I Speed of light on a moving platform

[jay t]

1. We have a moving platform traveling 700 miles per hour to the Right-Direction.
2. In the middle of the platform, we have a gun which fires a laser both at the front and back of the platform at the same time.
3. We have at both the front and back of the platform, a mechanical device. These devices once started, will walk the distance from the front or back of the platform (which ever one is started first) towards the middle where the gun was fired.
4. The devices are started once the bolt of light hits the front or back of the platform.

Will device #2 reach to the center first according to the computer doing the measurement?

 

[PeterDonis]

at the same time

At the same time in which frame? The frame of the platform, or the frame of the track on which it is moving?

 

[PeterDonis]

These devices once started, will walk the distance from the front or back of the platform (which ever one is started first) towards the middle where the gun was fired.

Walk at what speed?

 

[Orodruin]

At the same time in which frame? The frame of the platform, or the frame of the track on which it is moving?

Well, the particular events the quote referred to were colocated so in that case any frame.

Of course, the relativity of simultaneity does come into play when the OP wants to start the devices. I would suggest the OP to do the analysis (whatever the specification) in the platform rest frame. It will make everything more transparent.

 

[PeterDonis]

the relativity of simultaneity does come into play when the OP wants to start the devices

Yes, that's actually what I was thinking of. Consider my question suitably modified.

 

[jay t]

Thanks for responding guys.

OK guys.. honestly, i suck at physics.. I'm a computer programmer. So bear with me as i explain this in no-so-physics terms.

The devices (which moves back to the center) both walk at the same speed back to the center. At the center, we just check to see which device reaches back to the middle gun first. A computer is the one that is doing to checking to get rid of any human perception error.

The gun in the middle has one trigger, which fires a laser. The laser is split into 2 beams, via the use of mirrors. Therefore, the laser is fired from the gun in 2 directions at the same time.

Will the computer say that the device #2 is the winner? Thanks.

 

[DrGreg]

1. We have a moving platform traveling 700 miles per hour to the Right-Direction.

This could be rephrased as "there is something" (you haven't said what that is) "traveling 700 miles per hour to the left direction relative to the platform". The something never gets mentioned again so you might as well omit condition #1, as it is irrelevant.

 

[Janus]

Thanks for responding guys.

OK guys.. honestly, i suck at physics.. I'm a computer programmer. So bear with me as i explain this in no-so-physics terms.

The devices (which moves back to the center) both walk at the same speed back to the center. At the center, we just check to see which device reaches back to the middle gun first. A computer is the one that is doing to checking to get rid of any human perception error.

The gun in the middle has one trigger, which fires a laser. The laser is split into 2 beams, via the use of mirrors. Therefore, the laser is fired from the gun in 2 directions at the same time.

Will the computer say that the device #2 is the winner? Thanks.

No, both devices will reach the center at the same time.

For someone in the car, this will be because the lights both reach the ends of the car at the same time and the devices, traveling at the same speed relative to the car will take equal times to travel to the center.

For someone for which the car is moving to the right at 700 mph, the light going to the left will reach device 2 first and it will set off first. However they will not measure the two devices as moving at the same speed relative to the car; device 2 will be moving a bit slower (this is due to the way velocities add up), and the two devices will reach the center at the same time. In both cases the devices reach the center together, even though in one case one device starts off first.

 

[jay t]

This could be rephrased as "there is something" (you haven't said what that is) "traveling 700 miles per hour to the left direction relative to the platform". The something never gets mentioned again so you might as well omit condition #1, as it is irrelevant.

The something is the ground... The platform has wheels and is moving across the ground.

 

[jay t]

No, both devices will reach the center at the same time.

For someone in the car, this will be because the lights both reach the ends of the car at the same time and the devices, traveling at the same speed relative to the car will take equal times to travel to the center.

For someone for which the car is moving to the right at 700 mph, the light going to the left will reach device 2 first and it will set off first. However they will not measure the two devices as moving at the same speed relative to the car; device 2 will be moving a bit slower (this is due to the way velocities add up), and the two devices will reach the center at the same time. In both cases the devices reach the center together, even though in one case one device starts off first.

This is the part i do not understand about relativity.
So let's say we remove the mechanical device, and just simply have the computer measure which wall gets hit first.
You are saying that for the guy sitting on the platform, the computer will register that both platforms get hit the same time.
And
For the guy on the outside, the computer will register that the platform on which device#2 sits, will get hit first?

 

[Herman Trivilino]

So let's say we remove the mechanical device, and just simply have the computer measure which wall gets hit first.

Since the computer can't be present at both walls, you will need to specify how the signals will be sent from the walls to the computer so that the computer can make its determination. And this puts you back at the original problem, which is an attempt to determine the ordering of two spatially separated events. It turns out that if two spatially separated events are simultaneous in one reference frame there will be other reference frames where they are not, and by choosing the frame you can have them occur in either order. But those events will be causally disconnected, meaning that one of them can't possibly influence the other (because there's a maximum possible speed at which information can travel).

 

[davenn]

You are saying that for the guy sitting on the platform, the computer will register that both platforms get hit the same time.
And
For the guy on the outside, the computer will register that the platform on which device#2 sits, will get hit first?

yes and no ( if you are referring to the same on platform computer)

if the computer doing the measuring is on the platform in both instances, then it will always register that the end of the platform will get hit at the same time ( the motion of the platform is irrelevant)

for a guy and computer outside the platform, then device 2 end of the platform will get hit first by the laser

 

[PeterDonis]

The devices (which moves back to the center) both walk at the same speed back to the center.

The same speed relative to what? Relative to the platform, or relative to the track on which the platform is moving?

 

[nitsuj]

The something is the ground... The platform has wheels and is moving across the ground.

He means the question never refers to the "ground" frame again...in turn relative motion is not part of the question.

Janus came to the rescue with the "ground" and a very well worded answer.

 

[Nugatory]

You are saying that for the guy sitting on the platform, the computer will register that both platforms get hit the same time.
And
For the guy on the outside, the computer will register that the platform on which device#2 sits, will get hit first?

Yes. If you google for "Einstein train simultaneity" you will find many explanations (some of them here at Physics Forums) of the relativity of simultaneity. Jumping ahead to the result... In general, if you find that two spatially separated events happened at the same time, anyone moving relative to you will find that they did not happen at the same time. This is the key to most of the "paradoxes" of relativity as well as understanding time dilation, length contraction, and relativistic addition of velocities.

 

[jay t]

Thanks guys for all the responses.
I realize that however i phrase the question, you will all find things which i may have missed. Therefore I will explain it one very last time (while trying to plug all the relativity holes I may have missed). If this is still explain by relativity, then i will give up and just study the theory more until i get it. Here goes.

In the image below we have 3 Scenes. I will explain the scenes, then tell me where I am wrong in my logic.

Scene #1
We have a stationary laser gun, 7 miles away from a computer ( computer which measures the time that the laser hits it). Assume, that the laser takes 5 seconds to hit it. (please these times/scales/distances are not drawn to scale, so no nit-picking please ) .

Scene #2
We have the same scene, only this time, the Computer is mounted on a truck that is traveling away from the laser-gun at 4600MPH. My assumption here is that the laser will take a longer time to hit. Let's say this takes 7 seconds (not drawn to scale please). The laser is first fired at the same 7 miles mark. The truck began moving at 4600mph the very same time the trigger was pulled. (please no talk about the 0-4600mph acceleration.. remember this is a thought experiment..)

All ok with the the first 2 scenes? good...

Scene #3
We have a crazy setup. A mobile truck with an attached laser-cart is speeding toward the start line. When both lasers are aligned with the start line, they are both fired at the same time. Assume that we have the firing of both laser-guns done at the same time. So we have

1) Guns fired at same point (7miles), and at the same time
2) Lasers traveling through the air (light is constant speed) uninterrupted by speed of truck or air resistance.
3) There is only one observer -> The computer whose sole purpose is to determine while laser hit first.

So from the computer's perspective, will both the yellow and the orange laser take the same time to hit? or, will the yellow laser have 5 seconds while the orange have 7 seconds (even though they are fired from the same distance and their end point is moving away from them)?

I guess another question that would explain the scenes above would be:
If i had a very long laser-gun with a barrel 30 miles long. And i fired a laser beam.
Now when the light burst is in the middle of the barrel, i push the barrel of the gun in the direction of the fired shot at the speed of light minus about 100 seconds (no nitpicking please.. just trying to understand my foolish question). When i push the barrel, the light particle's position relative to the run's barrel should have been at the dotted yellow circle's point. But instead it will be at a lesser distance.

I hope the 2 diagrams explains what I am trying to ask in Scene#3 above. If my logic is incorrect, then i give up. I'll have to hit those relativity books.

 

[jartsa]

I hope the 2 diagrams explains what I am trying to ask in Scene#3 above. If my logic is incorrect, then i give up. I'll have to hit those relativity books.

Laser from the yellow rectangle hits the thing that it hits at the same time as the laser from the orange rectangle hits he thing that it hits. And that is so according to anyone you ask.

And how do we know that answer? We imagine two laser pulses one micrometer apart and moving to the same direction. They stay one micrometer apart forever.

 

[Nugatory]

When i push the barrel, the light particle's position relative to the run's barrel should have been at the dotted yellow circle's point. But instead it will be at a lesser distance.

A lesser distance relative to who? According to an observer moving along with the gun barrel, the distance between the point where the light was emitted and where it is now will be one thing. According to an observer at rest relative to the gun before you pushed it up to near lightspeed the distance between the point where the light was emitted and emitted and where it is now will be something else. And yes, the first will turn out to be less than the second.

However, both observers will agree that the flash of light is moving at speed $c$. Both observers will agree that if the light was emitted and the gun was pushed at time zero then at time $T$ the distance between where the light was emitted (which is not where the left-hand end of the gun is at time $T$, except for the observer moving with the gun) and where it is now will be equal to $cT$.

What's going on here is that the speed of light is always $c$ for all observers even if they are moving relative to one another. The speed of both the light source and the detector is irrelevant; all that matters is the distance between where the source was at the moment of emission and where the detector is at the moment that the light reaches it. Divide that distance by $c$ and you'll get the travel time.

However, if two observers are moving relative to one another, they will agree about the speed of light but they generally will not agree about the distance (as in your gun example) so they will not agree about the travel time either. To make sense of this, you absolutely have to work through the relativity of simultaneity as we've already suggested - google for "Einstein train simultaneity" and see what you find.

 

[Herman Trivilino]

@jay t . To demonstrate the relativity of simultaneity your yellow and orange emitters must be separated along the line of relative motion. Otherwise they will reach the computer at the same time regardless of the computer's speed along that line. What you want to show (if you're demonstrating relativity of simultaneity with your thought experiment) is that the computer will reckon that the flashes left the emitter at certain times after it accounts for the delay due to the travel time of the flashes. If the computer reckons that the flashes were emitted simultaneously for one particular computer speed relative to the emissions, then there will be other speeds for which it will reckon that they weren't emitted simultaneously.

 

[jay t]

Laser from the yellow rectangle hits the thing that it hits at the same time as the laser from the orange rectangle hits he thing that it hits.

Just to confirm what you said. You say that from the perspective of the computer, both lasers will hit the same time?

 

[jay t]

A lesser distance relative to who? According to an observer moving along with the gun barrel, the distance between the point where the light was emitted and where it is now will be one thing. According to an observer at rest relative to the gun before you pushed it up to near lightspeed the distance between the point where the light was emitted and emitted and where it is now will be something else. And yes, the first will turn out to be less than the second.

In all the examples i post here here, I am only referring to one observer. In the case of the first image. The observer is the Computer waiting of the lasers to hit it. In the case of the 2nd image the observer is the computer sitting at the very end of the gun's barrel.

In the 2nd image, if the gun was never pushed, the computer knows that the laser should have exited the barred (thereby hitting the computer) at say 5 seconds. But because of the pushing action, it took say 7 seconds. Only the computer is the observer. This was only an example to further explain the first image.

Will you explanation change now that we know that the computer, and only the computer is the observer? The only thing the computer cares about is the duration time the light takes to hit after firing.

 

[jay t]

@jay t . To demonstrate the relativity of simultaneity your yellow and orange emitters must be separated along the line of relative motion. Otherwise they will reach the computer at the same time regardless of the computer's speed along that line. What you want to show (if you're demonstrating relativity of simultaneity with your thought experiment) is that the computer will reckon that the flashes left the emitter at certain times after it accounts for the delay due to the travel time of the flashes. If the computer reckons that the flashes were emitted simultaneously for one particular computer speed relative to the emissions, then there will be other speeds for which it will reckon that they weren't emitted simultaneously.

You see.. the physics is hard to understand for me. So that's why I broke down the image into 3 small easy-to-understand scenes. My goal was to get you guys to agree which scene was incorrect.

1) Is scene #1 correct logically? Yes/No. I say yes.
2) Is scene #2 correct logically? Yes/No. I say yes.

therefore if both scene #1 and scene #2 are correct, then in scene#3, both lasers should hit at the same time (ie 7 seconds) right?
If you can break down the logic and tell me in which scene did the logic go wrong that will be great.

you said that they will reach the computer at the same time regardless of the computer's speed along that line. According to the diagram above, will that be 5 seconds same time? or 7 seconds same time?

(note, there is only one observer i.e. the computer)

 

[Nugatory]

In the case of the 2nd image the observer is the computer sitting at the very end of the gun's barrel.

Assuming that you mean to attach the computer to the right-hand end of the barrel so that it moves with the barrel:
The left-hand end of the barrel will be somewhere at the moment that the flash of light is emitted. The computer at the right-hand end of the barrel will be somewhere when the flash of light reaches it. Take the distance between these two points, divide by $c$, and you will get the flight time because the light is moving at speed $c$ relative to the computer and the barrel. The movement of the barrel is irrelevant except to the extent that it affects where the computer is when the light reaches it.

The exact same logic applies if the computer is not attached to the end of the gun barrel, although the computer will be in a different place at the moment that light reaches it so the distance traveled and time in flight will be different.

 

[jay t]

Assuming that you mean to attach the computer to the right-hand end of the barrel so that it moves with the barrel:
The left-hand end of the barrel will be somewhere at the moment that the flash of light is emitted. The computer at the right-hand end of the barrel will be somewhere when the flash of light reaches it. Take the distance between these two points, divide by cc, and you will get the flight time because the light is moving at speed cc relative to the computer and the barrel. The movement of the barrel is irrelevant except to the extent that it affects where the computer is when the light reaches it.

Ok, so basically you are saying that just measuring the time it takes the light source to leave the end of the barrel (where the computer is attached) should not be done the way I am doing it. But I measured it the way I did, to show that movement of the computer does indeed affect the time it takes for the light to hit it.
Even you said -> the movement of the barrel does affect where the computer is when the light reaches it, therefore this affects the time it takes for the light reaches it right?

The barrel image was only meant to show that the movement affects the time it takes to hit it. I am sorry this question has gone on so long ok.. Its just that I had this discussion with someone before and he couldn't get what I was trying to show..

 

[Herman Trivilino]

If you can break down the logic and tell me in which scene did the logic go wrong that will be great.

In Post #17 @jartsa told you that you had it all correct.

you said that they will reach the computer at the same time regardless of the computer's speed along that line. According to the diagram above, will that be 5 seconds same time? or 7 seconds same time?

If you want your thought experiment to demonstrate that it's possible for the computer to determine that the same flashes are in some cases emitted simultaneously and and in other cases not simultaneously you'll have to alter it in the way I described in Post #19.

 

[jay t]

In Post #17 @jartsa told you that you had it all correct.

He did, but just one followup question i had was -> Will it be 5 seconds for both? or 7 seconds for both?
He just said that it will be the same...

 

[Nugatory]

He did, but just one followup question i had was -> Will it be 5 seconds for both? or 7 seconds for both?
He just said that it will be the same...

The first thing to get straight in your mind here is:
The problem with the cart moving to the right at 4600 mph and both the ground-mounted and cart-mouned lasers firing at the same time when they are lined up (if they weren't lined up saying "at the same time" would be very problematic, but they are lined up so it's OK) is the exact same problem as the cart and the cart-mounted laser standing still while the ground and the ground-mounted laser are moving to the left at 4600 mph.

Now, if the computer's clock is set to zero at the moment (according to an observer who is at rest relative to the cart) that the two lasers coincide, and if the cart is 930,000 miles long (as measured by that observer who is at rest relative to the cart), then the computer's clock will read five seconds when the light from both beams reaches it because that's how long it takes for light to travel 930,000 miles. The fact that the ground is moving to the left at 4600 mph is irrelevant.

 

[jartsa]

Just to confirm what you said. You say that from the perspective of the computer, both lasers will hit the same time?

Yes.

 

[jay t]

Now, if the computer's clock is set to zero at the moment (according to an observer who is at rest relative to the cart) that the two lasers coincide, and if the cart is 930,000 miles long (as measured by that observer who is at rest relative to the cart), then the computer's clock will read five seconds when the light from both beams reaches it because that's how long it takes for light to travel 930,000 miles. The fact that the ground is moving to the left at 4600 mph is irrelevant.

(The drawings are not drawn to scale...)
And yes, this is the heart of the confusion. You just said that they would both take 5 seconds according to the computer observer.
But in scene#1, it took 5 seconds, and in scene #2 it took 7 seconds. Logically, both Scene#1 and Scene#2 are correct (if the computerdistance/trucklength is drawn to scale).
So i just cannot get how in scene#3 that the computer would detect them both at 5 seconds, when clearly, the ground-mounted laser mimics scene #2 which takes 7 seconds. to arrive. Wouldnt it be logical to say that they both took 7 seconds?

 

[Nugatory]

(The drawings are not drawn to scale...)
And yes, this is the heart of the confusion. You just said that they would both take 5 seconds according to the computer observer.
But in scene#1, it took 5 seconds, and in scene #2 it took 7 seconds. Logically, both Scene#1 and Scene#2 are correct (if the computerdistance/trucklength is drawn to scale).
So i just cannot get how in scene#3 that the computer would detect them both at 5 seconds, when clearly, the ground-mounted laser mimics scene #2 which takes 7 seconds. to arrive. Wouldnt it be logical to say that they both took 7 seconds?

Notice how I carefully specified "according to an observer at rest relative to the cart" in describing the five-second result in scene 1? The description I gave is accurate for both scene 1 and scene 2 according to an observer at rest relative to the cart.

You are getting your seven-second result (by the way, something is wrong with your algebra because 4600 mph is nowhere near enough to make two seconds worth of difference) in scene 2 by considering things from the point of view of an observer who is moving relative to the cart. According to this observer, and compared with the description I gave for scenes a and 2::
1) The light travels more than 930,000 miles. You've spotted this effect.
2) The cart is not 930,000 miles long; it's bit shorter because of length contraction.
3) Because of time dilation, the clock attached to the computer on the cart is running slow compared with the clock the observer would use to see how much time the light spent in flight.
4) Because of relativity of simultaneity, the computer's clock does not read zero at the same time that the lasers fire (although it does according to the guy on the cart).

All of these effects combine so that the ground observer finds that the light spends more than five seconds in flight, yet both flashes arrive at the same time.

 

[jay t]

by the way, something is wrong with your algebra because 4600 mph is nowhere near enough to make two seconds worth of difference

As I have mentioned before. Nothing is drawn to scale. It should not matter if one took 5-seconds vs 7-seconds right? I could just called them X and Y if you prefer. The values do not matter. What matters is the fact that the values are different. The same goes for the length of the truck (L). And the same also goes for the distance of 7 miles (M) etc etc. I just gave them random values because It is easier for me to call numbers instead of X and Y etc. I know this must bug you physics guys.. so sorry, I will now try to use variable instead of numbers.

Notice how I carefully specified "according to an observer at rest relative to the cart" in describing the five-second result in scene 1? The description I gave is accurate for both scene 1 and scene 2 according to an observer at rest relative to the cart.

Thanks for your response.. But right now I almost feel like just giving up and memorizing these concepts without trying to understand it. :(
Because no matter how many times I say that there is only one observer which is the computer at the right end of the diagram, the answers here always insert an observer at rest relative to the cart. I see below that you said -->

You are getting your seven-second result in scene 2 by considering things from the point of view of an observer who is moving relative to the cart.

1. We are already clear that Scene#1 takes X seconds.

2. From the computer observer's point-of-view (not another observer standing relative to the cart... but only the computer), in Scene#2, will the computer-observer see the light hit at the same time as in Scene#1? I am sure you would say no. And why? Because the light had to travel an increased distance greater than 7Miles (or M + N).
Therefore we can both agree at the according to the computer observer (only the computer observer), the light from Scene#1 hit at X-seconds while the light from Scene#2 from the computer observer hit at Y-seconds. Two clearly different times. I think this is correct so far right? If i am wrong at this point, then please let me know. There is no point in explaining scene#3 if it is wrong at this point.

3. Now for scene#3 (assuming you agree with the prior 2 scenes).
And again, keeping in mind that we are talking about times from the computer observer's view (not someone relative to the cart).
Which of the options below will the single computer observer (currently traveling away from its rest point just like it did in Scene#2) detect?

A) Both lasers hit at X-seconds
B) Both lasers hit at Y-seconds
C) One laser hit at X-seconds while the other hit at Y-seconds
D) None of the above

To make the answer very simple without much back and forth,
The answer i choose is Option (B). Which one is yours?

(if at this point, the question i am asking is unclear, then.. i do not know how else to ask it)

 

[jartsa]

I guess another question that would explain the scenes above would be:
If i had a very long laser-gun with a barrel 30 miles long. And i fired a laser beam.
Now when the light burst is in the middle of the barrel, i push the barrel of the gun in the direction of the fired shot at the speed of light minus about 100 seconds (no nitpicking please.. just trying to understand my foolish question). When i push the barrel, the light particle's position relative to the run's barrel should have been at the dotted yellow circle's point. But instead it will be at a lesser distance.

Is there a computer-observer bolted on the right end of the barrel?

If so, then we are now talking about the question that is same as this: is it possible to delay the hit from a death-ray by starting to move away from the ray.

The answer to that problem is: Yes, the faster you move the the longer you live. Is it possible to delay the hit from a death-ray gun that hangs on you by a rope? Yes, but only the speed that you gained after the firing of the ray counts.

 

[jay t]

Is there a computer-observer bolted on the right end of the barrel?

If so, then we are now talking about the question that is same as this: is it possible to delay the hit from a death-ray by starting to move away from the ray.

The answer to that problem is: Yes, the faster you move the the longer you live.

Is it possible to delay the hit from a death-ray gun that hangs on you by a rope? Yes, but only the speed that you gained after the firing of the ray counts.

Thanks for understanding :D
1. Yes, you are correct. There is a computer bolted down to he right end of the barrel.
2. You are also correct by modifying the question to -> is it possible to delay the hit from a death-ray by starting to move away from the ray.

So this means that both yours and my understanding of that image is on the same page.
Now, is it possible that the understanding of this image here can be ported over to the first image (with the 3 scenes) together with the question i asked @Nugatory in post #31 ?

In scene#1, from the computer's perspective, it takes X seconds
In scene#2, from the computer's perspective, it takes Y seconds. It will not take X seconds, because the distance the light travels is longer. Therefore it takes Y, with Y being > greater than X)
Therefore,
In Scene#3, from the perspective of the computer observer bolted on the end of the moving truck, will:
A) Both lasers hit at X-seconds (as explained in post#31)
B) Both lasers hit at Y-seconds (as explained in post#31)
C) One laser hit at X-seconds while the other hit at Y-seconds (as explained in post#31)
D) None of the above

The answer i choose is Option (B). But i am curious to find out which one you guys choose.

 

[jartsa]

So this means that both yours and my understanding of that image is on the same page.
Now, is it possible that the understanding of this image here can be ported over to the first image (with the 3 scenes) together with the question i asked @Nugatory in post #31 ?

Scene-1: The computer says that the laser's travel takes 5 seconds from the orange rectangle, which is placed 7 miles away by jay t, or by the computer, no difference there.

Scene-2: The computer says that the laser's travel takes 7 seconds from the orange rectangle, which is placed 9 miles away by by jay t.

Scene-3: The computer says that both lasers' travels take 7 seconds from the two rectangles, which are 9 miles away when the lasers are fired.

(the computer has a shortened ruler if it moves, that's why it says the distance is 9 miles)

(numbers are not correct, numbers are for illustrative purposes only)

 

[Janus]

I guess another question that would explain the scenes above would be:
If i had a very long laser-gun with a barrel 30 miles long. And i fired a laser beam.
Now when the light burst is in the middle of the barrel, i push the barrel of the gun in the direction of the fired shot at the speed of light minus about 100 seconds (no nitpicking please.. just trying to understand my foolish question). When i push the barrel, the light particle's position relative to the run's barrel should have been at the dotted yellow circle's point. But instead it will be at a lesser distance.
View attachment 212333

There's a problem with this scenario. If you push the barrel from the left end, The impulse from that push can't travel through the barrel of the gun at a speed greater than the speed of light. The right end of the barrel won't have even moved before the light pulse reaches it. This is often a tripping point for people first learning Relativity, they assume that they can use objects of perfect rigidity in their examples, but Relativity prohibits the existence of perfectly rigid bodies.
Another thing they sometimes do is to "jump ahead" to scenarios that involve accelerations or sudden changes of velocities before they have fully come to terms with dealing with scenarios with uniform motion alone.

 

[Nugatory]

But right now I almost feel like just giving up and memorizing these concepts without trying to understand it. :(
Because no matter how many times I say that there is only one observer which is the computer at the right end of the diagram, the answers here always insert an observer at rest relative to the cart.

We are not introducing an observer at rest relative to the cart - you are. You introduced it in scene 1, where the cart and the computer are not moving relative to one another so the computer is clearly at rest relative to the cart. This computer is 930000 miles to the right of the point where the light is emitted, so of course it takes five seconds for the light to get from the point where it is emitted to where it is detected by the computer.

Now, in scene 2... is the computer supposed to be at rest relative to the ground or the cart? If it is at rest relative to the ground, is it located 930000 miles to the right of the point where the light flash is emitted, or is it located to be lined up with the point where the light flash reaches the right-hand end of the cart?

2. From the computer observer's point-of-view (not another observer standing relative to the cart... but only the computer), in Scene#2, will the computer-observer see the light hit at the same time as in Scene#1? I am sure you would say no. And why? Because the light had to travel an increased distance greater than 7Miles (or M + N).

If the computer is at rest relative to the cart, then scene 1 and scene 2 are the exact same problem; the ground is moving left at 4600 mph in scene 2 but that doesn't change anything. The computer will find that the both flashes of light reach it at the same time, after five seconds have passed.

If the computer is at rest relative to the ground and located to be lined up with the point where the two flashes of light reach the right-hand end of the cart, then the computer will find that the two flashes of light arrive at the same time, after very slightly more than five seconds have passed. (That's "very slightly", because 4600 mph is so slow compared to the speed of light that the effect is almost unnoticeable... I'll calculate and post the exact amount in a few minutes).

3. Now for scene#3 (assuming you agree with
And again, keeping in mind that we are talking about times from the computer observer's view (not someone relative to the cart)

Now I do not understand what you are asking. "Not someone relative to the cart" and "currently traveling away from its rest point" seem to be contradictory. If the computer is traveling away from its rest point then it is moving along with the cart so at rest relative to the cart. So let's be precise: is the computer at rest relative to the ground or relative to the cart?

Which of the options below will the single computer observer (currently traveling away from its rest point just like it did in Scene#2) detect?
A) Both lasers hit at X-seconds
B) Both lasers hit at Y-seconds
C) One laser hit at X-seconds while the other hit at Y-seconds
D) None of the above

With both lasers firing at time zero, the correct answer is:
If the computer is at rest relative to the cart, both lasers hit at the same time, after five seconds have passed.

If the computer is at rest relative to the ground, both lasers hit at the same time, after a bit more than five seconds have passed (I'll post the exact amount more when I have time, unless someone else reading this thread gets to the calculation first).

If we do this problem with two computers, one on the ground and one on the cart, they will disagree about the time that passed (one will say "five seconds" and the other will say "a bit more than five seconds") and they will both be right.

 

[jay t]

Scene-1: The computer says that the laser's travel takes 5 seconds from the orange rectangle, which is placed 7 miles away by jay t, or by the computer, no difference there.

Scene-2: The computer says that the laser's travel takes 7 seconds from the orange rectangle, which is placed 9 miles away by by jay t.

Scene-3: The computer says that both lasers' travels take 7 seconds from the two rectangles, which are 9 miles away when the lasers are fired.

(the computer has a shortened ruler if it moves, that's why it says the distance is 9 miles)

(numbers are not correct, numbers are for illustrative purposes only)

Thank you @jartsa
You seem to be the only one here who understand what I asked and translated my non-physics like language, and then answered in plain old English. So that i can understand.

From what you said, I got this:
scene1 took 5 seconds.
scene2 took 7 seconds.
in scene 3 both took 7 seconds.

Now i will try to frame the question in a way that the others get it as well.. Because i do have follow up questions.

 

[jay t]

We are not introducing an observer at rest relative to the cart - you are. You introduced it in scene 1, where the cart and the computer are not moving relative to one another so the computer is clearly at rest relative to the cart. This computer is 930000 miles to the right of the point where the light is emitted, so of course it takes five seconds for the light to get from the point where it is emitted to where it is detected by the computer.

Awesome! :D
We both agree that scene#1 takes 5 seconds. So from now on, I will not refer to it again.
I will also not mention scene #3 yet, because I do not think you understand what i meant in Scene#2.

So here is scene #2 again. I will not go to scene#3 unless we both agree with scene#2.

In Scene#2, it is exactly like Scene#1 --> BUT, we have this important difference ---> Which is this:
The very exact moment that the laser leaves the stationary laser-gun, the computer then begins to move away from the same stationary laser-gun at speed S. (if the speed S does not matter, then ignore it. The point I am trying to make is that the distance the light has to travel will be different than scene#1).

Since the distance that the light has to travel to hit this computer is longer than the distance it had to travel in scene#1, will the time taken for the laser to hit the computer be the same time taken in scene#1? or will it be greater than the time taken in scene#1?
NB: We are talking from the the perspective of the computer (the only observer). To the computer, will the will the time taken for the laser to hit be greater than scene#1?

 

[jay t]

There's a problem with this scenario. If you push the barrel from the left end, The impulse from that push can't travel through the barrel of the gun at a speed greater than the speed of light.

Ok, you might have a point here.. But like I said, this second image was only shown to better explain image#1 with the 3 scenes. If using the 2nd image was a bad example, then i apologize.. we can ignore it.
but before i truly give up image#2, will it make a difference in rigidity if I said that the barrel was pulled from the right end instead of being pushed from the left?

 

[Nugatory]

The very exact moment that the laser leaves the stationary laser-gun, the computer then begins to move away from the same stationary laser-gun at speed S. (if the speed S does not matter, then ignore it. The point I am trying to make is that the distance the light has to travel will be different than scene#1).

In scene 2, the distance the light has to travel to get from laser to computer is more than 930,000 miles for an observer who is at rest relative to the ground. However, it is still exactly 930,000 miles and five seconds for the computer.

One way of seeing this (and I mentioned this in post #27 above) is to consider that as far as the computer is concerned, scene 2 is the exact same problem as if the ground starts moving to the left instead of the cart and computer moving to the right.

Since the distance that the light has to travel to hit this computer is longer than the distance it had to travel in scene#1, will the time taken for the laser to hit the computer be the same time taken in scene#1? or will it be greater than the time taken in scene#1?
NB: We are talking from the the perspective of the computer (the only observer). To the computer, will the will the time taken for the laser to hit be greater than scene#1?

From the perspective of the computer, the distance in scene 2 is not longer than in scene 1, so the time is the same: five seconds.
(I don't know if you've noticed this... But when you say that the computer and cart are moving to the right at 4600 mph, you've introduced a second observer who is at rest relative to the ground. That 4600 mph speed is a speed according to that observer - as far as the computer is concerned it is always at rest relative to itself).

 

[Nugatory]

Thank you @jartsa
You seem to be the only one here who understand what I asked and translated my non-physics like language, and then answered in plain old English. So that i can understand.

Unfortunately, both of @jartsa's answers in this thread are either incorrect answers to the question you're asking in scenario 2 ("What is the time measured by the computer?") or they are correct answers to a different question than the one you are asking ("What is the time measured by an observer at rest relative to the ground and moving relative to the computer?").

 

[Dale]

NB: We are talking from the the perspective of the computer (the only observer).

Note that the computer is a non inertial observer, so none of the standard formulas apply in its frame. In fact, its frame is not even well defined.

 

[Nugatory]

Note that the computer is a non inertial observer, so none of the standard formulas apply in its frame. In fact, its frame is not even well defined.

This is true, but somewhere far back in this thread OP allowed that the acceleration would be instantaneous, so we can work in the post-acceleration inertial frame in which the cart and detector are at rest while the ground is moving to the left. The emission event has time and space coordinates zero in that frame as well as in the frame in which the ground is at rest.

I've been glossing over this issue because OP is still hung up over stuff that's more basic; we'd be making the same explanations if in scenario 2 the cart (rest length five light-seconds) was moving at a constant speed and its left-hand end aligned with the light emission at time zero,

 

[jay t]

In scene 2, the distance the light has to travel to get from laser to computer is more than 930,000 miles for an observer who is at rest relative to the ground. However, it is still exactly 930,000 miles and five seconds for the computer.

Thanks man.. I think i finally understand the part where we both disagree on.

How is it possible that the computer's perspective will still be the same time of 5 seconds in Scene#2? It (the computer) moved further away from the start point before the light beam reached it. And therefore the light-beam took a longer time to reach it.. right? Therefore this longer time will be extra time added to the 5 seconds right?
You saying that for the computer, it is still 930,000 is very confusing...

 

[Nugatory]

How is it possible that the computer's perspective will still be the same time of 5 seconds in Scene#2? It (the computer) moved further away from the start point before the light beam reached it. And therefore the light-beam took a longer time to reach it.. right? Therefore this longer time will be extra time added to the 5 seconds right?

From the computer's perspective, the computer is not moving away (it can't be moving relative to itself, can it?). The computer is sitting still while the ground is moving away to the left; and the ground moving to the left doesn't change the fact that the light was emitted 930,000 miles (more convenient to call that five light-seconds - a light-second is the distance light travels in one second, 186000 miles) away from the computer.

To believe that the computer is moving to the right requires adopting the perspective of someone who is at rest relative to the ground. From that point of view the distance traveled by the light is longer and it does take more time for the light to travel from source to computer. But that's not the perspective of the computer, which is what you're asking about.

Read the first paragraph of post 27 again.

 

[Ibix]

It (the computer) moved further away from the start point before the light beam reached it.

This is true in the rest frame of the ground, but not in the frame of the computer. In the frame of the computer it is not moving - the ground is moving.

This is why it is just as easy to pour a drink on a moving train (ignoring rocking due to mechanical imperfection) as it is sat at home. You simply treat the train as stationary and the landscape as moving. This is an application of the principle of relativity - the laws of physics are the same in all inertial reference frames.

 

[jay t]

From the computer's perspective, the computer is not moving away (it can't be moving relative to itself, can it?). The computer is sitting still while the ground is moving away to the left; and the ground moving to the left doesn't change the fact that the light was emitted 930,000 miles (more convenient to call that five light-seconds - a light-second is the distance light travels in one second, 186000 miles) away from the computer.

To believe that the computer is moving to the right requires adopting the perspective of someone who is at rest relative to the ground. From that point of view the distance traveled by the light is longer and it does take more time for the light to travel from source to computer. But that's not the perspective of the computer, which is what you're asking about.

Read the first paragraph of post 27 again.

But doesn't this go against observable experiment?

If both the clock at the Start and the clock at the EndComputer are synchronized. Then we know the following-->
A) We know there exists a scenario where light will take 5 seconds to travel from Start point to Computer (from the computer's perspective), with start-point being stationary and computer being stationary.

B) We know there exists a scenario where light will take 7 seconds to travel from Start point to Computer (from the computer's perspective), with start-point being stationary and computer being stationary.

C) Therefore we can deduce that there also exists a scenario where light will take 7 seconds to travel from Start point to Computer (from the computer's perspective), with start-point being stationary and computer starting to move from P1 and the light hitting it at P2.

If this experiment is conducted with some constant moving object (on a computer simulation), and with synchronized clocks, the computer will measure 7 seconds on the 3rd image. This scenario happens in real life right?

 

[Ibix]

If both the clock at the Start and the clock at the EndComputer are synchronized

Synchronised in which frame? The frame where the ground is at rest or the frame where the computer is at rest? They don't agree on what "synchronised" means.

 

[jartsa]

From the perspective of the computer, the distance in scene 2 is not longer than in scene 1

How do we know that? Because it says so in the picture?

The picture is not drawn from the perspective of the computer in the scene 2, at least I don't think so.

 

[Erland]

This might have been answered in some previous post, but l ask it anyway for clarification:

jay t, in your post no 16, scene 3, which distance is 7 miles? Is it

a) the distance between the laser gun on the truck and the computer, measured in the reference system of the ground, or
b) the distance between the laser gun on the truck and the computer, meusured in the reference system of the truck?

Those distances are different, because of length contraction.

Notice also that I here disregard the laser gun on the ground, which is not needed to answer the question, and it would only complicate things to involve it, since we then must specify times.