Is Time Dilation Misunderstood in Popular Science Videos?

[seb7]

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.

 

[mfb]

Which last bit exactly do you mean? At which point in the movie?

wouldnt 50% of c would cause 50% of time?

No. That's not even well-defined without referring to another observer.

 

[seb7]

When he talks about time dilation, 5min 15s, into the video. Is he correct?

 

[Herman Trivilino]

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.

 

[m4r35n357]

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.

 

[seb7]

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?

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?

 

[jbriggs444]

The speed of light is absolute, not relative.

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".

 

[mfb]

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.

 

[stevendaryl]

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.

 

[Herman Trivilino]

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.

 

[seb7]

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).

 

[Nugatory]

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

 

[Herman Trivilino]

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.

 

[seb7]

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.

 

[seb7]

thanks Mister T, and Nugatory for the explanation. I will ponder on this.

 

[Herman Trivilino]

Read the details of the analysis in the link Nugatory posted in Post #12.

 

[stevendaryl]

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.

 

[seb7]

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.

 

[Herman Trivilino]

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".

 

[Herman Trivilino]

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.

 

[stevendaryl]

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.

 

[FactChecker]

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)

 

[seb7]

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?

 

[Nugatory]

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

 

[Herman Trivilino]

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.

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 $\beta$.

 

[seb7]

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.

 

[jtbell]

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?

 

[Herman Trivilino]

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.

 

[Janus]

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

 

[Nugatory]

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.