# Measuring distances between a moving object and a stationary object using light

1. May 13, 2012

### peterspencers

Hi there
When measuring the distance between 2 objects, say, the earth and a spacecraft travelling very close to the speed of light. If I used a laser and a mirror to bounce a light beam between the 2 objects, I would take the time the light took to return to me, multiplied by the speed of light, to obtain my distance at the time the light pulse bounced back and was received.

If I performed this from both the earth and the spacecraft, and had synchronised the 2 measuring pulses to be activated at earths timeframe (so onboard the craft we use a special clock that can calculate earths timeframe by knowing how fast the craft is going) would the pulse released from earth and the one from the craft give the same distance measurement?

2. May 13, 2012

### Mentz114

You would use the round-trip time of the pulse times c and divide that by two. This is called the 'radar distance'.

If you also use Doppler to estimate the speed of the distant object, you can correct your radar measurement to get the distance.
I don't follow.

3. May 13, 2012

### yuiop

The answer is no. The spaceship will measure a shorter distance. If the pulse were sent simultaneously in the spaceship rest frame then the Earth would measure the shorter distance. It is only if the pulses are sent simultaneously in a reference frame that measures the Earth and spacecraft to equal speeds (but going in opposite directions), that the radar distances would be equal.

4. May 13, 2012

### peterspencers

I am confused as to how the 2 distance measurements are reverseable, given that the earth has had no acceleration present and no inertia, earth has not had its time dilated and therefore its spacetime will be different to that of the spacecrafts regardless of the reference frame. Also, by that logic, if the distances measured are reverseable by switching reference frames, surely the time dilation would be interchangable by switching referance frames, but surely that isnt the case as the returning ship 'will' find that more time has passed on earth? I apologise if this is wildly off key, I am very new to this concept and am trying to reconsile many confusing concepts, please bear with me.

5. May 13, 2012

### yuiop

Given that the Earth orbits around the Sun and the Sun orbits around the Galaxy, what makes you so sure that the Earth has no inertia?
Yes.
It is the case.
Everything changes when the spaceship turns around and returns to Earth. There is no single inertial reference frame where the spaceship is always at rest for the entire journey, but there is one for the Earth, so the situation is no longer symmetrical when the spaceship accelerates mid flight and returns to Earth.

6. May 14, 2012

### peterspencers

Ok, i think I'm nearly there, thankyou. Surely the spaceship turning round is also a reversable situation? As in the ships reference frame it is the earth that turns around and returns?

7. May 14, 2012

### yuiop

The ship is aware that it was the one that turned around because it can feel the acceleration or measure the acceleration with an accelerometer. This breaks the symmetry of the situation. The basic equations of SR assume an inertial reference frame, which is a frame in which no proper acceleration occurs. If the Earth and the rocket send timing signals regularly to each other, the spaceship sees an immediate increase in the frequency of the signals from Earth as it turns around while the Earth does not see an increase in the signal frequency from the rocket until much later. This is more evidence that the situation is non reversible.

8. May 14, 2012

### peterspencers

YES! Im with you, thankyou so much for taking the time to answer my questions so thoroughly. I am incredibly grateful, you have helped me immensely :)

9. May 15, 2012

### Austin0

Hi Your explanation here makes perfect sense. I have been trying to understand the Dolby -Gull presentation but have found it difficult

Consider an observer travelling on path $\gamma$
with proper time $\tau$. Define:
$\tau$+(x) ≡ (earliest possible) proper time at which a light ray
(technically, a null geodesic) leaving point x could intercept .
$\tau$-(x) ≡ (latest possible) proper time at which a light ray
(null geodesic) could leave $\gamma$, and still reach point x.

$\tau$(x) ≡ 1/2 ($\tau$+(x) + $\tau$-(x)) = ‘radar time’.

p(x) ≡ 1/2 ($\tau$+(x) - $\tau$-(x)) = ‘radar distance’.

I don't get their radar distance.
Why is the proper time going to the target subtracted from the return time?
Also I have found no reference to multiplying by c to obtain the distance. What am I missing?
I have searched but have found no, more elementary, treatment of radar time and its implementation.
Thanks

Last edited: May 15, 2012