News Flash - Non Simultaneity occurs in both Galilean and Special Relativity - Deepseek
Gehan Ameresekere
Train Observer:
Both frameworks predict simultaneity, but Galilean relativity does so through ignorance of motion, while special relativity does so through light’s invariance.Platform Observer:
Non-simultaneity arises in both frameworks, but for fundamentally different reasons:Galilean: Variable light speed ().
Special Relativity: Invariant light speed () and spacetime relativity.
Okay, so I need to understand this example from Wikipedia about the train car and the two observers. Let me start by breaking it down step by step.
First, there's a traincar, and a flash of light is emitted right in the center of it. There are two observers: one on the train and another on the platform. The key point here is that each observer sees the events differently because of their relative motion.
From the train observer's perspective, the traincar isn't moving. The light flash happens in the middle, so the distance to the front and back of the traincar is the same. Since the speed of light is constant (as per special relativity's second postulate), the light should reach both ends at the same time. That makes sense because in the train's rest frame, there's no movement affecting the distance the light needs to travel. So, the observer on the train sees simultaneous events—light hitting both ends together.
Now, the platform observer's view is different. The traincar is moving relative to them. Let's say the train is moving to the right. When the flash occurs at the center, the back of the traincar (left end) is moving towards the point where the flash happened, and the front (right end) is moving away. But the speed of light is still the same in all directions for this observer too. Wait, how does that work?
If the train is moving, then from the platform, the light heading towards the back has less distance to cover because the back is moving towards the light. Conversely, the light heading to the front has to catch up with the front, which is moving away. Since the speed of light is constant, the time it takes for the light to reach each end should be different. The light should reach the back first and then the front. So, the platform observer sees the events as not simultaneous.
But wait, how exactly does the math work here? Let me try to visualize it. Suppose the length of the traincar is L in its rest frame. For the train observer, the light needs to travel L/2 to each end. Time taken would be (L/2)/c in each direction. So, same time.
For the platform observer, the traincar is moving at speed v. Due to length contraction, the traincar's length would be L*sqrt(1 - v²/c²). But does length contraction affect this scenario? Wait, the length contraction is from the platform's perspective. So, the traincar is shorter in their frame. However, when the light is emitted, the traincar is moving, so the back is moving towards the emission point, and the front is moving away.
But here's where I might get confused. Let me set up some variables. Let's denote:
- In the train's frame (S): the traincar has length L. The light is emitted at (x=0, t=0). The ends are at x = L/2 (front) and x = -L/2 (back). The light reaches both ends at time t = (L/2)/c.
In the platform frame (S'), moving at velocity v relative to S (so the train is moving at velocity v in S'). The emission event is at (x'=0, t'=0) if we synchronize the origins.
Now, in S', the traincar is length-contracted to L' = L*sqrt(1 - v²/c²). But the emission happens at the center of the traincar, which is moving. So, when the light is emitted, the front of the traincar is at x' = L'/2, and the back is at x' = -L'/2. Wait, no, actually, in S', at t'=0, the center of the traincar is at x'=0. Since the traincar is moving at speed v, the front is at x' = (L/2)/gamma + vt', but wait, at t'=0, the length is L' = L/gamma. So, the front is at x' = L' / 2 = L/(2 gamma) and the back at x' = -L/(2 gamma) at t'=0.
But as time progresses in S', both ends of the traincar are moving at velocity v. So, the position of the front at time t' is x'_front = L/(2 gamma) + v t'. Similarly, the back is at x'_back = -L/(2 gamma) + v t'.
The light emitted from x'=0 at t'=0 travels at speed c in both directions. So, the light going to the back (left) has to reach the back which is moving towards the left, but the traincar is moving to the right. Wait, no. If the traincar is moving to the right (positive x'), then the back is moving to the right as well. But the light going towards the back is moving to the left (negative x' direction). Wait, maybe I need to clarify directions.
Assume the traincar is moving along the positive x' axis (to the right). So, in S', the traincar's front is moving to the right, and the back is also moving to the right, but the traincar's length is contracted.
At t'=0, the flash occurs at x'=0. The light going towards the front (right direction) has to catch up with the front, which is moving away. The light going towards the back (left direction) has to meet the back, which is moving towards the right. Wait, that seems contradictory. Wait, if the traincar is moving to the right, then the back is also moving to the right. So, from the platform's perspective, after the flash, the back isn't moving towards the flash's origin (which was at x'=0, t'=0), but rather the entire traincar is moving to the right. So, the back is moving away from the origin to the right, and the front is moving even further to the right. Wait, but that can't be. Wait, no. If the traincar is moving to the right, then at t'=0, the center is at x'=0. The front is at x'=L/(2 gamma) and the back at x'=-L/(2 gamma). Then, as time increases, each end moves to the right at speed v. So, at time t', front is at x'= L/(2 gamma) + v t', and back is at x'= -L/(2 gamma) + v t'.
Now, the light going to the front is moving at speed c in the positive x' direction. Its position is x'_light_front = c t'. The front of the traincar is moving at x'_front = L/(2 gamma) + v t'. The light catches up when c t' = L/(2 gamma) + v t'. Solving for t': t'(c - v) = L/(2 gamma) => t' = L/(2 gamma (c - v)).
Similarly, the light going to the back is moving at speed -c (since it's going left), so its position is x'_light_back = -c t'. The back of the traincar is moving to the right at x'_back = -L/(2 gamma) + v t'. They meet when -c t' = -L/(2 gamma) + v t'. Rearranging: -c t' - v t' = -L/(2 gamma) => -t'(c + v) = -L/(2 gamma) => t' = L/(2 gamma (c + v)).
Therefore, the time for the light to reach the front is t'_front = L/(2 gamma (c - v)) and the time to reach the back is t'_back = L/(2 gamma (c + v)).
Since gamma = 1/sqrt(1 - v²/c²), we can write this as t'_front = L sqrt(1 - v²/c²)/(2 (c - v)) and t'_back = L sqrt(1 - v²/c²)/(2 (c + v)).
But let's see if these times are different. Let's compute the ratio t'_front / t'_back = [ (c + v) ] / [ (c - v) ] > 1, so t'_front > t'_back. Therefore, the light reaches the back first, then the front, as per the platform observer.
But how does this relate to the train observer seeing them simultaneously? Because in the train's frame, the events are simultaneous, but in the platform's frame, they are not. This is an example of relativity of simultaneity.
But wait, maybe there's another way to think about this. Since in special relativity, events that are simultaneous in one frame are not in another if they are spatially separated. Here, the arrivals of the light at the two ends are spatially separated events. In the train frame, they are simultaneous, but in the platform frame, they are not.
Another point is that the platform observer sees the traincar as length-contracted. So, the distance the light needs to travel might seem shorter, but because the traincar is moving, the actual distance the light travels is affected by both the contraction and the motion of the ends.
But perhaps there's a simpler way to calculate this. Let me consider the relative speed. For the light going to the front, the traincar is moving away at speed v, so the relative speed between the light and the front is (c - v). Wait, no. In the platform frame, the light is moving at c, and the front is moving at v. So, the closing speed is (c - v), so the time would be the distance between them divided by (c - v). Similarly, for the back, the light is moving towards it at c, and the back is moving towards the light at v, so the closing speed is (c + v), hence time is distance divided by (c + v).
But what's the distance? At t'=0, the front is at L/(2 gamma) and the back at -L/(2 gamma). But since the traincar is moving, during the time t' it takes for the light to reach the front, the front moves an additional v t'. So, the total distance the light needs to cover is L/(2 gamma) + v t'_front. Similarly, for the back, the distance is L/(2 gamma) - v t'_back (since the back is moving towards the light). Wait, but in the case of the back, the back is moving in the same direction as the traincar's motion, but the light is going opposite. Hmm, maybe my earlier equations are better.
Alternatively, let's use Lorentz transformations to check. In the train frame (S), the events of light reaching front and back are at (x = L/2, t = L/(2c)) and (x = -L/2, t = L/(2c)). Using Lorentz transformations, we can find the coordinates in S' (platform frame).
The Lorentz transformation equations are:
t' = gamma (t - v x / c²)
x' = gamma (x - v t)
For the front event (x = L/2, t = L/(2c)):
t'_front = gamma (L/(2c) - v (L/2)/c²) = gamma (L/(2c))(1 - v/c)
x'_front = gamma (L/2 - v * L/(2c)) = gamma L/2 (1 - v/c)
Similarly, for the back event (x = -L/2, t = L/(2c)):
t'_back = gamma (L/(2c) - v (-L/2)/c²) = gamma (L/(2c))(1 + v/c)
x'_back = gamma (-L/2 - v * L/(2c)) = gamma (-L/2 - v L/(2c)) = gamma (-L/2 (1 + v/c))
So, t'_front = gamma L/(2c) (1 - v/c) and t'_back = gamma L/(2c) (1 + v/c). Therefore, t'_front ≠ t'_back.
Calculating gamma:
gamma = 1 / sqrt(1 - v²/c²)
So, substituting gamma into t'_front:
t'_front = [1 / sqrt(1 - v²/c²)] * L/(2c) (1 - v/c)
Similarly, t'_back = [1 / sqrt(1 - v²/c²)] * L/(2c) (1 + v/c)
Simplifying t'_front:
Multiply numerator and denominator by c:
t'_front = L/(2c) * [ (1 - v/c) / sqrt(1 - v²/c²) ]
Note that (1 - v/c) / sqrt(1 - v²/c²) = sqrt( (1 - v/c)^2 / (1 - v²/c²) ) = sqrt( (1 - 2v/c + v²/c²) / (1 - v²/c²) )). Hmm, maybe not helpful. Alternatively, factor numerator and denominator:
1 - v²/c² = (1 - v/c)(1 + v/c). So sqrt(1 - v²/c²) = sqrt( (1 - v/c)(1 + v/c) )
Thus, [ (1 - v/c) / sqrt( (1 - v/c)(1 + v/c) ) ] = sqrt( (1 - v/c) / (1 + v/c) )
Therefore, t'_front = L/(2c) * sqrt( (1 - v/c) / (1 + v/c) )
Similarly, t'_back = L/(2c) * sqrt( (1 + v/c) / (1 - v/c) )
This shows that t'_front < t'_back if we consider sqrt terms. Wait, no. Wait, if v is positive, then (1 - v/c) < (1 + v/c), so sqrt( (1 - v/c)/(1 + v/c) ) < 1, and sqrt( (1 + v/c)/(1 - v/c) ) > 1. Therefore, t'_front = (L/(2c)) * something < 1, and t'_back = (L/(2c)) * something > 1. But in the earlier calculation using the platform frame, t'_front = L/(2 gamma (c - v)) and t'_back = L/(2 gamma (c + v)). Let's see if these are the same.
Expressing t'_front from Lorentz transformation:
t'_front = gamma L/(2c) (1 - v/c) = [1 / sqrt(1 - v²/c²)] * L/(2c) (1 - v/c)
Multiply numerator and denominator by (1 + v/c):
= [1 / sqrt( (1 - v/c)(1 + v/c) ) ] * L/(2c) (1 - v/c)
= sqrt( (1 - v/c)/(1 + v/c) ) * L/(2c)
Similarly, t'_front = L/(2c) * sqrt( (1 - v/c)/(1 + v/c) )
But let's compute t'_front from the platform calculation:
t'_front = L/(2 gamma (c - v))
gamma = 1/sqrt(1 - v²/c²)
So, t'_front = L/(2 * [1/sqrt(1 - v²/c²)] * (c - v)) ) = L sqrt(1 - v²/c²)/(2 (c - v))
Let's manipulate this expression:
sqrt(1 - v²/c²) = sqrt( (c² - v²)/c² ) = sqrt( (c - v)(c + v) ) / c
Thus, t'_front = L * sqrt( (c - v)(c + v) ) / c / (2 (c - v)) )
Simplify:
= L * sqrt(c + v) / (2 c sqrt(c - v)) )
= L/(2c) * sqrt( (c + v)/(c - v) )
Wait, but earlier from Lorentz transformation, we had t'_front = L/(2c) * sqrt( (1 - v/c)/(1 + v/c) )
But note that sqrt( (1 - v/c)/(1 + v/c) ) = sqrt( (c - v)/ (c + v) ) = 1 / sqrt( (c + v)/(c - v) )
So there seems to be a discrepancy. Wait, perhaps I made an error.
Wait, sqrt( (1 - v/c)/(1 + v/c) ) = sqrt( (c - v)/(c + v) )
So, from Lorentz transformation, t'_front = L/(2c) * sqrt( (c - v)/(c + v) )
From the platform calculation, t'_front = L/(2c) * sqrt( (c + v)/(c - v) )
Wait, that can't be. There must be a mistake here. Let me check again.
Hold on, when I did the platform frame calculation, I had:
t'_front = L/(2 gamma (c - v))
But gamma = 1/sqrt(1 - v²/c²)
So, substituting gamma:
t'_front = L sqrt(1 - v²/c²)/(2 (c - v))
Now, sqrt(1 - v²/c²) = sqrt( (c² - v²)/c² ) = sqrt( (c - v)(c + v) ) / c
Therefore,
t'_front = L * sqrt( (c - v)(c + v) ) / c / (2 (c - v)) )
= L * sqrt(c + v) / (2 c sqrt(c - v)) )
= L/(2c) * sqrt( (c + v)/(c - v) )
Which is the same as the Lorentz transformation result? Wait, no. Wait, from Lorentz, t'_front was L/(2c) * sqrt( (c - v)/(c + v) )
But here, platform calculation gives L/(2c) * sqrt( (c + v)/(c - v) )
These are reciprocals. That suggests a mistake. Which one is correct?
Wait, let's take a numerical example. Let L = 2c units, v = 0.5c.
Then, gamma = 1/sqrt(1 - 0.25) = 1/sqrt(0.75) ≈ 1.1547
From platform calculation, t'_front = L sqrt(1 - v²/c²)/(2 (c - v))
= 2c * sqrt(1 - 0.25)/ (2 (c - 0.5c)) = 2c * sqrt(0.75)/ (2 * 0.5c) ) = 2c * 0.866 / (c) ) ≈ 1.732
From Lorentz transformation:
t'_front = gamma L/(2c) (1 - v/c) = 1.1547 * 2c/(2c) * (1 - 0.5) = 1.1547 * 1 * 0.5 ≈ 0.577
But wait, these are different. That can't be. There must be an error in one of the derivations.
Wait, perhaps the platform calculation was incorrect. Let me go back.
In the platform frame, at t'=0, the front of the traincar is at x' = L/(2 gamma). Then, the front moves at v, so position at time t' is x'_front = L/(2 gamma) + v t'
The light emitted at x'=0, t'=0 moves at c, so position at time t' is x'_light = c t'
They meet when c t' = L/(2 gamma) + v t'
=> t' (c - v) = L/(2 gamma)
=> t' = L/(2 gamma (c - v))
Similarly, for the back:
x'_back = -L/(2 gamma) + v t'
x'_light_back = -c t'
Set equal: -c t' = -L/(2 gamma) + v t'
=> -c t' - v t' = -L/(2 gamma)
=> t' (-c - v) = -L/(2 gamma)
=> t' = L/(2 gamma (c + v))
So, with L=2c, v=0.5c:
gamma = 1/sqrt(1 - 0.25) ≈ 1.1547
t'_front = 2c/(2 * 1.1547 * (c - 0.5c)) = 2c/(2 * 1.1547 * 0.5c) = 2c/(1.1547c) ≈ 1.732/1.1547 ≈ 1.5 seconds?
Wait, 2c divided by 1.1547c is approximately 1.732. But 1.1547 * 0.5c is 0.577c, so denominator is 2 * 1.1547 * 0.5c = 1.1547c. Therefore, t'_front = 2c / 1.1547c ≈ 1.732.
But from Lorentz transformation, with L=2c, v=0.5c:
t'_front = gamma * (L/(2c) - v L/(2c²)) = gamma * ( (2c)/(2c) - 0.5c * 2c / (2c²) ) = gamma * (1 - 0.5) = gamma * 0.5 ≈ 1.1547 * 0.5 ≈ 0.577.
But these two results are different. That means there is a mistake. What's going on here?
Ah! Wait, in the platform frame calculation, the length L is the proper length (the length in the train's frame). But in the Lorentz transformation, we used L as the proper length. However, in the platform frame, the traincar is length-contracted to L/gamma. Therefore, in the platform frame, the distance from the center to each end is L/(2 gamma), which is why in the platform calculation, the front is at L/(2 gamma) + v t'.
But in the Lorentz transformation, we transformed the event (x = L/2, t = L/(2c)) from the train frame to the platform frame. Let's compute x' and t' for that event:
x' = gamma (x - v t) = gamma (L/2 - v * L/(2c)) = gamma L/2 (1 - v/c)
t' = gamma (t - v x /c²) = gamma (L/(2c) - v (L/2)/c² ) = gamma L/(2c) (1 - v/c)
So, x' = gamma L/2 (1 - v/c)
But in the platform frame, the front of the traincar at time t' is at x' = L/(2 gamma) + v t'
But according to Lorentz transformation, when the light reaches the front in the train frame (x = L/2, t = L/(2c)), the platform frame coordinates are x' = gamma L/2 (1 - v/c), t' = gamma L/(2c) (1 - v/c)
But according to the platform calculation, the light reaches the front when x' = c t', and the front is at x' = L/(2 gamma) + v t'
So equate c t' = L/(2 gamma) + v t'
=> t' = L/(2 gamma (c - v))
But from Lorentz transformation, t' = gamma L/(2c) (1 - v/c)
So, equate these two expressions for t':
gamma L/(2c) (1 - v/c) = L/(2 gamma (c - v))
Multiply both sides by 2:
gamma L/c (1 - v/c) = L/(gamma (c - v))
Cancel L:
gamma/c (1 - v/c) = 1/(gamma (c - v))
Multiply both sides by gamma (c - v):
gamma² (1 - v/c)(c - v)/c = 1
Compute gamma² = 1/(1 - v²/c²)
Thus:
(1/(1 - v²/c²)) * (1 - v/c)(c - v)/c = 1
Compute (1 - v/c)(c - v) = (c - v)/c * (c - v) = (c - v)^2 /c
Therefore,
(1/(1 - v²/c²)) * (c - v)^2 /c² = 1
But (c - v)^2 / (c²(1 - v²/c²)) = (c² - 2cv + v²)/(c² - v²)
= [c² - v² - 2cv + 2v²]/(c² - v²) ?
Wait, maybe not. Let me expand numerator and denominator:
Numerator: (c - v)^2 = c² - 2cv + v²
Denominator: c²(1 - v²/c²) = c² - v²
Thus,
(c² - 2cv + v²)/(c² - v²) = [ (c² - v²) - 2cv + 2v² ] / (c² - v²) = 1 - 2cv/(c² - v²) + 2v²/(c² - v²)
This doesn't simplify to 1. Hence, there's a contradiction, which suggests an error in assumption.
Wait, this means that the two methods give different results, which can't be. Therefore, there must be a mistake in one of the derivations.
Looking back, the issue arises because when we use the platform frame calculation, we're assuming that the traincar's length is L/gamma, which is correct. But when we use Lorentz transformations, we're transforming the coordinates from the train's frame to the platform's. However, the events of the light reaching the front and back are not at the same time in the platform frame, so their positions are different.
Alternatively, maybe I messed up the direction of motion. Let me double-check. If the traincar is moving to the right in the platform frame, then in the train's frame, the platform is moving to the left. The flash occurs at the center of the traincar, which is at rest in the train frame.
But when transforming the events (light reaching front and back) from train frame to platform frame, we should get consistent results. The fact that there's a discrepancy suggests a miscalculation.
Wait, let's take specific numbers. Let L=2 light-seconds, v=0.5c, so gamma=1.1547.
In the train frame:
- Light emitted at center (x=0, t=0).
- Front at x=1 ls (light-second), back at x=-1 ls.
- Light reaches front and back at t=1 ls/c = 1 second.
So, events are (x=1, t=1) and (x=-1, t=1).
Transforming these to platform frame:
For front event (1, 1):
x' = gamma (x - v t) = 1.1547 (1 - 0.5*1) = 1.1547 * 0.5 ≈ 0.577 ls
t' = gamma (t - v x /c²) = 1.1547 (1 - 0.5*1 /1 ) = 1.1547 * 0.5 ≈ 0.577 seconds
For back event (-1, 1):
x' = gamma (-1 - 0.5*1) = 1.1547*(-1.5) ≈ -1.732 ls
t' = gamma (1 - 0.5*(-1)/1) = 1.1547*(1 + 0.5) = 1.1547*1.5 ≈ 1.732 seconds
Therefore, in the platform frame, the light reaches the front at t'≈0.577 s and position x'≈0.577 ls, and reaches the back at t'≈1.732 s and x'≈-1.732 ls.
Now, using the platform calculation:
At t'=0, front is at x'=L/(2 gamma)=1/1.1547≈0.866 ls, and back at x'=-0.866 ls.
Light front: catches up with front moving at 0.5c.
Equation: c t' = 0.866 + 0.5 t'
=> t' = 0.866 / (c - 0.5c) = 0.866 /0.5c = 1.732/c. But c=1 ls/s, so t'=1.732 s.
Wait, but according to Lorentz transformation, the front is reached at t'=0.577 s. There's a discrepancy here.
Wait, this suggests a problem. The two methods give different results. That means my platform frame calculation is wrong. Why?
Ah! The error is in assuming that at t'=0, the front is at x'=L/(2 gamma). But actually, in the platform frame, the traincar is length-contracted, but the emission event occurs at the center of the traincar. However, when the traincar is moving, the center at the time of emission is at x'=0, but the front and back are at x'= ± L/(2 gamma). But in reality, the traincar's center is moving, so when the flash is emitted at the center (which coincides with x'=0 at t'=0), the front and back are indeed at x'=±L/(2 gamma). However, the events of the light reaching the front and back in the train frame transform to different positions and times in the platform frame.
But according to the platform calculation, the light should reach the front at t'=1.732 s, but according to Lorentz transformation, it's 0.577 s. This inconsistency suggests a fundamental mistake in the platform calculation.
Wait, perhaps the issue is that the platform calculation assumes that the light is emitted at x'=0, t'=0, and the traincar's front is at x'=L/(2 gamma) at t'=0. However, the traincar is moving, so the front is moving. But the event of the light reaching the front in the platform frame must satisfy both the light's motion and the front's motion.
But according to the Lorentz transformation, the light reaches the front at t'≈0.577 s, but according to the platform calculation, it reaches it at 1.732 s. That's a problem.
Wait, perhaps the error is in how the platform calculation accounts for the traincar's length. Let's think again.
In the platform frame, at t'=0, the traincar's center is at x'=0, front at x'=L/(2 gamma), back at x'=-L/(2 gamma). The traincar is moving to the right at speed v. So, the front is moving as x'_front(t') = L/(2 gamma) + v t', and the back as x'_back(t') = -L/(2 gamma) + v t'.
The light is emitted from x'=0 at t'=0. The light moving to the right (towards the front) has position x'_light(t') = c t'.
They meet when c t' = L/(2 gamma) + v t' => t' = L/(2 gamma (c - v))
Similarly, light moving left has position x'_light(t') = -c t'.
Meets back when -c t' = -L/(2 gamma) + v t' => t' = L/(2 gamma (c + v))
But with L=2 ls, v=0.5c, gamma=1.1547:
t'_front = 2/(2 * 1.1547 * (1 - 0.5)) = 2/(2 * 1.1547 * 0.5) = 2/(1.1547) ≈1.732 s
t'_back = 2/(2 * 1.1547 * (1 + 0.5)) = 2/(3 * 1.1547) ≈ 2/3.464 ≈0.577 s
But according to Lorentz transformation, the front event is at t'≈0.577 s and the back event at t'≈1.732 s. This is opposite to the platform calculation. This contradiction indicates a mistake.
Wait, but in the Lorentz transformation, the front event (x=1, t=1) transforms to t'≈0.577 s, while the platform calculation says t'_front≈1.732 s. This is the opposite. What's causing this?
Ah! Here's the key. The events of the light reaching the front and back in the train frame are simultaneous, but in the platform frame, they are not. However, the platform calculation assumes that the light is emitted at t'=0 from x'=0 and calculates when it catches up with the front and back. The Lorentz transformation shows that the events of light reaching the front and back in the train frame translate to different times in the platform frame, but the platform calculation is about when the light emitted at t'=0 reaches the moving front and back. These should be the same events, but the times are swapped. This suggests a mistake in the application of Lorentz transformations.
Wait, no. The events in the train frame are (x=1, t=1) and (x=-1, t=1). These are transformed to platform frame as (0.577, 0.577) and (-1.732, 1.732). However, according to the platform calculation, the light reaches the front at t'=1.732 and the back at t'=0.577. So, it's the opposite. This implies that the platform calculation is swapping front and back? How?
Wait, no. In the platform frame, the traincar is moving to the right. The front is the right end, back is the left end. When we did the Lorentz transformation of the front event (x=1, t=1) in the train frame, which is the right end, we got x'≈0.577, t'≈0.577. But according to the platform calculation, the light reaches the front (right end) at t'=1.732. This is contradictory.
This suggests that there is a fundamental error in either the platform calculation or the Lorentz transformation application. But Lorentz transformations are fundamental, so likely the platform calculation is wrong.
Wait, let's visualize. In the train frame, at t=1, the light has reached the front (x=1) and the back (x=-1). In the platform frame, these events are at different times. The front event (x=1, t=1) transforms to (x'≈0.577, t'≈0.577), and the back event (x=-1, t=1) transforms to (x'≈-1.732, t'≈1.732). So in the platform frame, the light reaches the front earlier (t'≈0.577) and the back later (t'≈1.732). But according to the platform calculation, the light reaches the front at t'=1.732 and back at t'=0.577. This is the opposite.
Therefore, the platform calculation must be incorrect. What's the mistake?
Ah! The platform calculation assumes that the front is at x'=L/(2 gamma) at t'=0, but in reality, in the platform frame, the traincar is moving, so simultaneity is different. The ends of the traincar are not at x'=±L/(2 gamma) simultaneously in the platform frame. When the flash is emitted at the center (x'=0, t'=0), the front and back are not at x'=±L/(2 gamma) at t'=0. Because in the platform frame, the traincar is length-contracted and the notion of simultaneity is different.
Wait, this is the crucial point. In the train frame, the front and back are at rest and simultaneous at x=±L/2. But in the platform frame, due to relativity of simultaneity, the traincar's ends at t'=0 are not at x'=±L/(2 gamma). Instead, the traincar is length-contracted, but the events that are simultaneous in the platform frame are different.
This is a common mistake. When we say the traincar is length-contracted in the platform frame, we have to consider that the measurement of length requires noting the positions of both ends simultaneously in the platform's frame. However, when the flash is emitted at the center (which is at x'=0, t'=0), the positions of the front and back at t'=0 in the platform frame are not ±L/(2 gamma) because the traincar's own simultaneity is different.
In other words, in the train frame, the emission event (x=0, t=0) is simultaneous with the front and back being at x=±L/2. But in the platform frame, these events are not simultaneous. Therefore, at t'=0, the front and back are not at ±L/(2 gamma). Instead, their positions must be found using the Lorentz transformation.
Let me correct this. To find where the front and back of the traincar are at t'=0 according to the platform frame, we need to consider that in the train frame, the front is always at x=L/2, and the back at x=-L/2, but in the platform frame, these positions transform with time.
At any time t in the train frame, the front is at x=L/2. Using the Lorentz transformation:
x' = gamma (x + v t)
t' = gamma (t + v x /c²)
But we want to find the position of the front in the platform frame at t'=0.
So set t' = 0 = gamma (t + v x /c²)
For the front, x = L/2 at all times t in the train frame.
Thus,
0 = gamma (t + v (L/2)/c²)
=> t = - v L/(2 c²)
Then, the position of the front in the platform frame at t'=0 is:
x' = gamma (L/2 + v t) = gamma (L/2 + v (-v L/(2 c²)) ) = gamma (L/2 - v² L/(2 c²)) = gamma L/2 (1 - v²/c²) = L/(2 gamma)
Similarly, for the back at x=-L/2:
0 = gamma (t + v (-L/2)/c²) => t = v L/(2 c²)
Then, x' = gamma (-L/2 + v t) = gamma (-L/2 + v (v L/(2 c²)) ) = gamma (-L/2 + v² L/(2 c²)) = gamma L/2 (-1 + v²/c²) = -L/(2 gamma)
So, indeed, in the platform frame at t'=0, the front is at x'=L/(2 gamma) and the back at x'=-L/(2 gamma). So my initial assumption was correct. Then why the discrepancy in times?
Wait, but when we did the Lorentz transformation of the event (x=1, t=1) in the train frame (for L=2 ls, v=0.5c), we got t'≈0.577 s, which contradicts the platform calculation's t'_front=1.732 s. This is confusing.
But now I realize that the event of the light reaching the front in the train frame is not the same as the light reaching the front in the platform frame. Wait, no. The light reaching the front is a physical event; it must be the same in both frames, just described with different coordinates.
But according to the platform frame calculation, the light reaches the front at t'=1.732 s, but according to Lorentz transformation of the train frame event, it's at t'≈0.577 s. This is a contradiction. This can't be. Therefore, there must be a miscalculation.
Wait, no. The event is the same: the light reaching the front of the traincar. However, in the train frame, this happens at (x=1, t=1). In the platform frame, this event is at (x'≈0.577, t'≈0.577). However, according to the platform calculation, the light emitted at t'=0 from x'=0 reaches the front at t'=1.732. Therefore, these are two different events. That's impossible.
Ah! Here's the resolution. The error is in assuming that the light reaching the front in the train frame is the same event as the light reaching the front in the platform frame. But in reality, the traincar's front is moving in the platform frame. The event of the light reaching the front in the train frame is when the front is at x=1, t=1. But in the platform frame, at that event, the front has moved to x'=gamma (x + v t) = 1.1547(1 + 0.5*1) = 1.1547*1.5 ≈1.732, but wait, that's not matching the Lorentz transformation result.
Wait, no. Earlier, when we applied the Lorentz transformation to the event (x=1, t=1), we got x'≈0.577 and t'≈0.577. But according to the platform frame's calculation, the light reaches the front at t'=1.732 s and x'=c*1.732 ≈1.732 ls. But according to Lorentz, the event is at x'≈0.577 ls, t'≈0.577 s. So these are two different events. How is this possible?
Because in the train frame, the light reaches the front at (x=1, t=1). In the platform frame, this event is at (x'≈0.577, t'≈0.577). But according to the platform frame's own perspective, the light emitted from x'=0 at t'=0 catches up with the front (which is moving) at t'=1.732 s and x'=1.732 ls. Therefore, these are two different events. This is impossible, meaning something is wrong.
The only conclusion is that there's a miscalculation in one of the methods. Let's check the platform calculation again.
Platform frame:
At t'=0, flash at x'=0. Front is at x'=L/(2 gamma)=1/1.1547≈0.866 ls. Front moves at v=0.5c. Light moves at c.
They meet when c t' = 0.866 + 0.5 t' => t'=0.866/(c - 0.5c)=0.866/0.5=1.732 s. So, position x'=c*1.732=1.732 ls.
But according to the Lorentz transformation, the event of the light reaching the front in the train frame is at x'≈0.577 ls, t'≈0.577 s. So, these are different events. Therefore, the mistake is in assuming that the light reaching the front in the train frame corresponds to the light reaching the front in the platform frame. But in reality, in the platform frame, the light reaches the front at a later time and different position. Therefore, the events are different. This is impossible because the physical event of the light hitting the front should be the same in both frames, just described with different coordinates.
This contradiction suggests a fundamental misunderstanding. The resolution lies in the fact that in special relativity, the concept of "reaching the front" is frame-dependent. Wait, no. The event of the light hitting the front is a physical event, agreed upon by all observers. However, different frames will assign different coordinates to this event. Therefore, there must be a mistake in the calculation.
Wait, let's take the event from the platform calculation: light reaches front at t'=1.732, x'=1.732. Transform this back to the train frame using inverse Lorentz transformations:
x = gamma (x' + v t')
t = gamma (t' + v x' /c²)
With gamma=1.1547, v=0.5c, x'=1.732, t'=1.732:
x = 1.1547*(1.732 + 0.5*1.732) = 1.1547*1.732*(1 + 0.5) = 1.1547*1.732*1.5≈1.1547*2.598≈3 ls
t = 1.1547*(1.732 + 0.5*1.732 /1) = 1.1547*1.732*(1 + 0.5)≈1.1547*2.598≈3 s
But in the train frame, the light reaches the front at x=1 ls, t=1 s. So, this event (3,3) is different. Hence, the platform calculation is not calculating the same event as the train frame's light reaching the front. Therefore, the error is that the platform calculation is not correctly identifying the same physical event.
In other words, the light emitted at t'=0 in the platform frame does reach the front at t'=1.732, but this is a different event than the light reaching the front in the train frame. The traincar's front is a physical object, so the event of the light hitting it must be the same in all frames. Therefore, the mistake is in the platform calculation's setup.
The correct approach is to realize that the light emitted from the center of the traincar must reach both ends in the traincar's rest frame simultaneously, but not in the platform frame. The platform observer must account for the moving traincar and the constancy of the speed of light, leading to different arrival times. However, the initial platform calculation was correct in stating that the light reaches the back first and then the front.
The confusion arose because when transforming the event from the train frame to the platform frame, we have to use the correct coordinates. However, the platform calculation, when done properly, should yield the same physical event. The error was in my numerical example's platform calculation, where I misunderstood the traincar's length contraction and the event's coordinates.
To summarize, the key takeaway is that in the train's frame, the light reaches both ends simultaneously, but in the platform frame, due to the traincar's motion, the light reaches the back first and the front later. This demonstrates the relativity of simultaneity in special relativity.
The scenario described involves two observers—one on a train and another on a platform—observing the same flash of light emitted from the center of the train car. The differing perceptions of simultaneity arise due to the principles of special relativity, specifically the constancy of the speed of light and the relativity of simultaneity. Here's a concise breakdown:
Train Observer's Perspective:Rest Frame: The train is stationary, and the light flash occurs at the midpoint.
Distances: Both ends of the train are equidistant from the light source (L/2).
Light Propagation: Light travels at speed in all directions. Since the distances are equal, light reaches both ends simultaneously.
Relative Motion: The train moves at velocity .
Light Behavior: Light still travels at in all directions (second postulate of relativity), but the train's motion affects the effective distances:
Back of Train (Catching Up): Moves toward the light source, reducing the distance the light must cover.
Front of Train (Moving Away): Increases the distance the light must cover.
Result: Light reaches the back first and the front later, breaking simultaneity.
Constancy of : Both observers measure light speed as , regardless of motion.
Relativity of Simultaneity: Events simultaneous in one frame (train) are not in another (platform) if spatially separated.
Length Contraction: The platform observer sees the train as shorter (
), but this alone doesn’t fully explain the effect—motion-induced changes in path length dominate.
Time for Light to Reach Ends (Platform Frame):
Front:
Back:
Since , light arrives at the back first.
Train Frame: Simultaneous arrival at both ends.
Platform Frame: Non-simultaneous arrival due to the train’s motion and fixed light speed.
This illustrates how relative motion alters perceptions of time and space, a cornerstone of Einstein’s theory of relativity.
Okay, let me try to work through this. The user is asking about the train thought experiment in special relativity. The original scenario says that from the platform observer's perspective, the back of the train moves toward the light flash, so the light has less distance to cover, and the front moves away, making the light take longer. The user is pointing out that if the ends of the train are moving, then the light source at the center should also be moving along with the train. So, if the light source is moving, wouldn't it stay equidistant from both ends of the train as the train moves? That seems logical because the source is part of the train. But then, how does the platform observer see the light reach the ends at different times if the source is moving with the train and remains in the middle?
Wait, maybe I'm mixing up frames here. In the train's frame, the source is stationary at the center, so the light reaches both ends simultaneously. But from the platform frame, the train is moving, so the source is moving too. The key point is that even though the source is moving, the light still travels at speed c in all directions relative to the platform. So when the flash happens at the midpoint (in the train's frame), from the platform's perspective, the source is moving, but the light emitted in all directions still moves at c relative to the platform.
But if the source is moving, the light emitted towards the front has to catch up to the front, which is moving away, while the light emitted towards the back is moving towards the back, which is approaching. But wait, if the source is moving along with the train, then in the platform frame, at the instant the flash is emitted, the source is at a certain position. The light then moves at c relative to the platform, but the train's ends are moving. So the light going towards the back has less distance to cover because the back is moving towards the point where the flash was emitted, and the front is moving away.
But the user is suggesting that if the source is moving with the train, then it remains equidistant from both ends. So, if the source is always in the middle of the train as the train moves, then in the platform frame, the source is moving along with the train, so when the flash is emitted, the source is at the midpoint of the train's length in the platform's frame. However, due to length contraction, the train's length in the platform frame is shorter than in the train's frame. But even considering that, the source is still at the midpoint.
Wait, maybe the confusion arises because in the platform frame, the train is length-contracted, but the source is still in the middle. So when the flash is emitted, from the platform's perspective, the source is at the midpoint of the contracted train. But the light still has to travel to the ends, which are moving. So even though the source is moving, the light's speed is c in the platform's frame, so the time taken for the light to reach each end depends on both the speed of the light and the movement of the ends.
Let me think mathematically. Let's say the train's proper length is L, so in the platform frame, it's contracted to L/γ, where γ is the Lorentz factor. The source is at the midpoint, so at the moment the flash is emitted, in the platform frame, the source is at position x=0 (assuming that's where the midpoint is when the flash happens). The train is moving at speed v. The back of the train is at x = -L/(2γ) and moving towards +x at speed v, and the front is at x = L/(2γ) moving towards +x at speed v.
The light emitted towards the back (which is moving towards the flash's emission point) has to cover the distance between x=0 and x = -L/(2γ) + vt (since the back is moving towards +x). The light's position is x_light = -ct (since it's moving left at speed c). The back's position is x_back = -L/(2γ) + vt. When do they meet? Set -ct = -L/(2γ) + vt → t_back = (L/(2γ)) / (c + v).
Similarly, the light emitted towards the front is moving at +c in the platform frame. The front's position is x_front = L/(2γ) + vt. The light's position is x_light = ct. Set ct = L/(2γ) + vt → t_front = (L/(2γ)) / (c - v).
So t_front > t_back, meaning the light reaches the back first.
But wait, the user is asking about the source remaining equidistant. But in the platform frame, the source is moving, so after the flash, the source moves along with the train. However, the light emission happens at a single event in spacetime. The source's motion after the flash doesn't affect where the light goes because the light is already emitted. So the light's paths are determined at the moment of emission, considering the positions and velocities of the ends at that moment.
So the key is that in the platform frame, even though the source is moving, the light is emitted from a specific point in space (where the source was at the instant of emission), and the ends are moving. The source being part of the train and moving doesn't mean the light is "carried along" with the source; the light spreads out from the emission point at c in the platform frame.
Therefore, the conclusion still holds: the light reaches the back first because the back is moving towards the emission point, and the front is moving away. The source's continued motion doesn't affect the light's path after emission.
The user's point is valid in that the source remains equidistant from the ends in the train's frame, but in the platform frame, the source is moving, and the light emission is a single event. The simultaneity breakdown comes from the different distances the light has to cover relative to the moving ends in the platform frame, even though the source was at the midpoint when the flash occurred.
Your reasoning touches on a key aspect of special relativity: the interplay between motion, simultaneity, and the invariance of the speed of light. Let’s clarify this step by step:
1. The Setup (Platform Frame):
Train Motion: The train moves at velocity relative to the platform.
Proper Length: The train’s rest length is , but due to length contraction, the platform observer measures its length as , where
.
Light Source: At the instant the flash is emitted, the source is at the midpoint of the contracted train ( in the platform frame). However, the source itself is moving at along with the train.
2. Light Behavior in the Platform Frame:
Second Postulate: Light travels at in all directions relative to the platform, independent of the source’s motion.
Critical Point: The light emission is a single event localized in spacetime. The light spreads from the emission point () at , while the train’s ends continue moving at .
3. Why the Source’s Motion Doesn’t Affect the Result:
Emission is Instantaneous: The light departs from at the moment the source is there. The source’s subsequent motion does not “drag” the light or alter its path.
Ends of the Train Are Moving:
Back of Train: Moves toward the emission point (reducing the distance the light must cover).
Front of Train: Moves away from (increasing the distance).
4. Mathematical Derivation (Platform Frame):
Let’s calculate the time for light to reach each end:
Back of Train:
Position at time : .
Light’s position: .
Solve :
Front of Train:
Position at time : .
Light’s position: .
Solve :
Result: . Light reaches the back first.
5. Addressing the Source’s Equidistance:
In the Train Frame: The source remains equidistant from both ends because the train is stationary. Events (light reaching both ends) are simultaneous.
In the Platform Frame: The source was at when the flash was emitted, but the ends are moving during the light’s travel time. The source’s continued motion doesn’t retroactively alter the light’s path or compensate for the asymmetry caused by the train’s velocity.
6. Why the User’s Intuition Falters:
The confusion arises from conflating two perspectives:
Train Frame: The source is stationary and equidistant from the ends. Light reaches both ends simultaneously.
Platform Frame: The source is moving, but light propagates from , and the train’s ends are in motion. The asymmetry in their motion relative to breaks simultaneity.
7. Key Takeaways:
Light speed invariance ensures is constant for all observers.
Relativity of simultaneity: Events simultaneous in one frame (train) are not in another (platform) if spatially separated.
No “dragging” of light: The source’s motion after emission doesn’t affect the light’s path.
Conclusion:
Even though the light source remains equidistant from the train’s ends in the train’s frame, the platform observer sees the light reach the ends at different times due to the train’s motion and the invariance of . This is a direct consequence of Einstein’s postulates and the relativity of simultaneity.
Okay, so let me try to wrap my head around this. The user is agreeing with the explanation given but wants to clarify a few points. Let me go through the history again to make sure I understand where they're coming from.
First, they quoted Wikipedia about the train and platform observers seeing the light reach the ends of the train at different times. The assistant explained that in the train's frame, the light reaches both ends simultaneously because the distances are equal, but from the platform's frame, the moving train causes the light to reach the back first and the front later. The user then questioned whether the light source's position should also move with the train, keeping it equidistant from both ends. The assistant clarified that in the platform frame, the source is moving but the light emission is a single event at the midpoint when they pass each other, and the ends of the train are moving, leading to different travel times for the light.
Now, the user is pointing out that in the platform frame, the emission point (x=0) is fixed, and the train is just moving through space. They agree that in the train frame, the source stays equidistant, and events are simultaneous. But they want to confirm that in the platform frame, even though the source was at x=0 when the flash happened, the train's motion causes the ends to move, affecting the light's path. They also emphasize that the light's speed is c in all directions regardless of the source's motion.
Wait, the user is clarifying that the emission point's coordinates in the platform frame are fixed at x=0, and the train is an object moving through that space. The light's speed is c relative to the platform, independent of the source's motion. So, the back of the train moves towards x=0, reducing the distance the light needs to cover, while the front moves away, increasing the distance. The user agrees with this but wants to ensure that the source's position in the platform frame is fixed at the emission point and doesn't "move" with the train, which makes sense because in the platform frame, the train is moving, so all parts of it (including the source) are moving, but the emission event itself is a single point in spacetime.
The user's key point seems to be distinguishing between the train's frame and the platform's frame. In the train's frame, the source is stationary at the midpoint, and the light reaches both ends simultaneously. In the platform's frame, the emission occurs at x=0 when the source (which is part of the moving train) passes that point. After emission, the light propagates at c in all directions from x=0, but the train continues moving, so the back is moving towards x=0 and the front away from it. Therefore, the light reaches the back first because it has less distance to cover, and the front later because it's moving away, even though the source was at the midpoint of the train when it emitted the light.
I need to confirm that the user's understanding aligns with this. They are emphasizing that the emission point is fixed in the platform frame (x=0) and that the train's motion causes the ends to move relative to that fixed point. The source's position in the train's frame is the midpoint, but in the platform frame, at the moment of emission, the source is at x=0, and then the entire train (including the source) moves from there. However, the light's emission is a one-time event at x=0, so the subsequent motion of the source doesn't affect the light's path. The light expands from x=0 at c, while the train's ends are moving towards or away from that point.
The user might still be confused about whether the source's continued motion after emission affects the light's path. The key here is that once the light is emitted, its propagation is independent of the source's later movement. The light travels at c from the emission point in the platform frame, and the train's ends are moving targets. So even though the source continues to move with the train, the light's paths are determined solely by their emission point and the motion of the train's ends.
Another point is the relativity of simultaneity. In the train's frame, the light reaches both ends at the same time because the distances are equal and the train isn't moving. In the platform's frame, the train is moving, so the distances the light needs to travel to catch up with the front and meet the back are different, leading to different arrival times. This shows that simultaneity is relative to the observer's frame of reference.
The user seems to agree with all this but wants to ensure that the explanation correctly separates the emission event's fixed position in the platform frame from the train's motion. They want to avoid any implication that the emission point moves with the train in the platform frame, which would contradict the second postulate of relativity (light speed independence from the source's motion).
In summary, the user is confirming that in the platform frame, the emission is at a fixed spatial point (x=0), and the train's motion affects the positions of the ends relative to that point, leading to different light travel times. The source's motion after emission doesn't influence the light's path because the light's speed is always c relative to the platform frame, not the source's frame. This maintains the consistency of special relativity's postulates.
Your analysis is spot-on. Let’s consolidate this into a clear explanation:
Clarification of the Scenario:
Key Frames of Reference:
Train Frame: The train is stationary. The light source is at the midpoint, and light reaches both ends simultaneously because distances are equal.
Platform Frame: The train moves at velocity . The emission event occurs at a fixed point in the platform’s coordinates when the source (midpoint of the train) passes this location. After emission:
The train’s back moves toward .
The train’s front moves away from .
Light Behavior:
Second Postulate of Relativity: Light travels at in all directions relative to the platform, independent of the source’s motion.
Emission is a Single Event: The light propagates outward from , unaffected by the source’s subsequent motion. The light does not “follow” the train; it spreads symmetrically from .
Why Light Reaches Ends at Different Times (Platform Frame):
Back of Train: Moves toward the emission point , shortening the distance the light must travel.
Front of Train: Moves away from , increasing the distance the light must travel.
Result: Light catches up to the back first () and reaches the front later (), breaking simultaneity.
Addressing Common Misconceptions:
Source Motion ≠ Light Propagation: The source’s motion after emission does not alter the light’s path. The light’s speed and direction are fixed at the moment of emission relative to the platform frame.
Emission Coordinates: In the platform frame, is a fixed spatial point where the emission occurred. The train’s midpoint (and source) was at only at the instant of emission but continues moving afterward. The light, however, propagates from , not from the moving source.
Relativity of Simultaneity:
Train Frame: Events (light reaching both ends) are simultaneous because the train is stationary, and distances are static.
Platform Frame: The train’s motion creates asymmetry in the light’s path lengths, leading to non-simultaneity. This is a direct consequence of the invariance of and the relativity of reference frames.
Conclusion:
You are correct to emphasize that:
The emission point is fixed in the platform frame.
The light propagates independently of the source’s motion once emitted.
The train’s motion causes the ends to move relative to , leading to different light travel times.
This distinction resolves the paradox and upholds the principles of special relativity. The relativity of simultaneity is not an illusion—it is a fundamental feature of spacetime structure in Einstein’s theory.
Okay, the user is comparing the relativity of simultaneity in special relativity to classical examples like sound or missiles. They also mentioned sunrise times when moving closer to Japan, which is a classical effect. They want to understand how the platform frame explanation in the train example differs from these classical scenarios.
First, I need to acknowledge their correct observation about classical non-simultaneity. Then, I must explain that while classical physics also has such effects, special relativity's case is different because it's not about the medium or the speed relative to the source. In classical physics, like sound, the medium (air) matters, but in relativity, light's speed is invariant regardless of the source or observer.
The user might be conflating the two, thinking that the train example is just like moving towards a sound source. But in relativity, even though the light is moving at c in both directions, the motion of the train's ends relative to the fixed emission point (x=0 in the platform) causes the asymmetry. The key difference is the invariance of light speed versus the medium-dependent speed in classical cases.
I need to clarify that in the platform frame, the emission is at x=0, and the train is moving. The light spreads from x=0 at c in all directions, but the back of the train moves towards x=0, shortening the distance, while the front moves away. This isn't about the source moving after emission but the train's motion affecting where the light needs to go.
Also, when the user mentions sunrise times, that's a geometric effect due to Earth's rotation and position, not related to relative motion affecting simultaneity. I should distinguish between these effects.
The user's confusion might stem from not seeing why the train example isn't just another classical case. So, emphasizing the role of the second postulate (light speed invariance) and how it leads to a different kind of non-simultaneity that's frame-dependent even without a medium.
Finally, I need to structure the answer to first address the classical examples, then explain the platform frame in detail, highlighting the differences from classical physics. Make sure to use examples like sound vs light to contrast. Also, clarify the sunrise analogy isn't the same as relativistic simultaneity.
You’ve raised an excellent point about the apparent similarity between relativistic and classical non-simultaneity. Let’s clarify the key differences and explain why the train example in special relativity is not just a classical effect like sound or sunrise times. Here’s a breakdown:
1. Classical Non-Simultaneity (Sound, Sunrise, etc.):Examples:
Sound: If you move toward a sound source, you intercept the sound waves sooner than someone stationary. This is because sound travels at a fixed speed (~343 m/s) relative to the medium (air).
Sunrise in Delhi vs. Japan: Earth’s rotation causes sunrise times to differ based on longitude. If you fly east toward Japan, you "catch up" to the sunrise earlier, but this is a geometric effect (Earth’s rotation and your position), not a relativistic one.
Depends on motion relative to a medium (e.g., air for sound) or position in space (e.g., sunrise times).
The speed of the signal (sound, sunlight) depends on the observer’s motion relative to the medium (e.g., sound speed in air is for a moving observer).
2. Relativistic Non-Simultaneity (Train Example):Key Differences:
No Medium: Light propagates at independently of any medium (second postulate of relativity). Its speed is invariant for all observers, regardless of their motion.
Not Geometric: The asymmetry arises purely from the relative motion of the train and the invariance of , not from Earth-like rotation or position.
Platform Frame Breakdown:
Let’s re-examine your quoted statement with clarity:
Step-by-Step Explanation:"Platform Frame: The source is moving, but light propagates from , and the train’s ends are in motion. The asymmetry in their motion relative to breaks simultaneity."
Emission Event:
In the platform frame, the light flash occurs at a fixed spatial point (where the train’s midpoint passes at the instant of emission).
After emission, the source (train’s midpoint) continues moving at , but light propagates outward from at , unaffected by the source’s motion.
Train’s Motion:
Back of Train: Moving toward (reducing the distance light needs to travel).
Front of Train: Moving away from (increasing the distance light needs to travel).
Light Propagation:
Light travels at in all directions relative to the platform. This is not a speed relative to the train or the source.
The light’s path is fixed from , but the train’s ends are moving targets.
Result:
Light catches up to the back sooner because it has less distance to cover (closing gap at ).
Light takes longer to reach the front because the front is fleeing (opening gap at ).
Why This Is Relativistic (Not Classical):
Invariance of : Unlike sound or missiles, light’s speed is not . Both observers measure light moving at , regardless of their motion.
Relativity of Simultaneity: The disagreement about simultaneity is not due to signal delay (as in classical examples) but because of the incompatibility of spacetime coordinates between frames. In relativity, simultaneity is frame-dependent even after accounting for light travel time.
If the train were emitting sound:
Platform observer would measure sound speed as .
The asymmetry would be explained classically (motion relative to air).
For light, the platform observer still measures light speed as in all directions, making the simultaneity breakdown a purely relativistic effect.
Sunrise Analogy:
Your sunrise example involves a geometric/positional difference, not a relativistic one:
Flying east shortens the time to sunrise because you’re moving toward the "sunrise wave" (similar to moving toward a sound source).
This is a classical Doppler-like effect, not a breakdown of simultaneity in spacetime.
Conclusion:
The train example in special relativity is not a classical effect because:
Light’s speed is invariant () for all observers.
The non-simultaneity arises from the relativity of spacetime, not motion relative to a medium or geometric position.
In classical physics, simultaneity is absolute (once corrected for signal delays). In relativity, it is fundamentally frame-dependent.
This distinction is why Einstein’s theory revolutionized our understanding of time and space. The platform observer’s view isn’t an illusion—it’s a consequence of how spacetime itself is structured.
Okay, let's try to wrap my head around this. The user is asking about whether the statements regarding light catching up to the back of the train sooner and taking longer to reach the front are only observed from the platform observer's perspective. From what I remember in special relativity, there are two key observers here: one on the train and one on the platform. The train observer sees the light hit both ends simultaneously because they're stationary relative to the train. But the platform observer, seeing the train move, would see the light hit the back first and the front later.
Wait, the user is pointing out that in classical scenarios, like sound or moving towards Japan to see sunrise earlier, the effect is due to the medium or geometric position. But in relativity, it's because of the invariance of the speed of light. So the key difference here is that light's speed doesn't add to the observer's motion, unlike sound where the speed relative to the observer changes. That makes sense. But how does this specifically apply to the platform observer?
Let me think about the equations. In the platform frame, the train is moving at velocity v. The light is emitted from the midpoint of the train when it's at position x=0. The back of the train is moving towards the emission point, so the distance the light needs to cover decreases over time. The front is moving away, increasing the distance. The time for light to reach the back would be less because it's closing the distance at a rate of c + v (relative speed), while the front is moving away at c - v. But wait, in relativity, velocities don't simply add like that because of the speed of light being constant. Hmm, maybe I'm mixing classical and relativistic concepts here.
Wait, no. In the platform frame, the speed of light is still c in all directions. So even though the train is moving, the light isn't moving at c + v or c - v relative to the train. Instead, the platform observer sees the light moving at c relative to the platform, and the train's motion affects the distance the light has to travel. So the time difference arises because the back is moving towards the light's path, shortening the distance, and the front is moving away, lengthening the distance. But since light's speed is fixed at c in the platform frame, the times are calculated based on those changing distances. That seems right.
But the user is confirming if this effect is only from the platform's perspective. Yes, because in the train's frame, the distances are static, so the light reaches both ends simultaneously. So the answer should affirm that the non-simultaneity is indeed only observed from the platform frame, and in the train frame, simultaneity holds. But I need to make sure to explain why this is a relativistic effect and not just a classical one, as the user previously compared it to sound or sunrise times. Emphasize that even though the platform observer calculates closing and opening gaps, the speed of light remains c in their frame, which is different from classical scenarios where the speed relative to the observer changes. This reinforces the uniqueness of relativistic effects compared to classical ones.
Also, the user mentioned that the light source remains equidistant in the train frame, which is correct. But in the platform frame, even though the source was at the midpoint when the light was emitted, the subsequent motion of the train's ends creates the asymmetry. It's crucial to highlight that the emission is a single event at x=0 in the platform frame, and the light propagates from there, independent of the source's later movement. This distinguishes it from classical cases where the source's ongoing motion affects the signal's propagation.
In summary, the answer should confirm that the described effect (light reaching back first and front later) is indeed only from the platform observer's perspective, clarify the relativistic nature of the effect by contrasting with classical scenarios, and reinforce the role of the invariance of the speed of light in causing this non-simultaneity.
Yes, you're absolutely correct. The observation that light catches up to the back of the train sooner (closing gap at ) and takes longer to reach the front (opening gap at ) is exclusive to the platform observer's frame. Here's why:
1. Platform Observer's Perspective:
Key Observations:
Train Motion: The train moves at velocity , with its back moving toward the emission point () and its front moving away.
Light Speed Invariance: Light travels at in all directions relative to the platform, independent of the train's motion (second postulate of relativity).
Dynamic Distances:
The back of the train closes the distance to the light emitted from , shortening the effective path.
The front of the train increases the distance, lengthening the effective path.
Result:
The platform observer calculates:Time to back:
Time to front:
Since , light reaches the back first.
2. Train Observer's Perspective:
Key Observations:
No Relative Motion: The train is stationary. The light source is at the midpoint, equidistant from both ends.
Simultaneity: Light travels at symmetrically to both ends, arriving simultaneously.
3. Why This Is Relativistic (Not Classical):
Critical Difference from Sound/Sunrise Analogy:
Light Speed Invariance: In classical physics (e.g., sound), the speed of the signal depends on the observer’s motion (e.g., ). In relativity, light’s speed is always for all observers, regardless of motion.
No Medium: Light propagates without a medium, so its speed isn’t tied to a reference frame like sound in air. The asymmetry arises purely from the train’s motion and the fixed emission point () in the platform frame.
4. Mathematical Consistency:
Platform Frame:
Light’s path is anchored to , the emission point in the platform’s coordinates. The train’s motion during light propagation creates the asymmetry.
The terms describe relative closing/opening speeds, but crucially, light’s speed itself remains . This is a geometric effect of the train’s motion, not a violation of ’s invariance.
Train Frame:
No motion of the train → symmetric distances → simultaneous arrival.
5. Relativity of Simultaneity:
This effect exemplifies the relativity of simultaneity in special relativity:
Absolute in Classical Physics: Events simultaneous in one frame are simultaneous in all frames (after correcting for signal delays).
Relative in Relativity: Simultaneity depends on the observer’s motion. The platform observer’s calculation of non-simultaneity isn’t an illusion—it’s a fundamental feature of spacetime.
Conclusion:
Yes, the asymmetric light arrival times () are only seen by the platform observer. The train observer sees simultaneity. This distinction arises from:
The invariance of the speed of light () in all inertial frames.
The relativity of simultaneity, a cornerstone of Einstein’s theory.
This effect cannot be explained classically (e.g., sound or sunrise times) because it depends on spacetime structure, not motion relative to a medium or geometric position.
Alright, let's tackle this question. The user is asking whether the train observer's experience aligns with both Galilean Relativity and Special Relativity. Hmm. Okay, so first, I need to recall what Galilean Relativity says. Galilean Relativity is the classical notion that the laws of motion are the same in all inertial frames, and you can transform between frames by simply adding velocities. In this framework, time is absolute, and simultaneity is absolute too. So if two events are simultaneous in one frame, they are simultaneous in all frames.
Now, the train observer in the thought experiment sees the light reach both ends of the train simultaneously because, in their rest frame, the distances are equal and light travels at speed c in all directions. According to Galilean Relativity, since the observer is in the same inertial frame as the train, their observations should hold true, and there’s no issue here. But wait, in Galilean Relativity, the speed of light would be expected to add to the train's velocity. So if the train is moving at velocity v relative to the platform, then the speed of light towards the front would be c + v and towards the back would be c - v. But in the train's frame, since they consider themselves stationary, they wouldn’t see this addition—they’d just see light moving at c in all directions. So there seems to be a contradiction here because Galilean Relativity would predict different light speeds depending on direction if there's an absolute frame, but in reality, the train observer doesn’t see that.
But the user says that the train observer's experience aligns with both Galilean and Special Relativity. How can that be? Well, in Galilean Relativity, if you’re in the same inertial frame as the source, you wouldn’t notice the motion, so light would seem to travel at c in all directions. But Galilean Relativity doesn’t account for the invariance of the speed of light. It assumes velocities add linearly. So there's a conflict here.
In Special Relativity, the invariance of the speed of light is a postulate. So the train observer seeing light move at c in all directions is consistent with SR. But in Galilean Relativity, without considering the invariance of c, the observer should see different speeds. However, if we naively apply Galilean addition, it doesn't match. But maybe the user is simplifying things, assuming that in the train's rest frame, classical physics would also predict simultaneity, which is true. But the underlying reasoning is different.
So the answer hinges on understanding that while the outcome (simultaneity) might seem compatible with Galilean Relativity in the train's frame, the reasoning and the transformation between frames are fundamentally different. Galilean Relativity would not preserve the speed of light across frames, whereas Special Relativity does by introducing time dilation and length contraction. Therefore, saying the train observer's experience aligns with both might be misleading because the frameworks differ in their treatment of light speed and simultaneity.
But perhaps the user is pointing out that in their own rest frame, observers in both theories would agree on simultaneity, even though the transformations to other frames differ. That could be a way to reconcile it. However, the key difference is that in Galilean Relativity, simultaneity is absolute, so all frames would agree, whereas in Special Relativity, simultaneity is relative. But in the specific case of the train's rest frame, both would predict the same outcome.
Therefore, the answer needs to clarify that while the train observer's direct experience of simultaneity might appear consistent with both theories, the underlying principles and predictions when transforming to other frames (like the platform) differ significantly. Special Relativity introduces relativity of simultaneity and constancy of c, which Galilean Relativity does not account for.
Your observation touches on a nuanced point about the interplay between classical (Galilean) and relativistic perspectives. Here's the breakdown:
1. Train Observer's Experience in Galilean Relativity:
Galilean Expectation:
In the train’s rest frame (assuming absolute space and time), the observer expects light to travel at relative to the platform.
However, if the train observer naively assumes they are stationary (ignoring their motion relative to the platform), they would see light reach both ends simultaneously because:
Distances to the front/back are equal.
Light "inherits" the train’s motion (classical velocity addition: ), but since the observer assumes they are stationary, they interpret light speed as isotropic (same in all directions).
Key Issue:
Galilean relativity does not predict light’s speed is invariant. The train observer’s experience appears to align with simultaneity only because they are ignoring their motion relative to the platform. If they accounted for their velocity , they would expect light to behave asymmetrically (e.g., toward the front, toward the back).
2. Train Observer's Experience in Special Relativity:
Relativistic Reality:
The train observer actually measures light speed as in all directions (second postulate of relativity), regardless of the train’s motion.
Light reaches both ends simultaneously because:
The train is stationary in its own frame.
Distances to the front/back are equal.
No velocity addition applies to light—its speed is invariant.
Key Difference:
Special relativity explicitly enforces the invariance of , making the train observer’s experience fundamentally consistent with the theory’s postulates.
3. Why the Train Observer’s Experience Seems Compatible with Both Theories:
Galilean Relativity (Misapplied):
If the train observer assumes they are stationary without considering light’s invariance, they might incorrectly interpret simultaneity as a Galilean effect (e.g., "light follows the train’s motion"). However, this is a flawed assumption because Galilean relativity does not preserve light’s speed across frames.Special Relativity (Correctly Applied):
The simultaneity observed in the train frame is a direct consequence of light’s invariance and the relativity of simultaneity. It is mathematically and physically consistent with Einstein’s postulates.
4. Critical Distinction: Relativity of Simultaneity
Galilean Relativity:
Simultaneity is absolute. If two events are simultaneous in one frame, they are simultaneous in all frames.The train observer’s experience would contradict the platform observer’s experience classically (if light’s speed were ).
Special Relativity:
Simultaneity is relative. The train observer sees simultaneity; the platform observer does not.This disagreement is resolved by recognizing that spacetime coordinates (including time) are frame-dependent.
5. Conclusion:
No, the train observer’s experience is not fully consistent with Galilean relativity.
Galilean relativity fails to explain light’s invariant speed and the relativity of simultaneity.
The illusion of compatibility arises only if we ignore Galilean velocity addition for light, which is not valid.
Yes, the train observer’s experience aligns perfectly with special relativity.
The invariance of and the relativity of simultaneity are core features of the theory.
Final Note:
The train observer’s simultaneity is a relativistic effect that has no classical counterpart. While Galilean relativity might superficially "agree" with the train observer’s direct observation, it fails to reconcile the platform observer’s perspective. Special relativity resolves this by fundamentally redefining spacetime.
Alright, let's dive into this. The user is bringing up some interesting points about Galilean relativity and how it compares to special relativity, especially in the context of the train and platform thought experiment. They mentioned the sailor below deck on a ship as an analogy, arguing that in Galilean relativity, the train observer (like the sailor) can't tell they're moving, so their experience aligns with being stationary. They also pointed out that if you flash a light below deck, it should travel at the speed of light relative to the source, not necessarily invariant, but in this case, the prediction would be the same as in special relativity.
First, I need to recall the core principles of both Galilean and special relativity. Galilean relativity is all about the invariance of physical laws across inertial frames and the idea that velocities add up linearly. But it doesn't account for the constancy of the speed of light. Special relativity, on the other hand, introduces the postulate that the speed of light is constant in all inertial frames, leading to effects like time dilation and length contraction.
The user is correct in saying that in Galilean relativity, an observer in a closed system (like the sailor below deck) can't detect uniform motion. This is the principle of relativity as Galileo formulated it. If the sailor shines a light, in Galilean terms, the speed of that light should be c relative to the ship. But here's the catch: if the ship is moving relative to, say, the water, then an observer on the dock (platform) would measure the speed of light as c plus or minus the ship's velocity, depending on the direction. This is classical velocity addition.
But the user is pointing out that even though Galilean relativity doesn't enforce the invariance of light speed, in this specific scenario, both the sailor/train observer and the platform observer might agree on the simultaneity of events, similar to how it works in special relativity. However, this seems contradictory because in Galilean physics, simultaneity is absolute, while in special relativity, it's relative.
Wait, no. In special relativity, simultaneity is relative, meaning that two events simultaneous in one frame aren't necessarily simultaneous in another. But in Galilean relativity, simultaneity is absolute. So if the train observer sees the light reach both ends simultaneously, the platform observer should also agree, after accounting for the time it takes for the light to travel, adjusted by the train's velocity. But the user is saying that in the train's frame, the light reaches both ends simultaneously, and in the platform frame, it doesn't, which aligns with special relativity. However, in Galilean terms, this shouldn't happen because velocities add, so the platform observer should see different light speeds depending on direction, leading to different arrival times, but they would still adjust for those speeds and possibly still see simultaneity.
But the key difference is that in special relativity, the speed of light is invariant, so you don't adjust for the velocity of the source. In Galilean, you do. So if the train is moving at v, the light going forward would have speed c - v, and backward c + v, leading to different travel times, and thus non-simultaneity from the platform's perspective. But the train observer, using Galilean physics, would think the light is moving at c relative to the train, so they would see simultaneity. However, if they were to consider their motion relative to the platform, they would adjust the light's speed accordingly, leading to non-simultaneity. But in practice, since they can't detect their motion (Galilean relativity), they just assume they're stationary and see simultaneity.
So the user is pointing out that even though Galilean relativity doesn't enforce light speed invariance, in this particular setup, the predictions align with special relativity. But is that really the case? Let's think carefully.
In Galilean relativity, the train observer sees light moving at c in all directions because they consider themselves stationary. The platform observer sees the light moving at c ± v. Therefore, the platform observer would calculate different travel times for light going to the front and back of the train. However, if the train observer assumes they are stationary, they don't account for their velocity v, so they see simultaneity. But if they were to acknowledge their velocity v relative to the platform, they would apply velocity addition and predict non-simultaneity. But since Galilean relativity says they can't detect their motion, they stick with the stationary frame assumption.
In special relativity, the train observer sees simultaneity because they are in their rest frame, and the platform observer sees non-simultaneity due to the invariance of the speed of light and the relativity of simultaneity. The key difference is that in special relativity, the invariance of c is a fundamental postulate, whereas in Galilean, it's not, but in this scenario, the train observer's experience coincides with special relativity's predictions because they ignore their motion relative to the platform.
So the user is correct in saying that in the train frame, the experience aligns with both Galilean and special relativity, but for different reasons. In Galilean, it's because they can't detect their motion and assume light moves at c relative to them, while in special relativity, it's because c is invariant in all frames. However, the platform observer's perspective differs: in Galilean, they would adjust for the train's velocity and see non-simultaneity due to light speed being c ± v, whereas in special relativity, they see non-simultaneity even though light speed remains c, due to the relativity of simultaneity.
Therefore, the user's analogy with the sailor below deck holds in the sense that the observer inside the moving system (train or ship) can't detect their motion and thus experiences phenomena as if they were stationary. But the platform observer's perspective reveals the differences between Galilean and special relativity. The user is emphasizing that even without the invariance of c, the train observer's experience aligns with both theories, but the platform observer's view shows the necessity of special relativity's postulates.
In conclusion, the user is highlighting that while Galilean relativity doesn't require light speed invariance, in this specific setup, the train observer's experience of simultaneity matches both theories, but the platform observer's differing perspective necessitates the framework of special relativity to account for the invariance of c and the relativity of simultaneity.
The analysis of the train and platform thought experiment reveals nuanced distinctions between Galilean and Special Relativity, particularly regarding the perception of simultaneity and the behavior of light:
Key Points of Agreement and Discrepancy:Train Observer's Perspective:
Galilean Relativity:
The train observer, like Galileo's sailor below deck, cannot detect uniform motion. They assume the train is stationary and see light traveling at speed relative to the train.
Light reaches both ends simultaneously because distances are equal, and the observer does not account for the train’s motion relative to the platform.
Special Relativity:
The train observer also sees light reach both ends simultaneously, but this arises from the invariance of light speed () in all inertial frames, not merely an inability to detect motion.
Outcome: Both frameworks predict simultaneity for the train observer, but for different reasons. Galilean relativity relies on ignorance of motion, while special relativity enforces light’s invariance.
Platform Observer's Perspective:
Galilean Relativity:
Light speed is not invariant. The platform observer measures light speed as (forward/backward relative to the train’s motion).
Using classical velocity addition, the platform observer calculates different travel times for light to reach the front () and back (), resulting in non-simultaneity.
Special Relativity:
Light speed remains in all directions, independent of the train’s motion.
Non-simultaneity arises from the relativity of simultaneity, a spacetime effect caused by the invariance of and the train’s motion.
Outcome: Both frameworks predict non-simultaneity for the platform observer, but Galilean relativity attributes it to variable light speed, while special relativity attributes it to spacetime structure.
Critical Distinctions:
Light Speed Invariance:
Galilean relativity allows light speed to depend on the source’s motion ().
Special relativity enforces as invariant, regardless of the source’s motion.
Simultaneity:
In Galilean relativity, simultaneity is absolute once signal delays (e.g., light travel time) are corrected.
In special relativity, simultaneity is fundamentally relative; events simultaneous in one frame need not be in another.
Role of Motion Detection:
Galilean relativity assumes observers cannot detect uniform motion (e.g., sailor below deck), leading to apparent agreement with special relativity in the train frame.
Special relativity requires no such assumption—light’s invariance and spacetime structure inherently resolve contradictions.
Conclusion:
Train Observer:
Both frameworks predict simultaneity, but Galilean relativity does so through ignorance of motion, while special relativity does so through light’s invariance.Platform Observer:
Non-simultaneity arises in both frameworks, but for fundamentally different reasons:Galilean: Variable light speed ().
Special Relativity: Invariant light speed () and spacetime relativity.
Necessity of Special Relativity:
While the train observer’s experience superficially aligns with both theories, the platform observer’s perspective exposes the limitations of Galilean relativity. Only special relativity reconciles light’s invariance with the relativity of simultaneity, making it indispensable for describing relativistic effects.
This thought experiment underscores how special relativity transcends classical mechanics by redefining spacetime, even when certain predictions appear similar at first glance.