Recognitions:

Benefits of time dilation / length contraction pairing?

 Quote by neopolitan Note that I too have said that time dilation is correct as defined. I have also said that the type of time used in time dilation is a different type of time to the one we usually use (one is the time between ticks, the other is the number of ticks displayed).
This distinction seems meaningless--unless you are postulating some notion of absolute time, the only way we can talk about "time between ticks" is by looking at the "number of ticks displayed" on a clock (or pair of synchronized clocks) with some smaller interval between ticks; for example, if the time between ticks is 1 second, put another clock next to it that ticks once every millisecond, and find that the second clock elapses 1000 ticks between each tick of the first clock. The time between a clock's ticks is just like any other time interval, we measure it using the difference in reading between the end of the interval and the beginning (or more abstractly, we can define it using the difference between the time coordinates between the events of two successive ticks).
 Quote by neopolitan I accept that there has been some dancing around, which makes things confusing. But the bit I was trying to do, the bit I was focussed on was "using your alternate equation". I ignored the words following it, because you were taking something out of context from post #45
When I said "using your alternate equation" I really meant "using whatever alternate equation you wish to define", not specifically using the alternate equation I gave in post #45.
 Quote by neopolitan The alternate equation I was using is this: $$t' = t / \gamma$$ You want to know in which frame the events are colocated. Really, they don't need to be colocated in any frame.
Fine, but regardless of what events you choose, you still haven't given any coherent definition of what t and t' mean in terms of coordinates assigned to the events, or in terms of actual physical measurements involving those events. Does t represent the coordinate time interval between two events in the unprimed frame, and t' represent the coordinate time interval between the same two events in the unprimed frame? If not, what do t and t' mean? You can't write down equations and say they have any relevance to physics if you can't even define the physical meaning of the variables in the equations!
 Quote by neopolitan Perhaps if I rephrase it will make it easier. I want to know the relativistic effects on my buddy's measurements of time and space. I can use that to calculate what the speed of light is in his frame (and I am most interested in that speed when light moves parallel to the direction of his motion). I want to use the same sort of dimensions that we usually use to calculate a speed (a length which can be traversed and a readout of time elapsed). What effect does putting my buddy into motion have on his dimensions? The answer is that his lengths contract and the readouts of time elapsed are reduced - according to me. (And my lengths and my readouts of time elapsed are reduced - according to him.) Are we agreed on this?
Your words are too vague and would only make sense with specific types of elaborations. Does "his lengths contract ... according to me" mean that the length you assign to an object at rest in his frame is smaller than the length he assigns to that same object? In that case I would agree. Does "his lengths contract ... according to me" mean that the length you assign to an object at rest in your frame is smaller than the length he assigns to that same object? If so, the statement is wrong. Does "his lengths contract ... according to me" mean that the spatial distance between two arbitrary events as measured by you is smaller than the spatial distance between the same two events as measured by him? If so, this would be true in some cases but not in others.

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.
 Quote by neopolitan I regret posting exactly when I did. Rasalhague asks a fine question in the post before mine.
Don't worry, I intend to reply to it.

 Quote by JesseM The usual convention is that the unprimed t represents the time interval between two events on the worldline of a clock as measured in the clock's rest frame (so both events happen at the same spatial location in the unprimed frame, and t will be equal to the time interval as measured by the clock itself), whereas the primed t' represents the time interval between the same two events in a frame where the clock is moving (so the events happen at different locations in the primed frame). This is consistent with the convention for length contraction, where the unprimed L represents the distance between two ends of an object in the object's own rest frame, and the primed L' represents the distance between the ends of the same object in the frame where the object is in moving.
As someone new to SR, the use of primed and unprimed coordinates has been a source of some confusion for me. I think that’s partly because, to begin with, I didn’t know which details of a textbook explanation were going to turn out to be the significant ones, and which were accidental to a particular author’s presentation. At first, I got the impression that the convention was simply that primed frame = rest frame. But then I encountered examples such as: “Consider a rod at rest in frame S’ with one end at x’2 and the other end at x’2 [...]” (Tipler & Mosca: Physics for Scientists and Engineers, 5e, extended version, p. 1274). From this, I guessed that the desciding factor in whether to call a frame primed or unprimed was the direction it was moving, a primed frame being one that moves in the positive x direction with respect to the unprimed frame, hence the signs used in versions of the full Lorentz transformation such as this:

x' = gamma (x – ut)
t' = gamma (t – ux/c^2)

x = gamma (x' + ut')
t = gamma (t' + ux'/c^2)

I think that’s the traditional choice of signs, isn’t it? But I don’t know yet how rigid that convention is. The practice of using primed coordinates to represent a frame moving in the positive x direction agrees with their use in discussions of Euclidian rotation for a frame rotated in the positive (counterclockwise) direction. At least, that’s how Euclidian rotation was introduced in the first books where I met it. But in Spacetime Physics, Taylor and Wheeler do it the other way around, using primed coordinates for a negative (clockwise) rotation. I suppose, looking on the bright side, the advantage to having a variety of presentation methods is that it, eventually, you can see by comparing them which details are the physically significant ones, and which a matter of convention. But to begin with it’s a lot of information to take in.

Recognitions:
 Quote by Rasalhague If I try putting this in my own words, could you tell me if I've got it right? As I understand it, the time dilation formula takes as its input the time between two events in a frame where there’s no space between them, i.e. two events colocal in the unprimed frame. When you plug into it this time and the speed the primed frame is moving relative to the unprimed, the formula t’ = t * gamma tells you the time in the primed frame between the two events. The length contraction formula takes as its input the distance between two events that are simultaneous in the unprimed frame. When you plug into it this distance and the speed the primed frame is moving relative to the unprimed frame, the formula L’ = L/gamma tells you the distance in the primed frame between one of these events *and a third event*, which third event is simultaneous in the primed frame with the first, and in the unprimed frame is colocal with the second.
Yes, these look right to me. But to make the physical meaning of the second one a little more clear, you might point out that if the first event occurs along the worldline of the left end of an object at rest in the unprimed frame, and the second event occurs along the worldline of the same object's right end (so if the distance between the events is L in the unprimed frame, this must be the length of the object in the unprimed frame), that means that the third event also occurs along the worldline of the object's right end, so since the first and third event are simultaneous in the primed frame, L' must be the length of the object in the primed frame. The idea is that the "length" of an object in any frame is defined as the distance between its two ends at a single instant in that frame.
 Quote by Rasalhague More generally, I guess I’ve been trying to understand what exactly the asymetry is, and where it comes from: whether a physical difference between time and space, an accident of the kind of questions asked in textbooks, or of the way the idea is expressed, or if there’s something about our relationship to time and space that makes us want to ask a different question of each--that is, something that makes this pair of not-directly-analogous questions more useful to us than any other combination of the four possible questions illustrated in the diagram. Does that make any sense?
I guess I would say the usefulness of these two equation is that in physics it is typical to calculate dynamics by taking as "initial conditions" a spatial arrangement of objects at a single moment in time (along with the instantaneous velocities at that time), and then use the dynamical equations of physics to evolve that initial state forward through time. So, it's useful to know the set of spatial coordinates an object occupies at a single instant which is where the length contraction equation comes in handy, and it's useful to know how much a clock will have moved forward if you evolve the initial state forward by some particular amount of coordinate time, which is where the time dilation equation comes in handy. The "spatial analogue of time dilation" and the "temporal analogue of length contraction" equations would come in more handy if we took as our "initial state" a surface of constant x rather than a surface of constant t, and then evolved this state forward by increasing the x coordinate and looking at how things changed in successive surfaces of constant x. But this points to a real difference between how the laws of physics treat time and space; in a deterministic universe the laws of physics do allow you to determine what the physical state will be in later surfaces of constant t if you know the physical state at an earlier surface of constant t, but they don't allow you to determine the state of a surface of constant x if all you know is the state of some other surface of constant x. I guess this is also related to the fact that in SR the worldlines are timelike, so if we assume no worldline has a start or end (no worldlines starting or ending at singularities as in GR, and no particle worldlines ending because the particle annihilates with another particle as in quantum theory), then a given worldline will pierce every surface of constant t precisely once, while a worldline can have no intersections with a surface of constant x, or one pointlike intersection, or multiple pointlike intersections, or an infinite collection of points on that worldline can lie on a single surface of constant x.

Recognitions:
 Quote by Rasalhague As someone new to SR, the use of primed and unprimed coordinates has been a source of some confusion for me. I think that’s partly because, to begin with, I didn’t know which details of a textbook explanation were going to turn out to be the significant ones, and which were accidental to a particular author’s presentation. At first, I got the impression that the convention was simply that primed frame = rest frame. But then I encountered examples such as: “Consider a rod at rest in frame S’ with one end at x’2 and the other end at x’2 [...]” (Tipler & Mosca: Physics for Scientists and Engineers, 5e, extended version, p. 1274). From this, I guessed that the desciding factor in whether to call a frame primed or unprimed was the direction it was moving, a primed frame being one that moves in the positive x direction with respect to the unprimed frame, hence the signs used in versions of the full Lorentz transformation such as this: x' = gamma (x – ut) t' = gamma (t – ux/c^2) x = gamma (x' + ut') t = gamma (t' + ux'/c^2) I think that’s the traditional choice of signs, isn’t it? But I don’t know yet how rigid that convention is. The practice of using primed coordinates to represent a frame moving in the positive x direction agrees with their use in discussions of Euclidian rotation for a frame rotated in the positive (counterclockwise) direction. At least, that’s how Euclidian rotation was introduced in the first books where I met it. But in Spacetime Physics, Taylor and Wheeler do it the other way around, using primed coordinates for a negative (clockwise) rotation. I suppose, looking on the bright side, the advantage to having a variety of presentation methods is that it, eventually, you can see by comparing them which details are the physically significant ones, and which a matter of convention. But to begin with it’s a lot of information to take in.
There isn't really any absolute convention about primed and unprimed, they're just ways of differentiating two distinct frames, although most authors seem to follow the conventions that you said you were most used to above. But as long as you understand the physical relations of the two frames that's all that really matters. For example, if someone writes $$\Delta t' = \Delta t * \gamma$$ (the most common form I've seen), then you know $$\Delta t$$ must refer to two events which occur at the same position in the unprimed frame, like two readings on a clock at rest in the primed frame; if someone instead wrote $$\Delta t = \Delta t' * \gamma$$, then unless they just made a mistake, you'd know they intended to refer to the time intervals between two events which occur at the same position in the primed frame. Likewise, if someone wrote the following as the Lorentz transformation:

x = gamma (x' - vt')
t = gamma (t' - vx'/c^2)

Then although this is different from how they're usually written, you can infer that this is just a situation where it's assumed the origin of the unprimed frame is moving at velocity v along the x' axis of the primed frame.

 Remember a while back I talked about an apparatus I had. I have it and it is at rest relative to me. Associated with this apparatus are a length and a time measurement. I called these L and t. I give these to my buddy, and he sets off on a carriage with a speed of v (in a direction that is convenient so that the length I measured as L is parallel to the direction of motion). My buddy will, if he checks, find a length and time measurement of L and t. 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). 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. If confuses you, and god knows it confuses me, because you have to step back a bit from the intial t I had. So, let's do it another way. 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. 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. I will find that $$\Delta t' = \Delta t / \gamma$$ (<- this is my equation, this is not time dilation!) Now I know that when $$\Delta t$$ has elapsed in my frame, $$\Delta t'$$ elapses in his frame. 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 $$\Delta t$$: $$L / \Delta t = c = L' / \Delta t'$$ Does that help? cheers, neopolitan
 Wow, thanks for your answer Jesse. That's given me lots to think about! In everyday life we're more used to regarding future time as being what's unpredictable, so it's curious to think of "the state of a surface of constant x" as being more fundamentally impossible to deduce "if all you know is the state of some other surface of constant x". But suppose that, in a deterministic universe, you knew everything about a surface of constant x to some arbitrary degree of precision. You'd know the positions of particles in that surface and be able to say something about other surfaces of constant x by examining the forces operating on the particles in your surface. Admittedly there'd be multiple possible states of the rest of space that could be responsible for the state of your surface of constant x, but then you could likewise have different histories that lead to the same state for some surface of constant t. So is it something about the future specifically, and its predictability, that makes all the difference? If that's even a meaningful question... Meanwhile, less philosophically, just to check I've understood: is the case where a worldline has multiple pointlike intersections with a surface of constant x only possible for a particle for which there's no inertial frame in which the particle can be said to be at rest (i.e. its worldline isn't a straight line)? (The other two cases you mention--no intersections, one pointlike intersection--being possible for a particle which can be described as being at rest in some inertial frame.)

Recognitions:
 Quote by neopolitan 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.
 Quote by neopolitan 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.
 Quote by neopolitan 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?
 Quote by neopolitan 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.

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.
 Quote by neopolitan 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.
 Quote by neopolitan 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.
 Quote by neopolitan 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?

Recognitions:
 Quote by Rasalhague Wow, thanks for your answer Jesse. That's given me lots to think about! In everyday life we're more used to regarding future time as being what's unpredictable, so it's curious to think of "the state of a surface of constant x" as being more fundamentally impossible to deduce "if all you know is the state of some other surface of constant x". But suppose that, in a deterministic universe, you knew everything about a surface of constant x to some arbitrary degree of precision. You'd know the positions of particles in that surface and be able to say something about other surfaces of constant x by examining the forces operating on the particles in your surface.
That's an interesting point, but consider the fact that it's quite possible to have a surface of constant x such that of the particles that cross it (and many particles may never cross a particular surface of constant x at all), each one crosses it only at a single point in spacetime. In this case you'd only know the instantaneous velocity of each particle at its crossing point, but I don't see how this would allow you to deduce the force unless you also knew the instantaneous acceleration (and keep in mind that in deterministic theories like classical electromagnetism, merely knowing the position and instantaneous velocity of each particle in a surface of constant t, along with the direction and magnitude of force field vectors in space in that surface, is sufficient to allow you to predict what will happen at later times, you don't need to know the instantaneous accelerations). Also consider that in principle it would be possible to have a surface of constant x where no particles crossed it at any point, even though particles did exist in that universe--I suppose you could still be told the direction and magnitude of force field vectors in this otherwise empty surface, since force fields like the electromagnetic field are imagined to fill all of space, but I don't think this would allow you to deduce the complete history of every particle in the universe (it's possible I could be wrong about this since I haven't actually seen any discussions of this question, though).
 Quote by Rasalhague Meanwhile, less philosophically, just to check I've understood: is the case where a worldline has multiple pointlike intersections with a surface of constant x only possible for a particle for which there's no inertial frame in which the particle can be said to be at rest (i.e. its worldline isn't a straight line)? (The other two cases you mention--no intersections, one pointlike intersection--being possible for a particle which can be described as being at rest in some inertial frame.)
That's right, assuming of course that we're talking about the x-coordinate of an inertial reference frame.

 "Picking appropriate values of $$L$$ and $$\Delta t$$" was too vague. The rest of what you were saying was akin to "You can't park four tanks on the rubber dingy you're designing". Here's what I mean about picking appropriate values, pick any value of $$L$$, any value you like - in the real world you probably want a really big value, but this is hypothetical world, so it is not so important. Then pick the value of $$\Delta t$$ so that L/$$\Delta t = c$$. If you haven't picked a really big value of $$L$$, then $$/Delta t$$ will be pretty damn small so that it will be challenging to take two readings $$t_{o}$$ and $$t_{i}$$ where $$t_{i} - t_{o} = \Delta t$$ - but we are in hypothetical world. We have no argument about length contraction. But do you deny that when I use my readings from my buddy's clock, and take into account the motion that I know he has, that I will get a $$/Delta t'$$ which is shorter than mine? Do you deny that the extent to which it is shorter is the same as the extent to which L' is shorter than L (where these are given by standard length contraction)? I've described the events, they aren't simultaneous (and in fact, I don't care about simultaneity, I know the time readings on my buddy's clock are not simultaneous with the time readings on mine, the only thing I bother with, or need to bother with, is the extra time the second reading takes to get to me because he has moved during the time). Any discussion of simultaneity in this scenario is a distraction. If you must have some simultaneity, then try thinking that my seeing the time on my buddy's clock is simultaneous with my reading of the time on my clock, but even that is not necessary since I could use a splitframe camera and look at the results afterwards. The bottom line, from you Jesse, is "there is no other way to do it" when the question is "what is the benefit with time dilation". It seems you truly think there is no other option. You have a very long winded way to say it, but I don't think there is any other way to interpret your approach to the original question. And yes, I haven't forgotten the original question. cheers, neopolitan

Recognitions:
 Quote by neopolitan Here's what I mean about picking appropriate values, pick any value of $$L$$, any value you like - in the real world you probably want a really big value, but this is hypothetical world, so it is not so important. Then pick the value of $$\Delta t$$ so that L/$$\Delta t = c$$. If you haven't picked a really big value of $$L$$, then $$\Delta t$$ will be pretty damn small so that it will be challenging to take two readings $$t_{o}$$ and $$t_{i}$$ where $$t_{i} - t_{o} = \Delta t$$ - but we are in hypothetical world.
Are you just picking a value of $$\Delta t$$ out of thin air, with no connection to anything physical (so you could just as easily pick a $$\Delta t$$ such that L/$$\Delta t$$ = 5c or any number you wish), or is it supposed to represent the time interval between some specific pair of events, like $$t_{o}$$ representing the time a photon passes next to one end of an object of length L which is at rest in your frame, and $$t_{i}$$ representing the time that photon passes next to the other end of the same object?
 Quote by neopolitan We have no argument about length contraction. But do you deny that when I use my readings from my buddy's clock, and take into account the motion that I know he has, that I will get a $$/Delta t'$$ which is shorter than mine?
What does "use my readings from my buddy's clock, and take into account the motion that I know he has" mean? This is something I specifically asked you about in my last response to you (post 75):
 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.
Can you please answer this question? And if the answer is "no, that's not what I meant" could you please tell me exactly what you do mean, preferably giving some sort of numerical example? For example, suppose your buddy's clock is moving at 0.6c in your frame and is initially right next to your, and when they are next to each other both your clock and his read 0 seconds; then later when your clock reads 16 seconds, you look through your telescope and see his clock reading 8 seconds, and you also see that when his clock reads 8 seconds it's next to a mark on your ruler that's 6 light-seconds away from you (so in your frame the event must have 'really' happened 6 seconds before you saw it through your telescope). How would the phrase "use my readings from my buddy's clock, and take into account the motion that I know he has" apply to this specific example. What would $$\Delta t'$$ be, and what would $$\Delta t$$ be?
 Quote by neopolitan I've described the events
Have you? Where? Are the events just the two readings on your buddy's clock?
 Quote by neopolitan they aren't simultaneous (and in fact, I don't care about simultaneity, I know the time readings on my buddy's clock are not simultaneous with the time readings on mine, the only thing I bother with, or need to bother with, is the extra time the second reading takes to get to me because he has moved during the time). Any discussion of simultaneity in this scenario is a distraction.
 Quote by neopolitan The bottom line, from you Jesse, is "there is no other way to do it" when the question is "what is the benefit with time dilation".
No it isn't, because I don't even know what the physical meaning of the "it" that you want to do actually is, your posts are just too unclear for me to judge them right or wrong. It will help if you give clear yes-or-no answers to questions about what you're saying when I ask for them.

 JesseM, You fragment too much. It leads, inexorably, to loss of context. That's why I am not responding to your fragmenting. Look back in previous posts and I explained what I meant about taking readings on my buddy's clock. I made mention of a telescope. But you must have overlooked it in your apparent excitement to demolish any discussion (and I mean that, "discussion", not argument because to demolish an argument you have to make an effort to understand). You make an effort to understand, pose an unfragmented question which indicates that you have made the slightest effort to understand, rather than attack, and I will answer it. cheers, neopolitan

 Quote by JesseM it's also possible to come up with a "temporal analogue for the length contraction equation" which looks like the length contraction equation but with $$\Delta t$$ substituted for L (this is more difficult to state in words, but it's basically the time-interval in the primed frame between two surfaces of constant t in the unprimed frame which have a temporal distance of $$\Delta t$$ in the unprimed frame, which is analogous to how length contraction tells you the spatial distance in the primed frame between two worldlines of constant position in the unprimed frame which have a spatial separation of L in the unprimed frame).
Suppose Alice and Bob are each wearing a watch. Bob, moving in the positive x direction in Alice's rest frame, at 0.6c, passes Alice and they synchronise watches. Some time later, Alice looks at her watch and wonders, "What time does Bob's watch say at the moment which in Bob's rest frame is simultaneous with me asking this question?" The answer is given by $$t_{B} = \gamma t_{A}$$, the standard time dilation formula. If Alice's watch says 4, Bob's will say 5. "How about that," thinks Alice. "To Bob, for whom my watch is moving, it's running slow."

Now suppose that Alice wonders, "What time does Bob's watch say at the moment which in my rest frame is simultaneous with me asking this question?" The answer to this is given by $$t_{B} = \frac{t_{A}}{\gamma}$$. If Alice's watch says 5, Bob's will say 4. "Fancy," thinks Alice. "From my perspective, Bob's watch, which is moving relative to me, is running slow."

And, of course, Bob can ask the equivalent questions about the time on Alice's watch with identical results by virtue of the fact that the two frames don't agree on which events are simultaneous (except for those that happen in the same place, such as their synchronisation).

But isn't Alice's second question none other than this exotic "temporal analogue for the length contraction equation"? She wants to know "the time-interval in the primed frame" (the time shown by Bob's watch, indicating a time interval along Bob's worldline) "between two surfaces of constant t in the unprimed frame" (one being the one which Alice and Bob's worldlines intersected when they synchronised watches, the other being Alice's present when she looks at her watch) "which have a temporal distance of $$\Delta t$$ in the unprimed frame" (the time shown by Alice's watch when she looks at it and makes her query).

Is Alice's second question in any way less natural than the first, or a less useful thing to ask of time than of space? I'm puzzled as to how it can be, if it is, as Jesse said, "just a trivial reshuffling of the usual time dilation equation"?

Recognitions:
 Quote by neopolitan You make an effort to understand, pose an unfragmented question which indicates that you have made the slightest effort to understand, rather than attack, and I will answer it.
I have made an effort to understand, and in fact the questions above are pretty clearly requests to nail down the meaning of your statements by asking if they match up with the precise definitions that I have suggested, for example:
 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. Can you please answer this question? And if the answer is "no, that's not what I meant" could you please tell me exactly what you do mean, preferably giving some sort of numerical example? For example, suppose your buddy's clock is moving at 0.6c in your frame and is initially right next to your, and when they are next to each other both your clock and his read 0 seconds; then later when your clock reads 16 seconds, you look through your telescope and see his clock reading 8 seconds, and you also see that when his clock reads 8 seconds it's next to a mark on your ruler that's 6 light-seconds away from you (so in your frame the event must have 'really' happened 6 seconds before you saw it through your telescope). How would the phrase "use my readings from my buddy's clock, and take into account the motion that I know he has" apply to this specific example. What would $$\Delta t'$$ be, and what would $$\Delta t$$ be?
If you see this line of questioning as simply an "attack" rather than an attempt to actually understand in precise terms the meaning of your phrase "take into account how long it took" (and thereby to figure out the precise physical relationship between the two intervals $$\Delta t$$ and $$\Delta t'$$ which appear in your equation L/$$\Delta t$$ = c = L'/$$\Delta t'$$), then I suppose that means you are simply too mistrusting of my motives to ever be interested in the process of actual communication with me (and 'communication' necessarily requires a willingness to clarify what the other person doesn't understand, especially in a discussion of physics where precise definitions are needed), in which case I take it there is basically nothing I could do other than nodding my head and agreeing with all your statements (even when I don't really understand what they mean) that would make you want to continue the discussion.

Recognitions:
 Quote by Rasalhague Suppose Alice and Bob are each wearing a watch. Bob, moving in the positive x direction in Alice's rest frame, at 0.6c, passes Alice and they synchronise watches. Some time later, Alice looks at her watch and wonders, "What time does Bob's watch say at the moment which in Bob's rest frame is simultaneous with me asking this question?" The answer is given by $$t_{B} = \gamma t_{A}$$, the standard time dilation formula. If Alice's watch says 4, Bob's will say 5. "How about that," thinks Alice. "To Bob, for whom my watch is moving, it's running slow." Now suppose that Alice wonders, "What time does Bob's watch say at the moment which in my rest frame is simultaneous with me asking this question?" The answer to this is given by $$t_{B} = \frac{t_{A}}{\gamma}$$. If Alice's watch says 5, Bob's will say 4. "Fancy," thinks Alice. "From my perspective, Bob's watch, which is moving relative to me, is running slow."
Here you can use the time dilation formula too. If the time dilation formula is written $$\Delta t' = \Delta t * \gamma$$, then that' formula is comparing the amount of time that's elapsed on a clock (whose rest frame is labeled the unprimed frame) with the amount of time that's elapsed in a frame where the clock is moving, with "time elapsed" in that frame being based on that frame's definition of simultaneity (and with this second frame being labeled primed). So in your second example, the clock is Bob's and the second frame whose definition of simultaneity you're using is Alice's, so you can just treat Bob's frame as the unprimed frame in the standard time dilation equation and Alice's question will be the same as asking for the time elapsed in the primed frame, meaning you're just substituting $$t_A$$ and $$t_B$$ into the time dilation equation giving $$t_A = t_B * \gamma$$. Of course, if you wish to divide both sides by gamma, you can get back the formula $$t_{B} = \frac{t_{A}}{\gamma}$$ you wrote above, but this is just a reshuffling of the time dilation equation.

But you do raise a great point which I hadn't thought of before, which is that the "temporal analogue of the length contraction equation" can always be used to calculate things that you'd normally use the time dilation equation to calculate, provided you reverse the meaning of which frame is primed and which frame is unprimed. Let me give a numerical example similar to yours. Suppose Bob is moving away from Alice at 0.6c and that both their clocks read 0 when they crossed paths as you suggested. But instead of starting Bob's time interval when his clock reads 0 as in your example, suppose we were interested in the time interval on Bob's clock that started with the event of his clock reading $$t_{B1}$$ = 8 seconds, and ended with his clock reading $$t_{B2}$$ = 12 seconds, so the length of the interval in Bob's frame is $$\Delta t_B$$ = ($$t_{B2}$$ - $$t_{B1}$$) = 4 seconds. If Alice wanted to know the time interval $$\Delta t_A$$ between these same two events in her frame, which is equivalent to wanting to know the time interval between the event $$t_{A1}$$ on her clock which is simultaneous in her frame with $$t_{B1}$$ (in this case $$t_{A1}$$ = 10 seconds) and the event $$t_{A2}$$ on her clock which is simultaneous in her frame with $$t_{B2}$$ (in this case $$t_{A2}$$ = 15 seconds), then she would plug these two different time intervals into the time dilation equation $$\Delta t' = \Delta t * \gamma$$, treating Bob's frame as unprimed and her frame as primed, which gives $$\Delta t_A = \Delta t_B * \gamma$$. If she wanted to reverse this and figure out the time interval $$\Delta t_B$$ on Bob's clock between two events on $$t_{B1}$$ and $$t_{B2}$$ on his clock's worldline that are simultaneous in her frame with two events on her clock's worldline $$t_{A1}$$ and $$t_{A2}$$ that are the beginning and end of a time interval $$t_A$$ (in the example above she would start with times 10 seconds and 15 seconds on her clock and then try to figure out how much time had elapsed on Bob's clock between these moments in her frame), she'd just divide the time dilation equation by gamma so it gives $$\Delta t_B$$ as a function of $$\Delta t_A$$, i.e. $$\Delta t_B = \frac{\Delta t_A}{\gamma}$$.

On the other hand, the "temporal analogue of length contraction" $$\Delta t' = \Delta t / \gamma$$ would be telling her something conceptually different, assuming she continues to treat Bob's frame as unprimed and her frame as primed. Basically, it would be saying "if you use $$t_{A1}$$ to label the time on Alice's clock that's simultaneous in Bob's frame with Bob's clock reading $$t_{B1}$$, and you use $$t_{A2}$$ to label the time on Alice's clock that's simultaneous in Bob's frame with Bob's clock reading $$t_{B2}$$, then the time interval on Alice's clock ($$t_{A2} - t_{A1}$$) is related to the time interval on Bob's clock ($$t_{B2} - t_{B1}$$) by the formula $$(t_{A2} - t_{A1}) = (t_{B2} - t_{B1}) / \gamma$$. If we use the same numbers for $$t_{B1}$$ and $$t_{B2}$$ on Bob's clock as before, namely $$t_{B1}$$ = 8 seconds and $$t_{B2}$$ = 12 seconds, then in this case we'd have $$t_{A1}$$ = 8*0.8 = 6.4 seconds (I just multiplied 8 by 0.8 because I know both clocks read 0 when they were next to each other and Alice's clock is moving at 0.6c in Bob's frame, so the standard time dilation equation tells me her clock is slowed by a factor of 0.8 in his frame) and $$t_{A2}$$ = 12*0.8 = 9.6 seconds. So the equation $$(t_{A2} - t_{A1}) = (t_{B2} - t_{B1}) / \gamma$$ does work here, since $$(t_{A2} - t_{A1}$$ = 9.6 - 6.4 = 3.2, $$t_{B2} - t_{B1}$$ is still 4 seconds, and gamma is still 0.8. But you can see that the time interval in Alice's frame we're talking about now (3.2 seconds) is different than the time interval in Alice's frame we were talking about when we were using the usual time dilation equation (5 seconds). But, that's only because we were treating Alice's frame as the primed frame in both equations! If we reverse the labels and treat Bob's frame as primed and Alice's frame as unprimed, then the standard time dilation equation $$\Delta t' = \Delta t * \gamma$$ does tell you that when 3.2 seconds have elapsed on Alice's clock, 4 seconds of time have passed in Bob's frame (or equivalently, if you look at the readings on Bob's clock that are simultaneous in Bob's frame with the two readings on Alice's clock, the difference between these two readings on Bob's clock is 4 seconds).

So I guess if you take the time dilation equation and divide both sides by gamma to solve for the interval in the primed frame, this is really just equivalent to taking the "temporal equivalent of length contraction" equation and reversing which frame we call primed and which we call unprimed. To me there's still a little bit of a conceptual difference though, in the sense that normally I think of these equations as relating a clock time-interval to a coordinate time-interval, with unprimed normally being the clock time-interval. For instance, when I read the time dilation equation $$\Delta t' = \Delta t * \gamma$$, I find it most natural to think that $$\Delta t$$ represents the difference between two clock-readings on a clock at rest in the unprimed frame, and then $$\Delta t'$$ represents the difference between the coordinate times of these two readings in the primed frame. Of course, because a clock at rest in the primed frame will keep time with coordinate time in that frame, this is equivalent to imagining there's also a clock at rest in the primed frame, and saying $$\Delta t'$$ represents the difference between two readings on the primed clock that are simultaneous in the primed frame with the two readings on the unprimed clock that were mentioned earlier. The first way of stating it just makes the usefulness of the time dilation equation more intuitive to me; as I said before, physics is all about setting up a spacetime coordinate system and then using equations to figure out how the state of objects in space changes as the time-coordinate increases.

JesseM responding to Rasalhague:
 Quote by JesseM But you do raise a great point which I hadn't thought of before, which is that the "temporal analogue of the length contraction equation" can always be used to calculate things that you'd normally use the time dilation equation to calculate, provided you reverse the meaning of which frame is primed and which frame is unprimed. The first way of stating it just makes the usefulness of the time dilation equation more intuitive to me; as I said before, physics is all about setting up a spacetime coordinate system and then using equations to figure out how the state of objects in space changes as the time-coordinate increases.
Thanks, I think you've given an answer my original question. I think you have said this:

There is another way of approaching the relativistic effects other than time dilation-length contraction. That is to use "temporal analogue of the length contraction equation"-length contraction. However, using time dilation is more intuitive to you - and possibly also for the majority of people. That said, there is nothing inherently wrong with using a "temporal analogue of the length contraction equation" (although one must note that a different prime convention is required).

Rasalhague has shown me that instead of:

 What exactly is the greater utility of time dilation and length contraction equations which prevents the use of two contraction equations ...?

 What exactly is the greater utility of length contraction and time dilation equations which prevents the use of a length contraction equation and a temporal equivalent of the length contraction equation ...?
For that I thank you Rasalhague.

cheers,

neopolitan

Recognitions:
But such a broadly-worded question would almost certainly not have led me to the same realization. This illustrates why I keep asking you to answer specifics about what you are saying, and I don't really understand why you are unwilling to grant these requests--is it that you don't like my attitude, is it that your ideas are mostly intuitive so you're not sure what the answers should be yourself, or something else? I really think a willingness to delve into specifics could allow us to make progress on things like the meaning of "L/$$\Delta t$$ = c = L'/$$\Delta t'$$" which I have been unable to make sense of so far, just as the specifics of Rasalhague's question allowed for progress on the issue of the uses of the "temporal analogue of length contraction" equation, so I hope that even if you decide you are not interested in continuing this line of discussion for whatever reason, you will at least consider that there may be a lesson here about the value of engaging with nitty gritty details when talking physics.