# Why is velocity defined as 4-vector in relativity?

In classical field theory class my professor first defined velocity as $$v^{\mu} = \frac{dx^{\mu}}{dt}$$ where, $\frac{dx^{0}}{dt} = 1$

Then he said it doesn't transform like $${v^{'}}^{\mu} = L^{\mu}_{\nu} v^{\nu}$$ So this is not a proper definition of velocity.

What I don't understand is, why should it transform like a 4-vector? What's wrong with ${v^{'}}_0^{ \mu} = L^{\mu}_{\nu} L_0^{0}v_0^{\nu}$ i.e it is a part of tensor $$v_{\nu}^{\mu} = \frac{d x^{\mu}}{d x^{\nu}}$$ Zeroth column is the velocity.

Does this break any physical laws trivially? Is there some axiom or postulate which bars this?

Or is it just that once we start doing physics with this definition, we get wrong results?

P.S - We do this for an electric field vector where it is zeroth column of the electromagnetic tensor. So what stops us from defining velocity in the conventional way (by that I mean w.r.t time of moving frame as seen by an observer, i.e., $x^{0} = t$. The velocity is defined from position-time assigned to the frame in its position 4-vector) instead introducing proper time and all that.

• A crude measure of the utility of a concept in relativity is whether it can be talked about in a coordinate independent manner or not--that is to say as a geometric object existing on its own in the 4-dimensional spacetime regardless of the coordinate system we, later on, choose to describe it. And the mathematical way to ensure that the object we are talking about is such an independent geometric object is to ensure that it is a tensorial object. That is the reason we tend to insist on obtaining tensorial definitions of quantities we value. – Dvij Mankad Aug 18 '18 at 13:08

"It should transform like a four-vector under a Lorentz transformation" is a generalisation of several intuitions you typically have regarding how natural objects/tensors should behave in special relativity -- an obvious one is "no special status to any individual dimension, since space and time are inherently symmetric. That $dx^\mu/dx^0$ doesn't transform like a four-vector is obvious from the fact that it gives special preference to time.

The conventional way to define four-velocity in relativity is as $dx^\mu/ds$. Your 2-tensor idea is cute -- it is similar to the angle tensor generalised to four-dimensions -- but it doesn't satisfy the uses we have of the standard four-velocity (e.g. how would the four-momentum be defined? $m\,dx^\mu/dx^\nu$? That wouldn't be conserved.)

• I don't know enough of Special Relativity to prove/disprove your Conservation argument, at-least for now. Hopefully I should be able to understand it within a few weeks. Thinking about it, I think defining Kineic energy as a scalar might become quite difficult/impossible with my tensor. However I am not yet skilled enough to actually try it.But I got the first argument. – Indigo1729 Aug 18 '18 at 6:44
• @GauthamAP The conservation argument is easy to see -- for conservation of a tensor to hold, each component must be conserved, including, e.g. $m\,dx^0/dx^0=m$. But mass isn't conserved in relativity. Neither is, for example $m\,dx^1/dx^0=mv$ (without the Lorentz factor). – Abhimanyu Pallavi Sudhir Aug 18 '18 at 6:50
• @AbhimanyuPallaviSudhir Note that $m\ dx^\mu/dx^\nu$ is conserved; it is the trivial tensor $m\delta^\mu_\nu$, which is constant. – AccidentalFourierTransform Aug 18 '18 at 13:36
• @AccidentalFourierTransform No, it is not -- like I said, you can take simple examples, like $\mu,\nu=0$, or $\mu=i,\,\nu=0$, or $\mu=0,\,\nu=i$ -- these give $m$, $mv$ and $m/v$ respectively, none of which are conserved. – Abhimanyu Pallavi Sudhir Aug 18 '18 at 18:02
• Perhaps you're confusing conservation and invariance. But it's not invariant either, since $dx^\mu/dx^\nu$ clearly isn't the identity tensor (you seem to be taking some sort of partial derivative independent of the object's motion). – Abhimanyu Pallavi Sudhir Jan 6 '19 at 20:58

Unfortunately, $\frac{\partial x^\mu}{\partial x^\nu}=\delta^\mu_\nu$ contains no information about how a body is moving. The total-derivative definition $\frac{dx^\mu}{dx^\nu}$ doesn't really transform in a useful way, because it's a ratio of two components of the $4$-vector I motivate in the paragraph below.

However, the proper time $\tau$ lets us define a $4$-vector viz. $v^\mu=\frac{dx^\mu}{d\tau}$. The difference is that $\frac{d}{d\tau}$ commutes with Lorentz transformations but $\frac{d}{dt}$ doesn't, because the latter uses a coordinate specific to one coordinate system, whereas proper time is an invariant. Nor is there any real reason to shy away from this: by the chain rule, all we've done is rescale the "obvious" non-tensorial definition, so $v^i/v^0$ is still a $3$-velocity component (give or take a power of $c$, depending on your definitions).

• Why is $v_{\nu}^{\mu} = \delta_{\nu}^{\mu}$ ? Derivatives of spatial part w.r.t $x^{0}$ is non-zero for moving frames/particles. – Indigo1729 Aug 18 '18 at 6:42
• @GauthamAP I was thinking in terms of partial derivatives, so I've amended the answer. – J.G. Aug 18 '18 at 6:46

# Write it out, find your confusion

So I think you are fundamentally confused at what you're trying to do. Just as an exercise, I would recommend that you try to actually write out this matrix, because I think you only have some idea of what you want for the first column of it, but I am not sure you can fill out the rest of it. More specifically, I think you are confusing some sort of dependent and independent variables. Ignore $y,z$, just focus on $w=ct$, $x$. What are you expecting? $$\begin{bmatrix}1&?\\\beta&1?\end{bmatrix}$$ Because before relativity you had a path $\vec r(t)$ but in forming this derivative matrix you appear to be generalizing to some sort of field $\vec r(t, \vec r'(t))$ and if you're going to do that you will need partial derivatives and you may find that in expecting $\vec r=\vec r'$ this matrix suddenly just becomes a diagonal identity matrix.

Once you have your matrix, see what it does under a Lorentz boost. This is even easier in some cases. If you have come up with something you find special, pay careful attention to what happens when $\beta=0$. Is your matrix the zero matrix? If so, then under boosts it remains the zero matrix. Is your matrix the identity matrix? If so then under boosts it probably remains the identity matrix, although this might depend on what valence the tensor has, whether it is a [1, 1] tensor or a [0, 2] tensor...

I will tell you that there is at least an antisymmetric [0, 3]-tensor because there is a [0, 4] orientation tensor that you can reversibly form the product with. If you had another 4-vector you could rely upon, they would combine together into an antisymmetric [0,2] tensor that you could indeed write as a matrix. Angular momentum is such a [0,2] tensor for precisely this reason.

# Where I would start instead

But it also makes sense to take a step back, to try to see what is going on and what you need. We know that we want a 4-position, a set of points in space-time that represent a path of a particle. Call this $X=\{x^\mu(s)~|~s\in\mathbb R\}$. We used to choose $s$ as the time coordinate $t$, but that was kind of an arbitrary choice, that we made to have one fewer variable to worry about.

We then have the tangent vectors $$X'=\left\{{dx^\mu\over ds}~\Big|~s\in\mathbb R\right\}.$$ Now the tangent vectors you get here depend on something unphysical, which is the exact functional dependence of $x^\mu$ on this arbitrary path parameter $s$. However in some sense that's not important, for example if you want to calculate the velocity in your local reference frame, you would just divide through: in 1+1 dimensions $w,x$ this is just $v(s)=c~x'(s)/w'(s).$ In other words, if you had $x^\mu\big(f(s)\big)$ for some monotonically increasing $f$, this dependence works out by the chain rule to multiplying all coefficients by a scale factor $f'(s)$.

So in this way of viewing things, there is never any discrepancy. As long as everybody agrees on the same coordinate $s$ parameterizing the set of points $X$ , they will agree on the set of tangent vectors $X'$.

Now we come to your question, which is fundamentally about choosing $s=w$ so that $w'(s)=1.$ The key here is that we can do it for one reference frame at a time only. So $dx^\mu/ds$, being a difference between two infinitesimally close (based) 4-vectors, is also a (unbased) 4-vector. Under that Lorentz transformation its $w$-component has to change, and so you have to choose in what reference frame you are setting this component to $1$, and you potentially get to make different choices for every $s$ if you want to.

You can't get $w'=1$ in all of them without forcing everybody to use their own different basis-dependent scale factor for $s$. Which indeed you can totally do, it's just extra bookkeeping when you're doing a Lorentz transformation that you have to keep track of “oh, this is a scaley 4-vector and when I boost it I need to multiply by the scale factor ratio,” just like how you need to keep track of whether a vector is based or unbased for questions of how they change under spacetime translation.

So we said that to avoid these scaley vectors, you would want to choose for each $s$ what reference frame sees $w'=1$. We have collectively chosen that it is the reference frame where $x'=y'=z'=0$, as a physics community. But we also know that the 4-norm is preserved by Lorentz transformations and so this amounts to saying that the tangent vector should be normalized to be a unit vector, at least in the $(+---)$ metric (which is indirectly the biggest reason I prefer that metric convention).

I hope that clarifies the trade-offs that you face.

† To be "based" is to be origin dependent in the obvious way: spacetime translations change based 4-vectors but leave unbased 4-vectors alone.

$dx^\mu/dx^0$ gives time the special position, so the normal way of solving this is to introduce proper time $\tau$ that everyone agrees on. The 4-velocity vector is then $dx^\mu/d\tau$.

However, you apparently found another solution that also gives equality to the four dimensions. This tensor $v^\mu_{\,\,\nu}=dx^\mu/dx^\nu$ has some intriguing properties. First, it transforms as a proper tensor. Second, according to the chain rule, $v^\mu_{\,\,\nu}=1/v^\nu_{\,\,\mu}$, and it has unity on its diagonal. It is not of much use though. We can't define a 'magnitude' of velocity in this way, and the acceleration and momentum are hard to define. And when a particle is not moving in one direction, some components of that tensor blows up.

The electromagnetic tensor encapsulates two 3-vectors. But your tensor only encodes velocity. It might serve as some decoration to the theory, but never a tool in practice.