I don't understand the result of my Earth-muon calculations

[GregAshmore]

This post is based on problem 1-11 in Taylor-Wheeler.

The problem is very similar to the others in this section, which is mostly devoted to the invariance of the spacetime interval. The problems are structured so that three of the four required numbers can be determined. The missing number can then be easily calculated.

The problem is stated this way:

Muons have a half-life of 1.5 microseconds, measured with respect to a reference frame in which they are at rest. All the muons are produced at the same height (60 kilometers); all have the same speed; all travel straight down; none are lost to collisions with air molecules on the way down.

Approximately how long will it take these muons to reach the surface of Earth, as measured in the Earth frame? [2E-4 seconds, assuming near light speed.]

A muon survives for three half-lives. [4.5E-6 sec]

In the muon (rocket) frame, what is the space separation between the birth of the muon and its arrival at the surface of Earth? [0 meters]

What is the spacetime interval between the birth event and the arrival event? [4.5E-6 sec]

That is the end of the problem in Taylor-Wheeler.

Now, suppose we turn the problem around and have the Earth be the rocket. The muon measures 4.5E-6 sec on its clock. The Earth whizzes by, traveling a distance of, say, 2.5E-6 light seconds in the muon frame. What is the time in the Earth frame?

The spacetime interval is sqrt(4.5E-6^2 - 2.5E-6^2) = 2.0E-6 sec. Because the Earth distance is zero in its own (rocket) frame, the spacetime interval is the time in the Earth frame. Therefore, the time in the Earth frame is 2.0 microseconds.

Obviously, 2 microseconds is much less than the 200 microseconds observed on the Earth. Why the difference?

 

[JesseM]

This post is based on problem 1-11 in Taylor-Wheeler.

The problem is very similar to the others in this section, which is mostly devoted to the invariance of the spacetime interval. The problems are structured so that three of the four required numbers can be determined. The missing number can then be easily calculated.

The problem is stated this way:

Muons have a half-life of 1.5 microseconds, measured with respect to a reference frame in which they are at rest. All the muons are produced at the same height (60 kilometers); all have the same speed; all travel straight down; none are lost to collisions with air molecules on the way down.

Approximately how long will it take these muons to reach the surface of Earth, as measured in the Earth frame? [2E-4 seconds, assuming near light speed.]

A muon survives for three half-lives. [4.5E-6 sec]

In the muon (rocket) frame, what is the space separation between the birth of the muon and its arrival at the surface of Earth? [0 meters]

What is the spacetime interval between the birth event and the arrival event? [4.5E-6 sec]

That is the end of the problem in Taylor-Wheeler.

Now, suppose we turn the problem around and have the Earth be the rocket. The muon measures 4.5E-6 sec on its clock. The Earth whizzes by, traveling a distance of, say, 2.5E-6 light seconds in the muon frame. What is the time in the Earth frame?

The spacetime interval is sqrt(4.5E-6^2 - 2.5E-6^2) = 2.0E-6 sec. Because the Earth distance is zero in its own (rocket) frame, the spacetime interval is the time in the Earth frame. Therefore, the time in the Earth frame is 2.0 microseconds.

Obviously, 2 microseconds is much less than the 200 microseconds observed on the Earth. Why the difference?

Because to measure the aging of the muon in the Earth frame, two clocks are required to make local measurements of the time at the position the muon is created in the upper atmosphere and the time and the position the muon reaches the Earth. In the muon frame, each of these clocks individually ticks forward by only two microseconds in the time between the muon's creation and the muon colliding with the Earth, but the clock on Earth is ahead of the clock in the upper atmosphere by 198 microseconds, so if the clock in the upper atmosphere read 0 microseconds at the moment the muon was created next to it, then at that moment (according to the muon frame's definition of simultaneity) the Earth clock already reads 198 microseconds, so by the time the muon reaches the surface the upper atmosphere clock reads 2 microseconds and the Earth clock reads 200 microseconds.

 

[PeterDonis]

I find that the best way to dispel confusion with a relativity problem is usually to draw a spacetime diagram, so I've attached one.

The diagram is drawn in the Earth frame and is scaled in kilometers (vertical distance on the x-axis, time in "light-kilometers" on the y-axis). The red line is a light ray from x = 60 km height at time t = 0 (Earth time), to zero height at time t = 60 km. The blue line, just above the light ray, is the worldline of one of the muons; I've used different specific numbers than the Taylor-Wheeler problem uses so that the difference in lines is visible on the diagram, and this worldline crosses the y-axis (which is zero height) at t = 61 km, meaning the muon's velocity is 60/61. (This means, of course, that it takes many more than three half-lives to complete the journey with the numbers I'm using; but the principle involved is the same with the numbers actually given in the problem.) This gives a gamma factor of 61/11, so only 11 km of time elapses in the muon's frame as it falls from 60 km to zero height.

The magenta line is the "line of simultaneity" in the muon frame, at the event where the muon starts out at x = 60 km. As you can see, this line crosses the y-axis at about t = 59 km; the actual slope of the line is 60/61, the inverse of the slope of the muon's worldline, so it actually crosses the y-axis at t = 3600/61 = 59.0164 km (to 4 decimal places, at least). So in the muon's frame, the Earth is rushing towards it at v = (60/61)c, and only 1.9836 km of time elapses on the Earth's clock (which is gamma times the 11 km elapsed in the muon's frame, as expected); *but* that time elapses between time t = 59.0164 km and t = 61 km on Earth's clock (i.e., between the magenta line crossing the y-axis and the blue line crossing the y-axis), so when the muon arrives at zero height, Earth's clock reads 61 km, *not* 1.9836 km.

 

[PeterDonis]

Just for good measure, here's another spacetime diagram drawn in the muon's frame. The scaling is the same (km distance on x-axis, light-km of time on y-axis), and the lines are as follows:

The green line is the same light ray as in the previous diagram.

The y-axis is the muon's worldline, and the x-axis is the line of simultaneity at the time the muon is at 60 km height as measured in the Earth frame--in the muon's frame, as you can see from the diagram, the distance is the "length contracted" distance of 60*(11/61).

The cyan line is the Earth's worldline (which crosses the x-axis at the "length contracted" distance just given), and the blue line is the Earth's "line of simultaneity" when the muon is at 60 km height in the Earth frame--in other words, the blue line is the same line that appeared as the x-axis in the "Earth frame" diagram in my previous post (and of course the cyan line, the Earth's worldline, was the y-axis in that diagram).

The (x, y) coordinates where the blue line meets the cyan line (which is the point that appeared as the origin in the previous diagram) are x = - 3660/11, y = - 3600/11. These are, of course, just the right values such that the "length contracted" distance to the origin along the blue line is 60 km, and the "time dilated" proper time along the cyan line to where it meets the y-axis (at y = + 11 km, which is the elapsed proper time to fall in the muon's frame) is 61 km.

 

[Dale]

Because the Earth distance is zero in its own (rocket) frame

The distance in Earth's frame is not zero. The birth of the muon occurs at the top of the atmosphere and the decay occurs at the bottom. Those locations have different coordinates in the Earth frame.

 

[GregAshmore]

The distance in Earth's frame is not zero. The birth of the muon occurs at the top of the atmosphere and the decay occurs at the bottom. Those locations have different coordinates in the Earth frame.

If the travel distance of the muon is zero in its own frame when the problem is set up with the muon as the rocket, why isn't the Earth's travel distance zero in its own frame when the problem is set up with the Earth as the rocket?

 

[Dale]

The Earth's travel distance is zero in the Earth's frame (by definition), but aren't you interested in the spacetime interval for the muon? Nothing interesting happens to the Earth in this problem (i.e. there are not two different events identified on the Earth's worldline).

 

[GregAshmore]

The Earth's travel distance is zero in the Earth's frame (by definition), but aren't you interested in the spacetime interval for the muon? Nothing interesting happens to the Earth in this problem (i.e. there are not two different events identified on the Earth's worldline).

I'm trying to analyze the problem from the point of view of the muon, without prejudice, so to speak. That is, if I were sitting in place of the muon observing the Earth go by, how would I interpret what I see according to SR?

I measure a certain time on my clock. In that time the Earth travels a certain distance as measured in my frame. (We assume that I know this distance by some means, analogous to our means of knowing the distance traveled by the muon in Earth's frame.)

With those two pieces of data--time and distance in my frame--and the knowledge that the Earth's travel distance is zero in its own frame, I can calculate the time in the Earth's frame. That time is the proper time for the event.

AND...now I see what I did wrong.

The reason this approach is wrong in this particular case is that the conditions are not truly the mirror image (so to speak) of the original problem.

In the problem as originally stated, the time of the event (the muon's life span) is first measured while the muon is at rest in the frame of measurement. That time, the muon's own time, must always be the time assigned to the event in the muon's frame, no matter which other frame is the viewing frame.

To be true to the form of the original problem, my construction of the problem should have first measured the time of an event in the Earth's frame (but not the life span of the Earth--much too long for the muon). That proper time, together with a distance in the muon frame, would be used to calculate the time in the muon frame.

The way I constructed the problem, I took the time and distance in the muon's frame and calculated the proper time of the event. That's fine--but that event is not the same event as in the original problem.

Which means that I took the long hard road to get to what you were trying to tell me in your reply.

This reminds me once again of the phrase which I want on my tombstone: "He was slow--but he got it eventually."

 

[Dale]

To be true to the form of the original problem, my construction of the problem should have first measured the time of an event in the Earth's frame (but not the life span of the Earth--much too long for the muon). That proper time, together with a distance in the muon frame, would be used to calculate the time in the muon frame.

The way I constructed the problem, I took the time and distance in the muon's frame and calculated the proper time of the event. That's fine--but that event is not the same event as in the original problem.

Yes, exactly.

FYI, just for some vocabulary, an event is a point in spacetime. I.e. a single position and time. So the "birth" of the muon is an event, and the "death" of the muon is a second event. The path that the muon takes from birth to death is called the muon's worldline. If you think of a 0-dimensional point particle it traces out a 1-dimensional line in spacetime. The spacetime interval measured along the worldline is called the proper time.

 

[John232]

You assumed that the muon would measure its own time to be the same as how long it took for it to reach Earth by its time. If the muons half life was 1.5 and it lived for 4.5 relative to Earth then the muon would be experiencing its time to be going 3 times slower than the Earth clock.

 

[John232]

The Earth should read 4.5 secounds to the muon frame if the muon traveled the same speed. It would go 3 times slower just like the muon did.

 

[John232]

The Earth should read 4.5 secounds to the muon frame if the muon traveled the same speed. It would go 3 times slower just like the muon did.

4.5 micosecounds

 

[John232]

Say, you have t' = t sqrt(1-v^2/c^2)

If t' is the dialated time that is slower then you would have to take that as the normal half life of the muon. Say, it assumed it was at rest then it's half life would always be 1.5 microsecounds. Then t would be the amount of time it actually lived 4.5 microsecounds. Then with that you can calculate the true velocity of your muon.

The velocity turns out to be sqrt[(1-sqrt(1/3))/c^2]

The sqrt of 1/3 is greater than one so then the velocity is some sqrt of a negagitive number that is imaginary. Then the equation doesn't work out for a velocity that has a percent of c that is greater than 1.

 

[GregAshmore]

Yes, exactly.

FYI, just for some vocabulary, an event is a point in spacetime. I.e. a single position and time. So the "birth" of the muon is an event, and the "death" of the muon is a second event. The path that the muon takes from birth to death is called the muon's worldline. If you think of a 0-dimensional point particle it traces out a 1-dimensional line in spacetime. The spacetime interval measured along the worldline is called the proper time.

Okay. Proper use of terminology aids understanding and avoids confusion.

Before:

I took the time and distance in the muon's frame and calculated the proper time of the event. That's fine--but that event is not the same event as in the original problem.

After:

I took the time and position intervals between the two events in the muon's frame and calculated the proper time of the episode. That's fine--but those events are not the same events as in the original problem.

 

[John232]

NVM, My calculator lied lol, not use to the one that is on the computer.

 

[John232]

Once you get the velocity you should be able to insert it into the equation as a percentage of the speed of light then leaving the one or the other times blank and solving for it should give you the same times. The times are in porportion to each other, so if one time is 1 and the other time is 3 then it only means one experience 3 times as much time for every 1 secound the other experiences 3 secounds. Then the dialated time would have to be of lesser value. If a muon experienced 4.5 microsecounds then that is greater than 1.5 microsecounds, but the muon hasn't experienced more time. It has actually experienced 1/3 less time to have it's lifespan increase by 3 times the normal amount.

 

[yuiop]

This post is based on problem 1-11 in Taylor-Wheeler.

The problem is very similar to the others in this section, which is mostly devoted to the invariance of the spacetime interval. The problems are structured so that three of the four required numbers can be determined. The missing number can then be easily calculated.

The problem is stated this way:

Muons have a half-life of 1.5 microseconds, measured with respect to a reference frame in which they are at rest. All the muons are produced at the same height (60 kilometers); all have the same speed; all travel straight down; none are lost to collisions with air molecules on the way down.

Approximately how long will it take these muons to reach the surface of Earth, as measured in the Earth frame? [2E-4 seconds, assuming near light speed.]

A muon survives for three half-lives. [4.5E-6 sec]

The time dilation ratio is 0.0002/0.0000045 = 44.44444

This equates to a velocity of 0.999746843c

At a velocity of 0.999746843c the Earth travels a distance of 0.999746843c*0.0000045 = 1.34872454 km or 4.49886081 light seconds in the muon frame. This means the thickness of the atmosphere in the Earth frame must be 1.34872454*44.44444 = 59.9433069 km or 1.99949349E-4 light seconds (so TW have obviously rounded the atmosphere height or clocks times).

1.34872454 km is the length contracted thickness of the Earth's atmosphere according to the muon and this explains why the atmosphere goes past the muon in only 4.5 microseconds from the muon's point of view.

In the muon's frame the the atmosphere travels past it in 4.5 microseconds.

What is the spacetime interval between the birth event and the arrival event? [4.5E-6 sec]
That is the end of the problem in Taylor-Wheeler.

Now, suppose we turn the problem around and have the Earth be the rocket. The muon measures 4.5E-6 sec on its clock. The Earth whizzes by, traveling a distance of, say, 2.5E-6 light seconds in the muon frame. What is the time in the Earth frame?

At a velocity of 0.999746843c the Earth travels a distance of 0.999746843c*0.0000045 = 1.34872454 kms or 4.49886081E-6 light seconds. (Not 2.5E-6 ls).

The spacetime interval is sqrt(4.5E-6^2 - 2.5E-6^2) = 2.0E-6 sec. Because the Earth distance is zero in its own (rocket) frame, the spacetime interval is the time in the Earth frame. Therefore, the time in the Earth frame is 2.0 microseconds.

The correct spacetime interval calculated from the muon's point of view is sqrt(4.5E-6^2 - 0^2) = 4.5E-6 seconds.

The correct spacetime interval calculated from the Earth point of view is sqrt(2E-4^2 - 1.99949349E-4^2) = 4.5E-6 seconds.

(The spacetime interval is invariant and should be same measured from any frame.)

Obviously, 2 microseconds is much less than the 200 microseconds observed on the Earth. Why the difference?

See above.