goodabouthood
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It would be nice to see what the equations you posted in your last diagram look like with actual numbers.
goodabouthood said: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?
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.goodabouthood said:What is meant when you say everyone travels at the speed of light on the time dimension?
If the train is going at .5c, then the time difference between the events according to the train's FoR is:goodabouthood said: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?
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.goodabouthood said: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.
goodabouthood said:It would be nice to see what the equations you posted in your last diagram look like with actual numbers.
BruceW said: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.
goodabouthood said: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.
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.goodabouthood said: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.goodabouthood said:I meant that the front strike happens at the same time in both FoR.
ghwellsjr said: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.