Speed of light on a moving platform

[Erland]

You need to be more careful here and consider what length is being measured here. The distance in the truck’s frame between the position where the laser fires and the computer at rest will be smaller than 7 miles. The distance you are thinking of is the distance between where the laser has traveled to and the computer at the time the signal reaches the computer. This distance is not very important for the setup.

No, this is a misunderstanding. It is clear from what jay t writes in his reply to me and otherwise, that 7 miles is the distance measured in the ground's frame between the laserguns's and the computer's positions, at the moment when the laserguns fire (this is more precisely described in my post #68).
Since the yellow lasergun and the computer both move with the truck, i.e. they are at rest in the truck's frame (so there is no "computer at rest", if "rest" refers to the ground's frame), so the distance between the yellow lasergun and the computer is constant in the truck's frame, and this distance is certainly more than 7 miles - this distance, call it L, can be considered as a "rest length", and by length contraction, the corresponding length measured in the ground's frame (and this distance is measured in the ground's frame as I described it in my post #68) is shorter then L, so 7 < L, or L > 7. This also follows from the Lorentz transformation.

But you are right that one must be very careful when dealing with these matters, one must carefully state which lenghts, times, readings of instruments etc. one is talking about and how measurements should be done, otherwise, the risk that one goes astray is very great.
Always remember this, jay t!

 

[Dale]

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

I don't think so. The starting condition is two spatially separated events which are simultaneous in the pre acceleration inertial frame, and then he wants answers in the computers frame only. So I think the non inertial aspect is essential to the problem.

 

[Nugatory]

I don't think so. The starting condition is two spatially separated events which are simultaneous in the pre acceleration inertial frame, and then he wants answers in the computers frame only. So I think the non inertial aspect is essential to the problem.

There are enough different questions floating around in this thread that we may not be talking about the same one?

There are no simultaneous spatially separated events in the one I'm talking about: we have a light signal emitted at the event x=0,t=0 using an inertial frame in which the cart is at rest after completing its instantaneous acceleration. In that frame the right-hand end of the cart is at x=5 so the light and the cart meet at x=5,t=5.

 

[Dale]

There are enough different questions floating around in this thread that we may not be talking about the same one?

Very likely!

There are no simultaneous spatially separated events in the one I'm talking about:

I thought that the light pulse at the left and the computer starting its timer at the right were always taken to be simultaneous in the ground frame in all of his scenarios.

 

[Nugatory]

I thought that the light pulse at the left and the computer starting its timer at the right were always taken to be simultaneous in the ground frame in all of his scenarios.

Ah - got it - I was reading the synchronization to be in the post-acceleration frame (and I rather doubt that OP was aware of the complexities he was introducing by accelerating the cart instead of just having it approach from the left at constant speed.

 

[jay t]

Finally got a chance to respond.

What does this mean? In the cart frame, the cart is 5.2 light seconds long (which length contracts to 5 light seconds when measured in the ground frame). So it takes 5.2s for the laser to cross the gap. In this frame the ground is moving, so the distance between the ground laser and target is length contracted to about 4.8 light seconds. But the target is moving towards the lasers so the ground laser pulse has a shorter distance to travel so was hit after only 3.7s.

So the cart target will always be hit second. The time and distance between the hits will depend on whose clocks and rulers you use. I haven't shown it, but whether the targets were next to each other when the lasers fired will also depend on whose clocks you use.

Edit: I recommend using speeds of either 0.6c or 0.8c. Those give rational values for $\gamma$, which makes exercises much less irritating.

I read over your response over and over...
And, i am sure it is a good answer. It's just that I do not understand it very well. (man you guys have a way of speaking too much of the physics lingo without "bringing it down to the common man".

Are you saying that in Scene#3, the laser from the gun that was mounted on the cart will be hit second.
And that the laser from the gun that was planted on the ground will hit first?

 

[Nugatory]

Are you saying that in Scene#3, the laser from the gun that was mounted on the cart will be hit second.
And that the laser from the gun that was planted on the ground will hit first?

It depends on the exact specification of the problem, and although it may not be clear to you, you have not exactly specified either scene 2 or scene 3. To do it right, you must:
1) Never ever speak of anything moving or not moving unless you say what it moving relative to.
2) Never ever mention any speed without specifying what that speed is relative to.
3) Any time you say that two things happen "at the same time", be sure that either they also happened at the same place or that you specify which frame you're using when you say they happened at the same time.

These are needed just to get rid of any ambiguity in the problem description. But also:
4) Eliminate all accelerations from the scenario. Instead of imagining a cart starting at rest and accelerating to some speed, imagine two carts one at rest and another passing by that one at the final post-acceleration speed, with one end or the other (but not both - see #3 above) aligned with one or the other end of the first cart. This is particularly important if you're imagining instantaneous accelerations.

#4 is not really needed, but it simplifies the math no end without changing the basic physics. Some first-year college-level relativity classes never even consider problems involving aceleration, just because they're hard to deal with until you have the basics down.

 

[Erland]

man you guys have a way of speaking too much of the physics lingo without "bringing it down to the common man".

This is probably beacuse the ordinary language is not suffiently exact to describe and analyze this problem with all its subleties (which are often unexpected by the common man), so the "physics lingo" is probably necessary for this.

 

[Ibix]

I read over your response over and over...
And, i am sure it is a good answer. It's just that I do not understand it very well. (man you guys have a way of speaking too much of the physics lingo without "bringing it down to the common man".

Are you saying that in Scene#3, the laser from the gun that was mounted on the cart will be hit second.
And that the laser from the gun that was planted on the ground will hit first?

If you use rulers nailed to the ground with zero next to the ground laser and clocks nailed to the ground and zeroed at the moment the cart laser passes the ground laser then the readings on those rulers and clocks will be the numbers I called x and t where and when events happen. If, on the other hand, you use rulers and clocks nailed to the cart then the readings on those rulers and clocks will be the numbers I called x' and t' where and when the events happen.

For the scenario I understand you to be describing, the ground target will be hit first and the cart target second. All observers will agree on this. They will not, in general, agree on the distances or times between the firing and the hitting.

 

[Dale]

@jay t

There are really only 3 steps to any problem in special relativity

1) Pick some inertial frame.
2) Completely described the scenario relative to that inertial frame including the position and time of every important event.
3) Use the Lorentz transform to change the positions and times to some new inertial reference frame.

 

[Janus]

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.

View attachment 212332So 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)?

Let's see if I can shed some light on this(pun intended).
I'm going to take this scenario, with just a couple of modifications. 7 miles and 4600 mph are really too small of figures to work with when dealing with relativity. ( light would only take 0.000037634... sec to travel 7 miles and 4600 mph is such a small fraction of c,that the effect we are looking for would just be too small.
So let's make the distance 10 light seconds or ~3,000,000 km, and the relative velocity 0.6c (this particular value gives a bit nicer looking numbers in the example)
So to start off, we can use this diagram for Scene 1 and scene 2, with the difference being that for scene 2 the sensor/computer is moving to the right at 0.6c. The lined box is a ruler resting on the ground marked off by 1 ls segments.

Scene 1 is easy to analyze since both the laser gun and sensor are at rest with respect to each other. Thus we can say that the laser light takes 10 sec to reach the sensor, whether it is according to the ground or the computer.

Scene 2 is a bit different. First we have to establish from what frame this image is taken in. Since we are basically using the same image as we did for scene 1, we will assume that this image also taken from being at rest with respect to the ground and ruler. Thus when the moving computer is 10 ls away from the laser cannon(at the end of the ruler) the laser fires. If we ask how long it takes for the laser to reach the computer, we have to take into account the distance between cannon increases after the cannon is fired and the light has to catch up to the receding computer. It works out that, with a relative velocity of 0.6c, it would take the light 25 seconds to hit the computer. That is how long it will take according to someone on the ground.

You are interested in how long it will take according to the computer. The thing to keep in mind, it that as far as the computer is concerned, it is the cannon that is receding from it. Thus what happens to the cannon after it fires has no effect on how fast the light reaches the computer. All that counts is how far apart they were when the cannon fired. At first, you might think that since they are 10 ls apart when the cannon fires, 10 seconds should pass for the computer between firing and reception of the light. However, the distance between the two is only 10 ls according to someone measuring the distance while at rest with respect to the ruler on the ground. Relativity says that frame with relative motion to each other measure distances differently. This is commonly known as Length contraction. what this means is that the computer and ruler on the ground each measure the length of each other as being contracted. At 0.6c this contraction factor is 0.8. So while the laser cannon and the ground ruler itself measure the ground ruler as being 10 ls long, the computer would measure it as being only 8 ls long.
So according to the computer, when it is even with the end of the ruler, things look like this:

and it is only 8 ls from the cannon (Note that the computer box itself is not a square in this image. This is because the square you see it as above for scene 2, is because it is length contracted according to the ground frame. This also means that technically I shouldn't have used the same image for scene 1 and 2, as for in scene one the computer was at rest and would not have been shown length contracted to a square. I just didn't want to confuse you with drawing it that way before I brought up length contraction.)
From this image it might appear that we can now say that it takes 8 sec for the light from the cannon to reach the computer as measured by the computer. Alas, we can't do this either. The problem is that while this shows the moment that the computer reaches the end of the ruler according to the computer, it is not the moment the laser fires according to the computer. (This is where the Relativity of Simultaneity brought up by other rears its head.)

To figure out when the laser fires according to the computer, we need to ask how does the laser know when to fire? We know that it is when the computer reaches the end of the ruler, but how does the laser know when this happens. A signal can't be sent back from the end of the ruler, because it would get to the laser until after the computer is well past the end of the ruler. You might send a signal from carefully chosen point along the ruler so that the signal it reaches the laser cannon at the same moment as the computer reaches the end of the ruler, or you might use this method:
Assume the computer starts to the left of the cannon. At the moment it passes the cannon a clock at the cannon starts counting from zero. You know how long the ruler is and how fast the computer is moving relative to it, so you know it will take to reach the end of the ruler after passing the cannon. The laser clock waits until its ticked off this much time, and then fires the cannon. With the values we are using, this works out to 16 2/3 seconds. So when the clock at the laser cannon reads 16 2/3 sec, it fires the cannon.

But what does this mean as far as the computer is concerned? Now we have to consider time-dilation. Clocks in motion with respect to you tick slow compared to your own clock. At 0.6 c, this factor is again 0.8 (for every 1 sec ticked of by your clock, a clock moving at 0.6c relative to you only ticks off 0.8 sec.)
So if we give the computer its own clock that starts at zero when it passes the laser cannon, for every 1 sec that ticks off on the laser cannon clock, 1.25 sec tick off for the computer clock. (It is important to note that this is not what the computer would visually see happening to the laser clock if it were watching it in a telescope as they separated, As that would be due to a combination of the increasing distance the light has to travel and time dilation. Here we are only interested in the time dilation part.)
If the laser fires when its local clock reads 16 2/3 sec, and 1.25 sec pass for the computer clock according to the computer for every sec on the laser clock, then 20 5/6 sec have passed on the computer clock when the laser fires, and at .6 c it will be 12.5 ls from the laser when it fires and it will take 12.5 sec for the light to reach the computer.
This is how things will appear for the computer when the laser fires:

Notice that it is far past the end of the ruler.

Now let's consider scene 3. As before, in the ground frame, things look like this when the laser is fired:

If both laser fire as they pass each other, both laser will catch the computer in 25 sec just like in the ground frame of scene 3.
But as mentioned before, this view of the computer and its attached laser cannon are length contracted due to the relative motion between ground frame and computer.

So when we transition to computer frame, the distance between the computer and its attached laser cannon is the non-length contracted distance of 12.5 ls. And at the moment that the two laser cannons pass and they both fire, things look like this:

Note that the only difference between this and the computer frame image for scene 2 is the arm extension and extra laser cannon.
Again, since both cannon are equal distance from the computer when they fire, both lights reach the computer at the same time, 12.5 sec later.

One last point. Earlier, I gave an alternative method for how the laser cannon decides when to fire. Using that alternative method would not have resulted in any difference in the final answers.

 

[jay t]

If you use rulers nailed to the ground with zero next to the ground laser and clocks nailed to the ground and zeroed at the moment the cart laser passes the ground laser then the readings on those rulers and clocks will be the numbers I called x and t where and when events happen. If, on the other hand, you use rulers and clocks nailed to the cart then the readings on those rulers and clocks will be the numbers I called x' and t' where and when the events happen.

For the scenario I understand you to be describing, the ground target will be hit first and the cart target second. All observers will agree on this. They will not, in general, agree on the distances or times between the firing and the hitting.

Thanks man.
The ground cannon will hit first and the cart-cannon will hit second.
Before I ask a followup question. Do all you guys agree this to be the correct answer? The other guys were saying before they they would hit at the same time.
So I'm just making sure. I didnt see that anyone say your answer is incorrect. So i guess they all agree?

 

[jay t]

Note that the only difference between this and the computer frame image for scene 2 is the arm extension and extra laser cannon.
Again, since both cannon are equal distance from the computer when they fire, both lights reach the computer at the same time, 12.5 sec later.

One last point. Earlier, I gave an alternative method for how the laser cannon decides when to fire. Using that alternative method would not have resulted in any difference in the final answers.

Wow.. thank you for the extremely detailed response. I'll be going over it after my work time today. Just one thing before i look it over however, I see that your answer differs to to post #79. (or at least i think so?). Any way, will respond later after work (got a deadline to complete). I really like all the explanations here. I am learning a lot.

 

[Janus]

Wow.. thank you for the extremely detailed response. I'll be going over it after my work time today. Just one thing before i look it over however, I see that your answer differs to to post #79. (or at least i think so?). Any way, will respond later after work (got a deadline to complete). I really like all the explanations here. I am learning a lot.

Post #79 refers to a "ground target". Which to me suggests a target affixed to the ground and not moving with the cart. ( like in scene 1). If you had such a target at the end of the ruler, then yes, the light would reach it before reaching the computer cart according to both the ground and computer frame in scenes 2 and 3. But this is not what I understood you as asking. I assumed that the target for both lasers in scene 3 was the cart, and that is what I based my reply on.

 

[Ibix]

I agree with Janus - there's some confusion about exactly what your experiment is. Differences in our answers are because we're interpreting your question differently, not because any of our analyses are wrong.

 

[jay t]

Post #79 refers to a "ground target". Which to me suggests a target affixed to the ground and not moving with the cart. ( like in scene 1). If you had such a target at the end of the ruler, then yes, the light would reach it before reaching the computer cart according to both the ground and computer frame in scenes 2 and 3. But this is not what I understood you as asking. I assumed that the target for both lasers in scene 3 was the cart, and that is what I based my reply on.

Yes both lasers are fired toward the computer (the big square object on the right of the diagram). The only difference is that one laser was fired from the ground, and the other laser was fired from a gun mounted unto the same truck with the computer (the gray square on the right of the diagram). Anyway, will look at this more in a bit.

lol man.. i think i will just do a vid to explain this.. so much confusion.

 

[Janus]

Yes both lasers are fired toward the computer (the big square object on the right of the diagram). The only difference is that one laser was fired from the ground, and the other laser was fired from a gun mounted unto the same truck with the computer (the gray square on the right of the diagram). Anyway, will look at this more in a bit.

lol man.. i think i will just do a vid to explain this.. so much confusion.

Then I think we are on the same page in regards to my post and your question.