I Speed of light on a moving platform

[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.

 

[jay t]

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.

man...
Synchronized as in, they tell the exact same time. When clock one says 1:00, then clock two also says 1:00.
The 2 clocks are synchronized using Quantum Entanglement. When one clock starts, the other does at the exact same time.

 

[PeroK]

man...
Synchronized as in, they tell the exact same time. When clock one says 1:00, then clock two also says 1:00.
The 2 clocks are synchronized using Quantum Entanglement. When one clock starts, the other does at the exact same time.

The problem for you, however, is that synchronisation is relative; not absolute. This is one of the key elements of relativity that distinguishes it from classical physics. If both clocks are synchronised in their rest frame, they will not be synchronised in any other frame.

Essentially, you have jumped into try to solve a problem in SR without having learned the basics. You need to back up and learn SR from the start, building up your knowledge step by step.

 

[jay t]

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.

If i measure the distance (represented by the dotted line) from the front of the laser gun, to the other dotted line (the start of the computer) will this be (a)? or will this be (b) ?
ok I am not a physics guy, so I am not sure which option this would be. I am just literally measuring the distance from the gun to the computer..

 

[Ibix]

man...
Synchronized as in, they tell the exact same time. When clock one says 1:00, then clock two also says 1:00.

The two frames don't agree on when this is since "simultaneous" is not an absolute concept. You may wish to look up the "relativity of simultaneity".

The 2 clocks are synchronized using Quantum Entanglement. When one clock starts, the other does at the exact same time.

That doesn't help. If you lay out a complete procedure for synchronising clocks you will find that - quantum mechanical procedure or not - the clocks will not be correctly synchronised when viewed from another frame.

 

[Ibix]

I am just literally measuring the distance from the gun to the computer..

Are you using rulers at rest with respect to the computer/truck or at rest with respect to the ground? That is all that you are being asked here.

 

[Erland]

If i measure the distance (represented by the dotted line) from the front of the laser gun, to the other dotted line (the start of the computer) will this be (a)? or will this be (b) ?
ok I am not a physics guy, so I am not sure which option this would be. I am just literally measuring the distance from the gun to the computer..

Yet, you must choose one of the options, since they give different lengths. Perhaps the question you must ask yourself is:
In which reference system are the dotted lines fixed? Are they fixed relative to the ground or relative to the truck?

 

[jay t]

The problem for you, however, is that synchronisation is relative; not absolute. This is one of the key elements of relativity that distinguishes it from classical physics. If both clocks are synchronised in their rest frame, they will not be synchronised in any other frame.

Essentially, you have jumped into try to solve a problem in SR without having learned the basics. You need to back up and learn SR from the start, building up your knowledge step by step.

But honestly, synchronization can be a non-factor right?
We are not trying to actually measure the time they take to hit. We are trying to see whether or not the time they hit are either both the same, OR they are different.

1. We know they fire from the same distance.
2. And we know the time they both hit it (from the computer's perspective)
3. We also know they both fire at the same time.

Therefore, we just want to know if the time they both hit (on the computer's end) is the same or different.
I just mentioned 7seconds and 5seconds as arbitrary values.

So its basically this ->
1. Did they fire at the same time? Yes
2. Did they fire at the same distance away from the computer? Yes

3. Therefore, from the computer's perspective did they hit at the same time in scene#3?

So synchronization can be a non factor i think.

[Update] in post #60 i include a small margin of error

 

[jay t]

Yet, you must choose one of the options, since they give different lengths. Perhaps the question you must ask yourself is:
In which reference system are the dotted lines fixed? Are they fixed relative to the ground or relative to the truck?

The dotted line markers are fixed on the ground. The truck is racing down a track toward the lines. As soon as the yellow lasergun lines up with first dotted line (which also lines up with the orange laser gun) they both fire at the same time

 

[PeroK]

But honestly, synchronization can be a non-factor right?
We are not trying to actually measure the time they take to hit. We are trying to see whether or not the time they hit are either both the same, OR they are different.

1. We know they fire from the same distance.
2. And we know the time they both hit it (from the computer's perspective)
3. We also know they both fire at the same time.

Therefore, we just want to know if the time they both hit (on the computer's end) is the same or different.
I just mentioned 7seconds and 5seconds as arbitrary values.

So its basically this ->
1. Did they fire at the same time? Yes
2. Did they fire at the same distance away from the computer? Yes

3. Therefore, from the computer's perspective did they hit at the same time in scene#3?

So synchronization can be a non factor i think.

man ... Distance is relative; synchronisation is relative; "when" is relative; "the same time" is relative.

 

[jay t]

man ... Distance is relative; synchronisation is relative; "when" is relative; "the same time" is relative.

Can we say this?
They are the same time with a very tiny margin of error. The margin or error of atomic clocks are very very small. Will this work? Therefore, i am ok having a very tiny margin of error using the very best time synchronization equipment we have today. The calculations we are working with are 7seconds and 5 seconds. Therefore we have a nice margin to play with in terms of small margin of error. I am not expecting perfection. I'm just trying to see if 2 hit times are the "same" or not. (margin of error included)