# Understanding how to derive the relativistic-relative velocity formula

• I
• JD_PM
In summary: 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."The proper velocity is just the component of the velocity that is in the direction of the proper time.
JD_PM
TL;DR Summary
I want to understand the derivation of the relativistic-relative velocity

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

JD_PM said:
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##.

JD_PM said:
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.

JD_PM
PeterDonis said:
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##.

JD_PM said:
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:
$$v_{rel} = \frac{P(E_1 + E_2)}{E_1E_2 + P^2}$$

JD_PM and vanhees71
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.

JD_PM, PeroK and Ibix
vanhees71 said:
So we finally have
$$v_{\text{rel}}=\frac{\sqrt{(u_1 \cdot u_2)^2-1}}{u_1 \cdot u_2},$$
In terms of rapidity, since
$u_1 \cdot u_2=\cosh\theta_{rel}$
we have
\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*}

JD_PM, vanhees71 and PeroK
vanhees71 said:
##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...

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 particle as measured in a Lorentz-boosted frame.

JD_PM
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.

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##.

JD_PM
vanhees71 said:
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.

$$v_{rel} = \frac{|\vec u_1|}{\gamma_1}=\frac{|\vec u_1|}{u_1^{0}}$$

vanhees71
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

$$\Big( \hat \beta \hat \beta^T \Big)^{j}_{ \ \ k}= \hat \beta^j \hat \beta^k$$

JD_PM said:
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.

JD_PM

## 1. What is the relativistic-relative velocity formula?

The relativistic-relative velocity formula is a mathematical equation that describes the relationship between the velocities of two objects in motion relative to each other in the context of Einstein's theory of relativity. It takes into account the effects of time dilation and length contraction at high speeds.

## 2. How is the relativistic-relative velocity formula derived?

The relativistic-relative velocity formula is derived from the principles of special relativity, which state that the laws of physics should be the same for all observers in uniform motion. It involves using the Lorentz transformation equations to convert between the measurements of velocity in different reference frames.

## 3. Why is it important to understand the relativistic-relative velocity formula?

Understanding the relativistic-relative velocity formula is important for accurately predicting the behavior of objects moving at high speeds. It is also crucial for many modern technologies, such as GPS systems and particle accelerators, which rely on the principles of relativity.

## 4. Can the relativistic-relative velocity formula be applied to all objects?

Yes, the relativistic-relative velocity formula can be applied to all objects, regardless of their mass or size. However, its effects are only noticeable at speeds close to the speed of light, which is why it is often used in the context of particles and spacecraft traveling at high speeds.

## 5. Are there any real-world applications of the relativistic-relative velocity formula?

Yes, there are many real-world applications of the relativistic-relative velocity formula. As mentioned before, it is used in GPS systems to accurately calculate the positions of objects on Earth. It is also used in particle accelerators to understand the behavior of subatomic particles at high speeds.

• Special and General Relativity
Replies
1
Views
715
• Special and General Relativity
Replies
3
Views
907
• Special and General Relativity
Replies
49
Views
5K
• Special and General Relativity
Replies
16
Views
2K
• Special and General Relativity
Replies
17
Views
3K
• Special and General Relativity
Replies
9
Views
4K
Replies
4
Views
758
• High Energy, Nuclear, Particle Physics
Replies
14
Views
2K
• Special and General Relativity
Replies
12
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
1K