Questions on length contraction in train light clock experiment?

[helpmeplz!]

Hi guys, I've been thinking about this experiment for the last few hours and I have a few questions which I am deeply confused about.

1) In the light clock experiment, where light clock is set up so that light travels vertically, and the train carrying the clock travels horizontally, wouldn't there be length contraction? For a stationary observe outside the train that sees this train moving, he would conclude that since the light has to move a greater distance between the two mirrors, then the actual space and all objects have grown in size? I know this is not true, since lengths only shrink and they only shrink in the direction of motion, but what is wrong with my reasoning? If you say the only reason the distance is longer is because the train is moving horizontally, why is that reasoning enough to talk your way out of length changes but not of time dilation?

2) In the muon experiment, where muons fall to Earth before decaying, the explanation is that in the Earth's reference point, it's at rest and the muon is moving towards it at close to speed of light, therefore it's clock runs slower and so it makes it to Earth on time before decaying. However in the muon's reference frame, it's stationary and the Earth rushes up to meet it. And so the explanation of why it's able to hit the Earth on time is that the space between it and the Earth shrinks. But would it be correct to say that the space between Earth and the muon would shrink in Earth's reference frame too? So the muon would appear smaller in size to someone on Earth than if it was stationary?

3) In the same muon experiment, if we placed a light clock that shoots light parallel to Earth's velocity, then we would see that the light would travel a shorter distance in its route, since it would go towards the higher mirror just the same distance, but it only has to come down a short distance as the Earth comes up to meet it. But we also know from relativity that the observer has to see time in the train slowed down since it is stationary and the train is moving. But if the length has shrunk, then the light has to go a slower distance, so it would take less time between ticks..meaning time has sped up? Where am i going wrong?

Thanks in advance guys!

 

[Orodruin]

1) In the light clock experiment, where light clock is set up so that light travels vertically, and the train carrying the clock travels horizontally, wouldn't there be length contraction? For a stationary observe outside the train that sees this train moving, he would conclude that since the light has to move a greater distance between the two mirrors, then the actual space and all objects have grown in size? I know this is not true, since lengths only shrink and they only shrink in the direction of motion, but what is wrong with my reasoning? If you say the only reason the distance is longer is because the train is moving horizontally, why is that reasoning enough to talk your way out of length changes but not of time dilation?

Length contraction is concerned with how long the distance between the ends of an object is at a single given point in time. In the light clock example this is not the case, the clock has length zero in the direction of motion. The length that the light travels in the clock's rest frame is orthogonal to the direction of motion and therefore does not contract when transformed to a different frame.

However, since the mirrors are going to move in the ground rest frame, the light needs to move not only in the vertical direction, but also in the horizontal direction, in order to bounce between them. Thus, the vertical distance ##y## is the same while the horizontal distance ##x## increases. The total distance is of course given by Pythagoras' theorem ##\ell = \sqrt{x^2 + y^2} \geq y^2## which means that the time between the bounces will be larger in the ground frame.

2) In the muon experiment, where muons fall to Earth before decaying, the explanation is that in the Earth's reference point, it's at rest and the muon is moving towards it at close to speed of light, therefore it's clock runs slower and so it makes it to Earth on time before decaying. However in the muon's reference frame, it's stationary and the Earth rushes up to meet it. And so the explanation of why it's able to hit the Earth on time is that the space between it and the Earth shrinks. But would it be correct to say that the space between Earth and the muon would shrink in Earth's reference frame too? So the muon would appear smaller in size to someone on Earth than if it was stationary?

If the muon had a size, yes that would be the case. However, elementary particles are point-like and it does not make so much sense to talk about the size of a muon.

3) In the same muon experiment, if we placed a light clock that shoots light parallel to Earth's velocity, then we would see that the light would travel a shorter distance in its route, since it would go towards the higher mirror just the same distance, but it only has to come down a short distance as the Earth comes up to meet it. But we also know from relativity that the observer has to see time in the train slowed down since it is stationary and the train is moving. But if the length has shrunk, then the light has to go a slower distance, so it would take less time between ticks..meaning time has sped up? Where am i going wrong?

It is a bit difficult to decipher what you are trying to say, but I will try to answer according to my understanding of it. Basically you are wondering what would happen if you put the light clock on the train horizontally rather than vertically? As you can imagine, things now become a bit more complicated, which is the reason to put the clock vertically in the first place. In the train rest frame, things will look exactly the same simply because the clock is not moving relative to the train.
In the ground rest frame, the clock is now length contracted - and moving. As you can imagine, in the ground frame it will take the light a longer time to go from the back mirror to the front mirror than vice versa as in that direction the light also has to catch up with the speed of the train and in the other direction it will meet faster because of the speed of the train.
Let us assume that the speed of the train is ##v## and that the clock rest length is ##L_0##. The time for the light to complete a full cycle in the train rest frame is of course ##T' = 2L_0/c##. So what is the time for the light to complete a full cycle (going from back mirror, to the front mirror, and back to the back mirror) in the ground rest frame? To start with, the clock has length ##L = L_0/\gamma## in the ground rest frame. The time ##t_1## to reach the front mirror is going to be given by the equation
$$
vt_1 + L = c t_1.
$$
The left-hand side of the equation is the length that the light has to travel in order to reach the position of the front mirror at ##t_1##, the right-hand side is the length the light travels in time ##t_1##. Solving for ##t_1## we are left with
$$
t_1(c-v) = L \quad \Rightarrow \quad
t_1 = \frac{L}{(c-v)}.
$$
In the same way, the time ##t_2## to go from the front to the rear mirror can be computed to be
$$
t_2 = \frac{L}{c(c+v)}.
$$
Thus, the total time is
$$
T = t_1+t_2 = L\left( \frac{1}{c+v} + \frac{1}{c-v}\right) =
L \frac{2c}{c^2-v^2} = \frac{L_0}{\gamma} \frac{2\gamma^2}{c}
= \frac{2L_0 \gamma}{c} = \gamma T'.
$$
Again we have the same time dilation formula as in the case of the vertical light clock. The reason that it takes the light longer to go forward than back is related to the fact that those two events are not located at the same point in space in the train's rest frame - which is the crucial assumption of deriving the time dilation.

Edit: Fixed LaTeX error

 

[Mr-R]

If the muon had a size, yes that would be the case. However, elementary particles are point-like and it does not make so much sense to talk about the size of a muon.

Lets say its an object like a ball or something instead of a muon. What would be the case for it?

 

[A.T.]

So the muon would appear smaller in size to someone on Earth than if it was stationary?

The electric field of the muon is length contracted in the frame of the Earth.

http://en.wikipedia.org/wiki/Length_contraction

The ionization ability of electrically charged particles with large relative velocities is higher than expected. In pre-relativistic physics the ability should decrease at high velocities, because the time in which ionizing particles in motion can interact with the electrons of other atoms or molecules is diminished. Though in relativity, the higher-than-expected ionization ability can be explained by length contraction of the Coulomb field in frames in which the ionizing particles are moving, which increases their electrical field strength normal to the line of motion.

 

[Mr-R]

The electric field of the muon is length contracted in the frame of the Earth.

http://en.wikipedia.org/wiki/Length_contraction

ow I never thought of it like this. Thanks :thumbs:

 

[helpmeplz!]

Length contraction is concerned with how long the distance between the ends of an object is at a single given point in time. In the light clock example this is not the case, the clock has length zero in the direction of motion. The length that the light travels in the clock's rest frame is orthogonal to the direction of motion and therefore does not contract when transformed to a different frame.

However, since the mirrors are going to move in the ground rest frame, the light needs to move not only in the vertical direction, but also in the horizontal direction, in order to bounce between them. Thus, the vertical distance ##y## is the same while the horizontal distance ##x## increases. The total distance is of course given by Pythagoras' theorem ##\ell = \sqrt{x^2 + y^2} \geq y^2## which means that the time between the bounces will be larger in the ground frame.




If the muon had a size, yes that would be the case. However, elementary particles are point-like and it does not make so much sense to talk about the size of a muon.



It is a bit difficult to decipher what you are trying to say, but I will try to answer according to my understanding of it. Basically you are wondering what would happen if you put the light clock on the train horizontally rather than vertically? As you can imagine, things now become a bit more complicated, which is the reason to put the clock vertically in the first place. In the train rest frame, things will look exactly the same simply because the clock is not moving relative to the train.
In the ground rest frame, the clock is now length contracted - and moving. As you can imagine, in the ground frame it will take the light a longer time to go from the back mirror to the front mirror than vice versa as in that direction the light also has to catch up with the speed of the train and in the other direction it will meet faster because of the speed of the train.
Let us assume that the speed of the train is ##v## and that the clock rest length is ##L_0##. The time for the light to complete a full cycle in the train rest frame is of course ##T' = 2L_0/c##. So what is the time for the light to complete a full cycle (going from back mirror, to the front mirror, and back to the back mirror) in the ground rest frame? To start with, the clock has length ##L = L_0/\gamma## in the ground rest frame. The time ##t_1## to reach the front mirror is going to be given by the equation
$$
vt_1 + L = c t_1.
$$
The left-hand side of the equation is the length that the light has to travel in order to reach the position of the front mirror at ##t_1##, the right-hand side is the length the light travels in time ##t_1##. Solving for ##t_1## we are left with
$$
t_1(c-v) = L \quad \Rightarrow \quad
t_1 = \frac{L}{(c-v)}.
$$
In the same way, the time ##t_2## to go from the front to the rear mirror can be computed to be
$$
t_2 = \frac{L}{c(c+v)}.
$$
Thus, the total time is
$$
T = t_1+t_2 = L\left( \frac{1}{c+v} + \frac{1}{c-v}\right) =
L \frac{2c}{c^2-v^2} = \frac{L_0}{\gamma} \frac{2\gamma^2}{c}
= \frac{2L_0 \gamma}{c} = \gamma T'.
$$
Again we have the same time dilation formula as in the case of the vertical light clock. The reason that it takes the light longer to go forward than back is related to the fact that those two events are not located at the same point in space in the train's rest frame - which is the crucial assumption of deriving the time dilation.

Edit: Fixed LaTeX error

Thank you very much! So to measure the length of something you have to do it simultaneously? But couldn't you also measure the length of something passing by you by starting your timer when the front passes you and stopping it when the back passes you? And then if you know the speed, you know the length.

If so, then here's my question. Suppose the muon is moving at speed 0.9c, and its half life is 10 seconds. Then from its reference frame, it will have traveled 9 light seconds until it decays. Whereas from our reference frame, we perceive his clock as running slower, so wouldn't we find that he decayed in more than 10 seconds? For example 12 seconds of our clock might correspond to 10 seconds of his clock since his clock is ticking slower. Hence we would conclude that it traveled a greater distance, not a shorter one? Where am I going wrong?

Finally, I think i get what you wrote in number 3 but I was looking for an intuitive way of understanding length contraction with just a light clock and using time dilation. Thanks!

 

[Nugatory]

If so, then here's my question. Suppose the muon is moving at speed 0.9c, and its half life is 10 seconds. Then from its reference frame, it will have traveled 9 light seconds until it decays. Whereas from our reference frame, we perceive his clock as running slower, so wouldn't we find that he decayed in more than 10 seconds? For example 12 seconds of our clock might correspond to 10 seconds of his clock since his clock is ticking slower. Hence we would conclude that it traveled a greater distance, not a shorter one? Where am I going wrong?

(You don't really mean "half-life", you mean "lifetime". That's a digression here, I'm just pointing it out so you'll know why I'm using the word "lifetime")

You are mixing frames again. If the lifetime of a particle is ten seconds, that means that it will live for ten seconds in the frame in which is at rest. In a frame in which the particle is moving at .9c, it will live for more than ten seconds and therefore will move more than nine light-seconds before it decays. In the frame in which the particle is at rest, it will live for ten seconds, an object at rest in the other frame will approach the particle at a speed of .9c, and will move 9 light-seconds during the lifetime of the particle. So in one frame we have a distance of nine light-seconds covered in ten seconds, and in the other frame we have a distance more than nine light-seconds, covered in more than nine seconds.

 

[Orodruin]

Thank you very much! So to measure the length of something you have to do it simultaneously? But couldn't you also measure the length of something passing by you by starting your timer when the front passes you and stopping it when the back passes you? And then if you know the speed, you know the length.

Yes, if you know (or can measure) the velocity of an object, then you can measure its length in this way assuming that the velocity is constant. It will give the same result as measuring the distance between the ends at a given time. The latter is how we define the length, but it is relatively easy to come to the conclusion that you can measure it the other way too.

 

[helpmeplz!]

(You don't really mean "half-life", you mean "lifetime". That's a digression here, I'm just pointing it out so you'll know why I'm using the word "lifetime")

You are mixing frames again. If the lifetime of a particle is ten seconds, that means that it will live for ten seconds in the frame in which is at rest. In a frame in which the particle is moving at .9c, it will live for more than ten seconds and therefore will move more than nine light-seconds before it decays. In the frame in which the particle is at rest, it will live for ten seconds, an object at rest in the other frame will approach the particle at a speed of .9c, and will move 9 light-seconds during the lifetime of the particle. So in one frame we have a distance of nine light-seconds covered in ten seconds, and in the other frame we have a distance more than nine light-seconds, covered in more than nine seconds.

Thanks, how does this make sense in the muon experiment though? It does not seem like there is any length contraction in the frame where the muon is at rest. Hence the muon would never make it through to Earth before decaying. For instance using the numbers here, in the frame in which the particle is at rest, the object (seperation of it and the earth) will move 9 light seconds before it decays, and not more which is what we would need for it to make it to earth. I realize the actual numbers are different btw.

 

[Orodruin]

In the frame where the muon is at rest, the Earth atmosphere is length contracted.