Is Time Dilation Misunderstood in Popular Science Videos?

[Herman Trivilino]

So whos to say we can't travel beyond light speed, it just can't ever be observed.

Things that are not observable, at least in principle, are not science.

 

[Herman Trivilino]

Quote: (from http://scitation.aip.org/content/aapt/journal/ajp/54/6/10.1119/1.14557)

"A ‘‘space‐time angle’’ φ is defined by setting v=c(sin φ). This leads to a form of Lorentz transformations which uses simple real trigonometric functions and yields a graphic correlation of important relativistic quantities for particles [...]

In that scheme $\phi$ is not an angle in the spacetime of special relativity. It's an angle in Euclidean space. All of the trigonometric relations presented in that paper can be derived from this figure. Namely $\sec \phi=\gamma$, $\sin \phi=\beta$, and $\tan \phi=\beta \gamma$.

Note that the Pythagorean theorem is valid: $\gamma^2=(\beta \gamma)^2+1^2$.

In the case of a triangle on a spacetime diagram we'd instead have $\gamma^2 = (\beta \gamma)^2-1^2$.

 

[MachPrincipe]

In that scheme $\phi$ is not an angle in the spacetime of special relativity. It's an angle in Euclidean space. All of the trigonometric relations presented in that paper can be derived from this figure. Namely $\sec \phi=\gamma$, $\sin \phi=\beta$, and $\tan \phi=\beta \gamma$.

View attachment 92532

Note that the Pythagorean theorem is valid: $\gamma^2=(\beta \gamma)^2+1^2$.

In the case of a triangle on a spacetime diagram we'd instead have $\gamma^2 = (\beta \gamma)^2-1^2$.

Yes. The diagram would be euclidean, with +1 +1 +1 +1. The interpretation of relativity from this POV is referred to as Euclidean relativity.

 

[Herman Trivilino]

Yes. The diagram would be euclidean, with +1 +1 +1 +1.

The diagram I drew is in Euclidean 2-space.

The interpretation of relativity from this POV is referred to as Euclidean relativity.

No, it's the same relativity, it's just using an angle in Euclidean space to do an analysis of Einsteinian relativity.

You were discussing an angle between a space axis and a time axis when you brought up this angle $\phi$. I was merely pointing out that the two are unrelated.

 

[MachPrincipe]

The diagram I drew is in Euclidean 2-

.

Yes, you are right. I was thinking of the more general 4d spacetime.

No, it's the same relativity, it's just using an angle in Euclidean space to do an analysis of Einsteinian relativity.

Yes, I agree. It's similar to Lorentzian relativity being equivalent to Einsteinian relativity. Still, if you look at the entry of Wikipediaa on alternatives to special relativity, you will. find that some advocates believe there are indeed differences. I think that is due to a misinterpretation of their own ideas.

You were discussing an angle between a space axis and a time axis when you brought up this angle $\phi$. I was merely pointing out that the two are unrelated.

Yes, the angles are unrelated as the diagrams, Minkowski, Brheme, Euclidean are all different. I was just guiding to the poster to ideas that could find close to what he is thinking of. And adviced to master on Minkowski ones, to avoid misinterpretation of more unconventional views.

 

[seb7]

v=c(sin φ).

yes, this was part of my maths. Some good links there, the pdf and "Minkowski diagrams' very similar to my way of thinking - don't feel such an idiot now.

you seem to be having a problem with the fact that there is no absolute reference frame

I disagree, I totally understand this (the principles and the maths involved. I do often seem to get answers about frames that I wasn't questioning).
I'm trying to go beyond the maths, to fully understand what spacetime is as described by the maths and observations.

Things that are not observable, at least in principle, are not science.

NIce answer. I suppose I am trying to look through a smokescreen to find an explanation for all observations.

 

[phinds]

I disagree, I totally understand this (the principles and the maths involved.

Good, it's just that I didn't see that fully reflected in your posts.

I do often seem to get answers about frames that I wasn't questioning).

Yes, and I suspect that's because you don't fully reflect your understanding in your posts, as was the case this time and why I made my comment in the first place.

I'm trying to go beyond the maths, to fully understand what spacetime is as described by the maths and observations.NIce answer. I suppose I am trying to look through a smokescreen to find an explanation for all observations.

Yes, and that's likely the heart of your problem. You are trying to answer "why" questions. As has been discussed on this forum ad nauseum, science doesn't answer "why" questions, it answers "how" questions. The reason for this is that no matter how far down you drill in answering a "why" question, there's always another "why" question lurking behind it and you just get into interpretations and philosophy and away from science.

As expressed by one of the mentors:

At the bottom of every stack of "why" questions is another "why" question. They never end. This is true for everything ...

 

[m4r35n357]

Time dilation is characterized by the ratio $\gamma=\frac{12.5}{7.5}$.

Sorry to leave it so long before replying. This simple statement was more troubling to me than you probably thought, and it has just struck me as to why.

The ratios of the readings and therefore the rate of the coordinate clock to the traveler's clock, is constant and greater than one for the whole trip. This means that a traveler observing only the clock "under his nose" as he travels will see (sic) the time in the other frame going more quickly, contrary to the popular argument that clocks should appear to run slower in the alternate frame. Of course if he looks at any other clocks they will be moving 3 times slower or quicker depending on whether they are in front or behind (Doppler effect).

 

[MachPrincipe]

Doppler effect afects frequency of light and other waves. Clock which move relative to the observer will be all time dilated.

 

[Herman Trivilino]

Of course if he looks at any other clocks they will be moving 3 times slower or quicker depending on whether they are in front or behind (Doppler effect).

If I see your clocking running 3 times slower than mine it will advance 1 year for every 3 years that mine advances. If my clock advances 7.5 years during the interval between two events, and yours advances 12.5 years during the interval between those same two events, I would likewise conclude that your clock is running slow. The former effect is called the Doppler effect, the latter effect is called time dilation. They are in fact logically equivalent, meaning that each implies the other. Each party sees the other's clock as running slow.

When a frequency is measured it involves measuring the time that elapses between two events, for example the arrival of two consecutive signals. If you want to compare that time to the time that elapses between the creation of those two signals there are two effects to be considered. One is the change in the distance between the source and receiver that occurs between the two events. The other is time dilation.

 

[Herman Trivilino]

This means that a traveler observing only the clock "under his nose" as he travels will see (sic) the time in the other frame going more quickly, contrary to the popular argument that clocks should appear to run slower in the alternate frame.

No, he won't. The second sentence in my last post is backwards. When 7.5 hours elapse on the rocket clocks, to the rocket observer 4.5 hours elapse on the Earth clocks. But if the Earth clocks are synchronized with the destination clock in their rest frame, to the rocket observer the clock at the destination will be 8 hours ahead of Earth clocks.

 

[m4r35n357]

No, he won't. The second sentence in my last post is backwards. When 7.5 hours elapse on the rocket clocks, to the rocket observer 4.5 hours elapse on the Earth clocks. But if the Earth clocks are synchronized with the destination clock in their rest frame, to the rocket observer the clock at the destination will be 8 hours ahead of Earth clocks.

Not sure where the 4.5 comes from . . .
I thought we had agreed on the figures (I am referring to post #44); that my spaceship clock advances 7.5 units over the journey, and that the coordinate clock (which I read at both ends) advances 12.5 units.
In which case, the coordinate clock (the one at my coordinates) must run be seen to run more quickly by a factor of $\gamma$, not more slowly.

 

[Herman Trivilino]

Not sure where the 4.5 comes from . . .

Since the Earth observer measures the rocket clock to be running slow by a factor of $\gamma=\frac{5}{3}$, that is, the ratio $\frac{12.5}{7.5}$, then the rocket observer must measure Earth clocks as running slow by the same factor $\gamma=\frac{5}{3}=\frac{7.5}{4.5}$.

I thought we had agreed on the figures (I am referring to post #44); that my spaceship clock advances 7.5 units over the journey, and that the coordinate clock (which I read at both ends) advances 12.5 units.

That was in the frame of reference of the Earth observer.

In which case, the coordinate clock (the one at my coordinates) must run be seen to run more quickly by a factor of $\gamma$, not more slowly.

No, because the rocket observer must use two "Earth clocks" to compare to his own rocket clock. One is located on Earth and the other is located at the destination. I call them "Earth clocks" because they are synchronized in Earth's rest frame. The rocket observer will measure the clock located at the destination to be 8 years ahead of the clock located at Earth. When he arrives he will find the clock at the destination reading a time that is 12.5 years more than the clock on Earth read when he left, but that is not because he measures the clocks running faster than his clock, it's (mostly) because the two clocks are not synchronized in his frame.

 

[m4r35n357]

The rocket observer will measure the clock located at the destination to be 8 years ahead of the clock located at Earth.

Now there's a magical seemingly 8 out of nowhere, you really should say how you get to these numbers ;) All my numbers out are in the open in posts #43 and #44. There is no 6 (post #48), 4.5 (#71) or 8 (#73). I see nothing in your arguments that contradicts my assertion at the end of #72.

If I'm wrong I want to know, but by now I'm pretty sure it's right.

 

[Herman Trivilino]

The clock on Earth is synchronized with the clock at the destination. Both clocks are at rest relative to each other and separated by a proper distance of 10 light years. To an observer moving from Earth towards the destination at a speed of 0.8c, the clock at the destination will be (0.8)(10)=8 years ahead of the clock on Earth.

 

[PeterDonis]

To an observer moving from Earth towards the destination at a speed of 0.8c, the clock at the destination will be (0.8)(10)=8 years ahead of the clock on Earth.

But this is not a direct "measurement"; it's a calculation. Also, it's a synchronization, not a rate.

 

[Herman Trivilino]

Now there's a magical seemingly 8 out of nowhere, you really should say how you get to these numbers ;) All my numbers out are in the open in posts #43 and #44. There is no 6 (post #48), 4.5 (#71) or 8 (#73). I see nothing in your arguments that contradicts my assertion at the end of #72.

If I'm wrong I want to know, but by now I'm pretty sure it's right.

Ok, I've got a few minutes this morning and need a distraction to help avoid the sympathetic panic of finals week ...

First and foremost the 1st Postulate implies that if rocket clocks run slow compared to Earth clocks then Earth clocks run slow compared to rocket clocks.

Second, let's see where all the numbers come from. Recall that we already have the following, based on the statement of the scenario, the definition of $\gamma$, the choice of unprimed coordinates for Earth frame and primed coordinates for rocket frame, and the usual conventions regarding the relationship between the primed and unprimed coordinate systems. $x$ is in light years, $t$ is in years. Note that $c=1$.

$\beta=\frac{4}{5}$
$\gamma=\frac{5}{3}$
$\Delta x=10$
$\Delta t = 12.5$
$\Delta x' = 0$
$\Delta t' =7.5$

So, let's take a look at one of the Lorentz transformation equations: $$\Delta t'=\gamma(\Delta t-\beta \Delta x)$$
Substituting the values, we have
$$(7.5)=\frac{5}{3}(12.5-\frac{4}{5}10)$$
$$(7.5)=\frac{5}{3}(12.5-8.0)$$
$$(7.5)=\frac{5}{3}(4.5)$$

In the second from last line you see the 8.0 you asked about. In the last line you see the 4.5 you asked about.

Let's look at one more Lorentz transformation equation: $$\Delta x=\gamma(\Delta x'+\beta \Delta t')$$
$$(10)=\frac{5}{3}\Big(0+\frac{4}{5}(7.5)\Big)$$
$$(10)=\frac{5}{3}(6)$$
This last line contains the 6 you asked about.

I think I got 'em all.

 

[PeterDonis]

the 1st Postulate implies that if rocket clocks run slow compared to Earth clocks then Earth clocks run slow compared to rocket clocks.

Actually, that's not what the first postulate says. It says the laws of physics are the same in all inertial frames; but that's a statement about physical measurements, not about coordinate conventions, which is what "clocks running slow" refers to. A direct physical measurement would be something like the Doppler shift of light signals traveling between the rocket and Earth; the first postulate then says that the redshift of rocket signals observed on Earth must be the same as the redshift of Earth signals observed on the rocket (assuming that neither Earth nor rocket change their state of motion while the light is traveling).