what will be the events relative to the ground observer , a time-like or space like?

Adel Makram

yes correct

nitsuj

& classical physics has left the building. (not that I think this is incompatible with classical physics)

IIRC there is no information exchanged in that scenario, ie no "break" in causality / faster than c. Not sure I subscribe to the scenario you describe with changing spin. A NYSE trader would love that kinda communication speed advantage, let alone any wireless communications company. It is a "state" that the particle is in, I guess there can be a "clone" particle of sorts that share indefinite properties (specifically a state); as wiki says "...which is indefinite in terms of important factors such as position,[5] momentum, spin, polarization, etc."

"Entanglement" is misleading.

Really this is not much different than "distant star" scenario we were discussing above. With theory, we can measure, calculate values & make predictions.

I hope someone who knows more can correct me on this if wrong, but there is no change in state of either particle, merely the values become mostly definitive/determined.

ghwellsjr

Could you explain what the single and double back quotes mean?

Michael C

Without using any relativistic calculations it's easy enough to see that this is impossible. Just put a mirror at A.

But you seem to be claiming that relativistic calculations show this to be possible. In order to understand your argument I need more explanation. In the PDF you just attached, what exactly do you mean by t`<sub>a</sub> and AB` ? In what frame of reference are you doing your calculations?

Fredrik

Apologies if this has all been sorted out. I haven't read all the posts after my previous one.

It only makes sense to ask about the separation between A and B if they are events. If they are, then the separation is determined from the sign of
$$-(t_A-t_B)^2+(x_A-x_B)^2.$$ If we define
$$x=\begin{pmatrix}t_A-t_B\\ x_A-x_B\end{pmatrix},\qquad\eta=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}$$ we have
$$-(t_A-t_B)^2+(x_A-x_B)^2=x^T\eta x.$$ A Lorentz transformation is defined as a linear transformation that preserves ##x^T\eta x##, so
$$-(t_A-t_B)^2+(x_A-x_B)^2=x^T\eta x=x'^T\eta x'=-(t_A'-t_B')^2+(x_A'-x_B')^2.$$
So the separation is clearly the same in all inertial frames.

Here A and B have two different meanings in the same sentence. You need to make up your mind if you want them to be events on locations on the ground.

Adel Makram

I know that it is impossible from the diagram and the experiment setup. So the physics says it is impossible but the math says it is possible!

AB` means the distance between A & B in the train
ta` is the time when A received the signal from the source
All calculations are done in the train rest frame

Adel Makram

Δt` is the difference in time of receiving the light signals at A and B
Δt`` is the time that would have been taken by a light signal emitted from A to reach B

Thanks,,,

Fredrik

Only if you're doing the math wrong.

Have you read any of my posts? I've been telling you lots of times that there's no such thing as a distance between A and B in the train's rest frame, if A and B are events that are simultaneous in the ground's rest frame.

Michael C

OK. I imagine that v is the velocity of the train relative to the ground?

Please could you explain the reasoning behind the first step in your PDF, where you state that:

[itex]t`_{a} = \frac{(\frac{1}{2}AB`)-vt`a}{c}[/itex]

ghwellsjr

Then why is there a "v" in the calculations and in your diagrams?

The speed of the light source has no bearing on anything except maybe to determine its location at the time it emits a flash which means you could analyze your scenario with the location of the light source as a variable instead of the way you are doing it, correct? Then instead of something special happening at v/c=0.618, it would happen at some position of the light source, correct? And then you could tell us what that location is and you could show us what the something special is, correct? Why don't you do that?

John232

It happens when particles split apart, their brother particle continues to have a link with the other particle it was joined with. So, they are like blood brothers for life. I think the mention of the indefinite qualities was only referring to the uncertainty principle. But looking further on the same page, "When a measurement is made and it causes one member of such a pair to take on a definite value (e.g., clockwise spin), the other member of this entangled pair will at any subsequent time[7] be found to have taken the appropriately correlated value (e.g., counterclockwise spin)." This is saying that it will happen at any subsequent time. It takes on a definite value when one of the particles are measured, that without measurement would be uncertain. The spins of the two particles would always be the opposite of the other. Then it would only be a matter of turning a clockwise spin of particle into a 1 and a counterclockwise spin into a 0 to gain 1 bit of information that is FTL, more bits would require more particles spins being measured and the tricky part would be haveing them have the same common ancestor. So then it would not be very feasible technology, unless computers could be fed these particles over the internet, even then it wouldn't make much of a difference since light can travel around the world many times in a secound.

Adel Makram

v is the velocity of the train relative to the ground, yes. But it is also the velocity of a source put in the ground moving with -v relative to the train if the location of that source coincides with the point of the source in the train at the beginning of the experiment. This I missed to mentioned at the PDF. So the light emitted from a source in the ground corresponding to a point in the train near A than B.
This also gives the answer to your second question. That is when the ground source coincides with the midpoint of the train relative to the ground observer, their light signal would reach A and B at the same time relative to him. It must coincides too with the center of the train relative to the train observer ( because the world lines of the source and the midpoint of the train coincide in all FORs) So ta` is the time recorded by the train observer when the light strikes A and tb` has the same meaning for B

Adel Makram

v is the velocity of the train relative to the ground, yes. But it is also the velocity of a source put in the ground moving with -v relative to the train if the location of that source coincides with the point of the source in the train at the beginning of the experiment. That is I missed to mention in the PDF. So the light emitted from a source in the ground corresponding to a point in the train near A than B.

So when the ground source coincides with the midpoint of the train relative to the ground observer, their light signal would reach A and B at the same time relative to him. It must coincides too with the center of the train relative to the train observer ( because the world lines of the source and the midpoint of the train coincide in all FORs) So ta` is the time recorded by the train observer when the light strikes A and tb` has the same meaning for B

ghwellsjr

You only addressed my first question and you didn't even answer it:
My question was: why, since "all calculations are done in the train rest frame" do you need "v" in those calculations?
Isn't it the case that the source of light emits a single flash at an instant of time? If this is true, why do you care where the source moves to after the flash has occurred? Couldn't the flash be emitted by a flash bomb that ceases to exist once it is set off? And if this is true, why do you care about its state of motion prior to it being set off?

Please answer the rest of my questions:
I think if you actually do what I'm suggesting, that is, break your scenario into two parts, one for the train with the flash occurring at a specific point and one where the speed of the ground observer determines where that point is, you will discover that at your special speed of 0.618, the flash actually occurs at location A and for higher speeds, it occurs to the right of A. But I don't know, because I still cannot figure out the second part of your scenario.

You still have not told us where the train observer is located on the train or where the ground observer is located with respect to the light source.

Michael C

It's important to remember this: what the source of the light signal does after emitting the signal is irrelevant to the course of the light flash that has already been emitted. The light source could be on the train and the experiment would work just the same way: all that matters is the space-time location of the source at the moment when the signal is fired.

All that you are doing in saying that the source is on the ground is keeping a record in the ground frame of the spatial position of the event "light signal is fired". I'll call this "position p" in the ground frame.

You have to be very careful here. We can distinguish 4 events:

Event 1: light flash is emitted.
Event 2: the middle of the train coincides with "position p" in the ground frame.
Event 3: the light reaches A
Event 4: the light reaches B

In the ground frame, events 2, 3 and 4 are simultaneous. In the train frame, they aren't: they happen in the order 3,2,4.

The two world lines meet at event 2, but I don't see where this helps us in the train frame: in this frame (as remarked above) this is not simultaneous with either event 3 or 4.

Yes, but that doesn't explain your first equation.

Adel Makram

You are right I made my calculation based on Events 2 & 3 and Events 2 & 4 are simultaneously. So how the wrong math leads to the correct LT?

Thanks,,,

ghwellsjr

Please don't ignore my response to you on the previous page. And don't expect any of us to figure out what the solution to your problem is until you tell us what your problem is. We can't read your mind.