# Length Contraction and Time Dilation

Length Contraction and Time Dilation

Gamma= 1/sqrt(1-v^2/c^2) = 2 (Assumed)
We have L=L’/Gamma…………………………………….1
And dT= dT’ * Gamma ………………………………2

Let primed frame is that of Bob and unprimed that of Dave.
Let us define identity as an equation in which values on both sides of the equal sign are equal.
And a formula in which given values at right side are calculated and assigned to a symbol at left side of the equal sign.
Equation 1 is treated as a formula. Proper length L’ is worked on and assigned to L. Therefore, L is the length in the Bob’s frame as calculated by Dave. If L’ is 100 meters, Dave disagrees and says that the length is only 50 meters. This is proper length contraction.
This is not the case with equation 2. It is treated as an identity.
Proper time of Bob, say 5 years, is multiplied by Gamma. To retain the identity, dT is increased to 10. In this case Bob’s clock doesn’t run slow. It is Dave’s clock that is running fast.
If equation 2 is used as a formula then according to Dave, Bob’s clock is running fast. That is even if Bob measures 5 years in his clock, according to Dave, this same time is 10 years.
But as soon as we use the word ‘measure’ we must treat the equations as formulae and not as identities, because we are finding values.
So equation 1 is correct but equation 2 should be dT= dT’ / Gamma. Now Bob’s time as measured by Dave is less, IOW according to Dave, Bob’s clock runs slow.
This scenario can be viewed from another angle.
We keep rod and clock in the frame of which the measurements are to be made.
So we keep a clock/s and a rod of length L’ in Bob’s frame. For length we have,
x = (x’+vt’)*Gamma
x2– x1= ((x’2 – x’1 ) + v(t’2–t’1)) * Gamma
or L= L’*Gamma
In the above we take help of the space-time diagram. We keep the rod on the line parallel to x’-axis. On this axis t’2 = t’1
Similarly
t2 – t1 = ((t’2–t’1)+(x’2–x’1)*v/c^2)*Gamma
or dT = dT’ * Gamma
In the above we keep two synchronized clocks on the line parallel to the time axis of primed frame.
In my opinion, once we derive Lorentz equations, we should show some respect to these basic equation. We can have dispute with the concepts behind the derivation but once derived, they cannot be manipulated further as we like. Present equations 1 and 2 are convenient but mathematically inconsistent.
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Doc Al
Mentor
We have L=L’/Gamma…………………………………….1
And dT= dT’ * Gamma ………………………………2
Realize that these are special cases of the Lorentz transformations and are not true in general (for arbitrary distances and time intervals).

Let primed frame is that of Bob and unprimed that of Dave.
Let us define identity as an equation in which values on both sides of the equal sign are equal.
And a formula in which given values at right side are calculated and assigned to a symbol at left side of the equal sign.
Seems like a rather meaningless distinction.
This is not the case with equation 2. It is treated as an identity.
Proper time of Bob, say 5 years, is multiplied by Gamma. To retain the identity, dT is increased to 10. In this case Bob’s clock doesn’t run slow. It is Dave’s clock that is running fast.
Nope. dT' represents the proper time between events--events that take place at a fixed location in the primed frame. If that proper time is 5 years, then the moving observer (in the unprimed frame) will measure the time between those events as taking 10 years. So he (Dave) judges that the primed clock (Bob's) is running slow.
If equation 2 is used as a formula then according to Dave, Bob’s clock is running fast. That is even if Bob measures 5 years in his clock, according to Dave, this same time is 10 years.
But as soon as we use the word ‘measure’ we must treat the equations as formulae and not as identities, because we are finding values.
So equation 1 is correct but equation 2 should be dT= dT’ / Gamma. Now Bob’s time as measured by Dave is less, IOW according to Dave, Bob’s clock runs slow.
Nah... Apparently you don't know how the 'time dilation formula' applies. Both frames measure the other's clocks as running slow. In that formula, dT' represents a proper time; dT does not.

No prize for you. Try again!

DrGreg
Gold Member
Gamma= 1/sqrt(1-v^2/c^2) = 2 (Assumed)
We have L=L’/Gamma…………………………………….1
And dT= dT’ * Gamma ………………………………2

In my opinion, once we derive Lorentz equations, we should show some respect to these basic equation. We can have dispute with the concepts behind the derivation but once derived, they cannot be manipulated further as we like. Present equations 1 and 2 are convenient but mathematically inconsistent.

You have to recognise that (1) and (2) do not apply in all circumstances, but the Lorentz transformation does. For reference, it is, in your notation:

$$dT' = \gamma ( dT - vL / c^2)$$.........(3)
$$L' = \gamma ( L - v \, dT)$$..............(4)​

These equations can also be rearranged to give

$$dT = \gamma ( dT' + vL' / c^2)$$........(5)
$$L = \gamma ( L' + v \, dT')$$...............(6)​

Comparing (1) and (4), we see that (1) is true only in the special case where dT = 0.

Comparing (2) and (5), we see that (2) is true only in the special case where L' = 0.

You should carefully think through what it means for dT = 0 or for L' = 0.

The reason (1) and (2) are mathematically inconsistent is because it's not possible for both of the conditions dT = 0 and L' = 0 to be true simultaneously (except in the trivial case where everything is zero).

Seems like a rather meaningless distinction.
Not always. Sometimes it can make substantial difference.

Nope. dT' represents the proper time between events--events that take place at a fixed location in the primed frame. If that proper time is 5 years, then the moving observer (in the unprimed frame) will measure the time between those events as taking 10 years. So he (Dave) judges that the primed clock (Bob's) is running slow.

Kindly give your ‘Yes’ or ‘No’ to the following points and if ‘No’ please elucidate, without bringing in extraneous matter.
1. Bob goes and returns after 10 years as counted on Dave’s clock. Since Bob returns younger, time dilation effect is real. Both have clocks which count years. One complete rotation of ‘year needle’ means one year.
2. If there was no time dilation then needles of both the clocks would be found to have completed 10 rotations.
3. Keeping aside the notion of proper time, let us talk about real clocks. If time dilation effect is real, then the needle of Dave’s clock rotated 10 times. If Bob has returned younger by 5 years then the needle of Bob’s clock has rotated only 5 times.
4. If (3) is true then the equation has to be T = T’ / Gamma.

You have skirted all the points and just produced statements from the text book.

Doc Al
Mentor
1. Bob goes and returns after 10 years as counted on Dave’s clock. Since Bob returns younger, time dilation effect is real. Both have clocks which count years. One complete rotation of ‘year needle’ means one year.
You've just introduced acceleration into the mix, which you hadn't before. Kindly stick to one topic at a time.
2. If there was no time dilation then needles of both the clocks would be found to have completed 10 rotations.
3. Keeping aside the notion of proper time, let us talk about real clocks. If time dilation effect is real, then the needle of Dave’s clock rotated 10 times. If Bob has returned younger by 5 years then the needle of Bob’s clock has rotated only 5 times.
4. If (3) is true then the equation has to be T = T’ / Gamma.
Only Dave remains in a single inertial frame throughout this exercise and only he is justified in using the time dilation formula in a simple manner. And it works just fine: T = T'*Gamma. (T is the time according to Dave's clock; T' the proper time according to Bob's clock.)

I recommend that you get basic time dilation between non-accelerating inertial frames straight before worrying about 'twin paradox' issues.

You've just introduced acceleration into the mix, which you hadn't before. Kindly stick to one topic at a time.

Only Dave remains in a single inertial frame throughout this exercise and only he is justified in using the time dilation formula in a simple manner. And it works just fine: T = T'*Gamma. (T is the time according to Dave's clock; T' the proper time according to Bob's clock.)

Totally irrelevant to the points 1, 2, 3 and 4.

I recommend that you get basic time dilation between non-accelerating inertial frames straight before worrying about 'twin paradox' issues.

Points 1 to 4 have nothing to do with twin paradox. I am just comparing clocks.

Doc Al
Mentor
Points 1 to 4 have nothing to do with twin paradox. I am just comparing clocks.
Then rewrite your scenario without either clock accelerating. (You cannot have them 'go and return' without introducing acceleration and the twin paradox.)

Then rewrite your scenario without either clock accelerating. (You cannot have them 'go and return' without introducing acceleration and the twin paradox.)

I am not contradicting basic result of twin paradox. I am not disputing the accepted result that Bob will return younger. I only beg for the answer to the point 3. Point 4 can wait.

Doc Al
Mentor
I am not contradicting basic result of twin paradox. I am not disputing the accepted result that Bob will return younger. I only beg for the answer to the point 3. Point 4 can wait.
OK, what's the big deal?
3. Keeping aside the notion of proper time, let us talk about real clocks. If time dilation effect is real, then the needle of Dave’s clock rotated 10 times. If Bob has returned younger by 5 years then the needle of Bob’s clock has rotated only 5 times.
Yes: During Bob's round trip, if Bob's clock 'rotated' 5 times then Dave's non-accelerating clock rotated 10 times. Perfectly in accord with the standard 'time dilation' formula.

OK, what's the big deal?

Yes: During Bob's round trip, if Bob's clock 'rotated' 5 times then Dave's non-accelerating clock rotated 10 times. Perfectly in accord with the standard 'time dilation' formula.

This is bit confusing to me. According to Newtonian relativity, Bob’s time is 10 years and so also Dave’s. So 10 years is Bob’s proper time and according to SR, there is no change in proper time. So why should Bob’s clock run slowly in Bob’s frame?

Doc Al
Mentor
According to Newtonian relativity, Bob’s time is 10 years and so also Dave’s.
Lots of luck trying to understand time dilation using 'Newtonian relativity'.
So 10 years is Bob’s proper time
No, Bob's proper time--his 'wristwatch time'--is 5 years.
and according to SR, there is no change in proper time. So why should Bob’s clock run slowly in Bob’s frame?
Huh? Bob's clock doesn't run slowly in its own frame. (Whatever that might mean.)

Lots of luck trying to understand time dilation using 'Newtonian relativity'.

No, Bob's proper time--his 'wristwatch time'--is 5 years.

Huh? Bob's clock doesn't run slowly in its own frame. (Whatever that might mean.)

Doesn’t it mean that it is Dave’s clock that ran faster? (According to Bob).

Doc Al
Mentor
Doesn’t it mean that it is Dave’s clock that ran faster? (According to Bob).
Not really. Note that this is the 'twin paradox' and not simple time dilation. If you want to understand how the result comes about, there have been many threads here describing it in excruciating detail.

Even though at any point in their relative motion each sees the other's clock as running slow, relativity of simultaneity ends up making the total elapsed time greater on the clock that remains in a single inertial frame (Dave's clock).

JesseM
Doesn’t it mean that it is Dave’s clock that ran faster? (According to Bob).
Are you talking about the average rate over the entire trip, or the rate during some section of the trip? For averages it's true that Dave must have had a faster average rate of ticking over the entire trip regardless of what frame you choose, but if you're just talking about the rate of Dave vs. Bob during some section of the trip like the period when Bob was moving away from Dave, then different frames can disagree whose clock was ticking faster during a given section.

Are you talking about the average rate over the entire trip, or the rate during some section of the trip? For averages it's true that Dave must have had a faster average rate of ticking over the entire trip regardless of what frame you choose, but if you're just talking about the rate of Dave vs. Bob during some section of the trip like the period when Bob was moving away from Dave, then different frames can disagree whose clock was ticking faster during a given section.
I think both clocks tick at the same rate, namely one second per second. What is different though is that Bobs clock simply recorded less second due to the particular way he traveled in spacetime.

When two people go from A to B on two different paths their ODO meters are different not due to one meter going slower or faster but because one path is shorter than the other. The same applies for time, some paths between two events require less seconds than other paths, but it does not mean the clock goes slower.

I do not understand why people have no problem understanding this for ODO meters but fail to understand that for clocks.

JesseM
I think both clocks tick at the same rate, namely one second per second.
To define a "rate" you need to talk about the rate one thing is changing relative to some other thing, obviously it's true that each clock ticks at 1 second per second relative to itself, but that's pretty trivial. I was talking about the rate a clock is ticking relative to coordinate time in some choice of reference frame.
What is different though is that Bobs clock simply recorded less second due to the particular way he traveled in spacetime.
Sure, I agree that's a good way to think about, but it doesn't mean you can't also talk about rates.
When two people go from A to B on two different paths their ODO meters are different not due to one meter going slower or faster but because one path is shorter than the other.
But even in this situation you can talk about the rate that the odometer reading is increasing relative to the person's increase in x-coordinate in some Cartesian coordinate system, analogous to talking about the rate a clock's proper time is increasing relative to the t-coordinate of some frame. I developed this geometric analogy at some length in [post=2972720]this post[/post], if you or anyone else is interested. I do find that thinking in terms of the geometric analogy is very helpful in clarifying thoughts about issues relating to time dilation and the twin paradox.

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Are you talking about the average rate over the entire trip, or the rate during some section of the trip? For averages it's true that Dave must have had a faster average rate of ticking over the entire trip regardless of what frame you choose.

Instead of average I would like to call it total time of Dave. Needle of Dave’s clock has rotated 10 times. But as you told me before, this is due to the jump in Dave’s time as seen by Bob during his turn around. Though it is Bob’s perception that Dave’s clock jumped it should have really jumped without which Dave’s clock will not show 10 years. So will Dave observe this jump in his clock? If not why not?

but if you're just talking about the rate of Dave vs. Bob during some section of the trip like the period when Bob was moving away from Dave, then different frames can disagree whose clock was ticking faster during a given section.

Does this disagreement suggest that time dilation effect is apparent?

JesseM
Instead of average I would like to call it total time of Dave. Needle of Dave’s clock has rotated 10 times. But as you told me before, this is due to the jump in Dave’s time as seen by Bob during his turn around.
When did I tell you that? In these sorts of discussions I always emphasize that a non-inertial observer like Bob doesn't have a single frame that defines his "perspective", you can choose different non-inertial coordinate systems where Bob is at rest which say different things about the rate of Dave's clock relative to coordinate time, none defines what is "seen by" Bob any more than the others (if you're talking about the question of what Bob sees visually then this doesn't depend on your choice of frame, but in this case there is no "jump")
Does this disagreement suggest that time dilation effect is apparent?
Time dilation is a frame-dependent notion, like velocity. Would you say that velocity is "apparent" since different frames disagree on which of two objects has a greater velocity? If so then you can say the same about time dilation, if not you shouldn't, it depends on your definition of "apparent" I guess.

Time dilation is a frame-dependent notion, like velocity. Would you say that velocity is "apparent" since different frames disagree on which of two objects has a greater velocity? If so then you can say the same about time dilation, if not you shouldn't, it depends on your definition of "apparent" I guess.

By apparent I mean unreal. Effect that is seen or assumed to take place but in reality doesn’t take place. When you say that “…then different frames can disagree whose clock was ticking faster during a given section” it is clear that time dilation in this situation is apparent. It has to be apparent. There is a modified version of twin paradox. From a space station, Dave and Bob set on voyage in opposite directions and come back to the station. For an observer on the space station, they both return younger. Between Dave and Bob, effects of turn around are symmetrical and so these can be neglected. During their uniform motion, according to each, other’s clock runs slowly and on meeting they find no change in their relative age and so their measurements must be apparent.

JesseM
By apparent I mean unreal. Effect that is seen or assumed to take place but in reality doesn’t take place.
I don't know whether a coordinate-dependent fact counts as a "reality" in your book. Can you please answer my question about velocity? If I have a higher velocity than you in some observer's frame, would you call that higher frame-based velocity "unreal" or "apparent" since in some other frame it would be you that has the higher velocity than I?
There is a modified version of twin paradox. From a space station, Dave and Bob set on voyage in opposite directions and come back to the station. For an observer on the space station, they both return younger. Between Dave and Bob, effects of turn around are symmetrical and so these can be neglected. During their uniform motion, according to each, other’s clock runs slowly and on meeting they find no change in their relative age and so their measurements must be apparent.
No, if you are using the inertial frame where Dave is at rest during the return voyage, you need to take into account simultaneity and realize that in this frame Bob's turnaround happened long before Dave's, so at the moment Dave turns around Bob is considerably older than Dave. Then during the return voyage, Bob is aging more slowly but he had that "head start" so his age continues to be greater than Dave's, but Dave is gradually "catching up" due to his faster aging in this frame, and it works out just right that they are the same age when they meet, despite the fact that Bob was aging slower during the entire return voyage.

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By apparent I mean unreal. Effect that is seen or assumed to take place but in reality doesn’t take place. When you say that “…then different frames can disagree whose clock was ticking faster during a given section” it is clear that time dilation in this situation is apparent. It has to be apparent. There is a modified version of twin paradox. From a space station, Dave and Bob set on voyage in opposite directions and come back to the station. For an observer on the space station, they both return younger. Between Dave and Bob, effects of turn around are symmetrical and so these can be neglected. During their uniform motion, according to each, other’s clock runs slowly and on meeting they find no change in their relative age and so their measurements must be apparent.

One may use the word "apparent" if they so wish, however it doesn't change anything in reality. IOWs, it's nothing more or less than semantics, to use the word apparent. The fact of the matter, is that the light and the image conveyed is measureable. The image may well be a clock's time readout, and we may then determine its location and velocity when the EM captured the clock's readout. We could even imagine that both clocks transmit their readouts via digitally encoded EM bursts to the other (or all others).

In your modified twins scenario here, each twin records the other's clock ticking more slowly while inertial, and can tick faster while non inertial ... similarly as in the classic twins scenario case. The difference here, the faster ticking of the other moving clock "will be just fast enough" such that their clocks read the very same upon return. They must, since any inertial observer will hold their clocks read the same upon return, and in any reasonable theory one cannot dispute such invariants. To allow such disputes, would be to allow some folks to claim 2 colocated clocks (even if momentary) read what they do, while allowing others to claim they read something altogether different ... in which case the theory would be reduced to rediculous. So one thing that becomes very clear here, is that the relative readout of clocks "is completely geometric", and because we all share the same spacetime symmetry that makes it as such, the 2 clocks in your modified scenario must read the same upon return. It can be no other way if the special theory of relativity is a correct theory.

GrayGhost

I don't know whether a coordinate-dependent fact counts as a "reality" in your book. Can you please answer my question about velocity? If I have a higher velocity than you in some observer's frame, would you call that higher frame-based velocity "unreal" or "apparent" since in some other frame it would be you that has the higher velocity than I?.

That depends on the situation. Since basically velocity is relative, all velocity dependent effects must be relative and so apparent. But in real world we have to fix its value for many kinds of measurements and to do that we decide a reference frame. So I will say that your velocity is definitely apparent and your velocity and all other velocities become real (only for the purpose of calculations) when these are referred to some frame. Unfortunately intuition and not logic or mathematics can be applied in some situations. For example when a spaceship is set off from earth and is in uniform speed, astronaut has to assume intuitionally, that his ship is in actual motion and the earth is not going away from him. Same is the case with the passenger in a train. Please note that record of acceleration plays no part during consideration of uniform velocities. However it should be remembered that no relative velocity can be set up without acceleration of one or more bodies. In the above case we assume that we do not know past accelerations of the frames and so a need to fix a suitable reference frame. Till such a frame is fixed, results dependent on relative velocities have to be apparent. When SR is applied to experiments, we invariably define a reference frame; to extract reality from relativity. It is also to be noted that in nature all velocities have a reference frame.

No, if you are using the inertial frame where Dave is at rest during the return voyage, you need to take into account simultaneity and realize that in this frame Bob's turnaround happened long before Dave's, so at the moment Dave turns around Bob is considerably older than Dave. Then during the return voyage, Bob is aging more slowly but he had that "head start" so his age continues to be greater than Dave's, but Dave is gradually "catching up" due to his faster aging in this frame, and it works out just right that they are the same age when they meet, despite the fact that Bob was aging slower during the entire return voyage.

If Bob turns around before Dave, then Bob is younger than Dave and not older. In any case we need not bother about instantaneous ages. Turn around effect is reciprocal and so need not be taken into account. What is important is that during their uniform motion, each clock is running slow in other’s frame. In short time measurements are apparent.
If you are not satisfied, then simply do not allow them to return. Let there be two more space stations, equidistant from the base. Let Dave and Bob count number of rotations of their ‘year needles’. After reaching the destination, they can inform each other, number of rotations recorded. These will be equal.

and because we all share the same spacetime symmetry that makes it as such, the 2 clocks in your modified scenario must read the same upon return. It can be no other way if the special theory of relativity is a correct theory.

GrayGhost

Yes exactly! That is what SR says and that is what we get.

JesseM
So I will say that your velocity is definitely apparent and your velocity and all other velocities become real (only for the purpose of calculations) when these are referred to some frame.
Then you should say the same about time dilation.
If Bob turns around before Dave, then Bob is younger than Dave and not older.
Nope, not true. You said their motions were completely symmetrical in the third observer's frame, so obviously in this frame they are the same age when they turnaround, and all frames agree in their predictions about local events like what a particular clock reads at the moment something happens to it (like that it begins to accelerate). If we assume that in the frame of the third observer, Bob and Dave move away from each other at the same constant speed, accelerate briefly, and then move towards each other at the same constant speed, then in the inertial frame where Dave is at rest during the inbound leg of the trip after the acceleration, in this frame Dave's speed must have been greater than Bob's during the outbound leg of the trip prior to either one turning around. So, in this frame Dave is aging slower during the outbound leg, which explains why Dave and Bob are the same age at the moment they turn around in spite of the fact that Bob turned around at an earlier time in this frame.

Are you familiar with how to use the Lorentz transformation? If so I can give you a numerical example to demonstrate this if you like.
What is important is that during their uniform motion, each clock is running slow in other’s frame.
Yes, and similarly each one has a higher velocity in the other's frame. But you said velocity was "real" when referred to a frame, so why not say the same about time dilation?
If you are not satisfied, then simply do not allow them to return. Let there be two more space stations, equidistant from the base. Let Dave and Bob count number of rotations of their ‘year needles’. After reaching the destination, they can inform each other, number of rotations recorded. These will be equal.
But in the frame where either (let's say Dave again) at rest during the journey, they arrive at their destinations at different times, because in such a frame the stations themselves are moving, one moving towards the point where they departed from one another and the other moving away from that point. Naturally if Bob is racing to catch up with a station moving away from the point where they departed, he will take longer to do so, which means that even though Bob's clock was ticking slower in this frame, Bob is still the same age when he finally catches up with his station as Dave was when he caught up with his own station. Again I can give you a simple numerical example if you don't believe me.