Understanding how to derive the relativistic-relative velocity formula

  • I
  • Thread starter JD_PM
  • Start date
  • #1
734
68

Summary:

I want to understand the derivation of the relativistic-relative velocity

$$v_{\text{rel}}=\frac{1}{1-\vec{\beta}_1 \cdot \vec{\beta}_2} |\vec{\beta}_1-\vec{\beta}_2|.$$

Main Question or Discussion Point

I was studying how to derive the cross-section formula in the CoM frame from Mandl & Shaw QFT's book, and they state the following formula for the relative velocity (I'm going to use Vanhees71's notation though)

$$\omega_1 \omega_2 v_{rel} = [(p_1 p_2)^2 - m_1^2 m_2^2]^{1/2} \ \ \ \ (2)$$

Then the relative velocity in the CoM system follows:

$$v_{rel}=\frac{|\vec p_1|}{\omega_1}+\frac{|\vec p_2|}{\omega_2}=|\vec p_1|\frac{\omega_1+\omega_2}{\omega_1\omega_2} \ \ \ \ (3)$$

Then Vanhees71 pointed out that (2) is wrong.

At #9 Vanhees71 stated:

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$

'Your Eq. (2) is wrong. It's an invariant and thus you have the Minkowski product p_1 p_2 in the first term under the square root and not ##\vec{p}_1 \cdot \vec{p}_2##.

Eq. (3) cannot be right either, because the relative speed is covariantly defined as
$$v_{\text{rel}}=\frac{I}{p_1 p_2}=\frac{P\omega}{\omega_1 \omega_2+P^2}.$$
Indeed the general formula for the relative velocity, as derived in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Eq. (1.6.5), confirms the above formula since or collinear velocities ##\vec{\beta}_1=\vec{p}_1/E_1=\vec{p}/E_1## and ##\vec{\beta}_2=\vec{p}_2/E_2=-\vec{p}/E_2## the formula simplifies to
$$v_{\text{rel}}=\frac{1}{1-\vec{\beta}_1 \cdot \vec{\beta}_2} |\vec{\beta}_1-\vec{\beta}_2|.$$
In your formula (3) the factor in front of the "naive" formula for the relative velocity. Unfortunately this is wrong in some textbooks. I cannot check Mandl and Shaw, whether it's correct in there, but I guess not, given your Eq. (3).'

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$

I did not really understand why Mandl & Shaw formulas were wrong so I thought the best was going through Vanhees71's derivation first... And here we go! :)

The intention is to go through the whole derivation (Section 1.6, Relative Velocity), so let's start from the very beginning

Two particles have the following four-velocities with respect to an arbitrary inertial frame

$$u_1^{\mu} = \frac{1}{\sqrt{1-\vec \beta_1^2}} \ \begin{pmatrix}
1 \\
\vec \beta_1 \\
\end{pmatrix}, \ \ \ \
u_1^{\mu} = \frac{1}{\sqrt{1-\vec \beta_2^2}} \ \begin{pmatrix}
1 \\
\vec \beta_2 \\
\end{pmatrix}
\ \ \ \ (*)$$

Then the rotation-free Lorentz boost to the rest frame of particle 1 is given by

$$(\Lambda^{\mu}_{ \ \ \nu}) = \hat B( \vec \beta_1)=
\begin{pmatrix}
\gamma_1 & -\gamma_1 \vec \beta_1^T \\
\vec \beta_2 & 1_3 + (\gamma_1-1) \hat \beta_1 \hat \beta_1^T \\
\end{pmatrix} \ \ \ \ (**)$$

My first questions are:

1) I do not follow ##(*)##

I know what's the definition of proper velocity; it's simply the change of the spacetime coordinate ##x^{\mu}## per unit of proper time:

$$u^{\mu} = \frac{d x^{\mu}}{d \tau}$$

My point is that I'd expect to have the scalar factor multiplied by a ##4 \times 1## matrix instead of a ##2 \times 1## matrix.

2) I do not understand how to write down the rotation-free Lorentz boost as presented

I understand that the rotation-free Lorentz boost can be written in matrix form as follows

$$\begin{pmatrix}
\bar x^0 \\
\bar x^1 \\
\bar x^2 \\
\bar x^3 \\
\end{pmatrix}= \begin{pmatrix}
\gamma & -\gamma \beta & 0 & 0 \\
-\gamma \beta & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
x^0 \\
x^1 \\
x^2 \\
x^3 \\
\end{pmatrix}
\ \ \ \ (***)$$

But how can we go from Eq. (**) to Eq. (***)?

Sources: Introduction to Electrodynamics, Griffiths and Manuscript in Special Relativity Theory, Vanhees71

PS: please note I am not used to modern SRT notation so I may ask too many naive questions. I am also aware that Griffiths' Electrodynamics book is maybe not the best source to study SRT. If you have any book suggestions please feel free to share

Thank you :smile:
 

Answers and Replies

  • #2
PeterDonis
Mentor
Insights Author
2019 Award
29,719
8,973
I'd expect to have the scalar factor multiplied by a ##4 \times 1## matrix instead of a ##2 \times 1 matrix##.
It is a ##4 \times 1## matrix; the bottom three components are the three components of the 3-vectors ##\vec{\beta}_1## and ##\vec{\beta}_2##.

I understand that the rotation-free Lorentz boost can be written in matrix form as follows
That's assuming a boost in the ##x## direction. The other formula (**) is for a boost in an arbitrary direction. For the case you're studying, you need the latter, since the directions of the two velocity 3-vectors ##\vec{\beta}_1## and ##\vec{\beta}_2## might not be the same.
 
  • Like
Likes JD_PM
  • #3
PeterDonis
Mentor
Insights Author
2019 Award
29,719
8,973
The other formula (**)
Also, you wrote that formula down wrong. The ##\vec{\beta}_2## in the lower left of the matrix should be ##- \gamma_1 \vec{\beta}_1##.
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
13,506
5,992
Then the relative velocity in the CoM system follows:

$$v_{rel}=\frac{|\vec p_1|}{\omega_1}+\frac{|\vec p_2|}{\omega_2}=|\vec p_1|\frac{\omega_1+\omega_2}{\omega_1\omega_2} \ \ \ \ (3)$$

Eq. (3) cannot be right either, because the relative speed is covariantly defined as
$$v_{\text{rel}}=\frac{I}{p_1 p_2}=\frac{P\omega}{\omega_1 \omega_2+P^2}.$$
Equation (3) is for the separation speed in the CoM frame.

Assuming the particles are colliding, we can get the relative speed as follows. Let's take ##m_2## to be at rest. We need a boost of magnitude ##v_2## to get to the rest frame of ##m_2##, where ##v_2## is the speed of ##m_2##. Then we can apply the velocity transformation to ##m_1## to get the speed of ##m_1## in the rest frame of ##m_2##:
$$v_{rel} = \frac{v_1 + v_2}{1 + v_1v_2}$$
And using:
$$v_1 = \frac{|\vec{p_1}|}{E_1}, \ \ v_2 = \frac{|\vec{p_2}|}{E_2}$$
We get:
$$v_{rel} = \frac{(E_1|\vec{p_2}| + E_2|\vec{p_1}|)}{E_1E_2 + |\vec{p_2}||\vec{p_1}|}$$
Finally, if you take ##|\vec{p_1}| = |\vec{p_2}| = P## for the CoM frame, we get:
$$v_{rel} = \frac{P(E_1 + E_2)}{E_1E_2 + P^2}$$
 
  • Like
Likes JD_PM and vanhees71
  • #5
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,151
6,608
One should also note that if we are only after the magnitude of the relative velocity it's more efficient to formulate it in covariant form. By definition it's the velocity of one of the particles in the rest frame of the other. Let particle 2 be at rest. Then ##u_2=(1,0,0,0)## and ##v_{\text{rel}}=|\vec{u}_1|/\gamma_1##. Now ##u_1=(\gamma_1,\gamma_1 \vec{\beta})## and thus ##u_1 \cdot u_2=\gamma_1##. Further ##u_1 \cdot u_1=1=\gamma_1^2-|\vec{u}_1|^2## , from which ##|\vec{u}_1|=\sqrt{\gamma_1^2-1}=\sqrt{(u_1 \cdot u_2)^2-1}##. So we finally have
$$v_{\text{rel}}=\frac{\sqrt{(u_1 \cdot u_2)^2-1}}{u_1 \cdot u_2},$$
which then holds in any frame.
 
  • Like
Likes JD_PM, PeroK and Ibix
  • #6
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,633
891
So we finally have
[tex]
v_{\text{rel}}=\frac{\sqrt{(u_1 \cdot u_2)^2-1}}{u_1 \cdot u_2},
[/tex]
In terms of rapidity, since
[itex] u_1 \cdot u_2=\cosh\theta_{rel}[/itex]
we have
[tex]
\begin{align*}
|v_{\text{rel}}|
&=\frac{\sqrt{(u_1 \cdot u_2)^2-1}}{u_1 \cdot u_2}\\
&=\frac{\sqrt{(\cosh\theta_{rel})^2-1}}{\cosh\theta_{rel}}\\
&=\frac{|\sinh\theta_{rel}|}{\cosh\theta_{rel}}\\
&={\left|\tanh\theta_{rel}\right|}\\
\end{align*}
[/tex]
 
  • Like
Likes JD_PM, vanhees71 and PeroK
  • #7
734
68
##v_{\text{rel}}=|\vec{u}_1|/\gamma_1##
Naive question: why ##v_{\text{rel}}=|\vec{u}_1|/\gamma_1##?

I understand the relative-velocity formula PeroK wrote at #4, but how do you keep that gamma factor? They cancel out to me, as

$$v_{rel}=\frac{\gamma(dx -u_1 dt)}{\gamma (dt -u_1 dx)}$$

So I must be missing something...
 
  • #8
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,151
6,608
That's not ##v_{\text{rel}}## in the definition usually used. I gave the definition in #5 in a manifestly covariant form, and it's a scalar. Your expression is not a scalar. It's rather the not covariant three-velocity of a partical as measured in a Lorentz-boosted frame.
 
  • Like
Likes JD_PM
  • #9
734
68
Oh I see your point, thank you.

My issue is that I understand everything on your #5 but the reason why the relative velocity has the following form ##v_{\text{rel}}=|\vec{u}_1|/\gamma_1##.

Maybe I need more thinking.
 
  • #10
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,151
6,608
It's the magnitude of the three-velocity of particle 1 in the rest frame of paricle 2. In this frame thus ##v_{\text{rel}}=|\vec{u}_1|/u_1^0=|\vec{u}_1|/\gamma_1##.
 
  • Like
Likes JD_PM
  • #11
734
68
It's the magnitude of the three-velocity of particle 1 in the rest frame of paricle 2. In this frame thus ##v_{\text{rel}}=|\vec{u}_1|/u_1^0=|\vec{u}_1|/\gamma_1##.
Thank you vanhees71.

I've been reading (Griffiths, Introduction to Electrodynamics (latest edition) page 533) and I think I got it.

The (proper) velocity 4-vector ##u^{\mu}## is defined as follows

$$u^{\mu} := \frac{dx^{\mu}}{d \tau} \ \ \ \ (1)$$

Where ##dx## refers to covered distance as measured from the ground and ##d \tau## to proper time.

The three velocity vector then looks as follows

$$\vec u_1 = \gamma_1 \vec v_{rel} \ \ \ \ (2)$$

And the ##u^{0}## component reads as follows

$$u_1^{0}=\frac{d x^0}{d \tau}=\gamma_1 \ \ \ \ (3)$$

Thus combining (2) & (3) I get the magnitude of the three-velocity of particle 1 in the rest frame of particle 2.

$$v_{rel} = \frac{|\vec u_1|}{\gamma_1}=\frac{|\vec u_1|}{u_1^{0}}$$
 
  • Like
Likes vanhees71
  • #12
734
68
Alright, so regarding

$$(\Lambda^{\mu}_{ \ \ \nu}) = \hat B( \vec \beta_1)=
\begin{pmatrix}
\gamma_1 & -\gamma_1 \vec \beta_1^T \\
\vec \beta_2 & 1_3 + (\gamma_1-1) \hat \beta_1 \hat \beta_1^T \\
\end{pmatrix} \ \ \ \ (**)$$

What I do not understand is why ##\hat B_{22} = 1_3 + (\gamma_1-1) \hat \beta_1 \hat \beta_1^T##; specifically, what I do not see is why

$$(\gamma_1-1) \hat \beta_1 \hat \beta_1^T=
\begin{pmatrix}
\gamma_1 -1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix} \ \ \ \
$$

Where

$$\Big( \hat \beta \hat \beta^T \Big)^{j}_{ \ \ k}= \hat \beta^j \hat \beta^k$$
 
  • #13
DrGreg
Science Advisor
Gold Member
2,284
875
specifically, what I do not see is why

$$(\gamma_1-1) \hat \beta_1 \hat \beta_1^T=
\begin{pmatrix}
\gamma_1 -1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix} \ \ \ \
$$
That is true only in the special case when $$
\hat \beta_1 =
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix} \ \ \ \
$$
i.e. motion along the ##x##-axis.
 
  • Like
Likes JD_PM
Top