# Thread: On Einstein's explanation of the invariance of c

1. Originally Posted by Tach
Since you are measuring the SAME distance in both directions, light cannot travel ".1s in one direction" and ".2s in the opposite direction".

You flunked
Wrong again, Tach. I hope someday you will be able to comprehend the difference between the distance light travels, and the distance between clocks. I'm confident with enough dedication and willingness to learn, you'll get there. I'm not 100% sure of that, but I have high hopes for you.

Originally Posted by Tach
Nope, since this is based on the error at the previous point. You will need to take this class over. Or, even better, sit in your dark box and take a break from posting nonsense.
I'm counting on you. Don't let me down!

2. Originally Posted by Motor Daddy
Originally Posted by Tach
Since you are measuring the SAME distance in both directions, light cannot travel ".1s in one direction" and ".2s in the opposite direction".

You flunked
Wrong again, Tach. I hope someday you will be able to comprehend the difference between the distance light travels, and the distance between clocks. I'm confident with enough dedication and willingness to learn, you'll get there. I'm not 100% sure of that, but I have high hopes for you.

I'm counting on you. Don't let me down!
Don't go out of your dark box. Stay!

3. Just think, if you understand it, you are doing better than Einstein, because he certainly never understood it!

4. BTW, since this is second grade math, if you can't understand it, I'm gonna have to hold you back and repeat second grade again. You can't progress to third grade until you know this material.

5. Perhaps I can assist here. Take a look at wikipedia for the definition of the second:

"Since 1967, the second has been defined to be the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom".

This refers to the NIST caesium fountain clock, where lasers and a microwave cavity are employed to cause hyperfine transition - an electron “spin flip” which emits microwaves. There’s a peak frequency in the emitted light, which is measured by a detector. But note that frequency is defined as cycles per second, and the second isn't defined yet. Thus what we're really doing here is counting the incoming microwave peaks, and when we get to 9,192,631,770, we say that's a second. Hence the frequency is 9,192,631,770 Hertz by definition.

Then as MotorDaddy pointed out, we use the second along with light to define the metre:

"Since 1983 the metre has been defined as the distance travelled by light in vacuum in 1⁄299,792,458 of a second".

The hyperfine transition is electromagnetic in nature, as are the emitted microwaves. So if electromagnetic events/effects occur/propagate slower, this will alter the calibration of our rods and clocks. Ditto if they occur/propagate faster. We won't be able to tell because we're effectively using the motion of light to define the second and the metre, which we then use... to measure the motion of light. Hence we always measure the local speed of light in vacuo to be 299,792,458 m/s.

6. Originally Posted by Motor Daddy
BTW, since this is second grade math, if you can't understand it, I'm gonna have to hold you back and repeat second grade again. You can't progress to third grade until you know this material.
making your theory even more laughable, second grade math shows that you produced bunkum.

7. Originally Posted by Tach
making your theory even more laughable, second grade math shows that you produced bunkum.
Would you like it more complicated, so you appear more intelligent? Making things as simple as possible is the goal, remember?

"If you can't dazzle 'em with brilliance, baffle 'em with BS."

8. Originally Posted by Motor Daddy
Would you like it more complicated, so you appear more intelligent? Making things as simple as possible is the goal, remember?
No, the way you conceived your theory is just fine. Only second grade math is sufficient to prove that you produced bunkum.

"If you can't dazzle 'em with brilliance, baffle 'em with BS."
That's precisely what you've been doing.

9. Originally Posted by Tach
No, the way you conceived your theory is just fine. Only second grade math is sufficient to prove that you produced bunkum.

That's precisely what you've been doing.
So do you now understand that it can take you different amount of time to traverse a school bus when you are going a constant 60 MPH in your car, depending on the speed of the bus, and the direction of travel of the bus?

Did you ride the short bus?

10. Originally Posted by Motor Daddy
1. You synchronize two clocks as such that they read the same time, simultaneously, as they would in a single time zone.
...

2. The synchronized clocks are a distance apart from each other. You will measure the length of that distance between the clocks.

3. You send a light signal from one clock towards the other clock and record the time the light signal was sent.

4. You record the time the light signal arrives at the other clock.

5. You subtract the start time from the arrival time to find the amount of elapsed time that the light signal took to go from one clock to the other.

6. You perform steps 3-5 in the opposite direction.

7. You now have the one-way travel times in each direction. You know the speed of light is 299,792,458 m/s, by definition. You therefore know how far light traveled in each direction by simply multiplying the elapsed time by 299,792,458. So for example, if you measured .1 seconds in one direction, and you measured .2 seconds in the opposite direction, light traveled 29,979,245.8 meters in one direction, and 59,958,491.6 meters in the opposite direction.

8. 59,958,491.6-29,979,245.8=29,979,245.8 meters. That means the box traveled 29,979,245.8 meters in .3 seconds, its velocity is 99,930,819.33 m/s.

9. The box traveled 99,930,819.33 m/s for .2 seconds in one direction, so the box traveled 19,986,163.866 meters in .2 seconds, and the light traveled 59,958,491.6 meters in .2 seconds. That means the distance between the clocks is 59,958,491.6-19,986,163.866=39,972,327.734 meters.
1. This only produces relative synch, because you don't know the absolute velocity.

5. The time will include compensating for the movement of the frame.

7. This is what SR specifies in the synch definition, even though you don't agree with it. Multply the time by c, and don't expect anything but d/t = c.

9. Since you ignore time dilation effects, you aren't aware there are many combinations of clock separations and speeds that will indicate the same time.

11. MotorDuddy has convinced himself that he's got something right that no physicist in the last 105 years has gotten right.

Why would anyone think that they can convince him otherwise?

12. Again, there is NO way Einstein's methods can find the distance between the two clocks, so his argument is circular, and he therefore concludes light always takes the same amount of time each way. He is SADLY mistaken, as light travel time is the distance light travels, not the distance between clocks. Unless of course the clocks have an absolute zero velocity, which he has no way of knowing. His argument is purely circular. Mine is not. I know the distance between the clocks. I know the velocity of the box. I know the distance the box traveled, and I know the distance each light traveled in each direction.

13. This is what Einstein says about synchronization:

"[If at the point A of space there is a clock, an observer at A can determine the time values of events in the immediate proximity of A by finding the positions of the hands which are simultaneous with these events. If there is at the point B of space another clock in all respects resembling the one at A, it is possible for an observer at B to determine the time values of events in the immediate neighbourhood of B. But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an ``A time'' and a ``B time.'' We have not defined a common ``time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A.]"

If this is wrong, then it should be possible to compare the times of events at A and B without defining a common time, equivalent to a rest frame for an observer.
light travel time is the distance light travels, not the distance between clocks
Are you saying that, despite the metre being defined in terms of wavelengths of light, the distance between clocks can't be measured with light?

14. Originally Posted by arfa brane
Are you saying that, despite the metre being defined in terms of wavelengths of light, the distance between clocks can't be measured with light?
Of course not, I did just that in my example. I told you the distance between the clocks, using light travel times.

I am saying using Einstein's methods, there is no way of knowing the distance between the clocks. You must know the velocity of the clocks to measure the distance between them using light, and he doesn't know that.

15. Originally Posted by Motor Daddy
You must know the velocity of the clocks to measure the distance between them using light, and he doesn't know that.
Why does Einstein say this then:

"[with the help of certain imaginary physical experiments we have settled what is to be understood by synchronous stationary clocks located at different places, and have evidently obtained a definition of ``simultaneous,'' or ``synchronous,'' and of ``time.'' The ``time'' of an event is that which is given simultaneously with the event by a stationary clock located at the place of the event, this clock being synchronous, and indeed synchronous for all time determinations, with a specified stationary clock.]"

16. Originally Posted by arfa brane
Why does Einstein say this then:

"[with the help of certain imaginary physical experiments we have settled what is to be understood by synchronous stationary clocks located at different places, and have evidently obtained a definition of ``simultaneous,'' or ``synchronous,'' and of ``time.'' The ``time'' of an event is that which is given simultaneously with the event by a stationary clock located at the place of the event, this clock being synchronous, and indeed synchronous for all time determinations, with a specified stationary clock.]"
Stationary compared to what? Does he mean that the distance between the clocks is not changing? So what, that says nothing about their velocity, as they could be two cars traveling down the road, each doing 60 MPH, maintaining the same distance between them. Does that mean they have a zero velocity just because the distance remains the same??

He's full of it!

When he can tell me the velocity of the box, then and only then will he be able to tell me the distance between the clocks.

17. I don't see what the big problem is with "stationary clocks located at different places".

I assume this means the distance between them is constant, as in your example of two cars traveling at 60mph--but presumably in the same direction. Such a comoving system means the cars have zero relative velocity. You know what relative means, don't you?

So the answer to: Stationary compared to what?, is compared to each other, or compared to the constant distance between A and B.

18. Originally Posted by arfa brane
I don't see what the big problem is with "stationary clocks located at different places".

I assume this means the distance between them is constant, as in your example of two cars traveling at 60mph--but presumably in the same direction. Such a comoving system means the cars have zero relative velocity. You know what relative means, don't you?
Anyone can understand that if two cars maintain the same distance apart from each other there is no motion compared to the other. Is that supposed to be some kind of difficult subject to comprehend? We are talking about the motion of light, as compared to the motion of a meter stick. Do you understand the meter stick can have different velocities?

19. Originally Posted by Motor Daddy
We are talking about the motion of light, as compared to the motion of a meter stick.
But the metre is defined in terms of the 'motion' of light.

20. Originally Posted by arfa brane
But the metre is defined in terms of the 'motion' of light.
And if the distance between the meter stick and light is changing, as is the distance between two cars is changing, how can you say that light always takes the same amount of time to traverse the meter stick?? Do you not realize that the meter stick can be in motion, or that it can not have any motion??