# Relativistic relative velocity

Hi. I'm reading some quantum field theory and I'm a bit rusty in my relativistic kinematics. I stumbled across the formula

$$E_1E_2 v_{rel} = ((p_1p_2)^2 - m_1^2m_2^2)^{1/2}$$

where 1 and 2 are two collinearly colliding paritcles with their respective masses and $v_{rel}$ are their relative velocity. My question is; how is this relation derived?

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tom.stoer
what is the context? what do the indices 1 and 2 mean?

what is the context? what do the indices 1 and 2 mean?
Hi! 1 and 2 are two collinearly colliding paritcles with their respective masses and $v_{rel}$ are their relative velocity.

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ghwellsjr
Gold Member
Hi. I'm reading some quantum field theory and I'm a bit rusty in my relativistic kinematics. I stumbled across the formula

$$E_1E_2 v_{rel} = ((p_1p_2)^2 - m_1^2m_2^2)^{1/2}$$

where 1 and 2 are two collinearly colliding paritcles with their respective masses and $v_{rel}$ are their relative velocity. My question is; how is this relation derived?

Fredrik
Staff Emeritus
Gold Member
Hi. I'm reading some quantum field theory and I'm a bit rusty in my relativistic kinematics. I stumbled across the formula

$$E_1E_2 v_{rel} = ((p_1p_2)^2 - m_1^2m_2^2)^{1/2}$$

where 1 and 2 are two collinearly colliding paritcles with their respective masses and $v_{rel}$ are their relative velocity. My question is; how is this relation derived?
You should post a better reference. Name the book and the page number, and if possible, link directly to the page at Google Books.

ghwellsjr
Gold Member
Are you going to answer my question in post #4?

Are you going to answer my question in post #4?

Yeah of course. I thought that was clear from the title and the statement

"where 1 and 2 are two collinearly colliding paritcles with their respective masses and $v_{rel}$ are their relative velocity."

but yes, it is their relative velocity :)

jtbell
Mentor

"Du har enten kommet til en side som ikke kan visas, eller nådd grensen for hva du kan vise av denne boken." Anyway, you could try using v = p/E to find the velocities of the two particles, then apply the relativitistic "velocity addition" formula. I assume the units are such that c = 1.

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ghwellsjr
Gold Member
Yeah of course. I thought that was clear from the title and the statement

"where 1 and 2 are two collinearly colliding paritcles with their respective masses and $v_{rel}$ are their relative velocity."

but yes, it is their relative velocity :)
How are you defining and/or measuring their individual velocities?

Bill_K
E1E2vrel = ((p1p2)2−m21m22)1/2
It's a pretty formula, but I don't believe it. For slow velocities, the right hand side becomes imaginary.

robphy
Homework Helper
Gold Member
I think $p_1p_2$ is the dot-product of the two 4-momenta.
...in terms of components: $E_1 E_2 - \vec p_1 \cdot \vec p_2$, where the spatial dot-product is used.
...in terms of rapidities ["angles" in spacetime]: $m_1 m_2 \cosh(\theta_1-\theta_2) = m_1 m_2 \gamma_{12}$, where $\gamma_{12}=\frac{1}{\sqrt{1-v_{12}^2}}$ is in terms of $v_{12}=\tanh(\theta_1-\theta_2)$, the velocity of object-1 according to object-2, what I would call the "relative velocity" (see below).

So, the quantity under the radical sign on the right-hand side ( $((p_1p_2)^2 - m_1^2m_2^2)^{1/2}$ ) is non-negative, even for small velocities.

However, I think the formula in Mandl is incorrect for another reason.
(2nd ed) http://books.google.com/books?id=Ef4zDW1V2LkC&pg=PA129#v=onepage&q&f=false (p. 129, eq 8.9)
(1st ed) http://archive.org/details/IntroductionToQuantumFieldTheory (p. 185, eq 23)

The (proposed) equation in Mandl [for spatially-parallel 3-momenta according to us... i.e. the 4-momenta of the two particles and us are coplanar in spacetime]
$$E_1 E_2 v_{rel} \stackrel{?}{=} ((p_1p_2)^2 - m_1^2m_2^2)^{1/2}$$
translates into rapidities as \begin{align} (m_1\cosh\theta_1) (m_2\cosh\theta_2) v_{rel} &\stackrel{?}{=} ((m_1m_2\cosh(\theta_1-\theta_2))^2 - m_1^2m_2^2)^{1/2} \\ \cosh\theta_1 \cosh\theta_2 v_{rel} &\stackrel{?}{=} ((\cosh(\theta_1-\theta_2))^2 - 1)^{1/2} \\ \cosh\theta_1 \cosh\theta_2 v_{rel} &\stackrel{?}{=} \sinh(\theta_1-\theta_2) \\ v_{rel} &\stackrel{?}{=} \frac{\sinh(\theta_1-\theta_2)}{\cosh\theta_1 \cosh\theta_2 } \end{align}
However, I would have expected
$$v_{rel} \stackrel{expected}{=} \tanh(\theta_1-\theta_2) = \frac{\sinh(\theta_1-\theta_2)}{\cosh(\theta_1-\theta_2)} = \frac{\sinh(\theta_1-\theta_2)}{\cosh\theta_1\cosh\theta_2 - \sinh\theta_1\sinh\theta_2}$$
so that Mandl's formula should probably read
\begin{align} (E_1 E_2 - \vec p_1 \cdot \vec p_2)v_{rel} \stackrel{expected}{=} ((p_1p_2)^2 - m_1^2m_2^2)^{1/2} \\ (p_1 p_2)v_{rel} \stackrel{expected}{=} ((p_1p_2)^2 - m_1^2m_2^2)^{1/2} \end{align} in its simplest form.

The further clue that something is wrong with Mandl's formula is that
eq. 8.10a on p. 129, 2ed and eq. 24 on p. 185, 1ed
appears to describe "relative velocity" in the Galilean way as the difference of two velocities.
If there are special cases or approximations being taken, they are not obvious to me.

Did I make a mistake somewhere? in interpretation?

Bill_K
Very good! That looks right. (With the assumption included that v1 and v2 are collinear.)

robphy
Homework Helper
Gold Member
There must be more to this story because Weinberg discusses this in his Quantum Theory of Fields book: p.137 - p.139

p.139 ... it can take values as large as 2.

Aha!
I see what it is now. It's a terminology confusion.
Mandl's and Weinberg's "relative velocity" is what DaleSpam and others here at PF call "separation velocity"... literally v1-v2.

In terms of rapidities, Mandl's formula is:
\begin{align} E_1 E_2 v_{separation} &= ((p_1p_2)^2 - m_1^2m_2^2)^{1/2}\\ (m_1\cosh\theta_1) (m_2\cosh\theta_2) v_{separation} &= ((m_1m_2\cosh(\theta_1-\theta_2))^2 - m_1^2m_2^2)^{1/2} \\ \cosh\theta_1 \cosh\theta_2 v_{separation} &= ((\cosh(\theta_1-\theta_2))^2 - 1)^{1/2} \\ &= \sinh(\theta_1-\theta_2) \\ v_{separation} &= \frac{\sinh(\theta_1-\theta_2)}{\cosh\theta_1 \cosh\theta_2 } \\ &= \frac{\sinh\theta_1\cosh\theta_2-\sinh\theta_2\cosh\theta_1}{\cosh\theta_1 \cosh\theta_2 } \\ &= \tanh\theta_1-\tanh\theta_2 \end{align}

So, while the relative-velocity $v_{rel}=v_{12}=\tanh(\theta_1-\theta_2)$ is a scalar (a Lorentz-invariant quantity),
the separation-velocity $v_{sep}=v_{1}-v_{2}=\tanh\theta_1-\tanh\theta_2$ is not Lorentz invariant.
(As we know, of course, these two quantities are equal in the Galilean case, as well as Galilean-invariant.)

However, $\cosh\theta_1\cosh\theta_2 v_{sep}=\gamma_1\gamma_2(v_1-v_2)=\sinh(\theta_1-\theta_2)$ is a Lorentz-invariant, the "relative celerity". (See http://en.wikipedia.org/wiki/Proper_velocity )
Thus, $E_1 E_2 v_{sep}$ is a Lorentz-invariant... as Weinberg motivates.

Whew... hopefully this clears up the confusion, as well as answers the original poster.
[Ok, great... now back to grading.]

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Fredrik
Staff Emeritus
"Du har enten kommet til en side som ikke kan visas, eller nådd grensen for hva du kan vise av denne boken." 