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Normandy6644
2004-Sep-16, 01:46 AM
Ok guys, my brain is fried right now from a lot of different things going on, so I need some help. (as with all homework help, I need to be pushed in the right direction, not told the answer :wink: )


#1. Suppose an event in reference frame S, an event A occurs at position x=0, t=0. Suppose that this causes a light beam to travel at v=2c to position x=L, whereupon it causes event B. Show that if this were true there exists a reference frame S' moving relative to S with relative velocity v (v<c) in which the order of the events are reversed.

Ok, I used the Lorentz transformations to find the coordinates of event B in S' (event A is just x'=0, t'=0), but I don't know how to show that this means the events are reversed...


Imagine there are a species of space elephants who are 1 meter tall at birth and grow at a rate of 1m per year, in their own time frame. An observer at rest in some inertial reference frame observes two spaceships (A and B) both moving at v=(4/5)c travelling toward him in opposite directions. At time t=-(5/4) year, when each ship is 1 light year away from his position, an elephant is born in each one. At time t=0, the two ships pass by each other at his position at which point the elephants have grown to the same height.

#2. An observer on ship A would say that the clocks on B are slower, and hence the elephant is growing more slowly. Relative to the observer on A, will the elephants be the same height when the two ships pass? You must determine the coordinates of both births according to an observer on ship A as well as the heights of both elephants as measured by observer A at the moment when the ships pass.

I understand the problem. Basically, by finding the coordinates, you know how long it will take for the ships to pass, and you also know how fast the elephants grow, so you can calculate the heights of both. My instincts tell me they won't be the same height, but I can't seem to do the transformation correctly.


Suppose a cosmic ray is a proton of energy 10^20eV. How long would it take this proton to cross our galaxy as measure on the proton's wristwatch? (the diameter of the galaxy is 10^5 light years and the rest energy of the proton is 10^9eV. How many centuries would this trip take as observed in our Earth frame?

My only problem with this one is finding the velocity of the proton, because I keep getting c, which is bad because there should be some time dilation according to the proton as it moves across the galaxy.

I'm real stuck on this. I'm gonna keep working, but any help will be appreciated. :D

Normandy6644
2004-Sep-16, 02:31 AM
Ok, I got number 1. :D

Normandy6644
2004-Sep-16, 03:13 AM
Now we're down to only the second question...

Normandy6644
2004-Sep-16, 03:44 AM
Man am I the only one doing my homework around here? :lol:

Charlie in Dayton
2004-Sep-16, 04:10 AM
Depends...how are you splitting the grade?

chiaroscuro25
2004-Sep-16, 04:45 AM
Since this is your homework, only hints. :wink:

Actually, the second problem isn't completely well posed as stated. Consider the orientation of the elephants with respect to their motion; how would that affect the result? Assuming that that isn't an issue, think about the implications of what the observers on Earth see.

Eroica
2004-Sep-16, 07:52 AM
Suppose a cosmic ray is a proton of energy 10^20eV. How long would it take this proton to cross our galaxy as measure on the proton's wristwatch? (the diameter of the galaxy is 10^5 light years and the rest energy of the proton is 10^9eV. How many centuries would this trip take as observed in our Earth frame?

My only problem with this one is finding the velocity of the proton, because I keep getting c, which is bad because there should be some time dilation according to the proton as it moves across the galaxy.

How are you calculating the velocity? These are the two equations I used:

E = (mc) + (pc)

p = mv/(√[1-(v/c)]

And I got v = c too! ](*,)

I guess that means it would take zero seconds on the proton's wristwatch.

milli360
2004-Sep-16, 10:34 AM
chiaroscuro25:
Actually, the second problem isn't completely well posed as stated. Consider the orientation of the elephants with respect to their motion; how would that affect the result? Assuming that that isn't an issue, think about the implications of what the observers on Earth see.
I think that is the whole point of using "height" -- it's perpendicular to motion here on earth, and intuitively we'd expect the same in any illustration of the ship's passing. That's a good point, but I think it's a reasonable assumption that the height is perpendicular to the spaceship motion.

If sam5 were here, he'd be able to tell you the answer in a second. That's a hint--and, remember, instincts are lousy in this area. :)

Normandy6644
2004-Sep-16, 11:41 AM
Suppose a cosmic ray is a proton of energy 10^20eV. How long would it take this proton to cross our galaxy as measure on the proton's wristwatch? (the diameter of the galaxy is 10^5 light years and the rest energy of the proton is 10^9eV. How many centuries would this trip take as observed in our Earth frame?

My only problem with this one is finding the velocity of the proton, because I keep getting c, which is bad because there should be some time dilation according to the proton as it moves across the galaxy.

How are you calculating the velocity? These are the two equations I used:

E = (mc) + (pc)

p = mv/(√[1-(v/c)]

And I got v = c too! ](*,)

I guess that means it would take zero seconds on the proton's wristwatch.

That turned out to be okay. I found the gamma factor from KE=gamma*moc^2-moc^2, and so I divided the time it took in the earth's frame (920 centuries or 1.2E9 seconds I think) by gamma (10^10) and got 29 seconds, which apparently is right. :)

Wally
2004-Sep-16, 01:54 PM
Remember, at the moment of passing, everyone must agree. Since from your frame of reference, you see both elephants being of equal height (since they were born at the same time from your FoR), then each ship would as well. The difference will be in the perceived moment of birth. You saw them happening at the same time. I quarantee you, those on the ships did not!

Normandy6644
2004-Sep-16, 02:48 PM
If sam5 were here, he'd be able to tell you the answer in a second.

What, that these problems have no point because SR is flawed? :D

chiaroscuro25
2004-Sep-16, 03:21 PM
I think that is the whole point of using "height" -- it's perpendicular to motion here on earth, and intuitively we'd expect the same in any illustration of the ship's passing. That's a good point, but I think it's a reasonable assumption that the height is perpendicular to the spaceship motion.
I'm quite certain you are correct in that the posers of the problem meant you to assume that, but I'm not sure I'd call it 'reasonable'. Height is generally not perpendicular to motion on earth either, e.g. if you're walking on stairs, taking an elevator, sitting in an ascending/descending plane, sleeping in a train car, etc. Better to be precise in these cases.

When I said "Think about the implications of what the observers on Earth see.", I was hinting at what Wally said directly: all of them had better see the same height when they pass.

Regarding the third problem, physicists don't bother calculating the velocity because you just get v ~ c as you've seen. Instead, they calculate and state the Lorentz factor \gamma.

Eroica
2004-Sep-17, 04:45 PM
How are you calculating the velocity? These are the two equations I used:

E = (mc) + (pc)

p = mv/(√[1-(v/c)]

And I got v = c too! ](*,)

I just realized what I was doing wrong. I was forgetting to convert electron volts to joules! ](*,)

chiaroscuro25
2004-Sep-17, 05:06 PM
I just realized what I was doing wrong. I was forgetting to convert electron volts to joules! ](*,)Well, that certainly doesn't help overall, but you'll still get v ~ c.

ToSeek
2004-Sep-17, 05:07 PM
How are you calculating the velocity? These are the two equations I used:

E = (mc) + (pc)

p = mv/(√[1-(v/c)]

And I got v = c too! ](*,)

I just realized what I was doing wrong. I was forgetting to convert electron volts to joules! ](*,)

There is a tiny little difference. ;)

George
2004-Sep-17, 09:29 PM
I'm still a rookie at this but the elephant one looks pretty straight forward.

I don't know how much to not tell you since this is homework....Hmmmm
Try these questions.

1) 4/5 (or .8c) is nice as it gives you a gamma of......what?

2) At .8c, how much time, as seen by the middle man, will it take for them to get to him?

3) How much time will the spaceships see it will take them to get to the middle man?

4) In what frame of refrence are they really growing?

5) So how much will they grow based on question 4?


I suspect the Doppler issue is required when comparing the elephant sizes and times between the two ships.

Hope this helps as it would mean I am right for a change on SR. :)

Eroica
2004-Sep-18, 09:27 AM
Suppose a cosmic ray is a proton of energy 10^20eV. How long would it take this proton to cross our galaxy as measured on the proton's wristwatch? (the diameter of the galaxy is 10^5 light years and the rest energy of the proton is 10^9eV. How many centuries would this trip take as observed in our Earth frame?
I found the gamma factor from KE=gamma*moc^2-moc^2, and so I divided the time it took in the earth's frame (920 centuries or 1.2E9 seconds I think) by gamma (10^10) and got 29 seconds, which apparently is right. :)
I don't get this. If the proton is travelling at less than c, shouldn't it take more than 1000 centuries for it to traverse the galaxy as measured by Earth clocks?

Also, isn't gamma = 10^11?

My answer is that the trip takes just over 3 seconds as measured by the proton's wristwatch. :-k

Eroica
2004-Sep-18, 09:43 AM
Regarding the third problem, physicists don't bother calculating the velocity because you just get v ~ c as you've seen. Instead, they calculate and state the Lorentz factor \gamma.
But surely you still have to derive v from the Lorentz factor if you want to find out how long it will take the proton to traverse the galaxy as measured by Earth clocks? 0.999999999999999999995 light-years per year is ~ c, but not quite = c. :-k

Normandy6644
2004-Sep-18, 01:35 PM
Suppose a cosmic ray is a proton of energy 10^20eV. How long would it take this proton to cross our galaxy as measured on the proton's wristwatch? (the diameter of the galaxy is 10^5 light years and the rest energy of the proton is 10^9eV. How many centuries would this trip take as observed in our Earth frame?
I found the gamma factor from KE=gamma*moc^2-moc^2, and so I divided the time it took in the earth's frame (920 centuries or 1.2E9 seconds I think) by gamma (10^10) and got 29 seconds, which apparently is right. :)
I don't get this. If the proton is travelling at less than c, shouldn't it take more than 1000 centuries for it to traverse the galaxy as measured by Earth clocks?

Also, isn't gamma = 10^11?

My answer is that the trip takes just over 3 seconds as measured by the proton's wristwatch. :-k

It's either 1x10^11 or 10^10 (I made that same mistake too). Doing that you get 30s (well, 29 depending on how much you round).

Normandy6644
2004-Sep-18, 01:39 PM
Regarding the third problem, physicists don't bother calculating the velocity because you just get v ~ c as you've seen. Instead, they calculate and state the Lorentz factor \gamma.
But surely you still have to derive v from the Lorentz factor if you want to find out how long it will take the proton to traverse the galaxy as measured by Earth clocks? 0.999999999999999999995 light-years per year is ~ c, but not quite = c. :-k

You can it this way: We know the energy of the proton (10^20eV or something, whatever i said :) ). Use the equation KE=gamma*(rest energy)-(rest energy). We are also give that the rest energy is 10^9. Solve for gamma, get around 10^10. Now for the earth frame assume that the particle is travelling at about c (since it won't really affect the calculation). Th distance is 10^5 light years (convert this to meters) and then divide by c to get the time in seconds. Divide by gamma to get the time as measured by the proton.

Eroica
2004-Sep-18, 02:35 PM
It's either 1x10^11 or 10^10 (I made that same mistake too). Doing that you get 30s (well, 29 depending on how much you round).
What mistake?

(Kinetic Energy + Rest Energy)/(Rest Energy) = Gamma

But KE + RE = Total Energy = 10^20 eV

(10^20)/(10^9) = Gamma = 10^11.

Normandy6644
2004-Sep-18, 02:45 PM
It's either 1x10^11 or 10^10 (I made that same mistake too). Doing that you get 30s (well, 29 depending on how much you round).
What mistake?

(Kinetic Energy + Rest Energy)/(Rest Energy) = Gamma

But KE + RE = Total Energy = 10^20 eV

(10^20)/(10^9) = Gamma = 10^11.

What I meant by mistake was I originally calculated everything using 10^11, which isn't right (I have the solutions now, so i know for sure). We have (10^20+10^9)/10^9, which gives 10^10.

chiaroscuro25
2004-Sep-18, 06:33 PM
But surely you still have to derive v from the Lorentz factor if you want to find out how long it will take the proton to traverse the galaxy as measured by Earth clocks? 0.999999999999999999995 light-years per year is ~ c, but not quite = c. :-k
Well, if you feel the bizarre need for 21+ digit accuracy, then go ahead and calculate v. :roll: Of course, they rounded every other parameter in the problem to one digit, so it's quite pointless to do so.

(Does anyone even know of a calculator/program that will give you that many digits?)

Normandy6644
2004-Sep-18, 09:44 PM
(Does anyone even know of a calculator/program that will give you that many digits?)

Computer algebra systems generally will, i.e., maple or mathematica.

Eroica
2004-Sep-19, 08:04 AM
What I meant by mistake was I originally calculated everything using 10^11, which isn't right (I have the solutions now, so i know for sure). We have (10^20+10^9)/10^9, which gives 10^10.
I think I see the problem now. You're mixing up two different forms of notation, powers and exponentials:

10^1 = 10

10E1 = 100 (E1 means x10^1)

As the problem was originally worded, Gamma = 10^11, and I still insist that the correct answer is about 3 seconds.

Could someone else jump in here an arbitrate? Am I just being ornery?

George
2004-Sep-20, 11:55 PM
As the problem was originally worded, Gamma = 10^11, and I still insist that the correct answer is about 3 seconds.

Could someone else jump in here an arbitrate? Am I just being ornery?
The gamma looks right, but, I get a length contraction of the galaxy of 8.9e9 meters. This should yield about 30 seconds.

So you may be off a decimal place or is it me that is "ornery"? #-o

That proton needs to be on the other side of the galaxy, too. :)

chiaroscuro25
2004-Sep-21, 01:15 AM
I also get about 30 seconds. (Gamma is 1E11.)

Eroica
2004-Sep-21, 11:54 AM
Thanks, guys. Yes, I must have been misplacing a decimal point somewhere. Now I get 30 seconds or so.


I also get about 30 seconds. (Gamma is 1E11.)Then I was right about Gamma = 10^11. So how did Normandy get 30 seconds using Gamma = 10^10?

Fram
2004-Sep-21, 12:38 PM
Normandy6644 wrote:
chiaroscuro25 wrote:
(Does anyone even know of a calculator/program that will give you that many digits?)


Computer algebra systems generally will, i.e., maple or mathematica

Don't need to look that far: your Windows calculator in scientific mode goes way beyond 21 digits (32 digits, to be more precise, on my Windows 2000 version).

Hey, I'm giving useful information on an astronomy board. Man, I must be smart 8)

Francis

Normandy6644
2004-Sep-21, 01:26 PM
Thanks, guys. Yes, I must have been misplacing a decimal point somewhere. Now I get 30 seconds or so.


I also get about 30 seconds. (Gamma is 1E11.)Then I was right about Gamma = 10^11. So how did Normandy get 30 seconds using Gamma = 10^10?

Something seems to work somehow, though the gamma I got was either 10^10 or 1E11, which are the same.

George
2004-Sep-21, 02:28 PM
Something seems to work somehow, though the gamma I got was either 10^10 or 1E11, which are the same.
Oh?

chiaroscuro25
2004-Sep-21, 02:36 PM
Something seems to work somehow, though the gamma I got was either 10^10 or 1E11, which are the same.
No, I'm afraid not. 10E10 = 1E11, but that's not equal to 10^10 = 1E10. That's where your error must've arisen.

Normandy6644
2004-Sep-21, 02:39 PM
Something seems to work somehow, though the gamma I got was either 10^10 or 1E11, which are the same.
Oh?

Oops! Didn't mean to write it like that. I meant to say that if you use a single notation for the problem you get the same answer, whether you put it in terms of 1 or 10.

Normandy6644
2004-Sep-22, 09:08 PM
Here's another one (just for fun, since I already solved it 8) )


A 1kW light beam from a laser is used to levitate a solid aluminum sphere by focusing it on the sphere from below. What is the diameter of the sphere, assuming that it floats freely on the light beam? (Assume that the sphere is highly polished, so that all the light reflects directly back, and that the sphere is near the surface of the earth. Also assume that the photon energy is much less than the rest mass energy of the sphere).

Have fun. (there are others, but I want to work them out first)

Normandy6644
2004-Sep-23, 12:29 AM
Ok, here are the others.


The pi meson (rest energy mc^2=135MeV) decays into two photons. a) find the energy of the two photons if the pion is at rest when it decays. b) suppose that the pion has energy of 1GeV in the lab at the time when it decays. Find the energies of the two photons in the lab if one emerges in the direction in which the pion was travelling.


A particle called a D* with rest energy Mc^2=2010MeV is at rest when it disintegrates into a D meson (rest energy mc^2=1865MeV) and a pi meson (rest energy mc^2=135MeV). Compute the momentum of the pion.