4-velocities in different frames

Question

We have an observer in an inertial frame $S$ who measures a particle's 4-velocity as $U$. We then have another inertial frame $S'$ with $X'=\Lambda{X}$, where $\Lambda$ is a matrix representing a Lorentz transformation. (To be clear, $U=\frac{dX}{d\tau}$ where $X$ is the 4-vector representing the trajectory of a particle through spacetime as viewed from $S$, parametrised by the proper time $\tau$).

I am told that an observer in $S'$ will then measure $U'=\Lambda{U}$ but this is not proved. I've tried to do it the following way, but I don't know if I can use the product rule as such when matrices and vectors are involved.

\begin{equation*} U'=\frac{dX'}{d\tau}=\frac{d}{d\tau}\Lambda{X}=\frac{d\Lambda}{d\tau}X+\Lambda\frac{dX}{d\tau}=\Lambda\frac{dX}{d\tau}=\Lambda{U} \end{equation*}

since the Lorentz transformation matrix $\Lambda$ is independent of the proper time $\tau$. I feel like this is one of those proofs which seems to work too easily - importantly, it offers no insight as to why $\tau$ is a good parameter to use rather than the time $t$ observed in $S$. The text I'm using implies that if we used a 4-velocity such as $V=\frac{dX}{dt}$, things would be much messier. This is not apparent with my proof though, so could someone tell me if it's correct? If not, how do I instead go about proving $U'=\Lambda{U}$ given $X'=\Lambda{X}$?

Answer 1

I've tried to do it the following way, but I don't know if I can use the product rule as such when matrices and vectors are involved.

\begin{equation*} U'=\frac{dX'}{d\tau}=\frac{d}{d\tau}\Lambda{X}=\frac{d\Lambda}{d\tau}X+\Lambda\frac{dX}{d\tau}=\Lambda\frac{dX}{d\tau}=\Lambda{U} \end{equation*}

since the Lorentz transformation matrix $\Lambda$ is independent of the proper time $\tau$.

If you are uncomfortable with matrices and vectors pick an arbitrary frame of unit orthogonal vectors $\vec e_0,$ $\vec e_1,$ $\vec e_2,$ and $\vec e_3.$ Then $\vec X=\sum_\mu X^\mu \vec e_\mu.$ You can also pick a potentially different arbitrary frame of unit orthogonal vectors $\vec e'_0,$ $\vec e'_1,$ $\vec e'_2,$ and $\vec e'_3.$ Then $\vec X=\sum_\mu X'^\mu \vec e'_\mu.$ Now the Lorentz transformation is a transformation of coordinates it sends the four tuple of numbers $(X^0,X^1,X^2,X^3)$ to a possibly different four tuple of numbers $(X^{'0},X^{1'},X^{2'},X^{'3}).$

And to be totally totally clear, a vector is a geometric object that points in a particular direction, from one when-where to another when-where. And the four tuple is just another way of writing the vector, four instance you can write $X=(X^0,X^1,X^2,X^3)$ instead of writing $\vec X= X^0 \vec e_0+X^1 \vec e_1+X^2\vec e_2+X^3 \vec e_3.$ And you could write $(X^{'0},X^{1'},X^{2'},X^{'3})$ instead of writing $X^{'0}\vec e'_0+X^{1'}\vec e'_1+X^{2'}\vec e'_2+X^{'3}\vec e'_3.$

All you are doing is choosing a different basis to write the same vector as a 4 tuple of numbers. Different four tuples, same vector. And one advantage of this is exactly to avoid having to learn how to deal with vectors and matrices. Instead you can note that $X^{'\mu}=\sum_\nu\Lambda^{\mu}_{\nu}X^\nu,$ where everything in that equation is a scalar. Then you can differentiate away.

Now that is an equation for scalars and it is only related to differentiation of vectors since you choice of basis vectors literally pointed in the same direction and magnitude for each time and each place. Otherwise you do need need to learn how to differentiate vectors. But the product rule does apply.

I feel like this is one of those proofs which seems to work too easily

You could differentiate a vector on the xy basis compared to the $\theta, r$ basis of you want something more complicated, you should learn that before you move on the general relativity.

• importantly, it offers no insight as to why $\tau$ is a good parameter to use rather than the time $t$ observed in $S$.

Now that you know it is the same vector, just expressed in two different basis you can see that it would make no sense whatsoever to differentiate it with respect to some random basis as if it is better than any other basis. Really you could differentiate it with respect to any orthonormal basis and then make a unit length vector at the end and that is fine. Because you are finding the tangent to the worldline, and when you parameterize by proper time you are scaling the tangent to have unit length.

Again, you can go back to calculus in a flat Euclidean plane and figure out how to compute a unit tangent to a curve and one way is to parameterize it by arc length and then differentiate it with respect to arc length.

And that isn't complicated the derivative is just a difference of vectors divided by a scalar. When that scalar is the length (proper time) then you have simply made a unit vector pointing between them. That's all that is going on.

Answer 2

There are two different questions that are unrelated to each other.

The first one is how the 4-velocity is defined. By definition, velocities are tangent flows on a differential manifold, therefore derivatives must be taken with respect to the parameter you are using to describe the flow with. In the context of special relativity such parameter is the proper length $s$ (or the proper time $\tau$, the two differ by a constant factor, thus you can just reparametrise the entire set of equations to obtain the former from the latter). The standard time $t$ is a variable on the manifold, not anymore the parameter you describe the flows with, hence it has no rights to be promoted as derivative. Notice that this is unrelated to how "messy" (according to your terminology) the final equations will be. Things may be messy but still correct, or viceversa.

The second question is how these 4-velocities transform in different reference frames. Once you have the definition you can just take derivatives as you have done above (brute force method) or you could use more sophisticated approaches of vector calculus showing how components of tangent vectors transform amongst charts of a manifold once you fix an underlying coordinates transformation: the results will be the same.

Answer 3

Your proof is entirely correct. Relativity doesn't have to be difficult :)

To be clear, the steps in your proof are simply the definition of $U'$, the definition of $X'$, the product rule, the constancy of $\Lambda$, and the definition of $U$, respectively. There is nothing wrong with any of these steps.

As for why $\tau$ and not $t$: $\tau$ is defined based on the worldline in such a way that it doesn't vary between coordinate systems. That is, whether I'm working in $S$ or $S'$, a particular event on the particle's worldline will happen at a particular proper time $\tau$. On the other hand, it will happen at coordinate time $t$ in $S$ and at $t' \neq t$ in $S'$. So by using $\tau$ in the first place we don't have to worry about transforming $t \to t'$ anywhere.

To see how this simplifies things, try to reconstruct your proof using $V$ instead: $$ V' = \frac{\mathrm{d}X'}{\mathrm{d}t'} = \frac{\mathrm{d}}{\mathrm{d}t'} (\Lambda X) = \frac{\mathrm{d}\Lambda}{\mathrm{d}t'} X + \Lambda \frac{\mathrm{d}X}{\mathrm{d}t'} = \Lambda \frac{\mathrm{d}X}{\mathrm{d}t'} \neq \Lambda \frac{\mathrm{d}X}{\mathrm{d}t}. $$ If we try to convert $\Lambda \, \mathrm{d}X/\mathrm{d}t'$ into $S$-frame quantities, we do indeed get a mess. In fact, I can't write out the expression without using components and Einstein summation notation. We can write $$ \Lambda \frac{\mathrm{d}X}{\mathrm{d}t'} = \Lambda^{\mu'}{}_\mu \frac{\mathrm{d}x^\mu}{\mathrm{d}t'} = \Lambda^{\mu'}{}_\mu \frac{\partial x^\nu}{\partial t'} \frac{\mathrm{d}x^\mu}{\mathrm{d}x^\nu} = \Lambda^{\mu'}{}_\mu \Lambda^\nu{}_{t'} \frac{\mathrm{d}x^\mu}{\mathrm{d}x^\nu}, $$ since $x^\nu = \Lambda^\nu{}_{\nu'} x^{\nu'}$. That is, you need to combine not only the four components of $V = \mathrm{d}x^\mu/\mathrm{d}t$ but also twelve other derivatives in order to get $V'$.¹

Now if you did try to use $t$ rather than $t'$ everywhere you differentiated, whether or not you were differentiating $X$ or $X'$, everything would work out. But velocities so defined would only have nice properties (like unit norm) in one coordinate system, and the whole idea of relativity is that you can do physics in any coordinate system, so long as you are consistent.

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In the language of tensors, this is because $V$ and $V'$ are each just four components out of sixteen of a rank-2 tensor. In general, you need all a tensor's components in order to find any of them in a different frame. $U$ and $U'$, on the other hand, are rank-1 tensors (aka vectors), so if you have all four components of $U$, you can easily find any/all components of $U'$.