 # If we have a light ray x^(nu) with velocity c, what is c^0 (the time component)? cofak48 2022-08-12 Answered
If we have a light ray ${x}^{\mu }$ with velocity $c$, what is ${c}^{0}$ (the time component)?
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For a massless particle like a photon, you have to use a new "affine" parameter $\lambda$ instead of using proper time $\tau$ (because $d\tau =0$). This parameter $\lambda$ must be a scalar, that is an invariant, under Lorentz transformations.
You will define the 4-velocity of the photon as:
${u}^{\mu }=\frac{d{x}^{\mu }}{d\lambda }\phantom{\rule{thinmathspace}{0ex}}\left(\mu =0,1,2,3\right)$
Here ${x}^{0}=ct$, where $c$ is the speed of light, and ${x}^{1},{x}^{2},{x}^{3}$ are spatial coordinates:
You will have the property:
${u}_{\mu }{u}^{\mu }={g}_{\mu \nu }{u}^{\nu }{u}^{\mu }=0$
where $g$ is the diagonal matrix $\left(1,-1,-1,-1\right)$
This could be also simply written:
$\left({u}^{0}{\right)}^{2}-{\stackrel{\to }{u}}^{2}=0$
Maybe, it is better, depending on your problem, to use the photon momentum/energy quadrivector ${p}^{\mu }$, which satisfies the same equations : ${p}_{\mu }{p}^{\mu }=0$, that is: $\left({p}^{0}{\right)}^{2}-{\stackrel{\to }{p}}^{2}=0$, This is best fitted if you use momentum conservation relations with several particles. The ordinary 3-velocity is easily obtained from $p$ as $\stackrel{\to }{v}=\frac{\stackrel{\to }{p}}{{p}^{0}}$

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Four-velocity actually isn't well-defined for light. This is because four-velocity is the derivative of the position four-vector with respect to the proper time, i.e. the time in the moving object's rest frame. Since you can't pick an inertial rest frame for a beam of light, you can't take the derivative with respect to the proper time.
Mathematically, the proper time of light is zero, so to take the derivative of light's position four-vector with respect to proper time, you would be dividing by zero, returning infinity/undefined (for all components, including the time component).

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