I Are the Lorentz Transformations False?

[Leepappas]

TL;DR Summary

SR just doesn't make sense.

Consider the Lorentz transformations with c=1, and consider any point in space whose x coordinate isn't zero, starting from

$t_{inital }= t'_{inital }=0$

$t' =\gamma (t-xv)$
$t= \frac { t'}{\gamma} + xv$
$\Delta t' = t'-0$
$\Delta t = t-0$
Time dilation provides
$\Delta t' =\gamma \Delta t$
So
$\Delta t = \frac {\Delta t'}{\gamma} + xv$
Or
$\Delta t = \Delta t + xv$
Whick implies x=0 or v=0.
Since x isn't zero, v=0.
But the rulers are in relative motion therefore not(v=0).
Therefore v=0 and not (v=0).
This contradiction was arrived at by assuming the Lorentz transformations are true, therefore they are false.

 

[Orodruin]

Time dilation provides
#\Delta t'=\gamma \Delta t#

No, it doesn’t. Not in general. You need to work on your understanding of the requirements that apply for time dilation and length contraction formulas to apply.

Edit: Working on correcting your use of math delimiters would also be desirable.

Edit 2: To put things in context: Here

consider any point in spacetime whose x coordinate isn't zero.

you are essentially stating “pick any event such that the time dilation formula I am going to use does not apply”.

 

[PeroK]

TL;DR Summary: SR just doesn't make sense.

Consider the Lorentz transformations with c=1, and consider any point in spacetime whose x coordinate isn't zero.

$t' =\gamma (t-xv)$
Time dilation provides
$\Delta t'=\gamma \Delta t$

Clearly, those two statements cannot both be true.

If you assume the Lorentz transformation, can you identify the constraint for time dilation to apply?

 

[Ibix]

$\Delta t' = t'-0$
$\Delta t = t-0$

I assume that $\Delta t$ and $\Delta t'$ are meant to refer to the intervals measured between some pair of events using different frames. But note that you've specified that the start time in both frames is zero. The only event that has $t=t'=0$ is the origin, where also $x=x'=0$. So you are here implicitly assuming an x coordinate of zero, in contradiction to the preceding paragraph.

That's where your problems start.

 

[Leepappas]

I assume that $\Delta t$ and $\Delta t'$ are meant to refer to the intervals measured between some pair of events using different frames. But note that you've specified that the start time in both frames is zero. The only event that has $t=t'=0$ is the origin, where also $x=x'=0$. So you are here implicitly assuming an x coordinate of zero, in contradiction to the preceding paragraph.

That's where your problems start.

Yes they are measured with respect to a pair of events. The first event has SpaceTime coordinate
$(t,x)=(0,0)$

The second event has space-time coordinate
$(t,x)$.

 

[Ibix]

The second event has space-time coordinate
$(t,x)$.

Then either $x=0$ here too, or else the clock is not at rest in the unprimed frame and the time dilation formula you quoted does not apply.

 

[Orodruin]

I assume that $\Delta t$ and $\Delta t'$ are meant to refer to the intervals measured between some pair of events. But note that you've specified that the start time in both frames is zero. The only event that has $t=t'=0$ is the origin, where also $x=x'=0$. So you are here implicitly assuming an x coordinate of zero, in contradiction to the preceding paragraph.

That's where your problems start.

This is not a problem. The assumed $x\neq 0$ is for the other event, that which OP is Lorentz transforming. However, the problem is that $x=0$ is necessary for the quoted time dilation formula to apply.

 

[Orodruin]

Then either $x=0$ here too, or else the clock is not at rest in the unprimed frame and the time dilation formula you quoted does not apply.

Indeed

you are essentially stating “pick any event such that the time dilation formula I am going to use does not apply”.

 

[Leepappas]

This is not a problem. The assumed $x\neq 0$ is for the other event, that which OP is Lorentz transforming. However, the problem is that $x=0$ is necessary for the quoted time dilation formula to apply.

The time dilation formula is a general statement about how amounts of time measured by clocks at rest in one inertial frame disagree with clocks at rest in another inertial reference frame moving relatively to the first frame. It says nothing about where those clocks are located

 

[Ibix]

The time dilation formula is a general statement about how amounts of time measured by clocks at rest in one inertial frame disagree with clocks at rest in another inertial reference frame moving relatively to the first frame. It says nothing about where those clocks are located

But you said the clock was at $x=0$ at first and $x\neq 0$ later, so its $x$ position is changing and it's not at rest!

If you want to use $x=0$ as your start position then you must either choose $x=0$ as the end position or your clock is not at rest and you cannot use the time dilation formula you are using. You cannot have it both ways.

 

[Leepappas]

But you said it was at $x=0$ at first and $x\neq 0$ later, so it's not at rest!

If you want to use $x=0$ as your start position then you must either choose $x=0$ as the end position or your clock is not at rest and you cannot use the time dilation formula you are using. You cannot have it both ways.

The clock that makes the measurement is stationary in the unprimed frame.

 

[Ibix]

The clock that makes the measurement is stationary in the frame.

Then its $x$ coordinate does not change and $x=0$ at both events.

 

[Leepappas]

Then its $x$ coordinate does not change and $x=0$ at both events.

Let its x coordinate be at x=a, where $a\neq 0$

Let the clock at rest at $x=a$ be synchronized with a clock at rest at the origin of the unprimed frame. Clearly the x coordinate is not zero at both events.

 

[PeroK]

Let its x coordinate be at x=a, where $x\neq a$

Let the clock at rest at $x=a$ be synchronized with a clock at rest at the origin of the unprimed frame. Clearly the x coordinate is not zero at both events.

You're getting nowhere with this approach. The time dilation formula applies to the time shown on a given clock. It does not apply to coordinates generally.

The time dilation formula is a special case of the Lorentz transformation for a single clock: I.e. at a constant spatial location at one of your frames.

 

[Leepappas]

You're getting nowhere with this approach. The time dilation formula applies to the time shown on a given clock. It does not apply to coordinates generally.

The time dilation formula is a special case of the Lorentz transformation for a single clock: I.e. at a constant spatial location.

The clock located at $x=a$ does have a constant spatial location in the unprimed frame.

 

[Ibix]

Clearly the x coordinate is not zero at both events.

Sure, but the clocks aren't synchronised in the primed frame, and blindly applying the time dilation formula doesn't acknowledge that.

Look. The Lorentz transform for time is $t'=\gamma(t-vx)$. If there is an event at $(x,t)=(x_0,t_0)$ and another event at $(x,t)=(x_0+\Delta x, t_0+\Delta t)$ then the primed frame sees time coordinates $t'_0=\gamma(t_0-vx_0)$ and $t'_0+\Delta t'=\gamma(t_0+\Delta t-v(x_0+\Delta x))$ for the same events. Subtracting them gives you the time interval $\Delta t'=\gamma(\Delta t-v\Delta x)$. This only simplifies to the time dilation formula you used in the case that $\Delta x=0$.

You keep using the $\Delta x=0$ form while insisting $\Delta x\neq 0$.

The clock located at $x=a$ does have a constant spatial location in the unprimed frame.

And if you plug $x=a$ in as the location of the first event and the correct corresponding value of $t'$ then you will find no contradiction.

What you cannot do is use one reading on one clock and another reading on another clock at another location and then use a formula that expects the clocks to be at the same location.

 

[PeroK]

The clock located at $x=a$ does have a constant spatial location in the unprimed frame.

So, we have two events $x_1 = a$ at time $t_1$ and $x_2=a$ at time $t_2$. Using Lorentz:
$$t'_1=\gamma(t_1 - va), \ t'_2=\gamma(t_2 -va)$$Hence$$\Delta t' = t'_2-t'_1 = \gamma(t_2-t_1) = \gamma \Delta t$$

 

[Orodruin]

Let its x coordinate be at x=a, where $a\neq 0$

Let the clock at rest at $x=a$ be synchronized with a clock at rest at the origin of the unprimed frame.

Then it is not synchronized with the clock at the unprimed origin in the primed frame and hence the time dilation formula does not apply to your events. This is relativity of simultaneity.

 

[PeterDonis]

The time dilation formula is a general statement about how amounts of time measured by clocks at rest in one inertial frame disagree with clocks at rest in another inertial reference frame moving relatively to the first frame. It says nothing about where those clocks are located

This is all wrong. The only correct general formula is the Lorentz transformation. The time dilation formula is only applicable to special cases, and the scenario you are trying to analyze is not one of them. That is what other posters are trying to tell you. Please take heed.

 

[Dale]

TL;DR Summary: SR just doesn't make sense.

Time dilation provides

Time dilation is already in the Lorentz transform. If you are already using the Lorentz transform then it doesn’t make sense to also separately use the time dilation formula.

The Lorentz transform is the general equation that always applies to any inertial frame. The time dilation formula is a simplification that only applies in specific cases. Here you have used it incorrectly

 

[Ibix]

Here are some Minkowski diagrams to explain what's going on. All are drawn using the unprimed frame.

The first diagram shows a green line representing a clock at rest in this frame and located at $x=0$. It starts ticking at $t=0$ and stops at $t=\Delta t$ and only this portion of its worldline is marked. There are also two fine blue lines showing the lines of simultaneity in the primed frame, and a thicker blue line showing a clock measuring the elapsed time in the primed frame.

If you were to compute the interval (the "length") of the blue line you would find that it was $\gamma$ times greater than the interval along the green line. This is time dilation.

Time dilation, indeed, does not care where you place the clocks. In this second diagram, I've added a second clock at $x=a$, again ticking for the same interval in this frame. I've faded out the simultaneity lines from the first diagram and put in new simultaneity lines and a new blue clock. You can see (or calculate!) that it's the same length as the blue line in the first diagram - so again, this is time dilation.

So it doesn't matter whether you use a clock at $x=0$ or a clock at $x=a$, nor where you place the blue clock that's at rest in the primed frame. The problems start when you try to use the left clock for the start time and the right clock for the end time. In the third diagram I've shown the simultaneity lines in the primed frame associated with those events and added another blue clock showing a much shorter time.

This is why your calculation fails, @Leepappas. It's because, when you apply the time dilation formula the way you are doing, you are assuming that the blue clock line in this diagram is the same length as the ones in the first two diagrams. It manifestly isn't due, as Orodruin says, to the relativity of simultaneity.

Just for completeness, here's the diagram if you use the right clock for the start event and the left clock for the end - this time the blue clock line is so long it doesn't fit on my scale.

 

[Bosko]

TL;DR Summary: SR just doesn't make sense.

Let's derive the Lorentz's transformation . The speed $v$ and the Lorentz factor $\gamma=(\sqrt{1-v^2})^{-1}$ are constant.

We need to express the coordinates of the same point in two coordinate systems.
For me, the most convenient way, is to use scalar product with a normal vector in order to calculate a distance from a coordinate axis.

[1] If we know $(x,t)$ and we want to calculate $(x',t')$

$$
\begin{align}
&x'=\gamma\left<(1,-v),(x,t)\right>=\gamma(x-vt)\nonumber\\
&t'=\gamma\left<(-v,1),(x,t)\right>=\gamma(-vx+t)\nonumber\\
\end{align}
$$

Also ...
$$
\begin{align}
\Delta x'&=x'_1-x'_2\nonumber\\
&=\gamma(x_1-vt)-\gamma(x_2-vt)\nonumber\\
&=\gamma(x_1-\cancel{vt}-x_2+\cancel{vt})\nonumber\\
&=\gamma \Delta x \nonumber\\
\end{align}
$$
and
$$
\begin{align}
\Delta t'&=t'_1-t'_2\nonumber\\
&=\gamma(-vx+t_1)-\gamma(-vx+t_2)\nonumber\\
&=\gamma(-\cancel{vx}+t_1+\cancel{vx}-t_2)\nonumber\\
&=\gamma \Delta t \nonumber\\
\end{align}
$$

[2] If we know coordinates in the prime reference frame $(x',t')$ and we want to calculate $(x,t)$

$$
\begin{align}
&x=\gamma\left<(1,v),(x',t')\right>=\gamma(x'+vt)\nonumber\\
&t=\gamma\left<(v,1),(x',t')\right>=\gamma(vx'+t')\nonumber\\
\end{align}
$$

Also, similarly as in [1] ...
$$
\begin{align}
\Delta x&=x_1-x_2\nonumber\\
&=\gamma(x'_1+vt)-\gamma(x'_2+vt)\nonumber\\
&=\gamma(x'_1+\cancel{vt}-x'_2-\cancel{vt})\nonumber\\
&=\gamma \Delta x' \nonumber\\
\end{align}
$$
and
$$
\begin{align}
\Delta t&=t_1-t_2\nonumber\\
&=\gamma(vx+t'_1)-\gamma(vx+t'_2)\nonumber\\
&=\gamma(\cancel{vx}+t'_1-\cancel{vx}-t'_2)\nonumber\\
&=\gamma \Delta t' \nonumber\\
\end{align}
$$

 

[Leepappas]

Sure, but the clocks aren't synchronised in the primed frame, and blindly applying the time dilation formula doesn't acknowledge that.

Look. The Lorentz transform for time is $t'=\gamma(t-vx)$. If there is an event at $(x,t)=(x_0,t_0)$ and another event at $(x,t)=(x_0+\Delta x, t_0+\Delta t)$ then the primed frame sees time coordinates $t'_0=\gamma(t_0-vx_0)$ and $t'_0+\Delta t'=\gamma(t_0+\Delta t-v(x_0+\Delta x))$ for the same events. Subtracting them gives you the time interval $\Delta t'=\gamma(\Delta t-v\Delta x)$. This only simplifies to the time dilation formula you used in the case that $\Delta x=0$.

You keep using the $\Delta x=0$ form while insisting $\Delta x\neq 0$.

And if you plug $x=a$ in as the location of the first event and the correct corresponding value of $t'$ then you will find no contradiction.

What you cannot do is use one reading on one clock and another reading on another clock at another location and then use a formula that expects the clocks to be at the same location.

Event 1 at (0,0).
Event 2 at (a,t).

The primed frame sees time coordinates $t'_0=0$ and $t'_0+\Delta t'=\gamma(t_0+\Delta t-v(x_0+\Delta x))$ Subtracting them gives you the time interval $\Delta t'=\gamma(\Delta t-v\Delta x)$. I completely see this is not the simple time dilation formula.

Something about this is counterintuitive. Why in the world should the amount of passage of time in the primed frame depend upon $where$ a clock is in the unprimed frame? To the contrary, it should be independent of the location of any clock at rest in the unprimed frame. We can change the amount of time in the primed frame simply by changing position in the unprimed frame, that's ludicrous.

 

[PeterDonis]

Subtracting them gives you the time interval $\Delta t'=\gamma(\Delta t-v\Delta x)$.

Yes. And that's the Lorentz transformation formula. It is not the time dilation formula. You have just demonstrated why your logic in the OP of this thread is wrong.

 

[Sagittarius A-Star]

$t_{inital }= t'_{inital }=0$
$t' =\gamma (t-xv)$

From those two equations follows, even without the time-dilation formula:
$0=t'_{inital } =\gamma (t_{inital }-x_{inital }v)=\gamma (0 -x_{inital }v)$
$\Rightarrow$
$x_{inital }v=0$.

With

But the rulers are in relative motion therefore not(v=0).

$\Rightarrow$
$x_{inital }=0$.

Then with ...

The clock that makes the measurement is stationary in the unprimed frame.

... follows, that you assume, that the clock rests at
$x=x_{inital }=0$.

 

[Vanadium 50]

This looks a lot like your locked thread.

@Leepappas what's your goal here? Is it to convince us that you have found a contradiction in SR, like you tried in your old thread? Sorry...you won't. It is known to be internally consistent and it has been shown to match observation.

If it's to better understand SR, I would start by going back to that thread and re-read those responses so we can start from there. Why should people type essentially the same answers to essentially the same questions if they did so before?

 

[Leepappas]

This looks a lot like your locked thread.

@Leepappas what's your goal here? Is it to convince us that you have found a contradiction in SR, like you tried in your old thread? Sorry...you won't. It is known to be internally consistent and it has been shown to match observation.

If it's to better understand SR, I would start by going back to that thread and re-read those responses so we can start from there. Why should people type essentially the same answers to essentially the same questions if they did so before?

Since that old thread I have familiarized myself with the relativity of simultaneity and Minkowski world lines. This is a new thread with a new idea. I'm starting to suspect that it is internally consistent as you say, but I have my intuition which tells me it contradicts reality. Can you explain to me why on Earth the amount of passage of time in the primed frame should depend upon where in space a clock is located in the unprimed frame?

 

[Dale]

I have my intuition which tells me it contradicts reality

This is a very common feeling. Intuition is based on experience and relativistic speeds are outside of our experience. So we simply do not have intuition that is trustworthy in relativity.

Part of the reason that we do careful scientific experiments is precisely so that we can overcome our limitations and find out how nature behaves in regimes we don’t directly experience. Here is a summary of the experimental evidence:

https://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

Our intuition is wrong. Reality is relativistic.

Can you explain to me why on Earth the amount of passage of time in the primed frame should depend upon where in space a clock is located in the unprimed frame?

What? I don’t understand what you are asking. Can you write it with math?

 

[Leepappas]

Lee Pappas said:
Can you explain to me why on Earth the amount of passage of time in the primed frame should depend upon where in space a clock is located in the unprimed frame?

Dale said:
What? I don’t understand what you are asking. Can you write it with math?

Now in response to Dale's question I say:
While the equation $\Delta t'=\gamma \Delta t$ was merely distasteful, the equation

$$\Delta t'= \gamma (\Delta t - v\Delta x)$$

Is totally repugnant.

In the former equation, $\Delta t'$ isn't a function of the spatial coordinates of a point in the unprimed frame. In the latter equation $\Delta t'$ is a function of the coordinates of a point in the unprimed frame. That goes against all common sense.

 

[Vanadium 50]

but I have my intuition which tells me it contradicts reality.

The universe, alas, is not required to conform to your intuition. You're just not that important.

 

[Dale]

While the equation $\Delta t'=\gamma \Delta t$ was merely distasteful, the equation

$$\Delta t'= \gamma (\Delta t - v\Delta x)$$

Is totally repugnant.

In the former equation, $\Delta t'$ isn't a function of the coordinates of a point in the unprimed frame. In the latter equation $\Delta t'$ is a function of the coordinates of a point in the unprimed frame. That goes against all common sense.

Ok, thanks, that is helpful. The term you are objecting to is the relativity of simultaneity. It is indeed the most challenging concept in special relativity and the one that most directly disagrees with student’s intuition and “common sense”. Nevertheless, nature is relativistic.

What this term represents is not that the passage of time depends on location. It is that there is a location-dependent offset to synchronized clocks.

In other words, if you have synchronized two clocks in your frame then in my frame they are desynchronized by an amount given by that term. Both clocks tick at the same rate in my frame, but they are not synchronized in my frame despite being synchronized in yours.

 

[PeterDonis]

In the former equation, $\Delta t'$ isn't a function of the coordinates of a point in the unprimed frame.

Yes, it is. $t$ is a coordinate. Frames in relativity describe spacetime.

That goes against all common sense.

So does a lot of modern physics. If you think relativity goes against your common sense, try quantum mechanics.

However, relativity and QM are confirmed by many, many experiments; in some cases the confirmation goes to as much as 13 decimal places. So your common sense is outvoted. The only way to deal with it is to retrain your common sense; that is probably the hardest part of learning modern physics.

 

[Dale]

The thread is reopened after deleting a brief tangent about QM.

@Leepappas your personal feelings about relativity are not relevant. You can discuss the theory and experiments without editorializing. We understand that this is a difficult concept. And now you have a good summary of the experimental facts that establish it anyway. So let’s focus on the science and not opinions.

 

[Leepappas]

The thread is reopened after deleting a brief tangent about QM.

@Leepappas your personal feelings about relativity are not relevant. You can discuss the theory and experiments without editorializing. We understand that this is a difficult concept. And now you have a good summary of the experimental facts that establish it anyway. So let’s focus on the science and not opinions.

Fair enough.

 

[Leepappas]

If $t' = \gamma(t-vx)$ then space is two dimensional.
Point 1: (0,0)
Point 2: (a,t)
We have established that for these two points in spacetime we have
$\Delta t' = \gamma(\Delta t - v\Delta x)$

Therefore, for these two points
Point 1: (0,0)
Point 2: (a,0)
We have:
$0 = \gamma(0- va)$
Since $v \neq 0$ it follows that $a=0$.
The y and z coordinates are not constrained to be any particular value but the x coordinate is constrained to be equal to zero. Therefore space is two dimensional if the Lorentz transformations are true.

 

[Dale]

for these two points
Point 1: (0,0)
Point 2: (a,0)
We have:
$0 = \gamma(0- va)$

No. By the Lorentz transform we have:
$\Delta t’=\gamma(0-va)$

 

[Bosko]

TL;DR Summary: SR just doesn't make sense.

Consider the Lorentz transformations with c=1, and consider any point in space whose x coordinate isn't zero, starting from

$t_{inital }= t'_{inital }=0$

I will place the origin of the primed coordinate system in that point. So the initial conditions are
$t_0=t'_0=0$
and
$x_0 \neq x'_0=0$

$t' =\gamma (t-xv)$

You need to adapt this formula to the changed initial conditions.

Otherwise, for $t_0=t'_0=0$ it would be
$t'_0 =\gamma (t_0-x_0v)$
$0 =\gamma (0-x_0v)$
$0 =-x_0v$

Do you want to try with :
$$t' =\gamma (t-(x-x_0)v)$$

I drew the coordinate systems

Edit: Be careful because $t=0$ and $t'=0$ are not the same moment ( they are not simultaneous )
$t'=0$ is on $x'$ axis. When $x=0$ and $t=0$ (check drawing) $t' \gt 0$

 

[Leepappas]

No. By the Lorentz transform we have:
$\Delta t’=\gamma(0-va)$

If the change in $\Delta t$ is zero, then how can time pass in another inertial reference frame?

 

[Dale]

If the change in $\Delta t$ is zero, then how can time pass in another inertial reference frame?

The two events are simultaneous in one frame, but not in the other frame.

According to the primed frame the events happened at different times. According to the unprimed frame the primed clocks are offset.

 

[Orodruin]

Since that old thread I have familiarized myself with the relativity of simultaneity

The following tells me you have not understood the relativity of simultaneity:

If the change in $\Delta t$ is zero, then how can time pass in another inertial reference frame?

That $\Delta t’$ is non-zero for events with $\Delta t = 0$ is the relativity of simultaneity. Two events that are simultaneous in the unprimed frame are not simultaneous in the primed frame. What you have just done is to derive an expression for how much they differ.

 

[PeroK]

If the change in $\Delta t$ is zero, then how can time pass in another inertial reference frame?

The real problem.is that you've fundamentally set your mind against relativity. This thread is more about tackling that psychology than about science or mathematics.

 

[Ibix]

Since that old thread I have familiarized myself with the relativity of simultaneity and Minkowski world lines.

The key realisation is that the Lorentz transforms are very closely analogous to rotations in Euclidean space. Rotating your coordinate system on a Euclidean plane can't lead to contradictions because all you are doing is picking a different way of labelling points. You can be sloppy and do it wrong, but that's not the fault of the maths. Similarly, boosting your coordinate system can't lead to contradictions because all you are doing is changing your system for labelling events.

Minkowski diagrams can help with this because you can draw a diagram of some arbitrary scenario and superpose the coordinate grid of any frame you like. The diagram never changes, only the coordinate grid. So how can you have a contradiction? Then you can imagine stretching the diagram so that one of the other coordinate grids is now orthonormal. It's a smooth deformation (no points need to cross or anything nasty like that) so, again, how can there be a contradiction?

I found that visualisation helpful enough to have written some Javascript. It doesn't work very well on touch screens (they weren't as ubiquitous when I wrote this), but the buttons at the bottom of the page set up some simple scenarios. You can then choose another frame and it animates the boost.

 

[Orodruin]

The universe, alas, is not required to conform to your intuition. You're just not that important.

This is something that should be repeated and repeated and repeated and … well, you get the picture.

The bottom line is that anybody’s ”common sense” or ”intuition” is based on their general experience. In everyday life most people will not come into contact with relativistic effects and so they will feel unfamiliar or counter-intuitive. The Universe however is not restricted to abide by anyone’s ”common sense”. This is why in physics we make experiments to test how the Universe behaves in extreme situations. Regardless of whether someone likes it or not, the results of such tests are what will tell you how things work, not ”common sense” or thought experiments.

The good news is that intuition for the theory can be built by working with it to the point that it does seem familiar.

 

[Leepappas]

The following tells me you have not understood the relativity of simultaneity:

That $\Delta t’$ is non-zero for events with $\Delta t = 0$ is the relativity of simultaneity. Two events that are simultaneous in the unprimed frame are not simultaneous in the primed frame. What you have just done is to derive an expression for how much they differ.

That is counterintuitive and a contradiction. One moment in time in the unprimed frame, corresponds to an infinite number of moments in time in the primed frame. The relativity of simultaneity is what's wrong with special relativity.

 

[Orodruin]

The relativity of simultaneity is what's wrong with special relativity.

There is nothing conceptually wrong with relativity of simultaneity. Counterintuitive, perhaps, but that is an artefact of your intuition, which the Universe is not obliged to care about. Contradictory, no - the theory is consistent. As little as there is something wrong with saying that two points that have the same x-coordinate in one frame has different x-coordinates in a frame that is rotated relative to the first. So no, there is no actual logical fallacy.

You may not like it or be able to make sense out of it personally, but that’s how the world works as confirmed by countless experiments. You can choose not to believe it, but your beliefs would be in direct conflict with empirical data.

The idea that you have found a logical gap in the theory that has been missed by many thousands of physicists over the last 100 years is, quite frankly, ludicrous. In particular as what you say is a known fallacy among students that have difficulty grasping the relativity of simultaneity (and there are a lot of them - hence my forum signature).

 

[Leepappas]

The two events are simultaneous in one frame, but not in the other frame.

According to the primed frame the events happened at different times. According to the unprimed frame the primed clocks are offset.

Ok that's counterintuitive. One moment in time in the unprimed frame corresponds to two moments in time in the primed frame.

 

[Leepappas]

There is nothing conceptually wrong with relativity of simultaneity. Counterintuitive, perhaps, but that is an artefact of your intuition, which the Universe is not obliged to care about. Contradictory, no - the theory is consistent. As little as there is something wrong with saying that two points that have the same x-coordinate in one frame has different x-coordinates in a frame that is rotated relative to the first. So no, there is no actual logical fallacy.

You may not like it or be able to make sense out of it personally, but that’s how the world works as confirmed by countless experiments. You can choose not to believe it, but your beliefs would be in direct conflict with empirical data.

The idea that you have found a logical gap in the theory that has been missed by many thousands of physicists over the last 100 years is, quite frankly, ludicrous. In particular as what you say is a known fallacy among students that have difficulty grasping the relativity of simultaneity (and there are a lot of them - hence my forum signature).

Why can't you see SR is wrong?

 

[Orodruin]

Ok that's counterintuitive. One moment in time in the unprimed frame corresponds to two moments in time in the primed frame.

Again: Counterintuitive does not mean that the Universe needs to oblige by your intuition. Regardless of how counterintuitive, experiments that test these limits confirm it. The correct reaction is to realize you need better intuition.

 

[Orodruin]

Why can't see SR is wrong?

Because it has been confirmed as a better model than classical mechanics by countless experiments. That is what is used to judge, not what you happen to find intuitive.

 

[Dale]

One moment in time in the unprimed frame corresponds to two moments in time in the primed frame.

Yes. Also one position in space in the unprimed frame corresponds to multiple positions in space in the primed frame.

Although this is a challenging concept, there is nothing particularly new about it. That happens routinely with coordinate transforms in space. The only new thing is that it happens with time as well as space. This is not a problem or flaw with relativity, it is just a coordinate transform on spacetime instead of just space.