On Thu, Mar 6, 2014 at 12:11 PM, Edgar L. Owen <edgaro...@att.net> wrote:
> Jesse, > > I don't think this is correct. It is meaningless to try to TAKE THE FRAME > VIEW OF ALL FRAME VIEWS. That's not the correct way to look at it. > > What we do is to take all frame views of any ONE proper time correlation. > Every frame view will give one and only one EXACT answer of how close those > proper times are to being equal. Once that's done we have the whole > picture. We DO NOT HAVE TO TAKE FRAME VIEWS OF THOSE FRAME VIEWS because we > already have ALL the frame views of that one situation. > I don't know what you mean by "the frame view of all frame views". I agree that for a given pair of clocks A and B that are at rest relative to each other and synchronized in their rest frame, each frame has only ONE answer to how much clock A is ahead of B (a number which can be zero if the frame in question is their rest frame, and can also be negative if the frame in question sees clock A as having a time that's behind clock B's time). But if we want to LABEL each such frame by velocity (so we can do an integral or sum over frames with different velocities to take the average), then we must use some specific "reference frame", and label the velocity of every other frame relative to the reference frame. So for example if a given frame X has a velocity of 0.9c relative to a pair of clocks that are 2 light-year apart in their own rest frame, then in X they are out-of-sync by 2*0.9 = 1.8 years. If we use as our "reference frame" the rest frame of the clocks themselves, then X is labeled with v=0.9c, and thus our "amount out of sync as a function of v" function will have a value of 1.8 at v=0.9c. On the other hand, if we use as our "reference frame" a frame moving at 0.8c relative to the clocks, then frame X will have to be labeled with v=0.357c--it's still the same frame X, and it still has the same amount-out-of-sync of 1.8, but it just has a different velocity label. So using this reference frame, our "amount out of sync as a function of v" function will have a value of 1.8 at v=0.357c. Point is, depending on the reference frame we use to define the "v" of every other frame, our "amount out of sync as a function of v" function will look different, and thus if we integrate over that function to find some sort of "average" value for the amount the clocks are out of sync, or just do an average over a finite number of values of the function at regular intervals of v, then we'll get different answers depending on what reference frame we chose. Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
