Real life examples of simultaneity

[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.