I Time difference caused by length contraction

[Dale]

But in LET what is the rocket's frame? if the front and the back are not synchronized, how can we speak of a reference frame?

Reference frames still exist in any theory. There are two common meanings for reference frames:

First, reference frames are coordinate systems used to assign position and time numbers to events. That is part of the mathematical analysis. LET and BU use the exact same math for the coordinate transforms.

Second, reference frames are a system of physical clocks and rulers used to assign position and time numbers to events. That is part of the experimental setup. This is an experimental fact independent of any theory.

In both cases reference frames are every bit a part of LET as BU.

Again, the fact that you think they are not part of LET shows a misunderstanding of LET. Note, this comment is not said as someone promoting a stance on the BU vs LET debate (they are both valid interpretations), but as a professional physics educator. You misunderstand the math of how LET works. It does not work the way you claim. You have a vague grasp of the philosophy, but not the concrete math. So please take some instruction on the topic.

Question: Where does the velocities and accelerations difference between the rear and the front come from?

Here is an argument: The contraction is the cause that the rear and the front do not have the same proper acceleration, and the proper acceleration is felt in the reference frame of the rocket.

This is a fallacious argument. Just because two points on a rigid object have different velocities and accelerations does not imply that it is due to length contraction. In Newtonian physics two objects at rest in a rotating reference frame will have different proper accelerations and in an inertial frame will have different velocities. That doesn’t imply length contraction caused it. Relativistic acceleration is a hyperbolic rotation, so it shares this kinematical fact with Newtonian rotational motion.

To show that length contraction is the cause will require an actual derivation, your simple assertions and rhetorical questions are insufficient.

 

[PeterDonis]

You are violating forum rules here with this statement. You are attempting to argue the superiority or veracity of BU.

I am doing no such thing. And at this point your question has been thoroughly answered, and you have failed to respond to any of the valid answers you have been given, but continue to either repeat the same incorrect statements, or raise irrelevant issues. Enough is enough.

This thread is closed.

 

[Dale]

Sorry @PeterDonis I was working on the math before the closure (actually since yesterday once @externo finally specified what they were looking for) and so I wanted to post it anyway. It just took some time to do.

In Rindler (accelerating) coordinates the metric is $$ds^2 = -c^2 d\tau^2 =-\frac{a^2 x^2}{c^2} dt^2 + dx^2 + dy^2 + dz^2$$ where an observer at $x_0=c^2/a$ experiences a proper acceleration of $a$ and has undilated time. Time dilation is given by $\frac{1}{\gamma} = \frac{d\tau}{dt}$, so for a clock at rest in the Rindler frame $dx=dy=dz=0$ and we have $$\frac{1}{\gamma}=\frac{d\tau}{dt}=\sqrt{\frac{a^2 x^2}{c^4} - \frac{dx^2+dy^2+dz^2}{c^2 dt^2}}= \frac{a x}{c^2}$$ For a clock at $x=x_0+h$ compared to a clock at $x=x_0$ this gives a time dilation ratio of $$ 1+\frac{ah}{c^2}$$ where $ah$ is the pseudo-gravitational potential in the Rindler frame. Note that $h=0$ corresponds to the reference clock which is undilated and has $1/\gamma=1$. So $h$ functions as a parameter that identifies specific members of the Rindler congruence, e.g. a clock at $h=0$ or a clock at $h=5 \mathrm{\ m}$.

Note that this is not a first-order approximation, but is the actual time dilation in the accelerating Rindler frame. By the equivalence principle this is the relationship that holds locally in a real gravitational field over a local region sufficiently small that tidal gravity effects are negligible.

Now, we will transform from the accelerating Rindler frame to an inertial Minkowski frame. If $t=0$ is the moment that the rocket is momentarily at rest in the aether frame, then this transforms to the aether frame, so this is the theoretical LET derivation even if we find it difficult to determine the correct $t$ as a practical matter.

In the inertial frame the worldline of the clock at $h$ is given by $$(T,X,Y,Z)=\left(\left(\frac{c}{a}+\frac{h}{c}\right) \sinh\left( \frac{at}{c} \right),\left(\frac{c^2}{a}+h\right) \cosh\left( \frac{at}{c} \right),0,0\right)$$$$ = \left(T, \left( \frac{c^2}{a}+h \right) \sqrt{1+\frac{a^2 c^2 T^2}{\left( c^2 + a h\right)^2}} ,0,0 \right)$$ where $T$ is time in the inertial frame and $t$ is time in the Rindler frame.

The metric in the inertial frame is $ds^2=-c^2 dT^2 + dX^2 + dY^2 + dZ^2$. Calculating the time dilation for this case (here $dX\ne 0$) we get $$\frac{1}{\gamma}=\sqrt{\frac{(c^2+ah)^2}{(c^2+ah)^2+a^2 c^2 T^2}}$$
Now, we have the time dilation in the inertial frame at any time $T$ for the Rindler clock at $h$. This time dilation, as expected, depends on both $T$ and on $h$. As $T$ increases the clock gets slower, regardless of $h$. So, we can take the ratio of the time dilation for a clock at $h=0$ and a clock at some other $h$. When we do that we get $$\left(c+\frac{ah}{c}\right)\sqrt{\frac{c^2+a^2 T^2}{(c^2+ah)^2+a^2 c^2 T^2}}$$
Now, the issue is that this is quantitatively the wrong value. Importantly, at $T=0$ there is no time dilation with this approach and furthermore the "time dilation" is a function of time, which is not correct for gravitational time dilation. In fact, this value is not even approximately correct. If we do just a first order approximation in $h$ we get $$1+\frac{a^3}{c^4+a^2 c^2 T^2}T^2 h + O(h^2)$$ which does not match the correct expression even to first order. Thus this expression is conclusively ruled out by experiment. The primary reason this is wrong is the one identified and explained to you by @Ibix