# How can we measure relative velocity of fast objects when there is time dilation

First of all sorry about the question title. I find it very difficult to ask this in less than 150 characters. I am not a physicist and only have basic understanding of advanced physical concepts.

So here is the detailed question:

Scenario:

1. There are two people. PA and PB.
2. A and B stands in some point in space in close proximity with 0 accelaration (constant velocity)
3. PB starts to accelerate away from PA in a spaceship and reaches very high velocity (eventually reaches 0.5c)

Assumptions: (please correct if these are wrong)

1. PA measures that PB is travelling at 0.5c
2. PA finds that PB's clock is moving slower than his (and vice-versa)
3. PA finds that objects in PB's ship are moving in slow motion

Question: If PA finds that PB, his ship and everything else in it moving slower(relative to PA's own clock), how come PA is still able to measure PB's velocity as 0.5c using PA's clock? Shouldn't every atom in PB and his ship feel like moving slower to PA?

Does lorentz contraction play any part in this?

• Possible duplicate of How can time dilation be symmetric? – John Rennie Oct 12 '18 at 5:50
• @JohnRennie I don't think so. I am not asking about the symmetry of time dilation. The question is: how does an object(the spaceship) moving at 0.5c wrt to an inertial frame of reference appear/measured to be 0.5c when everything in it appears to move slower to an a stationary observer in the same inertial frame of reference? I am aware that time dilation due to relative velocity (without acceleration) is symmetric. – Alwin Tom Oct 12 '18 at 6:21
• The relative motion of the frames of reference causes the time dilation effects observed in the other frame. Cause -> Effect, not the other way around. – JMLCarter Oct 12 '18 at 6:28
• @JMLCarter Could you elaborate? (like I said, I am not very good with advanced physical concepts) So you are saying that time dilation is an effect of the measured relative velocity. ie the amount of time dilation experienced by PA is proportional to the measured velocity between PA and PB wrt to PA. That makes sense. – Alwin Tom Oct 12 '18 at 6:36
• If you measure the speed of me relative you, why should you care about what my clock says? You only have to use your clock and your rules. – md2perpe Oct 12 '18 at 20:07

I think there are some things you are confusing in your question. Time dialation is measured between the reference frame where $$P_a$$ Is still, and the reference frame where $$P_b$$ Is still, and the second reference frame is moving relative to the first at a constant velocity of $$0.5c$$. The “ship” is not moving in this second reference frame, because the second frame is chosen so that it goes along with the ship, the ship has $$0$$ velocity in this frame, so time dialation would not play a role when measuring its velocity relative to the $$P_a$$ frame. If any object is moving in the ship, so it is moving in the reference frame where $$P_b$$ is stationary, then time dialation and space contraction should be used when relating the motion measured in the “frame $$P_b$$” to the ones measured in the “frame $$P_a$$“, but other than that, everything is defined properly, and velocities are measured appropriately.

• Thank you for the answer. I do have one more question. Assuming that the ship has glass walls (transparent) and Pa has some mechanism to see the inside of this ship, how would Pa perceive a ball that is moving up and down inside the ship at a fixed rate? Does Pa feel like the ball is moving slower (when he compares its movement in relation to the ship's body) compared to Pb? – Alwin Tom Oct 12 '18 at 11:00
• Yes, in the vertical direction the ball would be seen slower. You can see that from simple time dialation $\Delta t_a=\frac {\Delta t_b}{\sqrt {1-\frac {v^2}{c^2}}}$, and because the motion is perpendicular to the relative velocity of the frames, the vertical distance is the same in both frames, call it D. The velocity in the B frame is $V_b=\frac D{\Delta t_b}$ so the vertical velocity as measured from $P_a$ is simply $V_a=\frac D{\Delta t_a}=\frac D{\Delta t_b}*\sqrt {1-\frac {v^2}{c^2}}=V_b*\sqrt {1-\frac {v^2}{c^2}}$, which is less than $V_b$ – Hugo V Oct 12 '18 at 13:29

Lets try to see how a measurement of the velocity of PB from PA can be made?

In astronomy a method that gives the velocity of the star with respect to the sun is by using the line spectra of known elements:

Analyzing the spectrum of a star can teach us all kinds of things in addition to its temperature. We can measure its detailed chemical composition as well as the pressure in its atmosphere. From the pressure, we get clues about its size. We can also measure its motion toward or away from us and estimate its rotation.

......

When we measure the spectrum of a star, we determine the wavelength of each of its lines. If the star is not moving with respect to the Sun, then the wavelength corresponding to each element will be the same as those we measure in a laboratory here on Earth. But if stars are moving toward or away from us, we must consider the Doppler effect (see The Doppler Effect). We should see all the spectral lines of moving stars shifted toward the red end of the spectrum if the star is moving away from us, or toward the blue (violet) end if it is moving toward us (Figure 2). The greater the shift, the faster the star is moving. Such motion, along the line of sight between the star and the observer, is called radial velocity and is usually measured in kilometers per second.

Let us suppose that the spaceship moving at .5c with respect to PA shines a specific line spectrum at intervals, which by its doppler shift will allow PA to know the velocity of PB. A predetermined definition of the interval of shining the spectrum would give the time dilation

PA finds that PB's clock is moving slower than his (and vice-versa)

PA observes that the signal with PB's clock are larger, because of the time of arrival of the prearranged light interval.

PA finds that objects in PB's ship are moving in slow motion

So? Suppose it was arranged that the signals would come every 1 hour, and PA observes them arriving at much larger intervals. This means that the clock of PB on the ship with respect to the clock of PA runs slower, consistent with the Lorenz transformations.

Question: If PA finds that PB, his ship and everything else in it moving slower(relative to PA's own clock), how come PA is still able to measure PB's velocity as 0.5c using PA's clock?

I described a measurement above, using known atomic spectral lines.

Shouldn't every atom in PB and his ship feel like moving slower to PA?

Sure, one trusts the mathematics of Lorenz transformations and does not need to measure every atom to know that time intervals on PB will be slower than for PA.

• What if the ship is revolving around Pa at 0.5c(perpendicular component of velocity). Will there be any doppler shift in this situation? – Alwin Tom Oct 13 '18 at 14:13
• if there is a relativistic component of the velocity vector on the line joining the ship to PA, yes. – anna v Oct 13 '18 at 14:28