-d\vec{x}^2}$$ from it.

When taking the derivative of $\tau$ with respect to $t$ I don't see how you can get an expression with second order differentials inside the square root. The only way I can think of to interpret this is to say that a small change in $\tau$ corresponds to a small change in both $t$ and $\vec{x}$, but this doesn't seem remotely rigorous at all.

Can someone help explain where this comes from?

$\endgroup$
  • $\begingroup$ Is that really how the book defined proper time? $\endgroup$ – G. Smith Jul 8 at 16:46
  • 1
    $\begingroup$ "$\vec{x}$ represents all the space dimensions of an event, so for example in 3 space dimensions it could actually represent $x^2+y^2+z^2$." This isn't right. The sum of the squared components is the square of magnitude of $\vec{x}$ which is written $|\vec{x}|^2$. The symbol with the arrow on it ($\vec{x}$) represents the vector. To write it in terms of components you use a compact notation like $(x, y, z)$ or unit vectors like $x\hat{i}+y\hat{j}+z\hat{k}$. But it is important to keep in mind that the magnitude is not the vector, just the size of the vector (without direction information). $\endgroup$ – dmckee Jul 8 at 17:04
  • 1
    $\begingroup$ @G.Smith The book defined it by trying to introduce the concept of a quantity that is the same in any two reference frames along the x axis $(x,t)$ and $x',t'$ by showing that $t^2-x^2$ and $t'^2-x'^2$ were the same when substituted into the Lorentz transformation equations for one dimension, and then trying to generalise it by showing $t^2-\vec{x}^2$ was invariant under coordinate rotations. A quick internet search shows a more rigorous definition - is the book just explaining things very badly? $\endgroup$ – IntegralPrime Jul 8 at 17:13
  • 1
    $\begingroup$ One common approach, rather than starting with differentials, is to define the proper time interval $\Delta\tau$ between two events in spaceetime with finite timelike separation as $$(\Delta \tau)^2 = (c \Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2,$$ after showing that this is invariant under Lorentz (or Poincare) transformations. Then going to the differential form is obvious. The point is that proper time is not a number at each point in spacetime, as in your book's first definition. It is a number measuring the interval between timelike-separated points in spacetime. $\endgroup$ – G. Smith Jul 8 at 18:00
  • 1
    $\begingroup$ You can’t get from one to the other, which is why you can’t define $\tau$ that way and then expect to take its differential to get what everyone agrees is $d\tau$. You have to start by thinking about deltas. $\endgroup$ – G. Smith Jul 8 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.