Similar to the Einstein train experiment

[Whatifitaint]

Please help with this. This is similar to the Einstein train experiment.

When C and C' are at the same place, lightning strikes at their location.

Both survive though. Assume another prime B' behind C' When B' is at the same location as C, the lightning is at (A,0,0) and (-A,0,0) in the C frame. These are simultaneous events. But, they are not simultaneous to B'. So, C says the lightning is at A and -A when B' and C are together but B' says the lightning cannot be at both A' and -A' (these are located at A and -A respectively when B' and C are together).

So, here is my problem. How can B' and C disagree, when they are at the same place, on the distance the lightning traveled along the positive x-axis and negative x-axis?

Thanks in advance.

 

[mfb]

So, here is my problem. How can B' and C disagree, when they are at the same place, on the distance the lightning traveled along the positive x-axis and negative x-axis?

Where is the problem? Different observers with different velocity disagree on the simultaneity of events and on their coordinate systems. Their axes are different in spacetime.

 

[Whatifitaint]

Where is the problem? Different observers with different velocity disagree on the simultaneity of events and on their coordinate systems. Their axes are different in spacetime.

I am not seeing the answer here.
B' and C are at the same place when C determines the lightning strike is a distance A from both B' and C. But, B' cannot agree with this because B' does not agree the A and -A events are simultaneous.

However, that mean B' and C must disagree on the distance the lightning strike has traveled when they are together. That means the light must be at 2 different locations along the common x-axis when B' and C are together. How can that be true?

Or, B' must say when they are together, the lightning is not at A and C must say it is at A. How does this work?

 

[mfb]

That means the light must be at 2 different locations along the common x-axis when B' and C are together.

There is no common x-axis for B' and C, as they are moving relative to each other.

 

[Whatifitaint]

There is no common x-axis for B' and C, as they are moving relative to each other.

I am a little confused. They do not share the same x coordinates for events, but they share the same x-axis.

Is this false?

 

[mfb]

I am a little confused. They do not share the same x coordinates for events, but they share the same x-axis.

Not in spacetime. The direction might be the same, but the scale and the simultaneity is not. Just look at a Minkowski diagram.

 

[ghwellsjr]

Please help with this. This is similar to the Einstein train experiment.

When C and C' are at the same place, lightning strikes at their location.

Both survive though. Assume another prime B' behind C' When B' is at the same location as C, the lightning is at (A,0,0) and (-A,0,0) in the C frame. These are simultaneous events. But, they are not simultaneous to B'. So, C says the lightning is at A and -A when B' and C are together but B' says the lightning cannot be at both A' and -A' (these are located at A and -A respectively when B' and C are together).

So, here is my problem. How can B' and C disagree, when they are at the same place, on the distance the lightning traveled along the positive x-axis and negative x-axis?

Thanks in advance.

Yes, diagrams help a lot in situations like this. Here's a diagram showing the rest frame of the ground observer C in blue. The black train locomotive is C' and the red caboose is B'. The train is traveling at 0.6c. As you said, when the black C' locomotive reaches the blue C ground observer (at the beginning of the scenario which is the origin of the frame), lightning strikes their location and emits two green light flashes in both directions:

When the red caboose, B', reaches the ground observer, C, we can note the Coordinate Positions of the progress of the two green light flashes in the ground frame and they have reached plus and minus 10 thousand feet. All three events are at a Coordinate Time of 10 microseconds.

Now let's transform the coordinates of all the events into the rest frame for the train:

We can see that the event of the red caboose coinciding with the blue ground observer occurs at the Coordinate Time of 12.5 microseconds but the events for the progress of the flashes occur at Coordinate Times of 20 and 5 microseconds making none of these events simultaneous.

Does this clear up all your confusion?

 

[Whatifitaint]

Yes, diagrams help a lot in situations like this. Here's a diagram showing the rest frame of the ground observer C in blue. The black train locomotive is C' and the red caboose is B'. The train is traveling at 0.6c. As you said, when the black C' locomotive reaches the blue C ground observer (at the beginning of the scenario which is the origin of the frame), lightning strikes their location and emits two green light flashes in both directions:

When the red caboose, B', reaches the ground observer, C, we can note the Coordinate Positions of the progress of the two green light flashes in the ground frame and they have reached plus and minus 10 thousand feet. All three events are at a Coordinate Time of 10 microseconds.

Now let's transform the coordinates of all the events into the rest frame for the train:

We can see that the event of the red caboose coinciding with the blue ground observer occurs at the Coordinate Time of 12.5 microseconds but the events for the progress of the flashes occur at Coordinate Times of 20 and 5 microseconds making none of these events simultaneous.

Does this clear up all your confusion?

You have a very nice way of putting this problem. So far, everything above is good.

When the red caboose, B', reaches the ground observer, C, how can they disagree on the distance the lightning is from their common location?

Initial conditions              | Lightning strike
                                C
                          B'    C'

After motion                               
 | Lightning location                                  | Lightning location
-A                        C                            A
                          B'    C'

So, when B' and C are at the same place, C claims they are equidistant to the lightning locations of A and -A. B', however, claims A calculated in its space-time occurs before -A in its space-time.

Therefore, for example, it is most likely that B' thinks in its time that the lightning strikes are further down the positive x-axis than event A. So, if we were to take the C' frame's A' and translate it to C space-time, say A2, we would find A2 > A. But, that means since A2 and A are in the same space-time, then when C and B' are together, the lightning is at 2 different C space-time positive x-axis locations.

But, this makes no sense, however I can't see what is false above.

 

[PAllen]

Therefore, for example, it is most likely that B' thinks in its time that the lightning strikes are further down the positive x-axis than event A. So, if we were to take the C' frame's A' and translate it to C space-time, say A2, we would find A2 > A. But, that means since A2 and A are in the same space-time, then when C and B' are together, the lightning is at 2 different C space-time positive x-axis locations.

But, this makes no sense, however I can't see what is false above.

Every event exists in every frame. A2 and A, as you've defined them, are simply different events. They are at different space time coordinates per B'/C', and they remain at different spacetime coordinates for C. Absurd would be if they were different for B'/C' but the same for C.

 

[ghwellsjr]

You have a very nice way of putting this problem. So far, everything above is good.

When the red caboose, B', reaches the ground observer, C, how can they disagree on the distance the lightning is from their common location?

Initial conditions              | Lightning strike
                                C
                          B'    C'

After motion                               
 | Lightning location                                  | Lightning location
-A                        C                            A
                          B'    C'

So, when B' and C are at the same place, C claims they are equidistant to the lightning locations of A and -A. B', however, claims A calculated in its space-time occurs before -A in its space-time.

Correct.

Therefore, for example, it is most likely that B' thinks in its time that the lightning strikes are further down the positive x-axis than event A.

Correct. Here's a spacetime diagram to show the new event A' that is simultaneous to when B' and C are at the same place in the rest frame for C':

So, if we were to take the C' frame's A' and translate it to C space-time, say A2, we would find A2 > A.

Correct. Here's the spacetime diagram we get when we transform the previous frame to the rest frame of C and we change the name of the A' event to A2:

But, that means since A2 and A are in the same space-time, then when C and B' are together, the lightning is at 2 different C space-time positive x-axis locations.

But, this makes no sense, however I can't see what is false above.

I think PAllen's explanation should cover it.

 

[Whatifitaint]

Every event exists in every frame. A2 and A, as you've defined them, are simply different events. They are at different space time coordinates per B'/C', and they remain at different spacetime coordinates for C. Absurd would be if they were different for B'/C' but the same for C.

I am not following you. I will say I think and then you come back and correct it so that it all makes sense.

- The C frame says the cos(0) lightning beam is at A when B' and C are colocated.
- The B' observer says the cos(0) lightning beam is at some A' when B' and C are colocated.
- All space-time coordinates map from one frame to the other.
- So, either Map(A',0,0,A'/c) = (A,0,0,A/c), Map(A',0,0,A'/c) > (A,0,0,A/c) or Map(A',0,0,A'/c) < (A,0,0,A/c), trichotomy property for light beams.
- B' believes there are cos'(\pie) and cos'(0) events that are simultaneous but different from C's simultaneous events.

1) If Map(A',0,0,A'/c) = (A,0,0,A/c), then apply the same argument to -A and then B' thinks -A and A are simultaneous, which is wrong.
2) If Map(A',0,0,A'/c) > (A,0,0,A/c), relativity concludes when B' and C are colocated, the cos(0) lightning beam is at two different C frame space-time locations since (A,0,0,A/c) is in the space-time of C and so is Map(A',0,0,A'/c), which can't be true for one light beam.
3) If Map(A',0,0,A'/c) < (A,0,0,A/c), relativity concludes when B' and C are colocated, the cos(0) lightning beam is at two different C frame space-time locations since (A,0,0,A/c) is in the space-time of C and so is Map(A',0,0,A'/c), which can't be true for one light beam.

 

[ghwellsjr]

I am not following you. I will say I think and then you come back and correct it so that it all makes sense.

- The C frame says the cos(0) lightning beam is at A when B' and C are colocated.
- The B' observer says the cos(0) lightning beam is at some A' when B' and C are colocated.
- All space-time coordinates map from one frame to the other.
- So, either Map(A',0,0,A'/c) = (A,0,0,A/c), Map(A',0,0,A'/c) > (A,0,0,A/c) or Map(A',0,0,A'/c) < (A,0,0,A/c), trichotomy property for light beams.
- B' believes there are cos'(\pie) and cos'(0) events that are simultaneous but different from C's simultaneous events.

1) If Map(A',0,0,A'/c) = (A,0,0,A/c), then apply the same argument to -A and then B' thinks -A and A are simultaneous, which is wrong.
2) If Map(A',0,0,A'/c) > (A,0,0,A/c), relativity concludes when B' and C are colocated, the cos(0) lightning beam is at two different C frame space-time locations since (A,0,0,A/c) is in the space-time of C and so is Map(A',0,0,A'/c), which can't be true for one light beam.
3) If Map(A',0,0,A'/c) < (A,0,0,A/c), relativity concludes when B' and C are colocated, the cos(0) lightning beam is at two different C frame space-time locations since (A,0,0,A/c) is in the space-time of C and so is Map(A',0,0,A'/c), which can't be true for one light beam.

I think your problem is that when you are going from A to -A, you are changing the sign of both the spatial term and the temporal term but the temporal term should be positive in both cases.

Also, I'm not sure what you mean by cos(0) and cos'(\pie). Do you just mean the light beam that is going to the right for cos(0) and the light beam going to the left for cos'(\pie)? Did you really mean pi instead of pie? Since cos(0) = 1 and cos(pi) = -1, why don't you just say that you are referring to the beam going to the right which has positive x values and the beam going to the left has negative x values? And remember, they both of positive t values in all cases.

If you just use the Lorentz Transformation process either with my specific numerical values or with your general symbolic variables and are careful to apply v as positive or negative depending on which way you are transforming (positive when going from the C frame to the B' frame and negative when going from the B' frame to the C frame) and make sure t is always positive then I think your problems will go away.

 

[Whatifitaint]

I think your problem is that when you are going from A to -A, you are changing the sign of both the spatial term and the temporal term but the temporal term should be positive in both cases.

Also, I'm not sure what you mean by cos(0) and cos'(\pie). Do you just mean the light beam that is going to the right for cos(0) and the light beam going to the left for cos'(\pie)? Did you really mean pi instead of pie? Since cos(0) = 1 and cos(pi) = -1, why don't you just say that you are referring to the beam going to the right which has positive x values and the beam going to the left has negative x values? And remember, they both of positive t values in all cases.

If you just use the Lorentz Transformation process either with my specific numerical values or with your general symbolic variables and are careful to apply v as positive or negative depending on which way you are transforming (positive when going from the C frame to the B' frame and negative when going from the B' frame to the C frame) and make sure t is always positive then I think your problems will go away.

Wow, misspelling I meant cos'(\apple pie), joke. Yes, cos(\pi). And, the direction of v is based on the motion of the frames not the light beams.

We are not communicating.

So, let's ignore -A for now which is a shorthand of writing (-A,0,0).

We know frame C says when B' and C are colocated, the cos(0) beam is located at (A,0,0) with t=A/c.

Now B' has an opinion where its cos(0) beam is located in its space-time. Let that be (A',0,0) with t'=A'/c. Then use the Lorentz Transformations on (A',0,0) with t'=A'/c, such that $x=(x'+vt')\gamma$. So, $x=(A' + vA'/c)\gamma$.

Either x=A, x<A or x>A.

If x ≠ A, then when B' and C are colocated, the lightning strike is located at A and also x which are both C space coordinates. One light beam, can't be at 2 different places in the coordinates of the C frame when B' and C are colocated.

Say x=A. Then apply the same argument to -A with -A' eliminating x2 ≠ -A leaving x2 = -A for the $cos(\pi)$ beam. But, A' and -A' are not simultaneous events in the C' frame, so that can't be true either.

Therefore, to solve this problem, we need some way such that B' thinks it sees simultaneous events in its frame just like C does, but we don't trigger an error in which one light beam is at 2 different places in the space-time of C when B' and C are colocated.

I can't think of a way.

 

[PAllen]

We know frame C says when B' and C are colocated, the cos(0) beam is located at (A,0,0) with t=A/c.

Now B' has an opinion where its cos(0) beam is located in its space-time. Let that be (A',0,0) with t'=A'/c.

Again these are two completely different events. Both of them exist for B'. Both of them exist for C. B' thinks the A' event is now and the A event is earlier. C thinks the A event is now and the A' event is later. Note, further, they both agree that A' is later than A (as required by causality since they are two different events in the history of the same light beam). They only differ on which is 'now'.

This all goes back to your seeming refusal to accept that two coincident observers (in different states of motion) can disagree on what 'now' is. Not only can they, but they must if they both use the same procedure to define it.

 

[Whatifitaint]

Again these are two completely different events. Both of them exist for B'. Both of them exist for C. B' thinks the A' event is now and the A event is earlier. C thinks the A event is now and the A' event is later. Note, further, they both agree that A' is later than A (as required by causality since they are two different events in the history of the same light beam). They only differ on which is 'now'.

This all goes back to your seeming refusal to accept that two coincident observers (in different states of motion) can disagree on what 'now' is. Not only can they, but they must if they both use the same procedure to define it.

Can you please instead of asserting I refuse to accept different now's, which I have not done, instead answer the question?

C claims A is the lightning location when B' and C are colocated.

What does B' claim, when B' and C are colocated, the location of the lightning is in the coordinates of the C frame?

Please answer this specific question since it is physical reality.

 

[PAllen]

Can you please instead of asserting I refuse to accept different now's, which I have not done, instead answer the question?

C claims A is the lightning location when B' and C are colocated.

What does B' claim, when B' and C are colocated, the location of the lightning is in the coordinates of the C frame?

Please answer this specific question since it is physical reality.

I've answered the question twice already. You refuse the answer and repeat the same mistake.

The answer is that B' thinks the signal is at A' when they are colocated. This is a different event than what C calls event A precisely because what B' and C mean by 'now' are two different things. Both A' and A have separate coordinates for both B' and C. You have even been given a beautiful picture of this by
ghwellsjr.

Despite coinciding, and agreeing on 'now' for the single event of coincidence, B' and C disagree on what 'now' is everywhere else.

 

[Whatifitaint]

I've answered the question twice already. You refuse the answer and repeat the same mistake.

The answer is that B' thinks the signal is at A' when they are colocated. This is a different event than what C calls event A precisely because what B' and C mean by 'now' are two different things. Both A' and A have separate coordinates for both B' and C. You have even been given a beautiful picture of this by
ghwellsjr.

Despite coinciding, and agreeing on 'now' for the single event of coincidence, B' and C disagree on what 'now' is everywhere else.

You are still not answering the question.

Try to translate A' to C coordinates.

Use the lorentz transformations.

Where does B' think the lightning strike is in the coordinates of C when C and B' are colocated.

We already know that C thinks it is A. What does B' think it is in C space-time?

That is the issue.

 

[PAllen]

You are still not answering the question.

Try to translate A' to C coordinates.

Use the lorentz transformations.

Where does B' think the lightning strike is in the coordinates of C when C and B' are colocated.

That is the issue.

See Gwellsjr's picture. He already did it. The answer is the same whoever does it.

 

[Whatifitaint]

See Gwellsjr's picture. He already did it. The answer is the same whoever does it.

I am not getting it.

We all agree C says A when B' and C are colocated.

Now, you have not given the view of B' for where the lightning is located in C coordinates when B' and C are colocated. That is a logical known for B' using lorentz transformations.

Let's put it down in writing. I want to see this using math and writing.

Surely this should be simple.

 

[PAllen]

I am not getting it.

We all agree C says A when B' and C are colocated.

Now, you have not given the view of B' for where the lightning is located in C coordinates when B' and C are colocated. That is a logical known for B' using lorentz transformations.

Let's put it down in writing. I want to see this using math and writing.

Surely this should be simple.

Gwellsjr put it in writing and a picture. He shows both A and A' per B', and A and A' (relabeled A2) per C. What part of this don't you get?

 

[PAllen]

Let that be (A',0,0) with t'=A'/c. Then use the Lorentz Transformations on (A',0,0) with t'=A'/c, such that $x=(x'+vt')\gamma$. So, $x=(A' + vA'/c)\gamma$.

Either x=A, x<A or x>A.

If x ≠ A, then when B' and C are colocated, the lightning strike is located at A and also x which are both C space coordinates. One light beam, can't be at 2 different places in the coordinates of the C frame when B' and C are colocated.

There is nothing wrong with your computation. There is everything wrong with your interpretation. your x above represents the distance (in C units) to the event in the beam's history that B' considers simultaneous with B'/c coincidence. 'A' refers to the distance in C units to the event in the beam's history that C considers simultaneous to the B'/C coincidence. How many way can I explain than these are two completely different events in the history of the beam, so there is problem with them having different distances from B'/C coincidence in both B' coordinates or C coordinates.

 

[Whatifitaint]

Gwellsjr put it in writing and a picture. He shows both A and A' per B', and A and A' (relabeled A2) per C. What part of this don't you get?

His diagram does not include the translation of A' to C coordinates.

I want this simply written down for its value.

Can you do this?

 

[PAllen]

His diagram does not include the translation of A' to C coordinates.

I want this simply written down for its value.

Can you do this?

Yes it does. It is at x=25,t=25.

 

[Whatifitaint]

Yes it does. It is at x=25,t=25.

Thanks, what does C say?

 

[PAllen]

Thanks, what does C say?

Yes, using the particular numbers for the scenario that Gwellsjr for the scenario.

 

[PAllen]

Let that be (A',0,0) with t'=A'/c. Then use the Lorentz Transformations on (A',0,0) with t'=A'/c, such that $x=(x'+vt')\gamma$. So, $x=(A' + vA'/c)\gamma$.

Either x=A, x<A or x>A.

If x ≠ A, then when B' and C are colocated, the lightning strike is located at A and also x which are both C space coordinates. One light beam, can't be at 2 different places in the coordinates of the C frame when B' and C are colocated.

There is nothing wrong with your computation. There is everything wrong with your interpretation. your x above represents the distance (in C units) to the event in the beam's history that B' considers simultaneous with B'/c coincidence. 'A' refers to the distance in C units to the event in the beam's history that C considers simultaneous to the B'/C coincidence. How many way can I explain than these are two completely different events in the history of the beam, so there is problem with them having different distances from B'/C coincidence in both B' coordinates or C coordinates.

You have not computed the t coordinate for A'. You have to transform both coordinates for any event. Then you would see that the transform of A' corresponds to a completely different time per C and does the event C calls A.

 

[Whatifitaint]

See Gwellsjr's picture. He already did it. The answer is the same whoever does it.

That's odd, we do not know how far apart B' and C are. How can this be decided?

Either way, say A is the view of the C frame.

Here is the question, does B' claim A' translates to A when B' and C are colocated as does C?

This is a yes or no.

 

[Whatifitaint]

You have not computed the t coordinate for A'. You have to transform both coordinates for any event. Then you would see that the transform of A' corresponds to a completely different time per C and does the event C calls A.

I do not know what the t coordinate for A' is. I know the t' coordinate is A'/c for A'.

And, you are not understanding. I want to see a 4D vector in the transformation for B' view. Can you say it is different from the A view vector of (A, 0, 0, A/c)?

This is yes or no.

 

[PAllen]

That's odd, we do not know how far apart B' and C are. How can this be decided?

Either way, say A is the view of the C frame.

Here is the question, does B' claim A' translates to A when B' and C are colocated as does C?

This is a yes or no.

No. Simple. Again.

 

[PAllen]

Yes it does. It is at x=25,t=25.

Thanks, what does C say?

That is what C says.

 

[PAllen]

Ok, here's a new set of numbers, slight variation because I have B' and C synch clocks when coincide. Suppose when B' and C are coincident, they synch clocks to 0. They are moving past each other at .6c. Let's use c=1 for easier numbers. At his moment of passing C announces: the beam emitted to the right when light struck when I passed your front is 'now' 20 units (it would be, say, light seconds) away. [edit: and C would know this because the lightning struck 20 seconds ago, by their clock] This means C is referring to the event on light's history that C labels as (x,t)=(20,0). B' would find that this moment on the light's history is given by:

t' = γ (t-vx/c^2) = (1.25) ( 0 - .6 * 20) = -15
x' = γ(x-vt) = (1.25) (20 - .6*0) = 25

So, if B' know the light was at (x',t') = (25,-15), then now, for B' it is at (x',t')=(40,0) just by light propagation. So these are the two relevant events per B' : B''s view of the event C calls 'where the light is now', and B's view of what he calls where the light is now. Two different events.

To see what C says, we run this backwards. First, transform (25,-15) back just for kicks:

t = γ (t' + vx/c^2) = (1.25) (-15 + .6 *25) = 0
x = γ (x' + vt') = (1.25) (25 + .6 * (-15)) = 20

as expected. Now, C's view of what B' calls where the light is now:

t = (1.25) (0 + .6 *40) = 30
x = (1.25) (40 + .6 *0) = 50

Note:

- This second event per C is consistent with light propagation for our c=1 convention
- Both B' and C agree on the time ordering (which came first) of these two events on the light's history

 

[ghwellsjr]

Ok, here's a new set of numbers, slight variation because I have B' and C synch clocks when coincide.

And here's a new set of diagrams.

Suppose when B' and C are coincident, they synch clocks to 0. They are moving past each other at .6c. Let's use c=1 for easier numbers. At his moment of passing C announces: the beam emitted to the right when light struck when I passed your front is 'now' 20 units (it would be, say, light seconds) away. [edit: and C would know this because the lightning struck 20 seconds ago, by their clock] This means C is referring to the event on light's history that C labels as (x,t)=(20,0).

B' would find that this moment on the light's history is given by:

t' = γ (t-vx/c^2) = (1.25) ( 0 - .6 * 20) = -15
x' = γ(x-vt) = (1.25) (20 - .6*0) = 25

So, if B' know the light was at (x',t') = (25,-15), then now, for B' it is at (x',t')=(40,0) just by light propagation.

So these are the two relevant events per B' : B''s view of the event C calls 'where the light is now', and B's view of what he calls where the light is now. Two different events.

To see what C says, we run this backwards. First, transform (25,-15) back just for kicks:

t = γ (t' + vx/c^2) = (1.25) (-15 + .6 *25) = 0
x = γ (x' + vt') = (1.25) (25 + .6 * (-15)) = 20

as expected.

Same as the first diagram.

Now, C's view of what B' calls where the light is now:

t = (1.25) (0 + .6 *40) = 30
x = (1.25) (40 + .6 *0) = 50

Note:

- This second event per C is consistent with light propagation for our c=1 convention
- Both B' and C agree on the time ordering (which came first) of these two events on the light's history

 

[Whatifitaint]

No. Simple. Again.

Thanks, please let me turn this into logical statements. Please explain why this logic that I learned is false.

Let statement P = B' and C are colocated.

Let statement Q = the lightning flash is located at A in the space-time of C.

Then, we have C claiming,

$Special relativity \Rightarrow P \rightarrow Q$.

Since you said no, we have B' claiming

$Special relativity \Rightarrow P \rightarrow \neg Q$.

This is called a contradiction.

Can you please show me where this is wrong?

 

[Whatifitaint]

Ok, here's a new set of numbers, slight variation because I have B' and C synch clocks when coincide.


Suppose when B' and C are coincident, they synch clocks to 0. They are moving past each other at .6c. Let's use c=1 for easier numbers. At his moment of passing C announces: the beam emitted to the right when light struck when I passed your front is 'now' 20 units (it would be, say, light seconds) away. [edit: and C would know this because the lightning struck 20 seconds ago, by their clock] This means C is referring to the event on light's history that C labels as (x,t)=(20,0). B' would find that this moment on the light's history is given by:

t' = γ (t-vx/c^2) = (1.25) ( 0 - .6 * 20) = -15
x' = γ(x-vt) = (1.25) (20 - .6*0) = 25

So, if B' know the light was at (x',t') = (25,-15), then now, for B' it is at (x',t')=(40,0) just by light propagation. So these are the two relevant events per B' : B''s view of the event C calls 'where the light is now', and B's view of what he calls where the light is now. Two different events.

To see what C says, we run this backwards. First, transform (25,-15) back just for kicks:

t = γ (t' + vx/c^2) = (1.25) (-15 + .6 *25) = 0
x = γ (x' + vt') = (1.25) (25 + .6 * (-15)) = 20

as expected. Now, C's view of what B' calls where the light is now:

t = (1.25) (0 + .6 *40) = 30
x = (1.25) (40 + .6 *0) = 50

Note:

- This second event per C is consistent with light propagation for our c=1 convention
- Both B' and C agree on the time ordering (which came first) of these two events on the light's history

I agree with everything above.

Even though this is a little different, it has the same issue in it.

C observer view of now and lightning at A when the clocks are synched,

t = γ (t' + vx/c^2) = (1.25) (-15 + .6 *25) = 0
x = γ (x' + vt') = (1.25) (25 + .6 * (-15)) = 20

B' frame calculations of the C observer view of now and lightning location in C frame coordinates when the clocks are synched.

t = (1.25) (0 + .6 *40) = 30
x = (1.25) (40 + .6 *0) = 50

It is correct, that C will see the lightning strike at x=50 when t=30 with t=30 from the new clock sync, which is in the future for C. That is perfectly true. But, that is not the correct now in the C frame.

What I am saying is that is it okay for SR to get the wrong answer for C frame "now" of the location of the lightning strike when the clocks are synched?

Again,

C frame "now" at the clock sync,
t = γ (t' + vx/c^2) = (1.25) (-15 + .6 *25) = 0
x = γ (x' + vt') = (1.25) (25 + .6 * (-15)) = 20

B' claim of C frame "now" at the clock sync
t = (1.25) (0 + .6 *40) = 30
x = (1.25) (40 + .6 *0) = 50

 

[PAllen]

Thanks, please let me turn this into logical statements. Please explain why this logic that I learned is false.

Let statement P = B' and C are colocated.

ok

Let statement Q = the lightning flash is located at A in the space-time of C.

Already a big mistake pointed out many times. This statement has no meaning. You need to add, for example, "located at A per C at Cs time t0". Otherwise it has exactly the content of "how high is up?".

Thus all the rest is completely meaningless.

Then, we have C claiming,

$Special relativity \Rightarrow P \rightarrow Q$.

This is independently false, as has also been pointed out multiple times. Two co-located observers disagree on where something else is unless their state of motion is the same. Here it is not, so P->Q is simple a totally false statement in SR.

Since you said no, we have B' claiming

$Special relativity \Rightarrow P \rightarrow \neg Q$.

This is called a contradiction.

Can you please show me where this is wrong?

Everywhere. Further, this has been all explained before.