How many times does the train pass?

[necronom]

Hi,

This is a complicated one to describe, but it's been driving me and a friend mad for a few days

Let's say a train is travelling round the Earth at almost the speed of light, and there is an observer on the train and another at the only train station.  Each observer counts how many times he passes the station (i.e. orbits the Earth).

 

The observer at the station says the counting will last for 1 minute, and he starts and stops the train.

Let's say the person at the station counted the train pass 400 times in the minute.

So far so good, but since the train is travelling close to the speed of light, time will move slower for the train passenger, so after the minute is up outside the train, he may have only felt like he travelled for two second.  Since the speed of light is constant for each observer, the train passenger will see the Earth passing by at almost the speed of light, so will see the station passing by at maybe 7.5 times a second.

If he only felt like he was in the train for 2 seconds he may have counted 15 stations, but the person at the station counted 400 passes.

Since the train MUST have passed the station a finite number of times, how many times did it pass?

We may be missing something obvious, but we can't think what it is.

Thanks for any info you can give.

Paul.

[insane_alien]

well, its both. the train will pass by 400 times in a station minute and say 800 times in a train minute.

 

depending on what minute you are using as a reference you'd have to translate that into a perios of time for the other observer.

[losfomot]

Hi,

This is a complicated one to describe, but it's been driving me and a friend mad for a few days

Let's say a train is travelling round the Earth at almost the speed of light, and there is an observer on the train and another at the only train station.  Each observer counts how many times he passes the station (i.e. orbits the Earth).

 

The observer at the station says the counting will last for 1 minute, and he starts and stops the train.

Let's say the person at the station counted the train pass 400 times in the minute.

So far so good, but since the train is travelling close to the speed of light, time will move slower for the train passenger, so after the minute is up outside the train, he may have only felt like he travelled for two second.  Since the speed of light is constant for each observer, the train passenger will see the Earth passing by at almost the speed of light, so will see the station passing by at maybe 7.5 times a second.

If he only felt like he was in the train for 2 seconds he may have counted 15 stations, but the person at the station counted 400 passes.

Since the train MUST have passed the station a finite number of times, how many times did it pass?

We may be missing something obvious, but we can't think what it is.

Thanks for any info you can give.

Paul.

 

A quick answer is that you are ignoring length contraction. The observer inside the train is orbiting a much smaller Earth than the observer in the train station experiences. If the station observer starts the train, observes 400 passes and then stops the train, then the observer inside the train will also count 400 passes.

[necronom]

A quick answer is that you are ignoring length contraction.  The observer inside the train is orbiting a much smaller Earth than the observer in the train station experiences.  If the station observer starts the train, observes 400 passes and then stops the train, then the observer inside the train will also count 400 passes.

 

 

 

 

I see.  I knew there was something we were missing.

 

So the actual distance he is travelling is much less for the train observer, than the person at the station perceives?  If not, then he would be going faster than the speed of light if he did 400 revolutions in two seconds.

 

 

I don't really understand how the distance can change, but I guess I should read up on the effect.  It's very strange.

 

 

Thanks for that.  I'll have to let my friend at work know tomorrow.

 

 

 

 

 

[swansont]

Yes, it is strange if you aren't used to it. losfomot is correct — both observers must agree on the number of times the train passed through the station. The train observer will reconcile the number as being due to the length contraction, while the station observer will say that the train's clock ran slow.

[androstan]

What if the two people decide that "number of train station passings" is their chosen unit of time? Reasoning that the train is undergoing uniform periodic motion, they both reasonably conclude that each cycle of the train is a good clock. After 400 cycles they compare results and find they agree exactly on the elapsed time, even though they were in relative motion wrt each other.

 

Put another way, perhaps they both use atomic clocks too. After coming together they find that their atomic clocks do not agree, but that their "train passing" clocks do. They agree that the train passing clock is a superior timekeeper.

 

This would contradict the particular physical interpretation of STR, that time itself dilates. If the theory only applies to particular clocks and not to others, then it isn't really a theory about time but rather it is a theory about those particular clocks. If the electromagnetic radiation emitted by an atom is redshifted when in motion then one person may call it "time dilation". Another person may simply call it a velocity-dependent redshift.

 

With respect to length contraction, that has to do with how length is measured for nonlocal inertial frames. If I'm holding a ruler, then it is my local frame and I can measure the length of any other colocal object by simply comparing them directly. For an object moving away from me I cannot simply hold up my ruler to it. I have to measure the length of the departing object remotely. The most sensible way to do this is with radar. Let's say I take two identical rulers and I send one (ruler B) hurtling off away from me while I keep mine (ruler A) at home. I wish to know the length of ruler B compared to ruler A. I need to measure the length of ruler A in a way that is comparable to ruler B. I send a light signal along ruler A and I count how many clock ticks it takes for the signal to traverse the ruler. I then send a signal to ruler B and count how many of B's clock ticks it takes for the signal to traverse the ruler. I count that it takes fewer of B's clock ticks to traverse ruler B than it takes A's clock ticks to traverse ruler A. Since the measured speed of light is constant, fewer clock ticks (smaller time) means smaller length/distance.

 

Looked at another way, if you (A) are standing stationary on the embankment and someone else (B) is running at +v, person B will necessarily count fewer clock ticks than A as s/he sends a light signal to the end of the train and back. This is perhaps more obvious than the previous example.

Edited September 28, 2010 by androstan

[Sisyphus]

 

 

This would contradict the particular physical interpretation of STR, that time itself dilates. If the theory only applies to particular clocks and not to others, then it isn't really a theory about time but rather it is a theory about those particular clocks.

 

So in the twin paradox, if I said the traveling twin aged 1 year while the stay at home twin aged 50 years, you would disagree and say that they both aged exactly one "traveling twin's journey?" Do you see the problem?

[swansont]

What if the two people decide that "number of train station passings" is their chosen unit of time? Reasoning that the train is undergoing uniform periodic motion, they both reasonably conclude that each cycle of the train is a good clock. After 400 cycles they compare results and find they agree exactly on the elapsed time, even though they were in relative motion wrt each other.

 

 

This is comparing a clock against itself, which is not a good way to evaluate a clock. All real clock measurements use two or (preferably) more different clocks.

[between3and26characterslon]

Hi,

This is a complicated one to describe, but it's been driving me and a friend mad for a few days

Let's say a train is travelling round the Earth at almost the speed of light, and there is an observer on the train and another at the only train station. Each observer counts how many times he passes the station (i.e. orbits the Earth).

 

The observer at the station says the counting will last for 1 minute, and he starts and stops the train.

Let's say the person at the station counted the train pass 400 times in the minute.

So far so good, but since the train is travelling close to the speed of light, time will move slower for the train passenger, so after the minute is up outside the train, he may have only felt like he travelled for two second. Since the speed of light is constant for each observer, the train passenger will see the Earth passing by at almost the speed of light, so will see the station passing by at maybe 7.5 times a second.

If he only felt like he was in the train for 2 seconds he may have counted 15 stations, but the person at the station counted 400 passes.

Since the train MUST have passed the station a finite number of times, how many times did it pass?

We may be missing something obvious, but we can't think what it is.

Thanks for any info you can give.

Paul.

 

We all know the scenario of the twins paradox and your question has similarities.

 

If one of the twins travels away from Earth at near the speed of light and travels 5 lightyears, turns around and travels 5 lightyears back at near the speed of light then he will only have aged 1 year whereas the stay at home twin will have aged 10 years.

 

So does that mean the travelling twin covered a distance of 10 lightyears in just 1 year, ie. he went at 10 times the speed of light? Yes

 

...and no.

 

From his perspective he will experience what he thinks is faster than light speed but, due to length contraction and time dilation he will still have measured the speed of light as constant inside his space ship. Also from the outside looking at his space ship we will see it going at not quite the speed of light.

 

So the people on the train might think they are going faster than light whereas in actual fact they are not, they will count say 200 passes per second for 2 seconds but the people at the station will count 400 per hour because 2 of their seconds = 1 of our hours.

[androstan]

So in the twin paradox, if I said the traveling twin aged 1 year while the stay at home twin aged 50 years, you would disagree and say that they both aged exactly one "traveling twin's journey?" Do you see the problem?

 

Again, if they come together and their atomic clocks disagree, it would seem better to use the "number of journeys" clock, since it is consistent.

 

This is comparing a clock against itself, which is not a good way to evaluate a clock. All real clock measurements use two or (preferably) more different clocks.

 

I mentioned comparing "train passings" with each person's atomic clock.

 

The point at issue between the two persons seems to be how to explain why cesium atoms emit light less frequently when in motion. The Cs atom absorbs radiation from its surroundings, is excited in some way, there is some internal processing, and the radiation is re emitted as the atom "settles back down".

 

If the atom moves as a longitudinal wave and the light moves as a transverse wave, they have perpendicular components. When the light encounters the atom they blend into a wave that has both longitudinal and transverse components, the resultant velocity of which still has to be c. So it would take the light signal more time to propagate through the atom and back out (re emitted) if the atom were moving due to the net propagation rate being the vector difference between the longitudinal (v) and transverse © components. So, the ratio between light's propagation speed "in space" (only transverse components) and its propagation speed through the longitudinal wave of the moving atom would be c/(c²-v²)^1/2, which would explain the particular form of the Lorentz factor.

[Sisyphus]

Again, if they come together and their atomic clocks disagree, it would seem better to use the "number of journeys" clock, since it is consistent.

 

No, it isn't consistent. That's the point. "It always takes me exactly the same amount of time to get to work: one trip."

 

In order to force the measurement of time into that standard, literally every other possible "clock" would have to be off by the same amount. The speed of light would change, etc.

[swansont]

 

The point at issue between the two persons seems to be how to explain why cesium atoms emit light less frequently when in motion. The Cs atom absorbs radiation from its surroundings, is excited in some way, there is some internal processing, and the radiation is re emitted as the atom "settles back down".

 

 

That's not how a Cesium atomic clock works. You excite them into a superposition of the hyperfine ground states, they oscillate between these two states, and you count the oscillations.

 

Anyway, the whole point of measurement is to have a standardized system. "One trip" doesn't do that.

[androstan]

No, it isn't consistent. That's the point. "It always takes me exactly the same amount of time to get to work: one trip."

 

Well when you compare different results you have to use one standard, one clock. If you're going to talk about 10 trips to work, you have to pick one of those trips as your standard to compare the others to. Unfortunately you can't because they're in the past, so you'll have to pick another standard. You could pick earth rotations, Cs clock oscillations, etc. In practice we use whatever is most convenient and easy, whatever's practical.

 

Let's say you go to work and back by the same exact path over and over, counting each time. Someone else goes there and back by a different path, also counting each time, and you both leave your starting points simultaneously starting from rest. Eventually you'll both arrive at work simultaneously, maybe on the 500th trip or the 10th trip or the 10⁸⁰ trip. If you took N trips while your friend took M trips, then by the "trip standard" all you can objectively say is that you took N/M times more/less trips than your friend. Whether this is because you moved faster or the path you took was shorter isn't known. You could both go back and lay down rulers to measure the path lengths, then you would know how much faster/slower you were moving than him.

 

You could both carry clocks with you and compare those results too. Should you believe the trip standard or the clock standard? Well, the trip standard will work fine if all you want to predict is when the next time you'll both meet at work again, assuming neither of you changes your habits. The trip standard works fine whether you and him are moving at the same speed or not. However, a change in habit (walk faster, take a different path) would constitute a new reference standard, and everything must be compared to a common reference standard.

 

The clock standard, on the other hand, does not work if you are not both moving at the same speed. You both notice that your clocks do not agree. You two will not be able to decide when you'll meet at work again since you don't know whose clock to believe. However, you reason that the clock ticks every time light is emitted. If the vector components of the light are perpendicular to the clock's velocity vector then the resultant will be the vector difference, and the time it takes for light to propagate into and back out of the clock will be increased. By using simple vector algebra you arrive at a transform that corrects your clock for differences in speed between you and your friend. Now you can both not only predict when you'll meet the other (according to either clock), you can also correct for changes in velocity.

 

In the absence of a theory about light propagation the "trip clock" is superior because it is consistent. Both persons count their trips and agree on the ratio of trips. They agree on when they'll meet again. It has the flaw of being ungeneral, i.e. a new standard has to be used every time there is a change in the physical situation. With a theory about light propagation clocks become superior because now we can reference everything to a single standard, the net propagation rate of signals ©.

[swansont]

 

In the absence of a theory about light propagation the "trip clock" is superior because it is consistent. Both persons count their trips and agree on the ratio of trips. They agree on when they'll meet again. It has the flaw of being ungeneral, i.e. a new standard has to be used every time there is a change in the physical situation. With a theory about light propagation clocks become superior because now we can reference everything to a single standard, the net propagation rate of signals ©.

(emphasis added)

 

And that's the deal-breaker.

[androstan]

(emphasis added)

 

And that's the deal-breaker.

 

 

Well, one method requires you to quantitatively reinterpret your measurements to do useful things with them but is generally applicable to observations so far. The other method doesn't require quantitative reinterpretation , but isn't as general. So it's arguable. One is more practical, the other contains fewer assumptions.

[Sisyphus]

To say the "trip" is the same quantity of time for both observers, they would have to be measuring different values of C (as well as differences in pretty much every other imaginable clock, from radioactive half lives to rate of thought), and consequently of all other fundamental constants. You would end up with totally different physics for each observer.

[androstan]

To say the "trip" is the same quantity of time for both observers, they would have to be measuring different values of C (as well as differences in pretty much every other imaginable clock, from radioactive half lives to rate of thought), and consequently of all other fundamental constants. You would end up with totally different physics for each observer.

 

Well, you have to pick a reference. For trips they will get different values of c. Let A and B approach their work place and bounce a radar signal off it. Let A approach at 1/2 and B approach at 1/4.

 

The time (in work trips) that A will meet back up with his/her pulse is given by:

 

T/2 = 1 - T

 

T = 2/3 of a trip

 

For B:

 

T/4 = 1 - T

 

T = 4/5 of a trip

 

A calculates c to be 1/2 and B calculates it to be 2/3.

 

So you can set your trip-distance to be 1 and use it as the standard, or you could set the speed of light to 1 and use that as your standard.

[swansont]

Well, you have to pick a reference.

 

And we find it exceedingly useful to have references that can be realized by anyone.

[Sisyphus]

 

 

So you can set your trip-distance to be 1 and use it as the standard, or you could set the speed of light to 1 and use that as your standard.

 

It's not that simple, because it isn't just C that changes. Everything has to change along with it. One example: pulling on a string with enough force to break it, measured by the acceleration of a mass. Except that if your unit of time is "trip," then you get a different values for acceleration, hence force, hence tensile strength, hence intermolecular forces, hence all electromagnetism. Depending whether you're pulling on the string on the train or not.

[androstan]

It's not that simple, because it isn't just C that changes. Everything has to change along with it. One example: pulling on a string with enough force to break it, measured by the acceleration of a mass. Except that if your unit of time is "trip," then you get a different values for acceleration, hence force, hence tensile strength, hence intermolecular forces, hence all electromagnetism. Depending whether you're pulling on the string on the train or not.

 

Someone on the train has a box full of blocks of a wide variety of sizes. S/he attaches each to a spring that is attached to the roof of the train and measures the spring's displacement for each block, recording each displacement. Then s/he attaches a string to the roof and ties a block to it to see if it'll break. Repeat until it does. Take an identical string and use a block with a slightly smaller spring displacement to see if the string will break. Repeat until you have identified the block with the smallest spring displacement that breaks the string.

 

I think that if this person gets off the train and does the experiment on the ground, they will get the same result. Or if s/he gets on a train going a different speed, s/he will get the same result.

 

Someone on the train has a box full of blocks of a wide variety of sizes. S/he attaches each to a spring that is attached to the roof of the train and measures the spring's displacement for each block, recording each displacement. Then s/he attaches a string to the roof and ties a block to it to see if it'll break. Repeat until it does. Take an identical string and use a block with a slightly smaller spring displacement to see if the string will break. Repeat until you have identified the block with the smallest spring displacement that breaks the string.

 

I think that if this person gets off the train and does the experiment on the ground, they will get the same result. Or if s/he gets on a train going a different speed, s/he will get the same result.

 

 

The experiment would be easier if you just tie the string to the roof and tie the other end to a spring. Pull on the spring until the string breaks, noting the spring's displacement from equilibrium. It will be the same whether you are on the moving train or not.

Edited September 30, 2010 by androstan

[Sisyphus]

Someone on the train has a box full of blocks of a wide variety of sizes. S/he attaches each to a spring that is attached to the roof of the train and measures the spring's displacement for each block, recording each displacement. Then s/he attaches a string to the roof and ties a block to it to see if it'll break. Repeat until it does. Take an identical string and use a block with a slightly smaller spring displacement to see if the string will break. Repeat until you have identified the block with the smallest spring displacement that breaks the string.

 

I think that if this person gets off the train and does the experiment on the ground, they will get the same result. Or if s/he gets on a train going a different speed, s/he will get the same result.

 

 

 

 

The experiment would be easier if you just tie the string to the roof and tie the other end to a spring. Pull on the spring until the string breaks, noting the spring's displacement from equilibrium. It will be the same whether you are on the moving train or not.

 

So you just have to say that f=ma is not true, then. Since the same net force, applied to the same mass, will accelerate it different amounts depending on whether or not its on train.

[androstan]

So you just have to say that f=ma is not true, then. Since the same net force, applied to the same mass, will accelerate it different amounts depending on whether or not its on train.

 

The results of the two experiments are identical, the spring displacement required to break the string is the same in any inertial frame.

[swansont]

 

I think that if this person gets off the train and does the experiment on the ground, they will get the same result. Or if s/he gets on a train going a different speed, s/he will get the same result.

 

 

I think you are ignoring the effect of the acceleration of the train.

[androstan]

I think you are ignoring the effect of the acceleration of the train.

 

We were talking about frames in uniform relative motion.

 

But either way, the results of the experiments are still the same. The spring displacement required to break the string is the same on an accelerating train as on an inertial train.

Edited September 30, 2010 by androstan

[Sisyphus]

The result of that experiment is the same. However, f=ma no longer holds, so you'll have to redefine what you're measuring (since that is the definition of force).