# Today Special Relativity is still alive

Let us add Lorenz transformation to our "story".

Code:
observer1
|(->[COLOR=YellowGreen]__.__.__.__.__[/COLOR](o)[COLOR=YellowGreen]__.__.__.__.__[/COLOR]<-)|
|(->[COLOR=YellowGreen]__.__.__.__.__[/COLOR](o)[COLOR=YellowGreen]__.__.__.__.__[/COLOR]<-)|
|(->[COLOR=YellowGreen]__.__.__.__.__[/COLOR](o)[COLOR=YellowGreen]__.__.__.__.__[/COLOR]<-)|
|(->[COLOR=YellowGreen]__.__.__.__.__[/COLOR](o)[COLOR=YellowGreen]__.__.__.__.__[/COLOR]<-)|

observer2
|(->[COLOR=YellowGreen]__.__.__.__.__[/COLOR](o)[COLOR=YellowGreen]__.__.__.__.__[/COLOR]<-)|
|(->[COLOR=Orange]___.___.___.___._[/COLOR](o)->[COLOR=teal].__.__.__[/COLOR]<-)|
|(->[COLOR=Red]____.____.____.____.[/COLOR](o)->[COLOR=Blue]_._._.[/COLOR]<-)|
|(->[COLOR=DarkRed]______.______.______.__[/COLOR](o)->[COLOR=Magenta]...[/COLOR]<-)|
So there is no difference between observer1 and observer2 because space scale shrinks like a rubber sheet without losing its density ( exactly as (x/1)*(1/x) = 1 ) in the direction of the movement of observer2, and expends in the opposite side of observer2.

These opposite states can be observed as Doppler Effect.

So in both cases the speed of light is invariant in both directions.

Last edited:

Related Other Physics Topics News on Phys.org
Is there any comment?

Lama said:
Is there any comment?
Looks like you shrink space in the direction of movement but you expand it in the opposite direction. Shouldn't it shrink in both directions?

Lama said:
Is there any comment?
Take a look at simultaneity as defined by Einstein: Events simultaneous wrt stationary frames are niot simultaneous wrt moving frames. AE uses the train station moving train gedunken to show the photons are not detected simultabeously and says that The observers in the moving frame must therefore come to he conclusion that the photons were not emitted simultaneously in the moving frame.

Take a sitiuation where two photons emitted simultaneously at A and B arrive at the midpoint M of the sources of A and B. Just as the photons arrive they are deflected a few wave lengths into the moving frame (heading to the B photon) by mirrors (-->\/<--) where they had entered side by side from the stationary frame. There is no way the moving frame observer is able to determine, 1). Which photon came from which source, and 2) which photon was emitted first as the observers only see simultaneously emitted photons. Einstein's simultaneity definition is not physics, it's a crock.

Try to get around this one using mrore than the definition itself.

AE took the simple minded case where the photons were detected sequentially when the B photon was detected first followed by the A photon. AE claimed the observers' "must, therefore come to the conclusion" that the emitted photons were not emitted simultaneously in the moving frame. From this he strings us along with the discarding of absolute time, time dilation and all the other silliness of SFR.

It is a wonder this sham has lasted as long as it has,

Special Relativity died last week sometime.

wespe said:
Looks like you shrink space in the direction of movement but you expand it in the opposite direction. Shouldn't it shrink in both directions?
geistkiesel and wespe,

Then how can you explain Doppler Effect?

Is there a correlation between Doppler Effect and Lorenz transformation?

If not then please show it, thank you.

Last edited:
Lama said:
geistkiesel and wespe,

Then how can you explain Doppler Effect?

Is there a correlation between Doppler Effect and Lorenz transformation?

If not then please show it, thank you.
Something tells me lama that you either didn't read or understand my post or you are ignoring me, all iof the above ok. But first things first, ok? No use going around in circles when it isn't necessary.

Einstein's gedunken experiment of photons from sources A and B ossucred just as the moving observer O' was at the midpoint iof the photon sources.The train moving along detects the B photon first as she is moving to the B source. Later the photon from A arrived. that the photons did not arrive simultaneoiusly in the moving frame Einstein stated that the observers in the moving frame "must" conclude the photons were not emitted simultaneously in the moving frame.

Code:
A _________________M__________________B
O'B       O'A
The diagram shows the B photon arriving before the A photon

If we modify the experiment and place two mirrors at M in the stationary frame and divert the photons to the moving frame when they arrive, the photons will be emitted simultaneously in the moving frame. This kills, absolute time, time dilation, shrinking and the whole nine yards of SR theory..

Code:
A-->\/<--B
____||_____
####### detectors in the moving frame.
If the photons arrive simultaneously in the moving frame, there is no Lorentz transformation, time dilation, shrinking, releative time or Special Relativity.

So before we disuss Lorentz you must tindicate if the photons arrive simultaneously in the moving frame where the detectors are. The photons have just been reflected by mirrors in the stationary frame. See the difference from the first example where the photons arrived sequentially?? or are you a believer? In the second we just let the photons go to the midpoint and reflect them into the moving frame. Simultaneous or not? It looks like it doesn't it?

Doc Al
Mentor
geistkiesel's mirror illusion

geistkiesel said:
Something tells me lama that you either didn't read or understand my post or you are ignoring me, all iof the above ok. But first things first, ok? No use going around in circles when it isn't necessary.
Talk about going around in circles!

On the off chance that someone doesn't spot geistkiesel's latest error, I explain it here (starting with post #41): https://www.physicsforums.com/showthread.php?p=248152

Last edited:
Janus
Staff Emeritus
Gold Member
Lama said:
geistkiesel and wespe,

Then how can you explain Doppler Effect?

Is there a correlation between Doppler Effect and Lorenz transformation?

If not then please show it, thank you.
Doppler shift can be explained simply through the invarience of the speed of light.

look at this first animation:

http://home.teleport.com/~parvey/doppler1.gif [Broken]

It shows a lightsource stationary between two observers. The different colored bands indicate different points of each light wave. Note that the waves expand outward as a circle at c and each wave is an equal distance from the previous one.

Now check out the second animation:

http://home.teleport.com/~parvey/doppler2.gif [Broken]

This shows the situation when the lightsource and observers are moving relative to each other. The view is from the point of the observers. Remember, according to the postulates of SR, the speed of light is invarient for all observers, which means that every observer must measure the speed of light as c relative to himself. Thus when the front of the wave is emitted it expands at c from the point of emission in a circle according to the observers. Now as the next part of the wave is emitted it also expands as a circle for the point of emission, but since the distance between the lightsource and the observer has nowchanged, it is emitted from a different point then the fornt of the wave was. Each succesive part of the wave is emitted from a slightly different position than the part before it. they get closer to the Blue observer and further from the Red observer. So what you get is a series of circular rings, each with its center shifted to the right in the image.

As a result, the waves are bunched up when they reach the blue observer and he sees a blue shift in the light(since the later parts of the waves were emitted when they were closer to the blue observer they follow closer behind the earlier parts of the waves). For the red observer the waves are stretched out and he sees a redshift.

Last edited by a moderator:
Janus said:
Now check out the second animation:

http://home.teleport.com/~parvey/doppler2.gif [Broken]
It is a beautiful animation of Doppler Eeffect, thank you.

So, if we move at the speed of light, then there are no waves infront of us for example:

Code:
Before the source of light moves
+-------------+
|+-----------+|
||+---------+||
|||+-------+|||
||||+-----+||||
A    |||||+---+|||||    B
*    ++++++ * ++++++    *
|||||+---+|||||
||||+-----+||||
|||+-------+|||
||+---------+||
|+-----------+|
+-------------+

When the source of light moves at the speed of light
+-------------+
| +-----------+
| | +---------+
| | | +-------+
| | | | +-----+
A    | | | | | +---+    B
*    + + + + + + * +->  *
| | | | | +---+
| | | | +-----+
| | | +-------+
| | +---------+
| +-----------+
+-------------+
But we also know that Lorenz trasformation cannot be ignored at the speed of light, so can you please draw the connection between Lorenz trasformation and Doppler Eeffect?

Is Lorenz trasformation takes place only in the source of light and only if it has a mass > 0 ?

Last edited by a moderator:

Janus said:
Doppler shift can be explained simply through the invarience of the speed of light.

look at this first animation:

http://home.teleport.com/~parvey/doppler1.gif [Broken]

It shows a lightsource stationary between two observers. The different colored bands indicate different points of each light wave. Note that the waves expand outward as a circle at c and each wave is an equal distance from the previous one.

Now check out the second animation:

http://home.teleport.com/~parvey/doppler2.gif [Broken]

This shows the situation when the lightsource and observers are moving relative to each other. The view is from the point of the observers. Remember, according to the postulates of SR, the speed of light is invarient for all observers, which means that every observer must measure the speed of light as c relative to himself. Thus when the front of the wave is emitted it expands at c from the point of emission in a circle according to the observers. Now as the next part of the wave is emitted it also expands as a circle for the point of emission, but since the distance between the lightsource and the observer has nowchanged, it is emitted from a different point then the fornt of the wave was. Each succesive part of the wave is emitted from a slightly different position than the part before it. they get closer to the Blue observer and further from the Red observer. So what you get is a series of circular rings, each with its center shifted to the right in the image.

As a result, the waves are bunched up when they reach the blue observer and he sees a blue shift in the light(since the later parts of the waves were emitted when they were closer to the blue observer they follow closer behind the earlier parts of the waves). For the red observer the waves are stretched out and he sees a redshift.
There are other doppler models. The best is the one based on the addition of the observes speed to the velocity of a wavelength of light. When the wave pass through the eye they pass through at v = 3x10^8m/s. or c..

Now as the observer moves against the stream the obsevers velocity is added. to detemine the realtive velocity of observer and wave front. Assume we have a wave length measure at 10^-8 meters. The the wave will pass the eye in 10^-8/3x10^8m/s = .333x 10^-16 seconds for a frequency of 1/.333x10^-16 = = 3.003x10^16/sec. When we add the observers velocity say.05c = .015x10^8 for a total relative velocity of 3.015x 10^8m/s.
To find the new frequency we divide the velocity by the wavelength to arrive at (3.015x10^8m/s)/(10^-8m) = 3.015X10^16/S.

If we did not consider the added velocity we would determine a wavelength of (3x10^8m/s)/(3.015x10^16/s) = .995x10^-8m, a shortened wave length that will always result in reading the velocity of light as c. here shortened by .005x10^-8. which we may get from the obseves velociy and frequency .015x10^8m/s)/3.015x10^16/s) = .0049x10^-8 which when added to the calulated wavelength with no relative observers velocity .995 + .0049 ~ 1x10^-8 meters.

Frequency gets one the redshift.

The red shift comes, not from the contracted wave length but from the increased frequency due to the added velocity of the observer.

Last edited by a moderator:

Thank you very much, wespe.

russ_watters
Mentor
geistkiesel said:
There are other doppler models. The best is the one based on the addition of the observes speed to the velocity of a wavelength of light.
...which doesn't work because we know C is invariant.

Alkatran
Homework Helper
geistkiesel said:
There are other doppler models. The best is the one based on the addition of the observes speed to the velocity of a wavelength of light. When the wave pass through the eye they pass through at v = 3x10^8m/s. or c..

Now as the observer moves against the stream the obsevers velocity is added. to detemine the realtive velocity of observer and wave front. Assume we have a wave length measure at 10^-8 meters. The the wave will pass the eye in 10^-8/3x10^8m/s = .333x 10^-16 seconds for a frequency of 1/.333x10^-16 = = 3.003x10^16/sec. When we add the observers velocity say.05c = .015x10^8 for a total relative velocity of 3.015x 10^8m/s.
To find the new frequency we divide the velocity by the wavelength to arrive at (3.015x10^8m/s)/(10^-8m) = 3.015X10^16/S.

If we did not consider the added velocity we would determine a wavelength of (3x10^8m/s)/(3.015x10^16/s) = .995x10^-8m, a shortened wave length that will always result in reading the velocity of light as c. here shortened by .005x10^-8. which we may get from the obseves velociy and frequency .015x10^8m/s)/3.015x10^16/s) = .0049x10^-8 which when added to the calulated wavelength with no relative observers velocity .995 + .0049 ~ 1x10^-8 meters.

Frequency gets one the redshift.

The red shift comes, not from the contracted wave length but from the increased frequency due to the added velocity of the observer.
The observer can only make measurements from his own frame. The observer is stationary in this frame. Light moves at c in all frames. Light moves at c relative to the observer.

The observer can't have an "added" velocity on the light because of that.

Alkatran said:
The observer can only make measurements from his own frame. The observer is stationary in this frame. Light moves at c in all frames. Light moves at c relative to the observer.

The observer can't have an "added" velocity on the light because of that.
A moving observer is moving. Train stations do not move or accelerate only in the shallow minds of people unable to see what reality is and anot e what insane mathematical aberrations try to make it. Trains and rocket ships do move and for your theoretical nonsense that assumes otherwise is, well it is just that insanity, total, insanity. like the logic in the the following: d;sadkaslm- x = = <sd>/(-owe*mm/lllcvma;lcv dls';.cf,v/s=[e4-34.

Your statement that moving observers don't move is stupidity and SR BS. Peace to you poor lost pilgrim.

jcsd
Gold Member
So, is the train station on the Earth which rotating and orbiting around the sun which in turn is orbiting around the centre of the galaxy which in turn has it's on movemnets. Clearly it is very arbitary to declare that train sations are always static.

Alkatran
Homework Helper
geistkiesel said:
A moving observer is moving. Train stations do not move or accelerate only in the shallow minds of people unable to see what reality is and anot e what insane mathematical aberrations try to make it. Trains and rocket ships do move and for your theoretical nonsense that assumes otherwise is, well it is just that insanity, total, insanity. like the logic in the the following: d;sadkaslm- x = = <sd>/(-owe*mm/lllcvma;lcv dls';.cf,v/s=[e4-34.

Your statement that moving observers don't move is stupidity and SR BS. Peace to you poor lost pilgrim.
You can't move relative to yourself. What makes a train station so special that it NEVER EVER moves? Last I checked they were on a planet that's spinning around.

I like how you completely proved me wrong by randomly typing on your keyboard .

Last edited:
Alkatran said:
You can't move relative to yourself. What makes a train station so special that it NEVER EVER moves? Last I checked they were on a planet that's spinning around.

I like how you completely proved me wrong by randomly typing on your keyboard .
Take some dancing instructions. You know what I mean when I say that trains move and that train statrions are stationary. One cannot reasonably and in conformity with the laws of physics stop and start frames as easily one thinks SR theory allows. :rofl:

So when you are doing a train and train station problem do you project the train station is moving with the vlocity of the earth in space as you hypothecized sin this post?? No you do not. Thereofere you shouldn't use that example you used, hence the need for dancing lessons. Do you also stop the train to dead in the water absolute v -= 0, no you do not. Your SR math sucks, not that it is yours, only the one you adopted. Get into gear, and like i saii take some dancing lessons, you are out of step as indicated by this post. You are headed in the right direction by dealing with questions of SR theory, You should just not assume SR is gospel until you prove it to yourself as opposed to having it proved to you by someone else. Do you see the difference? good.

Alkatran said:
The observer can only make measurements from his own frame. The observer is stationary in this frame. Light moves at c in all frames. Light moves at c relative to the observer.

The observer can't have an "added" velocity on the light because of that.
You mean the eart6h isn't moviang through space, or you can't be moving in a satellite in orbit? SR says you can't add velocities huh? SR is wornmg. Just do it you will like it. Read Grounded's posts especially his original post opening his thread.

russ_watters
Mentor
geistkiesel said:
You mean the eart6h isn't moviang through space, or you can't be moving in a satellite in orbit? SR says you can't add velocities huh? SR is wornmg. Just do it you will like it. Read Grounded's posts especially his original post opening his thread.
No one said anything about how you add the velocities of objects. Its the speed of light that is invariant.

Alkatran
Homework Helper
geistkiesel said:
Take some dancing instructions. You know what I mean when I say that trains move and that train statrions are stationary. One cannot reasonably and in conformity with the laws of physics stop and start frames as easily one thinks SR theory allows. :rofl:

So when you are doing a train and train station problem do you project the train station is moving with the vlocity of the earth in space as you hypothecized sin this post?? No you do not. Thereofere you shouldn't use that example you used, hence the need for dancing lessons. Do you also stop the train to dead in the water absolute v -= 0, no you do not. Your SR math sucks, not that it is yours, only the one you adopted. Get into gear, and like i saii take some dancing lessons, you are out of step as indicated by this post. You are headed in the right direction by dealing with questions of SR theory, You should just not assume SR is gospel until you prove it to yourself as opposed to having it proved to you by someone else. Do you see the difference? good.
You mean the eart6h isn't moviang through space, or you can't be moving in a satellite in orbit? SR says you can't add velocities huh? SR is wornmg. Just do it you will like it. Read Grounded's posts especially his original post opening his thread.
Alright, your obviously grasping here because you seem to deem it necessary to constantly insult the people arguing with you. Did I say that objects can't move? NO! I said they can't move "Relative to themselves". And by that I don't mean your arms can't move either, because your arm is an object too, and the cells it is made off, and the elements those are made off... ETC

And yes, if need be, the train is considered at 0 and everything else is moving around it. This might not work in general relativity (the rotating earth) since I only know special.