Time dilation from length contraction

[intervoxel]

Suppose there exists an absolute frame.
The observer is at rest.
an object moves with velocity v.
Suppose it contracts as expected.

Does this imply automatically that time dilation occurred?

 

[Naty1]

Who was it who said 'time and space will forever cease to be independent entities'...or something to that effect. Minkowski??

Anyway, the Lorentz transform works equally explaining 'ether' philosophy or special relativity.

 

[AdrianMay]

You got the "suppose" and "imply" in the wrong place. It actually goes like this:

Suppose the speed of light is the same for all inertial observers. That implies that time dilation and lorentz contraction must happen.

 

[harrylin]

Suppose there exists an absolute frame.
The observer is at rest.
an object moves with velocity v.
Suppose it contracts as expected.

Does this imply automatically that time dilation occurred?

Time dilation can only "occur" if there is a physical process going on, but your description has none. So let's say that your object is a clock. Then its clock rate will reduce as expected.

 

[intervoxel]

Time dilation can only "occur" if there is a physical process going on, but your description has none. So let's say that your object is a clock. Then its clock rate will reduce as expected.

Bingo! It means that, in a euclidian space, if I implement a mechanism that contracts objects in the direction of motion, time dilation comes for free. Is this correct?

 

[Dale]

No. You need some other postulates to get time dilation.

If you doubt this, then please try to mathematically derive time dilation from length contraction without any other assumptions.

 

[bcrowell]

No. You need some other postulates to get time dilation.

If you doubt this, then please try to mathematically derive time dilation from length contraction without any other assumptions.

OK, but the other postulates required are pretty mild: (1) homogeneity of spacetime, (2) isotropy of space, (3) relativity of motion, and (4) causality. See http://www.lightandmatter.com/html_books/0sn/ch07/ch07.html#Section7.2 . Length contraction eliminates case I in subsection 7.2.1, and causality kills off case II. All that's left is case III, which has time dilation.

 

[Dale]

OK, but the other postulates required are pretty mild

Agreed, they are mild postulates, but they are there.

I am just not sure what intervoxel is planning, and I have seen people try to make really silly claims by thinking that "all I need is X and Y comes automatically".

 

[intervoxel]

No. You need some other postulates to get time dilation.

If you doubt this, then please try to mathematically derive time dilation from length contraction without any other assumptions.

Suppose we have homogeneity and isotropy in this euclidian space.
The starship bears two mirror clocks and leaves the station where
the observer there remains. One of the clocks is oriented along the motion
path, the other, perpendicular to the first. When the ship is at rest, the
period of both clocks is $T=\frac{2d}{c}$, where $d$ is the distance between
the mirrors of both clocks. When the ship is moving with velocity $v$, the
expected period due to time dilation as seem by the observer at the station
will be

$T^{\prime }=\gamma T=\frac{2d}{c}\gamma$

$T^{\prime }=\frac{2d}{\sqrt{c^{2}-v^{2}}} \label{factor1}$

Let's check this result considering the two clocks separately. The
perpendicular one does not suffer length contraction but, seem from the
reference frame the photon makes a zig-zag path. So we have a right triangle

$d^{2}+x^{2}=c^{2}t^{2}\text{,}$

where $x$ is the distance along the motion path. Therefore

$x=\sqrt{c^{2}t^{2}-d^{2}}=vt\text{.}$

We can now calculate the time to the photon hit the other mirror

$t =\frac{d}{\sqrt{c^{2}-v^{2}}}$

$T =2t$

$T=\frac{2d}{\sqrt{c^{2}-v^{2}}} \label{factor2}$

The other clock is contracted ($d^{\prime }=d\,/\,\gamma$). The
time the photon leaving the left mirror (supposing motion to the right)
takes to hit the right mirror is $t_{1}=\frac{d}{\gamma (c-v)}$. The time to
this photon return to the left mirror is $t_{2}=\frac{d}{\gamma (c+v)}$. So
we have

$T =t_{1}+t_{2}$

$T =\frac{2cd}{\gamma (c^{2}-v^{2)}}$

$T=\frac{2d}{\sqrt{c^{2}-v^{2}}}\text{.} \label{factor3}$

The three expressions are similar, then we can conclude that, in this approach, time dilation
comes for free when we implement length contraction.

What about an onboard observer? Well, we can imagine the observer's body as
a digital computer controlled by a clock. This clock is also affected by the
motion, like the two other clocks. So in his point of view nothing changed,
no length contraction nor time dilation ever occurred in his own frame.

Where is the error after all?

 

[harrylin]

Bingo! It means that, in a euclidian space, if I implement a mechanism that contracts objects in the direction of motion, time dilation comes for free. Is this correct?

No, I interpreted "as expected" to mean, as expected by Lorentz - what we expect is Lorentz contraction. What other expectation could you have meant? A mechanism that contracts objects in motion by your implementation has nothing to do with that.

 

[harrylin]

Suppose we have homogeneity and isotropy in this euclidian space.
The starship bears two mirror clocks and leaves the station where
the observer there remains. One of the clocks is oriented along the motion
path, the other, perpendicular to the first. When the ship is at rest, the
period of both clocks is $T=\frac{2d}{c}$, where $d$ is the distance between
the mirrors of both clocks. When the ship is moving with velocity $v$, the
expected period due to time dilation as seem by the observer at the station
will be

$T^{\prime }=\gamma T=\frac{2d}{c}\gamma$

$T^{\prime }=\frac{2d}{\sqrt{c^{2}-v^{2}}} \label{factor1}$

Let's check this result considering the two clocks separately. The
perpendicular one does not suffer length contraction but, seem from the
reference frame the photon makes a zig-zag path. So we have a right triangle

$d^{2}+x^{2}=c^{2}t^{2}\text{,}$

where $x$ is the distance along the motion path. Therefore

$x=\sqrt{c^{2}t^{2}-d^{2}}=vt\text{.}$

We can now calculate the time to the photon hit the other mirror

$t =\frac{d}{\sqrt{c^{2}-v^{2}}}$

$T =2t$

$T=\frac{2d}{\sqrt{c^{2}-v^{2}}} \label{factor2}$

The other clock is contracted ($d^{\prime }=d\,/\,\gamma$). The
time the photon leaving the left mirror (supposing motion to the right)
takes to hit the right mirror is $t_{1}=\frac{d}{\gamma (c-v)}$. The time to
this photon return to the left mirror is $t_{2}=\frac{d}{\gamma (c+v)}$. So
we have

$T =t_{1}+t_{2}$

$T =\frac{2cd}{\gamma (c^{2}-v^{2)}}$

$T=\frac{2d}{\sqrt{c^{2}-v^{2}}}\text{.} \label{factor3}$

The three expressions are similar, then we can conclude that, in this approach, time dilation
comes for free when we implement length contraction.

What about an onboard observer? Well, we can imagine the observer's body as
a digital computer controlled by a clock. This clock is also affected by the
motion, like the two other clocks. So in his point of view nothing changed,
no length contraction nor time dilation ever occurred in his own frame.

Where is the error after all?

It's a good start, but too limited in scope. For example, what does that predict for a C14 clock or muons that reach the Earth's atmosphere?

 

[AdrianMay]

You've implictly used the universality of c, from which you can derive both TD and LC.

I find it instructive to use your two clocks in the two orientations, require that they TD by the same degree, and prove that LC is necessary for that.

 

[John232]

You can get the time dilation equation just by setting up two similar triangles that are the path of a photon, each triangle being another frame of reference with their own independant values for distance and time. The sides adjacent to the right angle would be the distance the object traveled and the distance light travels at rest to an observer. The side opposite the right angle (hypotenus) is the distance light is seen travel from a moveing frame of reference. It turns out that you end up getting a smaller value for time on the side of the triangle that the observer at rest uses for their own time. So then the object moveing time slows down since they measure their photons to move a shorter distance than an observer moveing at a different velocity.

If a distance contracted and the speed of light stayed the same then time dilation would have to occor in order for them to measure the same speed for light, since velocity equals distance divided by time.

I think it is wrong to say time slows down because the photon is seen to travel a longer distance since that line would have a higher value for time. But it happens a lot, and it kind of really makes you scratch your head on to what varieable should be which that each observer uses.

 

[Dale]

Suppose we have homogeneity and isotropy in this euclidian space.

These are some additional assumptions.

What about an onboard observer? Well, we can imagine the observer's body as
a digital computer controlled by a clock. This clock is also affected by the
motion, like the two other clocks. So in his point of view nothing changed,
no length contraction nor time dilation ever occurred in his own frame.

You need also the principle of relativity to make this conclusion. Otherwise you could assume that the light clock is seen to slow down while other clocks do not slow down.

I think that length contraction, homogeneity, isotropy, and the principle of relativity are sufficient to derive time dilation.

 

[intervoxel]

Thank you all for the enlightening discussion.