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Benefits of time dilation / length contraction pairing?

by neopolitan
Tags: contraction, dilation, length, pairing, time
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neopolitan
#73
Mar17-09, 11:38 PM
P: 645
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 [tex]L'=L / \gamma[/tex]. (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, [tex]t'_{o}[/tex] and [tex]t'_{i}[/tex] 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 [tex]\Delta t' = \Delta t / \gamma[/tex] (<- this is my equation, this is not time dilation!)

Now I know that when [tex]\Delta t[/tex] has elapsed in my frame, [tex]\Delta t'[/tex] 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 [tex]\Delta t[/tex]:

[tex]L / \Delta t = c = L' / \Delta t'[/tex]

Does that help?

cheers,

neopolitan
Rasalhague
#74
Mar18-09, 12:02 AM
P: 1,402
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.)
JesseM
#75
Mar18-09, 01:23 AM
Sci Advisor
P: 8,470
Quote Quote by neopolitan View Post
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 [tex]L_{cbb}[/tex] 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 [tex]L_{cbb}[/tex], 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 [tex]L_{cbb}[/tex] refers to the length of the apparatus in your frame when it's being carried by your buddy, and [tex]L'_{cbb}[/tex] 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 [tex]L_{cbb} = L'_{cbb} / \gamma[/tex], 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 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 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 [tex]L'_{cbb}[/tex] is exactly equal to my value L." Correct?
Quote 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 [tex]L'=L / \gamma[/tex]. (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, [tex]t'_{o}[/tex] and [tex]t'_{i}[/tex] 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 [tex]t'_{o}[/tex] and [tex]t'_{i}[/tex], 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 [tex]t_0[/tex] to another time [tex]t_1[/tex] 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 [tex]t_1[/tex] in your frame) - (his clock's readout at the event on the clock's worldline that occurs at time [tex]t_0[/tex] in your frame) is smaller than the difference [tex]t_1 - t_0[/tex] 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 [tex]t_0[/tex] and B happening at [tex]t_1[/tex] 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 Quote by neopolitan
I will find that [tex]\Delta t' = \Delta t / \gamma[/tex] (<- this is my equation, this is not time dilation!)
So, if according to your frame's definition of simultaneity, your clock's reading [tex]t_ 0[/tex] is simultaneous with your buddy's clock reading some time [tex]t'_0[/tex], and according to your frame's definition of simultaneity your clock's reading [tex]t_1[/tex] is simultaneous with your buddy's clock reading some time [tex]t'_1[/tex], and if [tex]\Delta t' = t'_1 - t'_0[/tex] and [tex]\Delta t = t_1 - t_0[/tex], then we get the equation [tex]\Delta t' = \Delta t / \gamma[/tex]. 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 [tex]\Delta t' = \Delta t * \gamma[/tex]. 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 [tex]\Delta t = \Delta t' * \gamma[/tex] using this convention. And of course, if we now divide both sides by gamma, we get back the equation you offered, [tex]\Delta t' = \Delta t / \gamma[/tex]. You can see that this is just a trivial reshuffling of the usual time dilation equation, not anything novel.
Quote Quote by neopolitan
Now I know that when [tex]\Delta t[/tex] has elapsed in my frame, [tex]\Delta t'[/tex] 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 [tex]t'_0[/tex] and B is simultaneous with [tex]t'_1[/tex]. Then would you agree that the time interval between these events in his frame is [tex]\Delta t' = t'_1 - t'_0[/tex]? And we also know that the time interval in your frame between the event of his clock reading [tex]t'_1[/tex] and the event of his clock reading [tex]t'_0[/tex] is related to this by [tex]\Delta t = \Delta t' * \gamma[/tex]. But that doesn't mean the time interval in your frame between A and B is [tex]\Delta t = \Delta t' * \gamma[/tex]!!! This is because although it's true that his frame's definition of simultaneity says that A is simultaneous with his clock reading [tex]t'_0[/tex] and B is simultaneous with his clock reading [tex]t'_1[/tex], 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 [tex]t'_0[/tex] and B may not be simultaneous with his clock reading [tex]t'_1[/tex], so knowing the time-interval in your frame between his clock reading [tex]t'_0[/tex] and his clock reading [tex]t'_1[/tex] 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 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 [tex]\Delta t[/tex]:

[tex]L / \Delta t = c = L' / \Delta t'[/tex]
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 [tex]\Delta t[/tex] and [tex]\Delta t'[/tex], are you going with the definition I suggested earlier where [tex]\Delta t'[/tex] is the difference between two clock readings [tex]t'_1[/tex] and [tex]t'_0[/tex] on your buddy's clock, and [tex]\Delta t[/tex] is the difference between two clock readings [tex]t_1[/tex] and [tex]t_0[/tex] on your clock, where you have picked the readings so that according to your frame's definition of simultaneity [tex]t_1[/tex] is simultaneous with [tex]t'_1[/tex] and [tex]t_0[/tex] is simultaneous with [tex]t'_0[/tex]? 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?
JesseM
#76
Mar18-09, 01:37 AM
Sci Advisor
P: 8,470
Quote Quote by Rasalhague View Post
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 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.
neopolitan
#77
Mar18-09, 05:04 AM
P: 645
"Picking appropriate values of [tex]L[/tex] and [tex]\Delta t[/tex]" 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 [tex]L[/tex], 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 [tex]\Delta t[/tex] so that L/[tex]\Delta t = c[/tex]. If you haven't picked a really big value of [tex]L[/tex], then [tex]/Delta t[/tex] will be pretty damn small so that it will be challenging to take two readings [tex]t_{o}[/tex] and [tex]t_{i}[/tex] where [tex]t_{i} - t_{o} = \Delta t[/tex] - 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 [tex]/Delta t'[/tex] 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
JesseM
#78
Mar18-09, 05:42 AM
Sci Advisor
P: 8,470
Quote Quote by neopolitan View Post
Here's what I mean about picking appropriate values, pick any value of [tex]L[/tex], 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 [tex]\Delta t[/tex] so that L/[tex]\Delta t = c[/tex]. If you haven't picked a really big value of [tex]L[/tex], then [tex]\Delta t[/tex] will be pretty damn small so that it will be challenging to take two readings [tex]t_{o}[/tex] and [tex]t_{i}[/tex] where [tex]t_{i} - t_{o} = \Delta t[/tex] - but we are in hypothetical world.
Are you just picking a value of [tex]\Delta t[/tex] out of thin air, with no connection to anything physical (so you could just as easily pick a [tex]\Delta t[/tex] such that L/[tex]\Delta t[/tex] = 5c or any number you wish), or is it supposed to represent the time interval between some specific pair of events, like [tex]t_{o}[/tex] representing the time a photon passes next to one end of an object of length L which is at rest in your frame, and [tex]t_{i}[/tex] representing the time that photon passes next to the other end of the same object?
Quote 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 [tex]/Delta t'[/tex] 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 [tex]t'_{o}[/tex] and [tex]t'_{i}[/tex], 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 [tex]\Delta t'[/tex] be, and what would [tex]\Delta t[/tex] be?
Quote Quote by neopolitan
I've described the events
Have you? Where? Are the events just the two readings on your buddy's clock?
Quote 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.
But if by "taking into account" the light delay, you mean taking the time on your clock (t=16 seconds in my example above) when you saw a reading on your buddy's clock (t'=8 seconds in the example) and then subtracting the distance/c that your buddy's clock was from you in your frame when it showed that reading (6 light-seconds/c = 6 seconds in the example) to get an earlier time on your clock (t=16-6=10 seconds), then this is physically equivalent (meaning you'll get the same answer for what the two clock readings would be) to just asking for the time on your clock that was simultaneous in your frame with the reading you saw on your buddy's clock (i.e. in the example, the reading of t'=8 seconds on your buddy's clock is simultaneous in your frame with the reading of t=10 seconds on your clock). If this is indeed what you meant by "taking into account", then do you agree that this is physically equivalent to my statement about simultaneity? Please answer yes or no.
Quote 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.
neopolitan
#79
Mar18-09, 06:08 AM
P: 645
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
Rasalhague
#80
Mar18-09, 09:29 AM
P: 1,402
Quote Quote by JesseM View Post
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 [tex]\Delta t[/tex] 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 [tex]\Delta t[/tex] 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 [tex]t_{B} = \gamma t_{A}[/tex], 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 [tex]t_{B} = \frac{t_{A}}{\gamma}[/tex]. 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 [tex]\Delta t[/tex] 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"?
JesseM
#81
Mar18-09, 01:12 PM
Sci Advisor
P: 8,470
Quote Quote by neopolitan View Post
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 [tex]t'_{o}[/tex] and [tex]t'_{i}[/tex], 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 [tex]\Delta t'[/tex] be, and what would [tex]\Delta t[/tex] 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 [tex]\Delta t[/tex] and [tex]\Delta t'[/tex] which appear in your equation L/[tex]\Delta t[/tex] = c = L'/[tex]\Delta t'[/tex]), 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.
JesseM
#82
Mar18-09, 04:23 PM
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P: 8,470
Quote Quote by Rasalhague View Post
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 [tex]t_{B} = \gamma t_{A}[/tex], 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 [tex]t_{B} = \frac{t_{A}}{\gamma}[/tex]. 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 [tex]\Delta t' = \Delta t * \gamma[/tex], 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 [tex]t_A[/tex] and [tex]t_B[/tex] into the time dilation equation giving [tex]t_A = t_B * \gamma[/tex]. Of course, if you wish to divide both sides by gamma, you can get back the formula [tex]t_{B} = \frac{t_{A}}{\gamma}[/tex] 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 [tex]t_{B1}[/tex] = 8 seconds, and ended with his clock reading [tex]t_{B2}[/tex] = 12 seconds, so the length of the interval in Bob's frame is [tex]\Delta t_B[/tex] = ([tex]t_{B2}[/tex] - [tex]t_{B1}[/tex]) = 4 seconds. If Alice wanted to know the time interval [tex]\Delta t_A[/tex] between these same two events in her frame, which is equivalent to wanting to know the time interval between the event [tex]t_{A1}[/tex] on her clock which is simultaneous in her frame with [tex]t_{B1}[/tex] (in this case [tex]t_{A1}[/tex] = 10 seconds) and the event [tex]t_{A2}[/tex] on her clock which is simultaneous in her frame with [tex]t_{B2}[/tex] (in this case [tex]t_{A2}[/tex] = 15 seconds), then she would plug these two different time intervals into the time dilation equation [tex]\Delta t' = \Delta t * \gamma[/tex], treating Bob's frame as unprimed and her frame as primed, which gives [tex]\Delta t_A = \Delta t_B * \gamma[/tex]. If she wanted to reverse this and figure out the time interval [tex]\Delta t_B[/tex] on Bob's clock between two events on [tex]t_{B1}[/tex] and [tex]t_{B2}[/tex] on his clock's worldline that are simultaneous in her frame with two events on her clock's worldline [tex]t_{A1}[/tex] and [tex]t_{A2}[/tex] that are the beginning and end of a time interval [tex]t_A[/tex] (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 [tex]\Delta t_B[/tex] as a function of [tex]\Delta t_A[/tex], i.e. [tex]\Delta t_B = \frac{\Delta t_A}{\gamma}[/tex].

On the other hand, the "temporal analogue of length contraction" [tex]\Delta t' = \Delta t / \gamma[/tex] 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 [tex]t_{A1}[/tex] to label the time on Alice's clock that's simultaneous in Bob's frame with Bob's clock reading [tex]t_{B1}[/tex], and you use [tex]t_{A2}[/tex] to label the time on Alice's clock that's simultaneous in Bob's frame with Bob's clock reading [tex]t_{B2}[/tex], then the time interval on Alice's clock ([tex]t_{A2} - t_{A1}[/tex]) is related to the time interval on Bob's clock ([tex]t_{B2} - t_{B1}[/tex]) by the formula [tex](t_{A2} - t_{A1}) = (t_{B2} - t_{B1}) / \gamma[/tex]. If we use the same numbers for [tex]t_{B1}[/tex] and [tex]t_{B2}[/tex] on Bob's clock as before, namely [tex]t_{B1}[/tex] = 8 seconds and [tex]t_{B2}[/tex] = 12 seconds, then in this case we'd have [tex]t_{A1}[/tex] = 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 [tex]t_{A2}[/tex] = 12*0.8 = 9.6 seconds. So the equation [tex](t_{A2} - t_{A1}) = (t_{B2} - t_{B1}) / \gamma[/tex] does work here, since [tex](t_{A2} - t_{A1}[/tex] = 9.6 - 6.4 = 3.2, [tex]t_{B2} - t_{B1}[/tex] 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 [tex]\Delta t' = \Delta t * \gamma[/tex] 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 [tex]\Delta t' = \Delta t * \gamma[/tex], I find it most natural to think that [tex]\Delta t[/tex] represents the difference between two clock-readings on a clock at rest in the unprimed frame, and then [tex]\Delta t'[/tex] 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 [tex]\Delta t'[/tex] 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.
neopolitan
#83
Mar18-09, 06:58 PM
P: 645
JesseM responding to Rasalhague:
Quote Quote by JesseM View Post
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.

<snip>

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 ...?
I should have asked:

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
JesseM
#84
Mar18-09, 07:46 PM
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P: 8,470
Quote Quote by neopolitan View Post
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).
Yes, although I hadn't actually realized that the "temporal equivalent of length contraction" equation could be used to answer exactly the same physical questions as the time dilation equation until Rasalhague asked that. And it was the specificity of the way he asked the question that made me realize this--he was asking about particular events on the worldlines of two physical clocks and stating in which frame an event on one clock's worldline was supposed to be simultaneous with a corresponding event on the other clock's worldline. You say that you "should have asked" this question:
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 ...?
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/[tex]\Delta t[/tex] = c = L'/[tex]\Delta t'[/tex]" 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.
neopolitan
#85
Mar18-09, 10:08 PM
P: 645
We were probably arguing at cross purposes, frustratingly enough for both of us. I thought you had taken that "temporal equivalent of length contraction" thing onboard a long time ago (it is in your diagram after all). So I was totally confused as to where you were coming from.

Since I thought you had understood the point and were still arguing it, it felt as if you were just trying to play games. That may have been a form of "tranference" (psychological term, relating to ascribing apparent motives of one person to another), since in real life I had a rather difficult person at work doing what I thought you were doing - playing dominance games through irrational argument.

I take your point about specifics. You may see that I have tried to be specific with figures in another thread.

May I ask why you had not come to the understanding that you just came to, when it seems that both Rasalhague and I did? This is not a hidden "you must be stupid" insult. I find you annoying, as you surely find me, but I don't find you stupid. What I am trying to do is see if you can identify, from the vantage point of someone who has only just came to this understanding, what prevents people from coming to this understanding naturally. Is there a block of some kind? If so, is it pedagogical or psychological?

(Clarification follows: I am distinguishing here between pedagogical and psychological, with a definition of "pedagogical" relating to how subjects are taught and "psychological" relating to the different ways in which people think and learn. Specific examples: "whole language" is a pedagogical method for teaching kids to read, moving away from phonics and instead recognition of whole words. As for "psychological", I am a visual, pattern identifying person which means that having a graph in front of me is more useful than a page of numbers. My visual, pattern identifying nature may lead me to link together all things that look the same (like all things with primes against them get grouped).)

This is the sort of discussion I really wanted back when I started the thread. Perhaps you might understand why I found the 80 or so posts in between frustrating, even if they were my own fault.

cheers,

neopolitan
JesseM
#86
Mar19-09, 03:12 AM
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P: 8,470
Quote Quote by neopolitan View Post
May I ask why you had not come to the understanding that you just came to, when it seems that both Rasalhague and I did? This is not a hidden "you must be stupid" insult. I find you annoying, as you surely find me, but I don't find you stupid. What I am trying to do is see if you can identify, from the vantage point of someone who has only just came to this understanding, what prevents people from coming to this understanding naturally. Is there a block of some kind? If so, is it pedagogical or psychological?
Sure, it basically comes from the way I had drawn it in that diagram I gave you, which was the first time I had even thought about the issue of a "temporal analogue for length contraction" (let's call it the TAFLC equation for short). Note that if we write the standard time dilation equation as [tex]\Delta t' = \Delta t * \gamma[/tex], I have no problem with the idea that you can divide both sides by gamma to get [tex]\Delta t = \Delta t' / \gamma[/tex] (call this the 'reversed time dilation equation'), which I think of conceptually as telling us the time elapsed on a clock at rest in the unprimed frame between two events on its worldline which we know are separated by a time-interval of [tex]\Delta t'[/tex] in the primed frame (I said basically the same thing about reshuffling the time dilation equation in post #61, the paragraph beginning with "Also"). But although this "reversed time dilation equation" looks exactly like the TAFLC equation [tex]\Delta t' = \Delta t / \gamma[/tex] except for the switch between primed and unprimed, I was mistakenly thinking that the physical meaning of [tex]\Delta t'[/tex] and [tex]\Delta t[/tex] in the TAFLC equation was totally different from either of the terms in the reversed time dilation equation. Again, the reason was how it was drawn in my diagram--I was thinking that [tex]\Delta t'[/tex] represented some weird notion of the temporal distance in the primed frame between two surfaces of simultaneity from the unprimed frame that crossed through readings on the worldline of the clock at rest in the unprimed frame which have a separation of [tex]\Delta t[/tex]. Superficially the notion of taking the temporal distance in the primed frame between two surfaces of constant t in the unprimed frame seems pretty weird and disconnected from anything physical (at least it did to me), an idea created only because it was analogous with taking the spatial distance in the primed frame between two worldlines of constant x in the unprimed frame, which is what length contraction is about.

What I had failed to realize was that if we imagine a physical clock at rest in the primed frame, then the "temporal distance between surfaces of constant t from the unprimed frame" just represents the difference [tex]\Delta t'[/tex] between the clock's readings at the two points where its worldline intersects these surfaces of constant t from the unprimed frame, and that if we then shift our perspective back to the unprimed frame, [tex]\Delta t[/tex] is now just the coordinate time between two readings on the primed clock, so now this is exactly like how I'd conceptualize the physical meaning of the terms in the reversed time dilation equation except with the roles of primed and unprimed reversed. So, this is one or two mental steps from what the TAFLC seemed to mean based on my diagram, and I didn't see the connection until I started working through a numerical example in response to Rasalhague's question. Also, it didn't help that I was used to conceptualizing the standard and reversed time dilation equations as relating a clock time-interval on a clock at rest in the unprimed frame with a coordinate time-interval in the primed frame, rather than normally thinking in terms of a clock at rest in the primed frame too. I was aware intellectually of the fact that the coordinate time in the primed frame between two events A and B could be rephrased in terms of readings on a physical clock at rest in the primed frame, specifically the difference between the reading that was simultaneous with A and the reading that was simultaneous with B according to the prime frame's definition of simultaneity. But that seemed like a more complicated way of thinking about the physical meaning of [tex]\Delta t'[/tex] (you can see it took me longer to write it out) so I usually just thought of it in terms of coordinate time.
neopolitan
#87
Mar19-09, 10:04 AM
P: 645
Quote Quote by JesseM View Post
Sure, it basically comes from the way I had drawn it in that diagram I gave you, which was the first time I had even thought about the issue of a "temporal analogue for length contraction" (let's call it the TAFLC equation for short). Note that if we write the standard time dilation equation as [tex]\Delta t' = \Delta t * \gamma[/tex], I have no problem with the idea that you can divide both sides by gamma to get [tex]\Delta t = \Delta t' / \gamma[/tex] (call this the 'reversed time dilation equation'), which I think of conceptually as telling us the time elapsed on a clock at rest in the unprimed frame between two events on its worldline which we know are separated by a time-interval of [tex]\Delta t'[/tex] in the primed frame (I said basically the same thing about reshuffling the time dilation equation in post #61, the paragraph beginning with "Also"). But although this "reversed time dilation equation" looks exactly like the TAFLC equation [tex]\Delta t' = \Delta t / \gamma[/tex] except for the switch between primed and unprimed, I was mistakenly thinking that the physical meaning of [tex]\Delta t'[/tex] and [tex]\Delta t[/tex] in the TAFLC equation was totally different from either of the terms in the reversed time dilation equation. Again, the reason was how it was drawn in my diagram--I was thinking that [tex]\Delta t'[/tex] represented some weird notion of the temporal distance in the primed frame between two surfaces of simultaneity from the unprimed frame that crossed through readings on the worldline of the clock at rest in the unprimed frame which have a separation of [tex]\Delta t[/tex]. Superficially the notion of taking the temporal distance in the primed frame between two surfaces of constant t in the unprimed frame seems pretty weird and disconnected from anything physical (at least it did to me), an idea created only because it was analogous with taking the spatial distance in the primed frame between two worldlines of constant x in the unprimed frame, which is what length contraction is about.

What I had failed to realize was that if we imagine a physical clock at rest in the primed frame, then the "temporal distance between surfaces of constant t from the unprimed frame" just represents the difference [tex]\Delta t'[/tex] between the clock's readings at the two points where its worldline intersects these surfaces of constant t from the unprimed frame, and that if we then shift our perspective back to the unprimed frame, [tex]\Delta t[/tex] is now just the coordinate time between two readings on the primed clock, so now this is exactly like how I'd conceptualize the physical meaning of the terms in the reversed time dilation equation except with the roles of primed and unprimed reversed. So, this is one or two mental steps from what the TAFLC seemed to mean based on my diagram, and I didn't see the connection until I started working through a numerical example in response to Rasalhague's question. Also, it didn't help that I was used to conceptualizing the standard and reversed time dilation equations as relating a clock time-interval on a clock at rest in the unprimed frame with a coordinate time-interval in the primed frame, rather than normally thinking in terms of a clock at rest in the primed frame too. I was aware intellectually of the fact that the coordinate time in the primed frame between two events A and B could be rephrased in terms of readings on a physical clock at rest in the primed frame, specifically the difference between the reading that was simultaneous with A and the reading that was simultaneous with B according to the prime frame's definition of simultaneity. But that seemed like a more complicated way of thinking about the physical meaning of [tex]\Delta t'[/tex] (you can see it took me longer to write it out) so I usually just thought of it in terms of coordinate time.
Thanks for that. With some things going on the background it took me some time to digest.

There is something which I find curious. It is a criticism of the pedagogy not of you nor of what time dilation is actually representing.

Note that you are "used to conceptualizing the standard and reversed time dilation equations as relating a clock time-interval on a clock at rest in the unprimed frame with a coordinate time-interval in the primed frame". It's quite a complex thing to internalise. When being taught, or trying to teach oneself, it is going to be a real uphill struggle to grasp that particular nature of the standard time dilation equation.

I certainly struggled with it and it was not helped that I have "back to fundamentals" sort of approach to mathematics which I applied to SR by reading a translation of Einstein's 1905 paper (I use the one at fourmilab.ch). I noted, and agonised over the fact that one of the standard equations is shown in mathematical form (length contraction) but the other is only given in words (on page 10) and this directly follows what was to me the far more intuitive equation - a form of "TAFLC" with [tex]\tau[/tex] instead of t'.

I do think there is a source of confusion there. I'd be willing to accept that it just me, but it seems there are many people with some problem or another with SR, which seems really odd. Why SR? Some are kooks for sure, but there are many people who seem to be otherwise able to maintain perfectly normal lives apart from an intuitive feeling that something is just not quite right about SR.

I won't say his name, and sadly he has probably passed away with cancer by now, but a professor in a city not far from where I lived up until recently was the lead lecturer for relativity at his university. He generously gave me many hours of his time to discuss my concerns and proposed solutions, and admitted that really, he didn't fully understand it. He did not stop me or explain that my concerns about time dilation were invalid, because he had never intuitively grasped the principles the standard way either. There is a fellow in southern Europe, another physics professor, albeit in a different field who expressed stronger views than I during our discussions that there was something amiss with SR. (I don't think SR is wrong but I do think it could be taught better.) A quantum physics professor in southern England also felt I was onto something with my arguments.

If professors of physics don't grasp SR properly, what chance do the average visitors to these forums have?

To make sure I am not presenting a biased account, I should clarify that at least four professors I corresponded with gave clear indications that they grasped SR well enough as taught (at least enough as to not be intrigued by my concerns), but sadly they had no time to go into it in depth with me. I had learned SR at university, read up on it, even going back to the original documents (Einstein and Feynman, Feynman because the light clock is sort of his). I had my uneasy feeling despite all this, and being told to go learn SR (again!) didn't really help.

Anyways, I've cast away a lot of my original stuff because I can now see that I was looking at the same thing as standard SR from a different perspective (I've not cast away everything, but I may in time cast away even the little that remains) and my deep-seated concerns that time dilation could actually be wrong were not justified.

However, if this feeling of there being something not quite right (which in my case were, as I said, deep-seated and may be equally concerning to others) is due to something as harmless as a pedagogical/psychological issue where some people intuitively think the way you do and others intuitively think another, yet both ways of thinking are completely valid, being just slightly different perspectives on the same thing, then it seems that there is some scope for improvement on how SR is taught.

I really do think that your suggestion a long time ago, when we had the discussion in which the diagram I posted here was central, was a good one.

You said that you would show your new students a similar diagram and explain that time dilation is not a TAFLC, and is not meant to be. I think you could go a little further and explain the physical significance of the actual TAFLC, and how it relates to length contraction so that c is invariant. That way, you would catch the people like me who feel that TAFLC is a useful equation and gently guide them towards a proper understanding of time dilation. At the same time, you would catch people like you, who go many years without grasping that there is any significance to a TAFLC.

Does that sound unreasonable?

cheers,

neopolitan

(I'm trying to get a lot into as few words as possible, it is late and it has been a long day. Sorry if there is anything which is hard to follow.)
JesseM
#88
Mar19-09, 10:03 PM
Sci Advisor
P: 8,470
Quote Quote by neopolitan View Post
Thanks for that. With some things going on the background it took me some time to digest.

There is something which I find curious. It is a criticism of the pedagogy not of you nor of what time dilation is actually representing.

Note that you are "used to conceptualizing the standard and reversed time dilation equations as relating a clock time-interval on a clock at rest in the unprimed frame with a coordinate time-interval in the primed frame". It's quite a complex thing to internalise. When being taught, or trying to teach oneself, it is going to be a real uphill struggle to grasp that particular nature of the standard time dilation equation.
But as I said to Rasalhague, part of why this may seem "natural" and not really that complex if you've already spent some time studying other theories of physics like Newtonian mechanics or QM is that when solving physics problems, the usual convention is to pick some initial conditions, representing a frozen instant in which you can visualize the arrangement of all the parts of the system in space at that instant, and then evolve them forward in (coordinate) time using the dynamical equations. Once you've picked the frame, the coordinate time of that frame is something that you just get used to thinking of as "time" for the sake of solving that problem, almost like the absolute time of Newton, although in the background of your mind you know that it's frame dependent. Put it this way--if everyone still believed in absolute time and space (and therefore absolute velocity), and we knew there was a time dilation effect where clocks in motion relative to absolute space "ran slow" in an absolute sense, then would you still find it complex or difficult to internalize the notion of a time dilation equation that tells you how much real time goes by when a certain amount of ticks go by on a moving clock? And wouldn't this be pretty much analogous to wanting to know how much real space is taken up by a moving ruler whose marks indicate it's a certain length, but which is shrunk in absolute terms because it's moving? And in this case if you treat the frame corresponding to absolute space and time as the primed frame, and the frame of the moving clock and moving ruler as the unprimed frame, you get the usual equations [tex]\Delta t' = \Delta t * \gamma[/tex] and [tex]\Delta x' = \Delta x / \gamma[/tex]. Of course you could also ask how many ticks go by on the moving clock given a certain amount of real time has passed, or what the rest length of a ruler is given that it's a certain length in real space, in this case you'd have to divide both sides of the first by gamma to get what I called the "reversed time dilation equation" [tex]\Delta t = \Delta t' /\gamma[/tex], and multiply both sides of the second by gamma to get [tex]\Delta x = \Delta x' * \gamma[/tex]. On the other hand, if you're thinking in terms of absolute space and time and treating the absolute frame as the primed one, then the meaning of the TAFLC equation [tex]\Delta t' = \Delta t /\gamma[/tex] seems less intuitive to me; I guess it would come out to something like "given that two events on the worldline of a clock at rest in absolute space are separated by a coordinate time of [tex]\Delta t[/tex] in a frame moving at velocity v relative to absolute space, how much real time (or clock time, given that the clock is not slowed-down) passes between those two events?"

Anyway, if you can see my point that the time dilation and length contraction equation seem fairly "natural" in a universe with absolute space and time, then maybe you can see why, once a physics student has gotten used to the idea of picking a coordinate system and then taking that system's space and time coordinates for granted for the purposes of actual calculating the dynamical behavior of physical systems, then it might seem equally natural to ask how much coordinate time goes by when a certain amount of ticks go by on a moving clock (moving relative to that coordinate system), or how much coordinate space is taken up by a moving ruler. That's the best way I can think of to explain why the equations make intuitive sense to me, but obviously it's subjective so not everyone would have the same intuitions.
Quote Quote by neopolitan
I do think there is a source of confusion there. I'd be willing to accept that it just me, but it seems there are many people with some problem or another with SR, which seems really odd. Why SR? Some are kooks for sure, but there are many people who seem to be otherwise able to maintain perfectly normal lives apart from an intuitive feeling that something is just not quite right about SR.
I've always thought that the main reason so many people have a problem with SR is because of the relativity of simultaneity, and what that might be taken to imply about the lack of any "objective" or "true" present, and therefore the lack of a real flow of time. In my experience--and I have seen a few physicists say the same thing--whenever people claim they have found a paradox in SR, the majority of the time it seems to come down to a failure to consider the relativity of simultaneity.
Quote Quote by neopolitan
I won't say his name, and sadly he has probably passed away with cancer by now, but a professor in a city not far from where I lived up until recently was the lead lecturer for relativity at his university. He generously gave me many hours of his time to discuss my concerns and proposed solutions, and admitted that really, he didn't fully understand it. He did not stop me or explain that my concerns about time dilation were invalid, because he had never intuitively grasped the principles the standard way either.
Is it possible that, like DaleSpam said above, he just found it simpler to use the full Lorentz transform to approach any problem which compares different frames? Since all the other distinct equations like the time dilation equation, the length contraction equation, the relativity of simultaneity equation, and the velocity addition equation are all derived from the Lorentz transform, I can see the appeal of just using that one set of equations rather than bothering to keep track of a bunch of different ones dealing with different quantities and concepts in SR.
Quote Quote by neopolitan
There is a fellow in southern Europe, another physics professor, albeit in a different field who expressed stronger views than I during our discussions that there was something amiss with SR. (I don't think SR is wrong but I do think it could be taught better.) A quantum physics professor in southern England also felt I was onto something with my arguments.
I have actually heard a few people working in quantum gravity who speculate that perhaps such a theory will restore a "true" flow of time and an objective present, Lee Smolin comes to mind for example. But I don't think this is a very common view.

Quote Quote by neopolitan
I really do think that your suggestion a long time ago, when we had the discussion in which the diagram I posted here was central, was a good one.

You said that you would show your new students a similar diagram and explain that time dilation is not a TAFLC, and is not meant to be.
Yes, if I was ever actually in a position to be teaching a class on SR, I'd be sure to do that! I don't want people reading this to get the impression that I'm a professor or anything... ;)
Quote Quote by neopolitan
I think you could go a little further and explain the physical significance of the actual TAFLC, and how it relates to length contraction so that c is invariant.
But when you say "how it relates to length contraction so that c is invariant", are you referring to the L/[tex]\Delta t[/tex] = c = L'/[tex]\Delta t'[/tex] argument? As I said that doesn't really make sense to me, because even if we assume that [tex]\Delta t[/tex] and [tex]\Delta t'[/tex] have the meaning given to them in the TAFLC equation, and L and L' have the usual meaning from the length contraction equation, I still don't see how it would make sense physically that L/[tex]\Delta t[/tex] and L/[tex]\Delta t'[/tex] could represent the "speed" of a single photon in two different frames if speed is given its usual interpretation as the distance covered in a certain interval of time. We can try going back to discussing this point if you want, or not if you don't want to get into it. As for the physical significance of the TAFLC, if we write it as [tex]\Delta t' = \Delta t / \gamma[/tex] I guess I would basically write it out as ""given that two events on the worldline of a clock at rest in the primed frame are separated by a coordinate time of [tex]\Delta t[/tex] in the unprimed frame, how much clock time (or coordinate time in the primed frame where the clock is at rest) passes between those two events?" Can you think of a more intuitive way to express the physical significance?

By the way, I'm about to go on a trip for a few days, so I probably won't be able to continue the discussion until next week sometime.
neopolitan
#89
Mar19-09, 11:44 PM
P: 645
Quote Quote by JesseM View Post
But when you say "how it relates to length contraction so that c is invariant", are you referring to the L/[tex]\Delta t[/tex] = c = L'/[tex]\Delta t'[/tex] argument? As I said that doesn't really make sense to me, because even if we assume that [tex]\Delta t[/tex] and [tex]\Delta t'[/tex] have the meaning given to them in the TAFLC equation, and L and L' have the usual meaning from the length contraction equation, I still don't see how it would make sense physically that L/[tex]\Delta t[/tex] and L/[tex]\Delta t'[/tex] could represent the "speed" of a single photon in two different frames if speed is given its usual interpretation as the distance covered in a certain interval of time. We can try going back to discussing this point if you want, or not if you don't want to get into it. As for the physical significance of the TAFLC, if we write it as [tex]\Delta t' = \Delta t / \gamma[/tex] I guess I would basically write it out as ""given that two events on the worldline of a clock at rest in the primed frame are separated by a coordinate time of [tex]\Delta t[/tex] in the unprimed frame, how much clock time (or coordinate time in the primed frame where the clock is at rest) passes between those two events?" Can you think of a more intuitive way to express the physical significance?
That definition is correct, although I would imagine a new student would need to be eased into it.

I can see why you can't make sense of L/t = c = L'/t'.

You specifically want to measure a time interval between two events in the primed frame and then compare that to a time inverval in the unprimed frame.

I wasn't doing that. I was saying that any time inverval in the primed frame between two events which are colocal in the primed frame, will be shorter in the unprimed frame than two analogous (but not the same) events in the unprimed frame. The half-life of one muon in the primed frame (viewed from the primed frame) will be the same as the half-life of a totally different muon in the unprimed frame (viewed from the unprimed frame. (Yes, I know half-lives are statistical, but using a gross misrepresentation here might still be instructive.)

What I am saying is that the half-life of the muon in the primed frame (viewed from the primed frame) will be less than the half-life of the muon in the primed frame (viewed from the unprimed frame).

In the example BobS raised earlier in the thread, a muon at a gamma of 29.3 had a measured life time of 64.4ms as opposed to a normal (gamma of 1) life time of 2.2ms.

In the experiment he refers to, I would call the measured lifetime t and I could use the gamma to calculate what the life time in muon's "rest frame" was (quotation marks because "rest frame" is a bit of a misnomer under the circumstances). I'd prime the rest frame of the muon and leave the laboratory rest frame unprimed. That would give me:

t' = t/gamma = 64.4ms / 29.3 = 2.2ms

If I had a different experiement, using light clocks, this is how I would be doing it.

At rest in the laboratory, my light clock has a tick time of 2.2ms. That makes the distance between mirrors ct/2 = 330km (giving a L = 660km, the total distance a photon travels between ticks).

Conceptually, put the light clock at a gamma factor of 29.3 (in reality, this would prove difficult).

I will measure, in the laboratory, that the time between ticks of the light clock is now 64.4ms.

This 64.6ms is the t which is equivalent to the t from the muon example. It is not equivalent to the t which I used in ct/2 = 330km (that t was 2.2ms).

What I do know is that, in the laboratory's frame, the photon in the light clock has not travelled 330km in 64.4ms. As you showed before (using time dilation) the photon has to travel much further from one mirror to the other mirror in one direction and a bit less in the other direction.

So the distance travelled between ticks (in the laboratory) is not the same L as before but rather ct where t = 64.4ms ... eg, 19320km.

This L, divided by this t = 19320km/64.4ms = 300000 km/s

The distance travelled in the rest frame of the light clock is the old L (330km) and the time a photon takes to travel between them and back again is the old t (2.2ms).

This L, divived by this t = 660km/2.2ms = 300000 km/s

If you want to use the clock in the laboratory you as your reference point, you have to do this:

While a photon in the laboratory moves between mirrors, travelling 660km in 2.2ms - what happens to a photon which is at gamma of 29.3?

If 2.2ms has elapsed in the laboratory, then a period of 2.2ms/29.3 = 75μs will have elapsed in the rest frame of the test clock (the one accelerated to gamma of 29.3). The test clock will not have ticked. In the rest frame of the test clock, the test clock's photon will only have travelled a distance of 22.5 km.

This time and this distance are t' and L'.

L'/t' = 22.5km/75μs = 300000 km/s

I hope this helps.

cheers,

neopolitan
Rasalhague
#90
Mar20-09, 01:51 AM
P: 1,402
Quote Quote by JesseM View Post
Anyway, if you can see my point that the time dilation and length contraction equation seem fairly "natural" in a universe with absolute space and time, then maybe you can see why, once a physics student has gotten used to the idea of picking a coordinate system and then taking that system's space and time coordinates for granted for the purposes of actual calculating the dynamical behavior of physical systems, then it might seem equally natural to ask how much coordinate time goes by when a certain amount of ticks go by on a moving clock (moving relative to that coordinate system), or how much coordinate space is taken up by a moving ruler. That's the best way I can think of to explain why the equations make intuitive sense to me, but obviously it's subjective so not everyone would have the same intuitions.
Maybe this is part of what confuses a novice like me with less experience of physics in general to draw on. Namely that, having been taught that there is no absolute space and time, we're then tacitly invited to pretend there is “for the sake of solving the problem”. But as a beginner, that leaves you wondering *how much* of the your intuition you're allowed to hang onto in this particular exercise. And unless it’s made explicit, you just can't tell, because the one thing you've been warned is that you can't trust your intuition when it comes to relativity. So I worry that I might make mistakes by being lulled by such a natural-seeming way of conceptualising it. Or, to put it another way, the "natural" way of treating one frame as preferential, for the sake of convenience, can sometimes feel to me as if it's bringing swarming after it all those apparent paradoxes that disappear only when you abandon certain intuitions, such as--especially--absolute simultaneity. But maybe when I'm more familiar with SR, that won't be so much of an issue.

I suppose "time dilation" and "length contraction" being just a shorthand for the full Lorentz transformation, of use in a special cases, the thing to be learnt is what those special cases are, and (on a more philosophical or abstract level) why a different special case is thus highlighted for time from the special case thus highlighted for space. Regarding which, I've found this a fascinating discussion.

Quote Quote by JesseM View Post
I've always thought that the main reason so many people have a problem with SR is because of the relativity of simultaneity, and what that might be taken to imply about the lack of any "objective" or "true" present, and therefore the lack of a real flow of time. In my experience--and I have seen a few physicists say the same thing--whenever people claim they have found a paradox in SR, the majority of the time it seems to come down to a failure to consider the relativity of simultaneity.
Definitely! I certainly found that when I learnt that simultaneity was relative too--although it's such a fiendishly counterintuitive concept--that was the moment when some of these bizarre ideas first started to fall into place. They're still very hard for me to understood, but they no longer feels an affront to reason! The other technique that often clears things up for me is to break the problem down and think of it in terms of events. That often helps to root out the false assumptions lurking in my brain.

Quote Quote by JesseM View Post
As for the physical significance of the TAFLC, if we write it as [tex]\Delta t' = \Delta t / \gamma[/tex] I guess I would basically write it out as "given that two events on the worldline of a clock at rest in the primed frame are separated by a coordinate time of [tex]\Delta t[/tex] in the unprimed frame, how much clock time (or coordinate time in the primed frame where the clock is at rest) passes between those two events?" Can you think of a more intuitive way to express the physical significance?
As in my example, I found it helpful to give the notional observers names, and make their circumstances perfectly symmetrical. That seemed to be the only way I could start get my head around which parts of the problem were significant features of spacetime geometry, and which were accidental details of the example. In the descriptions I'd encountered, I often found myself struggling to keep track over whether a particular author was using the primed/unprimed convention to represent some unique feature of a particular frame. Some introductions use unprimed as you describe, but others use it according to some other convention, or arbitrarily. And of course, where the problem is more elaborate and involves converting back and forth between frames, or where the direction of movement is significant, it's less obvious which frame is the more natural choice to be called unprimed.


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