neopolitan said:
But while my buddy in motion measures L and t, I will work out that, because he is motion, the length is contracted. I call that L', because I already have an L (unprimed is my frame so I making primed my buddy's frame).
You seem to be confused about what "unprimed is my frame so I am making primed my buddy's frame" means. The length of your buddy's apparatus is contracted when measured in
your frame, his apparatus is not contracted in his own frame, so it's totally wrong to call the contracted length L' if you just said your frame is unprimed! If your frame is unprimed, then any variable that refers to how something appears in your frame--like the coordinate distance between either ends of an apparatus at single instant of time in your frame--
must be unprimed, regardless of whether the physical object that you're measuring is at rest in your frame or not. Remember, physical objects aren't "in" one frame or another, different frames are just different ways of assigning coordinates to events associated with any object in the universe. And it's true that, as you say, "you already have an L" if you previously defined L to be the length of the same apparatus in your frame when it was at rest relative to you, but that just mean you need some different unprimed symbol to refer to the length of the apparatus in your frame once you've given it to your buddy and it's at rest relative to him, like L_{cbb} where "cbb" stands for "carried by buddy".
Perhaps this confusion about what quantities should be primed and what quantities should be unprimed is related to your (so far unexplained) belief that there is something "inconsistent" about the way the standard time dilation and length contraction equations are written?
And even if I changed your statement above to "I will work out that, because he is motion, the length is contracted. I call that L_{cbb}, because I already have an L", your statement would still be too vague, for exactly the same reason as the statement in your last post was too vague (I offered several possible clarifications so you could pick which one you meant, or offer a different clarification). If L_{cbb} refers to the length of the apparatus in
your frame when it's being carried by your buddy, and L'_{cbb} refers to the length your buddy measures the apparatus to be using his own ruler (which is equal to L, the length you measured the apparatus to be using your ruler
before you gave it to your buddy, when it was still at rest relative to you), then these will be related by the equation L_{cbb} = L'_{cbb} / \gamma, which is just the length contraction equation with slightly different notation. If on the other hand what your buddy "measures" is a distance of L' between two
events using his apparatus, then the distance you measure between the same two events will
not necessarily be L'/gamma, in fact it could even end up being
larger than L'. So you really need to
be specific about precisely what is being measured like I keep asking.
neopolitan said:
I will also work out that, because he is in motion, my buddy's clock will have slowed down. What reads on his clock will less than what I read on mine.
It is meaningless to compare two compare two clock readings unless A) the clocks are located at the same position at the moment you do the comparison, or B) you have specified which frame's definition of simultaneity you're using. Do you disagree? If not, which one are you talking about here? If it's B, and if you're using your own frame's definition of simultaneity, and if the clocks initially read the same time at some earlier moment in your frame, then I agree that at the later moment his clock will read less than yours. But again, you really need to be way more specific or you'll end up using inconsistent definitions in different statements and end up with conclusions that don't make any sense, as seems to be true of your "t' = t/gamma has to be true in order that L/t=c and L'/t'=c" argument.
neopolitan said:
Say I have two sets of the apparatus. I keep one, and give the other to my buddy. I know they are identical. I ask him to measure it lengthwise, he gets L and I get L.
"It" is too vague since you have two sets, but I assume you mean "I ask him to measure the apparatus at rest relative to him, while I measure the apparatus at rest relative to me, his value L'_{cbb} is exactly equal to my value L." Correct?
neopolitan said:
But if I compare my length to his length (and I can do this with lasers and time measurements in my frame), I will find that he is "confused". His length is actually L'=L / \gamma. (And yes, I know if he does the same thing, he will find that I am "confused".)
Time is a little more complex to describe, but equivalent to using lasers and time measurements in my frame. Using a very high quality telescope, I keep track of my buddy's apparatus, most specifically the clock. I note down two times on his clock, t'_{o} and t'_{i} along with the times that I make them (my times, my frame, unprimed). I have to take into account how long it took each of those displayed times on his clock to get to me.
And when you "take into account how long it took", you are using
your frame's measurement of the distance that his clock was from yours when it read each of these two times, and assuming that the light from his clock travels at c in
your frame, and subtracting distance/c from the time on your clock when you actually
saw these readings, is that correct? For example, if when your clock reads 10 seconds you look through your telescope and see his clock reading 6 seconds, and at this moment you see his clock is next to a mark that's 2-light seconds away from you on your ruler, then you'd say his clock "really" read 6 seconds at the moment your clock read 8 seconds, correct?
If that is what you mean by "take into account"--and please actually tell me yes or no if it's what you meant--then note that this is exactly the same as asking what times on your clock were
simultaneous with his clock reading t'_{o} and t'_{i}, using your own frame's definition of simultaneity. So note that although you didn't really respond to my list of possible clarifications, it appears that your meaning is exactly identical to the first one I offered, which I'll put in bold (in the original comment I was using unprimed to refer to the buddy's frame and primed to refer your frame, but since you appear to want to reverse that convention by making times on your buddy's clock primed, I'll change the quote to reflect the idea that times in your frame are unprimed and times in your buddy's are primed):
And what does "readouts of time elapsed are reduced ... according to me" mean? Does it mean that if you consider a time interval from some time t_0 to another time t_1 in your frame, then for a clock at rest in his frame, the difference (his clock's readout at the event on the clock's worldline that occurs at time t_1 in your frame) - (his clock's readout at the event on the clock's worldline that occurs at time t_0 in your frame) is smaller than the difference t_1 - t_0 which represents the size of the time interval in your frame? In that case I would agree. Does it mean that if you consider two arbitrary events A and B (like two events on the worldline of a photon), with A happening at t_0 and B happening at t_1 in your frame, then for a clock at rest in his frame, the difference (his clock's readout at the event on the clock's worldline that occurs simultaneously with B in your frame) - (his clock's readout at the event on the clock's worldline that occurs simultaneously with A in your frame) is smaller than the time interval between A and B in your frame? If so, this is exactly equivalent to the first one, so I'd agree with this too. But does it mean that if you consider the same two events A and B, then for a clock at rest in his frame, the difference (his clock's readout at the event on the clock's worldline that occurs simultaneously with B in his frame) - (his clock's readout at the event on the clock's worldline that occurs simultaneously with A in his frame) is smaller than the time interval between A and B in your frame? If so this is not true in general. And if the two events are events on the path of a photon that occur on either end of a measuring-rod of length L in his frame, then it is that last difference in his clock's readouts that you divide L by to get c.
neopolitan said:
I will find that \Delta t' = \Delta t / \gamma (<- this is my equation, this is not time dilation!)
So, if according to
your frame's definition of simultaneity, your clock's reading t_ 0 is simultaneous with your buddy's clock reading some time t'_0, and according to your frame's definition of simultaneity your clock's reading t_1 is simultaneous with your buddy's clock reading some time t'_1, and if \Delta t' = t'_1 - t'_0 and \Delta t = t_1 - t_0, then we get the equation \Delta t' = \Delta t / \gamma. Is that what you mean? If so, then yes, I agree, and as I said this is exactly equivalent to the statement from my earlier post that I bolded above. But
in this case you are simply confused if you think this is any different from the standard time dilation equation--it only looks different because you've reversed the meaning of primed and unprimed from the usual convention and then divided both sides by gamma! Normally, if we want to take two events on the worldline of a clock (in this case your buddy's) and then figure out the time interval between these events in a frame where the clock is moving (in this case yours--of course, figuring out the time interval between these events in your frame is exactly equivalent to figuring out which readings on your clock these two events are simultaneous with in your frame and then finding the difference between the two readings on your clock), the usual convention is to call the first frame unprimed and the second frame primed, in which case we get the time dilation equation \Delta t' = \Delta t * \gamma. You have simply adopted the opposite convention, calling the first frame primed and the second frame unprimed, so the time dilation equation would just have to be rewritten as \Delta t = \Delta t' * \gamma using this convention. And of course, if we now divide both sides by gamma, we get back the equation you offered, \Delta t' = \Delta t / \gamma. You can see that this is just a trivial reshuffling of the usual time dilation equation, not anything novel.
neopolitan said:
Now I know that when \Delta t has elapsed in my frame, \Delta t' elapses in his frame.
No you don't, not for an arbitrary pair of events! Say you pick two events A and B which
don't occur on the worldline of his clock (they may be two events on the worldline of a light beam for example), but such that
according to his frame's definition of simultaneity, A is simultaneous with t'_0 and B is simultaneous with t'_1. Then would you agree that the time interval between these events in his frame is \Delta t' = t'_1 - t'_0? And we also know that the time interval in
your frame between the event of his clock reading t'_1 and the event of his clock reading t'_0 is related to this by \Delta t = \Delta t' * \gamma.
But that doesn't mean the time interval in your frame between A and B is \Delta t = \Delta t' * \gamma! This is because although it's true that
his frame's definition of simultaneity says that A is simultaneous with his clock reading t'_0 and B is simultaneous with his clock reading t'_1,
your frame uses a different definition of simultaneity, so according to
your frame's definition of simultaneity A may
not be simultaneous with his clock reading t'_0 and B may not be simultaneous with his clock reading t'_1, so knowing the time-interval in your frame between his clock reading t'_0 and his clock reading t'_1 tells us
nothing about the time interval in your frame between A and B.
Do you understand and agree with all this? Please tell me yes or no.
neopolitan said:
It is not just ticks on clocks, or the time interval between two events - his time dimension is affected. And it is affected in the same way as his spatial dimension is affected. So any speed in his frame will be calculated using contracted length divided by shortened time which will give you the same result as using unaffected length divided by unaffected time. Picking appropriate values of L and \Delta t:
L / \Delta t = c = L' / \Delta t'
Nope, you
still are unable or unwilling to define what you are actually supposed to be measuring the length of and time-intervals between, "appropriate values" is hopelessly vague. Do L and L' represent the distance between a single pair of events on the worldline of a photon, as measured in each frame? Or are you measuring two separate photons with two separate apparatuses, so L is the distance between one pair of events as measured in your frame and L' is the distance between another pair as measured in your buddy's frame? Or is it something else entirely? And how about \Delta t and \Delta t', are you going with the definition I suggested earlier where \Delta t' is the difference between two clock readings t'_1 and t'_0 on your buddy's clock, and \Delta t is the difference between two clock readings t_1 and t_0 on your clock, where you have picked the readings so that according to
your frame's definition of simultaneity t_1 is simultaneous with t'_1 and t_0 is simultaneous with t'_0? If not, can you be specific about what events you are taking "deltas" between? And if so, are any of these events on the clocks' worldlines supposed to be simultaneous with events on the worldline of a photon in some frame?
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