I Two observers moving in opposite directions relative to each other

Sang-Hyeon Han

Hello guys. I have a question for one of postulates of relativity. Consider there are three observers (called A, B, and C) in x-direction only. A is at rest. B is moving to the left relative to A with velocity 0.7c. C is to the right relative to A with velocity 0.7c. Then when A sees B or C, they can not move faster than c. It's correct right? However, when B and C see each other, is it possible that their (relative) velocities are faster than c (maybe 1.4c) or not?

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.Scott

Homework Helper
No. When B and C look at each other, they will see the other moving away at a speed grater than 0.7c but less than c. It will be about 0.94c.

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Sang-Hyeon Han

No. When B and C look at each other, they will see the other moving away at a speed grater than 0.7c but less than c. It will be about 0.94c.
How can we get that value??

DaveC426913

Gold Member
How can we get that value??
Using the 'relativistic velocity addition' formula:

You will find that, no matter what values you use for the two spaceships, the receding velocity will not exceed c.

eg: If you set v=-0.7c and u' as 0.7c, the result will be 0.94c.

Note: this is not just for relativistic velocities. It will give accurate results at any speed (even stoned koala speed) it's just at - at anything less than relativistic velocities - the denominator becomes one plus (nearly) zero and we get the common v+u' velocity we know and love.

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Sang-Hyeon Han

Ahh I understand. many thanks guys!!

vanhees71

Gold Member
The relative speed of two objects is the speed of one object in the rest frame of the other object. It's best to work with manifestly covariant objects. In this case these are the proper four-velocities, which are in your case
$$u_A=\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad u_B=\frac{1}{\sqrt{1-0.7^2}} \begin{pmatrix}1 \\ -0.7 \\ 0 \\ 0 \end{pmatrix}, \quad u_C=\frac{1}{\sqrt{1-0.7^2}} \begin{pmatrix}1 \\ 0.7 \\ 0 \\ 0 \end{pmatrix}.$$
Now you don't need a Lorentz transformation to get the relative speed of each observer since you can get this with Minkowski products between these four-velocities.

For observer A, at rest in the computational frame, it's easy to see that you get the time-component of the four-velocities of B and C just by the Minkowski product of their four-velocities with $u_A$:
$$(\gamma_B)_A =u_A \cdot u_B=\frac{1}{\sqrt{1-0.7^2}}.$$
This is the $\gamma$ factor of B measured by A who is at rest in the computational frame. From the $\gamma$ factor you get back $\beta=|\vec{\beta}|=|\vec{v}/c|$ simply by
$$(\beta_B)_A=\sqrt{1-\frac{1}{(\gamma_B)_A^2}}=0.7.$$
That's trivial, but the magic of covariant treatments is that since it's working with invariants, it's a general valid formula, i.e., to get the speed of $B$ as measured by $C$ you simply calculate the $\gamma$ factor of $B$ as measured by $C$ via the Minkowski product of the four-velocities,
$$(\gamma_B)_C=u_C \cdot u_B=\frac{1}{1-0.7^2}(1+0.7^2)=149/51$$
and thus the relative speed
$$(\beta_B)_C=\sqrt{1-1/(\gamma_B)_C^2} \simeq 0.940.$$

"Two observers moving in opposite directions relative to each other"

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