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

Fredrik

Yes, it obviously doesn't matter if you observe something with your eyes or with a video camera.

You're right, but he wasn't talking about the distance in the train's frame between the points on the ground where A and B occur. That distance changes with speed as ##1/\gamma##, i.e. it gets shorter. He was talking about how the coordinate difference ##x_A-x_B## in the train's frame depends on the speed of the train. That difference changes with speed as ##\gamma##, i.e. it gets longer.

John232

In the orignal post it sounded like A and B where two seperate places outside of the train. So then if you put two objects at a location down the track they would appear to come closer together as the train moved down the track. I was trying to say that the track itself would become shorter from the frame of reference of the train while it was moveing, not longer.

Adel Makram

In this occasion, I was not talking about the distance Δx` between places where A & B occur, but about the distance between the 2 train ends A and B. Your calculation is right that the Δt`=[itex]\gamma[/itex]vd/c[itex]^{2}[/itex]. OK, so can this time difference allows a timelike separation between A and B if a signal would take off from A to reach B?
In other words, if the train observer emits a light signal from A once A happens, can that signal reaches B before or at the same time when B happens?

ghwellsjr

Adel, you have indeed discovered a special speed, but it's not really significant. Here's how I would explain it:

If you use a distance in the ground frame of 1 (using compatible units like seconds and light-seconds and where c=1 and speeds are expressed as beta which is v/c) and you set the first event at the origin (all components equal zero), then as you increase beta from 0 to almost 1, the value of the reciprocal of gamma goes from 1 to almost 0 and the absolute value of the time coordinate of the transformed event goes from 0 to almost infinity. At some particular value of beta, these other two values will become equal. This value of beta is 0.618034 and the reciprocal of gamma is 0.786151 and the time component is -0.786151.

But what does this mean? Nothing at all.

In order to compare a length of an object in one frame to its length in another frame, you have to do it when the events at both ends of the object have the same time component. Clearly, in your discovery, those two events do not have the same time component (one end is at the origin with time equal to zero and the other end has a time component of -0.786151) and so the length component, which will be equal to gamma or 1.272020, is not the correct length to use. What you have to do is pick another value of time for the second event in the original frame such that the value of the time component of transformed event is zero. This is very easy to do when you normalize the problem as I have, you just set the time component to the value of beta. So if you set the time component for the second event to 0.618034 (with x equal to 1) then the transformed time component is 0 and the distance component is 0.786151 (the reciprocal of gamma). Of course, if you wanted to use any other distance, you would multiply beta by the distance to get the correct time component to use and then the transformed distance would be the original distance divided by gamma.

So the bottom line is that in order to see what length anything is in any frame, you have to make sure the time components at both ends are equal. The corollary to this is that if you want to see what a time interval is in any frame, you have to make sure the spatial components are the same at both events. Now it should be pretty obvious that there cannot be a single pair of events for which both of these conditions are satisfied, that is, you cannot use the same two events in any frame to know both the length and time intervals in any other frame.

Adel Makram

Please refer to the attached diagram

Suppose that a light source is put near the A end of the train so as to be seen simultaneously at A & B for a ground observer. It will be definitely seen at A before B for the train observer. Now can this time difference allows a timelike separation?

http://www.physicsforums.com/attachm...1&d=1330703748

Attached Thumbnails

 

Michael C

If two events occur simultaneously in a certain frame, but not at the same location, then their separation is space-like in all frames. Neither event is in the light cone of the other event. If a light signal is emitted at event A, there is no way that it can arrive at the location of B before B happens.

In order to make sense of your example, we need to be clear as to whether we are talking about an event or the location of this event in a particular frame. By using "A" and "B" both for events and for "the 2 train ends", we end up in confusion.

nitsuj

lol too hippy, well in that context your scenario is too trippy.

the measure of length & time (dimensions) are defined by c, why do you mention the speed of sound? If you want to define "reality" as a measure of sound then I'd say yes. But that is a stupid definition.

In physics I find it is better defined as everything that is physical, from the perspective of measurement. "Fundamental Interaction" is limited by c (i assume). In "reality" spacetime is a continuum of fundamental interaction.

Adel Makram

http://www.physicsforums.com/attachm...1&d=1330705974

Here is my calculation

This can be also considered as simple derivation of Lorentz transformation up to the equation (i)

Now no worries to assign A and B to whether events or train ends as long as we are talking from the perspective of the train rest frame

the second image is more clear and accurate

http://www.physicsforums.com/attachm...1&d=1330706663

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ghwellsjr

Would you please do a little more work and type your calculations into a response instead of expecting all of us to scrutinize your attachments. It is very difficult to read what you have written and frankly, I'm not motivated. You also need to explain every detail of what you are doing. It might make sense in your mind, but remember, we can't read your mind.

Michael C

No worries? Maybe you need to think that over.

I can't decipher all that, but I get the impression that there is still confusion between these two things:

- the distance between the two ends of the train in a certain frame
- the distance between the locations where the two events occur in a certain frame.

If you could type your argument in a post, complete with explanations of the steps, it would be easier to see what is going on.

ghwellsjr

You're rehashing the same situation you brought up in Is the order of the events reversed here? and we're having the same problem in this thread that we had in that thread--you aren't precise and clear in what you are presenting and we can't tell what you are talking about.

Adel Makram

I know that I was in hurry. I would like to apologize for that,,, Sure I will take my time to write it in a more clear form before the next post
Thank you

ghwellsjr

Adel, I'm going to try to explain what time-like and space-like mean, and why if two events have one type of likeness in one frame, they have the same type of likeness in all frames. Again, I'm going to use compatible units where the value of c=1 and I'm going to align the first event at the origin. Then we can use the Lorentz transformation in a simple way to understand these terms.

First off, time-like simply means that the time coordinate is greater than the spatial coordinate. Space-like simply means that the spatial coordinate is greater than the time coordinate. Light-like simply means that both coordinates are equal.

So if we look at the simplified Lorentz transform, we see:

t' = γ(t-xβ)
x' = γ(x-tβ)

What we want to do is compare the values of these two equations for different values of β, where β can range from -1 to +1 (but not including those values). So we compare like the following where the calculation of the new time coordinate, t', is on the left and the calculation of the new space coordinate, x', is on the right (the question mark can be replaced with =, <, or > depending on our evaluation):

γ(t-xβ) ? γ(x-tβ)

Now we note that γ is a function of β but for all values of β, γ is a positive number equal to 1 or greater than 1:

γ = 1/√(1-β²)

Therefore, we can remove γ from our conditional statement without effecting the evaluation of the condition:

t-xβ ? x-tβ

Next we rearrange the terms so that the t factors are on the left and the x factors are on the right:

t+tβ ? x+xβ

Now we factor out the common terms:

t(1+β) ? x(1+β)

Now since β can have a range of -1 to +1 but not including those numbers, the factor 1+β has a range of 0 to 2, not including those nuimbers.

Therefore we can divide out the (1+β) factor without changing the evaluation of the condition:

t ? x

What does this mean? It means what everyone has been telling you:

If t>x in one frame, t'>x' in any other frame, no matter what the value of β is.
If t=x in one frame, t'=x' in any other frame, no matter what the value of β is.
If t<x in one frame, t'<x' in any other frame, no matter what the value of β is.

So no amount of coming up with specific scenarios, especially ill-defined ones, can produce an example that will violate the conclusion that whatever likeness two events have in one frame, they have the same likeness in any other frame.

It's kind of a waste of time to try to decipher your scenarios and point out where your confusion is so that you will accept our critiques of what you are doing. My single most important advice to you is to learn what an event is, how to define a scenario in ONE frame, not two like you have been doing, and then use the Lorentz Transform to see what that scenario looks like in another frame.

John232

Someone could take two particles that are entangled, and then change the spin of one of them, and the other particle would change its spin instantly faster than the speed of light. Then your "reality" would depend on how fast of a transmission you observed it with. Someone observing it with a string of thread would see it differently than someone with a camera, or even with someone with an entanglement experiment.

Michael C

Adel, I'm going to stick to the train frame and give some names to things so we can be clear about what we are discussing:

IN THE TRAIN FRAME
A and B are the two ends of the train.
C is the point somewhere between them where there is a light source.
p is the event "a light signal is emitted at C".
q is the event "the light signal from C reaches A".
r is the event "the light signal from C reaches B".

What you seem to be claiming is that it could be possible to send a light signal from A after event q that will reach B before event r. Am I correct?

ghwellsjr

Are you doing this for his original scenario in post #1 or for the one in post #22? Or are they supposed to be the same or related in some way?

Adel Makram

Thank u for your care to clarify things for me. But my scenario is different from just assigning events to different coordinates in different frames. I posted here a clear detailed diagram and calculation based on my idea. Even from the diagram it is clear that A and B are still spacelike for the train observer because the source is put near to A than B which means that Signal must reaches B in no later than a signal would have taken to reach B coming from A. But how about the math! I see no contradictory to insert ∆t``= (AB`)/c and equating it with ∆t`= (AB`v/c2)/(1-(v2/c2)) can you? In other words, if A and B are spacelike for the train observer, why the math of LT allows the time difference between receiving signal at both ends to equal the time difference a light signal would take to reaches B from A?

Attached Files

Let.pdf (516.1 KB, 8 views)