# [ntp:questions] ntpd: time reset problem

Unruh unruh-spam at physics.ubc.ca
Wed Sep 16 16:32:37 UTC 2009

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>Unruh wrote:
>>> IIRC, the round trip time approaches 40 min when they are
>>> farthest apart?  With the worst case velocity difference
>>> approaching 121K mph (54KmpS)?  Thats approaching 65km
>>> difference in distance between when a message is sent,
>>> and the response is sent?
>>
>> So what? The signal goes from Mars to earth-- 20 min. On earth the clock
>> timestamps the receipt and the sending of the ntp packet. -- typically .00001 sec.
>> between those events. During that time the earth moves say 5 m.
>> Then going back the packet
>> takes 20min-5/3*10^8= 20min-.000000016 sec. Ie, the outward and inward
>> time delays are the same to .016usec. Who cares what the earth does
>> before or after it receives the packet? That the earth happens to be
>> 60000 km closer to mars when the packet arrives at mars is irrelevant.

>You are (for once) wrong here Unruh:

>To make it simple we can assume that the Earth is standing still and
>only Mars is moving (towards Earth):

Ah yes. relativity bites you in the tail again. This is exactly the
situation Einstein anaysed in 1905 to show that synchronization of time
depends on the frame of reference.
To figure out the time on mars, you need to assume that mars is at rest
not the earth.

>ntpd on Mars sends a request at (local) time t1, it is received and
>returned by Earth ot times t2, t3 (which are effectively the same), then
>received back on Mars at t4, right? This is standard ntp after all!

>However, due to the way Mars is getting closer all the time, the return
>path happens to be ~60000 km shorter, so (t4-t3) will be 2 ms less than
>(t2-t1).

Yup, and thus you find that IF you use the earth's point of view, you
are correct. But surely if you want the time on mars, yo uwant Mars's
point of view.

>This bias is something I'm sure Dave Mills' interplanetary version of
>ntp knows about, and can therefore correct for.

Nothing to correct. Mars at rest is the correct point of view for mars
time. There IS a correction that is needed and that is the timedilation
effect. Ie, the procedure will transfer earth time to mars, but since
earth is moving with respect to mars, earth time goes more slowly than
mars time, and that effect does need correction. Mind you it is not much
(at your assumption of 60Km/s relative velocity, that effect is .02PPM
difference in rate. It is quite possible that that is unmeasureable, but
it does make about .01 sec difference in a year.
)

>> You are probably trying to analyse things from the viewpoint of the
>> earth. That is called the "synchronization" effect in special
>> relativity. Yes, the midpoint of the receive/send time on earth and the
>> midpoint of the send/receive time on mars are mot the same according to
>> the earth bound observer, but they are to the Mars observer which is
>> what counts. (In special relativity things which are synchronized to one
>> observer, or not to a moving observer)

>IMHO, this is simply wrong.

Well, your humble opinion in this case does not accord with the way the
world works.

>Terje

>--
>- <Terje.Mathisen at tmsw.no>
>"almost all programming can be viewed as an exercise in caching"

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