# Thread: Special Relativistic time dilation and length contraction

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

2. 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.

3. 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.

4. Like the one you warned, promised, and threatened us you would post two years ago?

5. Originally Posted by Pete
Like the one you warned, promised, and threatened us you would post two years ago?
Indeed. It seems that CANGAS has learned how to use a proxy.

6. Originally Posted by Pete
Like the one you warned, promised, and threatened us you would post two years ago?

Why is "CANGAS" so important to you?

7. Originally Posted by BenTheMan
Indeed. It seems that CANGAS has learned how to use a proxy.

Why is "CANGAS" so important to you?

8. 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):
1. $A(v_3) = A(v_2) A(v_1) + {{v_2}\over{v1}} A(v_2) A(v_1) - {{v_2}\over{v1}} {{A(v_2)}\over{A(v_1)}}$
2. $v_3 A(v_3) = A(v_2) v_1 A(v_1) + v_2 A(v_2) A(v_1) = (v2 + v1) A(v_2) A(v_1)$
3. ${{1}\over{v3}} \left( A(v_3) - {{1}\over{A(v_3)}} \right) = {{1}\over{v_2}} \left( A(v_2) - {{1}\over{A(v_2)}} \right) A(v_1) + A(v_2) {{1}\over{v1}} \left( A(v_1) - {{1}\over{A(v_1)}} \right)$
4. $A(v_3) = {{v_1}\over{v_2}} A(v_1) A(v_2) - {{v_1}\over{v_2}} {{A(v_1)}\over{A(v_2)}} + A(v_2) A(v_1)$
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)

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
or
c²(Δτ/Δt)² = c² - (Δx/Δt)² = c² - (v
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
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/i...2no6p31_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...entum_equation
http://en.wikipedia.org/wiki/Noether's_theorem

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

10. Mod Note

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

11. Originally Posted by camilus
I would gladly post a derivation of the time-dilation formula if anyone would be interested.

12. Originally Posted by Uno Hoo
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...

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

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