# Thread: The burn mark problem

1. Originally Posted by Jack_
True and O does not need a clock.
Hooray! You do understand that time is involved!

Now, try this:
SR says that according to O', a certain time elapses between O meeting O', and O' meeting the burn mark, a time equal to d/cλ.
Once more, O can know this without needing a clock.

True or False?

2. Originally Posted by Pete
Come on, Jack, I know you can answer this.
Let's make it a true/false question.
The race begins at the starting line, and ends at the finish line.
When the race is run, the starting point of the race is still at the starting line.
True or false?
I want two observers in relative motion at the start position.

When the race is done each frame calculates the distance based on its start point to the end.

Cept, each frame diverges by vt.

You tell me which is correct. Thre is no correct absolute answer under SR.

3. Originally Posted by Jack_
Yesm but time under SR has skips and jumps. I was getting around those.
Well, "skips and jumps" is an awfully vague and inaccurate way of desribing it.
But regardless, there's no "getting around" what SR says about time. SR doesn't rely on clocks, SR works on time.

What do you think a second is? It is a smalll motion of the earth. I will help you.

60 seconds = 1 minute
60 minute = 1 hour.
24 hours = one day.

365 days = 1 year, the time it takes for one earth orbit around the sun.
Yeah, I know dude. I just think it's funny that you're using motion as a clock, while still swearing black and blue that there's nothing like a clock in your scenario.

4. Originally Posted by Pete
Hooray! You do understand that time is involved!

Now, try this:
SR says that according to O', a certain time elapses between O meeting O', and O' meeting the burn mark, a time equal to d/cλ.
Once more, O can know this without needing a clock.

True or False?
Yes, O will claim the time on its clocks will be d/cλ for O' to meet BM.

Let me continue, and O' will claim the time on its clocks will be d/c.

Yet, t' = ( t - vx/c²)λ under SR.

Are you getting the problem yet?

5. Originally Posted by Pete
Well, "skips and jumps" is an awfully vague and inaccurate way of desribing it.
But regardless, there's no "getting around" what SR says about time. SR doesn't rely on clocks, SR works on time.

Yeah, I know dude. I just think it's funny that you're using motion as a clock, while still swearing black and blue that there's nothing like a clock in your scenario.
Stop whining.

You are agreeing there is no time only motion.

6. Originally Posted by Pete
Well, "skips and jumps" is an awfully vague and inaccurate way of desribing it.
But regardless, there's no "getting around" what SR says about time. SR doesn't rely on clocks, SR works on time.

Yeah, I know dude. I just think it's funny that you're using motion as a clock, while still swearing black and blue that there's nothing like a clock in your scenario.
Let's get back on task of the multiple light emissions.

7. Originally Posted by Jack_
I want two observers in relative motion at the start position.

When the race is done each frame calculates the distance based on its start point to the end.

Cept, each frame diverges by vt.

You tell me which is correct. Thre is no correct absolute answer under SR.
That's right, Jack. SR says there is no correct absolute answer, that it is relative, frame dependant.

But I want to know what you say. Where did the race begin?
When someone says "The race started at the starting line, of course!" Do you think they're right, or are you going to tell them they're wrong?

8. Originally Posted by Jack_
You are agreeing there is no time only motion.
No, Jack. I'm laughing at you because you insist there are no clocks in your scenario, only motions... but you then happily use motion as a clock.

So, Jack, do you think that the motion of O' from O to the burn mark might serve as a clock?

9. Regarding time and motion:
Time, distance, and motion form a triad. Any two imply the third.

d = vt
v = d/t
t = d/v

10. Originally Posted by Jack_
Yes, O will claim the time on its clocks will be d/cλ for O' to meet BM.
Let me continue, and O' will claim the time on its clocks will be d/c.
Wrong, Jack.
O says it takes a time of d/c for O' to go from O to the burn mark, remember?
And SR tells us that O' says it takes a time of d/cλ between O' passing and the burn mark passing.

Yet, t' = (t - vx/c²)λ under SR.
Almost. In this case v is in the opposite direction to the positive x direction, so t' = (t + vx/c²)λ

Let's see where that takes us. The burn mark meets O' at:

$t = d/c$
$x = -vd/c$

Now let's find t':

[tex]t' = \gamma(t + \frac{vx}{c^2})
$t' = \gamma(d/c - \frac{v^2d}{c^3})$
$t' = \gamma \frac{d}{c}(1 - \frac{v^2}{c^2})$
$t' = \gamma \frac{d}{c}(1/\gamma^2)$
$t' = \frac{d}{c\gamma}$

What do you know?
t' = d/cλ, just like we said.

11. Originally Posted by Jack_
Yes, if you do not understand SR is actually a theory of multiple emission points, then you would not get anywhere with me.
Special relativity is a theory about the underlying symmetry of the space-time metric. Would you care to explain precisely what 'a theory of multiple emission points' is?

Originally Posted by Jack_
OH, say, since you are so brilliant, at least in your own view, why don't you explain the components of LT specifically and explain exactly what each term means and does.
I wouldn't say I'm 'brilliant'. I'd say I'm better than most but those things are not synonymous. what do you mean by the 'components of LT'? Lorentz transformations are defined as those which leave the space-time interval invariant. While you can represent them as a matrix acting on some kind of vector space which is equated with the metric in some way this is not fundamental to the Lorentz group itself and the components of the matrices depend on the representation you use. For instance, SO(3,1) can be represented by 4x4 matrices which act on the 4x4 metric $\eta$ but you can also represent the same transformations in terms of SL(2,C) matrices. This is because SL(2,C) ~ Spin(3,1), which is the double covering group of the SO(3,1) Lorentz group. Each representation has different components.

I would say that demonstrates a pretty decent understanding of this area of mathematical physics and demonstrates you're considerably less familiar with this stuff than you would like people to believe. Hence I think its all the more laughable you try to throw the insult of egotism at me with such comments as "since you are so brilliant, at least in your own view". Being better than you doesn't make me brilliant, you're setting the bar pretty low.

12. I would say that time becomes an issue due to direction. In the two slit experiment with an observer light has a straight path. Without an observer, light has a bent path. I call it the stacking problem. They can't both equal the same time, and yet only the flow of movement was altered. Even in a vacuum, there would be uncontrolled flow of movement. It's tiny, but there. I suppose there's no easy solution to it. Time is the best that there is to offer.

13. Originally Posted by Pete
That's right, Jack. SR says there is no correct absolute answer, that it is relative, frame dependant.

But I want to know what you say. Where did the race begin?
When someone says "The race started at the starting line, of course!" Do you think they're right, or are you going to tell them they're wrong?
I am OK with co-location for the start point.

And, no the light path cannot be relative.

Light moves through space at one speed c regardless of the motion light source. It forms one light path and not an infinite collection of them with an infinite number of distinct light emission points.

14. Originally Posted by Pete
Regarding time and motion:
Time, distance, and motion form a triad. Any two imply the third.

d = vt
v = d/t
t = d/v
v does not depend on a clock, it just happens, You are forcing a model of think onto that motion using a clock.

15. Originally Posted by Pete
Wrong, Jack.
O says it takes a time of d/c for O' to go from O to the burn mark, remember?
And SR tells us that O' says it takes a time of d/cλ between O' passing and the burn mark passing.

Almost. In this case v is in the opposite direction to the positive x direction, so t' = (t + vx/c²)λ

Let's see where that takes us. The burn mark meets O' at:

$t = d/c$
$x = -vd/c$

Now let's find t':

[tex]t' = \gamma(t + \frac{vx}{c^2})
$t' = \gamma(d/c - \frac{v^2d}{c^3})$
$t' = \gamma \frac{d}{c}(1 - \frac{v^2}{c^2})$
$t' = \gamma \frac{d}{c}(1/\gamma^2)$
$t' = \frac{d}{c\gamma}$

What do you know?
t' = d/cλ, just like we said.
I am OK with your calculation.

It is consistent with the fact that the light beam in O' starts at O' and ends up a distance d/λ from the B< which has been my point all along.
What do you know?

However, when O' moves to the BM, the O from has the light beam d + (v/c)d down the x-axis.

Note the disagreement.

I have a different way to put it. Let me see what you do with this.

Assume the standard configuration using O and O'
Each agree in their own frames they will start a clock at co-location and stop the experiment when the time reaches an agreed upon d/c where d is some chosen agreed value between the two.
When the experiment ends, they mark a measuring rod in their own frames only describing the positions and distances.

Here is what O marks.
BM-------(v/c)d-------O>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>d

Here is what O' marks.
O'>>>>(v/c)d>>>>O>>>>>>>>>>>>>>>>>>>>>>>>>d

They get back together after the experiment with their rods in the same frame and compare them.
BM--------(v/c)d--------O>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>d..Measurin g rod of O
O'>>>>>(v/cd>>>>>O>>>>>>>>>>>>>>>>>>>>>>>>>d................ ..............Measuring rod of O'

By matching up O and O', they find two different light paths.

All is consistent with SR with the following conditions met in each frame.
1) Light is measured c in each frame
2) The light emission point in the frame is located at the observer in the frame.
3) O and O' are separated by their relative motion caused by v. While light moved d, the frames diverged by (v/c)d
4) Light travels d in ct.= c(d/c)

Hence, SR is the theory of two different light paths.

16. Originally Posted by AlphaNumeric
Special relativity is a theory about the underlying symmetry of the space-time metric. Would you care to explain precisely what 'a theory of multiple emission points' is?

I wouldn't say I'm 'brilliant'. I'd say I'm better than most but those things are not synonymous. what do you mean by the 'components of LT'? Lorentz transformations are defined as those which leave the space-time interval invariant. While you can represent them as a matrix acting on some kind of vector space which is equated with the metric in some way this is not fundamental to the Lorentz group itself and the components of the matrices depend on the representation you use. For instance, SO(3,1) can be represented by 4x4 matrices which act on the 4x4 metric $\eta$ but you can also represent the same transformations in terms of SL(2,C) matrices. This is because SL(2,C) ~ Spin(3,1), which is the double covering group of the SO(3,1) Lorentz group. Each representation has different components.

I would say that demonstrates a pretty decent understanding of this area of mathematical physics and demonstrates you're considerably less familiar with this stuff than you would like people to believe. Hence I think its all the more laughable you try to throw the insult of egotism at me with such comments as "since you are so brilliant, at least in your own view". Being better than you doesn't make me brilliant, you're setting the bar pretty low.
You say very little.

x' = ( x - vt )λ

x' = xλ - vtλ

What is xλ and what is vtλ. Pleae be specific on their meaning and why they are in the equation.

Then expain the subtraction.

17. Originally Posted by Jack_
I am OK with co-location for the start point.
And, no the light path cannot be relative.
So you agree that when the race ends, the location of the start is still at the starting line?
And you declare that this is the only true answer?

18. Originally Posted by Jack_
You say very little.
Don't be so fucking lazy, Jack.
If you pay attention to what peple say and put in the effort required to understand, you'll learn a lot and get a lot smarter.

On the other hand, if you're happy with how smart you are and much you know know now, feel free to keep your head in that bucket.

19. Originally Posted by Jack_
v does not depend on a clock, it just happens, You are forcing a model of think onto that motion using a clock.
Again with the clocks?

Shit on a fucking shingle, Jack, how can you be so thick? I've seen Tellytubbies episodes with more to say.

Time, Jack. Not clocks. Time doesn't need clocks.

Time, distance, and motion form a triad. Any two imply the third.

d = vt
v = d/t
t = d/v

Jack,
Are you claiming that there is no such thing as time?
Do you think that you can form a meaningful mathematical model of reality without time?
If so, then SR is the least of your worries. You should instead work on proving that Newtonian mechanics is inconsistent.

20. Originally Posted by Pete
So you agree that when the race ends, the location of the start is still at the starting line?
And you declare that this is the only true answer?

What? The starting line is a member of each frame. Remember? We had two frames co-locate at the starting line.

Now, ae you claiming you know specifically where this line is in space?

Otherwise, you have no choice but to agree, each frame has a startline.