SR Question [Archive] - Physics Forums

StarThrower

Apr7-04, 01:24 PM

What exactly allows you to apply Euclidean geometry to spacetime in this case? (I am sincerely asking this because I don't know the answer.)

Answer: The Pythagorean Theorem

Keep in mind that the time dilation formula of the special theory of relativity is derived by assuming that the pythagorean theorem is true in the kind of reference frame we have in my question here.

By stipulation, the photons are departing from the atom at right angles to one another in the rest frame of the atom, which is an inertial reference frame by stipulation. The distance between them at any moment in time is found using the pythagorean theorem. We actually have an isosceles triangle here. The hypotenuse of which is greater than either leg.

However, if the relative speed of the photons is c, as was suggested by Jdavel, then you can prove that the triangle is equilateral. Thence, the emission angle of the photons is 60 degrees.

But it was stipulated that the emission angle is 90 degrees.

Kind regards,

StarThrower

P.S. The very first theorem of Euclid, was to prove that an equilateral triangle could be constructed. Here is a link to Euclid's first theorem:

Euclid 's First Theorem (http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI1.html)

In Euclid's eleventh proposition, he proves that a right angle can be constructed:

Euclid Book One Theorem Eleven (http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI11.html)

Severian596

Apr7-04, 02:08 PM

StarThrower

Apr7-04, 02:21 PM

First Issue:
I posed this question to StarThrower in post #92 on this page of the following thread:
http://www.physicsforums.com/showthread.php?t=17024&page=7&pp=15

I proposed that the photons would never observe each other and therefore could never determine how fast they were traveling relative to each other. An observer (in this case the atom) at point C could certainly propose how fast they were traveling with respect to himself, or how fast they were traveling with respect to each other with respect to himself. I'm intrigued about the legitimate answer.

Second issue:
Flat space is only one kind of space. As you say, "By stipulation, the photons are departing from the atom at right angles to one another in the rest frame of the atom." But you cannot necessarily assume that the photon is restricted to Euclidian 3-space. The Pythagorean theorem applies to flat space. For curved spaces you must use a modified version of the theorem that takes into account the metric of the curvature of space.

http://mathworld.wolfram.com/Distance.html

The metric is a (sometimes complex) matrix that tells you HOW to measure the distance between any two points in a specific locale of the space.

Let me reiterate:

Keep in mind that the time dilation formula of the special theory of relativity is derived by assuming that the pythagorean theorem is true in the kind of reference frame we have in my question here.

All kinds of mathematical nonsense is resting on the incorrect assumption that speed of light is c in any inertial reference frame. Space is three dimensional Euclidean.

From the site quoted we have:

In Euclidean three-space, the distance between points A=(x1,y1,z1) and B=(x2,y2,z2) is:

D(A,B) = \sqrt{ (x2-x1)^2 + (y2-y1)^2+(z2-z1)^2 }

You cannot bend the vacuum.

Kind regards,

StarThrower

Chen

Apr7-04, 02:28 PM

All of that kind of nonsense is predicated on the incorrect assumption that speed of light is c in any inertial reference frame. Space is three dimensional Euclidean.

You cannot bend the vacuum.
You're in the wrong profession, mate. You should have become a comedian.

How come light, which travels at straight lines, can and does bend around the sun, allowing us to see objects which are actually behind it? Or are you going to suggest that light does not travel at a straight line now?

Severian596

Apr7-04, 02:30 PM

Space is three dimensional Euclidean. You cannot bend the vacuum.

How do you define gravity in 3-space?

StarThrower

Apr7-04, 02:33 PM

You're in the wrong profession, mate. You should have become a comedian.

How come light, which travels at straight lines, can and does bend around the sun, allowing us to see objects which are actually behind it? Or are you going to suggest that light does not travel at a straight line now?

For light to travel in a curved path, it must experience a force. Otherwise, by Newton's First Law of Motion (Which is roughly Aristotle's law of Inertia), the photon would continue to move in a straight line at a constant speed.

BUT YOU ARE TELLING ME THE PHOTON IS IN A GRAVITATIONAL FIELD. Therefore, something is interacting with the photon to exert a force on the photon. Hence the photon isn't in an inertial reference frame anymore.

Kind regards,

StarThrower

Severian596

Apr7-04, 02:38 PM

For light to travel in a curved path, it must experience a force. Otherwise, by Newton's First Law of Motion (Which is roughly Aristotle's law of Inertia), the photon would continue to move in a straight line at a constant speed.

BUT YOU ARE TELLING ME THE PHOTON IS IN A GRAVITATIONAL FIELD. Therefore, something is interacting with the photon to exert a force on the photon. Hence the photon isn't in an inertial reference frame anymore.

Kind regards,

StarThrower

Like I said define the force of gravity in FLAT 3-space.

StarThrower

Apr7-04, 02:40 PM

How do you define gravity in 3-space?

\vec{F} =- GM_1 M_2 \frac{\vec{R}}{|\vec{R}|^3}

Severian596

Apr7-04, 02:43 PM

\vec{F} =- \frac{GM_1 M_2}{|vec{R}|^2}

Indeed it was. According to this equation when you wave your hand from left to right (specifically when you modify R in the equation above), you instantaneously influence atoms on jupiter, the moon...etc, because your hand exerts a gravitational force on them. Are you comfortable with this?

turin

Apr7-04, 04:22 PM

Space-time does NOT have a Euclidean geometry; it has a Euclidean topology. Since the pathagorean theorem is a geometrical theorem for Euclidean geometry, it does not apply to space-time.

Hurkyl

Apr7-04, 04:36 PM

What is the relative speed of the photons?

Undefined, since, as we've pointed out to you before, photons don't have rest frames.

By Euclidean geometry, if the relative speed of the photons is c, then the two photons form an equilateral triangle with the atom, whose area is increasing in time. The interior angles of an equilateral triangle are 60 degrees. However, the emission angle was stipulated to be 90 degrees. This is a contradiction.

Allow me to remind you that the difference of two velocity vectors in a given reference frame is not, in general, give you the relative velocity of the said objects in SR.

BUT YOU ARE TELLING ME THE PHOTON IS IN A GRAVITATIONAL FIELD. Therefore, something is interacting with the photon to exert a force on the photon. Hence the photon isn't in an inertial reference frame anymore.

According to GR, the photon is travelling in a straight line, period. And, of course, photons never have inertial rest frames.

Tom Mattson

Apr7-04, 05:14 PM

After all the time people have spent showing you that a reference frame moving at c is ill-defined in SR, I can't believe we're seeing another "what does motion look like from the photon's point of view?" thread.

By Euclidean geometry, if the relative speed of the photons is c, then the two photons form an equilateral triangle with the atom, whose area is increasing in time. The interior angles of an equilateral triangle are 60 degrees. However, the emission angle was stipulated to be 90 degrees. This is a contradiction.

So what??

The angle is stipulated to be 90 degrees in the lab frame. Even in Galilean relativity, the angle would not be 90 degrees in any other frame. In fact, it is always possible to transform to a frame in which both photons are moving along the same line! It's called the "center of momentum" frame.

Severian596

Apr7-04, 05:24 PM

Undefined, since, as we've pointed out to you before, photons don't have rest frames.

Hurkyl, is this because in order to determine the relative velocity of two objects they must be in inertial rest frame(s) (IRF)?

Let me pose a sub-light question (god forbit)...two objects A and B speed away from point C. At time t_{0}= 0 the distance between any two of these three points (A, B, and C) can be considered zero. A and B are traveling at 0.7c wrt point C.

A
^
|
|
|
|
C---------->B

How do we determine how fast A and B are travelling with respect to each other? Usually in SR this kind of thing is stated up front (S' is traveling at speed v with respect to S).

StarThrower asserts (as he usually does) that we can state how fast A and B are traveling with respect to each other if we just use our observations from our own frame of reference and apply [you know who's] theorem. I don't believe this is correct. If we concentrate on sub-light speeds for this problem perhaps we can get "real" SR answers instead of having to listen to StarThrower assert something else about absolute time and space...

jdavel

Apr7-04, 05:33 PM

Prove that the relative speed of the two photons is c if you can.

I can prove it. But I can't prove it to you, because your mind is closed to the possibility that I can. So I'm not going to waste my time.

A few days ago I showed you the source of all your confusion here: you don't understand what the variables used in SR mean. You think you do. But you don't.

Hurkyl

Apr7-04, 06:00 PM

Hurkyl, is this because in order to determine the relative velocity of two objects they must be in inertial rest frame(s) (IRF)?

Yes, I do interpret "relative velocity" to mean the velocity of one object in the rest frame of the other object.

russ_watters

Apr7-04, 08:51 PM

All kinds of mathematical nonsense is resting on the incorrect assumption that speed of light is c in any inertial reference frame. On what do you base this assertion? And don't say the Pythagorean theorem - I want emperical evidence, for example, give me just one (confirmed) test of the speed of light that shows a variation from C.

Hurkyl

Apr7-04, 10:27 PM

Hurkyl, is this because in order to determine the relative velocity of two objects they must be in inertial rest frame(s) (IRF)?

Let me pose a sub-light question (god forbit)...two objects A and B speed away from point C. At time the distance between any two of these three points (A, B, and C) can be considered zero. A and B are traveling at 0.7c wrt point C.

A
^
|
|
|
|
C---------->B

How do we determine how fast A and B are travelling with respect to each other? Usually in SR this kind of thing is stated up front (S' is traveling at speed v with respect to S).

StarThrower asserts (as he usually does) that we can state how fast A and B are traveling with respect to each other if we just use our observations from our own frame of reference and apply [you know who's] theorem. I don't believe this is correct. If we concentrate on sub-light speeds for this problem perhaps we can get "real" SR answers instead of having to listen to StarThrower assert something else about absolute time and space...

Ok, let us put coordinates on this system in order to solve it. (Things are much clearer with coordinates, which is probably one of the reasons why anti-SR crackpots tend to dislike them)

In the rest frame of C, let us put the event where A, B, and C are all at the same place at the origin of the coordinate system. Let the coordinate axes x, y, and z be right, up, and out of the page respectively, and let t be the coordinate time.

The worldlines of A and B are given parametrically by (being sloppy and parametrizing by t):

A: (t, 0, 0.7ct, 0)
B: (t, 0.7ct, 0, 0)

Letting t', x', y', and z' be the coordinates in A's reference frame, the Lorentz transform from C's frame to A's frame gives the relations:

x' = x
y' = γ(y - 0.7ct)
z' = z
t' = γ(t - 0.7y/c)

where γ := 1/√(1-0.72) = 1/√0.51

So, the worldlines are given parametrically in A's frame by (yes, I mean to parametrize by t and not t'):

A: (γ(t - 0.7 (0.7ct)/c), 0, γ(0.7ct - 0.7ct), 0)
A: (γ 0.51 t, 0, 0, 0)
B: (γ(t - 0.7 0/c), 0.7ct, γ(0-0.7ct), 0)
B: (γt, 0.7ct, -γ0.7ct, 0)

(I only did A as a sanity check)

So, we have that the coordinate velocity of B in A's frame is given by
v = (0.7c √0.51, -0.7c, 0)

So, unless I've made a mistake along the way, B's speed relative to A is 0.7 c √1.51.

jdavel

Apr7-04, 11:38 PM

If we concentrate on sub-light speeds for this problem perhaps we can get "real" SR answers instead of having to listen to StarThrower assert something else about absolute time and space...

Ok, Hurkyl just did it. But StarThrower isn't going read that post. He wouldn't have clue what Hurkyl was talking about, because he doesn't understand what the variables mean.

You can even do a little better than what Hurkyl did. Start the problem using an unspecified sub-light speed v. Then once you find the relative speed, let v go to c as a limiting value, and watch the relative speed go to c. But StarThrower wouldn't understand that either!

ramcg1

Apr7-04, 11:42 PM

Severian596

Apr7-04, 11:50 PM

Thank you very much, Hurkyl! This is my first full-fledged 4-space Lorentz transformation. Usually they factor out the y and z to make it more friendly, so I worked along side you to see if I got the same result.

My work agrees up until the transformation of C to A. I'm working out of Hogg's publication and am using his Lorentz Transformation notation. I cut it out of the PDF and posted it here (http://copperplug.no-ip.org/homesite/Spacetime_files/Lorentz.gif)...it's probably important to explain my discrepancy.

To transform from y to y', I did the following:

t = t
y = 0.7ct
\beta_y = 0.7
\beta^{2} = 0.7^{2}

Therefore the transformation should go:

y' = t(-\gamma \beta_{y}) + 0 + y(1+\frac{(\gamma -1)(\beta^{2}_{y})}{\beta^{2}}) + 0
y' = (t)(-\gamma 0.7) + (0.7ct)(1+\frac{(\gamma -1)(0.7^{2})}{0.7^{2}})

Eventually I simplify this to the following (I resubstituted y to compare it to yours):

Mine was: y' = \gamma(y-0.7t)
Yours was: y' = \gamma(y-0.7ct)

I don't have a c term in the final result where you do...where did I go wrong?

ramcg1

Apr7-04, 11:53 PM

jdavel

Apr8-04, 12:04 AM

ramcg1 said:

"If the part of the universe the two photons are passing though is at rest with respect to the source atom...."

How would you know whether it was or not. Michelson & Morley tried pretty hard to know, but they never did!

"....then neither photon will have suffered the effects of SR."

What does that mean?

"Detectors at a known distance from the source atom would register simultaneous arrivals."

True

"Thus a logical person would deduce that the relative velocity of the two photons were square root of 2 times 299792458 meters per second...."

Logical? Not if they were aware of SR.

"SR is irrelevent to the calculation as you are not trying to observe a photon travelling between the two."

Not true. SR is relevant to any calculation involving motion.

Hurkyl

Apr8-04, 06:35 AM

I don't have a c term in the final result where you do...where did I go wrong?

It is often cleaner to require that all 4 axes have the same units, so instead of the t axis you have a ct axis, and your 4-vectors are (ct, x, y, z). If you use ct for the 0-th coordinate when multiplying by the matrix you linked, you will get my equation, so I presume that's the thing.

Thus a logical person would deduce that the relative velocity of the two photons were square root of 2 times 299792458 meters per second or approx: 423970560 meters per second.

No, a logical person would deduce that (in this particular reference frame), the distance between the photons is increasing at a rate of 42390560 m/s.

Severian596

Apr8-04, 09:05 AM

It is often cleaner to require that all 4 axes have the same units, so instead of the t axis you have a ct axis, and your 4-vectors are (ct, x, y, z). If you use ct for the 0-th coordinate when multiplying by the matrix you linked, you will get my equation, so I presume that's the thing.

Excellent! That makes me feel pretty good about it. Normally I see the 0th coordinate as ct and didn't catch the fact that you changed it from t to ct.

StarThrower

Apr8-04, 09:21 AM

Indeed it was. According to this equation when you wave your hand from left to right (specifically when you modify R in the equation above), you instantaneously influence atoms on jupiter, the moon...etc, because your hand exerts a gravitational force on them. Are you comfortable with this?

It is interesting to postulate that Newton's law of gravity is literally correct. Then, we don't need a 'field theory' of gravity. Fields are spooky too. If one takes action at a distance seriously, and the law of inertia, some rather strange conclusions can be reached e.g. one particle can be in more than one place simultaneously.

At any rate, there is still a third option which doesn't involve the notion of action at a distance or the notion of 'field'... gravitons.

At any rate, I do have a personal desire to contemplate the notion of one particle being in more than one place simultaneously, but that is neither here nor there.

Kind regards,

StarThrower

StarThrower

Apr8-04, 09:26 AM

On what do you base this assertion? And don't say the Pythagorean theorem - I want emperical evidence, for example, give me just one (confirmed) test of the speed of light that shows a variation from C.

Give me one experiment that should have revealed a speed of light grossly different from c=299792458 m/s, which resulted in measurement c instead.

Kind regards,

StarThrower

Chen

Apr8-04, 09:38 AM

Ever heard of the Michelson-Morley experiment?

Severian596

Apr8-04, 09:46 AM

It is interesting to postulate that Newton's law [...] neither here nor there.

The reason I ask the initial question was because I wanted you to consider contemplating it. If you won't be swayed by emperical evidence, and you won't be swayed by accepted theory, and you sure won't be swayed by our input, put current theory on your list of things to contemplate. There's a reason we've progressed further than Newton.

Namely this is experimental evidence (which led to the development of theories which then sufficiently predicted and/or explained the data), but your most frustrating trait is you take the stance "I can think it, therefore it is." I believe the Physical sciences differ from Philosophy because physical sciences seek evidencial proof. "Common sense," and "what any logical person would conclude," come crashing down in the face of valid, correct experimental data.

StarThrower

Apr8-04, 09:52 AM

After all the time people have spent showing you that a reference frame moving at c is ill-defined in SR, I can't believe we're seeing another "what does motion look like from the photon's point of view?" thread.

So what??

The angle is stipulated to be 90 degrees in the lab frame. Even in Galilean relativity, the angle would not be 90 degrees in any other frame. In fact, it is always possible to transform to a frame in which both photons are moving along the same line! It's called the "center of momentum" frame.

Tom, first of all not a single physicist on this planet has proved to me that a reference frame traveling along with a photon isn't an inertial reference frame. The reason being of course, that it is (provided the photon isn't being subjected to a force).

That being said, the burden now falls squarely on the shoulders of the relativists to prove the impossible. Namely that a frame at which a photon is at rest, is non-inertial, in the case where the speed of this particular photon is constant in some IRF.

It's not that the reference frame is ill-defined in SR, that involves a bit of logico/deductive confusion on your part, or handwaiving, I'm not sure. Also, I am a bit confused as to why you wrote that in Galilean relativity it wouldn't be 90 degrees in any other frame. Perhaps you can elaborate on that?

Kind regards,

StarThrower

StarThrower

Apr8-04, 10:06 AM

Ever heard of the Michelson-Morley experiment?

Yes, indeed I have. The first thing to say, is when I did a derivation for the 'interferometer formula' using the S and S` frames found in a textbook of mine, if I do recall, I spotted an error. Yes I do recall that.

At any rate, if a photon must leave a source at speed c=299792458, regardless of what kind of reference frame the photon is in, then the Michelson Morely results are explained without using SR.

Regards,

StarThrower

Severian596

Apr8-04, 10:14 AM

At any rate, if a photon must leave a source at speed c=299792458, regardless of what kind of reference frame the photon is in, then the Michelson Morely results are explained without using SR.

SR is the theory that asserted 'a photon must leave its source at speed c regardless of its reference frame'. Michelson/Morely exhibited this behavior of the assertion experimentally. Therefore the Michelson/Morely results are explained by special relativity, not without using them.

The fact is you need to educate yourself on the topic. Period. Or you will never learn the answers to the questions you ask.

StarThrower

Apr8-04, 10:30 AM

Ok, let us put coordinates on this system in order to solve it. (Things are much clearer with coordinates, which is probably one of the reasons why anti-SR crackpots tend to dislike them)

In the rest frame of C, let us put the event where A, B, and C are all at the same place at the origin of the coordinate system. Let the coordinate axes x, y, and z be right, up, and out of the page respectively, and let t be the coordinate time.

The worldlines of A and B are given parametrically by (being sloppy and parametrizing by t):

A: (t, 0, 0.7ct, 0)
B: (t, 0.7ct, 0, 0)

Letting t', x', y', and z' be the coordinates in A's reference frame, the Lorentz transform from C's frame to A's frame gives the relations:

x' = x
y' = γ(y - 0.7ct)
z' = z
t' = γ(t - 0.7y/c)

where γ := 1/√(1-0.72) = 1/√0.51

So, the worldlines are given parametrically in A's frame by (yes, I mean to parametrize by t and not t'):

A: (γ(t - 0.7 (0.7ct)/c), 0, γ(0.7ct - 0.7ct), 0)
A: (γ 0.51 t, 0, 0, 0)
B: (γ(t - 0.7 0/c), 0.7ct, γ(0-0.7ct), 0)
B: (γt, 0.7ct, -γ0.7ct, 0)

(I only did A as a sanity check)

So, we have that the coordinate velocity of B in A's frame is given by
v = (0.7c √0.51, -0.7c, 0)

So, unless I've made a mistake along the way, B's speed relative to A is 0.7 c √1.51.

No, you didn't make a visible mistake along the way, and the reason you didn't is because the initial speed was taken as .7c. Let it be c, and redo the calculation.

Kind regards,

StarThrower

Here, I will start you off:

GOAL: Compute the relative speed of B and A.

As Hurkyl did, start off viewing the motion of A,B in the rest frame of C.

Let there be a clock at rest (anywhere in this frame, it doesn't matter), whose readings are denoted by the letter t. At the beginning of this event, the clock reads zero. As time goes foward, this clock's readings increase numerically. And let us presume that the clock ticks so fast, that whenever point A or point B changes coordinates in this frame, this clock has changed readings. Furthermore, let the tick rate of this clock be constant. This having been said...

At any moment in time t in this frame, the y coordinate of point A is ct on the y axis, x coordinate = 0, and z coordinate =0. Simultaneously in this frame, the x coordinate of B in this frame is ct, y coordinate =0, and z coordinate =0.

Or using coordinates so as to not be accused of being an anti-SR crackpot, we have:

A: (t, 0, ct, 0)
B: (t, ct, 0, 0)

Now, we are interested in viewing the motion of B in the rest frame of A. Let a coordinate system have been set up, whose origin is the point A. Let the coordinate axes x`, y`, and z` be right, up, and out of the page respectively, and let t` be the coordinate time. The motion of B is going to be diagonally down, and to the right, placing B in the fourth quadrant of this frame FA, once t`>0.

There is a clock at rest in this frame, whose rest rate matches that of the other clock. In fact, this clock is constructed identically to the other clock. When the event begins, this clock reads zero too. Hence, if t=0 then t`=0.

The Lorentz transformations from rest frame C, to rest frame A give:

x' = x
y' = γ(y - ct)
z' = z
t' = γ(t - y/c)

where γ := 1/√(1-1) = 1/0

Oops, I guess I can't use the Lorentz tranformations. Let me try the Galilean transformations instead... etc. etc.

Lets check out the location of B in FA, when t` = 1.

We have a clock at rest in frame A, which we shall call clock A. In the time it took clock A to go from zero, to 1, the point C moved downwards, onto the negative y` axis. This distance down it moved (in this rest frame FA), is given by ct`. Now, the y coordinate of A in reference frame C, at any moment in time according to clock C is ct, and the y` coordinate of C in reference frame A according to clock A is -ct`. Separation distance must be equivalent, hence we must have |-ct`| = |ct| which implies t=t`. In other words, the current coordinates of point C in frame A are: (1, 0, -c, 0), and clock C also reads 1.

Using the galilean tranformation we have t = t`, hence t = 1.

In this time, the point B has moved onto the positive x axis of frame C, at an x coordinate of ct or ct` (take your pick). This places point B in the fourth quadrant of the x`,y` plane, at a location with the following coordinates in frame A:

(t`,x`,y`,z`) = (1,c,c,0)

At this point in the argument c is in meters, because c = 299792458 m/s, but t` = one second in frame A= one second in frame C = t, hence the distance in meters is 299792458 meters.

The hypotenuse of this triangle can be computed using the pythagorean theorem. It corresponds to the distance travelled by point B in frame A, in the time it took clock A to go from zero to one. Using the pythagorean theorem we have:

D^2 = c^2 + c^2 = 2c^2

From which we infer that the distance traveled is:

D = (\sqrt{2}) c = (\sqrt{2}) (299792458) meters

Hence, the speed of B relative to A, in the rest frame of A is:

|\vec{V}| = (\sqrt{2}) (299792458) meters/second

Because it did so in one tick of clock A, and I recently stipulated that the unit of clock A is the second.

Let objects A,B be photons. Thus, the relative speed of these two photons exceeds the speed of light in the rest frame of A. Thus, if the fundamental postulate of the special theory of relativity is true, then FA is not an inertial reference frame. FC is an inertial reference frame by stipulation, and FA is moving relative to FC at a constant speed. Therefore, FA is an inertial reference frame. Therefore, FA is and isn't an inertial reference frame, which is impossible. Therefore, the fundamental postulate of SR is false.

Kind regards,

StarThrower

StarThrower

Apr8-04, 11:20 AM

SR is the theory that asserted 'a photon must leave its source at speed c regardless of its reference frame'. Michelson/Morely exhibited this behavior of the assertion experimentally. Therefore the Michelson/Morely results are explained by special relativity, not without using them.

The fact is you need to educate yourself on the topic. Period. Or you will never learn the answers to the questions you ask.

This is incorrect.

Proof:

(forthcoming)

Kind regards,

StarThrower

jdavel

Apr8-04, 11:32 AM

StarThrower said: "We are viewing the motion of A,B in the rest frame of C."

That's your problem!

With Galileo's relativity, you can calculate the relative speed of A and B in the rest frame of C or B or A and always get the same result, namely sqrt(2)*c

With Einstein's relativity, you can calculate the relative speed of A and B in the rest frame of A or B and always get the same result, namely c. But you'll get a different result in the rest frame of C, namely sqrt(2)*c.

But the postulate that says photons travel at c in all inertial frames means just that. Any measurement of a photon's speed made in an inertial frame, will always give the result c. In your setup there are two photons (A and B) moving at c in the rest frame of C, just like the postulate says there should be. If it were physically possible (but it's not) to get a measuring apparatus into the rest frame of B, then B would vanish (photons don't exist at rest) and you would measrue the speed of A as c.

By the way, I don't understand why you made the photons go at a right angle to eachother. Why not simplify the problem and have the atom emit them in opposite directions. That way, showing that each photon is moving through the other's frame at speed c (in other words that their relative speed is c) might be easy enough for you to understand. :wink:

jdavel

Apr8-04, 12:08 PM

StarThrower said: "There is a clock at rest in this frame, whose rest rate matches that of the other clock."

Relativity says this is impossible! You can't disprove a theory by DESCRIBING an experiment that would disprove the theory, if the theory says the experiment is impossible. You have to DO the experiment. In fact, it wouldn't even matter what the result of doing this experiment was. Just being able to do it would disprove relativity.

StarThrower

Apr8-04, 12:25 PM

StarThrower

Apr8-04, 12:37 PM

Chen

Apr8-04, 12:41 PM

You cannot divide by zero, therefore you cannot use the Lorentz transformations here.
You also cannot perform this in reality (something you cannot seem to grasp for some reason), so what seems to be the problem? Again, why should physical equational yield sensible answers for non-physical situations? Just to cater to your whim?

StarThrower

Apr8-04, 12:48 PM

You also cannot perform this in reality (something you cannot seem to grasp for some reason), so what seems to be the problem? Again, why should physical equational yield sensible answers for non-physical situations? Just to cater to your whim?

My whole point, is that you can perform this experiment in reality. You can fire two photons at right angles to one another simultaneously. According to the theory of relativity, these two photons cannot move relative to each other. That is impossible. Hence, SR contradicts.

Kind regards,

StarThrower

Chen

Apr8-04, 12:56 PM

My whole point, is that you can perform this experiment in reality. You can fire two photons at right angles to one another simultaneously. According to the theory of relativity, these two photons cannot move relative to each other. That is impossible. Hence, SR contradicts.
By saying this you assume that "a reference frame traveling along with a photon isn't an inertial reference frame". You have still not proven this. You told us you think it's right, you told us no one has proven otherwise, but you have never actually proved that it's correct. And until you do, nothing we can tell you will make any difference.

Now you can either frown upon us and say that "a reference frame traveling along with a photon is inertial until proven otherwise", in which case we can end the discussion right here, or you can actually attempt to prove this statement.

It's like I'd come to some mathematicians and tell them "I believe 1 + 1 = 3, and therefore (1 + 1) + (1 + 1) = 6 and your math contradicts itself."

StarThrower

Apr8-04, 01:01 PM

By saying this you assume that "a reference frame traveling along with a photon isn't an inertial reference frame". You have still not proven this. You told us you think it's right, you told us no one has proven otherwise, but you have never actually proved that it's correct. And until you do, nothing we can tell you will make any difference.

Now you can either frown upon us and say that "a reference frame traveling along with a photon is inertial until proven otherwise", in which case we can end the discussion right here, or you can actually attempt to prove this statement.

It's like I'd come to some mathematicians and tell them "I believe 1 + 1 = 3, and therefore (1 + 1) + (1 + 1) = 6 and your math contradicts itself."

You have what I am saying backwards. I am saying the following:

If a photon isn't experiencing a force then it is in an inertial reference frame.
From this it follows that if the speed of a photon is constant in some inertial reference frame, then the rest frame of the photon is an inertial reference frame.

By the fundamental ASSUMPTION of SR, the speed of a photon in any inertial reference frame is c>0.

From that assumption the following contradiction can be reached using perfect binary logic:

In inertial reference frame X, c=0 and not (c=0), where c denotes the speed of light in frame X.

Therefore, the fundamental postulate of SR must be false, by deduction.

Kind regards,

StarThrower

Chen

Apr8-04, 01:09 PM

You have what I am saying backwards. I am saying the following:

If a photon isn't experiencing a force then it is in an inertial reference frame.
From this it follows that if the speed of a photon is constant in some inertial reference frame, then the rest frame of the photon is an inertial reference frame.
That is not allowed by SR. You are basically saying this:

Let us assume something that contradicts a specific theory. Let us attempt to analyze the assumed situation. We reach a contradiction, therefore the specific theory is wrong.

StarThrower

Apr8-04, 01:12 PM

That is not allowed by SR. You are basically saying this:

Let us assume something that contradicts a specific theory. Let us attempt to analyze the assumed situation. We reach a contradiction, therefore the specific theory is wrong.

No, that's not what I am saying. I've repeatedly said what I am saying, but I can say it again, no harm done. I am saying this:

If the fundamental postulate of SR is true, then there is at least one statement X, such that (X and not X). Therefore, the fundamental postulate of SR is false.

The statement X being reached in this example problem is this: c denotes the speed of light in inertial reference frame F, and c=0 and not (c=0).

Kind regards,

StarThrower

Chen

Apr8-04, 01:17 PM

If the fundamental postulate of SR is true, then there is at least one statement X, such that (X and not X). Therefore, the fundamental postulate of SR is false.

The statement X being reached in this example problem is this: c denotes the speed of light in inertial reference frame F, and c=0 and not (c=0).
Except that X does not conform with the theory of SR. Which is what we have been trying to tell you for the past... what, two weeks now?

StarThrower

Apr8-04, 01:19 PM

Except that X does not conform with the theory of SR. Which is what we have been trying to tell you for the past... what, two weeks now?

You don't understand the reductio absurdum.

What do you mean by, the word 'conform' in your sentence, and I quote

"statement X does not conform with the theory of SR."

What's that mean?

Kind regards,

StarThrower

Severian596

Apr8-04, 01:27 PM

No, that's not what I am saying. I've repeatedly said what I am saying, but I can say it again, no harm done. I am saying this:

If the fundamental postulate of SR is true, then there is at least one statement X, such that (X and not X). Therefore, the fundamental postulate of SR is false.

The statement X being reached in this example problem is this: c denotes the speed of light in inertial reference frame F, and c=0 and not (c=0).

Kind regards,

StarThrower

Let's cut through the rather poor wording. You should not begin this kind of argument by saying "if something is true, then that something is false." Instead:

(1) Special Relativity depends on the condition that the speed of light through a vacuum c is constant (remains unchanged) in any inertial frame of reference.
(2) There is at least one inertial frame of reference where the speed of light through a vacuum does NOT equal c.
(3) Therefore, based on Special Relativity's dependence on condition (1) and the existence of condition (2), Special Relativity is not consistent.

Now all you have to do (as was stated before) is prove that condition (2) does actually exist. If it does, it makes Special Relativity inconsistent. If it does not exist then Special Relativity is not proven inconsistent.

Now...how do you plan on [b]proving condition (2)?

Severian596

Apr8-04, 01:29 PM

In essence, StarThrower, you have constructed a theory. How do you plan on proving your theory?

..:: EDIT ::..

Thus far you have only been able to illustrate that if your theory is true SR is inconsistent. Perhaps we should just end this discussion and conclude that your theory, if it is ever proven, could bring the downfall of SR?

jdavel

Apr8-04, 01:37 PM

StarThrower said: "You cannot divide by zero, therefore you cannot use the Lorentz transformations to make a coordinate transformation here."

Do you have to divide by zero to find the derivative of a function? No, you find it by calculating the limit of a ratio as both the numerator and denominator approach zero.

So try this? Start with two particles traveling at speed v in the rest frame of C, and calculate the speed of A in the rest frame of B. That's the relative speed of A and B, right?

Now see if that relative speed has a finite limit as v approaches c.

It will, and it will be c.

Severian596

Apr8-04, 01:40 PM

I guess I should also say that you should be prepared for the consequences of your theory, too. You'll receive a lot of fame and attention if your theory proves true, but much of the fame and attention will be negative. People will expect you to present a lot of proof that your theory is true.

And I doubt you'll be able to wow them with, "Okay. Suppose the following scenario....[insert the scenario here]...SEE! Bam! Inconsistent."

*StarThrower dusts his hands and is quite pleased with himself*

Tom Mattson

Apr8-04, 02:22 PM

Tom, first of all not a single physicist on this planet has proved to me that a reference frame traveling along with a photon isn't an inertial reference frame. The reason being of course, that it is (provided the photon isn't being subjected to a force).

It has already been shown that motion is ill-defined if we take the speed of light to be the same in every frame. If you assume that, then it is not possible to put the a photon at the origin of a frame that is at rest. Jeez, instead of waiting for a physicist to prove it for you, just work it out yourself. It's not that hard.

That being said, the burden now falls squarely on the shoulders of the relativists to prove the impossible. Namely that a frame at which a photon is at rest, is non-inertial, in the case where the speed of this particular photon is constant in some IRF.

There is no frame in which a photon is at rest.

What you are doing here is no different that asking me to prove that all mermaids have tails. Since the class of mermaids is empty, there is nothing for me to prove.

Similarly, the class of reference frames in which the photon is at rest is empty, by assumption.

It's not that the reference frame is ill-defined in SR, that involves a bit of logico/deductive confusion on your part, or handwaiving, I'm not sure.

Actually, it is ill defined in SR. Starting from any frame that is moving with a speed that is less than light, a Lorentz boost to v=c results in the Lorentz factor g to having a zero denominator, which makes it *drumroll, please* undefined.

Come on, man, do a little work for yourself. Your insistence on having others prove everything for you when it is so easy to derive gives me the impression that you are here only to stir things up, rather than having any genuine interest in learning.

Also, I am a bit confused as to why you wrote that in Galilean relativity it wouldn't be 90 degrees in any other frame. Perhaps you can elaborate on that?

Again: Try doing a little work for yourself. It won't kill you.

Assume Galilean relativity. Let S be stationary and S' be moving with a velocity vi with respect to S (v>0). Let two objects, U1 and U2, be moving at the same speed, but right angles in S. Give them velocities u1=ui and u2uj, respectively.

Now transform them to frame S' according to Galilean relativity:

u1'=u1-v
u2'=u2-v

Substituting the velocities given above yields:

u1'=(u-v)i
u2'=-vi+uj

Take the dot product:

u1.u2=v2-uv,

which in general does not vanish.

Kind regards,

Right back atcha.

Severian596

Apr8-04, 02:47 PM

FINALLY, to sum things up, look at it this way:

Postulates of the Theory Special Relativity (SR):

(1) All laws of physics apply uniformly to all inertial frames of reference
(2) All inertial frames of reference measure the speed of light in a vacuum c to be the same value

Now consider StarThrower's Theory (S) that depends on the following conditions:
(S1) A photon is in an inertial reference frame
(S2) A photon measures c differently than at least one other inertial frame of reference

StarThrower's Theory asserts the following:
((S1) \bigwedge (S2)) \rightarrow (S) \rightarrow (\sim 2) \rightarrow (\sim SR)

So there you go!! All you have to prove is (S1) \bigwedge (S2) and you've got the theory of SR right where you want it.

jdavel

Apr8-04, 02:48 PM

StarThrower said: "By the fundamental ASSUMPTION of SR, the speed of a photon in any inertial reference frame is c>0."

Suppose Einstein had said this:

For any inertial reference frame in which photons in empty space and matter can be observed, all the photons will be observed to be moving at a speed c, and all the matter will be observed to be moving at speeds less than c. Since all observers are made of matter, nothing can be observed in a frame of reference where only photons might (or might not) exist. Thus, such a frame of reference (if it exists) is irrelevant to all of physics, and is not considered in this theory.

That wording would lead just as inevitably to the Lorentz transforms and the other predictions of SR. But isn't Einstein's wording more elegant? And really, everyone who understands the theory knows what he meant.

ramcg1

Apr8-04, 02:58 PM

"Thus a logical person would deduce that the relative velocity of the two photons were square root of 2 times 299792458 meters per second...."

Logical? Not if they were aware of SR.

"SR is irrelevent to the calculation as you are not trying to observe a photon travelling between the two."

Not true. SR is relevant to any calculation involving motion.

You only believe this because you were taught this.

It does not make it true.

Try this thought:

A photon arriving at an atom can either join that atom or continue on its journey. At each atom there is an interaction. Each time the photon moves on it moves on with the velocity c with respect to the atom it just interacted with. If that atom was travelling at a different velocity to the previous atom then SR is applied to its frequency and wavelength (it may also have to adjust its amplitude).

SR and GR are only descriptions of these interactions not the macro universe.

Have a good easter.

Tom Mattson

Apr8-04, 03:01 PM

A photon arriving at an atom can either join that atom or continue on its journey. At each atom there is an interaction. Each time the photn moves on it moves on with the velocity c with respect to the atom it just interacted with. If that atom was travelling at a different velocity to the previous atom then SR is applied to its frequency and wavelength (it may also have to adjust its amplitude).

SR and GR are only descriptions of these interactions not the macro universe.

???

The paragraph above does not in any way imply the conclusion.

jdavel

Apr8-04, 03:50 PM

StarThrower, how about this?

Define: io to be a frame of reference "in which the equations of mechanics hold good"

Let I be the set of io and all other frames of reference in (n>0) that move at constant speed with respect to io

Let C be the subset of I in which all photons travel at c

Let V be the subset of I in which at least one photon travels at a speed v, not equal to c.

Let P be the subset of I which are possible in the physical world

Would you agree that I is the union of C and V?

Would you agree that the intersection of C and V is the null set?

If so, can you use this formalism to explain what you think the postulate of relativity means?

I think it means this: (C is not the null set) AND (P=C)

Chen

Apr8-04, 04:29 PM

You don't understand the reductio absurdum.

What do you mean by, the word 'conform' in your sentence, and I quote

"statement X does not conform with the theory of SR."

What's that mean?
I'll make it easier to understand.

Let's say you have a theory which includes two statements, X and Y. These statements must be true for the theory to be correct. Now let's take a statement Z and prove that it is self-contradictory. What does it have to do with the theory at all? How exactly does it prove the theory to be wrong?!

Hurkyl

Apr8-04, 06:17 PM

By the way, I don't understand why you made the photons go at a right angle to eachother. Why not simplify the problem and have the atom emit them in opposite directions

The simplest version of his argument would go as follows:

"Consider the rest frame of the photon. The photon isn't moving at c in this frame, which is a contradiction!"

However, I would suppose that this doesn't sufficiently obfuscate his conceptual error, so he has to go and complicate the scenario in order for him to ignore the blatantly obvious.

jdavel

Apr8-04, 06:33 PM

However, I would suppose that this doesn't sufficiently obfuscate his conceptual error, so he has to go and complicate the scenario in order for him to ignore the blatantly obvious.

Sort of like the "Triplet Paradox" used by many of StarThrower's ilk!

This thread has become more about human nature than about physics. Who can hold out longer, a mob of arguers with the certainy based on 99 years of experiments and 99+% of the world's physics community agreeing with us or one anonymous player in cyber space with who knows what motives?

Anyway I give up, he's all yours. I come here to share what I think I know, and to learn from what the rest of you know. StarThrower doesn't want the former, and he hasn't got any of the latter!

MindWarrior

Apr8-04, 08:00 PM

The simplest version of his argument would go as follows:

"Consider the rest frame of the photon. The photon isn't moving at c in this frame, which is a contradiction!"

However, I would suppose that this doesn't sufficiently obfuscate his conceptual error, so he has to go and complicate the scenario in order for him to ignore the blatantly obvious.

If you look back at post 36, you will see something interesting.

At any moment in spacetime when t>0, the y coordinate of photon A is ct.
At any moment in spactime when t`>0, the y` coordinate of photon gun C is -ct`. Hence the following equations are true:

y = ct & y` = -ct`

From which it follows that

y/t = c & c = -y`/t`

From which it follows that:

y` = -y \frac{\t^\prime}{t}

The above transformation is correct. If we use the Lorentz transformation, we will have division by zero error.

Thus, starthrowers analysis is correct. The theory of special relativity self-contradicts.

If we finish off his analysis, the contradiction becomes clear. Photon B is moving through an inertial reference frame at a speed greater than c=299792458.

His argument begins as follows:

In the rest frame of C, the Pythagorean theorem gives the distance between photon A and photon B as a function of the spacetime coordinate t. We have:

H^2 = (ct)^2 + (ct)^2 = 2 (ct)^2

From which it follows that they hypotenuse (which is the distance between the photons) as a function of spacetime coordinate t is:

H(t) = \sqrt{2} ct

We can thus write H(t`) using the transformation above. H(t`) is the distance that photon B is from photon A, in terms of spacetime coordinate t`. The analysis will show that in the rest frame of photon A, the speed of photon B exceeds the speed of light. The rest frame of photon A is an inertial reference frame, since it is moving in a straight line at a constant speed in the rest frame of C, which is also inertial. Thus, a photon can move in an inertial reference frame at a speed greater than c.

Kind regards,

Star

Hurkyl

Apr8-04, 08:16 PM

Thus, starthrowers analysis is correct.

...

Kind regards,

Star

You know, if you're gonna pretend to be someone else, you should probably change your sig.

russ_watters

Apr8-04, 08:18 PM

You know, if you're gonna pretend to be someone else, you should probably change your sig.Lol, hang a left and enter the surreal.

Severian596

Apr8-04, 11:56 PM

ROTFL

I am not Severian, I am his evil twin. And I am here to tell you this discussion has degenerated into people pretending to be someone else so people will listen to...people.

HAHHAA!

outandbeyond2004

Apr9-04, 01:08 AM

I am reminded of Bill Clinton's asking for a definiton of the verb 'to be,' (or something else?) in a legal deposition somehow.

jdavel

Apr9-04, 10:44 AM

I am reminded of Bill Clinton's asking for a definiton of the verb 'to be,' (or something else?) in a legal deposition somehow.

outandbeyond,

Oh, it was even better than that! He said, "That depends on what your definition of the word "is" is."

And with that, I suggest that it's time for this thread to be locked!

ramcg1

Apr9-04, 02:13 PM

???

The paragraph above does not in any way imply the conclusion.

" A photon arriving at an atom can either join that atom or continue on its journey. At each atom there is an interaction. Each time the photon moves on it moves on with the velocity c with respect to the atom it just interacted with. If that atom was travelling at a different velocity to the previous atom then SR is applied to its frequency and wavelength (it may also have to adjust its amplitude).

SR and GR are only descriptions of these interactions not the macro universe."

It does if you step out of relativistic thinking and allow the velocity of the photon to change so that it has a new velocity with repect to the previous interaction.

Severian596

Apr9-04, 02:22 PM

It does if you step out of relativistic thinking and allow the velocity of the photon to change so that it has a new velocity with repect to the previous interaction.

Okay I'm having trouble with a couple points about this theory but I'll put them in your thread here: http://physicsforums.com/showthread.php?t=17809

see ya there...

russ_watters

Apr9-04, 04:01 PM

ROTFL

I am not Severian... Ehh, it doesn't matter anyway: this board is just a text-based video game and all of you are just simulations generated by the PF server for my amusement.

Prove me wrong. :tongue:

dgorman

Apr10-04, 01:29 AM

StarThrower,

You are a fool. And for the record, ten years at a community college don't qualify for "[studying] at a fine northeastern university."

You've attempted to apply philosophy to physics. And you've (no surprise here) done this incorrectly. Much as it bores me to respond to crackpots, I shall (for the benefit of the poor unfortunates who might possibly be persuaded by your drivel) debunk your logic.

In order to prove a system internally inconsistent, one must assume its postulates and arrive, preferably through formal logic, at a contradictory conclusion. You, however, must begin by asserting a proposition not defined by Special Relativity--meaning that you are attempting to prove SR inconsistent outside the system. That doesn't work.

For instance:

Given that (a), "God does not exist," (b) "Only God can create humans," and (c) "Humans exist," prove that the system is internally inconsistent.

One might begin thus:

1. a (given)
2. c & b
3. c -> ~a
QED.

The point is, ST, you must begin from the rules of the system. You haven't done so, and it appears that you cannot do so. Case in point:

You propose to illustrate that SR is internally inconsistent. However, you begin by supposing an intertial frame of reference in which a photon is at rest. You can't do that--the system of SR has no concept of such a reference frame.

You haven't "proven" anything.

Therefore you are a crackpot. QED.

Eyesaw

Apr10-04, 04:57 AM

If the speed of light is source independent then it is c relative to any observer moving at any velocity as long as the observers all referred to the vacuum frame to measure the flight path of the light. If inertial coordinates not at rest with respect to the vacuum
are used however, the speed of the light must be c + v and c-v with respect to the origin of the inertial observer because the space between the inertial observer and the photon becomes "mixed". This results in anisotropic effects inside different inertial frames because the detector of light from one direction inside an inertial frame would receive it with less energy than light from the opposite direction. So there should be different Doppler shifts of single photons inside an inertial frame, if their speeds are source independent. I.e., Einstein's two postulates contradict each other.

Only if light speed is source dependent can physics be the same inside different inertial frames. And I really think this is the case because waitresses in cafes, servers inside train cabins
moving at 52 miles per hour, and stewardesses on a 747 flying at
534 miles per hour can pour a cup of coffee right into a cup without
spilling a drop so it can't be all that difficult for light to move with a velocity dependent of its source.

Inertial coordinates are a system of coordinates, which can be represented by particles moving at a constant velocity through space, that are at rest with respect to each other. Physical interactions between these particles then should reveal the fact that they are at rest with respect to each other, else Nature's definition of the state of rest between particles would become more complicated. E.g., imagine a source and detector both moving at constant velocity represented by V or J or M or N or S or C or G or L or P or Q or any other letter of the alphabet. If light is independent of source velocity, the detector would think the source is moving relative to it even though they are at rest with respect to one another in the same inertial frame moving relative to a stationary vacuum, leading to an infinite number of definitions for rest. This would be unsatisfying on a philosophical level at least. And as mentioned previously, every day observations (events inside a car, a plane, or on an Earth moving at 30 km/sec through space - we should fall of the earth when we take a step
if we did not have the same motion as the Earth already) verify source dependence of events for particles, and since waves are the collective motion of particles in space- if they are source dependent then so are the waves they produce. So whether depicted as a wave or as a particle, the speed of photons should depend on the velocity of the source frame, i.e., it should be c+v and c-v in the two directions with respect to the vacuum.

Hurkyl

Apr10-04, 07:46 AM

as long as the observers all referred to the vacuum frame

How, pray tell, does one tell if they are in "the" vacuum frame? And why must there only be one?

I.e., Einstein's two postulates contradict each other.

I don't see how doppler shift causes a contradiction between the two postulates...

Only if light speed is source dependent can physics be the same inside different inertial frames.

You have it backwards. The only theory of light we have says that light speed is not source dependant. Thus, physics can only be the same inside different inertial frames if the speed of light is not source dependant.

You'll have a point once you can put forth a source dependant theory of electromagnetism that is consistent with experiment.

If light is independent of source velocity, the detector would think the source is moving relative to it

How so?

Tom Mattson

Apr10-04, 11:33 AM

" A photon arriving at an atom can either join that atom or continue on its journey. At each atom there is an interaction. Each time the photon moves on it moves on with the velocity c with respect to the atom it just interacted with. If that atom was travelling at a different velocity to the previous atom then SR is applied to its frequency and wavelength (it may also have to adjust its amplitude).

SR and GR are only descriptions of these interactions not the macro universe."

Tom: ???

The paragraph above does not in any way imply the conclusion.

ramcg1: It does if you step out of relativistic thinking and allow the velocity of the photon to change so that it has a new velocity with repect to the previous interaction.

No, it doesn't follow. I'm not talking relativity, I'm talking logic. The paragraph does not contain a single premise regarding "the macro universe", and yet somehow it appears in your conclusion.

OneEye

Apr12-04, 10:38 PM

Prove that the relative speed of the two photons is c if you can.

Regards,

StarThrower

P.S.

By Euclidean geometry, if the relative speed of the photons is c, then the two photons form an equilateral triangle with the atom, whose area is increasing in time. The interior angles of an equilateral triangle are 60 degrees. However, the emission angle was stipulated to be 90 degrees. This is a contradiction.

How can someone have such a math education and be so confused at the same time?

If two dogs leave the igloo at the same time, and travel at the same speed at right angles to each other then (according to the above logic), the dogs will form an equilateral triangle with the igloo, and thus will not be at right angles to each other!

Gadzooks! Geometry is broken! Pythagoras is busted! I and my anti-twin will immediately converge and annihilate each other, generating dark energy and forcing the galaxies apart!

The figure described by the igloo-and-dogs (or the atom and photons) is an isosceles triangle (not an equilateral) with sides ct, ct, and \sqrt { 2 } ct.

Let's make it a point to check this guy's math in the future, okay?

P.S. Someone else may already have made this comment. I confess to having been impatient enough not to read the entire thread.

outandbeyond2004

Apr12-04, 11:49 PM

1. StarThrower is diddling us.
2. He is very careless in thinking.
3. He is congenitally unable/averse to think logically.
4. His prejudicies would take enormous effort to eradicate.
5. Some other explanation may be possible.

I don't know which of these explanations are true. However I am now going to guess that this thread is not worth perusing any more. I did gain some insights, but they came at some cost, and I feel that any more, if any, insights would not be worthwhile.
The thread did provoke some thinking (maybe good mental exercise), also, but . . .

I will lurk a little while to see what response, if any, comes to this post.

geistkiesel

Jun6-04, 10:00 PM

An atom is at rest in an inertial reference frame. Suddenly, two photons are emitted from it, at right angles to one another.

What is the relative speed of the photons?

P.S. Assume they depart from the atom at 299792458 meters per second in the atomic frame, as has been found true by experiment.
The obvious 2c.

SR theory predicts that the relative velocity will be c. The reason for this prediciton is derived from the postulates of SR. AS you have described the experiment, a pure observation, so to speak, you have related the unambiguous experimental result that the photons are moving at the relative velocity of 2c. Any measuement of the velocity of any one photon must show an escape velocity of c with respect to the immobile atom of the photon's birth. To mesure otherwise would contradict the experimental results.
Whatever is measured of the moving photons, The elative velocity of the photons is determined by adding the absolute value of their respective speeds (we let c = 1) to get 1^2 + 1^2 = 2. The relative part we take from the vector directions.

The SR theorists must also respond that the photons were not emitted simultaneously as measured from any moving frame. So a frame very near the emitting atom moving perpendicular (or any other angle) to the subsequently emitted photons calculates the emited photons were not simultaneously emitted, and this even if the observer on the faster frame also measures the velocity of each photon as c with respect to the host emitting atom.
SR says to revise the experimental results to conform with theory. Obedience, faith and professional security considerations sum to determine the 'real to postulated transition scenario' equates to the most expedient, the wisest option among a host of options, "shucks, its even obvious", certainly the best of all possible worlds...........

We are here now where all experimental results measures a non-SR value. Alongs come SR to demonstrate that the measured experimental value, unambiguously determined, or measured, axiomatically demonstrates the relative velocity of 2c, must now be revised to be c. A demonstrated real 2c is equated to a theoretical postulated value of c. 2c = c. 2 =1. Real unergoes a transformation to a theoretically postulated variant value, where "postulated" means "formed in someone's mind".

Real is transformed to the postualtes of a dead man, Albert Einstein, graduate advisor to the host of current Special Relativity Theorists.

Does this forum have a "surrender at" window? It should.
Or is the surrender window built into the thread?

Tom Mattson

Jun6-04, 10:17 PM

The obvious 2c.

Silly wabbit, it wouldn't be 2c even if Galileo's relativity were correct. They're traveling at right angles to each other, not in antiparallel paths.

SR theory predicts that the relative velocity will be c.

No, it doesn't. StarThrower's whole point was that you could make predictions from the vantage point of the rest frame of the photon. In SR, such a frame is undefined. When he goes ahead and tries to transform to a frame moving at v=c to show a contradiction in SR, he uses the LT at a point outside its domain of definition.

The reason for this prediciton is derived from the postulates of SR.

No, it's derived from your misunderstanding of those postulates.

AS you have described the experiment, a pure observation, so to speak, you have related the unambiguous experimental result that the photons are moving at the relative velocity of 2c.

Any measuement of the velocity of any one photon must show an escape velocity of c with respect to the immobile atom of the photon's birth. To mesure otherwise would contradict the experimental results.

Does any relativity crackpot know the difference between a thought experiment and a real experiment?

Whatever is measured of the moving photons, The elative velocity of the photons is determined by adding the absolute value of their respective speeds (we let c = 1) to get 1^2 + 1^2 = 2. The relative part we take from the vector directions.

No, it isn't. Even in Galilean relativity, the relative speed is sqrt(2)c, not 2c.

The SR theorists must also respond that the photons were not emitted simultaneously as measured from any moving frame. So a frame very near the emitting atom moving perpendicular (or any other angle) to the subsequently emitted photons calculates the emited photons were not simultaneously emitted, and this even if the observer on the faster frame also measures the velocity of each photon as c with respect to the host emitting atom.

Yes, so what?

SR says to revise the experimental results to conform with theory. Obedience, faith and professional security considerations sum to determine the 'real to postulated transition scenario' equates to the most expedient, the wisest option among a host of options, "shucks, its even obvious", certainly the best of all possible worlds...........

We are here now where all experimental results measures a non-SR value. Alongs come SR to demonstrate that the measured experimental value, unambiguously determined, or measured, axiomatically demonstrates the relative velocity of 2c, must now be revised to be c. A demonstrated real 2c is equated to a theoretical postulated value of c. 2c = c. 2 =1. Real unergoes a transformation to a theoretically postulated variant value, where "postulated" means "formed in someone's mind".

Real is transformed to the postualtes of a dead man, Albert Einstein, graduate advisor to the host of current Special Relativity Theorists.

Sorry, I missed the part where real data was taken in a real experiment. Could you give me the reference please?

Does this forum have a "surrender at" window? It should.
Or is the surrender window built into the thread?

No, we don't have that, but we'll have one made, just so you can wave your white flag. :rofl: :rofl:

geistkiesel

Jun6-04, 10:44 PM

Undefined, since, as we've pointed out to you before, photons don't have rest frames.

Allow me to remind you that the difference of two velocity vectors in a given reference frame is not, in general, give you the relative velocity of the said objects in SR.

According to GR, the photon is travelling in a straight line, period. And, of course, photons never have inertial rest frames.

Whatever you say about rest frames is to drag SR thepry into a description of a physical event. The photons are moving from their host atom at the velocity as measured from the immobile, unambiguously located atom. The experimental results told you that the photons were moving the velocity c with respect to the host atom. Or pick some other point in the uncerse, what would you determine he velocity of c to be? I bet you will always determine c. This is fundamental to SR, tis the basic postulate. Don't forget he invariance of physical law in all inertial frames.

All you SR theorists, yopu grab this and that from here and thee and say "See, reality isn't real! Re;lity is whja is lurking around in the minds of SR therists". If "postuilate" is a mere mental construct, then the mind of the SRists takes pecedence over the noncurrupted reality of direct and unamiguous observation. Wow, thisnk of the concept.

Think also of the conscious and unconcious considerations of 'professional security matters of SRists iintuiting the inevitable collapse of their precious theory' . . . .

Tom Mattson

Jun7-04, 03:56 AM

Whatever you say about rest frames is to drag SR thepry into a description of a physical event. The photons are moving from their host atom at the velocity as measured from the immobile, unambiguously located atom.

That is no problem at all.

The experimental results told you that the photons were moving the velocity c with respect to the host atom. Or pick some other point in the uncerse, what would you determine he velocity of c to be? I bet you will always determine c. This is fundamental to SR, tis the basic postulate. Don't forget he invariance of physical law in all inertial frames.

Don't worry, we won't forget it.

All you SR theorists, yopu grab this and that from here and thee and say "See, reality isn't real! Re;lity is whja is lurking around in the minds of SR therists". If "postuilate" is a mere mental construct, then the mind of the SRists takes pecedence over the noncurrupted reality of direct and unamiguous observation. Wow, thisnk of the concept.

Think also of the conscious and unconcious considerations of 'professional security matters of SRists iintuiting the inevitable collapse of their precious theory' . . . .

You are seriously disturbed, if you think that.

You think we say, "Reality isn't real"????

We are the ones who refer to real experiments, while your crackpot kind refers to imagined experiments.

If that nonsensical rant is what you think, then you are hopeless.

edit: fixed bold font bracket

russ_watters

Jun7-04, 07:34 AM

All you SR theorists, yopu grab this and that from here and thee and say "See, reality isn't real! Re;lity is whja is lurking around in the minds of SR therists". If "postuilate" is a mere mental construct, then the mind of the SRists takes pecedence over the noncurrupted reality of direct and unamiguous observation. The trouble is quite simply that you can't accept reality at face value, probably because you just don't like it. "What is lurking around in the minds of SR theorists" has been proven beyond reasonable (even some unreasonable) doubt. It is reality.

geistkiesel

Jun7-04, 08:40 AM

That is no problem at all.

Don't worry, we won't forget it.

You are seriously disturbed, if you think that.

You think we say, "Reality isn't real"????

We are the ones who refer to real experiments, while your crackpot kind refers to imagined experiments.

If that nonsensical rant is what you think, then you are hopeless.

edit: fixed bold font bracket
What imagined experiments are you refering, Tom, to the imagined experiment described by Einstein in his book, "Relativity" page 25-27? The one he used to justify discarding the concept of simultaneity? This is the same one you memorized, how long ago was it anyway?

Yes, indeed, you say reality is not real, it must be mediated by the insanity of relativity theory. For isntance, in David's hypothetical of photon moving opposite each other from the host atom at the speed of c, that because the photons are moving at the speed of c, because that is what we measured the velocities to be, then the photons are moving away from each other, also at the speed of c.

You invoke a theory to corrupt reality, a memorized reality in the case of most SR theorists. You might say something about light not having a frame of reference, very well.

At what velocity, Tom does the radius of an expanding electromagnetic sphere travel in a stationary frame? It wouldn't do damage to the physics to have a photon pulse length giving say 10^6 photon thick as the thickness of the sphere would it?

The expanding sphere would still be a sphere, all the photons faithfully moving outward from the center of the sphere not all that unlike David's two photons moving outward from the central atom at the speed c, correct? Of course it is correct.

Our expanding sphere reaches the diameter of 2tc after t seconds, correct?
If we measure the diameter of the sphere in the stationary frame after 1 unit of time we can draw a circle on the floor of the stationary platform the exact diameter of our expanding sphere projected on to the floor of the stationary platform.

We do this a million time for the value of t = 1.

Next all the scientists gather together and ponder the question: If the radius is expanding at the rate of c and after 1 time unit the sphere diameter is 2c then how fast have we measured the relative velocities of photons on opposite sides of the expanding sphere to be?

After a few experimental runs we place clocks at a number of points around the circumference of the projected sphere, all at equal diameters, of course. Everytime we measure the diameter of the sphere we get the same number for a unit time t =1. When we compare the arrival times of points along the sphere directly opposite each other we get an expansion rate of 2c, correct? Is there any ambuguity here? I mean a measured expansion rate of oppositiely moving photons being 2c, a measured 2c, that is. No ambiguity, correct?

When we compare clock time of photons along a 1/2 circumference line of the sphere we measure the expansion rate of adjacent photons to be proportional to the angle theta between radius segments connecting the photons, correct? The expansion rate goes from 0 to 2c as the angle theta goes from 0 to pi, correct? All measured relative velocities, correct? Both Measured and calculated, correct?. At an angle of pi/2 the expansion rate would be what? I guess c, or half the rate of photons moving directly opposite each other.

But you say, David's photons don't behave in this way. Why was that again, just one more time?

It could be that Davids's photons just weren't "up to speed" so to speak. I would check those photons out if I were him, this would be really big news, if you know what I mean, his photons moving at different speeds than your photons and all?

It would be 2c, wouldn't it Tom. I mean 2c as measured in the stationary frame, but only c as modified by SR theory, right?

You might go over tht reality thing again Tom. You had a lot of "!!!", excuse me, I stand corrected, a lot of "???" to ponder.

By the way, I browzed through the relativity forum and noticed there wasn't the energy and intensity of discussions as we have here, as a result many of the discussions were trivial echoes of dogma, with few, if any spirirted challenges. No one wants to be exiled to the theory development forum, for shame, should that occur. A lot of pomposity and smugness down there though, an awful lot.

Tom Mattson

Jun7-04, 01:28 PM

What imagined experiments are you refering, Tom, to the imagined experiment described by Einstein in his book, "Relativity" page 25-27? The one he used to justify discarding the concept of simultaneity? This is the same one you memorized, how long ago was it anyway?

I was referring to the same imagined experiment that you were referring to: StarThrower's.

Yes, indeed, you say reality is not real, it must be mediated by the insanity of relativity theory. For isntance, in David's hypothetical of photon moving opposite each other from the host atom at the speed of c, that because the photons are moving at the speed of c, because that is what we measured the velocities to be, then the photons are moving away from each other, also at the speed of c.

Where did I say, "reality isn't real?"

You invoke a theory to corrupt reality, a memorized reality in the case of most SR theorists. You might say something about light not having a frame of reference, very well.

There's nothing corrupt about it: SR makes no predictions from the vantage point of a frame moving at c.

At what velocity, Tom does the radius of an expanding electromagnetic sphere travel in a stationary frame?

It travels at the speed of light.

It wouldn't do damage to the physics to have a photon pulse length giving say 10^6 photon thick as the thickness of the sphere would it?

The expanding sphere would still be a sphere, all the photons faithfully moving outward from the center of the sphere not all that unlike David's two photons moving outward from the central atom at the speed c, correct? Of course it is correct.

Our expanding sphere reaches the diameter of 2tc after t seconds, correct?
If we measure the diameter of the sphere in the stationary frame after 1 unit of time we can draw a circle on the floor of the stationary platform the exact diameter of our expanding sphere projected on to the floor of the stationary platform.

We do this a million time for the value of t = 1.

Next all the scientists gather together and ponder the question: If the radius is expanding at the rate of c and after 1 time unit the sphere diameter is 2c then how fast have we measured the relative velocities of photons on opposite sides of the expanding sphere to be?

After a few experimental runs we place clocks at a number of points around the circumference of the projected sphere, all at equal diameters, of course. Everytime we measure the diameter of the sphere we get the same number for a unit time t =1. When we compare the arrival times of points along the sphere directly opposite each other we get an expansion rate of 2c, correct? Is there any ambuguity here? I mean a measured expansion rate of oppositiely moving photons being 2c, a measured 2c, that is. No ambiguity, correct?

Small correction: The rate of separation between the two points is sqrt(2)c, not 2c. I said that before.

When we compare clock time of photons along a 1/2 circumference line of the sphere we measure the expansion rate of adjacent photons to be proportional to the angle theta between radius segments connecting the photons, correct? The expansion rate goes from 0 to 2c as the angle theta goes from 0 to pi, correct? All measured relative velocities, correct? Both Measured and calculated, correct?. At an angle of pi/2 the expansion rate would be what? I guess c, or half the rate of photons moving directly opposite each other.

OK

But you say, David's photons don't behave in this way. Why was that again, just one more time?

No, I did not say that. Your confusion arises from the fact that you think that the rate of separation of two photons in one frame is the speed of one photon relative to another. It isn't.

It could be that Davids's photons just weren't "up to speed" so to speak. I would check those photons out if I were him, this would be really big news, if you know what I mean, his photons moving at different speeds than your photons and all?

Since David doesn't have any real photons, it's all pretty academic, now isn't it?

edit: The imaginary photons are StarThrower's, not David's.

It would be 2c, wouldn't it Tom. I mean 2c as measured in the stationary frame, but only c as modified by SR theory, right?

I already told you that it would be sqrt(2)c. That is the prediction of both Galilean and Special Relativity. Relativistic effects don't occur until you start comparing data from different frames.

geistkiesel

Jun8-04, 01:29 AM

Silly wabbit, it wouldn't be 2c even if Galileo's relativity were correct. They're traveling at right angles to each other, not in antiparallel paths.

No, it doesn't. StarThrower's whole point was that you could make predictions from the vantage point of the rest frame of the photon. In SR, such a frame is undefined. When he goes ahead and tries to transform to a frame moving at v=c to show a contradiction in SR, he uses the LT at a point outside its domain of definition.

No, it's derived from your misunderstanding of those postulates.

Does any relativity crackpot know the difference between a thought experiment and a real experiment?

No, it isn't. Even in Galilean relativity, the relative speed is sqrt(2)c, not 2c.

Yes, so what?

Sorry, I missed the part where real data was taken in a real experiment. Could you give me the reference please?

No, we don't have that, but we'll have one made, just so you can wave your white flag. :rofl: :rofl:
M<y mistake. I saw something elose. I retract th post that is the subject of this post.

So sorry.
:cry:

geistkiesel

Jun8-04, 05:07 AM

geistkiesel

Jun8-04, 05:54 AM

Tom, In my expanding EM sphere with a radius expanding at c then we must have a diameter of 2c after 1 unit of time would we not? Or, say just measure ona radius leg and draw in the projected circle, it is all so simple. Show me how the sphere gets from a 2c diameter sphere with the radius expanding at C, to less than 2c, I think you said the (2)^1/2c. If ever there was a contradcition in terms does not it sound like a contradiction in terms? A c radius sphere after 1 second turns into a (2)^1/2c diameter sphere.

Where is there to look to get a straight answer on this? I apologize, I just don't believe it. Convince me or something. I am pliable, in my new found humilty after my carelessness in that post to you.
Show me., I'll listen.

Tom Mattson

Jun8-04, 11:40 AM

Tom, In my expanding EM sphere with a radius expanding at c then we must have a diameter of 2c after 1 unit of time would we not?

Looks like I've made a mistake, too. I thought you were still talking about points moving in perpendicular directions, in which case the seperation (in both Galileo's and Einstein's relativity) would be sqrt(2)c. Yes, points on opposite sides of the sphere would have a separation of 2c, in either theory.

But, as has been pointed out, the rate of change the separation between the two photons is not the same as the speed of one photon relative to the other. In Galileo's relativity, one photon sees the other moving away at 2c. SR makes no rigorous prediction, because v=c is outside the domain of definition of the Lorentz transformation.

geistkiesel

Jun9-04, 06:40 AM

Looks like I've made a mistake, too. I thought you were still talking about points moving in perpendicular directions, in which case the seperation (in both Galileo's and Einstein's relativity) would be sqrt(2)c. Yes, points on opposite sides of the sphere would have a separation of 2c, in either theory.

But, as has been pointed out, the rate of change the separation between the two photons is not the same as the speed of one photon relative to the other. In Galileo's relativity, one photon sees the other moving away at 2c. SR makes no rigorous prediction, because v=c is outside the domain of definition of the Lorentz transformation.

OK plot a circle in front of you with a diameter with arrows pointing outward, in the direction of expansion, just like we discussed. Now draw a radius rotated say 4 degrees from one of the radii already drawn. Connect this rotated radius tip to the radius tip farthest back (should look like a shortened diameter). What is the expansion rate of these radii?

If oppositely directed radii are expanding at a 2c rate then slightly off "direclty opposite" radii should conform to the same logic should it not? Aren't all radii expanding with the same logic even though the velocity of expansion, v(e) < c for nonparallel radii? Are we still outside the Lorentz transformaion tho v(e) < c? If so when, if ever , does SR enter this scenario?

Sorry about that other "expansion" sitiuation in the previous posts. I simply misread and was talking about oppositely expanding photons while everyone else in the room was talking about photons leaving the source 90 degrees apart. My apologies and for all my dissing that accompanied my mistake .

geistkiesel

Jun9-04, 06:49 AM

The trouble is quite simply that you can't accept reality at face value, probably because you just don't like it. "What is lurking around in the minds of SR theorists" has been proven beyond reasonable (even some unreasonable) doubt. It is reality.

OK. Focus on one photon of a short photon burst located at M. What is the rate of expansion of the radius of the expanding sphere as measured by the photon selected for scruiny? The answer is c is it not?

Tom Mattson

Jun10-04, 12:02 PM

OK plot a circle in front of you with a diameter with arrows pointing outward, in the direction of expansion, just like we discussed. Now draw a radius rotated say 4 degrees from one of the radii already drawn.

OK, so let the radius along the +y-axis be:

r1(t)=ctj,

and let the radius along the -y-axis be:

r2(t)=-ctj.

Then let a third radius be rotated from the first one by an angle θ, so it is:

r3(t)=(ct)sin(θ)i+(ct)cos(θ)j.

Connect this rotated radius tip to the radius tip farthest back (should look like a shortened diameter).

The vector joining r2 and r3 is:

r32(t)=r2(t)-r3(t)=
r32(t)=-(ct)sin(&theta)i-ct(cos(θ) +1)j.

What is the expansion rate of these radii?

The rate of change of this vector is:

r32'(t)=-c(sin(θ))i-c(cos(θ)+1)j

and the norm is:

|r'32(t)|=[2cos(θ)+2]1/2c, which for small angles is slightly less than 2c, as expected.

If oppositely directed radii are expanding at a 2c rate then slightly off "direclty opposite" radii should conform to the same logic should it not?

Yes, indeed it does.

Aren't all radii expanding with the same logic even though the velocity of expansion, v(e) < c for nonparallel radii? Are we still outside the Lorentz transformaion tho v(e) < c?

We would still be outside the domain of applicability of SR if we tried to Lorentz boost to one of the photons' rest frames, because SR says that photons don't have rest frames.

If so when, if ever , does SR enter this scenario?

It enters when we switch from one frame to another.

geistkiesel

Jun13-04, 07:24 PM

We would still be outside the domain of applicability of SR if we tried to Lorentz boost to one of the photons' rest frames, because SR says that photons don't have rest frames.

prove the statement above with more than 'because SR says . . .'

Regarding the question of when SR is applicable:

It enters when we switch from one frame to another.

Ditto here - prove the ,'switching frames' statement.

Tom Mattson

Jun13-04, 07:42 PM

prove the statement above with more than 'because SR says . . .'

It is outside the realm of applicability of SR because the LT is undefined at v=c.

Regarding the question of when SR is applicable:

Ditto here - prove the ,'switching frames' statement.

SR only enters when switching frames because relativistic effects only turn up when an observer looks at the other guy's rods and clocks.

Geistkiesel, why don't you crack open a book every once in a while? It would help you out immensely. This is such basic information that I can't even believe you're asking me for it.

And as to your first question: It is just plain silly.

The domain of applicability of SR is determined precisely by "what SR says".

geistkiesel

Jun16-04, 04:56 AM

It is outside the realm of applicability of SR because the LT is undefined at v=c.

SR only enters when switching frames because relativistic effects only turn up when an observer looks at the other guy's rods and clocks.

Geistkiesel, why don't you crack open a book every once in a while? It would help you out immensely. This is such basic information that I can't even believe you're asking me for it.

So SR comes into play, a physical dynamic, a reality 'turns up', "when an observer looks at the other guy's rods and clocks." AAAmazzzing!!

Why don't I crack a book so I can believe that I can make reality turn up by looking at someone's 'clocks and rods'? I know what SR says, Mattson. Why do not you see that? Deduce your own answer from the silliness you just posted in the guise of science. Grounded has you Tom_mattson, and you sense the end is near, don't you?

Someone in your life made a big mistake when they convinced you that you knew what you were talking about.

Tom Mattson

Jun16-04, 07:59 AM

Why don't I crack a book so I can believe that I can make reality turn up by looking at someone's 'clocks and rods'?

No, you should crack a book open so that you can see how SR fits into the "big picture" of physics.

I know what SR says, Mattson. Why do not you see that?

Uhhh, because you have yet to show that you understand anything about it, and you have been very clear about not wanting to change that.

Deduce your own answer from the silliness you just posted in the guise of science. Grounded has you Tom_mattson, and you sense the end is near, don't you?

This is just dumb. Neither you nor grounded understands electrodynamics well enough to see why the pro-SR arguments are correct. That's why you both continue to use simple-minded arguments about counting wavefronts and clocks on trains. Neither of you has the knowledge to look at Maxwell's equations and see the deep problems that physicists saw at the turn of the last century, problems that only find their solution in the Lorentz transformation. And I don't know about grounded, but as for you, you don't seem to want to change your state of ignorance, which is a pity.

Someone in your life made a big mistake when they convinced you that you knew what you were talking about.

Look in the mirror.

geistkiesel

Jun19-04, 07:28 AM

No, you should crack a book open so that you can see how SR fits into the "big picture" of physics.

Uhhh, because you have yet to show that you understand anything about it, and you have been very clear about not wanting to change that.

This is just dumb. Neither you nor grounded understands electrodynamics well enough to see why the pro-SR arguments are correct. That's why you both continue to use simple-minded arguments about counting wavefronts and clocks on trains. Neither of you has the knowledge to look at Maxwell's equations and see the deep problems that physicists saw at the turn of the last century, problems that only find their solution in the Lorentz transformation. And I don't know about grounded, but as for you, you don't seem to want to change your state of ignorance, which is a pity.

Look in the mirror.
Look in the mirror.|. rorrim eht ni kooL

Tom, what we need is not pro-SR arguments, we need pro-SR book burners.

So what? Tom, do you want to travel to the stars, the farthest stars I mean? Do want to leave your children and grandchildren with the ability to really explore the cosmos? Your answer is yes, of course.

Then Tom, you had better change your ways and take a close, close analytic lookat SR as this theory, believed by the public, bureaucrat and scientist alike, has the manifest effect of limiting space exploration to an insignificant number of inhabitants of the planet earth.

Tom, would you like to go, or is arguing theory more appealing to your wandering sense of exploration? Synchronize your watches everyone.

Convince yoursel Tom, then convince your colleagues

ahrkron

Jun19-04, 08:31 AM

So what? Tom, do you want to travel to the stars, the farthest stars I mean? Do want to leave your children and grandchildren with the ability to really explore the cosmos? Your answer is yes, of course.

Then Tom, you had better change your ways and take a close, close analytic lookat SR as this theory, believed by the public, bureaucrat and scientist alike, has the manifest effect of limiting space exploration to an insignificant number of inhabitants of the planet earth.

Regardless of how strongly Tom, you or I wish to "go to the stars", nature will follow the same patterns. The only thing we can do is learn what those patterns are and work within them.

And those patterns do include what SR predicts, so we better take it into consideration. You would be much better starting from it. But that's your choice.

Just be aware (as every person that works for real in experimental physics would be able to tell you) that rejecting SR is a poor choice.

geistkiesel

Jun19-04, 09:31 AM

Regardless of how strongly Tom, you or I wish to "go to the stars", nature will follow the same patterns. The only thing we can do is learn what those patterns are and work within them.

And those patterns do include what SR predicts, so we better take it into consideration. You would be much better starting from it. But that's your choice.

ahrkron: be advised Grounded, for instance, has made some seios progress even though he isn't attacking SR.

Do you want to go to the stars.

Just be aware (as every person that works for real in experimental physics would be able to tell you) that rejecting SR is a poor choice.[/QUOTE]
Scientific evolutionary theory and observation proves you wrong: There has never been an invariant theory.

Therefore, why sit around waiting for someoen else to discover the next evolved state step: , giant or baby.

Hurkyl

Jun19-04, 10:36 AM

And where in history has any progress been made by denying existing theory without strong theoretical or experimental justification?

Tom Mattson

Jun19-04, 03:17 PM

ahrkron: be advised Grounded, for instance, has made some seios progress

No, he isn't. He is just re-hashing the physics of the 19th century. And if we all did what he is doing, we'd all be making some serious anti-progress.

even though he isn't attacking SR.

I'll agree that he isn't attacking SR. He's simply denying it.

Do you want to go to the stars.

This is irrelevant! Despite your beliefs to the contrary, the outcome of real experiments cannot be changed as capriciously as those of thought experiments.

Scientific evolutionary theory and observation proves you wrong: There has never been an invariant theory.

That does not imply that it is correct to go backwards, which is in fact what you are advocating.

geistkiesel

Jun20-04, 09:01 PM

No, he [Grounded] isn't. He is just re-hashing the physics of the 19th century. And if we all did what he is doing, we'd all be making some serious anti-progress.
I'll agree that he isn't attacking SR. He's simply denying it.

This is irrelevant! Despite your beliefs to the contrary, the outcome of real experiments cannot be changed as capriciously as those of thought experiments.
Note:To investigate/re-hash what Grounded is saying is "anti-progress", investigating/re-hashing, is anti-progress?

Here is a simple thought experiment that has been discussed in various forms over the years.

1. In the Einstein stationary platform and moving train experiment an observer O' on the train located at M' on the the train heading to a photon source at B, arrives at the midpoint M of A and B, two light sources, just as the sources emit photons along the line of motion of the train. M, A and B are on a stationary frame. The observer determines her v' wrt M, and t'1 (the time from when she passed through M, say t'0) detects a photon emitted from B and later detects a photon from A at t'2. SR tells us the photons emitted in the stationary frame were not emitted simultaneously in the moving frame. This is a given.

From the information in front of you, t'0, t'1 and t'2, v', the spatial location of M' at t'0, or M'(t'0), determine if M'(t'0) was at the midpoint of the photons at t'1, when the B photon was detected on the moving platform. All values are moving frame values. You may use the given information that when M'(t'0) was colocated with M in the stationary platform that the photons were emitted from A and B simultaneously.

2. In the Michelson-Moeley experiment virtually all descriptions I have seen have the photons moving transverse to the photon moving 'parallel' with the motion of the source reflected at an angle, as if the photon were bounced ahead.However, the photon is directed, or should be directed, at an angle of 90 degrees wrt the transverse moving photon. Therefore, the reflected photon will arrive behind the moving photon and will pass through the half-silvered mirror above the point from which it was originally reflected downward.

In capturing the reflected photons in the eye piece (now moving parallel to each other) the photons upward moving parallel photons should be separated by an amount vt, the distance the sources moved during the time t, the time of roundtrip of the downward moving photon. Both photons interfer orthogonally before their final passage through the silvered mirror on their path to the eye piece. The silvered mirror is at 45 degrees to the direction of motion of the downward, tranverse, photon.

From this, calculate the optical path length difference expected from a given range of V, the velocity of the source for a.) any 24 hour period and, b). any one year period. To combine the calculations you may plot t(years) exponentilly using real hours for the 24 hour period.