The Relativity of Simultaneity

[arfa brane]

The speed of light is the same in all reference frames. It's the velocity of the reference frames that changes, and that causes one to measure the speed of light incorrectly, unless of course they account for their frame's velocity in the calculations.

If the first sentence above is correct, then the speed of light is independent of the velocity of a reference frame, so all the sentences after the first one are based on a misunderstanding of what the first sentence actually means. If you measure the speed of light in ANY reference frame, it's always the same speed, c.

Note that Einstein isn't initially concerned about proving that the speed of light is measured to be the same in all reference frames, but is concerned with the logical implications of the assumption. Today, we know it isn't based on conjecture, but on physics.

 

[Motor Daddy]

I've thought of a clearer explanation.

The train observer can't measure t_forward or t_rearward in seconds.
But, he can measure $$t_{forward} \over t_{rearward}$$, because that ratio is the same for ticks as it is for seconds:

$$\frac{t_{forward}}{t_{rearward}} = \frac{\gamma.t_{forward}'}{\gamma.t_{rearward}'} = \frac{t_{forward}'}{t_{rearward}'}$$

He can then use that ratio to calculate the ratio of his absolute velocity to c through this equation:

$$\frac {c-v}{c+v} = \frac {t_{rearward}}{t_{forward}}$$

$$\frac {c-v}{c+v} = \frac {t_{rearward}'}{t_{forward}'}$$

$$\frac{v}{c} = \frac {t_{forward}' - t_{rearward}'}{t_{forward}' + t_{rearward}'}$$

Just show me what you have.

 

[Motor Daddy]

If the first sentence above is correct, then the speed of light is independent of the velocity of a reference frame, so all the sentences after the first one are based on a misunderstanding of what the first sentence actually means. If you measure the speed of light in ANY reference frame, it's always the same speed, c.

There is one frame that Einstein measures light, and that is a zero velocity frame, because he always assumes his measurement frame to be at rest. But when a real zero velocity frame is present, such as the tracks in this case, it is very clear that Einstein is wrong. The light from a train that is moving .5c can't be 300,000,000 meters in front of the train after 1 second has elapsed, and at the same time be 450,000,000 meters from the line on the tracks it was emitted. Take your pick, did the light travel 1c, or 1.5c, it's not both.

Note that Einstein isn't initially concerned about proving that the speed of light is measured to be the same in all reference frames, but is concerned with the logical implications of the assumption. Today, we know it isn't based on conjecture, but on physics.

I'm not concerned about proving the speed of light either, it is defined, so you don't need to measure it. It is always 299,792,458 m/s, by definition.

 

[Pete]

Do you understand that the train observer can calculate his velocity if he has synchronized clocks, using this equation:

$$\frac{v}{c} = \frac {t_{forward}' - t_{rearward}'}{t_{forward}' + t_{rearward}'}$$

 

[Motor Daddy]

Do you understand that the train observer can calculate his velocity if he has synchronized clocks, using this equation:

$$\frac{v}{c} = \frac {t_{forward}' - t_{rearward}'}{t_{forward}' + t_{rearward}'}$$

The train observer has no way of knowing the conversion of ticks to seconds or sticks to meters. For that matter, he has no units for c, or v. He can not use units of meters or seconds because he doesn't know the conversion factor. If he did know the conversion factor, his numbers would be exactly the same as the embankment.

 

[Pete]

The train observer has no way of knowing the conversion of ticks to seconds or sticks to meters.

That's correct.

Do you understand the previous equation?

Do you understand that $$t_{forward} \over t_{rearward}$$ (in seconds/seconds) is the same as $$t_{forward}' \over t_{rearward}'$$ (in ticks/ticks)?

 

[Motor Daddy]

That's correct.

Do you understand the previous equation?

Do you understand that $$t_{forward} \over t_{rearward}$$ (in seconds/seconds) is the same as $$t_{forward}' \over t_{rearward}'$$ (in ticks/ticks)?

No, I don't see how you get seconds from ticks.

 

[arfa brane]

There is one frame that Einstein measures light, and that is a zero velocity frame, because he always assumes his measurement frame to be at rest.

I don't want to upset any progress that may have been made here, but that just isn't true.

Einstein makes measurements in a stationary frame only when clocks need to be synchronised. In Einstein's universe, you can only synchronise clocks if they area at rest relative to each other.

That's all it means; it doesn't mean that both clocks have zero velocity relative to another frame of reference, it means they have the same velocity relative to each other, namely zero velocity. That means they remain at a constant distance from each other, and light travels a constant distance in a constant time, from one to the other. That's really simple, but you don't seem to believe it's possible.

 

[Pete]

No, I don't see how you get seconds from ticks.

I don't get seconds from ticks.
I get a ratio from a ratio.

If the train observer knows that t_frontward' (in ticks)is twice as long as t_rearward' (in ticks), then he also knows that t_frontward (in seconds) is twice as long as t_rearward (in seconds).

Right?

 

[Motor Daddy]

I don't want to upset any progress that may have been made here, but that just isn't true.

Einstein makes measurements in a stationary frame only when clocks need to be synchronised. In Einstein's universe, you can only synchronise clocks if they area at rest relative to each other.

That's all it means; it doesn't mean that both clocks have zero velocity relative to another frame of reference, it means they have the same velocity relative to each other, namely zero velocity. That means they remain at a constant distance from each other, and light travels a constant distance in a constant time, from one to the other. That's really simple, but you don't seem to believe it's possible.

Another frame's velocity has nothing to do with that frame's velocity in space. Every frame has its own absolute velocity in space, including a zero velocity frame. The speed of light is relative to the zero velocity frame, because the zero velocity frame is the standard. The zero velocity frame is where light travel time is equal to distance in that frame, and by definition, a meter is a specific light travel time. v=(ct-l)/t

 

[Motor Daddy]

I don't get seconds from ticks.
I get a ratio from a ratio.

If the train observer knows that t_frontward' (in ticks)is twice as long as t_rearward' (in ticks), then he also knows that t_frontward (in seconds) is twice as long as t_rearward (in seconds).

Right?

Is the clock in the direction of motion dilated and the other direction the opposite? Or, better said, they both move in the same direction, but the dilation is different if you move one to the rear and one to the front?

 

[James R]

Motor Daddy:

Yes I do agree that it's very different than Einstein's. He got a little confused and made a mistake. Poor guy, I wish he were alive today so I could set him straight.

He didn't make a mistake, because all the experiments that have been done show he was right and you are wrong. Looks like you made the mistake.

The speed of light is the same in all reference frames.

The measured speed of light is the same in all reference frames - that's Einstein. You say the measured speed of light is different in every frame.

So you agree my method is self consistent and mathematically sound?

Yes. It's self-consistent and mathematically sound. It's the underlying assumption of absolute space and time that is completely and utterly wrong. Garbage in - garbage out.

Einstein's second postulate is wrong. Sorry, that's a fact.

Show me one piece of actual real-world experimental evidence that proves this supposed "fact" of yours.

You know you cannot, because there isn't any. All real-world experiments prove you wrong.

The speed of light is not dependent on the frames velocity. When will you understand that, James?

Don't patronise me. I understand your arguments just fine. They are based on a faulty starting premise, that's all.

Explain how you reconcile the light from the train being 450,000,000 meters from the line? That is 1.5c, James. Maybe you think the light travels a different velocity when you place the light on the track at the line? If there was a light race between the light on the track and the light on the train, which one would get to the 300,000,000 mark on the tracks first, James? Let's talk about this in such fine detail that we get to the bottom of it, shall we?

Fine. I'm not entirely clear on what this new scenario of yours involves. But let me take a stab at it. Tell me if I'm wrong.

At time t=t'=0, let's say, the train is moving along at 0.5c relative to the track. At that instant, somebody draws a mark on the side of the track, level with the front of the train. Now, at the same instant, a flash of light is fired off from the front of the train and travels in the same direction as the train. After it has travelled for 1 second, a person on the track places another mark at the location of the light pulse. We wish to determine the distance between the two marks and the distance between the second mark and the front of the train at the instant the second mark is made.

Now, you've already done the calculation in your favorite reference frame - the frame of the track. In that frame, light travels 300,000,000 m in 1 second, and the train travels 150,000,000 m in the same amount of time. So, the light (at the second mark), ends up 150,000,000 m in front of the train at the time the second mark is made. The distance between the two marks on the track is, of course, 300,000,000 m.

Next, we need to look at the situation in the train's frame. To do that, we must use the train's rulers and clocks, not the embankment's. In the train's frame, the train is stationary, and according to Einstein the speed of light in that frame is 300,000,000 m/s. But in the train's frame, the light does not take 1 second to travel to the second mark, because relativity tells us that the train's clocks tick at a different rate to the track clocks. If you do the correct relativistic calculation, the light takes 0.866 seconds to go between the first and second marks on the track, as measured using the train's clocks. The relevant time dilation factor at the train's speed is 1.15, and we have 1/1.15 = 0.866. In 0.866 seconds, the light travels approximately 259,800,000 m (at a speed of 300,000,000 m/s) between the two marks. But in the train's frame, the position on the track where the second mark is made has been approaching the front of the train at a speed of 0.5c for 0.866 seconds, so that it ends up in front of the train a distance of 259,800,000 - 129,900,000 m = 129,900,000 m.

Just to check, the distance between the front of the train and the second mark when the light hits it is 150,000,000 m in the embankment frame. The train sees the track/embankment moving at 0.5c, so distances on the track are length-contracted. Therefore, we expect the distance between the front of the train to be less than 150,000,000 m, by a factor of 1.15 (which is the same factor as for the time dilation effects). And we find that 150,000,000/129,900,000 = 1.15, exactly as expected. Also, comparing the distance that light travelled between the two marks, we have 300,000,000 m in the embankment frame, which took 1 second, and 259,800,000 in the train frame, which took 0.866 seconds. Calculating the speed of light in both frames using the given distances and times gives 300,000,000 m/s in both frames, as expected.

Now, I agreed with your that your mathematics is self-consistent given the assumption of absolute space and time. Will you now agree that Einstein's mathematics is consistent given his postulate of the constancy of the speed of light in all frames?

I certainly understand relative motion, but you seem to get relative motion confused with the speed of light compared to objects.

No. You keep wanting to use the speed of light in the embankment frame, and never in the train frame. So, you're not really ever working in the train frame at all.

Wrong, James, I've already shown the embankment to be at an absolute zero velocity. When light takes the same time to travel each direction the same distance in the frame, the absolute velocity is zero, and that means your illusions fall apart. You are living in Einstein's world of illusions. Why can't you see that?

Light always takes the same time to travel the same distance in each direction in any frame, whether on the embankment or in the train, so using your method you must conclude that the absolute velocity of every object is zero, and that means your illusions fall apart. You are living in Motor Daddy's world of illusions. Why can't you see that?

Ask your self honestly what your concept of "at rest" means. Tell me what you think "at rest" means. At rest compared to what?

We've been through this many times. "At rest" is measured in a reference frame. If I don't move relative to rulers that are stationary in a reference frame, then I'm at rest in that frame. Simple.

Right. IF the times are different. But in a single frame, they never are, as a matter of fact.

As a matter of fact you are wrong. You not understanding it doesn't make me wrong.

You're attempting to patronise me again. That won't work. I understand your arguments just fine. Your assumption about absolute space and time is wrong.

Of course the source stays at the center if it has an absolute zero velocity. If it doesn't have a zero velocity then it's no longer at the center. There goes your "light is always 300,000,000 meters from the source 1 second later.

You left out any mention of reference frames from your statement. Let me help you re-write it so it is correct:

"Of course the source stays at the centre if it is stationary in a reference frame. If it doesn't have zero velocity in a particular reference frame, then it doesn't stay at the centre."

So you do 2 tests. One you remain 10 meters from a light and test the time it takes light to reach you. The other test you are at 10 meters, and as soon as the light is emitted you move away from the light. You are telling me that the times are both the same.

Not at all. I agree that in that case it will take light longer to reach you when you're moving away.

That's what you say, James. So if the time is always the same at say 10 and 11 meters, then it's the same at 25 meters, and 698 meters. Do you honestly believe that?

No. That would be silly. What ever gave you that idea? Nothing I have said mentions anything of the kind. You're just setting up a straw man. Try responding to what I actually write, ok?

Light travels at the speed of light from the point in space it was emitted. If a wall on each side of the light moves, one closer to that point and the other farther from that point, then the two times will be different. It's not even debatable, James, it's a rock solid fact!

I agree. But unlike you, I also add that this applies in any reference frame. Also, it is important to note that in the train's frame, the walls of the train never move at all, which is why the times are always the same. The same is not true in the embankment frame, which sees the train walls move.

All frames are valid, as long as you acknowledge that frames are different because they are at a different velocity. You seem to think of frames as all having a velocity, just not your frame. That's absurd, James. There are an infinite amount of frames, all with a different velocity, and yes, there is a zero velocity frame, in which the speed of light is measured to be 299,792,458 m/s. You know why, because the speed of light is DEFINED!

The only velocity that makes any sense when you're talking about reference frames is relative velocity. That's why it's called "relativity". A frame doesn't have a velocity in isolation from the universe. It has a velocity with respect to something else, such as some other frame or object.

And yes, there are an infinite number of frames, and yes, some of them have zero velocity with respect to others. And in all of them the speed of light is 299,792,458 m/s, according to Einstein's second postulate.

No, as a matter of reality, Einstein's world is a world of illusions. My world is reality.

Repeated assertion doesn't prove your case. In fact, it does nothing to even advance your case.

 

[arfa brane]

Every frame has its own absolute velocity in space, including a zero velocity frame.

How do you tell if frame A has a zero absolute velocity?

Does it need you to find a reference point in another frame (let's call it B), or can you tell that frame A has zero velocity without referring to any other physical location? (The answer to that is, that of course you need to refer to another frame of reference, doh!).

The zero velocity frame is where light travel time is equal to distance in that frame, and by definition, a meter is a specific light travel time. v=(ct-l)/t

But a frame of reference doesn't define any distance. But two reference frames allow this; in fact distance requires two distinct points, so how do you tell that two different points remain at a fixed distance, and at zero or any other velocity, without reference to yet another point?

See, you don't seem to have anything like a good grasp of the fundamentals here. You can't even describe what a frame of reference is, and you can't say why or how a single point in space and time can even have zero "absolute" velocity. Velocity of any frame of reference is always relative to another frame of reference, so there cannot be any zero velocity except where the distance between two points remains constant. If A has a velocity relative to B, they must be moving together or apart along some direction of motion. If they have zero relative velocity, they stay at a constant distance (is that not explained very well?), Anyway, at a constant distance, light takes the same time to go from A to B as from B to A, bingo--synchronous events!

Einstein's theory is about how to determine there are such points with a fixed distance between them; it's also about how this "arrangement" is really the exception to a more general rule--everything is in motion.

 

[Pete]

Is the clock in the direction of motion dilated and the other direction the opposite? Or, better said, they both move in the same direction, but the dilation is different if you move one to the rear and one to the front?

They are both moving at the same speed as the train, so they both tick at the same rate.

They will tick at different rates while one is being moved along the train, which is why that syncronization method won't work (no matter how slowly it's moved).

But when they are in place, they both tick at the same rate.


So if the train observer can synchronize his clocks with each other, he can measure his velocity.

In the next post we'll see what happens to his velocity measurement with his current synchronization scheme.

 

[Motor Daddy]

Next, we need to look at the situation in the train's frame. To do that, we must use the train's rulers and clocks, not the embankment's. In the train's frame, the train is stationary, and according to Einstein the speed of light in that frame is 300,000,000 m/s. But in the train's frame, the light does not take 1 second to travel to the second mark, because relativity tells us that the train's clocks tick at a different rate to the track clocks. If you do the correct relativistic calculation, the light takes 0.866 seconds to go between the first and second marks on the track, as measured using the train's clocks.

Why are the train's clocks dilated when it is assumed to be at rest? If it's at rest, its clocks match the absolute zero velocity frame's clocks (the track's clocks).

By you saying the train is at rest, and that the speed of light is measured to be 299,792,458 m/s in the train frame, you are saying the train is at an absolute zero velocity.

Is that your story, James? Is the train at an absolute zero velocity or is it traveling at .5c?

So the train says it's at an absolute zero velocity, and the tracks say that they are at an absolute zero velocity. What is your test to determine which, if either are at an absolute zero velocity?

If you can consider the train to be at rest, and the tracks to be moving, can you also consider the train to be traveling any speed you want to, say 274,000 m/s? If not, why not?

Is there ever a situation where the tracks are moving AND the train is moving at the same time, according to Einstein??

 

[Motor Daddy]

So if the train observer can synchronize his clocks with each other, he can measure his velocity.

In the next post we'll see what happens to his velocity measurement with his current synchronization scheme.

No, he can't measure his velocity, because he can't determine the length of a meter and the duration of a second. He can measure in units of sticks and ticks, but has no conversion factor for either. And in order to know the length of the train he must first know the velocity of the train, and to do that he must first measure the speed of light in the train, independent of the train. But again, he can measure the speed of light in units of sticks and ticks, but in order to do that he must have defined a unit of measure "stick" to be a specific distance that light travels in a vacuum in a specific amount of the unit of measure "tick." In order to use a unit of measure "tick" he must first have defined the duration of a unit "tick."

 

[Pete]

OK, let's try a worked example.

For the moment, let's say the train observer has managed to synchronize the clocks at each end of the train with each other.

Let's see how he calculates his velocity.
When the train clocks read t'=0, a flash of light is sent from the front clock.
t = d/(c+v) = 0.625 ms later, the flash reaches the rear clock, which reads:
Measured rearward time = t/gamma = 0.5 ticks

When the train clocks read t'=0, a flash of light is also sent from the rear clock.
t = d/(c-v) = 2.5 ms later, the flash reaches the front clock, which reads:
Measured forward time = t/gamma = 2 ticks


(Measured rearward time) / (Measured forward time) = 0.5 ticks / 2 ticks = 0.25

So the train observer knows that the ratio of rearward light travel time to forward light travel time is 0.25.

Through a little algebra, it's not difficult to work out that:
v/c = (1 - ratio) / (1 + ratio) = 0.75 / 1.25 = 0.6

The train observer knows that c = 299792458m/s, so it's now trivial for him to work out his velocity:

v = 0.6c = 179875474.8 m/s

 

[Motor Daddy]

OK, let's try a worked example.

For the moment, let's say the train observer has managed to synchronize the clocks at each end of the train with each other.

Let's see how he calculates his velocity.
When the train clocks read t'=0, a flash of light is sent from the front clock.
t = d/(c+v) = 0.625 ms later, the flash reaches the rear clock, which reads:
Measured rearward time = t/gamma = 0.5 ticks

He is trying to calculate his velocity using the equation t = d/(c+v) ?

That's strange, there is a v in there. If he doesn't know his velocity how does he figure out what t is, again?

 

[wellwisher]

Say I was on earth and emitted a powerful light beam and let it travel for one light year to the left. At the same time the beam is pulsed, my friend starts his rocket and travels to the right at V=0.25C. He also travels for one year but in the opposite direction.

The distance I would measure to the light beam would be 1 light year. While the distance to my friend would be 0.25 light years. Therefore the distance between the light beam and my friend will be 1.25 light years. What will be the distances my friend will measure to me and to the light beam?

 

[Motor Daddy]

Say I was on earth and emitted a powerful light beam and let it travel for one light year to the left. At the same time the beam is pulsed, my friend starts his rocket and travels to the right at V=0.25C. He also travels for one year but in the opposite direction.

The distance I would measure to the light beam would be 1 light year. While the distance to my friend would be 0.25 light years. Therefore the distance between the light beam and my friend will be 1.25 light years. What will be the distances my friend will measure to me and to the light beam?

The clock has already stopped, the measurements are complete, and the results are in. The distance between your friend and the light is 374,740,572.5 meters, and the distance between you and your friend is 74,948,114.5 meters.

Their velocities were calculated in the absolute zero reference frame where distance and time are inseparable according to the definition of the meter, the second, and the speed of light. The absolute reference frame is inseparable from the definition of the speed of light.