Special Relativistic time dilation and length contraction

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James R

10-09-04, 02:02 AM

This thread is for those who want to learn how a couple of the well-known effects from Einstein's Special Theory of Relativity can be derived from first principles. The derivation of time dilation and length contraction presented here is based on a simple thought experiment.

We start with the postulates of Special Relativity:

1. The laws of physics take the same form in all inertial reference frames.
2. The speed of light is constant for all observers.

These are the only assumptions we will make in order to derive the time dilation and length contraction effects predicted by the theory. We will also need some algebra.

The situation that the derivation will be based on is as follows:

There are two cars, which we will call BLUE and GREEN, on a straight road. The green car travels at speed v relative to the blue car, in a straight line along the road. We will assume that v is less than the speed of light, c.

We are interested in two different reference frames, so we need two different coordinate systems. The blue car denotes distances measured along the road as x, and times as t. The green car denotes distances by a different coordinate, x', and times by t'. We make no assumptions initially about the relationship between t and t' or x and x'. That is, the time intervals or distances between any given pair of events may be different when measured by the blue car, as compared to when they are measured by the green car.

The blue car chooses its coordinate system such that all distances to other objects on the road are measured from the blue car itself. This means that, according to BLUE, its own x coordinate is always x=0. Similarly, the green car measures distances from its own location, so that the position of GREEN at all times is x'=0. We assume that the x and x' axes both point along the road in the same direction. We know nothing about the relative scales of the axes yet.

In the equations which follow, keep in mind that any quantity labelled with a prime (') is measured in the reference frame of the GREEN car, while any quantity without a prime is measured in the frame of the BLUE car.

A diagram of the situation is given below. The diagram shows both reference frames - the points of view of both the BLUE and the GREEN car.

Diagram attached (http://www.sciforums.com/attachment.php?attachmentid=3332&stc=1)

When the GREEN car reaches the point x=A, where A is some constant distance away from the blue car, the BLUE car emits a pulse of light in the direction of the green car. This pulse of light is represented as a pink blob in the diagram. We will call the emission time of the pulse t=t'=0. It is assumed that the clocks used by both cars to measure time are mechanically identical, tick at the same rate when at rest, and keep time accurately.

We will look at where the green car is when the light pulse hits it, and also consider the time it takes the light to travel between the cars, from the point of view of both the blue and the green cars.

First, we write down expressions for the positions of each of the cars and the light pulse in each reference frame. Here we use the basic constant-speed equation x = x0 + vt repeatedly, where x0 is the position of an object at t=0, and v is its velocity in the x direction. This equation is presumed to hold in both reference frames, due to the first postulate of Special Relativity.

BLUE frame (top diagram)

BLUE car: x(t) = 0 ........(1)
GREEN car: x(t) = A + vt ..(2)
light pulse: x(t) = ct ....(3)

In this frame, the blue car is at rest, the green car moves in the positive x direction at speed v, and the light pulse moves in the positive x direction at speed c. Note that, at t=0, the blue car and the light pulse are both at x=0, and the green car is at x=A.

GREEN frame (bottom diagram)

BLUE car: x'(t') = -B' - vt' .....(4)
GREEN car: x'(t') = 0 ............(5)
light pulse: x'(t') = -B' + ct' ..(6)

In this frame, the GREEN car is at rest. The BLUE car moves in the negative x' direction. We assume that the speed of the blue car as seen by the green car is the same as the speed of the green car as seen by the blue car, but in the opposite direction. Hence, the quantity v has the same magnitude in the GREEN frame as in the BLUE frame, and it needs no prime (') to distinguish it.

On the other hand, since we don't know anything about the distance scales yet, we cannot assume that the distance between GREEN and BLUE is the same in both frames. Therefore, in the GREEN frame, we designate the position at time t'=0 as x'=-B'. We don't know how the constant B' relates to A yet.

Finally, we use the second postulate of special relativity to say that in both frames, the speed of the light pulse must be the same, so c, like v, is the same in both sets of equations above, and c needs no prime. (Another way of saying this is that c=c' and v=v').

Also note that in equations (4), (5) and (6), both the distances and times are measured in the GREEN frame, which is why x' is expressed as a function of t' (and not t). Remember, we don't yet know if t' is the same as t or not.
------

We now consider the time of arrival of the light pulse at the green car.

BLUE frame

The pulse arrives when the x coordinates of the pulse and the green car are the same, which means, from equations (2) and (3):

A + vt = ct

Solving for t, we get:

Time of arrival of pulse in BLUE frame: t = A/(c - v) = .......(7)

Where is the green car when the pulse arrives? Plug the value of t from equation (7) into quation (2) we get:

Position of GREEN car when pulse arrives: x = A + v[A/(c - v)] = A/(1 - v/c) ....(8)

GREEN frame

In this frame, the pulse arrives when the x' coordinates of the pulse and the green car are the ame. From equations (5) and (6) we get:

0 = -B' + ct'

Solving for t':

Time of arrival of pulse in GREEN frame: t' = B'/c = .......(9)

Using equation (4):

Position of BLUE car when pulse arrives at the GREEN car: x' = -B' - vt' = -B' -vB'/c = -B'(1 + v/c)....(10)
------

Now, let's look at the DISTANCES between the two cars, both when the pulse is emitted and when it is received.

When the pulse is emitted, we have:

BLUE car: x = 0, x' = -B'
GREEN car: x = A, x' = 0

The distance between the cars is as follows:

In BLUE frame: d(0) = A - 0 = A ........(11)
In GREEN frame d'(0) = 0 - (-B') = B' ..(12)

When the pulse is received, we have:

BLUE car: x = 0, x' = -B'(1 + v/c), from equation (10), above.
GREEN car: x = A/(1 - v/c), x' = 0, from equation (8), above.

The distance between the cars is as follows:

In BLUE frame: d(t) = 0 - [-B'(1 + v/c)] = B'(1 + v/c) .......(13)
In GREEN frame d'(t') = A/(1 - v/c) - 0 = A/(1 - v/c) ...........(14)
------

We are now in a position to relate the length scales of the x and x' axes. We write:

d'(0) = g d(0), .....(15)
d'(t') = g d(t) ...(16)

where g is some constant to be determined. In words, these equations say "The distance d' between the two cars, measured in the GREEN frame, are equal to g times the distance d between the cars measured in the BLUE frame." They are taken at two specific times, corresponding to two specific events (when the light pulse is emitted and received). We assume that the scale relationship is the same at time t as it was at time 0.

Using (15), (11) and (12) we get:

B' = g A ....(17)

Also, from equations (13), (14) and (16) we get:

A/(1 - v/c) = g B'( 1 + v/c) ....(18)

Replacing B' in (18) with the expression given in (17) we get:

A/(1 - v/c) = g2 A (1 + v/c).

The constant A cancels out on both sides. Rearranging we get:

g2 = 1/[(1 - v/c)(1 + v/c)]

g2 = 1/[1 - (v/c)2]

g = 1/sqrt(1 - (v/c)2) ....(19)

Since v is less than c, g is a number greater than 1. g is the well-known relativistic "gamma" factor. Note that it depends only on the relative speed v between the objects in question (in this case the two cars), since c is a constant according to the second relativistic postulate.

-------
Length contraction

Re-writing equation (16), with the value of g from equation (19):

d'(t') = d(t) / sqrt(1 - (v/c)2) .... (20)

In other words, the distance between the cars, as measured by the GREEN car, is GREATER than the distance between the cars as measured by the blue car. Since we are talking about the same physical distance here, this must mean that the GREEN car's rulers, which the GREEN observer uses to measure the distance, are contracted compared to the BLUE car's rulers. Shorter rulers mean that GREEN measures a longer distance.

This is relativistic length contraction.
------

Time dilation

How far did the light pulse travel? It covered the distance between the cars. We need to take into account that the cars were already a certain distance apart when the pulse was emitted. According to each observer:

BLUE observer:

d(0) = A ..............(11)
d(t) = B'(1 + v/c) ....(13)
So distance travelled by pulse = B'(1 + v/c) - A .... (21)

Since the speed of the pulse was c, the time taken to cover the distance, as measured on A's clock is the result in (21) divided by c.

GREEN observer:

d'(0) = B' ............(12)
d'(t') = A/(1 - v/c) ..(14)
So distance travelled by pulse = A/(1 - v/c) - B' .... (22)

The speed in this frame is also c, so the the time taken in this frame is the result in (22) divided by c.

Using B' = g A, equation (21) reduces to:

distance(BLUE) = B'(1 + v/c) - A
= A [g(1 + v/c) - 1]

And equation (22) reduces to:

distance(GREEN) = A/(1 - v/c) - B'
= B' [1/g(1 - v/c) - 1]

Taking the ratio of equations (22) and (21), we get:

distance(GREEN) / distance(BLUE) = time(GREEN)/time(BLUE)

= B'[1/g(1 - v/c) - 1] / A [g(1 + v/c) - 1] ....(23)

Recall from (19) that

1/g2 = 1 - (v/c)2 = (1 + v/c)(1 - v/c)

Rearranging, we get:

1/g(1 - v/c) = g(1 + v/c) .... (24)

This allows us to cancel the terms in square brackets in equation (23), leaving:

time(GREEN) / time(BLUE) = B'/A = g

or

time(GREEN) = g time(BLUE)

This equation says that the travel time for the pulse, as indicated by the GREEN observer's clock, was longer than the travel time as measured on the BLUE observer's clock. In other words, the GREEN clock runs slow compared to the BLUE clock. This is relativistic time dilation.

© James R 2004

geistkiesel

10-19-04, 06:17 PM

James R

10-19-04, 09:29 PM

geistkiesel

10-20-04, 01:01 AM

geistkiesel:

I don't think I understand your objection. Another way to express the postulate would be:

2. In any inertial frame of reference, the speed of light is always c (a constant).

But this is just saying the same thing in a slightly different way.

You don't understand the objection that the postulate describing the motion of light being indenpendent of the source of the light, is corrupted by your postulate defining the velocity of light in all inertial frames?

As you have stated it the postulate would apply equally to ducks and snails, though you would allow for different reference frames to measure different relative velocities of the duck and frame. So I ask yiou again, is the speed of light measured differently than h espeed of ducks as seen in all reference frames?

If I travel in increments of 1 km/s up to 10000 km/s will I always measure the relative velocty of frame and photon the same?

You are saying, are you, that I will never get a different result if I measure the relative velocity of my frame moving Vn = 1km/sec that will not measure , 300000 - 1= 299999 km/sec, or of any of the masurements in the table below,With Ve = 0, Vc - Ve = C and Vn - Ve = > 0:
Light | Frame | Relative Velocity of
Speed | Speed | Frame and Photon
_Vc______|_____Vn____|____Vc - Vn_______________
300000 - 1 = 299999 km/sec
300000 - 1000 = 299000 km/sec
300000 - 10000 = 290000 km/sec

One objection is that the postulates you quoted above are corruptions of the postulate governing the independence of the motion of light and the source of the light. Is this unclear?

If the motion of light is independent of the velocity of any frame and/or source, how do you justify that a frame moving 3000 will measure the same relative velocity of frame and photon as one moving 30000 km/sec?

Your postulates are saying there is no such a thing as a relative velocity of light. How then do you explain that photons emitted from opposite ends of a frame in motion wrt Ve = 0 will measure different arrival times at the respective opposite ends of the frame as seen in he figure.

The firsts figure is showing the emission of the photns from the L and R emitter/absorbers simultaineously when triggered by the || .
|| ||
L|>________________________________<|R t = 0
The photons head for the opposite ends of the moving frame moving to the right. The photon are seen just as the photon from the R side reaches the L emitter/absorber, at a distance L - vt from the R emitter/absorber, while the L photon still has 2vt to travel before arriving at the R emitter/absorber.
L|----|<____________________________>____|----|R t = 1
The left moving photon travels the distance between R to L as = ct.
The right moving photons must add the distance 2vt + vt2 to cover, or the time t2 between emission and arrival at R.

You will see that left moving photon arrives at the left end iin less time time than the photon emitted in the direction of motio to the right. The times we are using are thje clocks on the moving frame.
The double vertical line above the L and R emitter/absorber pairs, n the first figure above, are mechanical switches on Ve that start the clocks simultaneously in the Vn frame.Thereafter, they pulse at regular intervals say 100/sec.
Should the observer interpret the data as if he were at rest and claim that the photon arriving at L was emitted first and that is why there is the descrepancy, or any other story he may want to claim. he is informed that the clock data show a longer time between moving from L to right than R to left by the amoun t2.
The observer on Vn is asleep throughout all of this and doesn't wake up until he arrives back at the Ve home station, and then he looks at the recorded L and R clock data showing the consistent t2 time added to the right moving photon.
The observer is also shown data for the test, where the frame is at rest wrt Ve, "really".
From this data the velocity v can be determined from the expression:

t2 = 2vt/(c - v)

Once t2 is determined, v may be calculated:
V = Ct2/(2t + t2)

The observe is asked if he ever felt any acceleration on the train when leaving the station to begin he tests?,"Sure" he said.
"Hear or feel the clickety clack of the wheels passing over the expansion spaces in the tracks?", 'Absolutely" he answered.
Having adnitted to thoroughly inspecting the setup for the guarantee of the simultaneous setting of the clocks and all other significant parameters the observer was asked if he could account for the added time other than the motion of the train. "Nope" he said.
"Of the 1000 times you were observer on the tests, how many of those times did the frame accelrate when leaving the station?" he was asked and then answered, "1000".
"Is an inertial fram with a mass of 6 x 10^ 27 grams equivalent to any other earth bound frame?". "Of course not" the observer said.
" Of the 1000 times you were in these tests how many instances of the Ve frames accelerating did yiou observe? "0",he replied.
"So, if we go by your observation you never observed a Ve > 0, or a Vn that did not first accelerate before reaching the uniform speed of the tests, right?? "Right", he said. . "We only pretend we are at rest so we can have a Special Relativity theory, some of these guys have been doing this for decdes. It would break their hearts to have to discard SR at this point. Just think of it, their entire professional career, it could be embarrassing.".
" Do you know if the Special Relativity Theorists know that McDonalds is hiring and they train you?" , "No" he replied, "I'll spread the word."
Whatever direction and velocity of the source of light the motion is independent of that source and there is no attribute of the motion of light, including direction and speed, that is functionally related to the source. Or once a photon, the photon knows nothing about sources, reference frames or anything other than maintaining a straight-line motion.

James R: you may have personl reasons to give my posts the short shrift, but could you please make an effort to answer the questions?
Thanks.
geistkiesel
10-19-2004

James R

10-20-04, 01:13 AM

geistkiesel:

Just a quick reply. I am short of time at the moment, and your posts require lengthy responses. That is why they will have to wait.

bertus

10-05-08, 11:07 PM

Admittedly I am a beginner here. I find a few errors in the reasoning.
First:
Because
1. The laws of physics take the same form in all inertial reference frames.
It is therefore impossible that the scales would be different. Since there is no absolute frame of reference Blue may as well be moving as Green and hence you would get the opposite relationship of the distances which is absurd and completes my proof. Further the light may as well proceed from Green towards Blue again leading to the opposite relation and hence your axis scaling is dependent on the reference frame a contradiction, or the direction of the light again a contradiction. The situation is symmetric. I think the equations may be valid given another interpretation. Further I suggest considering the distance the light has traveled in each frame and the time. That may lead to a reasonable interpretation. I also suggest the following thought experiment to illustrate your contradictions. A train goes through a tunnel at high speed. Can you shut both doors on it as it shrinks? What does the train observer have to say? Ok again new to this but think there are some errors.

BenTheMan

10-05-08, 11:31 PM

A train goes through a tunnel at high speed. Can you shut both doors on it as it shrinks? What does the train observer have to say?

This is a common thought experiment in special relativity and has a well known resolution (http://www.rdrop.com/~half/Creations/Puzzles/pole.and.barn/index.html). Basically, the doors closing in the tunnel is a simultaneous event in the tunnel's frame, but this doesn't necessarily hold in the train's reference frame.

Uno Hoo

10-06-08, 05:37 PM

Length contraction

In other words, the distance between the cars, as measured by the GREEN car, is GREATER than the distance between the cars as measured by the blue car. Since we are talking about the same physical distance here, this must mean that the GREEN car's rulers, which the GREEN observer uses to measure the distance, are contracted compared to the BLUE car's rulers. Shorter rulers mean that GREEN measures a longer distance.

This is relativistic length contraction.

This equation says that the travel time for the pulse, as indicated by the GREEN observer's clock, was longer than the travel time as measured on the BLUE observer's clock. In other words, the GREEN clock runs slow compared to the BLUE clock. This is relativistic time dilation.
© James R 2004

Well now. James R has bent over backwards to confuse a learner new to Special Relativity. Perhaps not intentionally.

My post here is not to be taken as my opinion, pro or con, about the validity of Einstein Special Relativity. However, I do wish for SR to be accurately taught in an understandable way so that anyone can clearly know what it claims or does not claim. With a clear insight, anyone can then accurately determine for themself if they believe the theory is true or false.

Re length contraction: someone familiar with Einstein's Principle of Relativity will quickly notice how James R's choice of language is an abysmal perversion of the correct understanding of Einstein Special Relativity length contraction. The Principle of Relativity states that any observer must consider himself as being stationary and must use only measuring devices stationary with respect to himself, (such as, but limited to, his stopwatch, his yardstick and his bathroom scales.). Therefore the observer will always use measuring devices which are not themselves contracted, dilated or changed in any other way due to a velocity. According to the Principle of Relativity there is a contraction, or dilation or other transformation only when there is a velocity between the observer and the observed object.. When there is no velocity between the observer and his measuring devices then he cannot observe any transformation(s) in said devices. James R's inscrutable phrasing strongly implies that the observer can see his yardstick as having contracted. Not so! The Green observer will never see his own yardstick as being contracted!

Re time dilation: Again James R has left us ineffably confused about which observer is seeing who's clock. If we guess that he is writing about the Green observer being active, the the Green clock will be seen by the Green observer as keeping time in the regular way (proper time). Everything moving with velocity with respect to the Green car will be seen by the Green observer to do everything slower than usual. The Green observer gazing over at the Blue clock will see the Blue clock keeping time slower than his own Green clock. James R's abominable language strongly implies that the observer can see his time as being dilated. Not so! The Green observer will never see his own clock as being dilated!

Uno Hoo

10-06-08, 05:54 PM

This is a common thought experiment in special relativity and has a well known resolution (http://www.rdrop.com/~half/Creations/Puzzles/pole.and.barn/index.html). Basically, the doors closing in the tunnel is a simultaneous event in the tunnel's frame, but this doesn't necessarily hold in the train's reference frame.

The url was delightfully colorful. One might even gush and call it dazzling. However I am baffled about what it was supposed to prove. The Special Relativity tunnel paradox looks to me to be as dangerous to the theory as before I clicked on the pretty url. There does not seem to be any reason why the doors' closings cannot be simultaneous in the train's reference frame. If they can be simultaneous in the train reference frame (even if accidental) then we can have (momentarily) intact doors and a busted theory.

Did you perhaps forget to say that I must first drink enough beer to get drunk and then I can believe that your Technicolor url will become believable?

D H

10-06-08, 06:50 PM

Keep it up, CANGAS. Ben isn't as forgiving a moderator as were Pete and James.

Saxion

10-06-08, 06:59 PM

Pete

10-06-08, 07:07 PM

How about latex, or could you not be arsed???
Check the post date.

For anyone here today thinking this is acceptable, is short of a delusion and a half, concerning the fact any ''beginner'' as James seems to target, can not understand his sloppy integrations.
I'm looking forward to your improved edition!

D H

10-06-08, 07:10 PM

How about latex, or could you not be arsed???
bertus necromanced a four year old thread. The ability to use the [tex] tags didn't exist until 2+ years after James wrote the original post.

So, to answer your question, no, he could not be arsed to use latex.

Saxion

10-06-08, 07:10 PM

What edition, friend.

I see math that would muddle up a stranger halfway to Dubai.

Saxion

10-06-08, 07:11 PM

bertus necromanced a four year old thread. The ability to use the [tex] tags didn't exist until 2+ years after James wrote the original post.

So, to answer your question, no, he could not be arsed to use latex.

Well, with his royal authority, he should be able to change.

No???

Saxion

10-06-08, 07:13 PM

I am saying {aboslutely} that he should have made this easier to learn, for the learners, as soon as he could, which he obviously has not.

Pete

10-06-08, 07:51 PM

The url was delightfully colorful. One might even gush and call it dazzling. However I am baffled about what it was supposed to prove.
It's an interesting effort to present the barn-and-pole story in a different way. While it is qualitatively correct, it is not intended to be rigorous, nor to prove anything - just to present a novel way to visualise the relative simultaneity involved. Personally, I prefer animations, numbers, and graphs of distance against time (http://webphysics.davidson.edu/Course_Material/Py230L/relativity/relativity-ex3.htm), backed by rigorous transformations of event coordinates between reference frames..

The Special Relativity tunnel paradox looks to me to be as dangerous to the theory as before I clicked on the pretty url. There does not seem to be any reason why the doors' closings cannot be simultaneous in the train's reference frame. If they can be simultaneous in the train reference frame (even if accidental) then we can have (momentarily) intact doors and a busted theory.
At what moment in the pole's rest frame can the doors be closed simultaneously with the pole inside the barn? In the pole's rest frame, the pole is always either in the way of at least one door, or outside the barn altogether.

Pete

10-06-08, 07:52 PM

I am saying {aboslutely} that he should have made this easier to learn, for the learners, as soon as he could, which he obviously has not.
I'm saying that if you're complaining about it, then you should {absolutely} do something about it.

James R

10-06-08, 08:02 PM

How about latex, or could you not be arsed???

LaTeX wasn't available on sciforums in 2004.

I invite you to repost it in LaTeX form. If you can be arsed, that is.

For anyone here today thinking this is acceptable, is short of a delusion and a half, concerning the fact any ''beginner'' as James seems to target, can not understand his sloppy integrations.

I don't understand your complaint. Please explain.

Uno Hoo

10-06-08, 09:59 PM

Dear Ben my friend,

I am only in the first stage of conducting the science experiment I invented and described in an earlier post: drinking beer while studying your url in the possibility that in a relaxed mental state with enhanced gullibility the url will seem to show the train- in -the- tunnel paradox being resolved. It is too early to notice the url having any results.

But I found something that you hid from us.

When we click on the goto url in your post we then see a page captioned The Pole and Barn Paradox . In the first sentence of the second paragraph we read a goto url: spacetime diagrams, calculations, and talk of simultaneity and multiple frames of reference.. When we click that goto, we see a page captioned The Pole-Barn Paradox. At the bottom of the page is another goto url: The Bug-Rivet Paradox. When we click on that url, we see a page captioned The Bug-Rivet Paradox..

On The Bug-Rivet Paradox page we read of a cute little paradox in the same genre as the train-tunnel paradox. When we read about the bug, the rivet and the relativity contraction we reach a moment of epiphany.

Gloryoski, Sandy! Look at what we found! A simple little relativity paradox without a resolution.

It's a shame, too, because the spectrum hued pole in your url was bitchin.

Pete

10-06-08, 10:42 PM

Changing the subject, CANGAS?
Keep looking, and perhaps start thinking. The Bug-rivet paradox does have a resolution... just not on that page.

You recall this thread?
Bug-rivet paradox

Uno Hoo

10-06-08, 10:47 PM

Dear Ben my dear friend:

A progress report on my science experiment.

I am a few wee drops past halfway and there is no measurable difference in the believability of your url. In fact, it is even becoming more difficult to understand. But, I have to point out that it it was very near to being impossible to understand to start. Sure was pretty.

I am thinking that it might be good to consider concentrating on the bug-and-rivet paradox.

Uno Hoo

10-07-08, 12:00 AM

My good Ben dearest friend,

I am now concluding my science experiment. I have reached a mental state of enhanced gullibility that is maximal with a concomitant minimal necessary ability to discern beans from used peanut butter and still operate complicated machinery like a computer and i have to break some bad news to you.

Your url is awful good looking with its rainbow pole but is a total wipe out for explaining the train and a tunnel paradox. Anybody happening upon your url and hoping that it will
A. Be understandable
B. And lay the train in a tunnel paradox away

is going to go away thinking that they cant get any satisfaction.

It will be much more productive for anyone desiring to get ready made relativity paradox (without thinking about the theory in detail themselves) satisfaction to look at your recommended bug n rivet url.

And of course there are some more Special Relativity paradoxes that cannot be resolved that you do not yet know about.

Pete

10-07-08, 12:22 AM

Like the one you warned, promised, and threatened us you would post two years ago?
1151576
1151573
:D

BenTheMan

10-07-08, 12:34 AM

Like the one you warned, promised, and threatened us you would post two years ago?
1151576
1151573
:D

Indeed. It seems that CANGAS has learned how to use a proxy.

Uno Hoo

10-08-08, 01:06 AM

Like the one you warned, promised, and threatened us you would post two years ago?
1151576
1151573
:D

Why is "CANGAS" so important to you?

Uno Hoo

10-08-08, 01:07 AM

Indeed. It seems that CANGAS has learned how to use a proxy.

Why is "CANGAS" so important to you?

rpenner

10-08-08, 05:13 AM

The following should look familiar...

What's this Theory of Relativity?
It's called relativity because all of the laws of physics look the same to the observer who is moving at the same inertial velocity as the watched experiment as to the observer who is in relative motion. It has rules that say that the precise manner in which the experiments (not the laws) look different when you and the experiment are not moving at the same velocity and the way it looks when you are at relative rest only depends on the relative motion between you and the experiment.

This is the principle of relativity. The modern theory of relativity differs from the Galilean theory of relativity in that clocks in relative motion and rulers parallel to their direction of relative motion cannot be made to agree with one another.

But that's crazy!
I would say "counter-intuitive" because our intuition is shaped by evolution over billions of years at scales and speeds where relativistic effects are far from obvious. Besides, if raw intuition was all that was needed to study the universe, Plato and Aristotle would have gotten around to finishing it.

I don't think that how intuitive a physical theory is should be a criteria in it's value, because the universe is under no obligation to be intuitively comprehensible. In fact, the original Newtonian theory of Universal Gravitation contradicts over 1000 years of intuition by resting on the axiom that terrestrial and celestial laws of motion are identical.

In the same way, Special Relativity is the well-tested assertion, in contradiction to Newton's conflict with Maxwell's equations, that the laws of physics are the same for light as for more familiar objects. It turns out that some violence has to be done to the assumptions that Newton, following Descartes, was working from, but Special Relativity is highly successful in being one set of laws for fast and slow.

It's still crazy -- it has to be mathematically complicated and inconsistent
Almost immediately, it was demonstrated to be not only mathematically consistent but simple by Minkowski, but somewhat simpler, as space-time, than Newtonian separate concepts of space and time. While the mathematics of the Poincaré group will never be introductory mathematics, most of the tools to understand special relativity, geometrically, algebraically and logically are developed in common high school courses.

Well, anyone could be wrong and Einstein is just one man
You will note at the end are some references and that many of these references are to "review articles" which are summaries of hundreds of research papers. Review articles are like textbooks of recent developments in science, and are frequently written to a wider audience than the source articles which they cite.

It's not "just Einstein" but every serious investigation of high speed objects which supports this model of the physical world.

That's just lies made up by the modern physics conspiracy. No evidence before Einstein exists for Special Relativity
While Einstein (1879-1955) was fortunate to be living at a time when classical physics was suffering under the load of experiments inconsistent with Newtonian mechanics, clear-cut evidence which favors Einstein over Newton was published before Einstein was born.

There are conflicts between Newton and experiment which was increasingly apparent in the nineteenth century. In fact, by 1859 we had enough experimental evidence to favor special relativity over Galilean relativity and Newtonian absolute space and time.

By 1859, Hippolyte Fizeau's experiment to measure the Augustin Fresnel's hypothetical ether drag was exactly consistent with a velocity addition formula of v_3 = {{v_1 + v_2}\over{1 + K v_1 v_2}} with K = 1/c² while Newton and Galileo would predict K = 0.

Lorentz and FitzGerald came up with equations which correctly relate observables like elapsed time and relative position for two distinct inertial observers, but didn't have physics understanding beyond phenomenology. Einstein proposed new candidate axioms of physics and demonstrated that not only were such axioms consistent with existing experimental results, but than they provided a basis to derive previous phenomenological results like the Lorentz-FitzGerald contraction or the Fresnel drag coefficients as first-class physical results. Finally, Minkowski demonstrated that there was a mathematical beauty behind the physical results, such that we no longer speak of Euclidean and absolute space and time, but only of Minkowski space-time.

While astonishing to some, these results were shown by von Ignatowsky and others to be very natural descriptions, provided we were willing to let Nature be our guide and not just rely on the authority of Newton, Aristotle and Euclid.

Wait, wait, wait -- not just any velocity addition formula can work! You have to start with physics, not just math created from thin air.
Well, you need math to create a framework in which to place your observations, but let us see.

On Time Dilation
Time Dilation, as was shown shortly after Einstein's 1905 papers, is a natural result which arises from testable intuitive statements of the nature of space and time and physics.

First of all, assuming God doesn't write a physics textbook, man will be forever ignorant of the actual mechanisms of the universe. Even if Professor Y comes up with the mechanistic ONE TRUE THEORY OF EVERYTHING, there will be no way to distinguish the universe of the mechanistic theory from another universe where everything conspires to act just like Y's mechanistic model. That's why physicists (as opposed to philosophers) use mathematical models to avoid talking about the mechanism and only the behavior. Our everyday experience of the universe has taught us a lot of everyday assumptions. The following 4 should be non-controversial when you neglect gravity. They are all statements about observed symmetries of the universe -- so all of them are falsifiable if you found a counter-example.

Four everyday assumptions

Let us assume the laws of physics are translationally invariant in space. Then it follows a statement about a experiment happening in an arbitrary place will work the same if we center our coordinate basis with it. This also implies that we can calculate what's happening in an arbitrary place and apply a translation transform to it, and the physics is the same.
\begin{array}{rclcr}x' & = & \mathbf{T}_x & + & x \\ y' & = & \mathbf{T}_y & + & y \\ z' & = & \mathbf{T}_z & + & z \end{array}

Let us assume the laws of physics are translationally invariant in time. Then it follows a statement about a experiment happening in an arbitrary time will work the same if we center our basis of "now" with it. This also implies that we can calculate what's happening in an arbitrary time and apply a translation transform to it, and the physics is the same.
t' = \mathbf{T}_t + t

Shorthand: \left( { t' \\ \mathbf{x}' } \right) = \mathbf{T} + \left( { t \\ \mathbf{x} } \right)

Let us assume the laws of physics are rotationally invariant. Then it follows a statement about a experiment oriented in an arbitrary direction will work the same if rotate our coordinate basis to be aligned with it. This also implies that we can calculate what's happening in an arbitrary aligned experiment and apply a rotation transform to it, and the physics is the same.
\begin{array}{rcllcrlcrl}
x' & = & \mathbf{R}_{xx} & x & + & \mathbf{R}_{xy} & y & + & \mathbf{R}_{xz} & z \\
y' & = & \mathbf{R}_{yx} & x & + & \mathbf{R}_{yy} & y & + & \mathbf{R}_{yz} & z \\
z' & = & \mathbf{R}_{zx} & x & + & \mathbf{R}_{zy} & y & + & \mathbf{R}_{zz} & z \end{array} Where R is a proper orthogonal matrix, which can be parameterized in various ways by 3 rotation angles.

Shorthand: \mathbf{x}' = \mathbf{R} \mathbf{x}

Let us assume the laws of physics are invariant with respect to inertial frame. Then it follows a statement about a experiment with a freely moving center of mass moving in an arbitrary direction will work the same if set up our coordinate basis to be co-moving with it with it. But clearly any corresponding change-of-frame transform must tie velocity, time and space together. Since we already assumed we are rotationally invariant and translationally invariant, let us work with v in the x direction and just coordinate differences.

\begin{array}{rcllcrlcrlcrlcrl}
\Delta x' & = & \mathbf{F}_{xx} & \Delta x & + & \mathbf{F}_{xy} & \Delta y & + & \mathbf{F}_{xz} & \Delta z & + & \mathbf{F}_{xt} & \Delta t & + & \mathbf{F}_{x1} \\
\Delta y' & = & \mathbf{F}_{yx} & \Delta x & + & \mathbf{F}_{yy} & \Delta y & + & \mathbf{F}_{yz} & \Delta z & + & \mathbf{F}_{yt} & \Delta t & + & \mathbf{F}_{y1} \\
\Delta z' & = & \mathbf{F}_{zx} & \Delta x & + & \mathbf{F}_{zy} & \Delta y & + & \mathbf{F}_{zz} & \Delta z & + & \mathbf{F}_{zt} & \Delta t & + & \mathbf{F}_{z1} \\
\Delta t' & = & \mathbf{F}_{tx} & \Delta x & + & \mathbf{F}_{ty} & \Delta y & + & \mathbf{F}_{tz} & \Delta z & + & \mathbf{F}_{tt} & \Delta t & + & \mathbf{F}_{t1}
\end{array} where F is a function of v, which we have agreed to consider in the x direction.

Shorthand: \left( { \Delta t' \\ \Delta \mathbf{x}' } \right) = \mathbf{F}_{\mathrm{inhomogeneous}} + \mathbf{F} \left( { \Delta t \\ \Delta \mathbf{x} } \right)

Limiting the form of the velocity transform

Since it makes no sense to talk about (Δx,Δy,Δz,Δt) = (0,0,0,0) which says that the two events happened in the same time and place in one frame transforming into other than (0,0,0,0) in the primed frame, it follows that (F_x1, F_y1, F_z1, F_t1) = (0,0,0,0). Thus F(v) represents a homogeneous transform.

If you think you know of a reason why a v in the x direction should involve displacements in the y or z direction, please let me know. I think that the rotational invariance we assumed earlier means that if v is in the +x direction, then it cannot have a reason to prefer +y or -y, and so the effect on y must be zero, and vice-versa, and the same for z.

Then F_yy = F_zz = 1 and F_xy = F_xz = F_yx = F_yz = F_yt = F_zx = F_zy = F_zt = F_ty = F_tz = 0. So we are over half done.

\begin{array}{rcllcrlcrlcrlcrl}
\Delta x' & = & \mathbf{F}_{xx} & \Delta x & & & & & & & + & \mathbf{F}_{xt} & \Delta t & & \\
\Delta y' & = & & & & & \Delta y & & & & & & & & \\
\Delta z' & = & & & & & & & & \Delta z & & & & & \\
\Delta t' & = & \mathbf{F}_{tx} & \Delta x & & & & & & & + & \mathbf{F}_{tt} & \Delta t & &
\end{array}

Since two events one frame which don't move at all have Δx = 0, but in the other frame Δx'/Δt' = v, then F_xt = v F_tt. Since if two events in one frame are connected by a particle moving at speed -v, and not moving in the other frame then Δx = -v Δt => Δx' = 0 = -F_xx v Δt + F_xt Δt => F_xt = v F_xx => F_xx = F_tt. Let's call that A(v).

\begin{array}{rcllcrlcrlcrlcrl}
\Delta x' & = & A(v) & \Delta x & & & & & & & + & v A(v) & \Delta t & & \\
\Delta y' & = & & & & & \Delta y & & & & & & & & \\
\Delta z' & = & & & & & & & & \Delta z & & & & & \\
\Delta t' & = & \mathbf{F}_{tx} & \Delta x & & & & & & & + & A(v) & \Delta t & &
\end{array}

For the same reason that length-contraction must be in the direction of movement, we expect two observers to experience the same relative time dilation. Since there are no preferred directions, then nothing but convention distinguished -x from +x and so nothing distinguished -v from +v and so we expect that the time dilation to be the same for two observers in relative motion, if there is any time dilation.

Consider a motionless clock. Two tick of the clock are separated by Δt, but Δx = 0.
so Δt' = A(v) Δt . Now let's move the clock at -v so it is motionless for Δx' = 0. So we want to solve Δt = A(v) Δt', Δx = - v Δt, and Δt'= F_tx Δx + A(v) Δt
So Δx = - v A(v) Δt', Δt' = F_tx Δx + A(v) A(v) Δt' and so
Δt' = - v F_tx A(v) Δt' + A(v) A(v) Δt' and so
F_tx = ( A(v)A(v) - 1 ) / ( v A(v) ) = (1/v) ( A(v) - 1/A(v))

\begin{array}{rcllcrlcrlcrlcrl}
\Delta x' & = & A(v) & \Delta x & & & & & & & + & v A(v) & \Delta t & & \\
\Delta y' & = & & & & & \Delta y & & & & & & & & \\
\Delta z' & = & & & & & & & & \Delta z & & & & & \\
\Delta t' & = & {{1}\over{v}} \left( A(v) - {{1}\over{A(v)}}\right) & \Delta x & & & & & & & + & A(v) & \Delta t & &
\end{array}

At this point both the Newtonian and the Relativist should be happy. The Newtonian assumes that A(v) is a constant with value 1, while the Relativist sees that our four initial assumptions do not yet force that choice. A(v), based on our four assumptions, is just a number and may yet turn out to be a non-constant function of v.

Working with the velocity transformation

Now with translations or rotations, they form a group. (A group is a mathematical way of talking about symmetries.) So that if you apply T1 and then T2, you get T3 which is also in the form of a translations. (Same for rotations.) This should be the same for two transforms related to velocity.

\begin{eqnarray}
\Delta x' & = & A(v_1) \Delta x + v A(v_1) \Delta t \\
\Delta y' & = & \Delta y \\
\Delta z' & = & \Delta z \\
\Delta t' & = & {{1}\over{v_1}} \left( A(v_1) - {{1}\over{A(v_1)}}\right) \Delta x + A(v_1) \Delta t
\end{eqnarray}

\begin{eqnarray}
\Delta x'' & = & A(v_2) \Delta x' + v A(v_2) \Delta t' \\
\Delta y'' & = & \Delta y' \\
\Delta z'' & = & \Delta z' \\
\Delta t'' & = & {{1}\over{v_2}} \left( A(v_2) - {{1}\over{A(v_2)}}\right) \Delta x' + A(v_2) \Delta t'
\end{eqnarray}

\begin{eqnarray}
\Delta x'' & = & A(v_3) \Delta x + v A(v_3) \Delta t \\
\, & = &
A(v_2) \left(
A(v_1) \Delta x + v1 A(v_1) \Delta t
\right)
+
v_2 A(v_2) \left(
{{1}\over{v_1}} \left(
A(v_1) - {{1}\over{A(v_1)}}
\right) \Delta x + A(v_1) \Delta t
\right) \\
\Delta y'' & = & \Delta y \\
\Delta z'' & = & \Delta z \\
\Delta t'' & = & {{1}\over{v_3}} \left( A(v_3) - {{1}\over{A(v_3)}}\right) \Delta x + A(v_3) \Delta t \\
\, & = & {{1}\over{v_2}} \left( A(v_2) - {{1}\over{A(v_2)}} \right) \left( A(v_1) \Delta x + v_1 A(v_1) \Delta t \right) + A(v_2) \left( {{1}\over{v1}} \left( A(v_1) - {{1}\over{A(v_1)}} \right) \Delta x + A(v_1) \Delta t \right)
\end{eqnarray}

When we equate our expressions of the double-primed coordinates in terms of the unprimed coordinates, we have the following relations in v and A(v):
A(v_3) = A(v_2) A(v_1) + {{v_2}\over{v1}} A(v_2) A(v_1) - {{v_2}\over{v1}} {{A(v_2)}\over{A(v_1)}}
v_3 A(v_3) = A(v_2) v_1 A(v_1) + v_2 A(v_2) A(v_1) = (v2 + v1) A(v_2) A(v_1)
{{1}\over{v3}} \left( A(v_3) - {{1}\over{A(v_3)}} \right) = {{1}\over{v_2}} \left( A(v_2) - {{1}\over{A(v_2)}} \right) A(v_1) + A(v_2) {{1}\over{v1}} \left( A(v_1) - {{1}\over{A(v_1)}} \right)
A(v_3) = {{v_1}\over{v_2}} A(v_1) A(v_2) - {{v_1}\over{v_2}} {{A(v_1)}\over{A(v_2)}} + A(v_2) A(v_1)
From equations 1 and 4, we have the important equality:
\begin{eqnarray} & & A(v_3) = A(v_2) A(v_1) + {{v_2}\over{v1}} A(v_2) A(v_1) - {{v_2}\over{v1}} {{A(v_2)}\over{A(v_1)}} \\
& = & A(v_3) = {{v_1}\over{v_2}} A(v_1) A(v_2) - {{v_1}\over{v_2}} {{A(v_1)}\over{A(v_2)}} + A(v_2) A(v_1) \end{eqnarray}
or
{{v_2}\over{v_1}} A(v_2) A(v_1) - {{v_2}\over{v_1}} {{A(v_2)}\over{A(v_1)}} = {{v_1}\over{v_2}} A(v_1) A(v_2) - {{v_1}\over{v_2}} {{A(v_1)}\over{A(v_2)}}
or
{{v_2 v_1}\over{v_1^2}} A(v_2) A(v_1) - {{v_2 v_1}\over{v_1^2}} {{A(v_2) A(v_1)}\over{A(v_1)^2}} = {{v_1 v_2}\over{v_2^2}} A(v_1) A(v_2) - {{v_1 v_2}\over{v_2^2}} {{A(v_1) A(v_2)}\over{A(v_2)^2}}
or, for generic v1 and v2,
{{1}\over{v_1^2}} \left( 1 - {{1}\over{A(v_1)^2}} \right) = {{1}\over{v_2^2}} \left( 1 - {{1}\over{A(v_2)^2}} \right)
But since this is true for any v, then there is some constant K = = {{1}\over{v^2}} \left( 1 - {{1}\over{A(v)^2}} \right) for all v. This means A can be written in the form A(v) = {{1}\over{\sqrt{1 - K v^2}}} ; Using the binomial theorem, we can show that when K v² << 1, A(v) is approximately: 1 + \frac{1}{2} K v^2 + \frac{3}{8} \left( K v^2 \right)^2 + \frac{5}{16} \left( K v^2 \right)^3 + \frac{35}{128} \left( K v^2 \right)^4 + \frac{63}{256} \left( K v^2 \right)^5 + \frac{231}{1024} \left( K v^2 \right)^6 + ... so the Newtonian will always appear correct as long as |v| is "small," or K << 1/v².

Only at high speed would there be evidence that K is not zero. (If K is zero, then A(v) = 1, just like the Newtonian assumed.)

(See Pal)

The generic velocity addition law
From equation 2 we see something that with a little algebraic reworking can become our velocity addition law from our four assumptions.
v_3 A(v_3) = (v_2 + v_1) A(v_2) A(v_1)
or
{{v_3}\over{ \sqrt{1 - K v_3^2} }} = {{v_2 + v_1}\over{\sqrt{1 - K v_2^2} \sqrt{1 - K v_1^2}}}
or
{{v_3^2}\over{ 1 - K v_3^2 }} = {{(v_2 + v_1)^2}\over{(1 - K v_2^2)(1 - K v_1^2)}}
or
v_3^2 = {{(v_2 + v_1)^2}\over{(1 - K v_2^2)(1 - K v_1^2)}} {{(1 - K v_2^2)(1 - K v_1^2)}\over{K (v_2 + v_1)^2 + (1 - K v_2^2)(1 - K v_1^2)}}
or
v_3^2 = {{(v_2 + v_1)^2}\over{K (v_2 + v_1)^2 + (1 - K v_2^2)(1 - K v_1^2)}}
or
v_3^2 = {{(v_2 + v_1)^2}\over{K v_2^2 + 2 K v_1 v_2 + K v_1^2 + 1 - K v_2^2 - K v_1^2 + K^2 v_1^2 v_2^2}}
or
v_3^2 = {{(v_2 + v_1)^2}\over{1 + 2 K v_1 v_2 + K^2 v_1^2 v_2^2}}
or
v_3 = {{v_2 + v_1}\over{1 + K v_1 v_2}}

Clearly if K is measured to be zero, then A(v) = 1 and there is no time dilation. However the results of an 1859 experiment (among thousands of others) are inconsistent with K = 0.

Fresnel and Fizeau's measured value of K
In the discredited dragged-ether theory of Fresnel, light is "slowed" and "dragged" by a transparent dielectric with dielectric constant n. It is slowed to V=c/n, but if the medium is moving at speed v, it is dragged and the measured speed is about U = c/n + v(1 - 1/n²). The amount of "ether dragging" by a moving dielectric is measured as the unexplained Frensel drag coefficient, 1 - 1/n². The result eventually help caused the downfall of the dragged-ether model, for where a physical medium can support a wide variety of waves, the phenomenon of dispersion shows that n is a function of wavelength, and so the Fresnel drag coefficient must also be a function of wavelength, and so there must be a different ether to drag for every wavelength of light.

But let's just take V=c/n the experimental velocity of light in stationary medium, and apply our generic velocity addition law to it and see how it predicts an observer moving relative to the medium measure its speed at.

v_3 = {{v_1 + v_2 }\over{1 + K v_1 v_2}} = {{v + V}\over{1 + K v V}} = {{v + { {c} \over {n} } } \over { 1 + K v c/n } }

if we assume K v c/n << 1, then we can use the binomial theorem to approximate v_3 as (v + {{c}\over{n}}) - (K v {{c}\over{n}}) (v + {{c}\over{n}}) + (K v {{c}\over{n}})^2 (v + {{c}\over{n}}) + ... \\
= {{c}\over{n}} + v - K v^2 {{c}\over{n}} - K v {{c^2}\over{n^2}} + K^2 v^3 {{c^2}\over{n^2}} + K^2 v^2 {{c^3}\over{n^3}} ... or, if you drop the terms which aren't linear in v, v_3 = {{c}\over{n}} + v (1 - K {{c^2}\over{n^2}})

If v_3 is close to the observed value U, then K c² = 1, or K = 1/c²

Epilogue
This was written to show that time dilation, by which I mean that observers who differ in velocity must have A(v) different than 1, is the only physical result if you accept the four assumptions. The other consequences of this idea have been well-developed. K is very close to zero in ordinary units, but we have thousands of experimental results which suggest that it is much closer to 1/c² than to zero.

This should show that if the four assumptions are good, then even experiments not involving light will show that c is an physically important speed in our universe. Further E = m_0 A(v) c² gives us E = m_0 c² + 1/2 m_0 v² + ... which has been used to relate Newton's approximate formula for kinetic energy to Einstein's Relativity.
See Mattingly.

According to Silagadze people have shown this to be true with various degrees of rigor since at least 1910.

Following Einstein, Minkowski (1908) showed that algebraically this was the same as saying space and time were not separate things, which is how all physicists work today. Both Length contraction and time dilation arise from treating space and time as separate things with separate meanings, which is the Newtonian and intuitionist view. In Minkowski space-time, these are trivial (boring!) effects.

It's still crazy -- besides you can't show that this works with anything other than light

Well it works everywhere we test it including for decaying muons and in Thomas precession. It's used in the design of modern electronics, the GPS system and of course particle accelerators.

What is Thomas precession?

It is the observation that a boost in the X-direction and a boost in the Y-direction do not combine into a boost in some third direction, but a boost in a third direction AND a rotation. And the rotation depends on the order of the boosts.

This could (and has been) expressed as a statement relating to the commutators of the Poincaré group (inhomogeneous Lorentz group, i.e. Special Relativity). See page 55 of Kim and Noz or any comparable text.

Historically, Thomas precession was observed as a mismatch between the predictions of non-relativistic electromagnetics and physical experiment, where non-relativistic electromagnetics predicted a precession rate twice what was measured. Thomas precession (which doesn't depend on the charge of the electron or the magnetic field of the nucleus) predicts a value -1/2 of the non-relativistic electromagnetic prediction, so since 1 - 1/2 = 1/2 this was Thomas' 1927 explanation of the experimental result.

I don't see E=mc², so you are pulling a fast one!
E=mc² is not synonymous with the theory of special relativity, but it is an important result of the theory.

You don't even mention energy or mass in those equations, so how can you get E=mc²?
Lets start with proper time, Δτ, between two events which are seen be an observer to happen at different times, Δt and different places, Δx. (Let us assume that these two events are on the world line of a massive particle moving slower than light, so we have finite proper time...)
c²(Δτ)² = c²(Δt)² - (Δx)²
or
c²(Δτ/Δt)² = c² - (Δx/Δt)² = c² - (v)²
or
Δt = Δτ/√(1 - (v/c)²) = γΔτ

Where for the first time we use γ = 1/√(1 - (v/c)²)

Now, lets say that this Δt and Δx were associated with a massive body, with invariant mass m. Then it follows from equation 1 that
m²c²(Δτ)² = m²c²(Δt)² - m²(Δx)²
or
m²c² = m²c²(Δt/Δτ)² - m²(Δx/Δτ)²
or
m²c² = m²c²(Δt/Δτ)² - m²(Δx/Δt)²(Δt/Δτ)²
or since we introduced the shorthand, γ,
m²c² = (mcγ)² - (mvγ)²
or
(mc²)² = (γmc²)² - (γmvc)²

Now,
γmc² = mc²/√(1 - (v/c)²) ≈ mc²(1 + ½(v/c)² + ... ) = mc² + ½mv² + ...
which has units of energy and the second term exactly corresponds to the Newtonian kinetic energy of a particle of mass m and moving at speed v, so because it changes pretty much like the Newtonian Energy lets say
E = γmc²
and
p = γmv (or p = Ev/c² which works better in the case of |v| = c and m = 0)
then our last result could be written as
(mc²)² = E² - (pc)²
which for a non-moving particle is E = mc²

So because (cΔt,Δx) can be described by all observers in all frames as being associated with a relativistic invariant cΔτ, so must (E = γmc², pc = γmvc) be associated with a relativistic invariant mc².

Which is just high school algebra. The interesting physics comes from the cases when E actually acts like energy and p acts like a momentum, which is to say they are conserved. Proving this requires Noether's theorem and a physically tested Lagrangian. Basically, if the physics don't change over time, then the Energy is expected to be conserved, and if the the physics don't care where in the universe you are, then the momentum is conserved.

This bites us a second time in quantum physics where canonical conjugates like time and energy or position and momentum are related by the uncertainty principle.

According to Lev Okun, who I have stolen from before, all you need for the physics of a free particle are:

E^2 - \mathbf{p}^2c^2 = m^2c^4
and
p = Ev/c²

So Relativity gives us Minkowski space-time, and space-time gives us relations between E and p and m and v. And we call E and p, Energy and Momentum because relativistic Lagrangians indicate they are conserved in the cases where Newtonian Lagrangians indicate that the Newtonian quantities ½mv² and mv are conserved.

Getting tired now... what about gravity?

If Newton was wrong about kinematics of fast-moving things, then logically, the original description of Universal Gravitation is suspect. Not only did Einstein correctly describe the action of gravity upon fast-moving objects, but he removed the feature Newton found philosophically objectionable about his own theory: Einstein's General Relativity is both in better agreement with experiment than Universal Gravitation and it is a local theory with no action-at-a-distance.

Will's famous reference article also states in axiomatic form what are the principles of General Relativity in simple terms. Let me simplify them one step more:

The trajectory of a freely falling “test” body (one not acted upon by such forces as electromagnetism and too small to be affected by tidal gravitational forces) is independent of its internal structure and composition.
The outcome of any local non-gravitational experiment is independent of the velocity of the freely-falling reference frame in which it is performed.
The outcome of any local non-gravitational experiment is independent of where and when in the universe it is performed.
The Einstein tensor, which is the description of the trace of the curvature of space-time such that setting the derivative equal to zero is equivalent to saying the curvature of space-time obeys the Bianchi identities, is proportional to the local energy-stress tensor, for which setting the derivative equal to zero implies local conservation of energy. Thus local conservation of energy implies that the Bianchi identities hold (or vice-versa).

References

H. Fizeau "Sur les hypothèses relatives à l'éther lumineux" Annales de chimie et de physique 57 385-403 (1859) http://gallica.bnf.fr/ark:/12148/bpt6k347981/f381.table
W. A. von Ignatowsky, “Einige allgemeine Bemerkungen zum Relativitatsprinzip,” Phys. Z. 11, 972-976 (1910).
Y.S. Kim, M.E. Noz Theory and Applications of the Poincaré Group (1986) http://books.google.com/books?id=Ok6dPuqc8XMC
L.B. Okun "The Concept of Mass" Physics Today 42(6) 31-36 (1989) http://www.physicstoday.org/vol-42/iss-6/vol42no6p31_36.pdf
N Ashby "Relativity in the Global Positioning System" Living Rev. Relativity 6, 1 (2003) http://www.livingreviews.org/lrr-2003-1
P.B. Pal "Nothing but Relativity" Eur. J. Phys. 24 315-319 (2003) http://arxiv.org/abs/physics/0302045
R.C. Henry "Teaching Special Relativity: Minkowski trumps Einstein" http://henry.pha.jhu.edu/henryMinkowski.pdf
D. Mattingly, "Modern Tests of Lorentz Invariance", Living Rev. Relativity 8, 5 (2005) http://www.livingreviews.org/lrr-2005-5
M. Montesinos, E. Flores "Symmetric energy-momentum tensor in Maxwell, Yang-Mills, and Proca theories obtained using only Noether's theorem" Rev. Mex. Fis. 52, 29-36 (2006) http://arxiv.org/abs/hep-th/0602190
K. Brown "2.12 Thomas Precession" in Reflections on Relativity http://www.mathpages.com/rr/s2-11/2-11.htm
C.M. Will, "The Confrontation between General Relativity and Experiment", Living Rev. Relativity 9, 3 (2006) http://www.livingreviews.org/lrr-2006-3
Z.K. Silagadze "Relativity without tears" http://arxiv.org/abs/0708.0929

http://en.wikipedia.org/wiki/Aether_drag_hypothesis
http://en.wikipedia.org/wiki/Fizeau_experiment
http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_energy-momentum_equation
http://en.wikipedia.org/wiki/Noether's_theorem

camilus

10-26-08, 09:41 PM

I would gladly post a derivation of the time-dilation formula if anyone would be interested.

BenTheMan

10-26-08, 11:44 PM

Mod Note

Uno Hoo has been banned one day for trolling in the physics forum.

Reiku

10-28-08, 03:44 PM

I would gladly post a derivation of the time-dilation formula if anyone would be interested.

Yes, go right ahead.

Zeno

11-01-08, 07:17 PM

And of course there are some more Special Relativity paradoxes that cannot be resolved that you do not yet know about.

And some that can't be resolved that they do know about...
http://www.sciforums.com/showthread.php?t=81532

rpenner

11-02-08, 03:14 AM

Zeno, you made mistakes on that thread that completely invalidate any point you were trying to make.

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