2 clocks -- Using orthogonal light path detectors in a space ship

[Ross B]

Hi

I can't see the error in this can someone please explain where I went wrong?

A man is in a spaceship traveling at a constant velocity

He makes 2 identical tubes of length L with a mirror at one end, tube a and tube b

He has a single light bulb. Next to the bulb is a detector. He carefully measures that the bulb and the detector are equidistant from the mirrors.

He places the non mirrored ends of the tubes at the globe, with the tubes perpendicular to each other see the figure

The detector detects if 2 flashes of light arrive at the detector at the same time. The bulb flashed for a brief instant and the detector detects if the flash arrives from each tube at the same time.

The analysis for tube “a” is

T = ((L – vt)/c) + ((L + vt)/c)

(L – vt)/c = is the time taken for the light to leave the light source and travel to the mirror.

(L + vt)/c = the time to do the return trip.

T = the time taken for the round trip

The analysis for tube “b” is

T = 2L/square root(c^2 - v^2)

But

2d/square root(c^2 - v^2) does not equal ((L – vt)/c) + ((L + vt)/c)

There are 4 identical space ships, traveling at the same velocity, wrt and external observer, east, north south and west. They all conduct exactly the same experiment, simultaneously and the observer in each spaceship all get the same result.

 

[Ibix]

If you are including the speed of the ship in the analysis then you are not working in the rest frame of the ship. What happens to the tube parallel to the motion of the ship in that case?

 

[Ross B]

I don't understand - is that not the correct analysis for the tube parallel (tube a)?

The analysis for tube “a” is

T = ((L – vt)/c) + ((L + vt)/c)

(L – vt)/c = is the time taken for the light to leave the light source and travel to the mirror.

(L + vt)/c = the time to do the return trip.

T = the time taken for the round trip

Say the globe flashes at t1. If the spaceship is moving at a constant velocity v won't the clock be in a different position in space when the flash arrives at the mirror say at t2?

 

[Ibix]

Say the globe flashes at t1. If the spaceship is moving at a constant velocity v won't the clock be in a different position in space when the flash arrives at the mirror say t2?

In a frame where theship is moving, yes. But in such a frame, what happens to the length of tube a?

 

[Ross B]

the length of tube a, according to the man at rest wrt tube a is L, as he measured it just before he made the tube?

all the external observer does is measure the velocity and direction of each space ship. They cannot see the experiment

 

[Dale]

2d/square root(c^2 - v^2) does not equal ((L – vt)/c) + ((L + vt)/c)

It does for v=0, which is the case in the ships reference frame.

You can use this fact to derive the length contraction formula in other frames.

 

[Ross B]

but v does not equal zero ...so I sort of fail to see yr point ?

I have no interest in length contraction, as the analysis is the analysis of the observer at rest wrt to the tube ? observer at rest wrt to the tube measured the tube at L just before he made the tube

As there are spaceships going east, north, south and west, according to the external observer, then all parties must agree that at least one spaceship v does not equal 0

 

[Ibix]

the length of tube a, according to the man at rest wrt tube a is L, as he measured it just before he made the tube?

So v is zero ("at rest with respect to") for the person making the measurements. As @Dale points out, your formulae reduce to the same thing in that case.

 

[Ross B]

but that is confusing. As there are 4 identical space ships, carrying out an identical experiment going north, south, east and west, all at v, simultaneously, they must all agree that at least one spaceship v does not equal zero. In which case for any party at rest wrt the tube to assume v = 0 is an erroneous assumption isn't it?

 

[Ibix]

No. They can each consider themselves at rest and the others to be moving. That's what the principle of relativity means, applied to this case.

 

[Dale]

but v does not equal zero ...so I sort of fail to see yr point ?

In the frame where the tubes are the same length v does equal zero. In other frames the tubes are not the same length.

I have no interest in length contraction, as the analysis is the analysis of the observer at rest wrt to the tube ?

Where v is zero.

 

[Ross B]

No. They can each consider themselves at rest and the others to be moving. That's what the principle of relativity means, applied to this case.

so if spaceship number one says I am at rest, your all moving, the others must concede they are in fact moving. In which case the results of the experiments in space ships 2 - 4 should not accord with the results of the experiment in spaceship 1?

 

[russ_watters]

What you are describing is similar to the Michelson Morley experiment and your error was their (failed) hypothesis, which led to the recognition that there was no aether or universal rest frame:
https://simple.wikipedia.org/wiki/Michelson–Morley_experiment

the length of tube a, according to the man at rest wrt tube a is L, as he measured it just before he made the tube?

all the external observer does is measure the velocity and direction of each space ship.

but v does not equal zero ...so I sort of fail to see yr point ?

I have no interest in length contraction, as the analysis is the analysis of the observer at rest wrt to the tube ? observer at rest wrt to the tube measured the tube at L just before he made the tube

As there are spaceships going east, north, south and west, according to the external observer, then all parties must agree that at least one spaceship v does not equal 0

Your basic error here is that you are mixing and matching frames inappropriately when doing the analysis. You can't measure the tube length in the stationary frame and then apply that length to a different measurement done in a moving frame. You must be consistent. And the easiest way to do that is for each ship to declare themselves stationary and then predict their experiment will show no motion.

If the other ships *could* observe each other's experiments, then you could have one ship monitor the experiment and see what results you would get -- but the catch is that they don't just have to measure the time remotely, they also have to measure the length of the tubes remotely. You cannot mix and match measurements in different frames.

 

[Dale]

the others must concede they are in fact moving

No, the others are free to use whatever frame they prefer. That is the whole point of the principle of relativity. There is no universal "in fact moving". There is only moving relative to a given frame.

 

[Ross B]

yep that's fine the guy in spaceship one prefers to assert he is at rest. so the others must accept they are not at rest. It is impossible for them all to claim to be at rest. And in this context what does "at rest" mean. If someone claims to be at rest, and at least one spaceship is not at rest, then does that mean the spaceship that is at rest is at rest in an absolute sense?

 

[Nugatory]

so if spaceship number one says I am at rest, your all moving, the others must concede they are in fact moving. In which case the results of the experiments in space ships 2 - 4 should not accord with the results of the experiment in spaceship 1?

You can analyze the problem using the frame in which spaceship #1 is at rest and the other three are moving. In that frame the tubes on spaceship #1 are the same length but the tubes on the other three ships are not.

Or you can analyze the problem using the frame in which spaceship number #2 is at rest and the other three are moving. In that frame the tubes on spaceship #2 are the same length but the tubes on the other three ships (including spaceship #1) are not.

Either way, if you do the analysis correctly you will calculate that on all four ships the two light flashes reach the globe together. No matter which frame we choose, we have three ships that are moving and have tubes of different length and one ship at rest and has tubes the same length.

 

[jbriggs444]

yep that's fine the guy in spaceship one prefers to assert he is at rest. so the others must accept they are not at rest.

The others are not obliged to do any such thing. All the measurements anyone guy in a spaceship can make is consistent with the idea that he is at rest. All of the measurements reported by the guys in other space ships are consistent with the idea that all the other spaceships are moving and that their measurements are affected by length contraction, time dilation and relativity of simultaneity.

It is impossible for them all to claim to be at rest.

On the contrary. Each one can claim to be at rest and all possible experiments are consistent with each such claim. What would be inconsistent with experiment would be for them to all claim to be at rest with respect to the same inertial frame.

 

[Janus]

yep that's fine the guy in spaceship one prefers to assert he is at rest. so the others must accept they are not at rest. It is impossible for them all to claim to be at rest. And in this context what does "at rest" mean. If someone claims to be at rest, and at least one spaceship is not at rest, then does that mean the spaceship that is at rest is at rest in an absolute sense?

Welcome to Einstein's universe. The fact is that all the rocket's have equal claim to being "at rest". Or put another way, there is no test or experiment that anyone of them could perform that would tell them that they are "moving". You seem to be operating under the assumption that the speed of light(in a vacuum) is c relative to one unique frame. This is not the case, the speed of light is c relative to whoever is measuring it.
So for example, If you and a friend were moving at 0.5c relative to each other, and at the moment you passed each other you both set off a flash bulb, Each of you would see yourself as remaining fixed with respect the centers of the expanding spheres if light, while seeing the other moving with respect to the center of the expanding light spheres.

This also means that each of your spaceships would measure the round trip time for the light, no matter which direction in is pointed as being 2L/c.

As already noted this was the gist of the M&M experiment. It was trying to measure the Earth's motion by this method. Instead of using two spaceship, they just did the experiment at two different times. This was to guard against the possibility of the Earth's motion by chance being zero when you did your experiment. Since the Earth travels in an orbit around the Sun, at a different time of the year, it would be moving at a different velocity than it was when you first made the experiment. What they found was that they got a null result both times. The experiment failed to detect any motion of the Earth either time.

This was not expected, and there were a number of attempts made to explain it. Einstein's Theory of Relativity came out as the accepted answer. Not only did it explain the result, but it made predictions of a number of effects that could be (and have been) tested by experiment or observation. And while many of the concepts in Relativity appear counter-intuitive( Many people have trouble with abandoning the comfortable idea of absolute space and time), it has proven to be very accurate in describing how the universe works.

 

[m4r35n357]

yep that's fine the guy in spaceship one prefers to assert he is at rest. so the others must accept they are not at rest. It is impossible for them all to claim to be at rest. And in this context what does "at rest" mean. If someone claims to be at rest, and at least one spaceship is not at rest, then does that mean the spaceship that is at rest is at rest in an absolute sense?

To misquote Janus badly, welcome to Galileo's universe. You do not need to invoke special relativity to have this discussion, here is how Galileo describes relativity and inertia nearly 400 years ago: "nor could you tell from any of them whether the ship was moving or standing still".

None of this is Einstein's fault ;)

 

[Ibix]

yep that's fine the guy in spaceship one prefers to assert he is at rest.

He doesn't have to claim to be at rest. But if he claims not to be at rest then he has to remember that his clocks are time dilated so don't measure time correctly, and his rulers (those that aren't perpendicular to his chosen direction of movement) are length contracted and don't measure distance correctly. It's a bit of a pain to do it but you can do. If you work it out correctly (remember the relativity of simultaneity) then your pulses will return simultaneously however you do it.

 

[Dale]

the guy in spaceship one prefers to assert he is at rest. so the others must accept they are not at rest.

No, they don't. You are missing the meaning of the principle of relativity.

It is impossible for them all to claim to be at rest

Incorrect. They may all claim to be at rest. There is no single frame where they are all at rest, but for each of them there is a frame where he is at rest. Each one may choose to use their own rest frame, regardless of what the others choose.

 

[Ross B]

You can analyze the problem using the frame in which spaceship #1 is at rest and the other three are moving. In that frame the tubes on spaceship #1 are the same length but the tubes on the other three ships are not.

So the guy in spaceship one (S1) claims to be at rest. But the guy on S3 also claims to be at rest, The guy in S1 looks out his window and see S3 speeding away from him. The guy in S3 does the same. The external observer thinks none of them are at rest. S2 thinks S1 and S3 are not at rest. S4 thinks S1 and S3 are not at rest.So an argument breaks out, who is at rest? They decide to resolve it with the experiment. Those at rest should get a result that the light flashes arrive at the same time. Those not at rest should see a light flash differential. What is the result?

According to S1 the tube in S3 has contracted, But S3 says he carefully measured it and it is definitely L long, how dare he accuse him of being incompetent. S3 accuses S1 also of faulty measurement, the reply is the same.

Alternatively S3 says how much too short is his tube and S1 says it is "x" short, so S3 assumes his ruler is inaccurate and cuts a new tube with an additional "X" on it

My understanding was the MM experiment was to determine if the "ether" existed not if the Earth was moving?

If they all get the same result isn't that an absurd result as patently at least one of them is moving and the error is with the experiment?

 

[Ross B]

So for example, If you and a friend were moving at 0.5c relative to each other, and at the moment you passed each other you both set off a flash bulb, Each of you would see yourself as remaining fixed with respect the centers of the expanding spheres if light, while seeing the other moving with respect to the center of the expanding light spheres.

I agree "position" is also relative

 

[Ross B]

another thing I don't understand is they say no experiment can be carried out by someone within a spaceship to determine if the spaceship is moving. In other words it is impossible to know, without reference to external factors, if the spaceship is moving or not.

If that is the case how does the tube know it is in a moving spaceship so that it knows to contract ?

It appears the tube does know so what factor is the tube measuring in order for it to know it is in a moving frame?

If the tube knows then surely it is possible to know?

 

[Ross B]

another thing I don't understand is they say the laws of physics are the same in all frames. But I would have difficulty explaining a self contracting tube?

does it affect the density of the material that the tube is constructed of?

 

[Dale]

Those at rest should get a result that the light flashes arrive at the same time. Those not at rest should see a light flash differential. What is the result?

They all get the result that the light flashes arrive at the same time. None of them see a light flash differential.

Those at rest should get a result that the light flashes arrive at the same time. Those not at rest should see a light flash differential. What is the result?

They all get the result that the light flashes arrive at the same time. None of them see a light flash differential.

how does the tube know it is in a moving spaceship so that it knows to contract ?

If you have a standard 8.5" by 11" piece of paper how does the paper know if it is 8.5" or if it is 11"?

what factor is the tube measuring in order for it to know it is in a moving frame?

What factor is the paper measuring in order for it to know it is landscape or portrait?

 

[Ross B]

If you have a standard 8.5" by 11" piece of paper how does the paper know if it is 8.5" or if it is 11"?

given numerous factors within the room (temperature, humidity etc etc ) the atoms in the paper react to the repulsive and attractive forces between the atoms which ultimately governs the length and width of the paper. But if I place that paper in a moving frame of reference those repulsive and attractive forces somehow know they are now in a moving frame of reference and adjust themselves accordingly resulting in a smaller piece of paper?

Does that mean if I am standing on a train station with train carriages whizzing by with all the doors open and I repetitively stick my hand in and pull it out of the moving carriages , my hand will shrink and elongate ?

 

[Ross B]

They all get the result that the light flashes arrive at the same time. None of them see a light flash differential.

assuming that is the case they must all,simultaneously, assume they are at rest. Yet they look out their window at the others and it appears that is not the case ?

what is the right answer?

Is the tube measuring some unknown factor to know that it must contract?

or does the clock measure some unknown factor to know it must run slower ?

 

[Janus]

given numerous factors within the room (temperature, humidity etc etc ) the atoms in the paper react to the repulsive and attractive forces between the atoms which ultimately governs the length and width of the paper. But if I place that paper in a moving frame of reference those repulsive and attractive forces somehow know they are now in a moving frame of reference and adjust themselves accordingly resulting in a smaller piece of paper?

You misunderstood. I'll try and explain. You have a sheet of paper on a table. As you look at it, it is 11 in high and 8.5 in wide. Now you walk around the table and look at it from another direction, so that it now appears to be 8.5 in tall and 11 in wide. How did the paper "know" to switch height for width?
The same type of thing is happening with length contraction. Nothing is physically compressing the tube that is moving, it is that according to the measurement made by someone which the tube has a relative motion, the tube is shorter than what someone at rest with respect to the tube would measure. (It goes the other way too. If you have two tubes, each 1 meter long as measured by someone at rest relative to each tube, and these tubes have a relative motion with respect to each other, Each person will measure the other tube as being shorter than his own.

Does that mean if I am standing on a train station with train carriages whizzing by with all the doors open and I repetitively stick my hand in and pull it out of the moving carriages , my hand will shrink and elongate ?

No, in order for you to measure length contraction in something it has to be moving with respect to you.

The main problem you are having here is that you trying to treat Relativistic effects as being the result of some outside influence that acts on objects that are "moving". But what Relativity is really about the is the nature of time and space itself and how we measure it. Observers in relative motion with respect to each other just measure time and space differently. For example if you have two clocks moving at 0.866c relative to each other, call them Clock A and Clock B, for every second Clock A ticks off, it will measure Clock B as ticking off only 1/2 second, but according to Clock B, it is clock A that only ticks off 1/2 sec for every second it ticks off itself.

 

[Dale]

given numerous factors within the room (temperature, humidity etc etc ) the atoms in the paper react to the repulsive and attractive forces between the atoms which ultimately governs the length and width of the paper.

I guess it was a poorly designed question. I was trying to get you to think about the geometry, not the material.

The point is that the same piece of paper is both 8.5" in one direction and 11" in another direction. The paper does not need to "know" anything to change from 8.5" to 11". Nothing changes about the paper, it only depends on which measurement you choose to make. Both are true.

what is the right answer?

They are all right. Each one is at rest and in its own reference frame. Each one explains the null result of all the experiments as being due to length contraction. Their own tubes have no length contraction (v=0) and the other tubes have just the right amount of length contraction to make the travel times equal.

Is the tube measuring some unknown factor to know that it must contract?

No, this cannot be the case. The same tube at the same time is both uncontracted in its own frame and contracted in frames where it is moving. Just like a piece of paper is both 8.5" in one direction, 11" in another. Nothing changes about the paper to make it different lengths nor does the paper need to react to some unknown factor to be measured as 8.5" or 11".

 

[russ_watters]

If that is the case how does the tube know it is in a moving spaceship so that it knows to contract ?

It appears the tube does know so what factor is the tube measuring in order for it to know it is in a moving frame?

If the tube knows then surely it is possible to know?

I guess it was a poorly designed question. I was trying to get you to think about the geometry, not the material.

The point is that the same piece of paper is both 8.5" in one direction and 11" in another direction. The paper does not need to "know" anything to change from 8.5" to 11". Nothing changes about the paper, it only depends on which measurement you choose to make. Both are true.

Last week I was in a boring meeting and my mind wandered a bit. The projector sitting on the table was angled-up, making the image look like a trapezoid on the screen. This annoyed me and I came up with an invention sure to make me rich: install a camera on the projector, which would then measure the shape of the image and enable the projector to auto-correct it (many include manual adjustments, but people never use them).

But then I realized that the *screen* looked flat/square to me and the *picture* angled/trapezoidal, but to the projector the *picture* looked flat/square and the *screen* looked angled/trapezoidal. So much for my invention...but who is right?

Both are right because both are reporting only what they themselves see and not making any claim about a universal reality that both must agree on. As others have said, that's the hard part of Relativity: recognizing that things we have considered to be universal realities all our lives - in most contexts* - really aren't.

Then I started thinking about laser rangefinders and I'm not going to say anything more until I patent it...

*This is actually less true than most people believe it to be and they know it even if they never make the connection. It clicks for some people when they start traveling long distances in an airplane, with the Earth rotating underneath them and start wondering "how far did I *really* travel?" The fact that there is no universal answer to that question is the "secret sauce" of Relativity...and Galileo knew that long before Einstein added his spin to it.

 

[jartsa]

another thing I don't understand is they say the laws of physics are the same in all frames. But I would have difficulty explaining a self contracting tube?

does it affect the density of the material that the tube is constructed of?

A moving tube is a short tube, a dense tube, and what else ... a slowly aging tube.So let's say we want to pump all the water on Earth through a tube whose diameter is 0.1 m in one second. We can do that because the very fast moving water becomes very dense water, right?Of course there are always some observers saying that the water in the tube is not moving. Those observers are also saying that the density of the water is the normal density of water, 1000 kg/m³

 

[jartsa]

Does that mean if I am standing on a train station with train carriages whizzing by with all the doors open and I repetitively stick my hand in and pull it out of the moving carriages , my hand will shrink and elongate ?

The train whizzes by and is contracted. While the hand is not whizzing by and is not contracted. This is very straightforward stuff

And as always there are some observers that say that the hand is whizzing by, and those observers say the hand that is whizzing by is contracted.

 

[Khashishi]

Ross B, I suggest you study the Michelson-Morley experiment, which is basically the same as your thought experiment.

Whenever you specify the velocity of something, you have to also specify what your frame of reference is, if it is not implied. It's incomplete to say that the train is moving at 100 km/hr. You have to say, the train is moving at 100 km/hr relative to the Earth's surface. If I am on the train, then I am also moving 100 km/hr relative to the Earth's surface. But I am moving at 0 km/hr relative to the train. And the train is moving at 0 km/hr relative to me. And the Earth's surface is moving at -100 km/hr relative to me. I am moving at 0 km/hr relative to me.

The length of something depends on the frame of reference. Since I am moving at 0 km/hr relative to me, I do not see myself as contracted. The train is moving at 0 km/hr, so it is also not contracted. The Earth is moving at -100 km/hr so it is contracted slightly in the direction of motion. If the Earth stops moving (because the train applies its brakes), the Earth becomes longer.

Does that mean if I am standing on a train station with train carriages whizzing by with all the doors open and I repetitively stick my hand in and pull it out of the moving carriages , my hand will shrink and elongate ?

No. You seem to think of a frame of reference as something that you can move your hand into and out of. A frame of reference isn't a physical thing, but a choice of coordinates. You can be inside the train and measure your motion relative to the Earth, or to the train, or to yourself, or to any point you choose.