# Extending a vector to a four-vector

In the Feynman Lectures on Physics Vol-II, Chapter 25, Section 1, there is a short discussion on four vectors. In this section Feynman briefly shows why we cannot extend the velocity vector to a four vector and gives the following reason.

The components of the velocity vector can be written as,

$$v_x=\frac{dx}{dt}\\ v_y=\frac{dy}{dt}\\ v_z=\frac{dz}{dt}\\$$

Now, we cannot include a time component to make this a four vector because the other components are expressed as a derivative with respect to time which is not an invariant under the Lorentz Transformation. However the momentum vector whose components are listed below:

$$p_x=\frac{m_0v_x}{\sqrt{1-\frac{v^2}{c^2}}}\\ p_y=\frac{m_0v_y}{\sqrt{1-\frac{v^2}{c^2}}}\\ p_z=\frac{m_0v_z}{\sqrt{1-\frac{v^2}{c^2}}}\\$$

is indeed a four vector with the time component being energy. How is this possible when clearly every component of the momentum vector can be written as,

$$p_\mu=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\frac{dx_\mu}{dt}$$

which is also expressed as a derivative with respect to time as was velocity (Here $$\mu = x,y,z$$)? How does the appearance of the scalar $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ change anything? Is there something I am missing? What is the criterion that we must check before deciding whether a vector can be extended to a four vector?

I have attached the lecture referred to above over here: https://www.feynmanlectures.caltech.edu/II_25.html

• $\gamma=(1-v^2)^{-1/2}$ is no more a constant than the individual components $v_x,v_y,v_z$ are. Commented Feb 18, 2021 at 15:51
• @jacob1729 'scalar' would be a better word to use. I have made the necessary changes but my question still remains. Commented Feb 18, 2021 at 16:38
• The components must transform correctly such that the 4d “dot product” is invariant. This is tedious to check but it’s trivial. See the first three sections of uspas.fnal.gov/materials/12MSU/Conrad_4vector.pdf . I might resubmit basically this as an answer .
– user288901
Commented Feb 18, 2021 at 18:29
• ...And because you’re considering one vector you would dot it with itself.
– user288901
Commented Feb 18, 2021 at 18:37

"Now, we cannot include a time component to make this a four vector because the other components are expressed as a derivative with respect to time which is not an invariant under the Lorentz Transformation."

This sentence is not entirely true. We can't make a four vector from things that aren't Lorentz covariant. We can, however, combine terms in a clever such that the four vector as a whole is covariant$$^\dagger$$. Note that I say covariant here and not invariant. Invariant means that something doesn't change under a Lorentz transformation while covariant means it changes in a particular way. For a four-vector to be covariant it has to obey $$V'\,^\mu=\sum_\nu{\Lambda^\mu}_\nu V^\nu$$ Where $$V^\nu$$ is the old vector, $$V'\,^\mu$$ is the new vector and $${\Lambda^\mu}_\nu$$ is the Lorentz transformation matrix. Written out in matrix form for the $$t,x$$ components it looks like this $$\pmatrix{V'\,^0\\V'\,^1}=\pmatrix{\gamma&-\beta\gamma\\-\beta\gamma&\gamma}\pmatrix{V^0\\V^1}$$

So it doesn't matter if I have time derivatives in my four vector components as long as the entire thing transforms like the equations above. Now I will show why the four momentum transforms properly. If I directly used $$p_\mu=m_0\gamma\frac{dx_mu}{dt}$$ it would be a huge pain. But it can be written more simply using $$\gamma=\frac{dt}{d\tau}$$. \begin{align} p_\mu&=m_0\gamma\frac{dx_\mu}{dt}\\ &=m_0\frac{dt}{d\tau}\frac{dx_\mu}{dt}\\ &=m_0\frac{d}{d\tau}x_\mu \end{align} Now we that $$\frac{d}{d\tau}$$ is invariant while $$x_\mu$$ transforms covariantly. So as a whole $$p_\mu$$ transforms covariantly.

$$\dagger$$ I might have confused covariant and contravariant. I will update later if I have confused covariant/contravariant.

• Thanks for your answer. I have a follow up question. Is it possible to introduce a time component to the velocity vector such that this hypothetical velocity four vector is covariant under the Lorentz Transformation? If yes then what it is the time component? If not then how can we formally prove this and how can I decide whether a four vector extension is possible or not for any given vector quantity? Commented Feb 18, 2021 at 16:59

So this is a common problem, about this ambiguity. Let me phrase it like this: if you are in 3D space you can take any vector and any scalar and combine them together and those are validly the components of a 4-vector. The problem is that if you choose the wrong things like, say, temperature and vorticity, then when you boost this into another frame, the scalar field is no longer a temperature anywhere in the new frame and the vorticity likewise. For this reason we sometimes say things like “a vector is anything that transforms like a vector.” It is not a circular as it sounds, the idea is that we know what transforming like a vector means due to knowing the coordinate transforms very well, but we need to find a thing which maintains its identity under those coordinate transforms. That identity has much more to do with us, the sorts of experiments that we want to perform, then with any numerical aspect of the system. Like in the temperature/vorticity case, it matters that I want to measure temperatures in these two different circumstances and I want a quantity that predicts those experiments even when Lorentz boosts are in the picture.

So a lot more of it is just figuring out the right interpretation.

In the case of 4-vectors from 3-vectors, these already transform under rotations and so boosts are all we have to “interpret.” So for instance I might take a vector $$\vec v$$ and adjoin its length, to produce a 4-vector with Minkowski length 0, a null vector. This is a valid 4-vector if we are treating it like a displacement of light, which follows such null-vectors, or a tangent-vector for a massless particle or so. But it won't work if our interpretation is trying to single out a massive particle moving at a certain velocity, say, because those follow timelike paths.

Now, if you have a timelike path in relativity, a world-line, we can describe it as some function $$s\mapsto(w,x,y,z)$$ where $$w=ct$$ per usual, and $$s$$ is some arbitrary number to parameterize the path.

Well, you want a velocity and so we need a difference between two adjacent points, $$\big( w(s+\delta s),x(s+\delta s),y(s+\delta s),z(s+\delta s) \big)-\big( w(s),x(s),y(s),z(s) \big)\approx \big( w'(s),x'(s),y'(s),z'(s) \big)~\delta s$$ and then the actual velocity has components $$v_x(s)=c~x'(s)/w'(s)$$ etc., because this measures how much it moved in space divided by how much it moved in time.

I haven't told you hardly anything about $$s$$ here except that it should be a continuous parameterization such that these derivatives are possible. In particular under the map $$s\mapsto s^3$$, say, you would get the same world-line! All I have done is construct a tangent vector to a 4D path. But there would be a very felicitous choice of $$s$$ if, say, $$\big( w'(s),x'(s),y'(s),z'(s), \big)$$ were related to its analogues in other coordinates by Lorentz transforms too, so that it was also interpretable as a 4-vector. Then this would be understandable as a “4-velocity.”

This comes down to a scaling by this $$\delta s,$$ which amounts to saying that everyone agrees on the Minkowski length of this vector: and if everyone agrees on some “velocity”-like scalar you can imagine fudging a time dimension into $$s$$ so that the 4-velocity length is the speed of light, in which case $$s$$ is called the “proper time” along the path and often denoted $$\tau$$. And in this particular case you get the result that you are interested in, the 4-velocity is $$(\gamma c, \gamma v_x, \gamma v_y, \gamma v_z)$$ in any particular coordinates.

Then finally we can come to Feynman's point, which is the interpretability of this expression, the first component is $$\delta w=\gamma c~\delta \tau$$ and thus $$\gamma=\mathrm dt/\mathrm d\tau$$ and these other components are e.g. $$\gamma v_x= (\mathrm dt/\mathrm d\tau)\cdot(\mathrm dx/\mathrm dt).$$ The interpretation is as a chain rule, it is that this is secretly a derivative of $$x(t(\tau)).$$ In other words, $$t$$ is not a properly “invariant” time coordinate as it belongs to the coordinate frame, but $$\tau$$ belongs to the path which is an independent geometric entity which everyone can “point to” in their equations.

One final note for people who might stumble on this with a related question... That the 4-momentum is the rest mass times the 4-velocity, is something that I regard as needing derivation. The 4-velocity itself “makes sense” per the above, and certainly multiplying it by the rest mass is “manifestly covariant,” but that it is the correct physical quantity to expect a conservation law on, that I find needs a little more “meat.” Probably my favorite derivation here would be to attach a massless mirror to the object and slow it down by firing packets of light at it, keeping in mind that for packets of light you know from classical electromagnetism that $$E=pc$$ and from the Lorentz transformation you know the relativistic Doppler shift. So you get nice integrals which allow you to derive that for example the kinetic energy is, if energy is conserved, $$(\gamma - 1) m_0 c^2.$$ And then the claim is, it doesn't matter how the energy was dumped into this thing when that is a valid procedure to extract it. just the fact that this is always in principle possible proves that this relation must hold generally.