Lorentz Transformation - Exponential factor, why not Proportional?

[russ_watters]

Since both clocks are stationary relative to each other and on the surface of the earth, they will agree on the distance between them: you can use either to measure the distance (and, by all means, compare the two).

 

[Gamish]

I think we understand just fine - its just that the relationship is established based on how we observe the universe to work: if the relationship observed is not proportional, you can't make it proportional by manipulating the equations. If you do, the equations will fail to accurately model what we see. They were not guessing. The Lorentz transformations (and Relativity), are derived. They weren't just pulled out of the air. And even if they were (there have been cases where equations were found via trial and error), that wouldn't change the answer to your question: either way, they still have to fit the data.

I'm afraid the answer to all of this won't be very satisfying: at some level it boils down to that's just the way it is.

OK, this is my conclusion thus far. MY little lorenz transformation (1-v/c) does make sense, but is is simply not what we observe in nature. There is as of now, no explanation of why the relavistic effects near c are not proportional to the % of c, which is the ONLY variable involved in relavistic effects.

My question is this. Let's say that I was the person who made the lorenz transformation, OK. I had 2 pieces of evidence to work with, that show the relavistic effects near c. These are the evidence.

1.At low speeds, there is little relativity, to the point that there is no need to calculate it.

2.When you reach c, the speed of light becomes constant. This means that time and space must contract to 0 when you reach c.

3.The only variable which effects space, time, and mass is your velocity.

4.You can never reach c.

So, with this evidence alone, I can establish that relavistic effects near c are dependant on your velocity. So, according to common sense, if I travel .5c, then my time will decrease to .5t
Now, let me clarify that in no way I am purposing that my equation will work, I am just trying to make a point. My question is how else was the lorenz transformation derived?

at some level it boils down to that's just the way it is.

How did we figure out that the relavisit effects are not proportional to the % of c? Certantly we did not look at something moving at .5c and say, lOOK, its length did not contract to .5L, relavistic effects are not proportional to v/c!

So how was it figured out? Hence, the equation was derived.

 

[Janus]

So how was it figured out? Hence, the equation was derived.

Galileo has already given you that derivation back on page two of this thread.

 

[yogi]

djavel: "You're just making this stuff up as you go along"

Take a look at Einstein's Theory of Relativity by Max Born. The problem is discussed on pages 96 - 132.

On page 131: "It is only a mean velocity during the path to and fro that is actually measured. The deviation of this from the velocity of light c in the ether is, however, a quantity of the second order [Beta = (v/c)^2] with respect to Beta and not open to observation."

And again On page 132 Born Says in dealing with terrestrial light sources of the first order effect Beta = v/c. "...That this result must always lead to a negative result follows from the fact that the true duration of motion of the light from one place to another is never measured but only the sum and difference of the trip there and the trip back over the same light path is found. For the reasons given above we thus see that the quantities of the first order always cancel out."

Also see Wisps post in the other tread. If you had actually read Zhang's books before you condemn them, you would know that Zhang is an ardent relativist.

djavel--- I suggest that if you read more you would know more - and perhaps with the maturity that comes with knowledge, you will find it advisable to use more guarded language

 

[Gamish]

Galileo has already given you that derivation back on page two of this thread.

Yes, he did explaine it, but I didn't fully understand it.
Maybe someone could further break it down, without getting too mathematical.

Im the master at time!

 

[JesseM]

Yes, he did explaine it, but I didn't fully understand it.
Maybe someone could further break it down, without getting too mathematical.

Im the master at time!

Did you understand the basic idea of what happens if you shine a flashlight at the floor on a moving train--namely, that a person on the train will see the light go straight down, while a person outside the train will see the light take a diagonal path from the point where the flashlight turned on to the point where the light hit the floor of the train? (since the train had moved a little since the flashlight turned on, from the external observer's point of view)

 

[Gamish]

Did you understand the basic idea of what happens if you shine a flashlight at the floor on a moving train--namely, that a person on the train will see the light go straight down, while a person outside the train will see the light take a diagonal path from the point where the flashlight turned on to the point where the light hit the floor of the train? (since the train had moved a little since the flashlight turned on, from the external observer's point of view)

Yes, I understood that, I just get a little confused with the mathamatics of it. Perhaps you can go on the expliane it...

 

[JesseM]

Yes, I understood that, I just get a little confused with the mathamatics of it. Perhaps you can go on the expliane it...

OK, from the point of view of the observer outside the train, if the light takes time t to travel diagonally from the flashlight to the bottom of the train, then if he sees the train as moving at velocity v, during that time he will see the train has moved sideways a distance vt. So if the flashlight was held a distance h above the train's floor, by the pythagorean theorem length of the diagonal path will be equal to squareroot(h^2 + v^2*t^2). [I tried to convert this to LaTeX but it messed everything up]. A basic assumption of relativity is that light must travel at c in all frames, so if we take this distance and divide it by the time t, we must get a velocity c:

$$\frac{\sqrt{h^2 + v^2t^2}}{t} = c$$

Based on this, we can solve for t:

$$\frac{h^2 + v^2t^2}{ t^2} = c^2$$

$$\frac{h^2}{t^2} = c^2 - v^2$$

$$\frac{t^2}{h^2} =\frac {1}{c^2 - v^2}$$

$$t^2 = \frac{h^2}{c^2 - v^2}$$

$$t = \frac{h}{\sqrt{c^2 - v^2}}$$

So, this is the time that the observer outside the train must see for the light to travel from the flashlight to the floor. But the observer on the train can't see it take the same amount of time, or he wouldn't see the light moving at c; since in his frame the light only goes a distance h, then the amount of time t' which he measures must satisfy

$$\frac{h}{t'} = c$$
or
$$t' = \frac{h}{c}.$$

So, let's find the ratio of the times each observer measured, or
$$\frac{t'}{t}$$
This would be
$$\frac{\sqrt{c^2 - v^2}}{c}$$
or
$$\sqrt{\frac{1}{c^2}}\sqrt{c^2 - v^2}$$
[another minor LaTex problem here, I tried to edit out that last square bracket but it won't go away]
or
$$\sqrt{1 - \frac{v^2}{c^2}}$$
So, this shows why if you assume the speed of light must be c in both frames (and if you also assume they both measure the same value for the height h--the only length contraction is along the axis of motion), the observer outside must see the clock of the observer on the train slow down by a factor of
$$\sqrt{1 - \frac{v^2}{c^2}}$$
where v is the velocity of the train.

If any steps of this are still unclear, please let me know and I'll try to clarify.

 

[banza]

an intuitive difference between proportional and exponential

If you plot y versus x on a graph and it is a straight line, the plot is proportional. It doesn't matter where you are on the Y-axis, the proportion (or rate of change) remains constant. If your plot curves upward at an accelerating rate, then it is exponential because the rate is increasing. How fast? It depends on where you are on the Y-axis. In other words, your "proportional" implies a constant rate irrespective of X or Y. Exponential implies that the rate changes depending on how much Y you have already accumulated. The more Y you have, the faster the rate changes.