This is the crux of your misunderstanding! You are taking the speed of light as measured it in one frame, and then you are applying it to another frame that is in motion. They measure the speed of light to be the same in both frames! You cannot say that the photon travels at a different relative speed because it measures the speed of light to be a certain speed in another frame. This is basically what you are doing. You say, it is this speed in one frame, therefore another frame measures a difference in relative speed. This is just the wrong way to think about it. It will be the speed of light in one frame, and it will be the speed of light in another frame that is in motion. These two frames can't say that the other frame has to have relative motion to the photon because of another frame measuring it to be one speed.
On the idea of time in physics-relativity
- Thread starter ash64449
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The speed of light in vacuum is the SAME in ALL frames. This is known fact for the last 108 years.
Here is a very good explanation for MMX. You need to learn math and SR in order to understand it.
No that is precisely what I am NOT doing I am simply say the speed of light will be measure as c in both frames!
Yes each frame measure the speed at c.
Don't you see that the RELATIVE speed between the ship and the light is not c for the stationary observer. That is right the relative speed between the ship and the light is not c for the stationary observer. It CAN'T be because the speed of light is always meausred as c.
That is exactly what I am doing. Saying the relative speed, for the stationary observer, between the ship and the light is less than c.
Exactly!!!!!! The MM experiment says nothing about relative speeds between frames it only says that the speed of light is constant for all frames.
If a ship moving at .9 c shoots a laser out the front of the ship what would a stationary observer see? He would see the beam moving at a relative velocity to the ship of .1 c. Why? because the light is always measured as c. So the relative velocity is .1c.
What possible alternative is there??? For a stationary observer, if the relative velocity between the ship and the light beam is c then the ship must not be moving. For a stationary observer if the ship is moving and the relative speed between the light and the ship is c the the speed of light is NOT c.
Both observers will say the light sphere expanded outward from the source at a uniform rate of c. Your assuming that a prefered frame that is at rest, like an absolute frame (like your frame of mind), is the only frame that says this. Both frames must be able to say this at the same time!
You must be going to the same "school" as Motor Daddy.
Say there is a space ship that has one headlight on the side of it. The ship is in motion and then the headlight sends a beam of light in a direction perpendicular to its direction of motion. The observer on the ship then measures the beam to travel a distance according to his own clock, d' = c t'. It then passes by a space station that then notices that the beam of light from the ship being sent perpendicular to its direction of motion travels at the speed of light, but then has a vector of the velocity of the ship. It travels at an angle, so that the beam moves forward in the direction of motion of the ship, and has a vector in this direction that is the same speed of the ship. It is then measured to travel at an angle according the stations clock that is at rest, d = c t. The ship is also measured to be traveling with a velocity according to the stations clock that is at rest, d = v t.
Then you have three distances, and then you can translate the distance the ship has moved down to the location where the beam of light has reached. This then forms a right triangle. You can then calculate the difference of the two clocks as they measure the single beam that is at the same location relative to both frames using the Pythagorean Theorem, $$ a^{2} + b^{2} = c^{2} $$
$$ ( v t )^{2} + ( c t' )^{2} = ( c t )^{2} $$
$$ c^{2}t'^{2} = (c^{2}t^{2} - v^{2}t^{2}) $$
$$ c^{2}t'^{2} = c^{2}t^{2} ( 1 - \frac{v^{2}}{c^{2}}) $$
$$ t'^{2} = t^{2}( 1 - \frac{v^{2}}{c^{2}}) $$
$$ t' = t sqrt{ 1 - \frac{v^{2}}{c^{2}} $$
So then the proper time, is the amount of time both observers would measure so that they both measure the same speed of light while they measure the same beam of light at the same location simultaneously.
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If the dog barks continuously, at some point of time, explosion of the star and barking of dog can be simultaneous.
Clock only measures 'local time'.
Yeah. I have said that about 50 times.
I have neither said nor implied anything about a prefered frame. You are just too stupid, apparently to realize that there are different inertial frames and what that implies. I thought that it just might be possible to educate you at least a tiny bit, but alas I realize that your delusion is too strong. I guess if you admitted that you are wrong then you would have to abandon this delusion that you are some superior mind that sees the truth that the rest of the world is not privy to. I'm done, I have finally accepted that you can't teach anything to a box of rocks.:shrug:
So all I can say is revel in your ignorance!
You know what they say, ignorance is bliss!
Which keeps you happy to our dismay.
Physical Mathematics likes to think , well things do what they do because of mathematics
Yet the truth is , without the objects mathematics has nothing on which to base anything theory of relativity on
Time is the study of the movement of any object(s) in space and why the object(s) moves in the first place
Yet the truth is , without the objects mathematics has nothing on which to base anything theory of relativity on
Time is the study of the movement of any object(s) in space and why the object(s) moves in the first place
I still think I would rather be a real scientist and study quantum mechanics.
If they happen at the exact same time according to the observer, the observer will say they were simultaneous. But with a little more information he will quickly realize that out where the star happened to explode, no dog was yet existing to bark at its demise. Therefore, simultaneity is relative, wouldn't you agree?
In measuring simultaneity, would it matter what time of day the clock reads, or how fast it's ticking? No. All that's necessary is to determine that no delay has elapsed between the two events. An actual clock or timebase wouldn't necessarily be required since this is a gating function with a simple positive or negative result.
Write4U
Valued Senior Member
This made it clear to me.
In the picture with stationary observer O, the train is not relevant at all. The scene depicts two light flashes with a stationary observer in the exact middle.
Now, removing O from any consideration
PL, are you saying that O' does not receive the flashes at the same time, due to the movement of the observer O'?
In the picture with the moving observer O', the train is only relevant as to speed of O'. As soon as the light flashes, O' s position changes before he receives the light, thus there will either be a non-simultaneous reception by O' or because of the constancy of SOL, O' may receive the light simultaneous, but each flash will present a Doppler shift to O'.
Seems to me the latter would occur. Can anyone confirm this?
In the picture with stationary observer O, the train is not relevant at all. The scene depicts two light flashes with a stationary observer in the exact middle.
Now, removing O from any consideration
PL, are you saying that O' does not receive the flashes at the same time, due to the movement of the observer O'?
In the picture with the moving observer O', the train is only relevant as to speed of O'. As soon as the light flashes, O' s position changes before he receives the light, thus there will either be a non-simultaneous reception by O' or because of the constancy of SOL, O' may receive the light simultaneous, but each flash will present a Doppler shift to O'.
Seems to me the latter would occur. Can anyone confirm this?
No, I am saying that it does receive the flashes at the same time, it is just the clocks of each observer read a different time when this event happens. This is because the beams in the MME arrive at the same time in the detector in the end of the experiment even when it is in motion, relative to both frames. Each frame just measures a different amount of time on his own clock for this to happen. So even though they measure different times on their own clock, the event still happens at the same time, but their clocks don't read the same time.
Like in my post #266, they both say the photon has traveled a certain distance, and they both say it has reached a location at the same time, but from solving the equation you find that when this event takes place, they don't agree on their own times by the ticks of their clocks. So then it is like the proof is asking what time does both clocks read when two different observers measure a beam to be at the same location at the same time.
They will not have a doppler shift relative to O' as observed from O'. They will only have a doppler shift relative to O from a beam in O'.
I will put it this way, if the MME is O' it will have the same results as when it is O. Relative motion doesn't alter the results of the MME. In Einsteins thought experiment it makes it sound like they don't reach at the same time, I am saying that they arrive at the same time, therefore time dilation must take place so that they do. Their clocks have to read differently in order for the beams to arrive at the same time from both frames. So they have to experience a different time inorder for the event to be simultaneous.
Neddy Bate
Valued Senior Member
As everyone has told you already, the train thought experiment is not quite the same thing as the MMX.
What I meant was that one cannot draw a diagram that describes your belief that light does not behave logically. Logic tells us that the moving observer moves toward the right light ray, thereby encountering it sooner. Logic also tells us that the moving observer moves away from the left light ray, thereby encountering it later. Have you tried drawing a diagram showing both light rays reaching the moving observer at the same time?
As a matter of fact they have, but I don't see how the differences are relavent to the issue. The MME showed that there is no aether.
I have read about Minkowski Diagrams but I have never actually worked any of them. Have you ever worked any Minkowski Diagrams ? Seemed like a smart guy, was just reading about him and he was Einsteins teacher.
Write4U
Valued Senior Member
One thing I am not clear about. The thought experiment proposes a simultaneous flash of light at two specific spacetime coordinates.
When the light flashes (simultaneously) does it make a difference if the train is in motion or if there is a train at all?
IMO, the relevancy of the train is that it moves the observer (O') who is ON the train, nothing else.
When the light flashes can this event have momentum? Seems to me a flash is
(a) stationary event at a specific spacetime coordinate.
(b) even in motion, it still occurs at a specific spacetime coordinate and is a stationary event.
Only the observer O' (on the moving train) has a relative movement to the light flashes. But that only creates a frequency shift, no? It does not affect when the light is perceived.
But allow me a introduce a condition which seems to be missing from all illustrations. No one has paid any attention to the distance each observer is removed from the points of the light flash.
If the train were 200' long, observer O' (in the middle of the train) would be 100' removed from both flashpoints.
However, observer O (by virtue of standing alongside the tracks) would be more than 100' removed from both flashpoints.
It seems reasonable to expect each observer to receive the light flashes at different times (even at SOL).
LOL, the introduction of a moving train is just to make the problem a little more difficult....:shrug:
When the light flashes (simultaneously) does it make a difference if the train is in motion or if there is a train at all?
IMO, the relevancy of the train is that it moves the observer (O') who is ON the train, nothing else.
When the light flashes can this event have momentum? Seems to me a flash is
(a) stationary event at a specific spacetime coordinate.
(b) even in motion, it still occurs at a specific spacetime coordinate and is a stationary event.
Only the observer O' (on the moving train) has a relative movement to the light flashes. But that only creates a frequency shift, no? It does not affect when the light is perceived.
But allow me a introduce a condition which seems to be missing from all illustrations. No one has paid any attention to the distance each observer is removed from the points of the light flash.
If the train were 200' long, observer O' (in the middle of the train) would be 100' removed from both flashpoints.
However, observer O (by virtue of standing alongside the tracks) would be more than 100' removed from both flashpoints.
It seems reasonable to expect each observer to receive the light flashes at different times (even at SOL).
LOL, the introduction of a moving train is just to make the problem a little more difficult....:shrug: