Velocity addition formula question

[MrBillyShears]

In the Galilean formula u=u'+v, the velocities are bold so we know they're vectors, so for instance if u'=(a,b), and v=(c,d), we say that u=(a+c,b+d). But, in SR u=(u'+v)/(1+vu'/c²), none of the velocities are shown in bold, at least not not the way I saw it written. So, if we used the same two dimensional hypothetical vectors as up top, in SR terms, how would you find the added velocity u? Would it be (a+c,b+d)/(1+(ac+bd)/c²), or would you have to break it up into the individual x,y,z components and solve it for each one dimension component.

 

[Nugatory]

Check out the "non-parallel velocities" section of http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html and the wikipedia article on relativistic velocity addition.

 

[Chestermiller]

In special relativity, velocities are 4 dimensional rather than 3 dimensional. If you are talking about the velocity addition formula for 4 velocities, then you recover the Galilean formula. But, in practice, you got to be careful how you do it, by referring the velocities to the same 4D coordinate system.

Chet

 

[Meir Achuz]

in SR u=(u'+v)/(1+vu'/c),

That formula is only for {\bf u'} parallel to {\bf v}.
There is another formula for {\bf u'} perpendicular to {\bf v}.

 

[Bill_K]

For the general case, in which v₁ and v₂ are not co-linear, velocity "addition" is not commutative. In other words, the relative velocity vector V_AB of particle A as viewed from particle B is different from the relative velocity V_BA of B as viewed from A. (Not just a minus sign!) This is because the Lorentz transformations associated with v₁ and v₂ do not commute.

It may also be thought of as an aberration effect. Although the magnitude of the relative velocity will be the same, if |v₁| ≠ |v₂| the direction of V will be different depending on who is viewing whom.

It's simpler to express the result not in terms of the v's but rather their γ factors. For the co-linear case it's easy to start with V = (v₁ + v₂)/(1 + v₁v₂) and derive Γ = γ₁ γ₂ (1 + v₁v₂). For the non-co-linear case this generalizes quite simply to Γ = γ₁ γ₂ (1 + v₁·v₂)

 

[yuiop]

In the Galilean formula u=u'+v, the velocities are bold so we know they're vectors, so for instance if u'=(a,b), and v=(c,d), we say that u=(a+c,b+d). But, in SR u=(u'+v)/(1+vu'/c²), none of the velocities are shown in bold, at least not not the way I saw it written. So, if we used the same two dimensional hypothetical vectors as up top, in SR terms, how would you find the added velocity u? Would it be (a+c,b+d)/(1+(ac+bd)/c²), or would you have to break it up into the individual x,y,z components and solve it for each one dimension component.

Generally when the velocities are not parallel you need the individual components as they transform in different ways. Let us say there is an object A moving with velocity v relative to some reference frame S. Another object B moves with velocity u' as measured in the rest frame of object A. What is the velocity (u) of object B as measured in frame S?

For convenience, we can align the x-axis with the velocity vector v and resolve velocity vector u' into components (u'x, u'y. u'z). (Note that when all measurements are made from a single inertial reference frame that we can resolve or add velocity components in the usual Euclidean way. For a real object the magnitude of the combined components should never be greater than c, as mentioned in the parallel thread.) We can now express u' in the form of a four vector as [a0, a1, a2, a3] = [γ(u'), γ(u')ux', γ(u')uy', γ(u')uz'] where γ(u') represents 1/√(1-(u')²) = 1/√(1-(ux²+uy²+uz²)). Now we can perform a Lorentz boost of -v using Matrix multiplication and obtain a transformed 4 velocity [b0, b1, b2, b3] = [γ(u), γ(u)ux, γ(u)uy, γ(u)uz]. The 3 velocity of B relative to S is then u = (ux,uy,uz) = (b1/b0, b2/b0, b3/b0). The boost is -v so that after the transformation the velocity of object A relative to S is positive. The magnitude of the transformed 3 velocity u is √(ux²+uy²+uz²).

N.B. The four velocities are usually expressed in a more compact form as γ(u')[1, u'x, u'y. u'z] and γ(u)[1, ux, uy. uz)] where c=1.

 

[Bill_K]

Generally when the velocities are not parallel you need the individual components as they transform in different ways.

It's easier by far to use matrices rather than writing out all those individual components. A Lorentz transformation along the x-axis

x' = γ(x - vt)
t' = γ(t - vx)

can be written as a 4x4 matrix

\left(\begin{array}{cccc}\gamma&-\gamma v&0&0\\-\gamma v&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)

The three spatial directions can then be combined into a 3-vector,

\left(\begin{array}{cc}\gamma&-\gamma \mathbf{v}\\-\gamma \mathbf{v}&\mathbf{I} + (\gamma - 1)\mathbf{\hat{v}} \mathbf{\hat{v}}\end{array}\right)

where $\mathbf{\hat{v}}$ is the unit vector in the v direction.

For the problem at hand, just write $\mathbf{v_1}$ and $\gamma_1$ in place of $\mathbf{v}$ and $\gamma$, and apply the transformation to the other velocity 4-vector

\left(\begin{array}{c}{\gamma_2}\\\gamma_2 \mathbf{v_2}\end{array}\right)

For example the time component of the result is, as stated earlier,

\Gamma = \gamma_1 \gamma_2 (1 - \mathbf{v_1 \cdot v_2})

EDIT: For completeness, here's the space components, which are not as pretty:

$- \gamma_1 \gamma_2 \mathbf{v_1} + \gamma_2 \mathbf{v_2} + \gamma_2 (\gamma_1 - 1) \mathbf{\hat{v_1}} (\mathbf{\hat{v_1} \cdot \mathbf{v_2}})$

Note that the space components are not symmetric (or even antisymmetric!) under the interchange $\mathbf{v_1} \leftrightarrow \mathbf{v_2}$, which illustrates the important point that the "addition" of velocities is not commutative.

 

[Meir Achuz]

The simple answer for u' perpendicular to v is
{\bf u_\perp}=\frac{\bf u'_\perp}{\gamma(1+{\bf u'\cdot v}}.