Is Time Dilation Misunderstood in Popular Science Videos?

  • #51
Mister T said:
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?
That is the point at which my brain screams "enough!" and I lose the will to continue ;)
I'm sure if I chose to pursue it (and I have done in the past) I would agree with you. Thing is, I have a scaled spacetime diagram in front of me, and I see (sic) nothing on it which corresponds to a distance of 6ly. I just don't get what use it is to me that's all, and utility is what motivates me to work things out.
To labour the point, I know that space does not really contract, so at best it is just a calculation resulting from division of two numbers, and not even an optical illusion, since in reality it would actually appear to stretch in front of me and contract behind me.
 
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  • #52
m4r35n357 said:
Thing is, I have a scaled spacetime diagram in front of me, and I see (sic) nothing on it which corresponds to a distance of 6ly. I just don't get what use it is to me that's all, and utility is what motivates me to work things out.

That's because on your spacetime diagram you're focusing on two events separated by an interval that is timelike. You can find a triangle on there with a hypotenuse of 7.5 y. One leg of that triangle is 10 ly, the other is 12.5 y. Time dilation is characterized by the ratio ##\gamma=\frac{12.5}{7.5}##.
To labour the point, I know that space does not really contract, so at best it is just a calculation resulting from division of two numbers,

Suppose you and your friend are separated on a train by a distance of 7.5 light microseconds. The train zips through a tunnel at a speed of 0.8 c. At the same time you each reach out a window and place a mark on the tunnel wall. Later, when you're at rest relative to the tunnel, you measure the distance between the marks to be 12.5 light microseconds. Drawing this on a spacetime diagram you have a triangle with a hypotenuse of 7.5 light microseconds. One leg of that triangle is 10 microseconds, the other is 12.5 light microseconds. Length contraction is characterized by the ratio ##\gamma=\frac{12.5}{7.5}##.

If space doesn't really contract, how is it that the distance between the marks is 12.5 when you measure it with a tape measure, yet when you use that same tape to measure the distance between you and your friend you get 7.5?

This time we're focused on two events separated by an interval that is spacelike.
 
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  • #53
Mister T said:
This time we're focused on two events separated by an interval that is spacelike.
[been guitaring this week!]
Yep, different scenario altogether. Your "time dilation" in the previous scenario was just the difference between two invariants (which I am fine with in principle), but the "space contraction" was coordinate dependent (it went away when relative motion ceased).

If your latest scenario is a "dual" of the previous one then I can conceive of calling the difference between two proper distances measured at various times "space contraction". I just don't find it a particularly instructive or satisfying concept (my choice).

As I have stated over and over, IMO what we can see/measure/observe trumps all.
 
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  • #54
Mister T said:
If space doesn't really contract, how is it tha
isnt this just caused by the time dilation?

We can consider any object with velocity to also be at rest, so..
Two ships traveling side by side, already moving at 0.8c. One of the ships then accelerates again attempting to reach +0.8c from the ship they left behind. Does the ship behind observe them achieve another 0.8c?
 
  • #55
seb7 said:
isnt this just caused by the time dilation?

We can consider any object with velocity to also be at rest, so..
Two ships traveling side by side, already moving at 0.8c. One of the ships then accelerates again attempting to reach +0.8c from the ship they left behind. Does the ship behind observe them achieve another 0.8c?
Yes, but relative to someone in the frame of reference in which you state that they both are originally traveling at .8c, the first one continues at .8c and the second one gets up to something like .96c (I'm making up the number but it's in the ball-park). SO, the second one is not achieving ANOTHER .8c, he's just moving away from the first one at .8c In exactly the same way as the first one is moving away from the "fixed" observer at .8c.

If the first ship were to attempt to move away from the fixed observer at "another" .8c, he could not do so, he could only approach c asymptotically.

It is your choice of the word "another" that is confusing you about the scenarios.
 
  • #56
so 0.96c(est) from the original fixed position, but is +0.8c beyond the ship. So whos to say we can't travel beyond light speed, it just can't ever be observed. Could it be light itself is infinite but can only be observed at light speed. Might help explain some of the weird effects of it apparently knowing the future.

----

This is how I think of spacetime, which is why when I saw the video, it made me rethink. Does this not work with the calculations? let me know if I'm just plain wrong.

I see four dimensions. I think of space as 3 dimensions, time as a 4 dimension. Time is like 90 degrees from space, but always 0 degrees in front of us. Every object (including us) are always moving through these 4 dimensions (spacetime) at the speed of light, it's that speed that is fixed, never changes.
ie. when we think we are still, we are moving through time at lightspeed. When a moving object changes direction, we are also changing the angle we were moving through time, this angle now being different to other objects traveling at different speeds. The time angle still appears to us to be front, but its now a different angle to other objects moving through time. This angle directly relates to the time and distance dilation observed.
This seems to work for all the calculations I've done.
 
  • #57
seb7 said:
so 0.96c(est) from the original fixed position, but is +0.8c beyond the ship. So whos to say we can't travel beyond light speed, it just can't ever be observed.
That's ridiculous.You keep ignoring reference frames. Beyond the speed of light RELATIVE TO WHAT?

Could it be light itself is infinite but can only be observed at light speed.
No it could not.

Might help explain some of the weird effects of it apparently knowing the future.
If you think you can know the future, you're off into crackpot land.

This is how I think of spacetime
spacetime does not care how we think of it, it just does what it does and the math tells the tale.

which is why when I saw the video, it made me rethink. Does this not work with the calculations? let me know if I'm just plain wrong.
I'm not sure what you are talking about but based on your other statements I'd hazard a guess that you are just plain wrong.
I see four dimensions. I think of space as 3 dimensions, time as a 4 dimension. Time is like 90 degrees from space, but always 0 degrees in front of us. Every object (including us) are always moving through these 4 dimensions (spacetime) at the speed of light, it's that speed that is fixed, never changes.
ie. when we think we are still, we are moving through time at lightspeed. When a moving object changes direction, we are also changing the angle we were moving through time, this angle now being different to other objects traveling at different speeds. The time angle still appears to us to be front, but its now a different angle to other objects moving through time. This angle directly relates to the time and distance dilation observed.
This seems to work for all the calculations I've done.
I don't follow you so can't comment but it sounds wrong. What does it even MEAN to have time at an angle? If you're talking about space-time diagrams, then I would suggest that you learn the standard terminology, then we can converse with you about them.
 
  • #58
@seb7 you seem to be having a problem with the fact that there is no absolute reference frame. Velocity is only meaningful in terms of comparison with another object. That is the why it's called the theory of "relativity" and that part of it was known well before Einstein.
 
  • #59
seb7 said:
isnt this just caused by the time dilation?

Not really. Time dilation and length contraction are both consequences of the postulates. Starting from them you can derive time dilation and then use that to derive length contraction. That's the usual route, but it's not the only way to do it.

We can consider any object with velocity to also be at rest, so.. Two ships traveling side by side, already moving at 0.8c.

When you say "already moving" what does that mean? As you said, you could consider them to be at rest.

One of the ships then accelerates again attempting to reach +0.8c from the ship they left behind. Does the ship behind observe them achieve another 0.8c?

You're saying that the second ship is "attempting" to move at speed 0.8c relative to the first. If he succeeds then yes, each ship will see the other as moving at speed 0.8c.

Look at the figure in Post #25: This is a view from the frame of reference in which both ships were previously moving at speed 0.8c relative to A. What we are calling the "first ship" is B, and it's at rest in this frame of reference. The "second ship" is C, and it has succeeded in moving at speed 0.8c relative to the first ship, B. Note that B will see the distance between A and C increase at a rate of 1.6c.

Relativity.png


Now let's look at things from the frame of reference of A. In this frame both ships were originally moving at speed 0.8c. Afterwards, the second ship C has succeeded in pulling away from the first ship B at speed 0.8c.

Relativity 2.png
 
  • #60
I see four dimensions. I think of space as 3 dimensions, time as a 4 dimension. Time is like 90 degrees from space, but always 0 degrees in front of us. Every object (including us) are always moving through these 4 dimensions (spacetime) at the speed of light, it's that speed that is fixed, never changes.

Reference https://www.physicsforums.com/threads/time-dilation-reasoning.844227/page-3

You can check this PDF: http://www.thequantummachine.com/Quantummachine_4thspeed.pdf

For a peer-reviewed source, which is essentialy the same concept but applied in an ellaborate and formal way:

Representation of relativistic quantities by trigonometric functions, http://scitation.aip.org/content/contributor/AU0018829 , Am. J. Phys. 54, 536 (1986); http://dx.doi.org/10.1119/1.14557.

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 and for corresponding de Broglie waves. A number of relativistic relationships is obtained by the use of common trigonometric identities and formulas."

As adviced to you, first learn carefully about Minkowski diagrams. When you master them, you can follow with other non-standard diagrams (Bhreme, for example), and then revisit the links above.
 
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  • #61
seb7 said:
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.
 
  • #62
MachPrincipe said:
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##.

phi.png


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##.
 
  • #63
Mister T said:
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.
 
  • #64
MachPrincipe said:
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.
 
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  • #65
Mister T said:
The diagram I drew is in Euclidean 2-
.

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

Mister T said:
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.

Mister T said:
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.
 
  • #66
MachPrincipe said:
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.

phinds said:
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.

Mister T said:
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.
 
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  • #67
seb7 said:
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 ...
 
  • #68
Mister T said:
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).
 
  • #69
Doppler effect afects frequency of light and other waves. Clock which move relative to the observer will be all time dilated.
 
  • #70
m4r35n357 said:
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.
 
  • #71
m4r35n357 said:
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.
 
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  • #72
Mister T said:
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.
 
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  • #73
m4r35n357 said:
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.
 
  • #74
Mister T said:
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.
 
  • #75
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.
 
  • #76
Mister T said:
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.
 
  • #77
m4r35n357 said:
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.
 
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  • #78
Mister T said:
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).
 
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