Time Dilation Formula: Clarifying Confusion

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The discussion centers on the correct formula for time dilation, with participants debating whether it is t' = γt or t' = t/γ. The consensus leans towards t' = t/γ, as it aligns with the observation that a moving clock appears to run slower compared to a stationary observer's clock. This leads to the conclusion that less time elapses on the moving clock, reinforcing the concept of time dilation. Additionally, the conversation touches on the terminology used to describe time dilation and the relativity of time perception between different observers. Ultimately, the complexities of time dilation and its implications in relativity are acknowledged, highlighting the nuanced understanding required in these discussions.
  • #61
matheinste said:
Dilation, no matter who uses it, always means making larger. Time dilation in all the references we have quoted always means the same thing.

For Adams, Freund, Lerner, Petkov, Schröder, Weinert, Taylor/Wheeler and Tipler/Mosca, it means dilating the total number of units. For you, Lawden and Anthony it means dilating the size of each unit and thus contracting the total number of units. It's as if doctors had one term "pupil dilation", but some doctors used it to mean that a patient's pupils are bigger than normal, while others used it to mean that normal is bigger than a patient's pupils (i.e. the opposite of how their colleagues understand the term), claiming that it makes no difference because in either case dilation refers to something being larger!

matheinste said:
The resting observer sees the other clock running slow. Some refernces translate this to "the moving observer sees his proper time projected onto the resting frame coordinates as being increased, running fast, more seconds passed", it still means the same, it just expresses it differently.

Since "observer" and "clock" are each resting in some frame and each moving in another (observer being a colloquial short-hand in this context for "intertial reference frame", and any inertial reference frame being populated by its own notional clocks), we need to pick some standard to say what is being compared to what. If we have a particular example, the standard is chosen for us by the details of the example, by which time interval we're given and which we need to calculate. In the most general case, what is there to break the symmetry and escape reasoning round in circles? The only thing I can think of here is that we have our input, the data we know, as our standard, then the formula gives us some output expressed in terms of that standard. This notion of input and output introduces a natural way of ordering the pair of time intervals, and this allows us to talk meaningfully about which is to be made bigger by which operation. Since we're calling this action time dilation, something relating to time must have been made bigger by it. If \Delta t' = \gamma \; \Delta t is time dilation, then a total is dilated, as we're told by Adams, Freund, Lerner, Petkov, Schröder, Weinert, Taylor/Wheeler and Tipler/Mosca. But if, like Lawden, we use a similar expression--"a moving clock [...] will appear [...] to have its rate reduced--to describe the inverse formula, and call this time dilation, then presumably it's the size of each unit that's been dilated, because it isn't the total: that's got smaller.

matheinste said:
Moving clocks run slow, very loosely, says all that needs to be said.

But every clock at rest in some inertial frame is moving in another, the physical situation being perfectly symmetrical. So if dilation is to have any meaning at all, there must be some convention as to what it refers to. Otherwise, why not call it time distortion and save awkward questions. Sure, we could dodge the question by switching our definition of what dilation refers to whenever we want to change from using one of these reciprocal expressions to the other, so as to disguise the fact that they're reciprocal and not identical, but that's hardly a recipe for clarity.
 
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  • #62
Its not that complicated.

ANY inertial observer will reckon that ANY clock moving inertially with respect to him is running slow. This may be expressed in other ways but, however it is expresed, it means this and the fact that it does mean this can be inferred from the given scenario. That's all there is to it.

So with regards to our overall discussion, yes, different authors do seem to illustrate time dilation diferently, but however they do it, they are describing the same effect. I can explain in more detail but for me it is long and winding road to express the two views completely unambiguously because although they are in a way reciprocal the use of that word in this context can cause more confusion.

Matheinste.
 
  • #63
matheinste said:
Its not that complicated.

ANY inertial observer will reckon that ANY clock moving inertially with respect to him is running slow. This may be expressed in other ways but, however it is expresed, it means this and the fact that it does mean this can be inferred from the given scenario. That's all there is to it.

Sure, it's not complicated when we leave the dilation out of it. But wasn't this a discussion about the complications relating to people's varying uses of that word?

matheinste said:
So with regards to our overall discussion, yes, different authors do seem to illustrate time dilation diferently, but however they do it, they are describing the same effect. I can explain in more detail but for me it is long and winding road to express the two views completely unambiguously because although they are in a way reciprocal the use of that word in this context can cause more confusion.

There are two functions, each the inverse of the other, and some people call one of them time dilation, and some call the other time dilation. What one function dilates, the other contracts, so they don't both dilate time in the same sense. Take my "pupil dilation" analogy. You could argue that each set of doctors is "describing the same effect", but they're using the word dilation in opposite ways to describe that effect; one doctor's dilation is another's contraction: not a healthy situation!
 
  • #64
matheinste said:
Its not that complicated.

ANY inertial observer will reckon that ANY clock moving inertially with respect to him is running slow. This may be expressed in other ways but, however it is expresed, it means this and the fact that it does mean this can be inferred from the given scenario. That's all there is to it.

Matheinste.

It is all very well to make a statement like that, but if the number of units can in one view increase and in the other decrease, in one sense the moving clock reads more time has passed and in the other that less time has passed, for do we not reckon time by the number of units of time passing rather than by the size of them?

So does the clock slow because each tick takes longer?

Another interesting fact is that however one measures it the total duration of whatever we are measuring is the same, moving or not.
The number of seconds multiplied by the length of one second gives the same total whether it is proper time or co-ordinate time. The difference is that the unit of measurement changes: take the muon experiment referred to earlier where we have 2.2 microseconds proper time and approximately 65 microseconds co-ordinate time and the conversion is made applying the Lorentz factor which was 29.4

So in which way is it slowing?

Grimble.
 
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  • #65
Rasalhague said:
Sure, it's not complicated when we leave the dilation out of it. But wasn't this a discussion about the complications relating to people's varying uses of that word?



There are two functions, each the inverse of the other, and some people call one of them time dilation, and some call the other time dilation. What one function dilates, the other contracts, so they don't both dilate time in the same sense. Take my "pupil dilation" analogy. You could argue that each set of doctors is "describing the same effect", but they're using the word dilation in opposite ways to describe that effect; one doctor's dilation is another's contraction: not a healthy situation!

Its no complicated even with time dilation.

Longer period and more ticks are reciprocal. But they are obtained by using reciprocal scenarios. They "cancel out" to produce the same effect. There is only ONE effect. We are using TWO, opposite ways of describing the same effect.

In one scenario the "STATIONARY" observer is saying "according to MY reckoning the other guy's clock is showing LESS ticks than mine and so his basic period is LONGER". In the other scenario the "MOVING" observer is saying " the other guy is saying that according to HIS reckoning my clock is showing less ticks than his and so according to HIS reckoning his clock must be showing MORE ticks than mine and so according to HIS reckoning his clock's basic period is SHORTER" . They are both correct. They are both describing the same effect. Compared to stationary observer the moving observer's clock is running slow.

Notice that in the second scenario the moving guy is saying, not what he himself sees, but what the stationary observer sees and making a correct inference from this. This is because, in fact, the stationary observer does see his own clock as running faster than the other.

Ita bit convoluted but that seems to be how it works out.

I much prefer the much simpler description based on simulatnaous clock readings.

Matheinste
 
  • #66
matheinste said:
In one scenario the "STATIONARY" observer is saying "according to MY reckoning the other guy's clock is showing LESS ticks than mine and so his basic period is LONGER". In the other scenario the "MOVING" observer is saying " the other guy is saying that according to HIS reckoning my clock is showing less ticks than his and so according to HIS reckoning his clock must be showing MORE ticks than mine and so according to HIS reckoning his clock's basic period is SHORTER" . They are both correct. They are both describing the same effect. Compared to stationary observer the moving observer's clock is running slow.

It's like saying "more than" means the same as "less than" because if x < y then y > x. It's true that these particular inequalities say the same thing--and that we could get by perfectly well with one or the other sign on all occasions--but that doesn't mean that we can disregard the difference between "more than" and "less than" because they're "opposite ways of describing the same effect". No matter how we look at it, x < y is the opposite of x > y.

Here what you've done is to apply the two mutually inconsistent definitions of time dilation to the same situation, and to make this work, you've had to switch your standard of comparison when you start talking from the other perspective. But if we apply a consistent standard of comparison, the only way to make it work is to also use a single, consistent defnition of time dilation. Much less fraught!

When I compared x < y and x > y, all I changed was the inequality. I did this to demonstrate the effect of changing the inequality. If I'd changed the inequality and simultaneously switched variables, then all I'd have done would have been to disguise the genuine difference between "more than" and "less than". That's why, in post #52, I referred each definition to the same arbitrarily chosen standard.
 
  • #67
It's as if someone were to argue that inflation and deflation both "describe the same effect" since the inflation of one currency is equivalent to the deflation of another. But see what we did there? We can only argue that they're equivalent by changing which currency we refer to whenever we switch from calling the phenomenon inflation to deflation or vice-versa. If we consistently refer to one currency, and consistently define inflation as rising prices, there's no way we can claim that inflation means the same as deflation (with respect to that same currency).
 
  • #68
Rasalhague said:
It's as if someone were to argue that inflation and deflation both "describe the same effect" since the inflation of one currency is equivalent to the deflation of another. But see what we did there? We can only argue that they're equivalent by changing which currency we refer to whenever we switch from calling the phenomenon inflation to deflation or vice-versa. If we consistently refer to one currency, and consistently define inflation as rising prices, there's no way we can claim that inflation means the same as deflation (with respect to that same currency).

No it is not. You miss the point. Both scenarios, illustrations, explanations do describe the same effect, but they describe the same phenomena whereby "moving" clocks appear to be running slow compared to a "stationary" observers clock as observed by the "stationary" observer. One says it explicitly that way, the other says "moving" observers reckon "stationary" observers will see the "stationary" clock running fast compared to their, the "moving" observers, own clock. The first uses extended basic time periods the other uses the extended passage of time or more ticks as part of their explnation. The outcome is the same and the rest of SR which is greatly dependent on the phenomena is, obviously, unaffected by which is used otherwise there would be great disagreement at the later stages of teaching the theory.

Its a shame to end our discusiion without agreement and it is probably as frustrating for you as it is for me. Among the many possible problems the two most likely ones are that I lack the verbal skills to get my point across or I am incorrect in my interpretations. In either case, mea culpa.

Its been an interesting exercise and I have learned from it.

Matheinste.
 
  • #69
matheinste said:
No it is not. You miss the point. Both scenarios, illustrations, explanations do describe the same effect, but they describe the same phenomena whereby "moving" clocks appear to be running slow compared to a "stationary" observers clock as observed by the "stationary" observer. One says it explicitly that way, the other says "moving" observers reckon "stationary" observers will see the "stationary" clock running fast compared to their, the "moving" observers, own clock.

My point is that in order to make the two conflicting definitions correctly describe the same situation, you've had to switch your standard of comparison. In my analogy, we define inflation to mean that one can buy less given a fixed total of money (prices get bigger). Of course, this also means that a fixed total of the affected currency can be bought for less of another currency (price gets smaller). But then we'd have redefined our original currency as a comodity, and taken a different currency as our standard against which to compare it. To see what difference the different definitions make, we need to apply them to the same standard.

So we might say that the inflation of one currency is equiavalent to the deflation of another, but that doesn't make inflation the same thing as deflation. Yes, your two verbal summaries of time "distortion" describe the same situtation. Whether both of these verbal summaries are characterised as time dilation or time contraction depends on whether we're referring to units or total. But as soon as we specify what kind of value is already known and what kind of value, relative to that, we want to calculate from it (a bigger one or a smaller one), then the difference in terminology becomes apparent, because then the definition shared by Adams, Freund, Lerner, Petkov, Schröder, Taylor & Wheeler, Tipler & Mosca leads to a dilated value being called "dilated time" (the dilation being explicit, a dilation of the quantity given), while the definition shared by Lawden and Anthony leads to a contracted value being called "dilated time" (the dilation being implict, a dilation of units).

Luckily, so long as the relevant details are known, and the Lorentz transformation and how to apply it are understood, there shouldn't be any disagreement over results. So in that sense it's not catastrophic if people take dilation to refer to different aspects of the same situation. The main problem that I see with such conflicting definitions is that it can be distracting for people trying to learn the subject.

matheinste said:
Its been an interesting exercise and I have learned from it.

Me too! Thanks for your patience.
 
  • #70
Thank you Gentlemen for an illuminating discussion, but what are your answers to the original subject of this thread; "Which is the correct formula for time Dilation?"

Rasalhague said:
Not if, like Adams, Freund, Lerner, Petkov, Schröder and Taylor & Wheeler, we call the following operation time dilation and refer to \Delta t&#039;_{1} as "dilated time":

(1) \; \Delta t&#039;_{1} = \gamma \Delta t_{1}

But yes if, like Anthony and Lawden, we call the following operation time dilation and refer to \Delta t&#039;_{2} as "dilated time":

(2) \; \Delta t&#039;_{2} = \gamma^{-1} \Delta t_{2}
This seems to be the nearest you have come to answering, stating that there are differing opinions; but as those opinions are in direct conflict where does this leave one?

And as Einstein derives his formula, http://www.bartleby.com/173/12.html" , in this passage:

“Let us now consider a seconds-clock which is permanently situated at the origin (x' = 0) of K'. t' = 0 and t' = 1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:
t = 0
and
E5.GIF

which is t = \frac {t^&#039;}{\sqrt{1 - \frac{v^2}{c^2}}} with {t^&#039;} set to 1

Which gives us t = \gamma {t^&#039;} as Einstein's formula

Or option (2) in the quote above...

So would I be right in deducing that (2) is, in fact, the correct formula??

:confused::confused::confused:Grimble
 
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  • #71
It depends on exactly what one means by \Delta t and \Delta t^{\prime}.
 
  • #72
Grimble said:
So would I be right in deducing that (2) is, in fact, the correct formula??

Sorry if I went off topic there...

As I said in post #52, I only used the prime symbol in those examples to indicate output, i.e. the value that we're using the formula to calculate. Here Einstein uses a different convention. In this particular formula, he labels his input t' (that's the given value, one second) and his output t (that's the value he's using the formula to compute). So the formula he uses is of type (1) in my scheme.

It's not a matter of either being correct or incorrect, as such, and (as we'd expect!) Einstein uses the correct formula for his purpose. They're each appropriate to a different task. The only disagreement is over which of them would be called time dilation (and which its inverse), and why... Does Einstein use the term "time dilation" (Zeitdilation)? If not, we can't be sure whether he'd label this particular formula "the time-dilation formula". That said, he does characterise his result as "a somewhat larger time", so like the majority of the writers I looked at (Adams, Freund, Lerner, Petkov, Schröder, Weinert, Taylor & Wheeler, Tipler & Mosca), presumably Einstein would regard time dilation as an operation that takes a smaller interval of time (a smaller total of seconds) and produces a larger interval of time (a larger total of seconds). The result of Einstein's calculation is a larger number. The above-named authors called the formula that produces a larger number "time dilation", and they called its result (the larger number) "dilated time".

This contasts with Lawden who presents the inverse of the formula Einstein uses here and calls that time dilation, and it contrasts with Anthony who called the smaller total produced by \Delta t&#039; = \gamma^{-1} \; \Delta t "dilated time".
 
  • #73
Should we not take Einstein's own usage as the convention in this case?

He referred to the stationary system as K and the moving system as K'; and his time t was that of the stationary system and t' that of the moving system as transformed by the Lorentz equations.
 
  • #74
...if the number of units can in one view increase and in the other decrease, in one sense the moving clock reads more time has passed and in the other that less time has passed, for do we not reckon time by the number of units of time passing rather than by the size of them?

So in what way does the clock slow?

Another interesting consideration is that however one measures it the total duration of whatever we are measuring is the same, moving or not.
The number of seconds multiplied by the length of one second gives the same total whether it is proper time or co-ordinate time. The difference is, that the unit of measurement changes: take the muon experiment referred to earlier where we have 2.2 microseconds proper time and approximately 65 microseconds co-ordinate time and the conversion is made applying the Lorentz factor which was 29.4

So in which way is it slowing?

And surely, whichever way we calculate it, the co-ordinate seconds have to be smaller than proper seconds, in the same way that co-ordinate metres are smaller than proper metres.

For how else can the speed of light in the moving system - measured in co-ordinate units from the stationary system - still be c?

If the transformed lengths are contracted and the times are dilated how can c = \frac{d}{t} = \frac {d^&#039;}{t^&#039;}

Grimble.
 
  • #75
Grimble said:
Should we not take Einstein's own usage as the convention in this case?

He referred to the stationary system as K and the moving system as K'; and his time t was that of the stationary system and t' that of the moving system as transformed by the Lorentz equations.

In section 11 of the book you linked to, Einstein uses t' to denote a time which results from applying the Lorentz transformation to a time labelled t. But in section 12, he uses t to denote a time which results from applying the Lorentz transformation to a time labelled t'. So if there is a convention here, it's not defined relative to the Lorentz transformation, which fortunately does have a conventional form.

http://www.bartleby.com/173/11.html
http://www.bartleby.com/173/12.html

Einstein doesn't use the term "moving system" in either of these sections, as far as I can see. (I haven't read the whole book, so maybe it's used elsewhere.) But I don't know how the terms "moving system" and "stationary system" could be used to determine a convention for which frame to call K', since these terms are just as arbitrary, given that each system is moving relative to the other, and an observer at rest in either system will observe the same effect (slowness) in a clock at rest in the other system.
 
  • #76
Rasalhague said:
In section 11 of the book you linked to, Einstein uses t' to denote a time which results from applying the Lorentz transformation to a time labelled t. But in section 12, he uses t to denote a time which results from applying the Lorentz transformation to a time labelled t'. So if there is a convention here, it's not defined relative to the Lorentz transformation, which fortunately does have a conventional form.

http://www.bartleby.com/173/11.html
http://www.bartleby.com/173/12.html

Einstein doesn't use the term "moving system" in either of these sections, as far as I can see. (I haven't read the whole book, so maybe it's used elsewhere.) But I don't know how the terms "moving system" and "stationary system" could be used to determine a convention for which frame to call K', since these terms are just as arbitrary, given that each system is moving relative to the other, and an observer at rest in either system will observe the same effect (slowness) in a clock at rest in the other system.

But in chapter 11 he writes:
A co-ordinate system K then corresponds to the embankment, and a co-ordinate system K' to the train.
and K' is the moving system in Figure 2.

Unfortunately it is confusing until we realize that what he is saying with relation to the Lorentz transformations is that the time in an inertial frame of reference (IFoR) is proper time and we apply the Lorentz transformation to convert it to co-ordinate time, the time perceived in another inertial frame of reference moving at a constant velocity with respect to the first.
If we then compare the resultant co-ordinate time with with the proper time in the 2nd IFoR we, not surprisingly, find that the relationship is the Lorentz Factor.
I believe we can change labels like system A and system B to either IFoR but, because they are fundamental to the Lorentz transformation equations, we need to have a solid convention for what they are ferring to.
 
  • #77
Grimble said:
But in chapter 11 he writes:

"A co-ordinate system K then corresponds to the embankment, and a co-ordinate system K' to the train."

and K' is the moving system in Figure 2.

Okay, so K' is shown moving in Fig. 2, while K is at rest with respect to the page. This K' is what's known as the "rest frame" of the train (that's to say the frame where the train is at rest). In Chapter 12, K' is the label used for the rest frame of the clock. The equations of the Lorentz transformation in Chapter 11 tell us how to convert intervals of time and distances of space, as measured with respect to K, the embankment's rest frame, into the corresponding times and distances with respect to K', the train's rest frame.

It's natural for us to think of the embankment as stationary and the train as moving, but this is an artifact of the scenario. Taylor and Wheeler take a similar approach; they begin by defining one coordinate system as "the lab(oratory) frame" and another as "the rocket frame", meaning the rest frame of a laboratory and the rest frame of a rocket which they use in their thought experiments. Similarly with muons, it's natural for us to think of the muon's rest frame as moving relative to the earth. But in all of these scanarios, the effects due to special relativity are entirely symmetric. No frame is moving or stationary in an absolute sense, only relative to something.

Grimble said:
Unfortunately it is confusing until we realize that what he is saying with relation to the Lorentz transformations is that the time in an inertial frame of reference (IFoR) is proper time and we apply the Lorentz transformation to convert it to co-ordinate time, the time perceived in another inertial frame of reference moving at a constant velocity with respect to the first.
If we then compare the resultant co-ordinate time with with the proper time in the 2nd IFoR we, not surprisingly, find that the relationship is the Lorentz Factor.
I believe we can change labels like system A and system B to either IFoR but, because they are fundamental to the Lorentz transformation equations, we need to have a solid convention for what they are ferring to.

I've been struggling with these issues myself as I've been learning about sepcial relativity, and I certainly don't have all the answers... I tend to prefer standard, general, abstract terms that will work in any scenario, and which don't hide the symmetry (rest frame, proper time, spacetime interval, etc.). But some people prefer to begin with more concrete sounding names like "laboratory frame" and "rocket frame" (those are Taylor and Wheeler's examples), or you could think of the train's rest frame and the embankment's rest frame in Einstein's example.

Proper time is a good term. It's traditionally represented by the Greek letter \tau (tau). In special relativity, proper time is a timelike spacetime interval between two events. It's invariant, which means that it's the same in all inertial reference frames. It's equal to the coordinate time in the unique inertial frame of reference where the events happen in the same place as each other. It's unambiguous as long as we define which pair of events we're talking about the proper time between. In Einstein's example, both events lie on the worldline of a certain clock. The first event is the clock showing 0 seconds, and the other event is the clock showing 1 seconds. These events happen in the same place in the clock's rest frame (where, by definitio, the clock is at rest), so the coordinate time in that frame is equal to the proper time between these events. In any other frame, the coordinate time will be greater than the proper time. We find the coordinate time between these events using t_{coordinate} = \gamma \; \tau, which is just the time equation from the Lotentz transformation with x = 0.
 
  • #78
Grimble said:
...if the number of units can in one view increase and in the other decrease, in one sense the moving clock reads more time has passed and in the other that less time has passed, for do we not reckon time by the number of units of time passing rather than by the size of them?

So in what way does the clock slow?

Yes, time is measured by the number of units, which is why it seemed more natural to me to call time dilation the process whereby a larger number of units is derived from a smaller one.

A "moving" clock runs slow in the sense that it shows less time passing (a smaller number of units) than a "stationary" clock. To an observer who considers themselves stationary, a clock moving relative to this observer will show a smaller number of units as having passed in any given interval than a clock which is stationary relative to the obsever.

The same scenario from a more symmetrical perspective: imagine two clocks, A and B, moving relative to each other at some constant velocity. The proper time between two events on clock A's worldline is always less than the proper time between two events on the worldline of clock B if the following two conditions hold. (1) The first event on clock A's worldline is simultaneous in clock B's rest frame with the first event on clock B's worldline; (2) the second event on clock A's worldline is simultaneous in clock B's rest frame with the second event on clock B's worldline.

Since A and B are arbitrary labels, this will still be true if you swap them.

Events are points in spacetime. In this example, they represent a clock showing a particular time. An object's worldline is its trajectory (the path it takes) through spacetime. In special relativity, which deals with objects moving constant velocity relative to intertial reference frames, worldlines are straight lines.

Another name for the longer of these two intervals of time (the proper time shown by clock B in this example) is the coordinate time in clock B's rest frame of the proper time between the stated pair of events on clock A's worldline.

Best read in with one eye on a spacetime diagram!
 
  • #79
Grimble said:
Another interesting consideration is that however one measures it the total duration of whatever we are measuring is the same, moving or not.
The number of seconds multiplied by the length of one second gives the same total whether it is proper time or co-ordinate time. The difference is, that the unit of measurement changes: take the muon experiment referred to earlier where we have 2.2 microseconds proper time and approximately 65 microseconds co-ordinate time and the conversion is made applying the Lorentz factor which was 29.4

So in which way is it slowing?

I don't understand your first point here. We can conceive of the calculation as making the total bigger, or equivalently as making the units smaller, but not both, because that would be like multiplying by gamma squared, wouldn't it? Or if you made the total of one clock bigger, and the units of the clock you're comparing it to smaller, then that would be like making them both bigger, or both smaller, i.e. multiplying by gamma and dividing by gamma, i.e. no change. But there is a change, the calculation converts 2.2 into 65.

We're talking about the interval of time between two events on the muon's worldline, its creation and its annihilation. This interval is the muon's lifetime. The muon lasts 2.2 microseconds in its own rest frame and 65 microseconds in the rest frame of the lab. The time between events that happen in the same place is the proper time between them (a timelike spacetime interval). Since the muon's birth and death happen in the same place in the muon's rest frame, the proper time between them is 2.2 microseconds. In the lab's rest frame, the coordinate time of this interval is 2.2 * 29.4 = approx. 65 microseconds. If we think of the muon as a clock, we can say informally that the muon is running slow in comparison to a clock at rest in the lab's rest frame in the sense that the muon only counts to 2.2 microseconds while the lab clock counts to 65. From the perspective of an observer at rest with respect to the lab, a 2.2-microsecond muon is slow to disappear; it lasts for 65 seconds.

Grimble said:
And surely, whichever way we calculate it, the co-ordinate seconds have to be smaller than proper seconds, in the same way that co-ordinate metres are smaller than proper metres.

For how else can the speed of light in the moving system - measured in co-ordinate units from the stationary system - still be c?

If the transformed lengths are contracted and the times are dilated how can c = \frac{d}{t} = \frac {d^&#039;}{t^&#039;}

Grimble.

You're right. If we conceptualise coordinate seconds as being smaller than proper seconds, and coordinate meters as smaller than proper meters, there will be more of either corresponding to a given proper time (timelike spacetime interval) or a given proper distance (spacelike spacetime interval). Time dilation, as most of the textbooks understand it, refers to the calculation of the coordinate time of the proper time between a given pair of events. To calculate this, we multiply the proper time by gamma. Likewise, to find the coordinate distance of the proper distance between two events, we multiply the proper distance by gamma. This latter procedure is not the same thing as "length contraction", for which the inverse equation is required.
 
  • #80
Rasalhague said:
I don't understand your first point here. We can conceive of the calculation as making the total bigger, or equivalently as making the units smaller, but not both, because that would be like multiplying by gamma squared, wouldn't it? Or if you made the total of one clock bigger, and the units of the clock you're comparing it to smaller, then that would be like making them both bigger, or both smaller, i.e. multiplying by gamma and dividing by gamma, i.e. no change. But there is a change, the calculation converts 2.2 into 65.
I'm sorry, I haven't explained it well but in the following quote you say that Freund et al. take "dilated time" (co-ordinate units?) to mean an expanded total (of reduced units).
Rasalhague said:
My point is just that Freund, like Adams, Lerner, Petkov and Schröder, takes "dilated time" to mean an expanded total (of reduced units). So for all of these authors, dilation seems to refer to the quantity of units, the total, rather than--as I thought you originally suggested--the size of individual units. If these authors had taken dilation to refer to the size of units, then surely they'd have used the label "dilated time" for the interval made up of a reduced quantity of these dilated units, wouldn't they?

And what I am saying is that if the co-ordinate time is an expanded number of reduced units, and if the 'expansion' and 'reduction' are both according to the Lorentz factor, then the only conclusion is that the total duration remains constant. It is precisely the number of units and the size of the units that change.

We're talking about the interval of time between two events on the muon's worldline, its creation and its annihilation. This interval is the muon's lifetime. The muon lasts 2.2 microseconds in its own rest frame and 65 microseconds in the rest frame of the lab. The time between events that happen in the same place is the proper time between them (a timelike spacetime interval). Since the muon's birth and death happen in the same place in the muon's rest frame, the proper time between them is 2.2 microseconds. In the lab's rest frame, the coordinate time of this interval is 2.2 * 29.4 = approx. 65 microseconds. If we think of the muon as a clock, we can say informally that the muon is running slow in comparison to a clock at rest in the lab's rest frame in the sense that the muon only counts to 2.2 microseconds while the lab clock counts to 65. From the perspective of an observer at rest with respect to the lab, a 2.2-microsecond muon is slow to disappear; it lasts for 65 seconds.
Yes, the 'clock' by which the muon's lifetime is measured is the muon's lifetime and that is occurring in the muon's rest frame(2.2 microseconds - proper time); the 'lab' time here (approx. 65 microseconds) is that time 'observed' from the "lab's" rest frame; and so it is in co-ordinate time; i.e. it is in co-ordinate microseconds not proper seconds.
But wherever and however it is measured, the duration of the muon's lifetime is and can only be 2.2 microseconds proper time, that is surely a physical constant. We are only discussing how that one fixed interval is measured in different circumstances.

So 2.2 proper microseconds are equal in duration to approx. 65 co-ordinate microseconds?:rolleyes:


You have been a tremendous help so far, my friend,(if I may call you that?) and I see much clearer now and appreciate your patience, and following your suggestion I am putting all that I have learned into a new diagram that I hope will pull all these different threads together. I should be able to post it later today - it is certainly helping me to see things more clearly and I hope I am not too presumptious if I ask you to view it?

Thanks, once again for your help and guidance.

Grimble:confused:
 
  • #81
Grimble said:
I'm sorry, I haven't explained it well but in the following quote you say that Freund et al. take "dilated time" (co-ordinate units?) to mean an expanded total (of reduced units).

I see what you mean. I got confused and contradicted myself. What I wrote above is one way of expressing the idea that these two time intervals (proper time and coordinate time) represent the same timelike spacetime interval. I suppose I was imagining proper time as being like one ruler, and the coordinate time as being like a previously identical ruler that's been squashed so that its units are closer together (smaller) which means that more of them will fit into the length of the original ruler representing proper time. Expanded (dilated) total of reduced (shrunk, contracted) units. When I objected to this idea in the later post, I think I must have had a different metaphor in mind. Sorry about that.

Grimble said:
And what I am saying is that if the co-ordinate time is an expanded number of reduced units, and if the 'expansion' and 'reduction' are both according to the Lorentz factor, then the only conclusion is that the total duration remains constant. It is precisely the number of units and the size of the units that change.

That "sameness" is the way this particular view expresses the idea that the coordinate time here is the time component of the same spacetime separation (between two events) as the proper time is the invariant interval of.

Another common way of thinking about it is in terms of vectors in spacetime. The separation is a displacement vector in spacetime. Its interval is its magnitude. The magnitude ||Q - P|| of a vector in Euclidean space from point P to point Q is invariant when you rotate the coordinate axes, even though its x, y and z components change. Similarly, the interval of a displacement vector between two points (called events) in Minkowski spacetime is invariant when you switch between frames moving at a constant velocity relative to each other.

If there exists some frame in which the events happen in the same place (which is to say they happen along the possible worldline of some object), it's possible to represent the separation between them as a vector parallel to the time axis (vertical according to the usual convention). In this frame, the vector has only a time component and no spatial component, so its time component (its coordinate time) is equal to its proper time. This is like the way that a vector in a cartesian coordinate system, in Euclidean space, which is parallel to the y-axis has only a y component, and so its y component is equal to its magnitude (length). If we rotate our cartesian coordinate system in the xy plane, this vector will have some x component too, and so it's y component will be less than its magnitude so that the magnitude is still equal to sqrt(x^2 + y^2).

In Minkowski spacetime, the principle is similar, except that the "rotation" is hyperbolic: it takes the tip of the vector along a hyperbola rather than a circlular path. The faster the new frame is moving relative to the original frame where the events happened in the same place, the further the vector slopes, approaching the lightcone in the limit as the speed approaches c. This is because the invariant interval is equal to the absolute value of sqrt(t^2 - x^2), or in all 4 dimensions |sqrt(t^2 - x^2 - y^2 - z^2)|. A consequence of this is that, as the separation vector slopes and acquires a spatial component, its time component (coordinate time) must increase for the interval to remain invariant (in contrast to the y component of the cartesian vector which had to decrease). Hence coordinate time is always greater than proper time in any frame where the events happen in different locations.

Grimble said:
Yes, the 'clock' by which the muon's lifetime is measured is the muon's lifetime and that is occurring in the muon's rest frame(2.2 microseconds - proper time); the 'lab' time here (approx. 65 microseconds) is that time 'observed' from the "lab's" rest frame; and so it is in co-ordinate time; i.e. it is in co-ordinate microseconds not proper seconds.
But wherever and however it is measured, the duration of the muon's lifetime is and can only be 2.2 microseconds proper time, that is surely a physical constant. We are only discussing how that one fixed interval is measured in different circumstances.

So 2.2 proper microseconds are equal in duration to approx. 65 co-ordinate microseconds?:rolleyes:

That's right. Another scenario involving muons which is often used in introductions to special relativity is the example of muons created in the upper atmosphere by cosmic rays. A muon travels towards the earth. In the Earth's rest frame, the muon lasts a long time and travels a great distance, all the way from the upper atmosphere to the ground. In the muon's rest frame, the muon only lasts a couple of microseconds, but the Earth's atmosphere is length-contracted. So the speed with which the ground approaches in the muon in the muon's rest frame is the same as the speed with which the muon approaches the ground in the Earth's rest frame.

Grimble said:
You have been a tremendous help so far, my friend,(if I may call you that?) and I see much clearer now and appreciate your patience, and following your suggestion I am putting all that I have learned into a new diagram that I hope will pull all these different threads together. I should be able to post it later today - it is certainly helping me to see things more clearly and I hope I am not too presumptious if I ask you to view it?

Thanks, once again for your help and guidance.

Grimble:confused:

You're welcome. I hope I haven't added to your confusion ;-)
 
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  • #82
Hello again, my friend; well here it is:
http://img41.imageshack.us/img41/5448/specialrelativitydiagra.jpg
A Special Relativity diagram to demonstrate the relationship between two Inertial Frames of Reference moving with a constant relative velocity.
A and B represent a single axis (time or length) of Minkowski spacetime for each of the two IFoRs. They are drawn against a common background representing Proper Units
An observer at rest within each IFoR will be experiencing proper units (length and time) within that system; as shewn by the horizontal lines, labelled A and B.
But from each IFoR, the other frame's axis -- rotated according to their relative velocity -- will be reckoned in co-ordinate units, as shewn by the perpendicular projections from the coloured diagonals onto the observer's own axis.
The diagram is drawn to scale to represent two IFoRs with a constant relative velocity = 0.6c, giving \gamma = 1.25 and \frac{1}{\gamma} = 0.8

From this we can see exactly what Einstein was saying in http://www.bartleby.com/173/12.html" , when he writes:
But the metre-rod is moving with the velocity v relative to K. It therefore follows that the length of a rigid metre-rod moving in the direction of its length with a velocity v is
M2.GIF


of a metre.

For the metre rod moving with the velocity 0.6c relative to B would be represented in the diagram by the number 1 on the red diagonal and we can see the projection onto B's x-axis (as it would be in this scenario) where it would be the green 1 co-ordinate unit.
And this agrees with {x^&#039;} = \frac {x}{\gamma}
i.e. {x^&#039;} = 0.8x

Similarly, in the second part of that same chapter, Einstein writes:
**Let us now consider a seconds-clock which is permanently situated at the origin (x' = 0) of K'. t' = 0 and t' = 1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:
t = 0
and
E5.GIF


As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but
M5.GIF


seconds, i.e. a somewhat larger time.

And again we can see just how this works, for this time he is converting the time from the observer's frame, t (proper time units) into the observed time {t^&#039;} (co-ordinate time units) which would be to take the blue, 1 proper unit and project it upwards onto the red diagonal line or, indeed, one could read it off the green co-ordinate scale on B's axis.

Not surprisingly this agrees with Einstein's own equation:
t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} or
t = \gamma({t^&#039;}) = 1.25 co-ordinate units

I am fairly certain that this diagram meets all the conditions, references and relationships between the elements that compose SR as described in this thread.

For instance, the co-ordinate units are greater in number but reduced in size.

The principal problem that is brought to light here is that there are far more elements than at first appear.
Consider if you will:

1) We start with a solitary IFoR where the measurements are all, by definition, in proper units.

2) We add a 2nd IFoR moving at a constant velocity with respect to the first: both are measured in proper units and their times are identical and synchronous.

3) They then observe one another and upon doing so we find that the observed frames are rotated with respect to their observers, as shewn, but their units are still proper units.

4) When observing the rotated frames, their proper units are projected vertically onto the observer's frame of reference, being there-by transformed into co-ordinate units.

5) So we have the axis of each IFoR and its rotation, both measured in Proper units and its projection onto the other's axis measured in co-ordinate units.

6) And as all this is matched reciprocally by the other IFoR we have this duplicated giving 4 measures in proper units and two in co-ordinate units.

Is it any wonder that we become confused when trying to deal with this using only primed and unprimed symbols? :cry:
And this is with it all reduced to two dimensions...:rolleyes:


This exercise has certainly helped me to understand how it all fits together – I only hope that I have understood it correctly? o:)

Many thanks, once again to Rasalhague, and Matheinste whose patience and explanations have been a great help. :rolleyes:
 
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  • #83
Looks good to me!

The conventional Minkowski spacetime diagram takes a slightly different approach to representing the same information. Here are a couple of examples.

The top diagram in my first attachment, the attachment labelled "parallels", shows which lines are parallel with which. The lower diagram in the same attachment shows the spacetime intervals between the origin and various events (represented by points). As you can see, in this kind of diagram, the intervals aren't exactly represented as the Euclidean distance between points since the hypotenuse of each of those right triangles corresponds to a shorter spacetime interval than the edge that lies along the axis, contrary to what we'd expect from Pythagoras. This is because the equation for the spacetime interval between two events is that of a hyperbola

||\mathbf{E}_{2} - \mathbf{E}_{1}|| = \sqrt[]{|\Delta t^{2} - \Delta x^{2}|}

rather than a circle like the equation that gives the distance between two points in Euclidean space:

||\mathbf{P}_{2} - \mathbf{P}_{1}|| = \sqrt[]{\Delta y^{2} + \Delta x^{2}}

One frame is represented with the vertical time axis and horizontal x axis. The axes of the other frame are inlined towards each other. The faster the frame shown with the inclined axes is traveling relative to the frame shown with orthogonal axes, the closer its axes will be to the dotted line at 45 degrees which represents the path of a photon travelling, of course, at c. The angle \beta represents the speed of either frame relative to the other, as a fraction of the speed of light: v/c.

Any vertical line is the potential worldline of an object at rest in the frame with orthogonal axes; events on those lines happen in the same place in that frame. Any horizontal line is a line of simultaneity in the frame with orthogonal axes; all events on that line will be happen at the same time as each other in that frame.

Similarly, lines parallel to the time axis of the frame with oblique axes represent potential worldlines of objects at rest in the frame with oblique axes; events on those lines happen in the same place in that frame. Lines parallel to the x-axis of the frame with oblique axes are lines of simultaneity in that frame; all events on those lines happen at the same time as each other in that frame.

There's nothing special about either frame. The only difference is that they're moving in opposite directions relative to each other. So we could just as well change our diagram and show the oblique frame with orthogonal axes, and the orthogonal frame with oblique axes. The only difference would be that the axes of our new oblique frame would have a negative slope because its going in the opposite direction to the other frame. So the oblique axes would appear in the 2nd and 4th quadrants, rather than the 1st and 3rd.

My other diagram, the one labelled "hyperbola" is meant to show how this relates to hyperbolic geometry. If we draw a curve through all points on a Minkowski diagram which stand for events with a timelike interval of 1 from the origin, the curve is one branch of a unit hyperbola. Consider one of these events, the one that lies on the axis labelled tau in this diagram. The proper time between the origin and this event is 1. The coordinate time of this interval with respect to the frame represented by the axes labelled here t and x is

\gamma = \frac{1}{\sqrt[]{1 - \beta^{2}}} = cosh\left(\phi \right)

where \phi is the "hyperbolic angle", the input of the hyperbolic trigonometric functions, which corresponds geometrically to the area shaded blue under the hyperbola here. Cosh, sinh and tanh are the hyperbolic cosine, sine and tangent respectively.

sinh\left(\phi \right) = \gamma \beta

and

tanh\left(\phi \right) = \frac{sinh\left(\phi \right)}{cosh\left(\phi \right)} = \beta

so \phi is the value of the inverse hyperbolic tangent of \beta:

\phi = artanh \left(\beta \right)

I gather there are other ways of depicting the Lorentz transformation in two dimensions, but this is the one I'm familiar with, and seems to be the one most commonly used in textbooks. Some more examples here:

http://hubpages.com/author/lgsims96/hot/

So far every source I've looked at has followed this comvention for which way round to depict the time and space axes, with the exception of James Callahan's The Geometryu of Spacetime which has them the other way around:

http://www.google.co.uk/books?id=UM...:+Spacetime+and+Geometry#v=onepage&q=&f=false
 

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  • #84
Is there a way to derive the time dilation formula without using light clocks? Specifically, without using Euclidean geometry, by simply using mechanics, maybe?
 
  • #85
collectedsoul said:
Is there a way to derive the time dilation formula without using light clocks? Specifically, without using Euclidean geometry, by simply using mechanics, maybe?

Hello, I'm not sure that I can answer this, but I am assuming that you find the derivations using light clocks,http://www.answers.com/topic/time-d...ce_of_time_dilation_due_to_relative_velocity" are missing something -- and you would be quite right! For they are nothing more than plain straightforward GALILEAN TRANSFORMATIONS. The only reference they make to SR is that they use the constant speed of light and so let the time lengthen.

And, taking the 2^{nd} example; {t_A} = {t_B}\sqrt{1 -\frac{v^2}{c^2}} may be rewritten as {t_B} = \frac{t_A}{\sqrt{1 -\frac{v^2}{c^2}}}

And this is the accepted formula for Time Dilation – BUT (and here lies the problem), none of it has anything to do with Special Relativity!
Everything done so far has been measured in Proper units; so all we have measured is a GALILEAN TRANSFORMATION! For in keeping the speed of light constant we have merely increased the time taken.

This is an example of what I would term the 'Slippery Slope of STR', by which I allude to the fact that, as space and time are no longer fixed scales, we have to be so very, very careful to be absolutely clear about what we are referring to at every step.

So let us now continue with the transformation into coordinate time:

1.The passage above brought us to the point where we had established the relationship between the longer path, seen when the clock is moving (relative to the observer), and that within the clock's own frame of reference. That is:
{t_B} = \frac{t_A}{\sqrt{1 -\frac{v^2}{c^2}}}
where both sides of the equation are in common units of measurement.

2.And we know from Einstein's first Postulate (the Principle of Relativity), that time in the A's frame of reference is identical to that in B's frame of reference.

3.So t_B in Proper units = t_A in Proper units,


4.And t_B in coordinate units = \frac{t_A}{\sqrt{1 -\frac{v^2}{c^2}}} in coordinate units


4.Giving us the conversion factor, \frac{1}{\sqrt{1 -\frac{v^2}{c^2}}} in coordinate units to convert from coordinate units to Proper units.

5.So giving us {t_B} = {t_A}\sqrt{1 -\frac{v^2}{c^2}}

or t = {t^&#039;}\sqrt{1 -\frac{v^2}{c^2}} with the more usual terms, as the corrected formula for Time Dilation.

6.And \frac{1}{\sqrt{1 -\frac{v^2}{c^2}}} for the Lorentz Transformation Factor.

The second common derivation that is often quoted is http://www.answers.com/topic/special-relativity#Time_dilation_and_length_contraction"

But the problem with this derivation is that we are using the clock in the unprimed system, but the moving clock is that in the primed system and the whole point of STR is to transform the Spacetime magnitudes of the moving clock into the unprimed system's frame of reference.

The clock in the unprimed system has no relevance to what we are doing.

No what we need to de here is to use the clock in the primed system. Where two consecutive ticks of this clock are characterized by
\Delta{x^&#039;} = 0

If we want to know the relation between the times between these ticks as measured in both systems, we can use the third equation --
\Delta{t}=\gamma(\Delta{t^&#039;} + \frac{v\Delta{x^&#039;}}{c^2})

and find \Delta {t} = \gamma\Delta{t^&#039;} which is the correct formula for Time Dilation.
 
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  • #86
Hi,

I've just recently learned about relativity by watching some videos and reading parts of some books and my understanding of it is very basic and quite shaky. It just seems so bizarre to me! So please bear with any errors on my part and correct me.

I think I understood what you said about the light clock derivation using Galilean relativity and simply keeping the speed of light constant and therefore stretching time to account for the increased distance. And this is where I had a doubt in the first place - isn't length being contracted in the moving frame wrt the stationary frame? If not the vertical length traveled by the light, then the horizontal length (is my reasoning in this regard valid)? If not at all, then in what cases are length contracted?

About your discussion of the coordinate units in comparison to proper units, are you referring in the first case to the relative measurements of a moving object wrt a stationary one, and in the second case to measurements within each frame itself? Then that makes sense to me.

I didn't follow you on the last bit though (not sure what an unprimed system is).

Coming back to the light clock derivation from your second link http://www.answers.com/topic/lorentz-factor#Derivation, I quote the premise:
Imagine two observers: the first, observer A, traveling at a constant speed v with respect to a second inertial reference frame in which observer B is stationary. A points a laser “upward” (perpendicular to the direction of travel). From B's perspective, the light is traveling at an angle.

When I asked the original qs., I was thinking of a case where A points the laser in the horizontal direction rather than vertical. Or simply the case where a stationary light source emits beams measured by two observers at a same distance x, one moving with velocity v and the other standing still. I was trying to find out the effect of light having the same measured speed by both observers using mechanics equations and getting really confused in the process. It'd be a big help if you could explain what is happening in this situation, and whether or not its possible to derive time dilation from this.

Thanks for your patience!
 
  • #87
collectedsoul said:
Hi,

I've just recently learned about relativity by watching some videos and reading parts of some books and my understanding of it is very basic and quite shaky. It just seems so bizarre to me! So please bear with any errors on my part and correct me.

Hello, and welcome to the mysteries of relativity! it is great to find someone else, like me, that has learned about it by their own efforts! - It can be done! I just try to make sure I understand each step along the way - the mystery is more about how it is described, different uses of the same terminology, rather than relativity being difficult in itself.

I think I understood what you said about the light clock derivation using Galilean relativity and simply keeping the speed of light constant and therefore stretching time to account for the increased distance. And this is where I had a doubt in the first place - isn't length being contracted in the moving frame wrt the stationary frame?

Yes.

If not the vertical length traveled by the light, then the horizontal length (is my reasoning in this regard valid)? If not at all, then in what cases are length contracted?

Yes, that had me wondering, but you are quite right it is just the horizontal length.

About your discussion of the coordinate units in comparison to proper units, are you referring in the first case to the relative measurements of a moving object wrt a stationary one, and in the second case to measurements within each frame itself? Then that makes sense to me.

Yes, Proper time or length are the measurements within each frame, co-ordinate units are those proper units, converted by the Lorentz Transformation Equations into what a moving observer would see.

I didn't follow you on the last bit though (not sure what an unprimed system is).

Yes, that had me fooled for quite a time, but all it means is the use of the quote ('). It turns out that t' is primed and t is unprimed

Coming back to the light clock derivation from your second link http://www.answers.com/topic/lorentz-factor#Derivation, I quote the premise:
Imagine two observers: the first, observer A, traveling at a constant speed v with respect to a second inertial reference frame in which observer B is stationary. A points a laser “upward” (perpendicular to the direction of travel). From B's perspective, the light is traveling at an angle.

When I asked the original qs., I was thinking of a case where A points the laser in the horizontal direction rather than vertical. Or simply the case where a stationary light source emits beams measured by two observers at a same distance x, one moving with velocity v and the other standing still. I was trying to find out the effect of light having the same measured speed by both observers using mechanics equations and getting really confused in the process. It'd be a big help if you could explain what is happening in this situation, and whether or not its possible to derive time dilation from this.

Now I am sure there are many (if not most) of the other contributors out there that can give a more detailed answer than I can but essentially, the problem we have to overcome is thinking that time and space are absolute quantities.
In the situation you describe, each observer will measure the same speed for the light in his frame of reference, the tricky point to understand is that if either were to measure the speed of the light in the other frame of reference he would get the same result! It is the units themselves that are not constant.

The diagram in post 82 may help.

Thanks for your patience!

And don't give up it gets easier but remains fascinating, Grimble.:biggrin:
 
  • #88
Oh I'm not about to give up, its far too interesting! Thanks for answering my questions but a couple remain unanswered...any help from anyone else would be welcome.

In the light clock case, if length is contracted, then how is the derivation valid? Because it assumes that the moving clock travels a distance vt in the horizontal, when it should be less than that.

Secondly, can someone please explain why I can't derive time dilation from it? And what actually is happening in terms of each frame? Here is the case again:
Stationary light source sends out beams in the horizontal direction received by two observers. A is moving with horizontal velocity v and B is stationary. Both are same distance x from the light source. Since both measure the same speed of light, is there a way to derive time dilation from this case. And, like I asked before, what exactly is happening to spacetime in each frame, AND relative to each other?
 
  • #89
collectedsoul said:
isn't length being contracted in the moving frame wrt the stationary frame? If not the vertical length traveled by the light, then the horizontal length (is my reasoning in this regard valid)? If not at all, then in what cases are length contracted?

Length is contracted in the direction of movement (and no other direction), but we don't need to worry about it in this derivation. That's because, using the notation of your link, vt_B and ct_B are distances measured wrt the frame where the light clock is moving, and ct_A is the same distance in either frame since length contraction only occurs in the direction of movement.

http://www.answers.com/topic/lorentz-factor#Derivation

So all distances in the derivation are consistent with respect to the frame where the clock is moving.

To see why distances must be the same in directions perpendicular to the direction of movement, imagine two vertical meter sticks moving past each other in a horizontal direction. Stick A has nails attached to either end. Stick B has balloons attached to either end. When they pass, the nails on stick A burst the balloons on stick B. If A was shorter or if B was shorter, the nails would miss. Each stick is moving relative to the other, so if one is shorter in the other's rest frame, the other should be shorter in the first one's rest frame. But the balloons either burst or don't burst; they can't do both. So neither stick can be shorter.

collectedsoul said:
When I asked the original qs., I was thinking of a case where A points the laser in the horizontal direction rather than vertical. Or simply the case where a stationary light source emits beams measured by two observers at a same distance x, one moving with velocity v and the other standing still. I was trying to find out the effect of light having the same measured speed by both observers using mechanics equations and getting really confused in the process. It'd be a big help if you could explain what is happening in this situation, and whether or not its possible to derive time dilation from this.

If A emits light in the direction of movement, then length contraction does come into play. In fact this is a typical scenario used by textbooks to derive the length contraction formula.

http://www.pa.msu.edu/courses/2000spring/PHY232/lectures/relativity/contraction.html
 

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