No postulate of light is violated in Galilean transformation.

[geistkiesel]

"... caught on to Galilean relativity ..."

I just finished the first page of the URL at the motivation of my personal mentor, Doc Al.

http://theory.uwinnipeg.ca/mod_tech/node134.html

The writer showed examples of adding velocities using non photon entities. In using the photon in order to show that postulates of the speed of light would be violated if the observer shines a light ahead of his frame moving at 3/4c and adding velocities as,

c + 3/4c = 7c/4

This surely would a bring a contradicition of the postulates of the constancy of the speed of light limited to a maximum of 4c/4.. But wait a minute. If A is moving to the right and B is moving to the left what is their relative velocity? It is,

v(A + B) = V(A) + v(B),

and yes, the velocities are added to determine their relative velocity, which is greater than either velocity alone. But the writer in our example says the light was shined ahead of the moving frame moving in the direction of the light flash. To find the realtive velocity of the frame and photon entities using a standard Galilean transformation we write the relative velocity as a subtraction, or as,

c - 3c/4 = c/4.

So we have no contradiction of a postulate of light regarding maximum speed allowed do we, or even of measurement of the speed of light? And boy it sure seems rational that the relative velocity of the photon and frame is their difference. Do you think the writer was trying to fool us?
I made a mistake above. The c + 3c/4 = 7c/4 doesnot vioate any postulate of lihjt. What we arediscussing here is the realtive velocity of frame and photons. An expanding velocity of 7c/4 is reasonable and rational.

Hey Doc Al, this Galilean stuff is great, thanks for turning me on. Now I can really begin to live!

 

[Janus]

I just finished the first page of the URL at the motivation of my personal mentor, Doc Al.

http://theory.uwinnipeg.ca/mod_tech/node134.html

The writer showed examples of adding velocities using non photon entities. In using the photon in order to show that postulates of the speed of light would be violated if the observer shines a light ahead of his frame moving at 3/4c and adding velocities as,

c + 3/4c = 7c/4

This surely would a bring a contradicition of the postulates of the constancy of the speed of light limited to a maximum of 4c/4.. But wait a minute. If A is moving to the right and B is moving to the left what is their relative velocity? It is,

v(A + B) = V(A) + v(B),

and yes, the velocities are added to determine their relative velocity, which is greater than either velocity alone. But the writer in our example says the light was shined ahead of the moving frame moving in the direction of the light flash. To find the realtive velocity of the frame and photon entities using a standard Galilean transformation we write the relative velocity as a subtraction, or as,

c - 3c/4 = c/4.

So we have no contradiction of a postulate of light regarding maximum speed allowed do we, or even of measurement of the speed of light? And boy it sure seems rational that the relative velocity of the photon and frame is their difference. Do you think the writer was trying to fool us?
I made a mistake above. The c + 3c/4 = 7c/4 doesnot vioate any postulate of lihjt. What we arediscussing here is the realtive velocity of frame and photons. An expanding velocity of 7c/4 is reasonable and rational.

Hey Doc Al, this Galilean stuff is great, thanks for turning me on. Now I can really begin to live!

Yes it does violate the postulate of light velocity because the postulate states that the speed of light is invarient. For the speed of light to be invarient the answer you get when you add c to any velocity will must be c. (not 7c/4 or even c/4)

The formula that perserves this, and is the correct formula for the addition of any velocities is

w = \frac{u+v}{1+\frac{uv}{c^{2}}}

 

[geistkiesel]

Yes it does violate the postulate of light velocity because the postulate states that the speed of light is invarient. For the speed of light to be invarient the answer you get when you add c to any velocity will must be c. (not 7c/4 or even c/4)

The formula that perserves this, and is the correct formula for the addition of any velocities is

w = \frac{u+v}{1+\frac{uv}{c^{2}}}

You missed the point. The velocity of light did not change. Nothing changed. We made a simple addition of velocities and did not vary c. When you add the observers velocity you don't change c you determine the relative velocity., simple isn't it?

We subtracted the observers velocity from C to get c/4 a reasonable relative velocity under the conditions given. Remember the link. The writer there stated the velocity of the observer and the light was c + v, when the true or correct addition is c - v. Similarly when the observer and light are moving in opposite directions the velocity is added, v(a + B) = v(A) + v(B). In that case the relative expansion velocity is 3c/4 + 4c/4 = 7c/4. I see no violation.

How do you add velocity? And why do ou do it your way? I opine you are thinking ahead, anticipating long memorized SR theory and aren't keeping track of the steps as you go rushing along. Tres dangereaux, n' cest pas?

Show me where c is not measured as c. Can you do this?

This is only an observation, but I see you giving me a mathematical formula and I give you physics arguments. Not necessarily that one is better than the other, it is just an observation. This is the error I see in this simple first step into Galilean Transformations, where light postulates are not violated.

You see Janus you have a long way to go to arrive at your fornmula, and quite frankly, just repeating formula mantras don't cut it any more around here. You have to prove it. Do you think I don't know of your ideas, or the ideas of SR?? Get real man.

 

[ram1024]

Geist: i made a discovery in the other thread. It's looking like for every "moving observer" case all the calculations were done as "stationary observer" with "stationary sources" so that's why they believed that light speed was invariant relative to the viewer.

such a simple mistake has set physics and astrosciences back 100 years :D

 

[eJavier]

So far, all experimental data confirms the postulate that the speed of light must be the same for every inertial observer. And that doesn't follows from your theory.

Edit: rephrased

 

[geistkiesel]

Geist: i made a discovery in the other thread. It's looking like for every "moving observer" case all the calculations were done as "stationary observer" with "stationary sources" so that's why they believed that light speed was invariant relative to the viewer.

such a simple mistake has set physics and astrosciences back 100 years :D

We see alike. Welcome pilgrim.

 

[ram1024]

i guess that's why Tom didn't come back with the data...

it's either that or something terrible happened to him and that would suck much, so i take the lesser "evil"

 

[wespe]

You missed the point. The velocity of light did not change. Nothing changed. We made a simple addition of velocities and did not vary c. When you add the observers velocity you don't change c you determine the relative velocity., simple isn't it?

You mean:
"When you add the observers velocity, you don't change c"
"When you add the observers velocity, you determine the relative velocity"

You are still thinking c is speed of light relative to aether or something (so it doesn't change). No, c is the relative speed with respect to every observer. In other words, after the addition/subtraction/calculation we should find c as the relative speed, in order to conform to experiment results. The formula Janus gave finds c, simple addition/subtraction does not.

 

[wespe]

for every "moving observer" case all the calculations were done as "stationary observer" with "stationary sources" so that's why they believed that light speed was invariant relative to the viewer.

Please carry out an experiment to measure the speed of light. You can't? OK, somebody else did. They found c. So it is not just calculated/believed, it was measured.

 

[geistkiesel]

So far, all experimental data confirms the postulate that the speed of light must be the same for every inertial observer. And that doesn't follows from your theory.

Edit: rephrased

I really don't have a theory. I referenced the link I was studying and I merely looked at the way the writer there handled addition of velocities. Your statement of experimental proof is valuless here unless you are able to discuss the experiments in detail. Can you do this, or are you just echoing SR dogma?

http://theory.uwinnipeg.ca/mod_tech/node134.html Here take a look for yourself. This is where I am at the present.

He stated that for an observer moving with velocty 3c/4 in the direction the observer shined his flaslight that Galilean transformation would result in 4c/4 + 3c/4 = 7c/4. I merely observed that he was adding velocities improperly. If automobiles are moving away from each other one adds the relative velocites to get maxim relative speed, right? maximum speed? If the automobiles are moving in the same direction one subtracts the velocities or 4c/4 - 3c/4 = 1c/4, which seemed like a rational and reasonable number, especially after I detected the error. If the observer was moving in the opposite direction from the light we add the velocities or 4c/4 + 3c/4 = 7c/4 which is a righteous expanding velocity form oppositely moving particles.

Some have objected that I a violated a postulate of light, that the speed of light is always measured as c. I was raised in a morally healthy family situation and I tell you that I did not violate, nor did I alter the speed of light one tiny fraction of a m/s, trust me.

This is very important: Measuring relative velocity with a photon and object does not vary the speed of light. The addition of velocities merely determines the relative velocity and the SR theorists must prove the contrary. You don't have to remind me that I am swimming upstream, I know it so, very clearly.

I know that SR theorists are going to complain about this and that, they already have, like yourself, what's your story? How did you fall into the SR trap?. Well let them, this thread is titled "No postulate of light is violated in Galilean Transformation." If people are gping to respond to the thread they should keep on track, don't yopu agree?. Just saying "No No that's not allowed by SR", doesn't cut through any chaff. The SR theorists are still blinded by the simple title as I sneak through their net. I would have used Guile and Cunning but they'e still in jail on some kind of technicality thing, like a mistake, or something like that, like a misunderstanding.

It should be a slam dunk easy as pie task to prove that what is said in this thread, by myself or any else does not conform to physical law.Those that want to trade witticisms, or preach the history if physics, as an offer of proof best have something substantial to say, other wise you will be directed, yes you, to the most brutal of task masters in this forum. Can you take it like a man, or a woman, I mean can you stand up to Doc Al when he really gets wound up and in your face? I can't, I couldn't, I had to surrender, finally.

What you need to do to get the most from this thread is to prove me wrong, or prove yourself right, [I'll take either one, though I do have a preference.] but keep your rote SR mantras to yourself, bitte schön. I don't mean to be mean and blunt, but time is skitting along. I mean it.

I have already been certified as "outre" if you know what I mean, so unless you have a uniquely interesting tag to place on me as a physical argument please be creative, OK? As the song goes, "I've got skin like iron and breath like kerosene, I wear my gun outside my pants for all the world to see." from "Poncho and Lefty", Willy N, et al.

 

[geistkiesel]

Educating Wespe: somebody has to do the hard work.

You mean:
"When you add the observers velocity, you don't change c"
"When you add the observers velocity, you determine the relative velocity"

You are still thinking c is speed of light relative to aether or something (so it doesn't change). No, c is the relative speed with respect to every observer. In other words, after the addition/subtraction/calculation we should find c as the relative speed, in order to conform to experiment results. The formula Janus gave finds c, simple addition/subtraction does not.

wespe, prove it. Just don't bark SR mantras at me OK? Not this time.

Please see where and what I was talking about. Here take a look, below, before you open your doodling fingers. Prove what you have to say. in the context of what is said. This is one step at a time, do you understand?. One step at a time. Intro to Galilean transformations 101..

Do you hionestly think I don't know what SR has to say about velocity addition? Do you honestly believe I didn't anticipate posts such as yours?
read and weep, wespe.
http://theory.uwinnipeg.ca/mod_tech/node134.html

 

[ram1024]

exactly as he says. moving observer has nothing to do with the speed of light because he is not TIED to the light.

Aether does not MOVE at the observer's velocity. there is no Aether... :D

Light will move at "c", Calculated or Measured (if you know what you're doing <wink>)

 

[wespe]

wespe, prove it.

Prove what? Which one is the exact statement you disagree?

c is the relative speed with respect to every observer

Do you mean prove this statement? Experiments prove this. The speed measured in experiments (c) is relative speed of light with respect to the experiment apparatus, of course. Do you disagree?

 

[Doc Al]

Galileo's secrets revealed!

Please see where and what I was talking about. Here take a look, below, before you open your doodling fingers. Prove what you have to say. in the context of what is said. This is one step at a time, do you understand?. One step at a time. Intro to Galilean transformations 101..

I'm glad to see you are finally learning something about Galilean tranformations! Good for you. Welcome to the wonderful world of frame swapping.

But you are still a century behind if you think Galilean transformations describe how velocities are added in the real world. (Of course they are a wonderful approximation for low speeds.) Using Galilean transformations to talk about light is just plain silly.

Do you hionestly think I don't know what SR has to say about velocity addition?

If you have secret knowledge of SR... why have you been hiding it so well in all your hundreds of posts? You sneaky devil!

 

[geistkiesel]

Prove what? Which one is the exact statement you disagree?


Do you mean prove this statement? Experiments prove this. The speed measured in experiments (c) is relative speed of light with respect to the experiment apparatus, of course. Do you disagree?

This is the link I am looking at. I am stepping through the pages and taking them as face value. So far all I have seen ar esimple statenments that say this or that.
http://theory.uwinnipeg.ca/mod_tech/node133.html

I hyaven't seen anything that insist I do anything that I agree has to be done. Sying it is oart of SR isn't enough. For instance the addition of velocities follows from the statements ergarding the popstiulates of SR that are derived, at least partilly, from simultaneity conisderations.. I don't see the magic in light phenomena that says it is of such a nature that I can't move with respect to it and measure my speed with respect to the light.

The "measurement of he speed of light" being equal in all frames comes partially from the measurement of light speed being c and partially from relativity considerations. Einstein tells us that relativity consideration imposes the restiction that we measure the speed of light the same from whatever frame measured. I just don't happen to see it thay way at this juncture.

other rhan the relativity statement being widely recognized in SR do you see any physical reason that we allow ourselves the addition of velocities when considering stampeding elephants and Ford Futuras, but not stampeding elephants and photons?

Soon we will be discussing the dilation of light. Read ahead in the link and see if the discussion there fits with everything you have learned regarding the propagation of light. There is a glaring contradiction from which the "dilationists" take off in wild abandon. Glaring!

 

[jcsd]

Geitskeissal everything in special relativity logically follows on from the 2 postulates (the invariance of the speed of light and the laws of physics in inertial reference frames) + pre-relativstic classical physics.

The postulates aren't derived from anything else but experiment, the failure of simultaneity at distance needn't be considered to arrive at the formula for the addition of velocities, as both are derived directly from the postulates of special relativity.

 

[wespe]

You quoted but didn't answer my questions...

I hyaven't seen anything that insist I do anything that I agree has to be done. Sying it is oart of SR isn't enough.

The "measurement of he speed of light" being equal in all frames comes partially from the measurement of light speed being c and partially from relativity considerations. Einstein tells us that relativity consideration imposes the restiction that we measure the speed of light the same from whatever frame measured. I just don't happen to see it thay way at this juncture.

I didn't say speed of light is constant because of SR. I said experiments and measurements say so. Look up any experiment that measured the speed of light. All find c [within experimental error]. Plus, MMX. (dont' say Miller, you never answered my post https://www.physicsforums.com/showpost.php?p=248717&postcount=241)

For instance the addition of velocities follows from the statements ergarding the popstiulates of SR that are derived, at least partilly, from simultaneity conisderations.. I don't see the magic in light phenomena that says it is of such a nature that I can't move with respect to it and measure my speed with respect to the light.

The special thing about light is that it has zero rest mass.

other rhan the relativity statement being widely recognized in SR do you see any physical reason that we allow ourselves the addition of velocities when considering stampeding elephants and Ford Futuras, but not stampeding elephants and photons?

Where did you get this idea? Velocity addition formula Janus gave applies to all velocities. Galilean formula is an appoximation for low speeds only.

 

[geistkiesel]

You quoted but didn't answer my questions...





I didn't say speed of light is constant because of SR. I said experiments and measurements say so. Look up any experiment that measured the speed of light. All find c [within experimental error]. Plus, MMX. (dont' say Miller, you never answered my post https://www.physicsforums.com/showpost.php?p=248717&postcount=241)


The special thing about light is that it has zero rest mass.



Where did you get this idea? Velocity addition formula Janus gave applies to all velocities. Galilean formula is an appoximation for low speeds only.

Janus is a robot. Hr repeats what somebody told him. He poves nothing. He attempts to prove nothing.

Eventually we will get to time dilation, so let's do that now. We will use he link for a model. It is supposed to be educational.
http://theory.uwinnipeg.ca/mod_tech/node135.html

In arriving at a time dilaion result the physics of light propagation is ignored.

___________________
              |
              |
              |
              |
              |
              |
_ _______|_________

The horizontal lines are mirrors reflecting a light pulse. Each pulse takes one second to go from floor to ceiling, or round trips take two seconds.This is a clock wih an invaiant time cycle.

Look again at the model.

Now if the clock i smoving to the right our model has th epulse that is recorded diffeent than he pulse in the stationary clock.

______________  _______________________________
           \                            /
             \                        /
               \                    /
                 \                /
                   \            /
                     \        /
                       \    /
        _ _______________\/____

Ther part of he wave front being measured here is not the same as the case of the stationary clock.

Question: Does the velocity of the source of light get added to the velocity of light?

Answwer: No.

Therefore the
model we are using is bogus.

Here is how the measurement should be conducted for a moving clock.

.
____________________
          |     |
          |     |
          |     |
          |     |
          |     |
          |     |
          |     |
          |     |
          |     |
          |     |
 _______|___ |_________

The two lines ,the right one "down" the left "up", should be supperimposed on each other and shown separate for instructional puposes. It is the frame that moves. Right? Isn't this the law of he propagation of light? Answer:yes.

Using th extended light ray as indicated in this linkis the corruption on the construction of time dilation, all followed so robotically by the masses of SR theorists that want to be.

Next post how to make a light clock that works. Stay tuned.

Responders, if any, please use the laws of physics in your posts.

 

[quantumdude]

Janus is a robot. Hr repeats what somebody told him. He poves nothing. He attempts to prove nothing.

Janus proved that you have no idea of what the postulate of the speed of light says, or what "invariant" means.

Eventually we will get to time dilation, so let's do that now. We will use he link for a model. It is supposed to be educational.
http://theory.uwinnipeg.ca/mod_tech/node135.html

In arriving at a time dilaion result the physics of light propagation is ignored.

This is so badly mistaken, I can hardly believe that anyone who claims to have studied any physics could say it. Time dilation comes from precisely a very careful consideration of the physics of light propagation. You've demonstrated over and over again that you don't understand how light propagates anyway, so I guess I shouldn't be too surprised at this.

Question: Does the velocity of the source of light get added to the velocity of light?

Answwer: No.

Therefore the
model we are using is bogus.

No, it's your logic that is bogus. The assumption you are tacitly making is that the Galilean velocity addition formula should hold. But it doesn't.

Responders, if any, please use the laws of physics in your posts.

Janus tried that, and you called him "robot". So what's the use?

 

[wespe]

Geistkiesel,
It turns out how light travels [direction]* is not totally independent from the source if the source is moving perpendicularly (this is something I didn't think clearly before, thanks). But, still, speed of light must remain the same, because otherwise you could send a message faster than light.

Suppose light source is stationary and a message is sent from A to B in 1 second.

_________B________
         |
         |
         |
         |
         |
         |
_________|_________ 
         A

Now, the light source is moving perpendicularly. According to the stationary observer, light is traveling diagonally, but it must still travel the same distance in 1 second:

_________________C
              
               
             /
            /
           /
          /
_________/________ 
        A     A'

distance from A to C is longer, so information can't be sent in 1 second. It should take more time (hence the time dilation)

According to the moving frame, light traveled perpendicularly, but it was C that moved to the old position B, so it still took 1 second:

_______  <-C________
         |
         |
         |
         |
         |
         |
_________|_________ 
         A'

Of course, when the first case is viewed from the moving frame, all of this is observed the other way, so both frames see each other's clocks running slower.

Next post how to make a light clock that works. Stay tuned.
Responders, if any, please use the laws of physics in your posts.

Ultimately physics is an experimental science.
see http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/twin.html#c0
How will you explain these results if time dilation is bogus?


edit: (*) I have thought of light as a photon particle above. There's also aberration of light which I don't know exactly. Also there's wave/particle duality but we enter Quantum theory there. So I'm not 100% sure about this.

 

[geistkiesel]

You quoted but didn't answer my questions...





I didn't say speed of light is constant because of SR. I said experiments and measurements say so. Look up any experiment that measured the speed of light. All find c [within experimental error]. Plus, MMX. (dont' say Miller, you never answered my post https://www.physicsforums.com/showpost.php?p=248717&postcount=241)


The special thing about light is that it has zero rest mass.



Where did you get this idea? Velocity addition formula Janus gave applies to all velocities. Galilean formula is an appoximation for low speeds only.

The speed of light is 3x10^8m/s sure enough, but who prevents one from moving with respect to the light beside SR theory? That is what some say that light has zero rest mass. I disagree for one reason alone. Light exhibits charasteristics indicating a complex polarization mechanism that I find difficult to see formed in a "rest mass = 0 fluidity" . There must be something more substantial, more to the question than a simple m = 0. Those that say m = 0 at v = o have they taken alll the QM aspects into consideration? Correctly? To your satisfaction, from your own investigation? Nope, you haven't, have ya? More work wespe, better get busy, times a wasting.

 

[geistkiesel]

Geitskeissal everything in special relativity logically follows on from the 2 postulates (the invariance of the speed of light and the laws of physics in inertial reference frames) + pre-relativstic classical physics.

The postulates aren't derived from anything else but experiment, the failure of simultaneity at distance needn't be considered to arrive at the formula for the addition of velocities, as both are derived directly from the postulates of special relativity.

So everyhting follows logiaically from the two postulates. If physical law is invarinat in all inertial frames, where does simultabeity fit into that? The measure of the speed of light is hard coded into theory and by omitting the observers speed the measure of c in all frames is guarnateed as you say but how
By inserting time dilation and mass shrinking, two very big objections to Sr theory from where I sit. Ad guess what I ain't alone. Check out the other threads on the subject here in Theory Developemnt.

 

[quantumdude]

The speed of light is 3x10^8m/s sure enough, but who prevents one from
moving with respect to the light beside SR theory?

The "who" would be nature

That is what some say that light has zero rest mass. I disagree for one reason alone. Light exhibits charasteristics indicating a complex polarization mechanism that I find difficult to see formed in a "rest mass = 0 fluidity" .

What the bloody blazes is "rest mass = 0 fluidity"?

There must be something more substantial, more to the question than a simple m = 0. Those that say m = 0 at v = o have they taken alll the QM aspects into consideration? Correctly?

Of course it has been done correctly. The fact that m=0 implies that there is no |J,M>=|1,0> state for the photon. If m is not zero, then there should be three states:

|J,M>={|1,1>,|1,0>,|1,-1>},

but as it is only 2 are observed. The synthesis of SR and QM is QED, and it has been worked out very completely.

 

[quantumdude]

So everyhting follows logiaically from the two postulates. If physical law is invarinat in all inertial frames, where does simultabeity fit into that?

It fits into it via the Lorentz transformations, which are derived from the invariance of Maxwell's equations. I posted a detailed explanation of this some time ago. You acknowledged it, but clearly did not read or absorb it beyond a superficial level.

The measure of the speed of light is hard coded into theory

Not so. The measure of a Lorentz-invariant speed is hard coded into the theory. It is Maxwell's equations that tell us that this speed must be c.

and by omitting the observers speed the measure of c in all frames is guarnateed as you say but how

We don't "omit" the observer's speed. In the observer's frame, he is at rest.

By inserting time dilation

Time dilation is not inserted, it is derived.

and mass shrinking,

"Mass shrinking" is not part of SR. In fact, the mass of a particle is an invariant, as it is the norm of the 4-momentum.

two very big objections to Sr theory from where I sit. Ad guess what I ain't alone. Check out the other threads on the subject here in Theory Developemnt.

"From where you sit", eh? Then it's time for all of you to get up and move to a new chair, from which you can see the blackboard better.

 

[ram1024]

We don't "omit" the observer's speed. In the observer's frame, he is at rest

which means the speed of ALL objects in the universe acquire the observer's velocity, LIGHT INCLUDED.

why would light behave differently? and don't say the data says so, because the data does NOT include the addition of observer motion.

 

[geistkiesel]

Janus proved that you have no idea of what the postulate of the speed of light says, or what "invariant" means.

Janus proved nothing . He merely quoted gospel and verse.

This is so badly mistaken, I can hardly believe that anyone who claims to have studied any physics could say it.

"I know I am but what are you?"

Time dilation comes from precisely a very careful consideration of the physics of light propagation. You've demonstrated over and over again that you don't understand how light propagates anyway, so I guess I shouldn't be too surprised at this.

No I guess you won't be surprised at this.

When light is reflected between two parallel mirrors that are stationary each reflection follows the exct same path, i.e. one line. Of course, there is always side radiation, but you know of what I speak.

When the mirrors move sideways in a direction parallel with the mirrors, the reflected light does not change [position. Remember, the source of the light, like each reflection from the mirror adds no light velocity in the direction of motion of the mirrors. The light, therefore, remains fixed like a perfect z-axis, stationary, absolute zero velocity. A perfect and absolute frame of refer3enhce.The reflections continues up and down. Eventually the mirrors will come to an end and unless some adjustments are made the left side of the apparatus will simply crash into the z-axis light beam.

Measuring side reflections is an arbitrary piece of physics sloth. The mirror system can measure where each returning relfection is located along the mirror surface, therefore the pulses can be cycled to start over on the right hand side of the apparatus, depending on the measured velocity. It is really an engineering problem tom.

The diection of motion of the apparatus wrt to the z-axis can be detemined in a number of ways, one I just mentioned. Another, for the x-axis is with a modified ram1024 buoy.Don't ever try to tell me tom_mattson that I cannot use the z-axis I just created and use it as an absolute reference frame.

     0      |         o          |
      1        |<-------o----->     |
       2       |------>    o   ------>|
        3              ------>o<------|
         0
          1
            2
              3

The 0 level is when the photon is emitted from 'o' to the left and right. Reflectors are situated extending as shown. In 1 the left reflector has closed on the left moving photon. The right reflector already extended is now two units to the right instead of one unit as when the photon was emitted. 2 shows the left photon reflected to the right as the right photon has just reached the right reflector. The next cycle has the photons arriving simultaneously at o where, as Yogi Berra said, "It's dejavu all over again."

As long as the reflection cycle is sufficiently high detector counting arrivals at the various reflectors can maintain a count and hence keep a, solid track of velocity of the appaatus with respect to the absolute point in space designated as the source of the photons. Neat huh? Basic optical engineering 101.

Notice, that inspite of claims

1. SR,
2. Galilean transformations,
3. Lorentzian transformations,
4. simultaneity,
5. time dilation
6. c measured as c in all inertial frames (with time dilation of course)
7. and length contraction

It all works as designed.

You SR theorists have clouded your brilliant minds with theoretical molasses. Just think where we would all be if the simplicity of physics had been working for the past hundred years instead of that SR crap.

It doesn't work you say? Prove it.

I mean prove it.

Don't send the robot Janus over here with his mantras, or you tom. Show it won't work in an experiment. This is all that will convince me. You are wasting your breath and good finger muscle by attempting to convince me otherwise. I read your galilean post tom and in fact I was impressed which I believe I mentioned. The only thing is it doesn't get us there from here.Those watching may be impressed by your singular and collective brilliances, but I need an experiment. The one shown here is simple enough isn't it? Do this one. Our mutual uncle can afford it. Dig into his pockets.

I understand very well that a good physicist is necessaily a good proposal writer, right?

The physics is extremely simple. The speed of light is independent of the motion of the source. Therefore the the moving apparatus can and does keep a running track of the velocity of the light with respect to the inertial frame. Heresey? Get used to it.

I did learn something I didn't know earlier. That simultaneity is not operational at distances. What distances? What needed adjustment that you discarded or diosconrived your precious simultaneity? It certainly wasn't at the generosity of SR theorists was it? Does that mean we get absolute time back? Wow, just think of it! Or is that only at distance also?

No, it's your logic that is bogus. The assumption you are tacitly making is that the Galilean velocity addition formula should hold. But it doesn't.

It will hold fpor me. I asked mr. galilean, very politely, if you can believe that!

Janus tried that, and you called him "robot". So what's the use?

Being a robot doesn't make you a bad person Tom.

 

[Alkatran]

I think it's funny that you deman experiments to prove SR, but you have none to back yourself.

 

[geistkiesel]

what is this?

Tom, somethiong completely different. You are the math guru so you figure it out. I found the following in a website' But before that I had developed an expression from an analysis of the Einstein train gedunken. The train moving right detects the B photon coming from the front then the A photon from the rear. Using t1 as the time the B photon was detected measured from the instant the O' observer was at the midpoint of A and B sources in the stationary frame (t2 the time the photons arrived simultabeously at the midpoint of A and B in the stationary frame) and t3 the time of detection of the A photon by O' in the moviong frame as t3, I derived an expression that

t3 = t1(C + V)/(C - V)

from the following:

A_________________M__________________B
                 t0 t2
____________A______O'_____B_______A_____
                          t1<-------------          
 -------------------------------->t3

t0 the time the O' obsever was at M and the time the photons emitted at A and B. t1 the time B was detected by the moving O'. t2 the time the A and B photons arrived simultaneously at M from A and B. and t3 he time the A photon detected by O'.
From symmetry considerations A and B are equidistant = t1v from M when B is detected. The frame is moving at velocity v. t1 measured from t0 = 0. Therefore the A photon to be detected at t3 it must travel
c(t3-t1) = 2vt1 + v(t3 - t1)
ct3 - ct1 = 2vt1 = Vt3 - V t1
t3(C - v) = t1(C + v)
t3 = t1(C + v)/ (C - v).
Look at the below which I got from the link
http://www.mathpages.com/rr/s4-08/4-08.htm

So you the dude with all the math geewhizz, so gee whizz me with the derived expression and the below which comes from the loss of simultaneity with distance. por favor, bitte schön, please.

I f I do it igt would take much miore time than you. Dod Al knows about, but what can I say, Doc and Al and me, well oil and water, as they say.


Letting C = dS/dT denote the speed of light with respect to these rotating non-inertial coordinates, we therefore have C = 1 ± v, where again the sign depends on the direction of the light relative to the direction of rotation of the disk.

Does this analysis lead to some kind of paradox? It indicates that the non-inertial "speed of light" with respect to these rotating coordinates is not equal to 1, and in fact the ratio of the speeds in the two directions is (1+v)/(1-v), but of course this doesn't conflict with special relativity, because these are not inertial coordinates (due to their rotation). However, suppose we increase Rd and decrease w in proportion so that the rim speed v remains constant. The above formulas still apply for arbitrarily large Rd and small angular speed w, and yet the speed ratio remains the same, (1+v)/(1-v). Does this conflict with special relativity in the limit as the radius goes to infinity and the angular speed of the rim goes to zero? Clearly not, since we saw in Section 2.7 that if t1 and t2 denote the travel times for light pulses circling the disk in opposite directions, as measured by a clock at a fixed point on the rim, so that t2/t1 = (1+v)/(1-v), then we have t2/t1 - 1 = f /p, where f is the angular travel of the disk during the transit of light. In other words, the observed ratio of travel times around the rim always differs from 1 by an amount proportional to the angular travel of the disk during the transit of light. Thus the net acceleration (change of velocity) of the rim observer during the measurement remains in constant proportion to the measured anisotropy of the transit times.

 

[ram1024]

That didn't make any sense when I first read it, but I think I figured out what you mean.

Suppose you are in a spaceship, and some stars ahead look stationary. Then you fire your thrusters for 10 seconds and then you are inertial again. Now you feel you are at rest, but it seems the stars acquired a speed towards you. You meant this, right? OK, but how much speed? Say v. Then you fire the thrusters again for 10 seconds. Now stars seem to have gained some more speed. But how much speed, the same v? No. You repeat this until relative speed of stars is 0.9999...c, then you can't make relative speed exceed c no matter how long you fire your thrusters. Now think that light's relative speed was already c to begin with... And how does this look from an outside observer? Separation speed does look more than c, but that's not what you see in your spaceship. SR effects explain this. See post#2 for the speed addition formula.

you've ALMOST "got" it, wespe. Except that for "the laws of physics" to be the same for "all inertial bodies", a relative speed of 0.9999...c = the rocket stationary and the star moving towards me at 0.9999...c. I can still accelerate just fine thankyouverymuch :D

the reason Einstein has trouble with "speeds never exceeding lightspeed" is from a stationary observer standpoint with his "lightspeed never changes for the observer" postulate that means that objects can only come in towards an observer at light speed or go away from the observer at that speed.

but why suppose the observer is stationary in the first place? if he IS stationary then he has NO problem continuing to accelerate even at a relative separation of 0.9999...c.

let's not even talk about light and assume two spaceships are traveling towards each other at 0.9999...c. if we take one as stationary, the other must be covering the relative distance twice as fast 1.998...c or whatever. F=MA doesn't include am object's current velocity, indeed it doesn't matter because according to the "laws of physics" postulate same force, same mass, same acceleration, no matter the speed.

 

[wespe]

let's not even talk about light and assume two spaceships are traveling towards each other at 0.9999...c.

If you mean a third frame measures each spaceship's speed as 0.9999..c, ok,

if we take one as stationary, the other must be covering the relative distance twice as fast 1.998...c or whatever.

yes, if you mean as seen in the third frame. Not so as seen from the spaceships! It's still below c. See again #2 for the formula.

F=MA doesn't include am object's current velocity, indeed it doesn't matter because according to the "laws of physics" postulate same force, same mass, same acceleration, no matter the speed.

Well, the relativistic mass does increase with speed according to a third frame.
[You know that in particle accelerators, they can't accelerate particles to or beyond c, because mass does increase with speed]

 

[ram1024]

If you mean a third frame measures each spaceship's speed as 0.9999..c, ok,

yes of course, the measurements must be made according to something that is defined "stationary"

yes, if you mean as seen in the third frame. Not so as seen from the spaceships! It's still below c. See again #2 for the formula.

nope. take either ship as stationary, the opposing ship MUST cover twice the relative distance in the same amount of time. there is no "perspective" shift in this direct translation. the "third observer" would be in the same reference frame AS the "stationary rocket"

Well, the relativistic mass does increase with speed according to a third frame.
[You know that in particle accelerators, they can't acelerate particles to or beyond c, because mass does increase with speed]

a well-jumped-to conclusion. I'm not going to assume that people know HOW to calculate light speed anymore at this point :D

 

[wespe]

yes of course, the measurements must be made according to something that is defined "stationary"

What I had in mind was:

A->0.99c 0.99c <-B
____________________

the speeds are measured in the ____ frame. But in A or B frame, their relative speed is still below c.

nope. take either ship as stationary, the opposing ship MUST cover twice the relative distance in the same amount of time. there is no "perspective" shift in this direct translation. the "third observer" would be in the same reference frame AS the "stationary rocket"

You mean it is like this?

A <-B
____________

then where does "twice the distance" come from?

a well-jumped-to conclusion. I'm not going to assume that people know HOW to calculate light speed anymore at this point :D

what?

Anyway, in any case, relative speed between two objects as measured by one of the objects can never exceed c. The relative speed between the lab frame and particles don't exceed c (this is a fact), which supports what I said.

 

[ram1024]

no i mean THIS

A <- <-B

you can't forget that speed is defined taking a distance over a time. if A covers .9999x300,000km towards B in one second and B travels .9999x300,000km towards A in one second then if A is defined as stationary, B travels .9999x300,000km + .9999x300,000km in one second towards A.

simply so.

Anyway, in any case, relative speed between two objects as measured by one of the objects can never exceed c. The relative speed between the lab frame and particles don't exceed c (this is a fact), which supports what I said

as stated above you only believe so because you're completely negating one "observer's" velocity instead of adding it to everything else in the universe when defining that observer as "stationary"

and as far as particle accelerators go, if we took a linear accelerator capable of accelerating a particle to .9999c and then launched that accelerator into space with a relative velocity of .9999c to the earth, then shot a particle down the tube <pointing away from earth> how fast would it be going <from Earth's frame> ?

and don't bother using the "formula", i want you to tell me LOGICALLY how you feel about the situation. parroting someone elses formulas does not demonstrate you know what you're talking about...

 

[geistkiesel]

The "who" would be nature

What the bloody blazes is "rest mass = 0 fluidity"?

It is something i made up on the spur of the moment. What I meant is that having no rest mass puts the light as some kind of wispy, waviness, which would be very difficult to maintain any kind of complex structute that is observed such as having the dual time variant characteristic of "vertical" vs. "Horizontal" polarization potential. Richard Feynman describes light as coming in two types "V" and "H" to explain the polarization. What RF is not considering is that light is one type. The time history of any photon, or what ever it is that gets polarized, goes as ...VHVHVHVHVHVHVHVHVH..., when it isn't V then it is H etc. However, it can only be one or the other, so what happens to V when H is observed and vice verse? Looking at the polarization function abstractly we describe it as Y(10f), which we undertsand to mean that an observed 1 means iobserved, 0 unobserved, or nonlocal. ergo the time history is as I just described. Y(10f) Y(01f) Y(10f) Y(01f) and so on, with f the frequency of the state generation, where polarization is defined as a state of the particle, photon. I know it is heresy, but that's the way it is, otherwise you get RF's two types of light, if that is the current standard model.
http://frontiernet.net/~mgh1/

Sooner or later the QM world is going to be reviewing the most basic Stern-Gerlag experiments. I suggest the link and a look at Feynman's "Lectures on Physics" Vol III ch 5, an absolute must. If you aren't completely familiar with this chapter, Tom you "ain't". What more can I say. It will change your life.

Do you know what I mean by "fluidity"? aas inconsistent with the model just presented?

Of course it has been done correctly. The fact that m=0 implies that there is no |J,M>=|1,0> state for the photon. If m is not zero, then there should be three states:

|J,M>={|1,1>,|1,0>,|1,-1>},

but as it is only 2 are observed. The synthesis of SR and QM is QED, and it has been worked out very completely.

I am sure it has. Check out the link.an "Lectures"

 

[geistkiesel]

Geist: i made a discovery in the other thread. It's looking like for every "moving observer" case all the calculations were done as "stationary observer" with "stationary sources" so that's why they believed that light speed was invariant relative to the viewer.

such a simple mistake has set physics and astrosciences back 100 years :D

It seems ram1024, that someione is cheating. Is there someone we can call about this?

 

[geistkiesel]

We don't "omit" the observer's speed. In the observer's frame, he is at rest.

Why does the moving observer always consider herself at rest when measuring the speed of light? Convenience I will surmise.


"

Mass shrinking" is not part of SR. In fact, the mass of a particle is an invariant, as it is the norm of the 4-momentum.

Would length contraction sound better than mass shrinking?

"From where you sit", eh? Then it's time for all of you to get up and move to a new chair, from which you can see the blackboard better.

Some of my more pleasant moments in school was to be able to ask whoever was chalking up the board at the moment, "Where did you get that 2?"
Hey tom, where did you . . .aw fagida abowdit!

 

[wespe]

no i mean THIS

A <- <-B

What the heck is this? Why is the arrow on the right side of A? The arrows are supposed to show a velocity vector. Do you mean like this?

<-A <-B
________

If so, their speeds are in the same direction according to the third observer, and then they are stationary wrt each other.

you can't forget that speed is defined taking a distance over a time. if A covers .9999x300,000km towards B in one second and B travels .9999x300,000km towards A in one second then if A is defined as stationary, B travels .9999x300,000km + .9999x300,000km in one second towards A.
simply so.

In which frame are all these measurements made? If it's the Earth frame, the diagrams below fit your description.

A <-B
_______

or

A-> B
_______

A does approach B at .9999c per second, and B does approach A at .9999c per second, all according to the Earth frame. But they don't approach at nearly 2c.

You didn't like this one, but again:
A-> <-B
________

According to the Earth frame, they approach each other at nearly 2c per second (separation speed). But that doesn't mean they will see each other approaching at nearly 2c. According to SR, their relative speed will still be below c.

as stated above you only believe so because you're completely negating one "observer's" velocity instead of adding it to everything else in the universe when defining that observer as "stationary"

that didn't make any sense. I'm not "defining" observer stationary. Any observer IS stationary wrt itself, what's so difficult to understand?

and as far as particle accelerators go, if we took a linear accelerator capable of accelerating a particle to .9999c and then launched that accelerator into space with a relative velocity of .9999c to the earth, then shot a particle down the tube <pointing away from earth> how fast would it be going <from Earth's frame> ?

A little over 0.9999c but still below c, according to SR. How can you think you can predict the result of such an experiment while you sit on your a##.

and don't bother using the "formula", i want you to tell me LOGICALLY how you feel about the situation. parroting someone elses formulas does not demonstrate you know what you're talking about...

How I FEEL? Nature doesn't have to conform to your or my feelings. If you want to know how nature acts, look at the experiments. You totally frustrated me now with your ignorance and arrogance and incapability to learn. Bye, Ram.

 

[geistkiesel]

I think it's funny that you deman experiments to prove SR, but you have none to back yourself.

I gave at the office and didn't keep enough for expeiments, hence uncle is going to pop for this one. Do you mind? Is this all you have to offer for the post I just completed? Whoooooosh!

You are very helpful, thank you for sharing all that with us.

 

[geistkiesel]

I think it's funny that you deman experiments to prove SR, but you have none to back yourself.

I just demanded the experiment one I just described in the post above. It does not back SR, it dumps SR in the trash bin. What are you talking about?

 

[quantumdude]

Why does the moving observer always consider herself at rest when measuring the speed of light? Convenience I will surmise.

Not only does she consider herself at rest when measuring the speed of light, she considers herself at rest period, because she is inertial.

"
Would length contraction sound better than mass shrinking?

Well, since length contraction is actually a prediction of the theory, while "mass shrinking" is a stupid figment of some retard's imagination, yeah, I guess it would sound better!.

 

[Doc Al]

yes,geistkiesel, distance = speed x time: So what?

Tom, somethiong completely different. You are the math guru so you figure it out. I found the following in a website' But before that I had developed an expression from an analysis of the Einstein train gedunken. The train moving right detects the B photon coming from the front then the A photon from the rear. Using t1 as the time the B photon was detected measured from the instant the O' observer was at the midpoint of A and B sources in the stationary frame (t2 the time the photons arrived simultabeously at the midpoint of A and B in the stationary frame) and t3 the time of detection of the A photon by O' in the moviong frame as t3, I derived an expression that

t3 = t1(C + V)/(C - V)

from the following:

A_________________M__________________B
                 t0 t2
____________A______O'_____B_______A_____
                          t1<-------------          
 -------------------------------->t3

t0 the time the O' obsever was at M and the time the photons emitted at A and B. t1 the time B was detected by the moving O'. t2 the time the A and B photons arrived simultaneously at M from A and B. and t3 he time the A photon detected by O'.
From symmetry considerations A and B are equidistant = t1v from M when B is detected. The frame is moving at velocity v. t1 measured from t0 = 0. Therefore the A photon to be detected at t3 it must travel
c(t3-t1) = 2vt1 + v(t3 - t1)
ct3 - ct1 = 2vt1 = Vt3 - V t1
t3(C - v) = t1(C + v)
t3 = t1(C + v)/ (C - v).

I've explained all this before, but you still don't get it. Yes, amazingly, your expression t3 = t1(c + v)/(c - v) is correct. (It's just d = vt.) But only if all the time measurements are made in the O frame, not in the O' frame:

t = 0 is the time of emission according to the O frame
t = t1 is the time that photon B hits O' according to the O frame
t = t3 is the time that photon A hits O' according to the O frame ​

That's the only way you can derive that result. It says nothing whatsoever about measurements made in the O' frame. It says nothing about "simultaneity". It's just a trivial expression relating times t1 and t3, both of which are O-clock times.

By the way, here's a simpler way to derive that result. Note that all measurements are O-frame measurements. Let L be the distance from the midpoint to either A or B. By the time the photon from B hits O', O' has moved towards B, so:
(1) ct1 + vt1 = (c + v)t1 = L
And by the time the photon from A hits O':
(2) ct3 = L + vt3 --> (c - v)t3 = L
Combine (1) and (2) and you're done. So what?

 

[ram1024]

A does approach B at .9999c per second, and B does approach A at .9999c per second, all according to the Earth frame. But they don't approach at nearly 2c.

You didn't like this one, but again:
A-> <-B
________

According to the Earth frame, they approach each other at nearly 2c per second (separation speed). But that doesn't mean they will see each other approaching at nearly 2c. According to SR, their relative speed will still be below c.

why are you so confused? A can travel east at 20 miles per hour. B can travel west at 20 miles per hour. both cars have a speed limit of 20 miles per hour.

you're telling me the maximum speed they can approach each other is 20 miles an hour, which is patently untrue given if they're 40 miles apart and they both drive towards each other they can collide in an hour.

by any definition if you take A as a stationary object, B now travels 40 miles an hour towards A.

"but light is different..."

BS.

 

[Doc Al]

A can travel east at 20 miles per hour. B can travel west at 20 miles per hour. both cars have a speed limit of 20 miles per hour.

Viewed from an observer at rest with the road, what you say is perfectly true. All speeds are given with respect to the road.

you're telling me the maximum speed they can approach each other is 20 miles an hour, which is patently untrue given if they're 40 miles apart and they both drive towards each other they can collide in an hour.

According to observers at rest with the road, the two cars approach each other at 40 miles per hour.

by any definition if you take A as a stationary object, B now travels 40 miles an hour towards A.

Ah, now you want to switch frames and view things from A's frame. Better do it right: use the relativistic addition of velocities formula. For low speeds only it does turn out that V_{B/A} = V_{B/road} + V_{road/A}.

"but light is different..."

Everything is different, not just light. Welcome to the 20th century!

 

[geistkiesel]

I've explained all this before, but you still don't get it. Yes, amazingly, your expression t3 = t1(c + v)/(c - v) is correct. (It's just d = vt.) But only if all the time measurements are made in the O frame, not in the O' frame:

t = 0 is the time of emission according to the O frame
t = t1 is the time that photon B hits O' according to the O frame
t = t3 is the time that photon A hits O' according to the O frame ​

No and nly if you invoke some unwanted and unwarranted SR theory. This gedunken precedes Sr and should, like AE ofered it, be able to stand on its own two 1/3meters.

No doc, the times are measured using the passenger wrist watches. I specifically instructed all the passenger to first get a WWV time hack and use only the watches they wear. Of course the stationary observers do the same, but we aren't comparing stationary and moving frame clocks here. C is the speed of light c = 3x 10^8m/s. v is measured with train watches and distance measuring devices. t3 and t1 are moving frame times. This is how I set up th experiment. If you want to conduct the same experiment using stationary frame times, go for it, but only your opinion insists on using stationary times, your opinion, only.

On what possible ground do you infer, that I infer, the times weren't moving frame times?

Also, at the velocities used here, you could never prove your SR time dilation SR stuff anyway, could you?. Gotcha Doc.

Anyway, some time in this experiment I want to do the impossible: Stop the train and accelerate the embankment.Uh, oh, I have to set v = 0, right? Then my expression becomes, t3 = t1(c + 0)/(c -0) or t3 = t1 , which is contradicted by measurement that t3 > t1, Hence the passengers conclude that the train ain't stopping for no SR theory.

That's the only way you can derive that result. It says nothing whatsoever about measurements made in the O' frame. It says nothing about "simultaneity". It's just a trivial expression relating times t1 and t3, both of which are O-clock times.

Doc the times are from watches on the passengers on he trains, OK? Why are you persisting to swap my times on me? This isn't a Hidden "frame swap" of which you are getting such a rep for inserting into people's posts, is it?

By the way, here's a simpler way to derive that result. Note that all measurements are O-frame measurements. Let L be the distance from the midpoint to either A or B. By the time the photon from B hits O', O' has moved towards B, so:
(1) ct1 + vt1 = (c + v)t1 = L
And by the time the photon from A hits O':
(2) ct3 = L + vt3 --> (c - v)t3 = L
Combine (1) and (2) and you're done. So what?

So what? Doc, check this out.
http://www.mathpages.com/rr/s4-08/4-08.htm
I thought I left you a message that I had found a twin expression in the link regarding simultaneity. I was actually asking youd to take a look at what I had found so you could "mentor" me on the comparison. Did you not get my message? Doc the expression is a crucial aspect of simultaneity, you had better look, before you leap.

 

[geistkiesel]

Viewed from an observer at rest with the road, what you say is perfectly true. All speeds are given with respect to the road.

I used radar instrumentation in both cars, where the return signal is translated into relative speed, as when the fuzz is setting behind the big sign waiting for you to speed through her trap, her gun measures your velocity respecto hers, when she is stationary. But moving radar can be used, even though it is translated into velocity wrt the ground. I know, a Texas Ranger clocked me at 92mph between Dallas and FT. Wrth and he only gave me a warning. He showed me the 92.

B]According to observers at rest with the road[/B], the two cars approach each other at 40 miles per hour.

[Ah, now you want to switch frames and view things from A's frame. Better do it right: use the relativistic addition of velocities formula. For low speeds only it does turn out that V_{B/A} = V_{B/road} + V_{road/A}.


Everything is different, not just light. Welcome to the 20th century!

But I don't want to use relativistic addition of velocities formulas.

 

[geistkiesel]

evenj if she accelerated?

Not only does she consider herself at rest when measuring the speed of light, she considers herself at rest period, because she is inertial.

If she measures her acceleration to .1c and then a partner on the stationary Earth frame sends her a light beam, that has a standard 10^-8m wavelength. She is heading into the oncoming beam. and measures the time it takes one lambda to pass her eye at 10^-8/3.3x10^8 = .3030 x 10^-16 sec., for a determined frequency of 3.3003 10^16/sec. If I hadn't used her velocity I would have calulated .3333x10^-16sec. for one wavelength to pass, for a frequency of 1/.333= 3.3003x10^16/sec. which is shorter than it physicall is measured. She calculates then .9090x10^8m for the wave length.

If she determines her relative velocity she calculates he wavelength as 3.3]3.3 = 10^-8m, he correct wavelength. She still gets her red shift , but she isn't imposing physical errors in measurement into the equation.
Calcukating the wavelength change by .3/3.3003 = .091x10^-8m. Add this to her calulated wavelength or ,9090 + .091 = 10^-8, the correct wavelength. I haven't changed the speed of light.



Well, since length contraction is actually a prediction of the theory, while "mass shrinking" is a stupid figment of some retard's imagination, yeah, I guess it would sound better!. [/QUOTE]

Ah yoo saeng i um ray tahded? Yes, but check with ram1024, you've only got another day of SR theorey left, unless he generously extends the deadline.

 

[Doc Al]

No and nly if you invoke some unwanted and unwarranted SR theory. This gedunken precedes Sr and should, like AE ofered it, be able to stand on its own two 1/3meters.

Show me what "unwanted and unwarranted" SR theory I invoked. It's just D = VxT, as seen by the O frame. That's all you're doing.

No doc, the times are measured using the passenger wrist watches. I specifically instructed all the passenger to first get a WWV time hack and use only the watches they wear. Of course the stationary observers do the same, but we aren't comparing stationary and moving frame clocks here. C is the speed of light c = 3x 10^8m/s. v is measured with train watches and distance measuring devices. t3 and t1 are moving frame times. This is how I set up th experiment. If you want to conduct the same experiment using stationary frame times, go for it, but only your opinion insists on using stationary times, your opinion, only.

Laughable. For one, I'll bet you haven't the foggiest notion of what the O' wristwatch time will read when that B photon hits him. It's certainly not what you call "t1". Go ahead, calculate it (in terms of L and v). I'll wait. (I'll give you a hint down below.)

Second, if you really meant that t1 and t3 are O' times, then your equation makes no sense. Remember you can't just ASSUME that O and O' clocks agree.

On what possible ground do you infer, that I infer, the times weren't moving frame times?

You are welcome to use the O' times, but then you'll have to use the O' frame. Can you handle it? But then your equation, which uses O-frame times, won't work any more. I'm probably giving you too much credit. You really don't have any idea what you're doing.

Also, at the velocities used here, you could never prove your SR time dilation SR stuff anyway, could you?. Gotcha Doc.

You still haven't graduated from Einstein's Train Gedanken academy yet. You are a long ways away from calculating time dilation.

Doc the times are from watches on the passengers on he trains, OK? Why are you persisting to swap my times on me? This isn't a Hidden "frame swap" of which you are getting such a rep for inserting into people's posts, is it?

If those times are O' watch times, then your equation makes no sense. You are the one "swapping frames": you try to use O' times in an equation derived using O times.

So what? Doc, check this out.
http://www.mathpages.com/rr/s4-08/4-08.htm
I thought I left you a message that I had found a twin expression in the link regarding simultaneity.

That's an article discussing relativity--too advanced for you. There is nothing wrong with your equation, it's just not what you think it is.

I'll give you a big hint. What time does the O frame say that photon B hits O'? For this, you can use your reasoning, since only O-frame measurements are involved. vt1 + ct1 = L, so t1 = L/(v + c). That's the time that the B photon hits O' according to the O clocks. Your assignment: figure out what the O' wristwatch reads when B photon hits him.

 

[Doc Al]

I used radar instrumentation in both cars, where the return signal is translated into relative speed, as when the fuzz is setting behind the big sign waiting for you to speed through her trap, her gun measures your velocity respecto hers, when she is stationary. But moving radar can be used, even though it is translated into velocity wrt the ground.

Right. Note that the speed is always measured with respect to the radar, then translated to be with respect to the ground. If the speeds were fast enough, that translation would require knowledge of SR.

But I don't want to use relativistic addition of velocities formulas.

I know, I know. It hurts so bad!
Pity that nature doesn't care what you want or don't want.

 

[geistkiesel]

She cares for me.

Right. Note that the speed is always measured with respect to the radar, then translated to be with respect to the ground. If the speeds were fast enough, that translation would require knowledge of SR.

I know, I know. It hurts so bad!
Pity that nature doesn't care what you want or don't want.

Nature cares what I want and don't want, that's why Mother Nature exposes herself so I can see how she dances, how she twirls and shows a flash or two of a well turned ankle, with just the slightest tinge of a blush on her cheeks. She is always there with crooked smile, but straight teeth, and a bent finger beckon to, slowly now, "come hither dear geistkiesel". She let's me spend the night with her, every beautiful night. And what a sweet dream it is.

 

[quantumdude]

Tom: We don't "omit" the observer's speed. In the observer's frame, he is at rest

Ram: which means the speed of ALL objects in the universe acquire the observer's velocity, LIGHT INCLUDED.

Wrong. I can't even imagine how you could end up at this ridiculous conclusion.

why would light behave differently?

Differently than what?

and don't say the data says so, because the data does NOT include the addition of observer motion.

The data says so. Deal with it.