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View Full Version : An issue I'm having with time dilation and other effects of speed


1MileCrash
Nov3-10, 08:02 PM
Here's my issue.

We will say that from our reference Object A is stationary and Object B is moving.

Object A witnesses Object B traveling 6/7 the speed of light, thus Object B's "onboard clock" is running half as fast as Object A's, though it appears normal to Object B. If Object B does this for 15 of "its" minutes, 30 of A's minutes have passed.

But this seems to violate the whole basic idea of special relativity, that know one really knows who is moving or who is stationary. If we know who's clock slowed than we know who moved, and that is impossible because there is no "absolute motion."

From Object B's reference, Object A is the one moving at 6/7 the speed of light, thus Object A's onboard clock should run half as fast as Object B's from it's reference.

If Object A's clock is twice as fast as Object B's clock from its reference, and object B's clock is twice as fast as Object A's from it's reference, if they "met up later" shouldn't their clocks read the same thing?

Please help me understand, from what I have learned so far is that there is essentially no objective difference between object A moving from object B and vice-versa.

DaleSpam
Nov3-10, 09:22 PM
if they "met up later" shouldn't their clocks read the same thing?This is called the twin paradox, and it cannot happen for two inertial observers.

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

ghwellsjr
Nov3-10, 09:54 PM
It depends on how they meet up again. If they both accelerated identically toward each other and then deccelerated identically to come to rest at the same location, then they will have aged the same amount, (assuming that they started at the same point at the same age), but in the classic Twin Paradox, only one accelerates, and that is the one who ages less. Be forewarned though, you have to be careful how you specify identical accelerations. The easiest way to illustrate and understand this is to specify the frame of reference where the two observers start off traveling in opposite directions at the same speed and at the same age, then they come back together traveling with identical accelerations and speeds. (When viewed this way, the two observers are always the same age.)

granpa
Nov4-10, 02:14 AM
its all to do with relativity of simultaneity.

ghwellsjr
Nov4-10, 02:40 AM
Yes, that's true if you wanted to transform 1MileCrash's original scenario into the frame I was proposing, but I was suggesting that we cheat and start over by defining a new scenario that is, in principle, the same as his, but with different details so that we don't have to do the hard work of transforming.