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 ?

  • 2
    $\begingroup$ 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
  • $\begingroup$ Let $v$ be $4$-vector. $$v=v^{i}\frac{\partial}{\partial u_{i}}$$ Linear coordinate transformation of basis vector: $$\frac{\partial}{\partial u_{i}}=\frac{\partial w_{k}}{\partial u_{i}} \frac{\partial}{\partial w_{k}}$$ $$v=v^{i}\frac{\partial w_{k}}{\partial u_{i}} \frac{\partial}{\partial w_{k}}$$ The vector basis is different, the vector component is different, but $v$ hasn't changed: $$v=v^{k} \frac{\partial}{\partial w_{k}}$$ $\endgroup$ – Cinaed Simson Feb 8 '20 at 8:41

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?


First of all, as knzhou pointed out in the comments, you made a mistake in the transformation law for the coordinates. You should have defined $x^{\mu}$ as a function of $y^{\nu}$, $x=x(y)$, and then have taken the derivative through chain differentiation:

$$ u^{\mu}_{(x)}(\tau)=\frac{dx^{\mu}}{d\tau}(\tau)=\frac{d}{d\tau}\ x^{\mu}(y(\tau))=\frac{\partial x^{\mu}}{\partial y^{\nu}}(y(\tau))\,\frac{dy^{\nu}}{d\tau}(\tau)=\frac{\partial x^{\mu}}{\partial y^{\nu}}(y(\tau))\,u^{\nu}_{(y)}(\tau) $$

It follows that the components of the velocity vector transform according to

$$ u^{\mu}_{(x)}(\tau)=\frac{\partial x^{\mu}}{\partial y^{\nu}}(y(\tau))\,u^{\nu}_{(y)}(\tau) $$

and thus they are not frame- (or coordinate-) independent. Indeed, their transformation law is the appropriate one for a vector and makes them covariant quantities, as opposed to invariant quantities.

However, if you define the velocity vector (i.e. not its components, but the actual vector) for any set of coordinates/for any frame as

$$ u|_{\tau}=u^{\mu}_{(x)}(\tau)\,\frac{\partial}{\partial x^{\mu}}\bigg|_{x(\tau)} $$

as you would do in differential geometry, then from the transformation law for the components it follows that

$$ u|_{\tau}=u^{\mu}_{(x)}(\tau)\,\frac{\partial}{\partial x^{\mu}}\bigg|_{x(\tau)}=u^{\nu}_{(y)}(\tau)\,\frac{\partial x^{\mu}}{\partial y^{\nu}}(y(\tau))\,\frac{\partial}{\partial x^{\mu}}\bigg|_{x(\tau)}=u^{\nu}_{(y)}(\tau)\,\frac{\partial}{\partial y^{\nu}}\bigg|_{y(\tau)}=u|_{\tau} $$

where I have used

$$ \frac{\partial x^{\mu}}{\partial y^{\nu}}(y)\,\frac{\partial}{\partial x^{\mu}}\bigg|_{x(y)}=\frac{\partial}{\partial y^{\nu}}\bigg|_{y} $$

i.e. the velocity vector is a well-defined geometrical quantity, independent of coordinates.

So the answer to your first question is: yes, but this actually holds for any frame, not limited to the inertial frames (observe that nowhere in the derivation given above I have assumed that $x$ and $y$ are coordinates associated to an inertial frame: $x=x(y)$ is a general coordinate transformation). Since the coordinates are arbitrary, the transformation law most assuredly also holds in general relativity, hence the answer to your third question is also: yes.

As for your second question, I believe that your teacher's statement should be modified to "we defined the 4-velocity by an expression which is the same in all frames, thus it is a quantity independent of frames". And of course, as you guessed, you must prove that said expression is the same in all frames, as we did above.


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

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


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



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

$${\eta}^{\mu} {\eta}_{\mu}=c^2$$

Whether the particle stands still (though this may sound strange) or goes at the speed of light (and any velocity in between these two extremes), it's velocity as defined above is always c. And thus absolute.

  • $\begingroup$ Wait what? Did you take a square root? $\endgroup$ – John Donne Nov 29 '17 at 20:27
  • $\begingroup$ Your last expression says that every component of the four-velocity is constant and equal to the speed of light. This is plain wrong. This answer merely shows (in a very confusing way) that the norm of the four-velocity is $c$, which doesn't answer the question. $\endgroup$ – Milan Feb 7 '20 at 23:33
  • $\begingroup$ Huh? Does the last expression say that every component is equal to c? It says the whole four-velocity is equal to c. What's so confusing? Don't you understand the answer? It's plain and simple as can be! $\endgroup$ – Deschele Schilder Feb 8 '20 at 7:04
  • $\begingroup$ @descheleschilder In physics, there is a difference between a (four-)vector and a scalar. They are two very different objects that can't just be compared by putting an equals to sign between them. Start here: en.wikipedia.org/wiki/Scalar_(physics) $\endgroup$ – Milan Feb 8 '20 at 19:46
  • $\begingroup$ @JohnDonne Where did I take a square root? $\endgroup$ – Deschele Schilder Feb 8 '20 at 23:03

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