Bah! I apologize for the poor grammer of my previous post. I also want to add that you may be able to purchase older editions of the same text as the content from say, the third to the fourth edition usually doesn't change that much. As for the problems, you should be able to copy them from the reserve text at the library.
Personally, I have found that the international editions tend to be of a lower quality than the more costly counterparts. For a text that I reference a lot, say Arfken, I would pay a bit more for the more study, hardbound, copy.
Perhaps I should have been a more careful with my wording. Mathematics is indeed a beautiful, and interesting area of study Insofar as science is concerned, it is merely a language.
No. You do not need a to be in possesion of a "shitload" of drugs to face felony charges in Florida. In general* possesion of any drug other than marijuana is a felony. If I recall correctly the amount of marijuana required for the state to pursue a felony charge is twenty grams, no matter though as it is much less than the five pounds that you have stated.
*Yes there is PTI etc..., however in most cases you will be charged/convicted with a felony. Each county will work in a slightly different manner, but in general my statement holds.
Not everybody in prison is a convicted felon. Some are awaiting the completion of their trial, others were convicted of misdemeanors.
Typically prisons house offenders who have been convicted of a felony. Jails house offenders awaiting trial, those convicted of misdemeanors offences and those .
Well, if you are using data from two different refernce frames, you can get some odd results.
As for muon decay, I have a bit of time so I am going to tell you what Tipler & Llewellyn have to say about it (pg.41-42 in the Modern Physics 4th ed.):
Muons decay according to N(t)=Noe^(-t/T) where No is the original number of muons at t = 0., N(t) is the number remaining at time t, and T is the mean lifetime (a proper time interval), which is about 2 microseconds for muons. Since muons are created high in the atmosphere, few muons should reach sea level. A typical muon moving with v =.998c would travel 600m in 2 microseconds. However, the lifetime of the muon measured in Earth's reference frame is increased due to time dilation by the factor gamma=1/(1-v^2/c^2)^1/2, which is 15 for this particular v. The mean lifetime measured in the Earth's reference frame is therefore 30 microseconds, and a muon with speed v travels about 9000 m in this time. From the muon's frame, it lives only 2 microseconds, byt the atmosphere is rushing past it at speed v. This distance of 9000 m in Earth's frame is thus contracted to only 600m in the muon's frame.
It is easy to distinguish experimentally between the classical and relativistic predictions of the obersvations of muons at sea level. Suppose we observe 10^8 muons at an altitude of 9000m in some time interval. How many would we expect to observe at sea level in the same time interval? According to the nonrelativistic prediction, the time it takes for these muons to travel 9000m is (9000m)/0.998c = (approx.) 30 microseconds, which is 15 lifetimes. Substituting No = 10^8 and t = 15T into the equation for N(t) we obtain N = (10^8)e^-15 = 30.6. We would expect all but about 31 of the original muons to decay before reaching sea level.
According to the relativistic prediction, Earth must travel only the contracted distance of 600m in the rest frame of the muon. This takes only 2 microseconds = 1T. Therefore the number of muons expected at sea level is N = (10^8)e^-1 = 3.68*10^7. Thus relativity predicts that we would observe 36.8 million muons in the same time interval. Experiments of this type have confirmed the relativistic predictions.
The last line is rather important. The effects of special relativity are very real and very observable provided you have the proper equipment.
You still fail to take into account length contraction! As mrzyx suggested, google for "Muon Decay" and "special relativity" and I believe that you questions shall be answered. If you are a university student, I suggest that you go to your physics library and check out a text on special relativity, I have heard French's text is reasonably good, although I have no personal experience with it. I will, however, suggest that you avoid Kogut's text like the plague.
I Understand you can't reach LS, was just using the easiest example. But even in that case, even considering H&K's flying clocks I still ask how much food you pack for the trip.. do I pack for.7 years or 7 years?
You would pack enough food for.7 years. That is the strange thing about special relativity, in one twin's reference frame, seven years have elapsed, in the other twin's frame, only.7 years have elapsed.
With regards to the question of velocity, I am rather certain that in the ship's frame of reference it is stationary. The ship sees earth2 moving twoards it at v =.99c.
I am not sure I follow you on transmitting information backwards in time. What we observe in the night sky is from the past but does not represent current time/locations for what we see ( unless you extrapolate based on your observations for how far away the object is... and your extrapolations are limited in acuracy by the N > 2 problem ).
If gravity is instentaneous then it would allow for concurrent communication even though our observations based on light transmission would indicate what we are communicating with would be in the past.
I believe I understand what the parent was trying to say, however the only way I am able to adequatly explain this is with a Minkowski diagram. I suggest you try googling it.
On a side note it is this reason I have always had trouble with the whole time dilation effect of light speed. In that same example if you left here in a ship traveling at light speed you would arrive on the parallel earth in 7 years time, or in the year 2011 on both planets. Its 2004 there now and in seven years you would be there in 2011. Many people seem to think you would get there in 2004.
If you have twins, one set here and one set there where one makes the trip then according to relativity the twin that makes the trip will be the same age as when they left thus from their perspective making the trip instantly while 7 years would pass for the twins that remained on their respective planets.
Actually, due to the fact that the ship (and you) have mass, you cannot travel at v=c, however you can approach c. For the sake of example, let v=.99c. In that case, the trip would not be insantaneous,it would take t' = t*[(1-v^2/c^2)]^1/2. Take t = 7 years and we find t' =.7 years. As for your concerns, time dilation is an
observable effect. It is quite real.
Slashdot Mirror
Bah! I apologize for the poor grammer of my previous post. I also want to add that you may be able to purchase older editions of the same text as the content from say, the third to the fourth edition usually doesn't change that much. As for the problems, you should be able to copy them from the reserve text at the library.
Personally, I have found that the international editions tend to be of a lower quality than the more costly counterparts. For a text that I reference a lot, say Arfken, I would pay a bit more for the more study, hardbound, copy.
Perhaps I should have been a more careful with my wording. Mathematics is indeed a beautiful, and interesting area of study Insofar as science is concerned, it is merely a language.
Mathematics is merely the language of science. Nothing more, nothing less.
Geek books you say? Well look no further than Introduction to Quantum Mechanics and Introduction to Electrodynamics, both authored by David Griffiths. If you are feeling particularly ambitious there is always the phone book.
No. You do not need a to be in possesion of a "shitload" of drugs to face felony charges in Florida. In general* possesion of any drug other than marijuana is a felony. If I recall correctly the amount of marijuana required for the state to pursue a felony charge is twenty grams, no matter though as it is much less than the five pounds that you have stated. *Yes there is PTI etc..., however in most cases you will be charged/convicted with a felony. Each county will work in a slightly different manner, but in general my statement holds.
Typically prisons house offenders who have been convicted of a felony. Jails house offenders awaiting trial, those convicted of misdemeanors offences and those .
As for muon decay, I have a bit of time so I am going to tell you what Tipler & Llewellyn have to say about it (pg.41-42 in the Modern Physics 4th ed.):
Muons decay according to N(t)=Noe^(-t/T) where No is the original number of muons at t = 0., N(t) is the number remaining at time t, and T is the mean lifetime (a proper time interval), which is about 2 microseconds for muons. Since muons are created high in the atmosphere, few muons should reach sea level. A typical muon moving with v = .998c would travel 600m in 2 microseconds. However, the lifetime of the muon measured in Earth's reference frame is increased due to time dilation by the factor gamma=1/(1-v^2/c^2)^1/2, which is 15 for this particular v. The mean lifetime measured in the Earth's reference frame is therefore 30 microseconds, and a muon with speed v travels about 9000 m in this time. From the muon's frame, it lives only 2 microseconds, byt the atmosphere is rushing past it at speed v. This distance of 9000 m in Earth's frame is thus contracted to only 600m in the muon's frame.
It is easy to distinguish experimentally between the classical and relativistic predictions of the obersvations of muons at sea level. Suppose we observe 10^8 muons at an altitude of 9000m in some time interval. How many would we expect to observe at sea level in the same time interval? According to the nonrelativistic prediction, the time it takes for these muons to travel 9000m is (9000m)/0.998c = (approx.) 30 microseconds, which is 15 lifetimes. Substituting No = 10^8 and t = 15T into the equation for N(t) we obtain N = (10^8)e^-15 = 30.6. We would expect all but about 31 of the original muons to decay before reaching sea level.
According to the relativistic prediction, Earth must travel only the contracted distance of 600m in the rest frame of the muon. This takes only 2 microseconds = 1T. Therefore the number of muons expected at sea level is N = (10^8)e^-1 = 3.68*10^7. Thus relativity predicts that we would observe 36.8 million muons in the same time interval. Experiments of this type have confirmed the relativistic predictions.
The last line is rather important. The effects of special relativity are very real and very observable provided you have the proper equipment.
You still fail to take into account length contraction! As mrzyx suggested, google for "Muon Decay" and "special relativity" and I believe that you questions shall be answered. If you are a university student, I suggest that you go to your physics library and check out a text on special relativity, I have heard French's text is reasonably good, although I have no personal experience with it. I will, however, suggest that you avoid Kogut's text like the plague.
You would pack enough food for .7 years. That is the strange thing about special relativity, in one twin's reference frame, seven years have elapsed, in the other twin's frame, only .7 years have elapsed.
With regards to the question of velocity, I am rather certain that in the ship's frame of reference it is stationary. The ship sees earth2 moving twoards it at v = .99c.
I believe I understand what the parent was trying to say, however the only way I am able to adequatly explain this is with a Minkowski diagram. I suggest you try googling it.
On a side note it is this reason I have always had trouble with the whole time dilation effect of light speed. In that same example if you left here in a ship traveling at light speed you would arrive on the parallel earth in 7 years time, or in the year 2011 on both planets. Its 2004 there now and in seven years you would be there in 2011. Many people seem to think you would get there in 2004. If you have twins, one set here and one set there where one makes the trip then according to relativity the twin that makes the trip will be the same age as when they left thus from their perspective making the trip instantly while 7 years would pass for the twins that remained on their respective planets.
Actually, due to the fact that the ship (and you) have mass, you cannot travel at v=c, however you can approach c. For the sake of example, let v=.99c. In that case, the trip would not be insantaneous,it would take t' = t*[(1-v^2/c^2)]^1/2. Take t = 7 years and we find t' = .7 years. As for your concerns, time dilation is an
observable effect. It is quite real.