Speed of light for different observers

[sisoev]

You need to have two paths with a phase difference arriving at the same detector in order to get interference. There is only one path for each detector, so no interference.

OK, I know that, but there are two slits for each detector.
If we repeat numerous times the experiment will we see the same interference on both detectors?

About the red light shift:
I don't understand very well the frequency of a photon and I'd appreciate some help here.
If the photon travels same distance for A and B in my experiment and it does it for the same time, how that would change its frequency for detector B (you already agreed in a previous post that the photon will hit detector B with less energy)
Would you explain, please?

No, why would the color have anything to do with talking about simultaneity? The doors can be opened simultaneously regardless of their colors. I think that this whole line of thought is based on the misconception that relativity is about appearances.

I think you did not understand the "purple door" experiment.
If we set an experiment to show simultaneity we set that simultaneity between the opening of a blue and red doors.
If you say that you saw the purple door to open later that would only prove that your observation of the event was wrong because there is no purple door in the experiment setting.
If we don't take in consideration the properties of the objects in one experiment we cannot get the right results from that experiment.

Note that in the first question from above I agreed with you that the photon traveled same distance for A and B. I agreed with the "purple door" for the sake of the conversation, but the experiment is set with the full length of the truck and just because we see it differently doesn't mean that the light didn't travel the full truck length

 

[Dale]

OK, I know that, but there are two slits for each detector.
If we repeat numerous times the experiment will we see the same interference on both detectors?

OK, if there is a two-slit apparatus between the source and the detectors then that gives you the two paths required for interference. The fringes will be different because the frequency is different for each detector due to Doppler shift.

About the red light shift:
I don't understand very well the frequency of a photon and I'd appreciate some help here.
If the photon travels same distance for A and B in my experiment and it does it for the same time, how that would change its frequency for detector B (you already agreed in a previous post that the photon will hit detector B with less energy)
Would you explain, please?

Do you understand the Doppler shift? If so, that is all this frequency shift is. The distance traveled doesn't matter, only the relative speed between the source and the detector.

http://en.wikipedia.org/wiki/Doppler_effect
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop2.html
http://www.fourmilab.ch/cship/doppler.html

I think you did not understand the "purple door" experiment.
If we set an experiment to show simultaneity we set that simultaneity between the opening of a blue and red doors.
If you say that you saw the purple door to open later that would only prove that your observation of the event was wrong because there is no purple door in the experiment setting.

So what? We know that the speed of light is finite. We account for that in determining simultaneity.

Note that in the first question from above I agreed with you that the photon traveled same distance for A and B. I agreed with the "purple door" for the sake of the conversation, but the experiment is set with the full length of the truck and just because we see it differently doesn't mean that the light didn't travel the full truck length

You are simply factually wrong on this point. Not only are you wrong according to SR, but you are also wrong according to Newtonian mechanics.

Let a light pulse be emitted from the source at x=0, t=0. Then the equation of motion for the light pulse is x=ct. If the truck is of length L then the equation of motion for the detector is x=L-vt. That is two equations in two unknowns so we can easily solve to obtain x=L/(1+v/c) and t=L/(c+v). Clearly x is smaller than L for any v>0.

Please stop repeating this error.

 

[sisoev]

OK, if there is a two-slit apparatus between the source and the detectors then that gives you the two paths required for interference. The fringes will be different because the frequency is different for each detector due to Doppler shift.

Thank You.

Do you understand the Doppler shift? If so, that is all this frequency shift is. The distance traveled doesn't matter, only the relative speed between the source and the detector.

http://en.wikipedia.org/wiki/Doppler_effect
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop2.html
http://www.fourmilab.ch/cship/doppler.html

I understand very well the Doppler shift.
It is because every next successive wave is approaching us later because of the increased distance (red shift)
I cannot connect it with the photon frequency though, since the photon is one wave.
The moment it hits the detector it stop existing for it as wave = one successive wave.
That's why I said that I don't understand very well the photon frequency, when compared to light frequency, especially in connection with Doppler shift.

So what? We know that the speed of light is finite. We account for that in determining simultaneity.

And does that turns the door from purple to blue?

You are simply factually wrong on this point. Not only are you wrong according to SR, but you are also wrong according to Newtonian mechanics.

Let a light pulse be emitted from the source at x=0, t=0. Then the equation of motion for the light pulse is x=ct. If the truck is of length L then the equation of motion for the detector is x=L-vt. That is two equations in two unknowns so we can easily solve to obtain x=L/(1+v/c) and t=L/(c+v). Clearly x is smaller than L for any v>0.

Please stop repeating this error.

I'm trying very hard, DaleSpam (to stop repeating the errors)
My thousands apologies for me being that annoying.

I don't understand why we have to make calculations if all parameters are given in the experiment.
It is irrelevant what we observe, from which coordinate system and from which point in it we observe it.
The light will travel the full length of the truck.

By the way, why no one attempted to use length contraction in explaining this experiment?

 

[PAllen]

By the way, why no one attempted to use length contraction in explaining this experiment?

What??! See my long post on a numeric example of the OP. It used length contraction as part of the comparison of different frame's measurements.

 

[sisoev]

What??! See my long post on a numeric example of the OP. It used length contraction as part of the comparison of different frame's measurements.

Sorry, Pallen.
I missed it.
I'll read it tomorrow.
It is already 00:14 here.
Time for bed.

Thank You guys.
Good day and good night :)

 

[Dale]

the photon is one wave.

Where did you get that idea? It is not generally correct.

I don't understand why we have to make calculations if all parameters are given in the experiment.

It is often possible to make qualitative predictions without making any calculations, but quantitative predictions always require calculations. The quantitative predictions can also be used to confirm or correct the qualitative predictions.

The light will travel the full length of the truck.

No it won't, as I have both explained and derived. Again, please stop repeating this same error.

By the way, why no one attempted to use length contraction in explaining this experiment?

I mentioned it earlier. But length contraction is a second order effect, and you are not even correctly analyzing the first order (Newtonian) effects. Once you learn Newton then we can start working on Einstein.

 

[harrylin]

[..] I don't understand why we have to make calculations if all parameters are given in the experiment.
It is irrelevant what we observe, from which coordinate system and from which point in it we observe it. [..]

In my experience - and probably most others who studied physics - without also doing calculations (and/or graphical constructions), I could not get a full and thorough understanding of such complex matters. Consequently, I'm afraid that you cannot either.

 

[sisoev]

Where did you get that idea? It is not generally correct.

The above is answer to my understanding that the photon represents one wave.
Perhaps my terminology is wrong again
As I understand it, according to complementarity principle a photon can behave as either particle or wave, but not both at the same time.
Why would we say that the photon is not one successive wave, since it "dies" as wave the moment it exhibits its particle behavior by hitting the detector.

It is often possible to make qualitative predictions without making any calculations, but quantitative predictions always require calculations. The quantitative predictions can also be used to confirm or correct the qualitative predictions.

Fully agree with you.
In our case we don't have to predict the path of the light for observer A. It is given; from one side of the truck to the other.
I account the fact that for a third observer the light travels from the place of emission to the detector, but in the same time I account the other fact, that for observer A the point of emission moved with the truck.
For A we are moving and for us A is moving.
In this case, should we not compare both light paths as in stationary frames?

No it won't, as I have both explained and derived. Again, please stop repeating this same error.

OK, it is equally annoying for me as it is for you, to go in a loop.
I consider the possibility that I'm a little slow in understanding, but do you think that your explanation is good enough for my question.
I understand where the problem is, and I'll try later to put the question in a better way.

I mentioned it earlier. But length contraction is a second order effect, and you are not even correctly analyzing the first order (Newtonian) effects. Once you learn Newton then we can start working on Einstein.

Kind of agree on this

 

[sisoev]

In my experience - and probably most others who studied physics - without also doing calculations (and/or graphical constructions), I could not get a full and thorough understanding of such complex matters. Consequently, I'm afraid that you cannot either.

Ha-ha
Obviously I have a problem, otherwise I wouldn't be here, asking "stupid" questions ;)

 

[Dale]

The above is answer to my understanding that the photon represents one wave.
Perhaps my terminology is wrong again
As I understand it, according to complementarity principle a photon can behave as either particle or wave, but not both at the same time.
Why would we say that the photon is not one successive wave, since it "dies" as wave the moment it exhibits its particle behavior by hitting the detector.

If you have a detector which detects the frequency of the photon then you will measure a Doppler shift. Personally, I think you need to focus on classical physics and not even worry about quantum effects for now. Once your grasp on classical physics is firm then learning QM is feasible.

In our case we don't have to predict the path of the light for observer A. It is given; from one side of the truck to the other.

It is given that it goes from the source at the time of emission to the detector at the time of detection. That distance is not equal to the length of the truck in any frame where the truck is moving, as explained and derived.

do you think that your explanation is good enough for my question.

Yes, I explained it multiple times, pointed out the error in your reasoning, and derived the correct result. That seems to be a complete explanation.

Before you pursue this line of discussion any further, spend a little effort of your own. Sit down and take a close look at each step of my brief derivation. Either the derivation is right or it is wrong. If it is right then you have no choice but to acknowledge that the distance traveled by the light is not equal to the length of the truck in any frame where the truck is moving. If it is wrong, then point out the specific error, and post the correct derivation which supports your claim quantitatively.

 

[Agerhell]

It is not possible to determine or measure the one-way speed of a light pulse or a photon. We can only measure the round-trip time it takes for a light pulse to start from a source, traverse to a mirror, reflect off the mirror and traverse back to the source. This is experimental evidence for the universal constant value of the speed of light.

In Einstein's Special Relativity, a Frame of Reference is defined in which the two halves of the trajectory of the aforementioned experiment are assigned equal times. This is Einstein's second postulate. Read his 1905 paper.

I think that the GPS-system is very much based upon knowing the one-way speed of light... By knowing where the satellites were when they send their signals and the one-way speed of light your gps can tell you where your are... I am not really a GPS-expert though...

If you want simultaneity to be universal, then you need to promote the Lorentz Ether Theory in which there is a preferred reference frame, although no one knows where it is.[...]
You have two choices: you can claim that there exists an absolute ether rest state in which the speed of light is exclusively constant in all directions and for which times, distances, and simultaneities are absolute, or you can claim that all rest states are equally valid and define times, distances and simultaneities relatively according to the definition of anyone of those rest states.

I am guessing that the GPS-system is actually treating the frame of the centre of the Earth as a preferred frame... Even though the Earth moves by 30 km per second around the sun, this is not something that have to be accounted for by the GPS-recievers... If you assume that the velocity of light is the same in all directions with respect to the centre of the Earth but not with respect to any other inertial frame you get the correct results... Both for atomic clocks and the GPS-system, treating the centre of the Earth as a preferred frame, seem to give the correct results, in agreement with observations.

 

[harrylin]

I think that the GPS-system is very much based upon knowing the one-way speed of light... By knowing where the satellites were when they send their signals and the one-way speed of light your gps can tell you where your are... I am not really a GPS-expert though...

It is very much based on assuming the one-way speed of light, just as gwellsjr explained. After setting the clocks according, one simply measures what one first assumed.

I am guessing that the GPS-system is actually treating the frame of the centre of the Earth as a preferred frame... [..]

It treats the ECI frame as an inertial frame (which is not exactly true, but good enough), as defined in classical physics and SR: in any inertial frame the laws of physics are supposed to work. Thus GPS should also work on Mars, using the centre of Mars frame. Note that also corrections are used for gravitational potential.Harald

 

[PAllen]

It is very much based on assuming the one-way speed of light, just as gwellsjr explained. After setting the clocks according, one simply measures what one first assumed.




Harald

Ok I've had enough of this overly strong claim. What about slow clock transport rather than light based clock synchronization? That this comes out the same as light synchronization is a strong physical statement (that uniquely specifies only LET theories that are completely indistinguishable from SR can be true; they must effect all possible physical laws the same way as a spacetime symmetr).

Anyway, by normal standards of logic, slow clock transport is independent of any signal speed, and combined with distance measurement based on e.g. rulers, allows measurement of one way lightspeed.

 

[Dale]

I think that in any theory with a one-way speed of light not equal to c that the net time dilation effect would not asymptotically approach 0 as v approached 0. Whether that has some deep significance I am less sure.

 

[harrylin]

Ok I've had enough of this overly strong claim. What about slow clock transport rather than light based clock synchronization? That this comes out the same as light synchronization is a strong physical statement (that uniquely specifies only LET theories that are completely indistinguishable from SR can be true; they must effect all possible physical laws the same way as a spacetime symmetr).

Anyway, by normal standards of logic, slow clock transport is independent of any signal speed, and combined with distance measurement based on e.g. rulers, allows measurement of one way lightspeed.

Sorry, I don't see the problem here.

First, this discussion relates to the fact that according to SR, for any inertial frame that we may choose, we "assume that the clocks can be adjusted in such a way that the propagation velocity of every light ray in vacuum - measured by means of these clocks - becomes everywhere equal to a universal constant c, provided that the coordinate system is not accelerated".

And slow clock transport doesn't affect that assumption: it doesn't in any way indicate anything "true" or "absolute" about the clock synchronization that we happened to choose. As you certainly know, we can pick any inertial reference frame wrt which we define the speed of light to be equal in both directions; that includes frames in which all our "slowly moving" clocks are moving very fast in one direction, and in which the (closing) speed of light relatively to us is very different in different directions. Harald

 

[sisoev]

If you have a detector which detects the frequency of the photon then you will measure a Doppler shift. Personally, I think you need to focus on classical physics and not even worry about quantum effects for now. Once your grasp on classical physics is firm then learning QM is feasible.

Well, that's why I asked for your help about the photon frequency.
I know how the Doppler shift works and depends either on the moving source or the moving observer. For Doppler shift of the light we need few successive waves.
It appears that it is not the same for a photon.

I really need to do some reading :)

It is given that it goes from the source at the time of emission to the detector at the time of detection. That distance is not equal to the length of the truck in any frame where the truck is moving, as explained and derived.

.........
.......

Before you pursue this line of discussion any further, spend a little effort of your own. Sit down and take a close look at each step of my brief derivation. Either the derivation is right or it is wrong. If it is right then you have no choice but to acknowledge that the distance traveled by the light is not equal to the length of the truck in any frame where the truck is moving. If it is wrong, then point out the specific error, and post the correct derivation which supports your claim quantitatively.

I promise to read and come with new question or acceptance.

Just one question to give me direction; if we put light source in a middle of a moving truck and detectors on both sides, the path of the light in both directions will be different, right?

Would that mean:
1) both detectors will measure different frequency
or
2) light travels different distance and that distance is traveled:
a) for the same time
b) for different time

3)does both lengths together give the full length of the truck?

[EDIT] the last question is some how incorrect to me...

Thank You, DaleSpam :)

 

[PAllen]

You [Harrylin] said:

"It is very much based on assuming the one-way speed of light, just as gwellsjr explained. After setting the clocks according, one simply measures what one first assumed. "

I disagree. SR can be derived from several different sets of assumptions, then confirmed with measurements. One can derive SR making no assumptions at all about the speed of light, let alone constant one way speed. And though hard in practice, in principle one can measure the one way speed of light without implicitly assuming it.

 

[Dale]

if we put light source in a middle of a moving truck and detectors on both sides, the path of the light in both directions will be different, right?

Yes.

Would that mean:
1) both detectors will measure different frequency

No.

2) light travels different distance and that distance is traveled:
a) for the same time
b) for different time

Different time. Remember, light travels at c, so the fact that the distances are different implies that the times are different also.

3)does both lengths together give the full length of the truck?

I don't think so, but I would have to do the math to be sure.

 

[ghwellsjr]

It is very much based on assuming the one-way speed of light, just as gwellsjr explained. After setting the clocks according, one simply measures what one first assumed.



Harald

Ok I've had enough of this overly strong claim. What about slow clock transport rather than light based clock synchronization? That this comes out the same as light synchronization is a strong physical statement (that uniquely specifies only LET theories that are completely indistinguishable from SR can be true; they must effect all possible physical laws the same way as a spacetime symmetr).

Anyway, by normal standards of logic, slow clock transport is independent of any signal speed, and combined with distance measurement based on e.g. rulers, allows measurement of one way lightspeed.

You don't have to use light to synchronize two remotely located clocks, you can use the slow transport of anything as long as it travels at the same constant speed in both directions. We know that moving a clock from one location and back results in a loss of time compared to the clock that remained stationary so you would have to have three clocks, two that remained stationary and one that made a round trip. Then you could adjust the remote clock according to one-half the loss of time to synchronize the two stationary clocks. This would also require communication from the local clock to the remote clock. It would take a very long time to do this. Why not just use light signals?

But as I say, you could do this with fast moving clocks or fast moving missiles, anything to establish the same time for light or anything to travel both halves of the round-trip in the same time, whatever that time may be.

This will allow you to then establish a Frame of Reference, just like Einstein proposed, but the time on the remote clocks are relative to the definition of that one Frame of Reference. Another set of clocks moving with respect to the first set will have a different definition of time for a second Frame of Reference and they are both just as valid as the other even though they will now disagree on the other one's definition of the one-way speed of light.

This issue is no different than other null experiments to establish a preferred reference frame or to establish an absolute ether rest state. Every inertial state appears the same as any other (they all look like an absolute ether rest state) but they disagree with each other about the definition of the time it takes for light to travel from one point to another where those two points are comoving in relation to the inertial state.

 

[harrylin]

You [Harrylin] said:

"It is very much based on assuming the one-way speed of light, just as gwellsjr explained. After setting the clocks according, one simply measures what one first assumed. "

I disagree. SR can be derived from several different sets of assumptions, then confirmed with measurements. One can derive SR making no assumptions at all about the speed of light, let alone constant one way speed. And though hard in practice, in principle one can measure the one way speed of light without implicitly assuming it.

Perhaps my reply to Agerhell (he didn't respond yet) wasn't clear enough? But that should not be read on its own, for my reply refers to the explanation that gwellsjr gave in post #23 and the discussion of GPS by Agerhell. "It" refers to GPS, not SR. See my further clarification in post #75 and also gwellsjr's elaboration in post #79.

Summarized in a single soundbite: relativity of simultaneity implies relativity of closing speed.Harald

 

[Agerhell]

Perhaps my reply to Agerhell (he didn't respond yet) wasn't clear enough? But that should not be read on its own, for my reply refers to the explanation that gwellsjr gave in post #23 and the discussion of GPS by Agerhell. "It" refers to GPS, not SR. See my further clarification in post #75 and also gwellsjr's elaboration in post #79.

What am I supposed to respond to?

It is very much based on assuming the one-way speed of light, just as gwellsjr explained. After setting the clocks according, one simply measures what one first assumed.

Well if you assume wrong, your GPS will basically tell you that you are at a location that you are not at. Is that the kind of measuring you are referring to? Or are you one of them standing on-a-train-bouncing-light-at-mirrors-looking-at-the-trainstations-clock-kind-of-people?

It treats the ECI frame as an inertial frame (which is not exactly true, but good enough), as defined in classical physics and SR: in any inertial frame the laws of physics are supposed to work. Thus GPS should also work on Mars, using the centre of Mars frame. Note that also corrections are used for gravitational potential.

Well maybe, but if you for an earthbound location imagine that the speed of light at your location is c in relation to some other inertial reference frame than that of the centre of the earth, your up for problems...

 

[harrylin]

[..] Well if you assume wrong, your GPS will basically tell you that you are at a location that you are not at. [..]

No, if you choose another inertial frame the calculations will only become much more complex for GPS, so that GPS may take too much time to compute your correct location. It's similar to choosing the centre-of-mass frame in particle physics, it's mere convenience.

 

[Agerhell]

No, if you choose another inertial frame the calculations will only become much more complex for GPS, so that GPS may take too much time to compute your correct location. It's similar to choosing the centre-of-mass frame in particle physics, it's mere convenience.

No, if you assume that the speed of light on Earth is c in relation to some other inertial reference frame, for instance that of the sun, you will get the wrong answer. Imagine that you are positioned such that you can see one gps sattelite in a position that is directly in front of the Earth in its orbit around the sun and one satellite directy behind the Earth in its orbit around the sun. Now if you assume that the signal from the satellites travels with a velocity of c in relation to the centre of the Earth you will be able to correctly position a surface in between of the satellites where you must be positioned (you need three satellites to know where on that surface you are). If you instead suppose the velocity of light is c in relation to the sun you will position that surface incorrectly because the Earth is traveling at roughly 30 km per second around the sun. You will think that you are closer to the satellite that is positioned behind the Earth in its orbit around the sun and at a greater distance from the satellite that is in front of the Earth in its orbits than you really are.

I can not really understand how you can think otherwise.

 

[harrylin]

No, if you assume that the speed of light on Earth is c in relation to some other inertial reference frame, for instance that of the sun, you will get the wrong answer. Imagine that you are positioned such that you can see one gps sattelite in a position that is directly in front of the Earth in its orbit around the sun and one satellite directy behind the Earth in its orbit around the sun. Now if you assume that the signal from the satellites travels with a velocity of c in relation to the centre of the Earth you will be able to correctly position a surface in between of the satellites where you must be positioned (you need three satellites to know where on that surface you are). If you instead suppose the velocity of light is c in relation to the sun you will position that surface incorrectly because the Earth is traveling at roughly 30 km per second around the sun. You will think that you are closer to the satellite that is positioned behind the Earth in its orbit around the sun and at a greater distance from the satellite that is in front of the Earth in its orbits than you really are.

I can not really understand how you can think otherwise.

The paradox that you sketch is just a misunderstanding of Einstein's train example: a different standard inertial reference system uses a different synchronization of clocks, thus making the speed of light isotropically c in that frame. So it's very simple: if the ECI frame would be preferred for the laws of physics as opposed to the solar frame, then relativity theory would be wrong.

PS: note also that the speed of light "on earth" is not c in GPS calculations; instead the speed of light is made c wrt to the ECI frame, in which the surface of the Earth is rotating. GPS corrects for the speed of the surface of the Earth.Harald

 

[Agerhell]

The paradox that you sketch is just a misunderstanding of Einstein's train example: a different standard inertial reference system uses a different synchronization of clocks, thus making the speed of light isotropically c in that frame. So it's very simple: if the ECI frame would be preferred for the laws of physics as opposed to the solar frame, then relativity theory would be wrong.

You are saying that the GPS-system would work equally well if we assumed the velocity of light to be c with respect to the sun and somehow did something with the clocks onboard all of the gps-satellites so that you would still get the right position at all times no matter where on Earth you are positioned? You are going to have to refer me to some proof of that before I believe you...

PS: note also that the speed of light "on earth" is not c in GPS calculations; instead the speed of light is made c wrt to the ECI frame, in which the surface of the Earth is rotating. GPS corrects for the speed of the surface of the Earth.

I have not been claiming anything ells than that to get the correct result from the GPS system you have to assume that the velocity of light on Earth is c with respect to the centre of the earth. That is also what the ECI frame is based upon, so I do not know who you are arguing with right now...

If you have signals from three satellites you basically know your position at a certain time. The fact that the Earth is rotating is not much of a problem.

 

[sisoev]

Originally Posted by sisoev
3)does both lengths together give the full length of the truck?

I don't think so, but I would have to do the math to be sure.

Do you think that if we do the round trip for the light, we will get the length?

 

[Dale]

Do you think that if we do the round trip for the light, we will get the length?

Both lengths together gives:
\frac{L}{1-v^2/c^2}
The round trip gives twice that.

 

[sisoev]

Before this thread veers completely into new directions, I thought I would put some numbers and clarifications of the OP experiment. Refer to the picture in the OP.

We have a truck moving fast. At the moment its front passes a light, it flashes. The back of the truck receives the signal at some time and place. A ground observer, B, is standing where the back of the truck passes as it receives the signal. The question was how do they both measure the speed of light to be the same?

We need to add a lot of specificity. Assume some time in the past, we collected identical tape measures and clocks. Now we describe how B and A (a truck traveling scientist) set up to do their measurements.

B) B places a clock where the signal will be received as the back of the truck passes; rolls out tape measure to where the emitter is (carrying a clock) and places clock next to emitter. Assistant stands at this clock to record when the emitter flashes. Scientist walks back to reception point, to await the truck's arrival (where he will record the reception event on his clock).

A) Scientist enters truck with tape measure and two clocks, and waits in back of truck as it gets up to speed. Then leaves one clock at back of truck, carries other to front as he measures the truck. An assistant waits at the front to record emission time according this clock at front of truck. Scientist walks to back of truck to record reception time.

Ok, now for some numbers. All units are light seconds, speed of light is 1 (light second / second), and we imagine we have all the time in the world to walk back and forth many light seconds.

A measures his truck length as 100, the speed of the ground going by as .6 (c). For him, the light took 100 seconds to reach the back and lightspeed is 1.

Now it gets interesting. If everything has been properly set up so that events happen as described at the beginning, then B will measure the distance between emitter and reception point as 50. This is a combination of seeing the truck as length 80, and the fact that the truck will travel 30 between emission and reception. B will measure the time between emission and absorption as 50, getting c for the speed of light. The discrepancy between B's 50 seconds and A's 100 is a combination B seeing A's clock running only 80% as fast as his, but also seeing a large discrepancy between the front and back clocks on the truck. According to B, A's front clock is set 60 seconds ahead of the back clock. So the 100 seconds measured by A is 'really' only 40 of A's seconds. Then since A's clock is only running 80% the rate of B's, this 40 of A's seconds correspond to 50 of B's.

Thank You, for the thorugh explanation, Palen.
However, you got one thing wrong;
the flash is ON the truck
and one thing not taken in consideration;
what if the graphics represent the contracted length of the truck
We should see it contracted from our point anyway.

 

[PAllen]

Thank You, for the thorugh explanation, Palen.
However, you got one thing wrong;
the flash is ON the truck
and one thing not taken in consideration;
what if the graphics represent the contracted length of the truck
We should see it contracted from our point anyway.

The flash being on the truck makes no difference whatsoever. (What matters is the 'event' of light emission. The motion of the source only matters for doppler effects, which were not part of the original scenario, and completely irrelevant to it).

If the graphics represent the contractel length, the just multiply most numbers by 5/4 (e.g. you don't multipl .6c by 5/4).

In a nushell, both these observations are completely irrelevant to the physics of the scenario.

 

[sisoev]

Both lengths together gives:
\frac{L}{1-v^2/c^2}
The round trip gives twice that.

As you probably noticed from the OP, I'm not physicist or mathematician, but I got the same result with a simple graphics.
So, why do we get this result?
Don't you think that it has something to do with not placing the emission point where it belongs, at the source point?

In general, both blue and red shifted lights are "old" lights; the source is either closer or further away from us.
With "old" light we cannot measure distance to the source unless we use its frequency in our calculation. Correct?

Not putting the emission at the place of the source is not in any favour for the length contraction, because I could easily set the experiment with the contracted length of the truck and we would still have different length of the light for A and B.