Why do we differentiate a 4 vector with respect to proper time to obtain 4-velocity?

Question

The coordinates of an event in spacetime are given by the 4-vector $(ct, \mathbf{r})$, where $\mathbf{r}$ is the spacial coordinates of the event. This 4-vector can be seen as 4-displacement of a worldline from the defined origin of the reference frame we're in at time $t$.

It seems sensible that $\frac{d}{dt}(ct,\mathbf{r})$ should give 4-velocity of the worldline, but instead everything I've read has stated that we differentiate with respect to the worldline's proper time $\tau$ instead, and yet I so far haven't seen any explanation as to why. This answer here on the Stack Exchange simply says we do it because it maintains the Lorentz invariant. However, why would proper time be invariant under the Lorentz transformation and other times wouldn't?

Consider $\mathbf{x^\mu}=(ct,x,y,z)^T$, which I differentiate with respect to time $t$ to get $\mathbf{v}=(c,v_x,v_y,v_z).$ Let's check if this is Lorentz invariant:

$$ \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} c\\ u_x\\ u_y\\ u_z \end{bmatrix} =\begin{bmatrix} c\gamma-\beta\gamma u_x\\ -\beta\gamma c+\gamma u_x\\ u_y\\ u_z \end{bmatrix} =\mathbf{v'} $$

$$ (c\gamma-\beta\gamma u_x)^2-(-\beta\gamma c+\gamma u_x)^2=c^2\gamma^2+\beta^2\gamma^2u_x^2+\beta^2\gamma^2c^2+\gamma^2u_x^2 =c^2\gamma^2(1+\beta^2)-u_x^2\gamma^2(1+\beta^2)=c^2-u_x^2 \\ \therefore \mathbf{v'}\cdot\mathbf{v'}=c^2-u_x^2-u_y^2-u_z^2=\mathbf{v}\cdot\mathbf{v} $$

Therefore, $\mathbf{v}$ is Lorentz invariant. Why then, do we reject it as the velocity 4-vector?

Answer 1

why would proper time be invariant under the Lorentz transformation and other times wouldn't?

A worldline is specified by giving the four coordinates $t,x,y,z$ as functions of some other parameter $\lambda$. Each value of $\lambda$ specifies one point on the worldline, and the functions $t(\lambda),x(\lambda),y(\lambda),z(\lambda)$ are the coordinates of that point. The proper time $\tau(\lambda)$ at any point along the worldline is given by solving $$ \dot\tau^2=c^2\dot t^2-(\dot x^2+\dot y^2+\dot z^2), \tag{1} $$ where an overhead dot denotes a derivative with respect to $\lambda$. By definition, a Lorentz transformation is a transformation of $(t,x,y,z)$ that leaves the right-hand side of (1) invariant, so the proper time $\tau$ is invariant under Lorentz transformations by construction. The coordinates are not.

The calculation used in the OP to check whether $\mathbf{v}=d/dt\,(ct,x,y,z)$ is a 4-vector is not a valid check, because it assumes that $\mathbf{v}$ is a 4-vector. To determine whether or not $\mathbf{v}$ is a 4-vector, we can express $\mathbf{v}$ in terms of coordinates, apply a Lorentz transformation to the coordinates, and then see what happens to $\mathbf{v}$. When we do that, the problem becomes apparent: the first component of $\mathbf{v}$ is $$ \frac{d}{dt} ct = c, \tag{2} $$ which is independent of coordinates. Therefore, a coordinate transformation (specifically a Lorentz transformation) can't change the first component of $\mathbf{v}$ at all, so $\mathbf{v}$ cannot be a 4-vector.

On the other hand, the proper time $\tau$ is invariant under coordinate transformations (by construction), including Lorentz transformations, so the quantity $$ \frac{d}{d\tau}(ct,x,y,z) \tag{3} $$ transforms just like the quantity $(ct,x,y,z)$. This is why (3) is a 4-vector.

Answer 2

The proper time is defined using the invariant interval: $$ ds^2= (c dt)^2-(dx^2+dy^2+dz^2) $$ The proper time is the time indicated by a clock at rest in the frame of the observer, so $(dx^2+dy^2+dz^2)=0$, which makes $ds=d\tau$ in the frame where the clock is at an obvious invariant. Dividing a $4$-vector by an invariant scalar produces another $4$-vector, thus guaranteeing that the $4$-velocity defined as $$ u^{\mu}=\frac{d}{d\tau}x^{\mu} $$ has the right transformation properties under Lorentz.

Answer 3

When you do this:

Consider $\mathbf{x^\mu}=(ct,x,y,z)^T$, which I differentiate with respect to time $t$ to get $\mathbf{v}=(c,v_x,v_y,v_z).$ Let's check if this is Lorentz invariant:

$$ \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} c\\ u_x\\ u_y\\ u_z \end{bmatrix} =\begin{bmatrix} c\gamma-\beta\gamma u_x\\ -\beta\gamma c+\gamma u_x\\ u_y\\ u_z \end{bmatrix} =\mathbf{v'} $$

$$ (c\gamma-\beta\gamma u_x)^2-(-\beta\gamma c+\gamma u_x)^2=c^2\gamma^2+\beta^2\gamma^2u_x^2+\beta^2\gamma^2c^2+\gamma^2u_x^2 =c^2\gamma^2(1+\beta^2)-u_x^2\gamma^2(1+\beta^2)=c^2-u_x^2 \\ \therefore \mathbf{v'}\cdot\mathbf{v'}=c^2-u_x^2-u_y^2-u_z^2=\mathbf{v}\cdot\mathbf{v} $$

Therefore, $\mathbf{v}$ is Lorentz invariant. Why then, do we reject it as the velocity 4-vector?

you actually only show that the Lorentz transformation preserves the pseudoscalar product. You do not show that $\mathbf{v}$ is Lorentz invariant.

Suppose that a particle follows $\mathbf{x} = (ct, 0, 0, 0).$ Then in this frame you get $\frac{d}{dt}\mathbf{x} = (c, 0, 0, 0).$ In the primed frame you have $\mathbf{x}' = (ct', -vt', 0, 0)$ so $\frac{d}{dt'}\mathbf{x}' = (c, -v, 0, 0).$ But this is not the Lorentz transformation of $\mathbf{x}$ which is $(\gamma c, -\gamma v, 0, 0):$ $$ \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} c\\ 0\\ 0\\ 0 \end{bmatrix} =\begin{bmatrix} \gamma c \\ -\gamma\beta c \\ 0\\ 0 \end{bmatrix} =\begin{bmatrix} \gamma c \\ -\gamma v \\ 0\\ 0 \end{bmatrix} $$

Answer 4

You are forgetting that you have obtained this four dimensional velocity vector as a derivative of the worldline with respect to the coordinate $t$. So when you are changing the ccorfinate system, you are forced to change also the parameter $t$ you are differentiating with respect to. That is going to cause the problems that interfere with Lorentz invariance.