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Please note that my question is not a duplicate, it is not about the speed of light, my question is only technical about the four velocity vector for light, its definedeness, value and constantness.

I have read these questions:

What is the time component of velocity of a light ray?

Where Izhov says:

Four-velocity actually isn't well-defined for light.

And where ClassicStyle says in a comment:

The four velocity of light is perfectly well defined. You just can't use proper time to parameterize the world line. Four velocity is just the tangent vector to a world line

Are components of the velocity of light equal to $c$?

https://en.wikipedia.org/wiki/Four-vector

The four-velocity defined here using the proper time of an object does not exist for world lines for objects such as photons travelling at the speed of light

Why is light affected by time dilations in space-time curvatures

Where Safesphere says in a comment:

The magnitude of the 4-velocity of light is always zero (see my comment above).

The (always) non-zero time component of the 4-velocity of light does NOT mean that light moves in time. To calculate the 4-velocity of light, we have to use a different affine parameter instead of proper time, because the proper time of light is always zero.

Now this is confusing. Light must have a four velocity vector, but it seems to be either well defined or not, and it seems to have a magnitude of 0 or c and it seems to be always constant or not.

Questions:

  1. Which one is right, is the four velocity of light well defined or not?

  2. Is the magnitude of the four velocity vector for light always constant?

  3. Is the magnitude 0 or c?

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So in relativity we have these things called four-vectors. Let $a$ be a four-vector, then in any given coordinate systems it has four components: $a^w$ in the time-direction $w=ct$, $a^{x,y,z}$ in the spatial direction.

Its squared-magnitude is defined by:$$a_\mu a^\mu = (a^w)^2 - (a^x)^2 - (a^y)^2 - (a^z)^2.$$ When this is negative we say that the four-vector is “space-like” or when it is positive we say that the four-vector is “time-like,” and when it is zero we say that it is null or “light-like.” And then if it's space-like we can either take the normal square root and get an imaginary number, or take $\sqrt{-a_\mu a^\mu}$ to get a positive number which we can call the magnitude; if it is time-like then the normal square root $\sqrt{a_\mu a^\mu}$ suffices to get a magnitude from a squared-magnitude. Like how rotations preserve vector lengths, Lorentz transformations preserve squared-magnitudes.

Four-velocity is one of these four-vectors. If we choose the right coordinate system so that your four-velocity lies along, say, the $z$-axis, then for particles it is defined as the vector $$v^w = c \cosh \phi,\\ v^{x,y}=0,\\ v^z = c\sinh \phi,$$ for some number $\phi$, and it corresponds to something moving with speed $c\tanh\phi.$ These functions, if you have not encountered them yet, are the hyperbolic functions.

Plugging this into the above equation gives $$v_\mu v^\mu = c^2(\cosh^2\phi - \sinh^2 \phi) = c^2,$$so that the four-velocity is always normalized to a constant value, it is timelike with magnitude $c$. Indeed you could imagine that it is the non-four-vector tangent vector to the particle's motion in spacetime, $(c, 0, 0, c\tanh\phi)$, but that has been normalized to have a constant magnitude $c$, so that Lorentz transformations can preserve this length.

Now whether you take this normalized value as a part of the definition of four-velocity, that is an aesthetic opinion rather than something which the mathematics forces on you. But there is a reason to be concerned.

See, you cannot normalize a light-like tangent vector into a normal four-velocity, because its squared magnitude is zero. So if you have the point in spacetime $(w,x,y,z) = (ct_0, x_0, y_0, z_0)$ and you add a little time $dt$ to this, a light ray moving from that point in the $z$-direction is now at point $(ct_0 + c~dt, x_0, y_0, z_0 + c~dt)$ and the difference between those two points is a four-vector, $$T^w = c~dt, T^{x,y} = 0, T^z = c~dt.$$ However we would find that $T_\mu T^\mu = 0$ and there is nothing you can multiply it by in order to give it constant magnitude $c$.

You might respond to this fact by saying “this is disastrous! let us say that light has no four-velocity!”—or you might instead respond by saying “ok, but this is actually a blessing in disguise, it does not mean that no normalizations are possible so much as all normalizations are trivial, I am free to choose whichever I want!”. Both responses have some merit. I personally tend towards the first one, for the following reason: my gut choice for the second is to normalize a lightlike tangent vector as $(c, 0, 0, c)$, so that the time component is constant. But, there is a problem with this normalization: the Lorentz transform will not preserve it. It will properly transform the vector, but I will have to renormalize it in the new context, too. I do not like that aspect very much, personally.

With that said there are occasionally reasons to do it. The easiest would be if you were thinking about what the universe looks like when you move through it: you would draw null rays in from all of the stars you can see at this very second, towards your face: and then it might be nice to project this all as if it had come from a sphere of fixed radius $R$ that you happen to be at the center of: a “celestial sphere.” And that is essentially the same normalization that I described above when I fixed the time component to a fixed value $c$. You might then perform a Lorentz boost in some direction, so the sphere gets mapped to some different sphere, and you might project the new sphere back onto a new sphere of fixed radius $R$, finding that the stars have all appeared to shift in the sky as a result of my boost (more specifically: they all appear to have shifted towards the direction that I was accelerating in). A similar argument about the light that I emit would suggest that it has also all crowded in that direction, leading to a well-known phenomenon called relativistic beaming where something which emits light preferentially emits it in the direction that it is travelling as it travels faster and faster.

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  • $\begingroup$ Thank you, is the magnitude for light (if it is defined) always constant or not, and if constant, then is it c or 0? $\endgroup$ – Árpád Szendrei May 26 at 15:46
  • $\begingroup$ @ÁrpádSzendrei the magnitude of the tangent vectors for light are always constant and that constant is zero; it is null rather than timelike or spacelike. Similarly one can have a null 4-displacement between two events; this means "one is objectively before the other but I can choose coordinates to shrink the time gap arbitrarily close to zero, similarly they are objectively not at the same place but I can choose coordinates to shrink the distance between them arbitrarily close to zero—indeed, these both shrink together." $\endgroup$ – CR Drost May 26 at 17:07
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From "A First Course in General Relativity":

2.3 The four velocity

A particularly important vector is the four-velocity of a world line. ... In our four-geometry we define the four-velocity $\vec U$ to be a vector tangent to the world line of the particle, and of such a length that it stretches one unit of time in that particle's frame.

The immediate problem for the case of a photon is that it does not have a frame. Schutz makes this explicit here:

2.7 Photons

No four-velocity. Photons move on null lines, so, for a photon path,

$$\mbox{d}\vec x \cdot \mbox{d}\vec x = 0$$

Therefore $\mbox{d}\tau$ is zero and Eq. (2.31) $[\vec U = \mbox{d}\vec x / \mbox{d}\tau]$ shows that the four-velocity cannot be defined. Another way of saying the same thing is to note that there is no frame in which light is at rest (the second postulate of SR), so there is no MCRF for a photon. Thus, no $\vec e_0$ in any frame will be tangent to a photon's world line.

Note carefully that it is still possible to find vectors tangent to a photon's path (which, being a straight line, has the same tangent everywhere): $\mbox{d}\vec x$ is one. The problem is finding a tangent of unit magnitude, since they all have vanishing magnitude.

So, by the above, the answer to your first question is: the four-velocity is not defined for photons.

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Short answer:

  • If the term "four-velocity" is used in the strict sense of $d x^\mu/d\tau$ where $\tau$ is the object's proper time, then four-velocity is undefined for light because the elapsed proper time is always zero ($d\tau=0$) along a lightlike worldline.

  • If the term "four-velocity" is used in the generalized sense of $dx^\mu/d\lambda$ where $\lambda$ is an affine parameter that increases monotonically along the lightlike worldline, then the four-velocity is perfectly well-defined for light.

So the four-velocity is either undefined for light or well-defined for light, depending on what the speaker/writer means by "four-velocity."

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  • $\begingroup$ Thank you, is the magnitude (if it is defined) always constant or not, and if constant, then is it c or 0? $\endgroup$ – Árpád Szendrei May 26 at 15:40
  • $\begingroup$ The four-velocity in the generalized sense (using an affine parameter $\lambda$) is zero for a lightlike worldline, even though the components $dx^\mu/d\lambda$ are not all zero. Using overhead dots to denote derivatives with respect to $\lambda$, the equation for the proper time $\tau(\lambda)$ along any worldline is $\dot\tau^2=\dot t^2-(\dot x^2+\dot y^2+\dot z^2)$ in flat spacetime with $c=1$. The right-hand side is the magnitude of the generalized 4-velocity. Lightlike worldlines are defined by the condition $\dot\tau=0$. $\endgroup$ – Chiral Anomaly May 26 at 18:54

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