Tutorial: Relativity - what is a reference frame?

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Good explanation, and good of you to help the non-technical people understand this. It's free tutoring--who can beat that?

 

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Excellent tutorial James. Good work!

 

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James, Can you elaborate a little more on how you would go about determining that a stick in your frame is 1 meter, using light (since a meter is defined by light travel time), while keeping in mind that light travels independently of objects? Also, speaking strictly about light and time, in what reference frame are the sticks laid out for the light sphere??

James, It's clear that you can be stationary compared to the bus while remaining seated on the bus, and also be traveling with respect to the road when the bus is in motion compared to the road. Nobody disputes that!! Get a grip on yourself.

 

K. So you are at a constant velocity in a ship, not accelerating. You send a light signal from the rear of the ship and find where the light hits after 1/299792458 of a second. The distance the light traveled in 1/299792458 of a second in that frame is a meter, correct? Now, you fire your rocket engine and accelerate for a duration of time and then turn off the engine and are once again at a constant velocity. You then perform the test again, sending a signal from the rear of the ship and find where the light is after 1/299792458 of a second. Are you saying that in both tests the light will be at the same point along the sticks after 1/299792458 of a second has elapsed?

 

The set of sticks nailed to the floor of the ship. The same set of sticks in both tests in which you marked on the sticks where light was after 1/299792458 of a second. You know, THE STICKS!

 

Yes, it will always be the same. That's just a small part of the beauty of Relativity.

 

Motor Daddy

Then you do understand the meaning of "Frame of Reference"? In this case the frame of reference is the bus, not the road. If the bus was traveling very near lightspeed you would still measure the speed of light and the distance it travels/time it takes exactly the same as if the bus was stationary to the road(paint over the windows and you could not tell whether you were stationary or moving very fast), but the effects you experience(speed of time, length in the direction of travel etc.)would be different from effects the people standing beside the road measure IN YOUR FRAME, and that is Relativity. IE what you experience within your frame is always the same as measured within that frame, but is relative to all other frames. This is a simple fact, get used to it.

Grumpy

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You know, I have a nagging suspicion you only know that because I've written almost exactly the same thing many times in various relativity related threads here over the last few years.

 

If you set the sign convention on the velocity such that

\( \begin{eqnarray} t' &=& \gamma (t \,-\, \frac{v}{c^{2}} x) \\ x' &=& \gamma (x \,-\, vt) \,, \end{eqnarray} \)​

then the inverse equations are

\( \begin{eqnarray} t &=& \gamma (t' \,+\, \frac{v}{c^{2}} x') \\ x &=& \gamma (x' \,+\, vt') \,, \end{eqnarray} \)​

Also it's a bit odd that you write

because the numerator \(t \,-\, vx\) is only correct if you apply the common convention of setting \(c \,=\, 1\), which makes it rather odd that you leave \(c\) explicitly in the denominator.

Also,

actually, if I ignore the weird \(c \,=\, 1\)? thing I mentioned just above, that should be

\( \frac{x \,-\, vt}{1 \,-\, \frac{v^2}{c^2}} \,-\, \frac{u(t\,-\,vx)}{1\,-\,\frac{v^2}{c^2}} \)​

(without the square roots).

Obviously anyone can make silly mistakes, but it doesn't look much like you really understand what you're doing.

 

So Motor Daddy, a reference frame is simply ''another point of view.'' The only thing in this universe that does not vary from one reference frame to another, is of course the speed of light - always gets measured moving at the speed of light.

 

I'm just observing that your post seems to follow a format that I've used many times (the Lorentz transformation is..., notice that \(x' \,=\, 0 \,\Rightarrow\, x \,=\, vt\)..., prove something that should be obvious to anyone who already understands relativity...).

I'm not flattered so much as amused. I wouldn't normally bother pointing something like this out (if people are reading and learning from my posts, then great!), except that you have something of a history of writing posts that consist mostly of a collage of stuff you read and didn't understand elsewhere, and you've even been caught out in [POST=2552811]blatant plagiarism[/POST] before.

Your last few posts roughly look like you're trying to show that the composition of two Lorentz boosts is still a Lorentz boost and that the velocity of the resulting boost is given by the Lorentz velocity addition formula (\(w \,=\, \frac{u \,+\, v}{1 \,+\, \frac{uv}{c^{2}}}\)), except you got that wrong too and it doesn't seem to be a useful addition to this thread or have any relation to the comment you were replying to.

 

I don't think I know you. I have no idea what you'd be like in person. But many people here are familiar with your online persona and patterns in your posting behaviour.

And whether or not you're the latest incarnation of reiku, you should keep in mind it is damn obvious to everyone when you use well known equations that you don't understand the meaning of.

 

If you assume the laws of physics are translationally invariant in space and time, and assume the laws of physics are rotationally invariant so that the directions we call X, Y and Z are arbitrary choices, and you assume the laws of physics are invariant with respect to inertial frame such that any inertial body is just as good as any other inertial body as the standard of "motionless," then one is compelled to deal with the geometry of space and time. It is then a matter of experiment, not opinion, as to what geometry our universe has.
The possibilities (that don't involve rotation or motion other than in the X direction) are parametrized in a generalization of the Galilean transform.
\(\begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex}\Delta x' \\ \rule{0pt}{20ex}\Delta y' \\ \rule{0pt}{20ex}\Delta z' \end{pmatrix} = \begin{pmatrix}\rule{0pt}{20ex}\frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex}\frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex}0 & 0 & 1 & 0 \\ \rule{0pt}{20ex}0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t\\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix}\)

This transform converts coordinate differences on a motionlessness world-line in one coordinate system to motion in another:
\( \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t\\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} \Delta t \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} \Delta t \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} v \Delta t' \\ \rule{0pt}{20ex}0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}\)
This transform also converts motion in one coordinate system to motionless in this specific case:
\( \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t\\ \rule{0pt}{20ex} -v \Delta t \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} \Delta t - \frac{K v^2}{\sqrt{1 - K v^2}} \Delta t \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} \Delta t - \frac{v}{\sqrt{1 - K v^2}} \Delta t \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex}0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}\)
And we see it is self-consistent if we compose two different transforms and see that the result is also a transform:
\(\begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K u^2}} & \frac{K u}{\sqrt{1 - K u^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{u}{\sqrt{1 - K u^2}} & \frac{1}{\sqrt{1 - K u^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}} & \frac{K u + K v}{1 + K u v} \times \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{u + v}{1 + K u v} \times \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}} & \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K w^2}} & \frac{K w}{\sqrt{1 - K w^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{w}{\sqrt{1 - K w^2}} & \frac{1}{\sqrt{1 - K w^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \)
Since \(w = \frac{u + v}{1 + K u v} \, \wedge \, \left| K u^2 \right| < 1 \, \wedge \, \left| K v^2 \right| < 1 \; \Rightarrow \; \frac{1}{\sqrt{1 - K w^2}} = \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}}\)

So the generalization of the Galilean transformation gives us the self-consistent generalization of the velocity composition law. \(w = \frac{u + v}{1 + K u v}\). We also have three different conservation laws (isometries) depending on the sign on K.
If K < 0, then \(\left( \Delta t \right)^2 + \left| K \right| \left( \left( \Delta x \right)^2 + \left( \Delta y \right)^2 + \left( \Delta z \right)^2 \right) = \left( \Delta t' \right)^2 + \left| K \right| \left( \left( \Delta x' \right)^2 + \left( \Delta y' \right)^2 + \left( \Delta z' \right)^2 \right) \). This is elliptical geometry.
If K > 0, then \(\left( \Delta t \right)^2 - \left| K \right| \left( \left( \Delta x \right)^2 + \left( \Delta y \right)^2 + \left( \Delta z \right)^2 \right) = \left( \Delta t' \right)^2 - \left| K \right| \left( \left( \Delta x' \right)^2 + \left( \Delta y' \right)^2 + \left( \Delta z' \right)^2 \right) \). This is hyperbolic geometry.
If K = 0, then \( \left( \Delta x \right)^2 + \left( \Delta y \right)^2 + \left( \Delta z \right)^2 = \left( \Delta x' \right)^2 + \left( \Delta y' \right)^2 + \left( \Delta z' \right)^2 \) and \(\Delta t = \Delta t'\) This is Euclidean geometry of space and an unconnected absolute time.

All physical experiment (since 1859 when testing became precise enough) favors \(K = c^{\tiny -2}\) over \(K = 0\) which is why we study the Lorentz transformations, Einstein velocity composition laws, special relativistic interval \( c^2 \left( \Delta t \right)^2 - \left( \Delta x \right)^2 - \left( \Delta y \right)^2 - \left( \Delta z \right)^2\) and hyperbolic geometry in modern physics classes. It also unified the mechanics of material bodies with the phenomena of electromagnetism.

 

I know what a frame of reference is, so stop pretending you are teaching me something, because you're not.

I am still waiting on an answer from James. If you care to answer the question then take a stab at it. You have sticks nailed to the floor in your ship. What is your process of determining a "meter?"

 

Impossible, James. Don't you know what you are saying?

The more you make increases or decreases in the ship's velocity by accelerating for a duration of time, and then retesting, the more your point of impact of light is going to change where you find it along your sticks 1/299792458 of a second later.

If the signal is sent from the rear of the ship, the greater the ship's velocity in the forward direction (x direction, we'll say) the closer to the rear of the ship along the sticks the light will impact, if tested for the same 1/299792458 of a second each test.