# One clock on Earth, one on Mars

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1. Jan 3, 2012

Let's say we have two clocks that are synchronous here on Earth.

One of the clocks is instantly transported to Mars.

What time does the clock on Mars read? Is it ticking slower or faster than the one on Earth?

I want to talk about SR here and not the gravity factor of GR.

Let's say the clocks were synchronized at time 0 and when time was equal to 1 second one clock arrived on Mars and the other stayed on Earth.

Does the clock tick depend on the speed of mars?

2. Jan 3, 2012

### bm0p700f

The weaker gravitational potential on the surface of mars due to its increased distance from the sun and the planets lower mass will cause the clok to run slower. Mars also orbits at a lower velocity relative to earth and roatates at a slower rate. So for all these reason the clock on Mars will run a bit faster than the one on earth. Not by much though.

3. Jan 3, 2012

If we ignored the gravity factor, how would we figure out the time?

I really want this example to be based on SR and not on GR.

Would we have to choose a frame of reference such as the Earth?

4. Jan 3, 2012

### Snip3r

is this the time on earth clock or mars clock? If earth, the clock on mars will show less than 1 second else earth clock shows more than 1 second

5. Jan 3, 2012

### MikeLizzi

You mean faster don't you?
The deeper you are in a gravitational well, the slower your clock. So the guy whose clock is not as deep (Mars) has the faster running clock.

6. Jan 3, 2012

### Passionflower

What do you mean by instantly?

7. Jan 3, 2012

### bm0p700f

t$^{'}$ = t$_{0}$/sqrt(1-(v$^{2}$/c$^{2}$))

Is what you need to caluate time dialation using SR. However becuase Mars rotates your velocity is constantly changing therefore the ammoun t of time dialation you experience changes. I am not sure if the roatation effects cancel out as I have not done the sums.

However the effect of gravity is more significant I think and therefore you need

t$^{'}$ = $\frac{t_{0}}{\sqrt{1-\frac{2gR}{c^{2}}}}$

Last edited: Jan 3, 2012
8. Jan 3, 2012

### Jorrie

For what you asked, it is better to consider a case where identical atomic clocks are situated on Earth and on Mars. Also consider both to be rate-compensated for the different gravitational potentials that they find themselves in, but that they are not rate-compensated for the differences in orbital speeds. Then, since Mars has a lower orbital speed than Earth, one would expect the Mars clock to slowly gain time on the Earth clock.

How to actually measure such a difference is another story.

9. Jan 3, 2012

### ghwellsjr

If you want to ignore the relativity effects of gravity then I think your best bet for understanding the SR effects of these two clocks is to use a reference frame in which the Sun is at rest and the Earth and Mars are in motion around the Sun. We can also ignore the surface speed of these planets because it is so much smaller than the motion around the Sun.

The Earth travels a little faster than Mars does around the sun which means that a clock on Earth will run slightly slower than a clock on Mars.

You start with two identical clocks on Earth set to zero. Then one of these is accelerated very quickly and decelerated to bring it to rest on Mars within one second (we don't want to do it instantly because that would violate SR). During that one second, of course the clock on the Earth will advance by one second but the clock going to Mars will run so slowly that it will only advance by a very small fraction of a second.

So right off the bat, the clock on Mars is behind the clock on Earth by almost a second. But then the clock on Earth runs slightly slower then the one on Mars, so if we wait long enough the clock on Mars will eventually catch up to the clock on Earth and from then on the clock on Earth will remain behind the clock on Mars.

10. Jan 3, 2012

### Snip3r

but if my FoR is earth, does mars clock continue to run slow? For a moment consider this assume earth and mars are in the same orbit (not concentric circles, just to avoid confusion and i think this has no effect)and earth as FoR. Now when clock B is instantaneously transported to mars it shows less than 1 second after reaching mars(clock A on earth shows 1 sec). Now as the speed of mars is less, according to earth clock B is returning. isn't this similar to twin paradox? so clock B continue to tick slower even after reaching mars?what am i missing?

11. Jan 4, 2012

### LuckyNate

just a quick reminder...what time the clocks both read depends on the observer's distance from the clocks, because the light from the clock's 'face' take time to reach the observer...there is no one frame of reference from which to determine simultaneity.
if you were equidistant from both clocks that will give the closest to an accurate reading but still the time between any two clocks not moving together in the same referential frame will deviate in their readings

12. Jan 4, 2012

### Snip3r

no, the clock is not actually looked at. the time on a clock, given its velocity is calculated using equations.

13. Jan 4, 2012

### LuckyNate

Then what is the discussion about? get out your calculator and do the math if you know how, i thought you wanted to discuss sr effects.
You can use a light clock (a photon bouncing between two mirrors) for thought experiments, this gives a better visualization in the head as you can imagine the zig zag line traced out by the photon in the light clock as the two mirrors travel around with the surface of each planet. Whichever clock has the widest angle of reflection is moving faster through space and therefore more time is passing between each 'tick'.
As far as calculating the amount of difference between the clock i imagine that doing the calculation would be challenging and exciting but I am no mathematician, and the only way to confirm your results would be to double check your work, and then do an experiment to see if your results bear out, a daunting task to say the least but, I'll bet that that the real students of physics out there can answer this accurately for extra credit in one of their courses.

14. Jan 4, 2012

### ghwellsjr

You can analyze any problem using any FoR and you will always get the same answers (that matter). If my analysis shows that the clock on Earth runs slower than the clock on Mars, how can any other analysis show otherwise?

The problem with using the Earth as a FoR is that it is not inertial and you have very complicated patterns of motion for Mars relative to Earth but if you're thinking that since the Earth is at rest in this non-inertial FoR and that Mars is always in motion and therefore the only one experiencing time dilation then you should consider this:

Let's think about the Sun and the Earth. Since the Earth is always in motion relative to the inertial rest frame of the Sun, it experiences a non-reciprocal time dilation that the Sun does not experience. This is an example of the twin paradox which is similar to the one that Einstein introduced in his famous 1905 paper introducing Special Relativity. Since the Earth is not inertial, you cannot use the Earth as a FoR and say that the Sun is in orbit around the Earth and conclude that the time dilation is reciprocal and applies to the Sun and not to the Earth.

So in a similar way, you cannot use the non-inertial rest frame of the Earth and say that the time dilation is all on Mars.
I'm afraid you've lost me here. If Earth and Mars are in the same orbit (the same distance from the Sun), then they will experience the same time dilation and their clocks will remain about 1 second apart.
The speed of mars is less than what? If you are using the Earth as a FoR, then the speed of mars is more than earth since earth is at rest. Please explain what you are thinking of here.

Then you say "according to earth clock B is returning"--but from where? I have no idea what you are proposing here. But whatever it is, would it be the same if you had concentric orbits instead of the planets being in the same orbit?

Clock B ticks slower (practically stops) during its high-speed one-second trip to Mars but as soon as it comes to rest on Mars, its tick rate goes almost back to normal but faster, not slower, than the Earth clock.

I'm not sure what you are missing because I'm missing an understanding of what you are proposing.

15. Jan 4, 2012

### ghwellsjr

You are correct that looking at distant clocks will result in differences in times from any concept of simultaneity, but we can rely on the fact that in this case, approximately every two Earth years, Mars will be at its closest approach to Earth and we can compare the two clocks to see how they change every two years and we will find that the Mars clock will gain on the Earth clock. Of course, we can't do this in reality because this analysis ignores the effects of gravity.

16. Jan 4, 2012

### Snip3r

i had this in mind. as mars revolves slowly in earth FoR i considered clock B as coming back to earth.and what is the below comment i dont get how

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17. Jan 4, 2012

### ghwellsjr

But if you move Mars to the same orbit as Earth, then it won't travel slower than the Earth like it does at its orbital distance from the Sun. It will travel at the same speed and remain the same fixed distance from Earth. In the non-inertial rest frame of the Earth, Mars is also at rest so the two clocks would tick at exactly the same rate.

Did you think maybe that Mars orbits the Sun slower than the Earth because it is smaller?

18. Jan 4, 2012

### Snip3r

yes you are correct i forgot the elementary aspect
GMm/r$^{2}$=mV$^{2}$/r velocity depends on r and m cancels out but imagine this what if mars had some kind of propulsion mechanism which keeps it velocity below the velocity of earth?now what happens in earth's FoR? and previous diagram holds

19. Jan 4, 2012

### ghwellsjr

It's always challenging to figure out what happens in a non-inertial FoR which is what earth's FoR is. Why do you want to torture yourself like that? Just work it out in the Sun's inertial FoR like your diagram shows and it will be super simple.

Are you are proposing that the Earth and Mars are in the same orbit but Mars has a rocket that slows it down (which would then actually take it out of Earth orbit, but we'll overlook that) with the idea that Mars will slowly get closer to the Earth until they meet? And you're wondering if this is like the Twin Paradox? Sort of, the clock that went to Mars and returned will have less time on it but mainly because it got most of that time difference during its high speed trip to Mars. On the slow speed return, even if it is traveling slower than earth, it would never be enough to overcome the initial one-second loss of time.

But to be precise, in the Twin Paradox, to make the solution obviously simple, one of the twins remains inertial. We usually call that the Earth twin but we have to pretend that the Earth is inertial and not orbiting the Sun. If both twins experience acceleration, then we have to carefully consider the velocity profile of each twin to determine which one is younger when they reunite.

20. Jan 4, 2012

### Snip3r

yes you are correct the problem becomes trivial when sun is chosen as the frame
yes!
precisely...
you got me!!
but what i also assumed is all these happen in a duration within which the path traced by the earth is almost a straight line so no acceleration and qualifies as inertial frame so whats wrong here?