The Relativity of Simultaneity

[wellwisher]

I always liked the idea of an absolute zero reference which is the same for all references just like C. It always seemed to be a logical consequence of C being the same in all references. This seems necesary to keep the laws of physics the same in all references.

When we include matter and matter/energy interactions, if there was no zero reference, matter could have variable velocity when it interacted with C. This should make the potential for matter/energy transitions different for each reference since going from C to V1 or C to V2 is different.

If we have an absolute zero reference, matter/energy transitions would always go from C to 0, regardless of reference. What that would do is add energy to all references, with the energy being unique to each reference, but after the fact in terms of the laws of physics. This explains the basis for relativistic mass, entropy and the uncertainty principle using only the assumption of zero reference.

Relativistic mass, increases as V approaches C. Since C needs to touch zero reference for mass/energy transitions, the closer to C the reference goes, the more output we get when C goes all the way to 0. This energy will be called relativistic mass and will create uncertainty as well as provide energy for entropy. It may even be a possible source for dark energy/matter.

 

[AlexG]

the absolute zero reference frame

Does not exist. Which is why you're totally wrong.

 

[Motor Daddy]

Does not exist. Which is why you're totally wrong.

Then you're in the same boat as Pete, up the creek without a paddle.

The standard meter and second are defined in the absolute zero reference frame, according to the constant speed of light. That doesn't change unless you change the definitions, but then you changed the speed of light if you changed the definitions of a meter and a second, and you're still in the same boat, just with different numbers.

Pete's problem is that he has disassociated his frame from the standard meter and second by using time dilation and length contraction. Those are both conversions from the standard meter and second. So you either disassociate and be up the creek without a paddle, or you convert and agree with the absolute reference frame, which means there is no relativity of simultaneity.

Take your pick, Pete's boat or my boat. My boat has standards, Pete's doesn't. He will chase his tail in circles.

oh, by the way, if he does decide to define his own speed of light in his frame using the defined units sticks and ticks then he's back to square one, in which in order to communicate to me, he will have to tell me what conversion factor to use for sticks and ticks to meters and seconds, at which time his numbers will agree with the zero velocity frame of the embankment, and in doing so he will be forced to admit that a relativity of simultaneity doesn't exist.

 

[AlexG]

Then you're in the same boat as Pete, up the creek without a paddle.

That would be the boat floating on over 100 years of experimental verification.

What floats your boat, besides your unsubstantiated and unproven assertions?

 

[AlphaNumeric]

So you either disassociate and be up the creek without a paddle, or you convert and agree with the absolute reference frame, which means there is no relativity of simultaneity.

Take your pick, Pete's boat or my boat. My boat has standards, Pete's doesn't. He will chase his tail in circles.

Your boat's standards must involve ignorance and denial of reality. The predictions of relativity are born out every single day to millions of people due to its use in many now common place technologies. Your claims simply do not square with reality. You can whine and complain all you like but the issue is ultimately how Nature behaves and it doesn't work as you claim. Whether or not you accept the 'philosophy' of special relativity you cannot deny its applicability and accuracy.

Unless of course you're just utterly detached from reality and a deluded wack job.

 

[Dywyddyr]

Unless of course you're just utterly detached from reality and a deluded wack job.

By George! I think you've found the solution!

 

[Pete]

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

No. That's my calculation.

This is the train observer's calculation:

0.5 ticks / 2 ticks = 0.25

v/c = (1 - 0.25) / (1 + 0.25) = 0.75 / 1.25 = 0.6

v = 0.6c = 179875474.8 m/s

 

[Pete]

Motor Daddy, this is really simple.

If the train observer has synchronized clocks, then he can measure that light takes four times longer to go through the ship forward than backward.

And the only way that can happen is if the ship is moving at 0.6c.

 

[James R]

Motor Daddy:

Why are the train's clocks dilated when it is assumed to be at rest?

We're comparing the train's clocks with the embankment's clocks here. An observer on the train does not think the train's clocks are dilated in any way. The train observer says the embankment clocks are dilated, because in the train's frame the embankment moves and the train is at rest. And the reverse applies to the embankment observer, who says his own clocks are fine and the train clocks are dilated.

If it's at rest, its clocks match the absolute zero velocity frame's clocks (the track's clocks).

The train isn't at rest with respect to the tracks. It is travelling at 0.5c along the tracks. Or, equivalently, the train is at rest and the tracks are travelling backwards at 0.5c relative to the train.

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.

No. The train is always at rest in its own frame. How many times do I need to say that? Every object is at rest in its own frame. If you understand what a reference frame is, as you say you do, then this one is a no-brainer.

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

The train travels at 0.5c relative to the tracks. That's the only thing that matters.

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?

There is no absolute zero velocity. Only relative zero velocity. And the train is not at zero velocity relative to the tracks. It's at 0.5c.

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?

Yes, you can work in any frame you like, including on in which the train travels at 274,000 m/s with respect to that particular frame. But the question you asked concerned how far ahead of the train the light was in the embankment frame or the train frame, so I worked in the two frames you asked about. Obviously.

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

Yes. Just pick any frame that is not stationary relative to the train and the tracks. For example, use a frame travelling in the direction of the train at 0.1c. In that frame, the tracks are travelling backwards at 0.1c, and the train is travelling forwards at less than 0.5c.

The standard meter and second are defined in the absolute zero reference frame, according to the constant speed of light.

No they aren't. There's no reference to any absolute zero frame in the definition of the metre and the second. A brief internet search will quickly confirm that for you.

You must use the definitions that actual physicists use. You can't just make up your own and pretend they are real.

oh, by the way, if he does decide to define his own speed of light in his frame using the defined units sticks and ticks then he's back to square one, in which in order to communicate to me, he will have to tell me what conversion factor to use for sticks and ticks to meters and seconds, at which time his numbers will agree with the zero velocity frame of the embankment, and in doing so he will be forced to admit that a relativity of simultaneity doesn't exist.

Not true. The Lorentz transformations give a precise translation between spacetime coordinates in any two frames. They allow us to convert lengths and times between frames. But when you've done the conversions, you find that lengths and times are relative (i.e. frame-dependent - different in each frame). There's no problem with that. Relativity is completely self-consistent. Unless you can show that it isn't, that is.

 

[Motor Daddy]

Motor Daddy, this is really simple.

If the train observer has synchronized clocks, then he can measure that light takes four times longer to go through the ship forward than backward.

And the only way that can happen is if the ship is moving at 0.6c.

Ok. So he understands that the train must be in motion because it takes light a different amount of time to travel the same length of the train, depending on which way he measures it. What does he think the train is in motion relative to?

 

[Pete]

Ok. So he understands that the train must be in motion because it takes light a different amount of time to travel the same length of the train, depending on which way he measures it.

Yes! At least, he would if he could synchronize his clocks.
But can he do that?

What does he think the train is in motion relative to?

We've had that discussion already. [post=2751811]Remember?[/post]

What is his concept of "at rest?" At rest compared to what?

At rest means that light takes the same time to go from front to back as back to front.

So he bases his concept of "at rest" on light, not on objects such as tracks etc...

In other words, if he thought the train was riding on tracks, he couldn't measure the train's velocity by measuring the distance the train traveled down the tracks and get a correct answer unless he knew the tracks to have a zero velocity. So he realizes that the tracks could have a velocity and the train could have a separate velocity, correct?

Yes

So since he bases his concept of "at rest" on light, does he realize that IF the train were to have a velocity that he would not measure the speed of light to be c in the train?

Yes.

Noted. Continue.

 

[Motor Daddy]

Yes! At least, he would if he could synchronize his clocks.
But can he do that?

He can, but maybe with a slight degree of inaccuracy. The point being, even if slightly inaccurate, the absolute motion is detected. So he knows his train is in motion, he just might have a degree of inaccuracy due to his failure to syn'c two clocks with 100% accuracy.

Would you suggest he throws all concepts and measurements away because of a possible slight inaccuracy? Would you throw away your watch and all concepts of time if you thought it was slightly inaccurate over the course of 1,000 years?

We've had that discussion already. [post=2751811]Remember?[/post]

Yes, just reiterating my point.

 

[Pete]

He can, but maybe with a slight degree of inaccuracy. The point being, even if slightly inaccurate, the absolute motion is detected. So he knows his train is in motion, he just might have a degree of inaccuracy due to his failure to syn'c two clocks with 100% accuracy.

Let's see the numbers.

Two clocks are synchronized at the back of the train.
One of the clocks is slowly moved to the front of the train.

The train is moving at 179875474.8 m/s. (gamma=1.25)
Let's say that when the clock is moved, it is moving at 179875474.9 m/s. (gamma=1.2500000004)

The train is 299792.458 m long, so it takes 2997924.58 seconds for the clock to reach the front of the train.

While the clock is being moved the 299792.458 m length of the train:

• 2997924.58 seconds elapse.
• The clock that remained at the rear of the train elapses 2,398,339,664 ticks.
• The clock being moved elapses 2,398,339,663.25 ticks.

So we know (but the train observer does not) that the front clock is 0.75 ticks behind the rear clock.


Now, he does the measurement.

• The rear-to-front measurement should be 2 ticks, but because the front clock is 0.75ms behind, the actual measurement is 1.25 ticks.
• The front-to-rear measurement should be 0.5 ticks, but because the front clock is 0.75ms behind, the actual measurement is 1.25 ticks.

Because of the synchronization failure, the train observer has measured the same time for light to travel each way.


Should he conclude that he is at rest?
Is there a better synchronization method?

 

[Motor Daddy]

No. That's my calculation.

This is the train observer's calculation:

0.5 ticks / 2 ticks = 0.25

v/c = (1 - 0.25) / (1 + 0.25) = 0.75 / 1.25 = 0.6

v = 0.6c = 179875474.8 m/s

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

You say the train observers calculations are .5 ticks and 2 ticks. But how did you know the .5 and 2 ticks without using v in the equation??

Show me how you know the ticks without using v?

 

[Pete]

You say the train observers calculations are .5 ticks and 2 ticks. But how did you know the .5 and 2 ticks without using v in the equation??

Show me how you know the ticks without using v?

We know the train's velocity, so we can how long it takes light to go through the train, and we know how many ticks elapse on the clocks.
It's right there in the post you quoted.

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

 

[Motor Daddy]

We know the train's velocity, so we can how long it takes light to go through the train, and we know how many ticks elapse on the clocks.
It's right there in the post you quoted.

Let me get this straight:

The train observer measures the ticks from front to rear and rear to front.

The clocks tick .5 ticks in one direction and 2 ticks in the other direction.

Then what?

 

[Pete]

Let me get this straight:

The train observer measures the ticks from front to rear and rear to front.

If he had synchronized clocks, that's what he'd measure.

The clocks tick .5 ticks in one direction and 2 ticks in the other direction.

Then what?

Then he knows that the light travel time (whether in ticks or in seconds) is four times longer in one direction than the other.

So he knows that:
$$\frac{d/(c-v)}{d/(c+v)} = 4$$

$$\frac{c+v}{c-v} = 4$$

$$c+v = 4(c-v)$$

$$c+v = 4c - 4v$$

$$5v = 3c$$

$$v = 0.6c$$

v = 179875474.8 m/s

 

[Motor Daddy]

Should he conclude that he is at rest?

No

Is there a better synchronization method?

You've already shown that if the clocks are in fact absolutely sync'd that the numbers are the same as in the embankment. So according to you, Einstein's entire theory of the relativity of simultaneity is based on unsync'd clocks.

He knows the relativity of simultaneity is non existent, but he can't find a way to sync clocks, correct?

 

[Motor Daddy]

If he had synchronized clocks, that's what he'd measure.

So he hasn't actually measured the ticks yet, because he has no sync'd clocks. He doesn't know how many ticks elapse, nor has he measured the speed of light or the length of the train or his rulers. YOU know the information only in reference to the tracks, but he doesn't know anything yet.

Then he knows that the light travel time (whether in ticks or in seconds) is four times longer in one direction than the other.

So he knows that:
$$\frac{d/(c-v)}{d/(c+v)} = 4$$

$$\frac{c+v}{c-v} = 4$$

$$c+v = 4(c-v)$$

$$c+v = 4c - 4v$$

$$5v = 3c$$

$$v = 0.6c$$

v = 179875474.8 m/s

But he hasn't actually measured anything yet. That's your numbers, not his. So far your numbers match the embankment's numbers, so there is no relativity of simultaneity at this point.

 

[Pete]

But he hasn't actually measured anything yet. That's your numbers, not his.

Yes he has, he measures the ticks.

Obviously, I have to calculate what he would measure.

So he hasn't actually measured the ticks yet, because he has no sync'd clocks.

He does make a measurement, and he does know the speed of light, because he knows the rules of this mathematical world.

You've already shown that if the clocks are in fact absolutely sync'd that the numbers are the same as in the embankment. So according to you, Einstein's entire theory of the relativity of simultaneity is based on unsync'd clocks.

He knows the relativity of simultaneity is non existent, but he can't find a way to sync clocks, correct?

All correct.