# Is relativity of simultaneity measurable?

#### Pete

##### It's not rocket surgery
Registered Senior Member
The first 90 posts of this thread were a sidetrack culled from another thread: [post=3068386]Basic Special Relativity Question[/post].

The sidetrack originated when Tach said that while the relativity of simultaneity is real, it can't be measured even in principle. I maintain that it can in principle be measured, but that I doubt the feasibility in practice.

How? Let's see the details. You waste so much time with repeating the same question
You could just answer it. Then the time wouldn't be wasted.
Do you think RoS is physically meaningful? Surely not, since you think it can't be measured. But then why do you think it is real?
you could have used all this time figuring out the details of the experiment. This should be an incentive, since no one else, to date , has come up with an experimental setup. You do not seem worried that there is no "RoS test" section here, so go for it!
There's not much to think about, it's a pretty trivial experiment in principle.
But, it's not practical. You're not going to get a precision synchronizable clock running on a rail at 100km/s.

Try this:
Place two clocks 1000 km apart. Synchronize them using Einstein synchronization.
Beside each clock is a rail, on which runs another clock (two rails, two clocks, 1000km apart). The rails are long enough that the clocks on the rails can be accelerated to 100km/s, then synchronized with each other using Einstein synchronization.

The clocks on rails are precisely accelerated and synchronized, so that they pass the stationary clocks at the same instant (as measured by the stationary clocks.)

As the moving clocks pass the stationary clocks, each clock records its current time.
If the experiment is a success, the stationary clocks will record the same time, the moving clocks will record a difference of approximately 1 microsecond.

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There's not much to think about, it's a pretty trivial experiment in principle.
But, it's not practical. You're not going to get a precision synchronizable clock running on a rail at 100km/s.

Try this:
Place two clocks 1000 km apart. Synchronize them using Einstein synchronization.
Beside each clock is a rail, on which runs another clock (two rails, two clocks, 1000km apart). The rails are long enough that the clocks on the rails can be accelerated to 100km/s, then synchronized with each other using Einstein synchronization.

The clocks on rails are precisely accelerated and synchronized, so that they pass the stationary clocks at the same instant (as measured by the stationary clocks.)

As the moving clocks pass the stationary clocks, each clock records its current time.
If the experiment is a success, the stationary clocks will record the same time, the moving clocks will record a difference of approximately 1 microsecond.

1. For an extra 20 points, the positions of the clocks on rails will not match the positions of the "stationary clocks". Why?
2. For extra 30 points, since the positions of the clocks don't match, how do you compare them?
3. For an extra 100 points: accelerated clocks cannot be e-synched if they are spatially separated (as in your setup), you have tried to e-synch them and now, that you accelerated them, they are out of synch. What do you do?

the positions of the clocks on rails will not match the positions of the "stationary clocks".
I suspect you're thinking of length contraction. Please clarify.

since the positions of the clocks don't match, how do you compare them?
You compare them as they pass each other.
One rail clock passes one ground clock. The times on both these clocks are recorded as they pass.
The other rail clock passes the other ground clock. The times on both these clocks are recorded as they pass.

Tach said:
accelerated clocks cannot be e-synched if they are spatially separated (as in your setup), you have tried to e-synch them and now, that you accelerated them, they are out of synch.
They are accelerated to equal and constant speed, then synchronized using Einstein synchronization.

I suspect you're thinking of length contraction. Please clarify.

You compare them as they pass each other.
One rail clock passes one ground clock. The times on both these clocks are recorded as they pass.

The ground clock disagrees with the rail clock. Now what?

The other rail clock passes the other ground clock. The times on both these clocks are recorded as they pass.

The other pair of clocks disagree as well. Now what?

In addition, the clocks on the ground disagree as to when the ends of the endpoints of the segment made by the rail clocks coincides with their respective positions on the ground. Now what?

Pete,

If you do these things, you need to think them through, not throw them over the fence.

The ground clock disagrees with the rail clock. Now what?

The other pair of clocks disagree as well. Now what?
If the experiment succeeds, the two ground clocks record the same time, meaning the passing-events were simultaneous in the ground frame.
If RoS is measured, then the two rail clocks record different times (about 1 microsecond difference), meaning the events were not simultaneous in the rail clock frame.

If the experiment succeeds, the two ground clocks record the same time, meaning the passing-events were simultaneous in the ground frame.

But you already know that this is not true due to length contraction.

But you already know that this is not true due to length contraction.

We know it is true, because that's how we set it up.
We didn't set up the rail clocks to be exactly 1000km apart in the rail clock frame, we set them up to pass the ground clocks simultaneously in the ground frame.

We know it is true, because that's how we set it up.
We didn't set up the rail clocks to be exactly 1000km apart in the rail clock frame, we set them up to pass the ground clocks simultaneously in the ground frame.

Ok, so you decreased the distance between the clocks in the ground frame such that it matches the contracted distance between the moving clocks. Now the positions of the endpoints agree in the ground frame but they disagree in the rail car frame because the distance has been shrunk by you. So, now the positions don't match in the rail car frame. What now? How do you compare the clock readings?

Ok, so you decreased the distance between the clocks in the ground frame such that it matches the contracted distance between the moving clocks. Now the positions of the endpoints agree in the ground frame but they disagree in the rail car frame because the distance has been shrunk by you. So, now the positions don't match in the rail car frame. What now? How do you compare the clock readings?
I don't understand your objection. It's as if you think the clocks don't pass each other in the rail clock frame.
They do pass each other, but the two pairs don't pass at the same time... which is the whole point.

I don't understand your objection. It's as if you think the clocks don't pass each other in the rail clock frame.
They do pass each other, but the two pairs don't pass at the same time... which is the whole point.

The point is that you can only compare clocks that are co-located. Your setup disallows that. You didn't think things through and you are trying to make it up as you go. Case and point: when the left hand clocks (ground and railcar) are lined up, the righthand clocks are lined up in the ground frame (through your preshrinking of their distance in the ground frame) but they aren't lined up in the railcar frame, so you cannot have a valid measurement.

The point is that you can only compare clocks that are co-located.
We can also compare recorded times.
Your setup disallows that.
Of course it allows it. The clocks pass each other. The times are recorded.

Case [in] point: when the left hand clocks (ground and railcar) are lined up, the righthand clocks are lined up in the ground frame (through your preshrinking of their distance in the ground frame) but they aren't lined up in the railcar frame,
That's the whole point of the exercise.
They line up simultaneously in the ground frame.
They line up non-simultaneously in the rail clock frame.

That's the whole point of the exercise.
They line up simultaneously in the ground frame.
They line up non-simultaneously in the rail clock frame.

...but you cannot measure that because you can't compare the clocks. That is the whole point, you can't get a valid measurement.
Tell you what, write up the equations describing the experiment and you will find out your errors by yourself (or, as before, I will help you out in finding them). While you are at it, you might want to think about also coming up with the setup for the actual puzzle, not the dumbed down version, i.e. how do you measure the $$(x^"_k,y^"_k,t^"_k)$$ sets in the "rod bending".

...but you cannot measure that because you can't compare the clocks.
We compare the recorded times.
The ground clocks record the same time. They show that the events are simultaneous.
The rail clocks record different times. They show that the events are not simultaneous.

We compare the recorded times.
The ground clocks record the same time. They show that the events are simultaneous.
The rail clocks record different times. They show that the events are not simultaneous.

Tell you what, write up the equations describing the experiment and you will find out your errors by yourself (or, as before, I will help you out in finding them). While you are at it, you might want to think about also coming up with the setup for the actual puzzle, not the dumbed down version, i.e. how do you measure the $$(x^"_k,y^"_k,t^"_k)$$ sets in the "rod bending".

No, I don't think so.
I've described a clear and simple way of measuring Relativity of Simultaneity, which you said was theoretically impossible.
When you acknowledge that RoS is a theoretically measurable effect, then we'll continue. Otherwise, we're done.

You could just answer it. Then the time wouldn't be wasted.
Do you think RoS is physically meaningful? Surely not, since you think it can't be measured. But then why do you think it is real?

There's not much to think about, it's a pretty trivial experiment in principle.
But, it's not practical. You're not going to get a precision synchronizable clock running on a rail at 100km/s.

Try this:
Place two clocks 1000 km apart. Synchronize them using Einstein synchronization.
Beside each clock is a rail, on which runs another clock (two rails, two clocks, 1000km apart). The rails are long enough that the clocks on the rails can be accelerated to 100km/s, then synchronized with each other using Einstein synchronization.

The clocks on rails are precisely accelerated and synchronized, so that they pass the stationary clocks at the same instant (as measured by the stationary clocks.)

As the moving clocks pass the stationary clocks, each clock records its current time.
If the experiment is a success, the stationary clocks will record the same time, the moving clocks will record a difference of approximately 1 microsecond.

The above will not work because the math disproves your claims.
Let $$S$$ be the frame of the moving 1000km (!!!) rail car with the clocks at both ends and $$S'$$ the frame of the rails. The $$S'$$ clocks are situated $$L=1000km$$ apart. we will label the clocks A and B:

$$t'_A=\gamma(t_A+vx_A/c^2)$$
$$t'_B=\gamma(t_B+vx_B/c^2)$$

The $$S$$ clocks are synchronized , so:

$$t_A=t_B=t_0$$

$$x_A=0$$
$$x_B=L$$

so:

$$t'_A=\gamma t_0$$
$$t'_B=\gamma t_0+\gamma vL/c^2$$

Therefore:

$$t'_B-t'_A=\gamma vL/c^2$$

No surprise here. But:

$$x'_A=\gamma(x_A+vt_A)=\gamma vt_0$$
$$x'_B=\gamma(x_B+vt_B)=\gamma vt_0+\gamma L$$

So, your scenario has a flaw, the clocks on the rail car do not line up with the clocks on the ground, the clock at B in the railcar is measured by the platform as being way past to the right of the corresponding ground clock. No wonder the ground clock has accumulated the extra time, the railcar clock has passed it by quite a bit. The larger the relative speed, the larger the missalignment, both spatial and temporal

So, your scenario has a flaw, the clocks on the rail car do not line up with the clocks on the ground

That is not a flaw, that is RoS. The misalignment is an important part of it. You cannot expect Pete to create an experiment in which the two pairs of clocks line up simultaneously in both frames. If that were possible, there would not be any RoS to measure.

That is not a flaw, that is RoS. The misalignment is an important part of it. If that were possible, there would not be any RoS to measure.

Sorry but you are wrong, the setup is flawed.

You cannot expect Pete to create an experiment in which the two pairs of clocks line up simultaneously in both frames.

They are not lined up in any frame.