Real life examples of simultaneity

[goodabouthood]

http://en.wikipedia.org/wiki/File:Relativity_of_Simultaneity.svg

Take that image for example. It says the red reference frame is moving at .28c relative to green reference frame. Now why goes the graph take on that shape for the red reference frame?

 

[BruceW]

These graphs show us how the position and time of some event are interchangeable depending on the reference frame used.

In the original (green) reference frame, an event might be given as (x,t) and in the new (red) reference frame, the same event might be denoted by (x',t'). So what we're saying is that the spatial coordinate and the time for an event are not absolute. Each can change depending on the speed of the reference frame used.

But one thing that is absolute is the quantity x^2 - c^2t^2. In other words, x^2 - c^2t^2 = x'^2 - c^2t'^2. (This is true because we are not changing the origin of the reference frame). The shape that the red graph takes is such that this quantity is the same when calculated by either graph.

 

[goodabouthood]

These graphs show us how the position and time of some event are interchangeable depending on the reference frame used.

In the original (green) reference frame, an event might be given as (x,t) and in the new (red) reference frame, the same event might be denoted by (x',t'). So what we're saying is that the spatial coordinate and the time for an event are not absolute. Each can change depending on the speed of the reference frame used.

But one thing that is absolute is the quantity x^2 - c^2t^2. In other words, x^2 - c^2t^2 = x'^2 - c^2t'^2. (This is true because we are not changing the origin of the reference frame). The shape that the red graph takes is such that this quantity is the same when calculated by either graph.

Well now I am understanding what it means for events to be simultaneous but I am still having a bit of trouble understanding exactly why this happens.

I am trying to visualize in my head what each frame of reference would look like for the train example.

I'm finding lately that sometimes I feel I understand it and other times I am losing it. I do understand the concept that events happen at different times for different reference frames I just have a hard time understanding why this is.

I know it has to do with the speed of light and with the movement of one reference train relative to another I just haven't pieced it all together in my mind.

 

[goodabouthood]

I am pretty much saying I would like to see a visual reference of both the train and observers coordinates.

 

[bobc2]

I am pretty much saying I would like to see a visual reference of both the train and observers coordinates.

Here are the coordinates again for both observers. The black coordinates are the rest frame of the guy on the platform and the blue coordinates are the frame of reference for the guy sitting in the center of the train passenger car. Circles mark different events of interest.

The sketch in the lower right corner shows just black and blue coordinates without the clutter of the other details. It makes it clear how the blue coordinate grid lines are skewed.

 

[bobc2]

Thanks.

I am still wondering why the other coordinate system takes on the position it does. I see as the coordinate system approaches the speed of light it closes in on each other I just don't understand why.

Historically, it was just discovered (theoretically with Maxwell's equations and experimentally with Michelson-Morely experiments) that the speed of light was the same for all observers, no matter what their velocity. It turned out that the only way that could be possible was if the coodinates for different observers moving at different velocities relative to some reference system are rotated in the manner we've been showing in the space-time diagrams. And along with this was the discovery that the laws of physics are the same in all of these different coordinate systems. All of this is of course what special relativity is about.

So this peculiar way nature has of orienting the different 3-D cross-section views of the universe for different observers is something that physics has discovered. So far, no one has come up with a reason for this. Some may maintain that the coordinates are that way in order to make the speed of light constant for everyone and so that the laws of physics would be the same for everyone. That sounds kind of like mother nature said, "I think I'll make a 4-dimensional universe populated by 4-dimensional objects in a way that guarantees a constant speed of light and a set of uniform laws of physics." No one really knows why the universe is the way it is.

It also would help to see the numbers for both coordinate systems.

I will show you those numbers in my next post. But, first you should try to understand the sketches below. I am using high school level algebra to derive the Lorentz transformation equation that is used to compute time dilation.

It is very useful to use what we call a symmetric space-time diagram. You have two guys (red and blue) flying off in their rockets at relativistic speeds in opposite directions. They are moving with the same speeds, one to the left and one to the right with reference to the usual black rest coordinates. By having them going at the same speeds, you can easily compare the distances and times along each of their respective axes (to compare numerical times and distances with black you must use hyperbolic calibration curves--which we will do in a later post).

And by the way, you can always turn a problem for two observers moving with respect to each other into a symmetric diagram of this type. Just add in a rest system whose X4 axis bisects the angle between the two moving observers and display the space-time diagram as shown here.

Note below that when blue is at station 9 (you could label it as either a time or a distance along X4) he is in the 3-D cross-section of the 4-D universe that includes the red guy at station 8. And when the red guy is at station 9, the blue guy in his (red's) simultaneous 3-D space at station 8. (How can the red guy and blue guy each be at two different places at the same time?)

 

[goodabouthood]

I really appreciate the diagrams you are making. I can now see that the act of when the lighting hits and when the observer sees it are two different events in the frame of reference. Am I correct in saying that?

Also in regards to the last diagram you posted, is there anyway you could show me an example with actual numbers filled in for the variables in the equations? I think that would help me a lot.

Thanks.

 

[bobc2]

I really appreciate the diagrams you are making. I can now see that the act of when the lighting hits and when the observer sees it are two different events in the frame of reference. Am I correct in saying that?

You certainly are.

Also in regards to the last diagram you posted, is there anyway you could show me an example with actual numbers filled in for the variables in the equations? I think that would help me a lot.

You want numbers for the train example?

 

[BruceW]

I really appreciate the diagrams you are making. I can now see that the act of when the lighting hits and when the observer sees it are two different events in the frame of reference. Am I correct in saying that?
...
I am trying to visualize in my head what each frame of reference would look like for the train example.

Yes! The lightning hitting ground and observer seeing the light from that strike are two different events.

For the train example, the only difference between the frames of reference is the relative speed of the two frames of reference. So the origin and coordinate axes are the same in both cases, but the frames have a relative motion. One frame measures the two lightning strikes to have happened at the same time in different places, so the other frame must measure the strikes to have happened at different times.

This may seem weird because the origin and axes of the two reference frames are identical, but in special relativity, the relative speed of the reference frames also decides where/when events happen.

 

[bobc2]

Yes! The lightning hitting ground and observer seeing the light from that strike are two different events.

For the train example, the only difference between the frames of reference is the relative speed of the two frames of reference. So the origin and coordinate axes are the same in both cases, but the frames have a relative motion. One frame measures the two lightning strikes to have happened at the same time in different places, so the other frame must measure the strikes to have happened at different times.

This may seem weird because the origin and axes of the two reference frames are identical, but in special relativity, the relative speed of the reference frames also decides where/when events happen.

Nice observations, BruceW. Thanks.

 

[goodabouthood]

It would be nice to see what the equations you posted in your last diagram look like with actual numbers.

 

[goodabouthood]

I have a couple more questions as well.

What is meant when you say everyone travels at the speed of light on the time dimension?

Also, let us go back to the train example. I understand that the events of the lightning bolts take place at different times for each frame of reference but I still can't exactly understand why this happens.

Let say the observer on the platform sees them simultaneously at t=0. Now let say that the train passing is going at .5c. Would that mean the lightning bolt will hit a half second latter in the train's frame of reference?

The thing that is bothering me in my head is that I imagine the train situation and I imagine the observer on the platform and the observer in the train passing each other when the bolts hit simultaneously for the observer on the platform. What I don't understand is why they can't hit simultaneously for the train FoR. I know it's because the train of reference is in motion I just don't understand why it happens.

Thanks.

 

[LaurieAG]

I can now see that the act of when the lighting hits and when the observer sees it are two different events in the frame of reference. Am I correct in saying that?

I was on the phone to a friend during a thunderstorm and there was a huge flash of lightning. We both said wow at the same time and, as I was only 2 miles away from the strike and my friend was 7 miles away, we both heard the bang at my location after 2 seconds or so and then we both heard the same bang at his location after another 5 seconds. As we were both on landlines we decided to continue our conversation after the storm had passed.

I went outside approximately 5 minutes later, the air was still and thick with humidity, and took several photos from different angles of what looked like smoke coming from the direction of the strike. The smoke was actually coming from a chimney but the photos revealed a dark straight line leading to the tip of the chimney in both photos.

 

[BruceW]

What is meant when you say everyone travels at the speed of light on the time dimension?

The velocity of an object through spacetime is called its four-velocity. The equivalent magnitude of this vector is equal to the speed of light for all objects with mass.
If we take a reference frame fixed to the earth, then all humans are moving mostly in one direction through spacetime. If we call this direction the 'time dimension', then clearly, all humans are moving at nearly the speed of light through the 'time dimension'.

Let say the observer on the platform sees them simultaneously at t=0. Now let say that the train passing is going at .5c. Would that mean the lightning bolt will hit a half second latter in the train's frame of reference?

If the train is going at .5c, then the time difference between the events according to the train's FoR is:
\frac{\sqrt{3}}{3c} \Delta x
Where \Delta x is the distance between the two strikes according to the platform's FoR. So clearly, it also depends on how far apart the two events were.

The thing that is bothering me in my head is that I imagine the train situation and I imagine the observer on the platform and the observer in the train passing each other when the bolts hit simultaneously for the observer on the platform. What I don't understand is why they can't hit simultaneously for the train FoR. I know it's because the train of reference is in motion I just don't understand why it happens.

From the platform's FoR, for the beams of light from each strike to reach the train at the same time, one strike would have to happen earlier, so its light could 'catch up' with the train. So if the events were simultaneous for the train's FoR, they can't be simultaneous for the platform's FoR and vice versa.

To be honest, making sense of it is the tricky part. The easy part is simply accepting that the laws of special relativity are better than Newton's laws, and then using special relativity to make predictions instead of using Newton's mechanics.

 

[bobc2]

It would be nice to see what the equations you posted in your last diagram look like with actual numbers.

I started to post another sketch with the numbers. But then I remembered that ghwellsjr has already done a nice job of presenting calculations in his posts #54 and #75. His post #75 is a particularly detailed discussion using numbers he calculated directly from the Lorentz transformations. You'll find his post #75 example on page 5. (good job, ghwellsjr)

 

[BruceW]

Yep. His post #75 is very good. Lorentz transforms are one of the first things to learn when you start learning special relativity. So its one of the things I recommend to start practising to get familiar with if you want to learn special relativity. My advice: As soon as you are given (or think up) some problem, make sure you define the events and what reference frame each of those events are given in.

goodabouthood, I can't remember what was said earlier in the thread, but are you learning relativity from online, or a book, or a teacher?

 

[BruceW]

From the platform's FoR, for the beams of light from each strike to reach the train at the same time, one strike would have to happen earlier, so its light could 'catch up' with the train. So if the events were simultaneous for the train's FoR, they can't be simultaneous for the platform's FoR and vice versa.

I have no idea what I was talking about here. Please ignore this bit. (Just goes to show my point about the 'meaning' being the difficult part).

Let me try again: Let's say the lightning strikes occur at both ends of the train simultaneously according to the platform's FoR. Someone on the platform would see the train moving to the right and the beams of light traveling toward each other at the same speed. So according to the person on the platform, the beams of light would meet somewhere in the left half of the train, not the centre of the train. (since the light doesn't care about the motion of the train around it).

Now in the train's FoR, the two beams of light must meet somewhere in the left half of the train. (Because that is what we found when we used the platform's FoR). And also we know that the lightning strikes occurred at both ends of the train. The people in the train know that light travels at the same speed from both ends of the train, so they must conclude that the strike on the right happened before the strike on the left, since the two beams of light meet in the left half of the train.

So that's an explanation using just the assumption that the speed of light is the same in either reference frame. Clearly, the explanation takes longer than simply using the Lorentz transforms, and in complicated situations, it gets even more important to use Lorentz transforms rather than explanations.

In other words, the principle of special relativity is that the speed of light is the same in all reference frames. And from this, we get the mathematics of special relativity. The mathematics is often easier to use rather than using the principle directly, which is why we often use the maths.

 

[goodabouthood]

Fair Enough.

I do need to brush up on my math to learn some of this stuff.

The thing that was getting to me was this:

The guy on the platform sees the bolts simultaneously. The guy on the train sees the front one strike first and the rear one strike later.

Now they both see the front one at the same time. In my head I try to imagine where the rear bolt is for the train at the time the platform FoR sees it. I guess that event just hasn't happened in the train's FoR yet. It's hard to picture that it's happening in one FoR but not the other.

Maybe I can make a picture in paint or something to better illustrate what I am saying.

 

[BruceW]

The guy on the platform sees the bolts simultaneously. The guy on the train sees the front one strike first and the rear one strike later.

Now they both see the front one at the same time.

These three statements cannot each be true if we are talking about one situation. In my example, the two people do not see the light from the front strike at the same time, because they happened to pass by each other at the time the strikes happened (according to the platform's FoR), not at the time when they saw the light from the strikes.

 

[goodabouthood]

I meant that the front strike happens at the same time in both FoR.

 

[ghwellsjr]

I meant that the front strike happens at the same time in both FoR.

The only way that could happen is if you set the origin of both Frames of Reference to be the event of the front strike.

EDIT: Actually, this isn't even technically correct. Simultaneity has to do with two events separated in space but having the same time coordinate as defined according to a single Frame of Reference. So talking about something happening at the same time in two different Frames of Reference really doesn't make any sense except for the fact that the origins of two Frames of Reference are locally coincident but it's misleading and shows misunderstanding to try to connect simultaneity between two Frames of Reference.

 

[BruceW]

I meant that the front strike happens at the same time in both FoR.

This can be done by specifying that the origins of the two reference frames coincide at the event of the front strike. It doesn't change any of the physics of the problem, since only the difference between events is actually important.

So we get the same answers as in my example before. (I just went through the maths to convince myself). It is tricky to get your head around at first (and it still takes me a while too). I would advise only imagining one frame of reference at a time. You can't really imagine them both as happening because they are two alternative (but equally valid) representations of the universe.

 

[goodabouthood]

Now we are ready to express our two events. Normally, I would use the nomenclature of [t,x,y,z] but since we have agreed to assign zeroes to y an z, I will use the shorthand nomenclature of [t,x]. So here are our two events for the lightning bolts (E1 is in front, E2 is behind:

E1=[0,+500]
E2=[0,-500]

The fact that they both have the same t coordinate means that they are simultaneous.

Now let's define the train Frame of Reference. In order to make things simple, we want to use the standard form so that we can easily use the Lorentz Transform and that means we want to use the same axis directions and units for distance and time and we want their origins to coincide. We will place the origin of the train at its midpoint.

Now we are ready to use the Lorentz Transform. We will use units such that the speed of light equals 1 which means that we are using nanoseconds for time and light nanoseconds (which equal one foot) for distances.

First we have to calculate gamma, γ, from this formula:
γ = 1/√(1-β²)
For β=0.6,
γ=1/√(1-0.6²)
γ=1/√(1-0.36)
γ=1/√(0.64)
γ=1/0.8
γ=1.25

Now the Lorentz Transform has two formulas, one for calculating the new t' coordinate and one for calculating the new x' coordinate from the old t and x coordinates. Here they are:
t'=γ(t-βx)
x'=γ(x-vt)

Since we are only interested in the time coordinate, we will do that calculation for each of our two events here:

t1'=1.25(0-0.6*500)
t1'=1.25(300)
t1'=375

t2'=1.25(0-0.6*-500)
t2'=1.25(-300)
t2'=-375

We can see right away that these two time coordinates are different so the events they go with are not simultaneous. In fact, as a sanity check, we can calculate the difference between them as 750 nanoseconds which is the same value we calculated in post #54 where we used BruceW's shortcut formulat and got 0.75 microseconds.

Show me how you would solve for x'=γ(x-vt) in this situation. Thanks.

 

[ghwellsjr]

I made a mistake here, the v should be β. Sorry about that. So the corrected formula is:

x'=γ(x-βt)

x1'=1.25(500-0.6*0)
x1'=1.25(500-0)
x1'=1.25(500)
x1'=625

x2'=1.25(-500-0.6*0)
x2'=1.25(-500-0)
x2'=1.25(-500)
x2'=-625

But now I see that I also screwed up the calculations for the times. They should be:

t1'=1.25(0-0.6*500)
t1'=1.25(-300)
t1'=-375

t2'=1.25(0-0.6*-500)
t2'=1.25(300)
t2'=375

So the two events in the train's rest frame are:

E1'=[-375,+625]
E2'=[+375,-625]

Thanks for asking about this, I hate to have lingering mistakes.

But just to make up for my screwups, I'll go into some more explanation:

The event of the front lightning strike in the train frame is [-375,+625]. Now this is not the time at which the train observer sees the lightning strike because it is located 625 feet or 625 light nanoseconds in front of him. Remember that he and the train are stationary in the train frame. So it will take 625 nanoseconds for the light to travel from the front of the train to his location which we set up to be at x=0 which means he will see it at 625-375 or 250 nanoseconds. This would be event [250,0] in the train frame.

Now the rear lightning strike also takes 625 nanoseconds to reach him but it occurred at +375 so he will see it at 625+375 or 1000 nanoseconds. This would be event [1000,0]. And the difference between these two events is also 750 nanoseconds. This is the same difference in the time components of the lightning strike events only because the train observer is midway between the two events.

 

[ghwellsjr]

Let's call the event in the train frame when the train observer sees the front flash E3' and the event when he sees the rear flash E4':

E3'=[250,0]
E4'=[1000,0]

Let's transform these into the ground frame. To do this, we must realize that β is now -0.6 because to the stationary train, the ground is going backwards. And we interchange the primed and unprimed terms:

t=γ(t'-βx')
x=γ(x'-βt')

t3=1.25(250-(-0.6*0))
t3=1.25(250-0)
t3=1.25(250)
t3=312.5

x3=1.25(0-(-0.6*250))
x3=1.25(0-(-150))
x3=1.25(150)
x3=187.5

E3=[312.5,187.5]

t4=1.25(1000-(-0.6*0))
t4=1.25(1000-0)
t4=1.25(1000)
t4=1250

x4=1.25(0-(-0.6*1000))
x4=1.25(0-(-600))
x4=1.25(600)
x4=750

E4=[1250,750]

So this means that in the ground frame, the train observer will be at location 187.5 at time 312.5 when he sees the front flash and he will be at location 750 at time 1250 when he sees the rear flash. Let's see if this makes sense.

Let's track what happens with regard to the front flash. At time 0, the front of the train is at 500 where the lightning strikes and both observers are at location 0. At time 312.5 the train observer sees the flash at location 187.5. Does this make sense? Well, the train is traveling at 0.6 so if we multiply the time 312.5 by 0.6, we do indeed get 187.5. In the meantime, the light has progressed from location 500 to location 187.5, a distance of 500-187.5=312.5 in 312.5 nanoseconds. So that works out, too.

Now let's do the same thing for the rear flash. At time 0, the rear of the train is at -500 where the lightning strikes and both observers are at location 0. At time 1250 the train observer sees the flash at location 750. We multiply the time 1250 by 0.6 and we get 750. The light has progressed from location -500 to location 750 in 1250 nanoseconds.

For completeness sake, we should note that the ground observer sees both flashes at time 500, because they occurred 500 feet in front of and behind him at time 0.

So here's a timeline of the front flash:

At time 0, flash occurs at 500 feet.
At time 312.5, train observer sees flash at 187.5 feet.
At time 500, ground observer sees flash at 0 feet.

And the timeline for the rear flash:

At time 0, flash occurs at -500 feet.
At time 500, ground observer sees flash at 0 feet.
At time 1250, train observer sees flash at 750 feet.

 

[BruceW]

This looks good to me.

You've got the spatial origin of each reference frame to coincide with a person, which makes calculating what each person 'sees' less complicated.

We could also have the origin to be different to the people, but this would give the same answers and would require more writing to calculate.