# A Test of Special Relativity

Discussion in 'Physics & Math' started by MacM, Jun 14, 2005.

1. ### funkstarratsknufValued Senior Member

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But that is exactly the point! All the coordinate points in the orbiter frame are pressed closer in the orbiter frame relative to the earth frame, and this is exactly why they both agree that the orbiter clock is ticking faster! This is why everonewill agree that the orbiter clock is ticking faster than the earth clock: No matter where you are, if you compare an orbiter meter to an earth meter, the orbiter meter will be shorter.
I've understood that all along, and never disputed it.
Ok, rephrase: Is the orbiter clock's reference frame the same as the earth clock's reference frame? Remember that we're assuming no relative velocity...

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3. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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by funkstar:

"But that is exactly the point! All the coordinate points in the orbiter frame are pressed closer in the orbiter frame relative to the earth frame, and this is exactly why they both agree that the orbiter clock is ticking faster! This is why everonewill agree that the orbiter clock is ticking faster than the earth clock: No matter where you are, if you compare an orbiter meter to an earth meter, the orbiter meter will be shorter."
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OK, here is where I am having trouble understanding what you mean. The coordinates
in an orbiter frame extend to infinity in all directions, just as the coordinates in an Earth frame. The GRID with coordinate points marked off on it in meters extends all the way to infinity in every direction also. A grid eminating from a specific location, on the
orbiter for instance, should seemingly portray a meter spacing that
continually gets RELATIVELY longer as the grid nears the Earth. An observer at this specific location, will measure a meter as a meter, but the meter gets relatively longer as it nears the Earth. An observer located on the Earth will measure a meter as a meter at his location on Earth, but the meter will get relatively shorter as the coordinates on the same grid are pressed closer together climbing out of the gravity well. In your explaination, you seem to imply there are only two meter lengths on the grid, not a progression of lengths as a meter rises and falls along the grid with gravitational potential. If you had said the OBSERVER on the orbiter would see the
meter on the Earth's surface as relatively longer than his, only the two lengths would need to be compared. But by using a grid marked off in meters, checkerboard style, it
is a progression of relative lengths of meters. Using the phrase 'pressed together' to represent a relatively longer meter in reduced gravity field is kind of counterintutive also. I tend to associate 'pressed together' with more intense gravity and a shorter meter as the meter falls in a gravity well, not a longer meter as it falls. I do realize
an object is 'stretched' or 'spaghettified' as it falls into a black hole as seen by a
distant observer, but saying it is pressed together as it escapes an accretion disk just doesn't sound right for some reason, hehe!
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by funkstar:

Ok, rephrase: Is the orbiter clock's reference frame the same as the earth clock's reference frame? Remember that we're assuming no relative velocity...
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You do seem to use unconvential phrases when talking physics. My explaination was
conventional and clear. No, the 'orbiter clock's reference frame' is not the same as the
'Earth clock's reference frame'. But the earth clock is in the orbiter clock's reference
frame and the orbiter clock is in the earth clock's reference frame.

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5. ### geistkieselValued Senior Member

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Let me try my hand at understanding the special theory of relativity as it applies in this instance. Using the twin aging scenario we see one twin going out and back at the SOL for 10 years round trip earth time. This is the one we refer to as the "younger". There is no question of "twin paradox" , this is another matter.

Any way frames contract in the direction of motion of the moving frame. Clock rates slow down but what of the motion of biological activity transverse to the direction of motion? Would the biolological activity moving transverse to the direction oif motion of the frame age more, perhaps the same as her brother back on earth, periodically anyway?

In other words would the body of the younger twin be slightlty younger in someplaces (and directions in those places) and slightly the same as twin bro in other places?

If we kept her constantly rotating could we zero out the aging gradient she would adopt if she just sat staring at the front of the front of the ship for 10 years earth travel time? But wouldn't this just lower her youngest possible and raise her oldest possible limits to aging? By rotating her for ten years I mean? Could she be satisfied with this? What do you think?
Geistkiesel​

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7. ### funkstarratsknufValued Senior Member

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That's not what I meant. The grid is directly related to the orbiter observer, and marks off his, and only his, meter. If the grid were to get spaced out, it would imply that the orbiter sees things getting contracted, which is the opposite of what happens.
Actually, there's only one meter length on the grid, and that is the meter of the orbiter observer. The distance between any two adjecent points on the grid will measure out a meter to the orbiter observer, and to him only.
I did, and I used the rubber sheet analogy to explain it.
I don't find it counterintuitive at all. In fact, since light is blueshifted falling into a gravity well, I think compression is a perfect analogy. I think the confusion arrives because you seem to think that the grid marks off the local meter. It doesn't, it only marks of the orbiter's meter, and compared to the local meter, going into the gravity well, the orbiter meter appears smaller and smaller in comparison.
Ok, perhaps we should just abandon that particular analogy, because we don't agree on what it says. The important thing is whether you've understood why gravitational redshift makes the orbiter clock tick faster than the earth clock.
When I say that two observers are in the same frame, I mean that they agree on spacetime. E.g. in SR, two observers would be (qualitatively) in the same frame, if they have no relative velocity, and would agree on the time and order of events, lengths etc. But it doesn't matter, just so long as you understand that they have different frames, and that these two frames don't coincide: They don't agree on time (and therefore length).

[EDIT: There's another reason we should abandon the analogy - it's probably wrong. I'm no expert on GR, and I think I've misunderstood this particular effect...]

[EDIT 2: Yep, I was wrong.]

Last edited: Jun 27, 2005
8. ### wesmorrisNerd Overlord - we(s):1 of NValued Senior Member

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I'm just thinking about the geometry of space-time:

The materials that comprise your body occupy space time. It isn't all in the same point in space time.

Since they are not all at the same point, each atom travels in its own reference frame through space time.

To my primitive comprehension of the issue, this implies directly that no two points of interest (like an atom) are truly simultaneous. It's a matter of the accuracy of your clock as to whether or not you can discern the difference in time between any two points.

Maybe I've got it all wrong, which is why I was asking for input.

9. ### geistkieselValued Senior Member

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2,471
I would only say, in general that no two events are simultanoeus if occuring separate from each other. When two events, like the motions of two phtons arrive some place they arrive simuiltanously,. though if arriving in separate locations in general the events would not be simultanous if measured from another frame moving with respect to the one where the events occured.
SRT places much emphasis on what observers see, or think they perceive. when observing events occurring in two separate reference frames.

Personally I conclude they, the SRTists, are missing the NOW that is occuring even though we know we cannot measuire the suimultaneous events because two lightning strikes that hit the griound at the same time means that the lightning bolt closest to the observer is going to be oberved first, but this doesn't mean the bolts of lightning didn't strike the ground at the same time in every body's frame of reference. The moving observer can verify this with some trivial data analysis.
Geistkiesel​