Does time dilation occur due to the speed limit of light or c?

[Grasshopper]

Oh one other thing: these discussions about what the meter is are not pedantic. As you learn more about this topic (speaking from personal experience), you’ll find that the rabbit hole goes much deeper than you might think. And with respect to the meter specifically, when Einstein was first making his arguments about SR, he carefully looked at what it means to measure lengths or time intervals. And the more you learn, you’ll see it was for good reason.

As I said, the rabbit hole goes deeper than one might think (or as a physics professor told me, there are “subtleties” to SR that aren’t clear at first glance).

Good luck in your searching!

 

[Sagittarius A-Star]

pedantic

Being pedantic is required in math and physics.

 

[anuttarasammyak]

Thanks all, but that doesn't answer my question. What I am asking is if time dilation is caused by the finite speed of light or the speed in which we are moving relative to another observer?

I would like to show you the third choice : time dilation of another body is caused by its speed to us who are at rest in an IFR.

Do you say in your latter choice that our time pace depends on whether another one observe us or not ? When v=0.99999c elementary particle is produced in CERN accelerator, does the particle become an observer and influence our pace of time ?

 

[anuttarasammyak]

What I am asking is if time dilation is caused by the finite speed of light

..Well, yes, in a sense that time dilation has gamma factor defined as
$$\gamma:=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
and if c were infinite, $\gamma$=1 for any finite v so no time dilation could take place.

 

[vanhees71]

In modern terms, the meter is defined in terms of the speed of light as @PeroK says, so you can't have a different speed of light.

You are correct that this definition scheme is new (2018) but the problem is not. The meter used to be defined as some fraction of the Earth's circumference. In such a system, if you change $c$ you have to change $\epsilon_0$, which changes the strength of the electromagnetic force, which changes the size of an atom (because they're held together by electrostatic forces), which changes the size of the Earth, which changes the definition of the meter in exactly the same way the modern definition would.

The new SI didn't change $c$ but kept it as it was defined since 1983.

The logic of the SI is driven by both the best theoretical knowledge about the natural laws we have today and the practical realization of the units as accurate as possible. That's why all starts with the definition of the second as a unit of time. It uses still the Cs hyperfine transition frequency as done since 1967:

The second, symbol s, is the SI unit of time. It is defined by taking the fixed numerical value of the caesium frequency $\Delta \nu_{\text{Cs}}$, the unperturbed ground-state hyperfine transition frequency of the caesium-133 atom, to be 9192631770 when expressed in the unit Hz, which is equal to s−1.

Then one uses the fact that to the best of our knowledge spacetime has to be described relativistically, and thus there is a universal limiting speed (having a priori nothing to do with the speed of light but it's always called the speed of light, because to the best of our empirical knowledge the em. field is massless, as discussed above) to define the metre as the unit of length:

The metre, symbol m, is the SI unit of length. It is defined by taking the fixed numerical value of the speed of light in vacuum c to be 299792458 when expressed in the unit $\text{m} \cdot \text{s}^{-1}$, where the second is defined in terms of the caesium frequency $\Delta \nu_{\text{Cs}}$.

This is all as before. The great achievement of the 2019 redefinition of the SI units however is that the fundamental natural constants (except Newtons gravitational constant $G$) have been given defined values once and for all. Thus the kg as the unit of mass is now pretty abstractly defined as

The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be $6.62607015 \cdot 10^{-34}$ when expressed in the unit J⋅s, which is equal to $\textbf{kg} \cdot m^2 \cdot s^{-1}$, where the metre and the second are defined in terms of $c$ and $\Delta \nu_{\text{Cs}}$.

The electromagnetic unit, the SI introduces for practical convenience (not so much for the convenience of theoretical physicists and students of electromagnetism though ;-)), an extra unit for electric charge or rather, for historical reasons, electrical current. It defines the elementary charge (the charge of a proton or the negative charge of an electron) to have a certain numerical value:

The ampere, symbol A, is the SI unit of electric current. It is defined by taking the fixed numerical value of the elementary charge e to be $1.602176634 \cdot 10^{−19}$ when expressed in the unit C, which is equal to A⋅s, where the second is defined in terms of $\Delta \nu_{\text{Cs}}$.

Now the pretty artificial constants $\mu_0$ and $\epsilon_0$ are to be empirically determined, and this brings the greatest deviations between the old and new definitions of the units due to the uncertainty inherited from the unertainty in the measurement of the finestructure constant $\alpha=e^2/(4 \pi \epsilon_0 \hbar c)$. While $e$, $\hbar=h/(2 \pi)$ and $c$ are by definition exact, the vacuum permittivity $\epsilon_0$ has to be measured, and the most accurate way is to measure $\alpha$, and this has an relative uncertainty (CODATA 2018) of $1.5 \cdot 10^{-10}$. Correspondingly also the vacuum permeability has an uncertainty of this order too, because it is related with $\epsilon_0$ due to $\mu_0 \epsilon_0=1/c^2$.

 

[PeterDonis]

The reason I took that approach this time is because I was getting shot down when asking the questions directly.

You were? Where? Not in this thread.

 

[PeterDonis]

I don't see why it needs to be consistent for a hypothetical situation

If the scenario you propose isn't consistent, how can anyone possibly reason about it? It would be like asking, "if 2 plus 2 were 5, what would the universe look like?" How is anyone supposed to answer that?

I don’t see how I could have asked the question without using a hypothetical situation

The point isn't to never propose a hypothetical situation; the point is that if you do, it has to be consistent so that people can reason about it. You already know that there are two consistent models--finite invariant speed (relativity) and infinite invariant speed (Newtonian physics)--so you could just ask, hypothetically, if our universe were described by Newtonian physics instead of relativity, would there be time dilation? (That question has already been answered in this thread anyway--the answer is no.)

Or, if you aren't sure about, say, whether the speed of light can be consistently modeled as different from the invariant speed, you could ask if such a model exists. (And then you would have gotten the answer that @Ibix gave you in post #2.) And then, if such a model exists, you could ask if there is time dilation in that model, and what it depends on. (The answer to that is yes, and it depends on the spacetime structure implied by the finite invariant speed, not the speed of light.)

Maybe I shouldn’t have used relative speeds, let’s convert them to m/s to be clear about what I meant.

Relative speeds are in m/s (or whatever unit of speed you are using). So I don't understand why you would need to "convert" them.

In the last two cases (where you said they would be the same), what I meant was ~300km/s when I said c and ~600km/s when I said 2c.

In these two models, you are proposing that the speed of light is different, not the invariant speed, correct?

 

[Jsauce]

The way I reason about it is with a thought experiment. All observers will measure c to be the same, no matter their velocity relative to the inertial of the universe. So, if you are in a ship moving at 0.5 c and you measure c you will get the same result as a “stationary” ship, but the photons can’t be translating at 1.5c just because you are observing them, this is where I imagine the time dilation adjustment comes in. The photon translates the same “distance” in the universe frame in the same amount of universal frame “time”. So instead of the photon having an absolute speed of 1.5c it is actually still c in the universe frame.

 

[PeroK]

The way I reason about it is with a thought experiment. All observers will measure c to be the same, no matter their velocity relative to the inertial of the universe. So, if you are in a ship moving at 0.5 c and you measure c you will get the same result as a “stationary” ship, but the photons can’t be translating at 1.5c just because you are observing them, this is where I imagine the time dilation adjustment comes in. The photon translates the same “distance” in the universe frame in the same amount of universal frame “time”. So instead of the photon having an absolute speed of 1.5c it is actually still c in the universe frame.

This is all wrong. There is no "universal" frame of reference.

 

[anuttarasammyak]

the photons can’t be translating at 1.5c just because you are observing them, this is where I imagine the time dilation adjustment comes in.

Not only time dilation but also Lorentz contraction of space length take place. Say light starts and goals with time and space difference of T and X
$$\frac{X}{T}=c$$
It holds for observation in any IFR say
$$\frac{X_n}{T_n}=c$$
where An means A observed value in IFR#n with any numbering of IFRs. We have both no ways and no needs to say which IFR is "the universal frame".

 

[Jsauce]

This is all wrong. There is no "universal" frame of reference.

Sorry for my imprecise language, but I was referring to asymptotically flat space time.

 

[jbriggs444]

Not only time dilation but also Lorentz contraction of space length take place. Say light starts and goals with time and space difference of T and X
$$\frac{X}{T}=c$$
It holds for observation in any IFR say
$$\frac{X_n}{T_n}=c$$
where An means A observed value in IFR#n with any numbering of IFRs.

This is correct but, perhaps, misleading. Just as you say, there is time dilation and length contraction. If we blindly apply the formula for length contraction, we might say "our measured displacement was $X$. His measured displacement must be greater": $$X_n = \frac{X}{\sqrt{1-\frac{v^2}{c^2}}}$$If we blindly apply the formula for time dilation, we might say "our measured duration was T. So his measured duration must be less": $$T_n = T \sqrt{1-\frac{v^2}{c^2}}$$
The above seemingly correct but actually wrong-headed calculation would lead us to believe that$$c_n = \frac{X_n}{T_n} = \frac{c}{1-\frac{v^2}{c^2}}$$
A correct way of doing the calculation would be to do the full Lorentz transformation on the starting and ending events and thereby include the relativity of simultaneity.

We have both no ways and no needs to say which IFR is "the universal frame".

Agreed.

 

[Mister T]

The way I reason about it is with a thought experiment. All observers will measure c to be the same, no matter their velocity relative to the inertial of the universe.

But there is no such thing as velocity relative to the inertial of the universe. Besides not making sense, it seems you are clinging to the notion the universe somehow has a rest frame and that all velocities can be measured relative to it. That's equivalent to the notion of velocity being absolute.

Syo, if you are in a ship moving at 0.5 c and you measure c you will get the same result as a “stationary” ship, but the photons can’t be translating at 1.5c just because you are observing them, this is where I imagine the time dilation adjustment comes in.

No need to imagine any such thing. You are simply hung up on the Newtonian notion of adding speeds to combine them. Adding them is not the correct way. When you correctly combine c and 0.5c you get c, not 1.5 c.