Will the velocity of light be the same in all reference frames?

[Herman Trivilino]

If the photon has eye, then?

Then it can't travel at speed c. Which was your premise. So your argument is self-contradictory.

 

[PAllen]

The relative velocity for two particles, where at least one particle (say particle 1) has non-zero mass is the velocity of the other particle in the rest frame of particle 1 (which is called the lab frame in scattering theory for historical reasons, when the accelerator experiments where fixed-target experiments). It's easy to formulate this in a manifestly covariant way. In the lab frame we have for the four-momenta (I use natural units, $c=1$)
$$p_1^*=(m_1,0)^T, \quad p_2^*=(E_2^*,\vec{p}_2^*) \; \Rightarrow \; v_{\text{rel}}=\frac{|\vec{p}_2^*|}{E_2^*}=\sqrt{E_2^{*2}-m_2^2}{E_2^*}.$$
Now we have for the Minkowski product in this frame and thus for any other frame
$$p_1^* \cdot p_2^*=p_1 \cdot p_2=m_1 E_2^* \; \Rightarrow \; E_2^*=\frac{p_1 \cdot p_2}{m_1},$$
and thus
$$E_2^{*2}-m_2^2=\frac{(p_1 \cdot p_2)^2}{m_1^2}-m_2^2=\frac{(p_1 \cdot p_2)^2-m_1^2 m_2^2}{m_1^2},$$
which leads to
$$v_{\text{rel}}=\frac{\sqrt{(p_1 \cdot p_2)^2-m_1^2 m_2^2}}{p_1 \cdot p_2}.$$
Now, if $m_1=m_2=0$ you have two possibilities: $p_1 \cdot p_2=0$; then the relative velocity is undefined because then $p_1 \propto p_2$ or $p_1 \cdot p_2 \neq 0$, and the relative velocity is 1. If $m_1 \neq 0$ but $m_2=0$ you have always $v_{\text{rel}}=1$.

Also note that $v_{\text{rel}}$ is only $|\vec{v}_1-\vec{v}_2|$ if $\vec{v}_1 \propto \vec{v}_2$. The general expression in an arbitrary frame is
$$v_{\text{rel}}=\frac{1}{1-\vec{v}_1 \cdot \vec{v}_2} \sqrt{(\vec{v}_1-\vec{v}_2)^2-(\vec{v}_1 \times \vec{v}_2)^2}.$$
For a derivation, see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Sect. 1.5 and 1.6.

Nice. This is eqivalent to the first formula in a post of mine a while ago: https://www.physicsforums.com/threads/generalized-velocity-addition-and-aberration-formulas.821733/ , where I note that the cross product term is a relativistic correction for orthogonal component of the relative motion, while the denominator not being 1 is a correction for the parallel component.

Also note that, as you are well aware, your statement " $v_{\text{rel}}$ is only $|\vec{v}_1-\vec{v}_2|$ if $\vec{v}_1 \propto \vec{v}_2$" neglects the denominator, which 'implements' parallel velocity addition. Thus, norm of vector difference isn't even true for the parallel case.

 

[Aswin Sasikumar 1729]

As asked, that question makes no sense. If two things are moving towards one another and there exists a coordinate system in which one of them is at rest, then they necessarily have a non-zero relative velocity using that coordinate system.

(If there is no such coordinate system, as is the case with flashes of light, then their relative velocity is undefined, although t

a

The relative velocity for two particles, where at least one particle (say particle 1) has non-zero mass is the velocity of the other particle in the rest frame of particle 1 (which is called the lab frame in scattering theory for historical reasons, when the accelerator experiments where fixed-target experiments). It's easy to formulate this in a manifestly covariant way. In the lab frame we have for the four-momenta (I use natural units, $c=1$)
$$p_1^*=(m_1,0)^T, \quad p_2^*=(E_2^*,\vec{p}_2^*) \; \Rightarrow \; v_{\text{rel}}=\frac{|\vec{p}_2^*|}{E_2^*}=\sqrt{E_2^{*2}-m_2^2}{E_2^*}.$$
Now we have for the Minkowski product in this frame and thus for any other frame
$$p_1^* \cdot p_2^*=p_1 \cdot p_2=m_1 E_2^* \; \Rightarrow \; E_2^*=\frac{p_1 \cdot p_2}{m_1},$$
and thus
$$E_2^{*2}-m_2^2=\frac{(p_1 \cdot p_2)^2}{m_1^2}-m_2^2=\frac{(p_1 \cdot p_2)^2-m_1^2 m_2^2}{m_1^2},$$
which leads to
$$v_{\text{rel}}=\frac{\sqrt{(p_1 \cdot p_2)^2-m_1^2 m_2^2}}{p_1 \cdot p_2}.$$
Now, if $m_1=m_2=0$ you have two possibilities: $p_1 \cdot p_2=0$; then the relative velocity is undefined because then $p_1 \propto p_2$ or $p_1 \cdot p_2 \neq 0$, and the relative velocity is 1. If $m_1 \neq 0$ but $m_2=0$ you have always $v_{\text{rel}}=1$.

Also note that $v_{\text{rel}}$ is only $|\vec{v}_1-\vec{v}_2|$ if $\vec{v}_1 \propto \vec{v}_2$. The general expression in an arbitrary frame is
$$v_{\text{rel}}=\frac{1}{1-\vec{v}_1 \cdot \vec{v}_2} \sqrt{(\vec{v}_1-\vec{v}_2)^2-(\vec{v}_1 \times \vec{v}_2)^2}.$$
For a derivation, see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Sect. 1.5 and 1.6.

whether these equations are dimensionally correct?

 

[Ibix]

a

whether these equations are dimensionally correct?

Yes. As @vanhees71 noted he is using natural units in which c=1, and therefore suppressing the factors of c, or including them in the units if you prefer to see it that way.

I do think there's a typo in the last part of the first equation - I think he should be dividing by $E^*_2$ rather than multiplying.

 

[vanhees71]

I do think there's a typo in the last part of the first equation - I think he should be dividing by $E^*_2$ rather than multiplying.

Thanks, I've corrected the typo.