# Given 2 objects in space, which is moving faster?

1. Aug 30, 2015

### zmqp

I have been unable to find a clear explanation to this basic question which is at the core of my lack of understanding the relativity. Everything I've found is either too general, or too technical. Please excuse my simplifications and lack of technical language.

As I understand it (in very simple terms), given two objects, the one which is moving faster relative to the other will experience time going by at a slower pace. What I don't understand is, given two objects in space, how can one be said to be moving faster or slower RELATIVE to the other? Wouldn't they both be moving at equal and opposite velocities RELATIVE to one another?

I don't understand. Can someone please shed some light, or direct me towards some reference material to help me with this? Thank you in advance.

2. Aug 30, 2015

### Orodruin

Staff Emeritus
This is not correct. There is no such thing as an object "moving faster" than another object - it all depends on the inertial frame. In particular, in the rest frame of object A, object B will be moving faster and vice versa.

Yes they would. What you might be missing is the relativity of simultaneity, which is what implies that each object would consider time to be moving more slowly for the other.

3. Aug 30, 2015

### mathman

4. Aug 30, 2015

### BvU

Hello zm, welcome to PF !

It is rather funny you are asking the one question that cannot be asked ! Not to worry, it happens a lot (to me too at some time) because we have our own built-in frame of reference. The key thing about relativity is exactly that all uniform motion is relative. No one can say "my frame of reference stands still" and back it up with a decisive experiment.

For objects A and B with their inertial reference frames a and b "slower or faster relative to the other" cannot be said. "slower or faster" cannot be said either. The only thing that holds up is something like "slower or faster relative to a third inertial reference frame c".

You'll be fine. Keep asking questions like this.

5. Sep 1, 2015

### vanhees71

The relative velocity of particle 2 to particle 1 is defined as the velocity of particle 2 in the restframe of particle 1.

The most easy way to obtain the relative velocity is to work with the four-velocities, which are four-vectors in Minkowski space,
$$u_1^{\mu}=\begin{pmatrix} \gamma_1 \\ \gamma_1 \vec{v}_1/c \end{pmatrix},\quad u_2^{\mu}=\begin{pmatrix} \gamma_2 \\ \gamma_2 \vec{v}_2/c \end{pmatrix}.$$
The Lorentz boost to the restframe of particle 1 is given by
$${\Lambda^{\mu}}_{\nu} = \begin{pmatrix} \gamma_1 & -\gamma_1 \vec{v}_1^T/c \\ -\gamma_1 \vec{v}_1/c & \mathbb{1}_3 +\frac{\gamma_1-1}{\vec{v}_1^2} \vec{v}_1 \otimes \vec{v}_1 \end{pmatrix}.$$
The four-velocity of particle 2 in the restframe of particle 1 is thus after some algebra given by
$$\tilde{u}_2^{\mu} = \frac{\Lambda^{\mu}}{\nu} u_2^{\mu} = \begin{pmatrix} \gamma_1 \gamma_2 (1-\vec{v}_1 \cdot \vec{v}_2/c^2) \\ (\vec{v}_2-\vec{v}_1)/c + (\gamma_1-1) [(\vec{v}_1 \cdot \vec{v}_2)/\vec{v}_1^2-1] \vec{v_1}/c \end{pmatrix}.$$
Finally the relative three-velocity is given by
$$\vec{v}_2^{(\text{rel})}=\frac{\tilde{\vec{u}}_1}{u_2^0} c=\frac{\vec{v}_2-\vec{v}_1}{\gamma_1(1-\vec{v}_1 \cdot \vec{v}_2/c^2)} - \frac{\gamma_1-1}{\gamma_1} \vec{v}_1.$$
The relative speed is easier to calculate directly from the temporal argument of $\tilde{u}_2^{0}=\gamma_2^{(\text{rel})}$
$$\left |\vec{v}_2^{(\text{rel})} \right|=c \sqrt{1-\frac{(c^2-\vec{v}_1^2)(c^2-\vec{v}_2^2)}{c^2-\vec{v}_1 \cdot \vec{v}_2}}.$$
As you see by exchanging the particle labels 1 and 2, the relative speed of particle 1 wrt. particle 2 is the same as the relative speed of particle 2 in the restframe of particle 1. Note, however that generally $$\vec{v}_1^{(\text{rel})} \neq -\vec{v}_2^{(\text{rel})}$$.