Is Time Dilation Misunderstood in Popular Science Videos?

seb7
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The video is pretty popular and being spread by science channels.
Is the last bit about the cause of time dilation correct?
I used to think like this, but came to the concoction its wrong, but seeing this video makes me wonder once again how could this be correct.

If the video is true then - wouldn't this mean everything isn't truly relative?
If the video is true then - wouldn't 50% of c would cause 50% of time? which does not match Einstein's equation on this.
 
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Which last bit exactly do you mean? At which point in the movie?
seb7 said:
wouldnt 50% of c would cause 50% of time?
No. That's not even well-defined without referring to another observer.
 
When he talks about time dilation, 5min 15s, into the video. Is he correct?
 
seb7 said:
If the video is true then - wouldn't this mean everything isn't truly relative?

The speed of light is absolute, not relative. That's what makes Einstein's relativity different from Galileo's. The notion that everything is relative was overturned by Einstein. That's what makes the theory revolutionary.

If the video is true then - wouldn't 50% of c would cause 50% of time? which does not match Einstein's equation on this.

Nowhere in that movie is that claim made.

In the movie an error is made at about the 6-minute mark. The narrator speaks of the traveler being closer to the speed of light than people on Earth. That's just not true, and violates both postulates. If someone had a speed closer to the speed of light than someone else, that would constitute a way to distinguish between their inertial reference frames. And it would mean that they would measure different values for the speed of light. The explanation of time dilation that follows from that idea therefore has subtle flaws. You couldn't, for example, use that argument to explain why clocks on Earth run slow relative to the traveler.

Other than that, the movie accurately portrays the two postulates and the different aging of the traveler compared to the person left on Earth.
 
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seb7 said:
If the video is true then - wouldn't 50% of c would cause 50% of time? which does not match Einstein's equation on this.
No, but I'm not surprised you asked the "50%" question because he didn't give any relativistic equations at all, just d = s x t.

OK, if this guy has piqued your interest in the subject then search the internet for introductory SR material, and come back with more specific question when you are ready. That video has got you as far as you can get with words, but science in not done with words; you need equations to understand what is "really" happening.
 
I understand Einstein's equations on time dilation, but this video talks about why, which doesn't seem correct to me.
Wouldn't his reasoning for the time dilation require a fixed ether frame?

Mister T said:
speed of light is absolute
yes, but only within its frame.

here's a question: Say we got on a ship traveling at 0.8c then while traveling we got on to its onboard shuttle and kicked it back to 0c, then the ship (still traveling at 0.8c), later went back to meet up with stationary shuttle.
When they compare clocks, which craft lost more time?
 
Mister T said:
The speed of light is absolute, not relative.
seb7 said:
yes, but only within its frame.
It would be more precise that the speed of light is both relative and invariant. "Invariant", meaning that the measured speed is the same, no matter what frame you use to measure it from. "Relative", meaning that the speed can only be measured relative to something else.

Since light has no frame, it is nonsense to say "within its frame".
 
seb7 said:
When he talks about time dilation, 5min 15s, into the video. Is he correct?
It is right in some directions. Without further explanation, I think it is at least a bit misleading.
You can look up "light clock", where the same argument is made in a more formal way.
 
seb7 said:
I understand Einstein's equations on time dilation, but this video talks about why, which doesn't seem correct to me.
Wouldn't his reasoning for the time dilation require a fixed ether frame?

No. What he's talking about when he talks about things slowing down because of speed is relative to ANY inertial rest frame. The simplest example to understand it is a light clock. A light clock consists of two parallel mirrors, with a pulse of light bouncing back and forth between the mirrors. The time for a complete circuit for the pulse of light is just T = 2D/c, where D is the distance between the mirrors. Now, if you set the two mirrors in motion (or alternatively, switch to a different rest frame in which the mirrors are already moving), then the time required for a complete circuit is longer. If we assume that the motion is perpendicular to the line connecting the mirrors, and that the distance between the mirrors is still D (a more thorough analysis is needed to conclude this, so for now, it's just an assumption), then for a pulse of light to travel from one mirror to the other and back at speed c would take a time equal to:

T' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} 2D/c

where v is the speed of the mirrors.

To see this: let \delta t be the time required to travel from one mirror to the other. During this time, the second mirror moves forward by an amount v \delta t. So the light, in traveling from one mirror to the other travels a distance D in one direction, and a distance v \delta t in the other. The total distance traveled, by Pythagoras, is: D_{total} = \sqrt{D^2 + v^2 \delta t^2}. If light always travels at speed c, then D_{total} = c \delta t. So we have: c \delta t = \sqrt{D^2 + v^2 \delta t^2}. Solving for \delta t gives \delta t = D/c \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. Double that for a round-trip.

So for a mirror in motion, the time for a complete circuit is longer, by a factor of \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. This is relative to the inertial reference frame in which v is measured. So the amount of slowing is different for different frames. There is no absolute frame relative to which the slowing is to be measured--time dilation works in any frame.
 
  • #10
seb7 said:
I understand Einstein's equations on time dilation, but this video talks about why, which doesn't seem correct to me.

I agree. I do not think the explanation is correct either. I explained why.

Wouldn't his reasoning for the time dilation require a fixed ether frame?

Part of it, yes. Part of the explanation is equivalent to the standard light clock analysis, which is correct.

here's a question: Say we got on a ship traveling at 0.8c then while traveling we got on to its onboard shuttle and kicked it back to 0c

0.8 c and 0 c relative to what? These speeds are relative, so you have to specify which frame of reference you're measuring the speed relative to. On the other hand, if you stated that a light beam was moving at speed c you wouldn't need to state which frame you're measuring the speed relative to, because every frame would measure it to be speed c. This is what I meant when I said that the speed of light is absolute.

, then the ship (still traveling at 0.8c), later went back to meet up with stationary shuttle. When they compare clocks, which craft lost more time?

You could analyze this situation in the following way. Let's choose a frame of reference where the shuttle is at rest after it leaves the ship, and measure every speed relative to this frame. Ignore the acceleration the shuttle must undergo to come to a stop. The ship departs from the shuttle at speed 0.8 c. It then turns around and comes back to meet up with the shuttle. The ship's clock would be behind the shuttle's.
 
  • #11
stevendaryl, yep, understand the mirror concept, and the resulting equation.
(I got the 50% thing wrong, as I was thinking in 1 dimension, not 2).

Using the video reasoning, I don't understand why an object with velocity has the slower time.
If a ship leaves earth, travels around at 0.8c and comes back, in both reference frames, the ship apparently aged more slowly - who decides which object with the set of mirrors had the faster velocity? (since object speed is relative).
 
  • #12
seb7 said:
Using the video reasoning, I don't understand why an object with velocity has the slower time.
If a ship leaves earth, travels around at 0.8c and comes back, in both reference frames, the ship apparently aged more slowly - who decides which object with the set of mirrors had the faster velocity? (since object speed is relative).

This is the classic twin paradox. Try: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
 
  • #13
seb7 said:
If a ship leaves earth, travels around at 0.8c and comes back, in both reference frames, the ship apparently aged more slowly - who decides which object with the set of mirrors had the faster velocity? (since object speed is relative).

In Post #4 I explained that you couldn't use the argument given in the video to explain why clocks on Earth run slow relative to the traveler. That's the part they got wrong.

In a proper analysis you can either look at things only from only Earth's frame of reference because it's the only inertial frame. The traveler's frame is not inertial because he has to turn around. Or, if you want to also understand things from the frames of reference of the traveler you have to take into account the change in reference frames when he switches directions. When you do that you find consistency. The traveler's clock is behind Earth's clocks upon the traveler's return.
 
  • #14
Mister T, I would expect the clocks to be the same (or close), since the shuttle had to accelerate at relative velocity of 0.8c away from the ship, the ship then (later) would have to accelerate beyond 0.8c to catch the shuttle, thus the ship dilation was even slower, but had further to travel.
 
  • #15
thanks Mister T, and Nugatory for the explanation. I will ponder on this.
 
  • #16
Read the details of the analysis in the link Nugatory posted in Post #12.
 
  • #17
seb7 said:
Using the video reasoning, I don't understand why an object with velocity has the slower time.
If a ship leaves earth, travels around at 0.8c and comes back, in both reference frames, the ship apparently aged more slowly - who decides which object with the set of mirrors had the faster velocity? (since object speed is relative).

A quantity such as "How old is the traveler in the ship when it gets back to earth?" is invariant, which means you can use any inertial frame you like to calculate the answer. If we use the inertial frame of the Earth, then we can reason that while the ship is traveling, its "light clock" is running slower, so the total number of complete circuits for a light pulse on the ship's light clock will be fewer than the number of complete circuits for the Earth's light clock. So less time (as measured by light clocks) will have passed for the ship. If you assume that biological processes are affected in the same way as light clocks, then the traveler will have aged less.

You can use any inertial reference frame to compute this answer. If you have some inertial reference frame that is traveling at speed v relative to the Earth in the same direction as the rocket, then according to this frame, the Earth's light clock is running slower than the rocket's during the outward journey, but the rocket's is running slower than the Earth's during the return journey. In this frame, the return journey takes longer, so the rocket ends up aging less.
 
  • #18
ok. got it. ship or Earth might move apart at 0.8c, but ship actually needs 1.6c to return, still making it the faster object. 0.8c to decelerate to a stop, another 0.8c to return.

but ... if photos/atoms are actually working like this 'light clock', wouldn't there be a direction which has less velocity than our current frame? though, I can't see how it could ever possibly be measured, probably the same issues when trying to measure one-way light.
 
  • #19
seb7 said:
ok. got it. ship or Earth might move apart at 0.8c, but ship actually needs 1.6c to return, still making it the faster object. 0.8c to decelerate to a stop, another 0.8c to return.

No! The ship's speed cannot exceed c in anyone's frame of refernece.

but ... if photos/atoms are actually working like this 'light clock', wouldn't there be a direction which has less velocity than our current frame?

When you say "less velocity" what do you mean? Velocity is not something that ship has unless you specify in which frame of reference that's occurring.

You are retaining the notion of a "true rest frame" in which the velocity of the ship can be measured. This is the same mistake the narrator makes in the movie when he says the ship's speed is "closer to c".
 
  • #20
stevendaryl said:
If you have some inertial reference frame that is traveling at speed v relative to the Earth in the same direction as the rocket, then according to this frame, the Earth's light clock is running slower than the rocket's during the outward journey, but the rocket's is running slower than the Earth's during the return journey. In this frame, the return journey takes longer, so the rocket ends up aging less.

It took me some effort to untangle that last sentence. In this frame, the rocket's clock will be observed to run slower during the return journey, so the rocket ends up aging less than Earth. In the observer's frame the return journey takes the same amount of time as the outbound journey.

I think I remember having seen this argument before, but I must have forgotten it somewhere along the way. Thanks for bringing it up.
 
  • #21
Mister T said:
It took me some effort to untangle that last sentence. In this frame, the rocket's clock will be observed to run slower during the return journey, so the rocket ends up aging less than Earth. In the observer's frame the return journey takes the same amount of time as the outbound journey.

I think I remember having seen this argument before, but I must have forgotten it somewhere along the way. Thanks for bringing it up.

For a rocket that goes out at constant velocity, turns around, and returns at constant velocity, there are three relevant frames:
  1. F_{earth}: the frame in which the Earth is at rest. In this frame, the rocket's clock runs slower than the Earth's clock, and the outward and return journeys are the same length.
  2. F_{out}: the frame in which the rocket is at rest during its outward journey. In this frame, the rocket's clock runs faster than the Earth's clock during the outward journey, but runs slower than the Earth's clock during the return journey. The return journey takes longer than the outward journey in this frame.
  3. F_{return}: the frame in which the rocket is at rest during its return journey. In this frame, the rocket's clock runs slower than the Earth's clock during the outward journey, but runs faster than the Earth's clock during the return journey. The outward journey takes longer than the return journey in this frame.
In all three frames, the rocket's clock advances less than the Earth's clock, in total.
 
  • #22
That youtube video is ok for motivating people to be interested in learning more. It tries to make everything very simple but is wrong in fundamental ways and can be misleading. If you are interested in relativity after one viewing of the video, it is time to move on without thinking any more about the video. There are many good books that are fairly heavy on the math, but if you are interested in more casual reading, check these out:
Mr Tompkins in Paperback by Gamow (first 4 chapters)
Relativity Visualized by Epstein
Relativity, The Special and the General Theory by Einstein (yes, the man himself)
 
  • #23
Mister T said:
No! The ship's speed cannot exceed c in anyone's frame of reference.
So a ship can go 0.8c, but can never stop and return at the same speed? because that would mean it adds up to relative speed difference of -1.6c?
 
  • #24
seb7 said:
So a ship can go 0.8c, but can never stop and return at the same speed? because that would mean it adds up to relative speed difference of -1.6c?

That is not right - the ship can go out at .8c and return at .8c, and there's no 1.6c involved when it does. Suppose two ships leave the Earth at the same time, traveling together in the same direction at the same direction at the same speed until one of them decides to turn around and return the earth, also at .8c relative to the earth.

One ship is moving at .8c to the left relative to the earth, the other ship is moving to the right at .8c relative to the earth, but neither ship is moving at 1.6c relative to the other. For this you'll need the relativistic formula for velocity addition ##(u+v)/(1+uv/c^2)##, and each ship's speed relative to the other turns out to be about .975c when you set u and v both to .8c

We have three equally valid and mutually consistent descriptions of the situation.
1) Outgoing spaceship is at rest. Earth is moving to the left at .8c, turnaround spaceship is moving to the left at .975c so will eventually return to earth.
2) Turnaround spaceship is at rest during its return leg. Earth is moving to the right towards it at .8c, outgoing spaceship is moving to the right at .975c
3) Earth is at rest. Turnaround spaceship is moving to the left at .8c, outgoing spaceship is moving to the right at .8c
 
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  • #25
seb7 said:
So a ship can go 0.8c, but can never stop and return at the same speed? because that would mean it adds up to relative speed difference of -1.6c?

No, as Nugatory points out, it doesn't work that way. Another way to see this is to imagine a space buoy planted at the turn-around point. The buoy is "B" in this figure, and it's at rest relative to Earth.

Relativity.png


A represents the ship after it has turned around and is headed back towards Earth. C represents what the ship would be doing had it not turned around but instead kept moving away from Earth. An observer on B would see both A and C departing at speeds of 0.8 c. But what about an observer aboard C?

To an observer aboard C, B would be moving away at speed 0.80 c. But A would be moving away at a speed of about 0.98 c.

Another way to calculate this is to use the relation $$\beta=\tanh \theta.$$ In this example ##\beta=0.80##. We don't double ##\beta## to get 1.6. Instead we double ##\theta## to get about 2.20, and then get about 0.98 for the corresponding value of ##\beta##.
 
  • #26
Yes, but onboad the ship, can't they do whatever speed they like? (Well, they think they can, as the time slow down makes them think they are traveling beyond c)

I don't have my calculation to hand, but i think 0.87c would cause a time dilation of around 0.5, thus the ship's navigation (when looking at time over distance) would show a speed of 1.74c.
 
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  • #27
When we calculate speed as distance / time, the distance and time must both be relative to the same inertial reference frame (IRF). If you get a speed greater than c, then you are probably measuring distance relative to one IRF, and time relative to a different IRF. How does your spaceship's navigation system measure time and distance, exactly?
 
  • #28
seb7 said:
Yes, but onboad the ship, can't they do whatever speed they like? (Well, they think they can, as the time slow down makes them think they are traveling beyond c)

On board the ship there's no experience of time dilation. But they do see the distance contracted.

I don't have my calculation to hand, but i think 0.87c would cause a time dilation of around 0.5, thus the ship's navigation (when looking at time over distance) would show a speed of 1.74c.

They would see the length contracted to half, so the speed is 0.87 c.

Observers on Earth would see the time dilated, so the speed is 0.87 c.
 
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  • #29
seb7 said:
Yes, but onboad the ship, can't they do whatever speed they like? (Well, they think they can, as the time slow down makes them think they are traveling beyond c)
No. Consider the following scenario. Your ship start's out with zero speed with respect to the Earth. It then starts moving at 1/10 the speed of light with respect to the Earth. It then drops a space buoy. (the buoy maintains the 1/10 c with respect to the Earth.) Then it accelerates until it is moving at 1/10 c with respect to this buoy. If he now measures his speed relative to the Earth, he has to us the formula given by Nugatory and gets an answer of 0.198c ( not 0.2c). If he drops another buoy and accelerates to 1/10 c with respect to it, he now is moving at 0.292 c. He can keep doing this forever, and he will never measure his speed with respect to the Earth as being anything but less than c.

So how is it that our ship can travel to a star 10 light years away at 0.8c and get there in 7.5 years by his clock, even though it should take 12.5 years to travel 10 light years at 0.8c? The reason is that the distance between Earth and Star that the Earth measures as 10 light years is only measured as being 6 light years by him. This is because of length contraction. For him, it is the Earth and star that are moving at 0.8c and the distance between them is length contracted. So according to him, the reason his clock only ticked off 7.5 years is because a distance of 6 light years is being crossed at 0.8c
 
  • #30
seb7 said:
Yes, but onboad the ship, can't they do whatever speed they like? (Well, they think they can, as the time slow down makes them think they are traveling beyond c)
I don't have my calculation to hand, but i think 0.87c would cause a time dilation of around 0.5, thus the ship's navigation (when looking at time over distance) would show a speed of 1.74c.

That video may be misleading you into focusing too much on length contraction and time dilation (in which case FactChecker's post #22 is really good advice, and I'd also consider "Spacetime Physics" by Taylor and Wheeler).

It's can be quite hard to understand relativity if you start from time dilation and length contraction without understanding how they arise. You may be better off starting with the Lorentz transformations and the relativity of simultaneity; understand these and time dilation and length contraction just naturally work.
 
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  • #31
Sorry but I can't restrain myself . . ;)

The ship's pilot is a trained, intelligent person, who understands relativity. He calculated the route before traveling, and knew full well that at 0.8c he would age 7.5 years over the course of the trip, even without checking his clock at the destination.

Describing this situation in terms of length contraction is merely inviting the division that seb7 cannot resist the urge to perform (if he knows to divide the numerator by gamma first then he also knows the spacetime interval).

Space does not magically contract at the whim of the traveler. It is merely an illusion. Here's how I would describe the situation:

The Doppler factor is 3, so over the whole journey the traveler sees the home clock advance by 2.5 years (1/3 of his own clock rate), and the destination clock advance by 22.5 years (3 times his own rate).

Before the journey the home clock reads (say) 0 years and the destination reads -10 years (they are synchronized). After the journey the home clock reads 0 years and the destination clock reads 10 years.

So, at the start of the journey, the destination clock jumps (in reality there would be finite acceleration so it wouldn't really be instantaneous) back by 2.5 years (from -10 to -12.5), and at the end of the journey the home clock jumps back 2.5 years (from 2.5 to 0).

Clear as mud? Maybe, but no sleight of hand with space "compressing" (I just hope I've got the maths right).
 
  • #33
m4r35n357 said:
But that is just wrong. Nobody would ever see (with their eyes) what is in those illustrations. The "real" situation is far more varied and interesting.
It is a dream where the speed of light is vastly reduced so that the relativity effects are evident in everyday life. Gamow was a famous scientist.
 
  • #34
FactChecker said:
It is a dream where the speed of light is vastly reduced so that the relativity effects are evident in everyday life.
I know that.

Please read my edited post, in particular the second link. Here is a quote:

"However, this picture is entirely wrong. It is wong because Gamov did not take the consequences of the finite speed of light into account. In everyday life we may to all intents and purposes pretend that the speed of light is infinite, since we are always concerned with much smaller velocities. But when a velocity close to the speed of light is involved, this is not justified any more. On the contrary, what we should see in this case would be substantially influenced by the fact that light propagates at a finite speed."
 
  • #35
m4r35n357 said:
I know that.

Please read my edited post, in particular the second link. Here is a quote:

"However, this picture is entirely wrong. It is wong because Gamov did not take the consequences of the finite speed of light into account. In everyday life we may to all intents and purposes pretend that the speed of light is infinite, since we are always concerned with much smaller velocities. But when a velocity close to the speed of light is involved, this is not justified any more. On the contrary, what we should see in this case would be substantially influenced by the fact that light propagates at a finite speed."
Yes. There is also a rotation. It is important in explaining some of the paradoxes. I am sure that Gamow was aware of that. So what? It takes about an hour to read the first 30 pages of the book. It is fun to read and educational, if not complete. The criticism that Gamow did not take into account that the speed of light is finite puzzles me. This book describes a dream where the speed of light is not only finite, it is small. Other than the rotation, I don't know what the complaint is.
 
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  • #36
FactChecker said:
Yes. There is also a rotation. It is important in explaining some of the paradoxes. I am sure that Gamow was aware of that. So what? It takes about an hour to read the first 30 pages of the book. It is fun to read and educational, if not complete.

More quotes:

"But it was not until 1924 that Anton Lampa [4] described the visual effects for the first time."

"And this notion was in fact illustrated by the physicist George Gamov in his 1940 book “Mr. Tompkins in Wonderland” "

So he must have known (or been ignorant of the facts, which I doubt), and yet he still wrote the book, and people are still being taught that this is what we "see" more than 90 years after it was proved false . . .
 
  • #37
m4r35n357 said:
More quotes:

"But it was not until 1924 that Anton Lampa [4] described the visual effects for the first time."

"And this notion was in fact illustrated by the physicist George Gamov in his 1940 book “Mr. Tompkins in Wonderland” "

So he must have known (or been ignorant of the facts, which I doubt), and yet he still wrote the book, and people are still being taught that this is what we "see" more than 90 years after it was proved false . . .
The rotation is the only effect I see mentioned in the link. Obviously Gamow took the finite speed of light into account. I can't see how the link could possibly imply otherwise.
 
  • #38
FactChecker said:
The rotation is the only effect I see mentioned in the link. Obviously Gamow took the finite speed of light into account. I can't see how the link could possibly imply otherwise.
Gamow's illustrations are based on a literal coordinate description of the scene. The simultaneity that underpins the coordinate description is not observable, since it is spacelike. We observe down light cones; we do not observe simultaneity. Terrell rotation is just one very specific example of several effects that Gamow omits (aberration, doppler, beaming).

Anyway, I think this thread is in danger of going off-topic, so I'll stop now as I am in danger of hiding post #31, which was my on-topic point.
 
  • #39
m4r35n357 said:
Gamow's illustrations are based on a literal coordinate description of the scene. The simultaneity that underpins the coordinate description is not observable, since it is spacelike. We observe down light cones; we do not observe simultaneity. Terrell rotation is just one very specific example of several effects that Gamow omits (aberration, doppler, beaming).
Yes. The illustrations in Mr Tompkins are not 100% complete and accurate. I don't think they were meant to be. But the OP video was so bad that alternatives were called for. In the first 30 seconds of that video, it gave rotating, orbiting planets as examples of inertial reference frames. Certainly that misconception would cause problems in the most elementary physics classes.
 
  • #40
m4r35n357 said:
The Doppler factor is 3, so over the whole journey the traveler sees the home clock advance by 2.5 years (1/3 of his own clock rate), and the destination clock advance by 22.5 years (3 times his own rate).

You criticized Gamow for using "sees" in the same way that you are using it. By the time the traveler arrives at the destination he will not have received all the signals yet, so he won't have yet "seen" the clock advance by that much. He'll have to wait for some time while the signal completes its journey. Also, when the ratio of frequencies is 3 the ratio you are using for clock rates is not ##\frac{1}{3}##, it's ##\frac{3}{5}##.

That ratio describes what people call time dilation. You may prefer to not call it that, but it is nevertheless not an illusion. Likewise for length contraction. If you like, you can think of those terms as describing a ratio that's not equal to one. The fact that people are using different terms than you to describe the same thing you are describing does not make what those terms are describing an illusion.
 
  • #41
Mister T said:
You criticized Gamow for using "sees" in the same way that you are using it. By the time the traveler arrives at the destination he will not have received all the signals yet, so he won't have yet "seen" the clock advance by that much. He'll have to wait for some time while the signal completes its journey. Also, when the ratio of frequencies is 3 the ratio you are using for clock rates is not ##\frac{1}{3}##, it's ##\frac{3}{5}##.

That ratio describes what people call time dilation. You may prefer to not call it that, but it is nevertheless not an illusion. Likewise for length contraction. If you like, you can think of those terms as describing a ratio that's not equal to one. The fact that people are using different terms than you to describe the same thing you are describing does not make what those terms are describing an illusion.
Thanks for taking the time to reply, but I don't think you are getting my point. The Doppler factor is 3 or 1/3 so that is the rate the the traveler sees (compared to his own clock). I prefer not to call it time dilation because it is not time dilation, it is the Doppler factor (you could say it "includes" time dilation, gamma (which is 3/5), or whatever). He sees all this happen in real time so he does not have to wait for anything.

I am using the word "sees" to mean precisely what it says. I might have made an error somewhere, but not where you are pointing to.

I think you might be suffering from Mr Gamow's "education" ;)

Try looking at this if you don't see my point, and the link inside it too. Drawing an accurately scaled spacetime diagram will help too.
 
  • #42
If errors were made in your analysis, please tell us what you think they were so that we can get them fixed. Then I might be able to understand what it is you're saying.

m4r35n357 said:
The ship's pilot is a trained, intelligent person, who understands relativity. He calculated the route before traveling, and knew full well that at 0.8c he would age 7.5 years over the course of the trip, even without checking his clock at the destination.

Can you show us how he did that? And explain what the calculation means. Like, for example, if using the Lorentz transformation, explain what the transformed and untransformed values mean, and what a comparison of them means.

The Doppler factor is 3, so over the whole journey the traveler sees the home clock advance by 2.5 years (1/3 of his own clock rate), and the destination clock advance by 22.5 years (3 times his own rate).

How does he "see" that the clock has advanced 2.5 years? Does he look at it with a telescope or does he "see" it some other way?
 
  • #43
Mister T said:
If errors were made in your analysis, please tell us what you think they were so that we can get them fixed. Then I might be able to understand what it is you're saying.
Can you show us how he did that? And explain what the calculation means. Like, for example, if using the Lorentz transformation, explain what the transformed and untransformed values mean, and what a comparison of them means.
How does he "see" that the clock has advanced 2.5 years? Does he look at it with a telescope or does he "see" it some other way?
1. I don't know of any, that's why I'm asking.
2. ##\tau^2 = t^2 - x^2 = x^2 / v^2 - x^2 = 100 * (1/0.64 - 1) = 56.25 = 7.5^2##
3. Yes, with a telescope
4. Goodnight, it's 11pm here and I have to work tomorrow . . . enjoy ;)
 
  • #44
OK I found my error, and had to get up early to correct it! I shall start afresh:

The ship's pilot is a trained individual who understands relativity. He planned the route before traveling, and knew full well (by calculating the spacetime interval as in #43) that at 0.8c he would age 7.5 years over the course of the trip, even without checking his clock at the destination.

The Doppler factor is 3, therefore he also knows in advance that over the whole journey the home clock will advance by 2.5 years (1/3 of his own clock rate), and the destination clock by 22.5 years (3 times his own rate).

Before the journey the home clock reads (say) 0 years and the destination reads -10 years (they are synchronized).

After the journey the home clock reads 2.5 years and the destination clock reads 12.5 years (the time it takes to travel 10 ly at 0.8c, and the part I forgot to take into consideration!).

QED

I shall now write out 100 times: CLOCKS DO NOT "JUMP" IN THE DOPPLER ANALYSIS!
 
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  • #45
ok, so the light clock explanation helps to get a point across; but I'm correct in saying it can not be the actual reason for time dilation?
(surely if this were correct then wouldn't electrons have an elliptic orbit, and some kind of zero rest frame would have to exist where electrons had a perfectly circular orbit.)

Mister T said:
They would see the length contracted to half, so the speed is 0.87 c.
meaning, as time halfs, distance to travel doubles?

If this were true, then what would happen if our ship circled the Earth 0.87c for a year (earth time). Both Earth and the ship occupants wouldn't both see a travel speed of 0.87c?
(ignoring gravitational effects on time)
 
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  • #46
seb7 said:
ok, so the light clock explanation helps to get a point across; but I'm correct in saying it can not be the actual reason for time dilation?
If you want to understand the "actual reason" for time dilation, there are two possibilities:
1) Learn about the relativity of simultaneity, as has been suggested above. When we say that clock clock A and clock B are both running at the same rate, we're really saying that if we look at clock A twice, and both times we look at clock B at the same time, the difference between the two readings from clock A will be the same as the difference between the two readings from clock B. Because of the relativity of simultaneity, different observers will mean different things by that all-important phrase "at the same time", and depending on ho wthey define "at the same time" may find one clock running slower than the other.

It's worth repeating that this phenomenon is not the differential aging that appears in the twin paradox.

2) Recognize that it's just the way the geometry of spacetime works. There are two possible geometries: in one of them the speed of light is the same for all observers and relativistic effects such as time dilation appear; in the other the speed of light is constant relative to some hypothetical fixed point somewhere in the universe, everyone else measures the speed of light to be the difference between ##c## and their motion relative to that fixed point. Experiments show that we live in the first kind of universe.

(surely if this were correct then wouldn't electrons have an elliptic orbit, and some kind of zero rest frame would have to exist where electrons had a perfectly circular orbit.)
It's off-topic here, but electrons don't have any sort of orbit at all. The idea that electrons are particles moving in circles or ellipses (or any trajectory at all) around the nucleus was abandoned more than three-quarters of a century ago.
 
  • #47
seb7 said:
ok, so the light clock explanation helps to get a point across; but I'm correct in saying it can not be the actual reason for time dilation?

The light clock demonstrates time dilation. Without a light clock you'd still have time dilation, so no, the light clock is not the reason for time dilation. The reason is that inertial reference frames are equivalent and the speed of light is the same in all of them.

(surely if this were correct then wouldn't electrons have an elliptic orbit, and some kind of zero rest frame would have to exist where electrons had a perfectly circular orbit.)

A hoop has a circular shape in its rest frame. The hoop is at rest relative to the observer. Another observer, moving relative to the first, could carry with him and exact duplicate of the hoop. Each would agree that their hoop is circular. Each could determine that the other's hoop is not circular.

meaning, as time halfs, distance to travel doubles?

No, you are talking about this as if all you need is one frame of reference to describe it. There are always (at least) two, otherwise time dilation and length contraction do not occur.

If this were true, then what would happen if our ship circled the Earth 0.87c for a year (earth time). Both Earth and the ship occupants wouldn't both see a travel speed of 0.87c? (ignoring gravitational effects on time)

At any instant, yes. Each would observe the other having a speed of 0.87 c.
 
  • #48
m4r35n357 said:
The ship's pilot is a trained individual who understands relativity. He planned the route before traveling, and knew full well (by calculating the spacetime interval as in #43) that at 0.8c he would age 7.5 years over the course of the trip, even without checking his clock at the destination.

m4r35n357 said:
2. ##\tau^2 = t^2 - x^2 = x^2 / v^2 - x^2 = 100 * (1/0.64 - 1) = 56.25 = 7.5^2##

So, we have ##\tau=7.5## and ##t=12.5##. But the time did not dilate from 7.5 to 12.5? How would you describe the relationship between those two amounts of time?

The traffic controller stationed on Earth is a trained individual who understands relativity, and knows full well (having done the same calculation) that at a speed of 0.8 c the clocks aboard the ship will register a time of 7.5 years upon arrival. He therefore reckons that the ship will travel a distance $$x'=vt=(0.8)(7.5)=6.$$ Previous experiments with light signals determined that the proper distance is ##L=10##.

So, we have ##L=10## and ##x'=6##. But the length did not contract from 10 to 6? How would you describe the relationship between those two amounts of distance?
 
  • #49
Mister T said:
So, we have ##\tau=7.5## and ##t=12.5##. But the time did not dilate from 7.5 to 12.5? How would you describe the relationship between those two amounts of time?

The traffic controller stationed on Earth is a trained individual who understands relativity, and knows full well (having done the same calculation) that at a speed of 0.8 c the clocks aboard the ship will register a time of 7.5 years upon arrival. He therefore reckons that the ship will travel a distance $$x'=vt=(0.8)(7.5)=6.$$ Previous experiments with light signals determined that the proper distance is ##L=10##.

So, we have ##L=10## and ##x'=6##. But the length did not contract from 10 to 6? How would you describe the relationship between those two amounts of distance?
Pardon the deleted post if you saw it, I was not happy with my initial answer.

On further reflection, I am just going to say that we agree with the relative times by doing things different ways. You are using a calculation of time dilation and I am using the Doppler shift (because I like to deal with things I can see). It looks like we are not going to agree on the approach, although we both get the same answers. In fact, since my calculation of space-time interval effectively contains gamma, we are not even using different mathematics.

As to length contraction, I maintain (unlike the ratio or difference between the two times) that there is nothing that is actually measurable at 6ly, so I have no interest or opinion on that particular relationship.

So, I have effectively come to the same conclusion as in my deleted post.
 
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  • #50
m4r35n357 said:
On further reflection, I am just going to say that we agree with the relative times by doing things different ways.

Yes. And I agree that your way is just as valid as mine.

In fact, since my calculation of space-time interval effectively contains gamma, we are not even using different mathematics.

The two methods are equivalent. I have no issues with either of them.

As to length contraction, I maintain (unlike the ratio or difference between the two times) that there is nothing that is actually measurable at 6ly, so I have no interest or opinion on that particular relationship.

If the ship had left a trail of bread crumbs and you were in another ship constrained to travel with the same velocity, is there no way for you to determine its length using only measurements taken from your ship?

If there's an airport runway on the surface of planet Earth, and you fly parallel to it at a speed of 0.8 c is there no way for you to determine its length using only measurements taken aboard your ship?
 
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