Photon Speed Addition: Can it Be FTL?

[jk22]

It is well known that relativistic speed addition is a non-commutative group (gyrogroup).

The addition of two orthogonal speeds is given by :

$\vec{u}\oplus_\perp\vec{v}=\vec{u}+\sqrt{1-u^2/c^2}\vec{v}$

Hence if $u$ is describing a photon with speed c then we can add whatever $v$ it always gives $\vec{u}$.

Does this mean in particular that a photon could have ftl components of its speed that are orthogonal to it's motion ?

 

[PeroK]

Does this mean in particular that a photon could have ftl components of its speed that are orthogonal to it's motion ?

No.

It's not clear why you would even think that. Consider $v + u$ instead.

 

[Vanadium 50]

Does this mean in particular that a photon could have ftl components of its speed that are orthogonal to it's motion ?

What does it even mean to have components of velocity (I know you said speed, but speed is a scalar and has no components) orthogonal to the direction of motion?

 

[jk22]

Yes velocity I meant. Due to the fact that in relativity speed cannot go beyond c, so the velocity could have kind of hidden components ? Like $\vec{c}=u\vec{e}_1+v\vec{e}_2$ but the vector addition is replaced by relativistic addition of velocities.

 

[Mark44]

If the photon is moving in a certain direction, why would there even be an orthogonal component of the velocity?

 

[DrStupid]

Like $\vec{c}=u\vec{e}_1+v\vec{e}_2$ but the vector addition is replaced by relativistic addition of velocities.

No it isn't. The relativistic "addition" of the velocities v and u actually is a transformation of the velocity v into a frame of reference moving with the velocity -u. It has nothing to do with the addition of vector components.

 

[PeroK]

If the photon is moving in a certain direction, why would there even be an orthogonal component of the velocity?

The idea is as follows. If a particle is moving at speed $u$ in some reference frame, and a second reference frame is moving at speed $v$ in an orthogonal direction wrt the first, then the speed of the particle in the second reference frame is given by the formula in the OP.

Of course, if $u = c$, then $u' = c$ independent of $v$, as is clear from the formula, as the gamma-like term is $0$.

Hence, even if you set $v > c$, you still get $u' = c$.

The question, therefore, is whether this allows the second reference frame to move at greater than $c$ with respect to the first. This, of course, is ruled out more fundamentally.

 

[chaksome]

I think when you make an assumption，you should always consider that whether you break the more basic rule.

 

[vanhees71]

Concerning the kinematics a photon is characterized by its four-momentum, which is light-like, i.e.,
$$p^{\mu}=\begin{pmatrix} |\vec{p}| \\ \vec{p} \end{pmatrix}.$$
Since a photon is a single-frequency mode of the em. field the classical pendant is the wave vector of a plane em. wave,
$$k^{\mu}=\begin{pmatrix} |\vec{k}| \\ \vec{k} \end{pmatrix}.$$
The relation is the Einstein-de Broglie relation $p^{\mu}=\hbar k^{\mu}$.

The dispersion relation of em. waves dictates this, and it's a Lorentz-invariant property,
$$p_{\mu} p^{\mu}=0,$$
i.e., it holds in any inertial frame.

 

[jk22]

The question is more mathematically posed :

It seemed to me that usual vector addition of velocities should be replaced by relativistic addition formula given above. But then, I lack of mathematical knowledge to understand for example what replaces a basis, since for example the decomposition in a basis $e_x,e_y,e_z$ would be : $\vec{v}=(v_x e_x\oplus_\perp v_y e_y)\oplus_\perp v_z e_z\neq v_x e_x\oplus_\perp (v_y e_y \oplus_\perp v_z e_z)$.

So that the decomposition where not unique, and a star shape or wreath should appear, for example 12 arms if we consider the velocities are in 3 dimensions ??

Since relativistic Velocity addition is not a group operation, all the notions of vector space are to be modified ?

 

[Nugatory]

Since relativistic Velocity addition is not a group operation, all the notions of vector space are to be modified ?

It's not that the notions of vector space need to be modified, it's that the velocity three-vectors don't form a vector space so you can't use these notions at all.

Easier, much easier, to work with four-vectors. You can always recover the velocity three-vector when you're done by projecting the the four-vector onto the three dimensional space you're interested in.

 

[jk22]

For example if I have 2 4-vectors $p_i^\mu=(E_i/c,\vec{p}_i)$ how do I add them ?
...
Ah I got it we just write $\vec{v}\gamma_v=\vec{v}_1\gamma_{v_1}+\vec{v}_2\gamma_{v_2}$ if we consider the same particle having those components by $\vec{p}=\vec{p_1}+\vec{p_2}$ with $\vec{p_i}=m\vec{v}_i\gamma_{v_i}$ ?

 

[vanhees71]

The four-vectors are, as their name says, vectors. Their components follow the usual rules for vector addition, multiplication with scalars etc. They transform under Lorentz transformations as
$$p^{\prime \mu}={L^{\mu}}_{\nu} p^{\nu}.$$
Take a boost in $x$ direction with velocity $v=\beta c$. Then you have
$$(L^{\mu \nu})=\begin{pmatrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\ 0& 0 & 0 &1 \end{pmatrix}.$$
Thus you have
$$\begin{pmatrix} E'/c \\ p^{\prime 1}\\ p^{\prime 2}\\ p^{\prime 3} \end{pmatrix} = \begin{pmatrix}\gamma(E/c-\beta p^1) \\ \gamma(p^1-\beta E/c) \\ p^2 \\ p^3 \end{pmatrix}.$$
The inverse transformation is simply given by setting $\beta$ to $-\beta$ in the transformation matrix, i.e.,
$$\begin{pmatrix} E/c \\ p^{1}\\ p^{2}\\ p^{3} \end{pmatrix} = \begin{pmatrix}\gamma(E'/c+\beta p^{\prime 1}) \\ \gamma(p^{\prime 1}+\beta E'/c) \\ p^{\prime 2} \\ p^{\prime 3 }\end{pmatrix}.$$
From this you get the velocity-addition law by comparing the three-velocities (which are NOT the spatial components of a four-vector) in the original to that in the primed reference frame:
$$u^1=\frac{c}{p^0} p^1=\frac{c}{\gamma(E'/c+\beta p^{\prime 1})}(\gamma p^{\prime 1}+\beta E'/c)=\frac{c(p^{\prime 1}+\beta E'/c)}{E'/c+\beta p^{\prime 1}}=\frac{u^{\prime 1}+v}{1+\beta u^{\prime 1}/c}.$$
In the last step I've devided both the numerator and denominator by $E'/c$.

For the other two components you get
$$u^2=\frac{c p^2}{p^0}=\frac{c p^{\prime 2}}{\gamma (E'/c+\beta p^{\prime 1})}=\frac{u^{\prime 2}}{\gamma (1+\beta u^{\prime 1}/c)}$$
and analgously for $u^3$.

As you see, the addition theorem for the $\vec{u}$'s is very complicated (note that it's not even commutative!), while the four-vector transformations are quite simply given by the linear Lorentz transformation law with the corresponding matrix.