My question is about 4-velocity but it is more general about my global comprehension of S.R.

In special relativity, we define the 4-velocity vector as ($\tau$ is the proper time) :

$$U=\frac{\partial x}{\partial \tau}$$

So, once I have chosen a frame $R$, it represent how the coordinates of the point change in $R$ when the proper time of the particle has changed of $d\tau$ : so the only frame dependence is on the upper part of the derivative, we always do the derivative according to $\tau$ no matter in which frame we are.

In my course, they say that this vector is "absolute" and doesn't depend on any frame.

But I have some questions about this.

In diff geometry, we define tensors as quantities that transform well, but here we have:

$$\partial_{\tau} x^{\beta}=\frac{\partial}{\partial \tau}\left(\frac{\partial x^{\beta}}{\partial y^\alpha}y^\alpha\right)=\frac{\partial^2 x^\beta}{\partial \tau \partial y^\alpha}y^\alpha+\frac{\partial x^{\beta}}{\partial y^\alpha}\frac{\partial y^\alpha}{\partial \tau}$$

So, the first term shouldn't be here to have a well defined quantity.

But here, we are focusing on inertial frame. So, they are linked with Lorentz boost that is a linear transformation. Thus the second derivative that is written above should be 0.

First question: From a math perspective, can we say that 4-velocity vector is indeed an absolute quantity because between inertial frames, the quantity transforms "well" as a tensor ?

Second question: in the course the teacher doesn't do such proof, he just says "we defined 4-velocity without referring a specific frame, thus it is a quantity independent of frames". I don't understand this, can we understand it is an absolute quantity without doing what I've written above?

Third question : In general relativity (that I just started to study), we are not focused on inertial frame only, we can do any change of coordinates. Thus is the 4-velocity still well defined ?

  • 1
    $\begingroup$ You made a mistake in the very first step: $x^\beta$ is not equal to $(\partial x^\beta / \partial y^\alpha) y^\alpha$. That's only true for linear transformations. Accordingly, your final result only makes sense for linear transformations. $\endgroup$ – knzhou Nov 29 '17 at 17:50

You seem to be interested in algebra, so I will try to give a more technical answer. I am not a matematician, so take it with a pinch of salt.

Lets say you have a scalar field $f$ defined in your spacetime. By this I mean that is a map from event in your spacetime $\mathcal{M}$ to, say, real numbers $f:\mathcal{M}\to\mathbb{R}$. This way $f$ is independent of your coordinate transformations: you can label spacetime as you wish, but it will still be the same spacetime.

Next, lets consider a world-line of your object. This can be thought of as a map from real numbers (proper time) to points on your spacetime $\bar{x}^\mu: \mathbb{R}\to\mathcal{M}$. Note that I make no assumption that $\bar{x}^\mu$ is a vector, instead it is simply a collection of functions that map proper time into particular coordinates (that you chose to address your spacetime).

We can now define $f\circ\bar{x}: \mathbb{R}\to\mathbb{R}$ (i.e. $\mathbb{R}\to\mathcal{M}\to\mathbb{R}$). Lets take a derivative of this function, and try to apply Leibniz rule.

$\frac{d}{d\tau}\left(f\circ\bar{x}(\tau)\right)=\frac{d}{d\tau}\left(f\left(\bar{x}^0(\tau),\bar{x}^1(\tau), \dots\right)\right)=\frac{d\bar{x}^\mu}{d\tau}\left(\partial_\mu f\right)\rvert_{@\bar{x}(\tau)}$

Note that there is still no assumption of $d\bar{x}^\mu/d\tau$ being a vector.

But now we consider this from point of view of differential geometry: $\frac{d}{d\tau}\left(f\circ\bar{x}(\tau)\right)$ is a scalar function - it cannot change due to change in coordinates. Also, you know how $\partial_\mu f$ transforms. It follows that $d\bar{x}^\mu/d\tau$ (note the full derivatives!) must transform as a vector. This works even if spacetime is not flat (in which case $\bar{x}^\mu$ is not a vector, but $d\bar{x}^\mu/d\tau$ is).

Does it answer your question?


We have to show that $${\eta}^{\mu}=\frac{{dx}^{\mu}}{d{\tau}}=\gamma{v}$$

is an invariant, i.e. it is the same in every inertial frame.


$${\eta}^0={\frac{dx^0}{dt}}=\frac{d(ct)}{(\frac{1}{\gamma})}dt=\gamma c,$$



from which it follows that


The last expression can be written as

$${\gamma}^2 c^2(1-{\frac{v^2}{c^2}})=\frac{{\gamma}^2c^2}{{\gamma}^2}=c^2,$$

so in every inertial frame


Whether the particle stands still (though this may sound strange) or goes at the speed of light, it's velocity as defined above is always c.

  • $\begingroup$ Wait what? Did you take a square root? $\endgroup$ – John Donne Nov 29 '17 at 20:27

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