The Relativity of Simultaneity

[Motor Daddy]

James, I have asked him several times for any evidence that the speed of light will change relative to the observers movement and he simply ignores the request. I think that he has realized at some point that he is wrong but is stubbornly refusing to admit it. It is really very simple, if MD is correct then the speed of light as measured from earth should vary continuously through the year as the direction of our orbital speed changes relative the stationary frame (wherever that is). MD knows this is not true so he just ignores the whole problem. Not to scientific...

I don't know that to be not true. Don't put words in my mouth. How do you explain your delusion that the light is 300,000,000 meters in front of the train after 1 second? This can be tested by putting a receiver at the 300,000,000 meter mark from the line on the tracks. The receiver receives the light at the 300,000,000 meter mark, and the train is half way between the receiver and the start line. The light is 150,000,000 meters in front of the train after 1 second. You're full of it.

 

[Pete]

Not exactly.

1. You say the clock was slowly moved to the front? Are you now saying that it will be in sync with the other clock when it gets there? If it moves faster will it still work? How about even faster? What is the limit to still being in sync when it gets there and being out of sync when it gets there, as far as the rate at which you move it there?

Good question. We'll find out when we do the numbers.
Or, there are other ways he could try to synchronize the clocks. Is there an particular method you have in mind?

2. You mention milliseconds. What is that?? How would he know such a thing when he doesn't know if his clocks are calibrated to the standard second?

He knows what milliseconds are, and he knows how velocity affects clocks, eg he knows the equation:

t_forward = time in milliseconds = gamma.(t_1b' - t_1a')​

He can't directly use that equation because he doesn't know the value of gamma, but he can indirectly use it to get a useable equation for v, as shown.

Note that the final equation for velocity only uses the direct readings from his clocks.

 

[Motor Daddy]

He may have knowledge of a millisecond, but he has no way to measure them. His clocks tick, that's it. He has no standard of time. For that matter he has no standard of distance either. Heck, he has no standards, so he can't measure the speed of light.

 

[Pete]

He doesn't measure milliseconds, they only appear as a step in the derivation of the velocity equation.
Look at the algebra.
Do you understand how the last equation is derived?
Do you see that it only uses values in ticks, not milliseconds?

 

[Motor Daddy]

No I don't agree with it.


You need to measure the distance light traveled from rear to front and front to rear, independent of the train. The distance light travels is not relative to the length of the train.

In other words, if the train has a velocity, and the train is say 10 meters in length, and it takes one second for light to travel the length of the train, the distance that light traveled is NOT 10 meters. Light traveled more than 10 meters. It had to, because the train had a velocity and light travels independently of the train.

 

[Pete]

Yes, that has all been accounted for in the algebra.
Which step/s in the derivation do you not agree with?

Good night.

 

[origin]

I don't know that to be not true.

Don't you think you find out then since that is the basis of your analysis, fercryinoutloud??? Rather incredible

 

[Motor Daddy]

Don't you think you find out then since that is the basis of your analysis, fercryinoutloud??? Rather incredible

What's incredible is that you think the light is 300,000,000 meters in front of the train. Prove it! I've shown the light is 150,000,000 meters in front of the train after 1 second. You show me how you figure it to be 300,000,000 in front of the train and 450,000,000 meters from the line after 1 second?

You can't do simple 2nd grade math.

 

[origin]

You can't do simple 2nd grade math.

Yuh huh!

Can't find any of that darned evidence to back up your claim - hmmm.

Now, I will shut up and let Pete continue.

 

[Motor Daddy]

Now the train observer will set up synchronized clocks at each end of the train, and use them to measure the train's velocity.

He proceeds like this:

• Two clocks are synchronized while together at the rear of the train.
• One clock is slowly moved to the front of the train.
• A flash of light is sent from the rear clock to the front clock.
• A flash of light is sent from the front clock to the rear clock.
• One distance measurement (in rulers) is taken
  • d' = the distance between the clocks
• Four time readings (in ticks) are taken:
  • t_1a' is the reading on the rear clock when the forward flash is sent.
  • t_1b' is the reading on the front clock when the forward flash is received.
  • t_2a' is the reading on the front clock when the rearward flash is sent.
  • t_2b' is the reading on the rear clock when the rearward flash is received.

The readings will be used to calculate the train velocity, after a little algebra:

• d / t_forward = c - v
• d / t_rearward = c + v
• therefore:
  $$\frac{c-v}{c+v} = \frac{t_{rearward}}{t_{forward}}$$
• t_forward = time in milliseconds = gamma.(t_1b' - t_1a')
• t_rearward = time in milliseconds = gamma.(t_2b' - t_2a')
• therefore:
  $$\frac{c-v}{c+v} = \frac{t_{2b}' - t_{2a}'}{t_{1b}' - t_{1a}'}$$
• rearranging:
  $$v = c \frac{(t_{1b}' - t_{1a}') - (t_{2b}' - t_{2a}')}{(t_{1b}' - t_{1a}') + (t_{2b}' - t_{2a}')}$$

OK?

My questions and objections:

I've already questioned the sync method.

The distance between the clocks is sticks, not meters, correct? So he has 374740 big sticks and 1 little stick. Would it be acceptable to make 2 equal length sticks, or 7 equal length sticks, or 10 equal length sticks, in order to have equal length sticks? There are 374741 sticks on the floor, 374740 equal in length, and 1 unequal to the other 374740. How do you reconcile that in your calculations?

Just to make sure we are on the same sheet of music, he has no clue of the length of a meter or how many sticks it takes to make a meter. A meter may be .3427 stick, or 7 sticks, or 6.29 sticks.

He also has no clue as to the duration of a second, or how many ticks elapse in the duration of a second, if even 1 tick. For all he knows a tick may be 2 seconds, or 30 seconds, or .1234 seconds.

What units do you plan on using for c, sticks/tick?

What units do you plan on using for v, sticks/tick?

The amount of ticks it takes for light to travel from the front to the rear, and then the rear to the front are absolute ticks, correct? In other words, if 10 ticks occur during the time it takes light to travel from rear to front, that is 10 absolute ticks, and nothing can be changed, because that was a reality. You can not multiply or divide the ticks by anything, because the ticks were counted, and that can't be changed.

You say

d / t_forward = c - v

So you would express this as sticks/ticks=sticks/ticks - sticks/ticks

Is that what you are saying?

In order to define a unit of measure "stick" using the speed of light, how many "ticks" does light travel in order to travel the length of a stick? In other words, a meter is defined as the length of the path traveled by light in vacuum in 1⁄299,792,458 of a second. What is your definition of a "stick", as far as the amount of ticks that light travels to travel the length of a stick?

Define the duration of a unit of measure "tick."

 

[Motor Daddy]

My recap of the situation:


The embankment has an absolute zero velocity.

The distance between the two points on the embankment (A and B) is 299,792.458 meters.

The embankment observer is at a midpoint between A and B.

The embankment observer was struck by the lights from A and B
simultaneously at .0005 seconds after 12:00:00.

The strikes occurred at A and B simultaneously.

The strikes occurred at exactly 12:00:00 at A and B.

The train is 299,792.458 meters in length

The train has an absolute velocity of 179,875,474.8 m/s

The train observer was struck by the light from B exactly 0.0003125 seconds after 12:00:00

The train observer was struck by the light from A exactly 0.00125 seconds after 12:00:00

The train observer doesn't know the length of a meter.

The train observer doesn't know the length of his sticks.

The train observer doesn't know the length of the train.

The train observer doesn't know the duration of a second.

The train observer doesn't know how many ticks equal 1 second.

The train observer doesn't know the velocity of the train.

The train observer doesn't know the speed of light in the train.

The train observer doesn't have sync'd clocks.

 

[AlexG]

Just a reminder to all engaged with MD, he does not accept the Lorentz transformation. Therefore, ALL his statements are based on there being an absolute frame of reference. MD does not consider any other situation.

 

[Pete]

My recap of the situation:


The embankment has an absolute zero velocity.

The distance between the two points on the embankment (A and B) is 299,792.458 meters.

The embankment observer is at a midpoint between A and B.

The embankment observer was struck by the lights from A and B
simultaneously at .0005 seconds after 12:00:00.

The strikes occurred at A and B simultaneously.

The strikes occurred at exactly 12:00:00 at A and B.

The train is 299,792.458 meters in length

The train has an absolute velocity of 179,875,474.8 m/s

The train observer was struck by the light from B exactly 0.0003125 seconds after 12:00:00

The train observer was struck by the light from A exactly 0.00125 seconds after 12:00:00

Correct.

The train observer doesn't know the length of a meter.

He knows that a metre is the distance light travels in 1/299792458 seconds.
He doesn't yet have a way of reproducing a metre on the train.

The train observer doesn't know the length of his sticks.

Not in metres, but he does know some things about their length.
He knows they are all the same length.
He knows that they are $$1/\sqrt{1-\frac{v^2}{c^2}}$$ metres long, where v is the train's velocity.

The train observer doesn't know the length of the train.

Not in metres. But he does know how many sticks long it is.

The train observer doesn't know the duration of a second.

He knows the standard definition, but again, as yet has no way to reproduce it on the train.

The train observer doesn't know how many ticks equal 1 second.

As for sticks, he knows the relationship between ticks, seconds, and train velocity.

The train observer doesn't know the velocity of the train.

The train observer doesn't know the speed of light in the train.

The train observer doesn't have sync'd clocks.

Correct. Must be frustrating for him.

 

[arfa brane]

Does he accept that a moving charge, say of an electron, is different to a stationary charge, and that the presence/absence of a magnetic field depends on relative motion of observers, of said charge?

The electric/magnetic "relativity" of charged particles kind of blows a big hole in absolute frames of reference, right there. But perhaps MD hasn't learned about electrons yet, or how they radiate energy when they accelerate?

 

[Pete]

My questions and objections:

I've already questioned the sync method.

A valid question. We will in fact find when we do the numbers that it doesn't work.
Have you thought of a reliable sync method?

The distance between the clocks is sticks, not meters, correct? So he has 374740 big sticks and 1 little stick. Would it be acceptable to make 2 equal length sticks, or 7 equal length sticks, or 10 equal length sticks, in order to have equal length sticks?

That would be acceptable. It makes no difference.

There are 374741 sticks on the floor, 374740 equal in length, and 1 unequal to the other 374740. How do you reconcile that in your calculations?

He is able to precisely cut a ruler to any desired fraction of its length.
The last ruler is exactly 0.5725 the length of the others.
Or maybe he has precise microscopic markings on his rulers, so he can read down to a millionth (billionth, trillionth, whatever) of a ruler.

Just to make sure we are on the same sheet of music, he has no clue of the length of a meter or how many sticks it takes to make a meter. A meter may be .3427 stick, or 7 sticks, or 6.29 sticks.

He knows how the length of a ruler and the length of a metre relates to the train's velocity.

He also has no clue as to the duration of a second, or how many ticks elapse in the duration of a second, if even 1 tick. For all he knows a tick may be 2 seconds, or 30 seconds, or .1234 seconds.

Same applies.
In particular, he knows that his rulers are less than a metre long, that his ticks are more than a second long, and the ratio of ticks to seconds is the same as metres to rulers.

What units do you plan on using for c, sticks/tick?
What units do you plan on using for v, sticks/tick?

m/s for both

The amount of ticks it takes for light to travel from the front to the rear, and then the rear to the front are absolute ticks, correct? In other words, if 10 ticks occur during the time it takes light to travel from rear to front, that is 10 absolute ticks, and nothing can be changed, because that was a reality. You can not multiply or divide the ticks by anything, because the ticks were counted, and that can't be changed.

That's correct.

You say:
d / t_forward = c - v

So you would express this as sticks/ticks=sticks/ticks - sticks/ticks

Is that what you are saying?

No, that's:
m / s = m/s - m/s

I'll generally use a dash (eg t') for any variable measured in ticks or rulers.

The following values are in metres, seconds, and m/s:

• d
• t_forward
• t_rearward
• c
• v

All others are in ticks and rulers.

Here are the equations again.
Work through them, and let me know where it's not clear what I'm thinking.

• d / t_forward = c - v
• d / t_rearward = c + v
• therefore:
  $$\frac{c-v}{c+v} = \frac{t_{rearward}}{t_{forward}}$$
• t_forward = time in milliseconds = gamma.(t_1b' - t_1a')
• t_rearward = time in milliseconds = gamma.(t_2b' - t_2a')
• therefore:
  $$\frac{c-v}{c+v} = \frac{t_{2b}' - t_{2a}'}{t_{1b}' - t_{1a}'}$$
• rearranging:
  $$v = c \frac{(t_{1b}' - t_{1a}') - (t_{2b}' - t_{2a}')}{(t_{1b}' - t_{1a}') + (t_{2b}' - t_{2a}')}$$

 

[Pete]

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:

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

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

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

 

[Pete]

I am of course happy to use your ideas of how the train observer could measure his ticks, rulers, or velocity.

 

[James R]

Motor Daddy:

What's incredible is that you think the light is 300,000,000 meters in front of the train. Prove it! I've shown the light is 150,000,000 meters in front of the train after 1 second.

Did you read any of my previous posts?

You haven't shown anything. You have assumed that light will be the same distance in front of the train in the train's reference frame as it will be in the embankment's frame. And you have zero evidence that that's a valid assumption.

Once again, you aren't taking reference frames seriously. You want to work in the frame of the embankment all the time, and it seems you have a mental block on picturing the frame of the train.

You show me how you figure it to be 300,000,000 in front of the train and 450,000,000 meters from the line after 1 second?

Sure. If Einstein is correct, then light travels at the same speed in every frame, including the train's frame. Therefore, light travels at 300,000,000 metres per second in the train's frame. Since the train is stationary in its own frame, light will move ahead of it at 300,000,000 m/s. So, after 1 second, the light will be 300,000,000 m ahead of the train.

You can't do simple 2nd grade math.

Can you see a problem with the maths in my last paragraph? It all seems simple enough to me.

 

[James R]

Motor Daddy:

You seem to have missed reading post #334, since there has been no reply from you concerning that post. I look forward to your response.

 

[Motor Daddy]

Yes. I agree. That's an equivalent way of putting it using your terminology. You agree that this is very different from Einstein's second postulate, don't you?

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 would have been overjoyed to finally understand the way it really works. He was a very confused individual. Long on BS and short on logic.

Einstein agrees that the speed of light is not determined by the speed of any object. But Einstein says the speed of light is the same in all reference frames. So, not only is it not determined by objects, it is not determined by your kind of "space" either.

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.

As a matter of reality, you will not. IF your assumption of absolute space were correct, then your conclusion would be correct, too. But you're wrong, as a matter of actual fact, verified by countless actual real-world experiments.

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

All of that is true in actual fact in the frame of the embankment. It is NOT, in fact, true in the frame of the the train. Or, to put it another way, the train never moves in its own frame, so the speed of light must be 299792458 metres per second in its frame, if Einstein's second postulate is correct. If Einstein's second postulate were wrong and you were correct, then the measured speed of light on the train would have to change.

Einstein's second postulate is wrong. Sorry, that's a fact. I know it hurts your feelings and you have emotional ties to the old guy's ways, but sometimes you have to let it go. A frame posses it's own velocity in space. The speed of light is not dependent on the frames velocity. When will you understand that, James?

It's not rubbish. It's just counterintuitive. Your gut feeling that it is wrong doesn't prove anything. That's what you don't seem to understand. You imagine how you think things ought to work, but they don't in fact work that way. Why do you get things wrong when you apply your common sense? Answer: you live in a "low speed" world, where nothing you encounter in your daily life ever moves at a reasonable fraction of the speed of light. So, in your everyday experience, relativistic effects are so tiny that you never notice them. But that doesn't mean they aren't there. And, moreover, they become very significant indeed when you deal with higher relative speeds.

It is rubbish. Pure unadulterated rubbish!

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?

No, I'm not wrong. The embankment always has zero velocity in its own reference frame. And the train always has zero velocity in its own reference frame. You have to start taking this reference frame stuff seriously if you want to understand relativity - even common-sense Newtonian relativity.

I certainly understand relative motion, but you seem to get relative motion confused with the speed of light compared to objects. You really need to do a gut check and ask yourself why you believe such nonsense.

In the embankment's frame, the train moves and the embankment is stationary. In the train's frame, the embankment moves and the train is stationary. And there's no physical experiment you can do that will tell you which frame is "absolutely" stationary. That's in spite of your claim that you can use light to measure the absolute speeds, because your assumption that you can do that is wrong as a matter of observed fact.

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?

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

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.

In a frame where the source is moving, you're right - the source is not at the centre of the light sphere 1 second later. But in a frame where the source is stationary, the source stays at the centre of the sphere at all times.

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.

What Einstein's postulate says is that the speed of light emitted from a source in the middle of your train is observed to move at speed c in both directions by somebody on the train. Therefore, the light wavefront is always the same distance from the source in both directions and the source remains stationary at the centre. The motion of the train is irrelevant.

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. 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?

According to YOUR postulate, things are very different. You say light moves at different speeds in the two directions, as measured in the train's frame. So, light spreads out from the source at different rates depending on its direction of travel and the source does not remain at the centre of the light sphere.

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!

Looking at things from the embankment frame, Einstein says that the source does NOT remain at the centre of the light sphere because light travels at c in both directions in the embankment frame and the source on the train is moving. And you agree with Einstein about the embankment picture, because in your mind the embankment frame is the only truly valid frame (provided it is stationary in "space").

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!

As a matter of reality, Einstein is right and you are wrong, as proven by countless experiments. Nothing you imagine or assert can change that.

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