1 / 38

Special Case: Proper Time

Special Case: Proper Time. LTE’s: x 1 = (x 2 + v*t 2 ) / [1-(v/c) 2 ] 1/2 t 1 = (t 2 + v*x 2 /c 2 ) / [1-(v/c) 2 ] 1/2 IF a time interval is measured using ONE CLOCK, then there is no problem with simultaneity - but the beginning and ending have to happen at the same place:

darva
Télécharger la présentation

Special Case: Proper Time

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Special Case: Proper Time LTE’s: x1 = (x2 + v*t2) / [1-(v/c)2]1/2 t1 = (t2 + v*x2/c2) / [1-(v/c)2]1/2 IF a time interval is measured using ONE CLOCK, then there is no problem with simultaneity - but the beginning and ending have to happen at the same place: x2 i = x2 f , so we get: tnon-proper = tproper / [1-(v/c)2]1/2,tNP > tP

  2. Special Case: Proper Length LTE’s: x1 = (x2 + v*t2) / [1-(v/c)2]1/2 t1 = (t2 + v*x2/c2) / [1-(v/c)2]1/2 IF a length is measured using NO CLOCKS, then there is no problem with simultaneity - but the beginning and ending have to happen at the same time in the other system: t2 i = t2 f , so we get: xproper= xnon-proper /[1-(v/c)2]1/2,xP > xNP

  3. Review of Equations: LTE’s: good for one EVENT (x,t) x1 = (x2 + v*t2) / [1-(v/c)2]1/2 t1 = (t2 + v*x2/c2) / [1-(v/c)2]1/2. Time Dilation: good for a time interval: tnon-proper = tproper / [1-(v/c)2]1/2,tNP > tP Length Contraction: good for space interval xproper= xnon-proper /[1-(v/c)2]1/2,xP > xNP

  4. LTE Velocity Transformation x1 = (x2 + v*t2) / [1-(v/c)2]1/2 t1 = (t2 + v*x2/c2) / [1-(v/c)2]1/2. v1 = x1 / t1 = (x1 f - x1 i) / (t1 f - t1 i) = {[(x2 f + v*t2 f) - (x2 i + v*t2 i) ] / [1-(v/c)2]1/2 } / {[(t2 f+v*x2 f/c2) -(t2 i +v*x2 i/c2)] / [1-(v/c)2]1/2} = [(x2 f-x2 i) + v*(t2 f-t2 i)] / [(t2 f-t2 i) + v*(x2 f-x2 i)/c2] now divide numerator and denominator by (t2f-t2i) = (v2+v)/(1+v*v2/c2) = v1

  5. LTE Velocity Transformation v1 = (v2+v) / (1+v*v2/c2) Note: If v and/or v2 << c, then get Galilean result NOTE:IF v2 = c, THEN v1= (c+v) / (1+vc/c2) = (c+v) / (1+v/c) = (c+v) / (1/c)*(c+v) = c . This result DOES AGREE with the Michelson-Morley experiment!

  6. Relativity Example #1 Question: Can there be a moving system that sees the explosion from a cannon shell happen before the cannon fires the shell? NOTE: If this is possible, we have a problem with cause and effect!

  7. Relativity Example #1 Answer: This situation entails two events: the cannon fires at x1e=0m at t1e=0s, and the cannon shell explodes at x2e at t2e, where x2e=vcst2e . [BOTH x2e AND t2e are greater than zero.] From the moving frame, we are suggesting t2s< t1s (shell explodes before cannon fires). Since we are dealing with two events, we need the LTE:

  8. Relativity Example #1 LTE: ts = (te +/- vxe/c2)/[1-(v/c)2]1/2 for the first event:ts1= (0 +/- 0)/[1-(v/c)2]1/2 = 0s. for the second event: ts2 < ts1 = 0s, so we need: 0s > ts2 = (te2 +/- vxe2/c2)/[1-(v/c)2]1/2 or te2 < vxe2 /c2, but xe2 = vcste2, so te2 < v*vcste2/c2, or v> c2/vcs The shell exploding before the cannon fires (the effect precedes the cause) is possible only if either the cannon shell, vcs, or the ship, v, can go faster than light!

  9. Speeds greater than c Because of the square root in the denominator, if we have a speed, v, greater than c we will get imaginary results! In the previous example, if we have a speed, v, greater than c we will mess up cause and effect. Our theory will have to eventually explain why we can’t have speeds of objects go faster than light if we are to have a consistent theory.

  10. Relativity Example #2 Question: A spaceship is 300 m long as measured by the spaceship captain. The spaceship is moving at v=0.8*c relative to the space station. How big is the spaceship as measured by the space station personnel?

  11. Relativity Example #2 Answer: Since the spaceship is at rest with respect to the captain of the spaceship, this length of 300 m is a proper length. The length as determined by the space station personnel is NOT a proper length. Therefore, we can use: Lnp < Lp , or Lnp = Lp* [1-(.8)2]1/2 = 300 m * 0.6 = 180 m

  12. Relativity Example #3 Question: A muon has a half life of 2.2 s in the lab. How fast are a beam of muons moving if they travel 1200 m on average before decaying?

  13. Relativity Example #3 Answer: Since the muons were at rest in the lab, the 2.2s is the proper time interval. Since the distance of travel is xearth =v*tearth the non-proper time interval, tearth, is (1200 m) / v. Since tnp > tp , (1200 m)/v = tearth = 2.2s / [1-(v/c)2]1/2 , or V = (1200m) * [1-(v/c)2]1/2 / 2.2s , or squaring both sides: [1-(v/c)2]*(1200 m/2.2s)2 = v2

  14. Relativity Example #3 [1-(v/c)2]*(1200 m/2.2s)2 = v2 the quantity (1200 m / 2.2 s ) can be simplified to: (1200 m / 2.2 s) = 5.45 x 108 m/s = 1.82 * c [1-(v/c)2]*(1.82*c)2 = v2 Now bring all the terms with v to the left: (1.82*c)2 = [1+(1.82)2]*v2 So we finally get: v = (1.82*c)/[1+(1.82)2]1/2 = .877*c

  15. Relativity Example #4 Question: A rocket moving toward the earth at a speed of 0.8*c fires a missile going at a speed of 0.95*c (relative to the rocket) directed toward the earth. How fast does the earth see the missile approaching?

  16. Relativity Example #4 Answer: v1 = (v2+v) / (1+v*v2/c2) vearth = (vmissile/rocket +/- vrocket) / [1 +/- vmr*vr/c2] use + sign in both since both rocket and missile are approaching the earth vearth = (0.95*c + 0.80*c) / [1 + 0.95*0.80] = 1.750*c / 1.760 = 0.994*c .

  17. Relativity Example #4 If the missile were fired away from the earth, we would have instead: vearth = (vmissile/rocket +/- vrocket) / [1 +/- vmr*vr/c2] vearth = (-0.95*c + 0.80*c) / [1 - 0.95*0.80] = -0.150*c / 0.240 = -0.625*c . The minus sign means the missile is headed away from the earth.

  18. Relativity Example #4 • If the missile approaching the earth (first case) then fires a laser pulse at the earth, how fast does the earth measure for the speed of the laser pulse? • This problem is a giveaway, since everybody always measures the speed of light (that is what a laser pulse is) as c = 3 x 108 m/s !

  19. Relativity • Note: for all these equations to work, the speed of one system with respect to another must be less than c (v < c). The equations will break down if v>c, since there is the square root of [1 - (v/c)2] in the denominator! • To be a good theory, then, the theory must explain WHY v<c ! • If x and t change between systems, does m? See the introduction to the Relativity 2 computer homework to see whether m changes as x and t do.

  20. Determining mass in different systems As was demonstrated in the Relativity 2 computer homework introduction, the determination of mass of an object depends on the system. The “best” mass is the mass determined in a frame in which the mass is at rest (call this mass mo). m = mo / [1-(v/c)2]1/2, m(v) > mo tnon-proper = tproper / [1-(v/c)2]1/2,tNP > tP xproper= xnon-proper /[1-(v/c)2]1/2,xP > xNP

  21. Kinetic Energy KE = Workdone = Force thru distance and if mass is constant, then dp = m*dv, so

  22. Relativistic Kinetic Energy However, if mass is not constant, then we need to do something different: Note that dp = d(mv) = vdm + mdv, so

  23. Relativistic Energy Since the integral on the previous slide looked hard, we will turn “tricky”: consider m = mo/[1-(v/c)2]1/2, or m2[1-v2/c2] = mo2, or m2c2 - m2v2 = mo2c2 ; now take d( ) of each side: 2mc2dm - 2mv2dm - 2m2vdv = 0 , or c2dm = v2dm + mvdv !

  24. Relativistic Energy c2dm = v2dm + mvdv so we can now write:

  25. Relativistic Energy If we start from rest, then we can have for the kinetic energy: KE = mc2 - moc2. Recall the m = m(v), so we can also write: KE(v) = moc2 ({1/[1-(v/c)2]1/2 } - 1). This does not look at all like KE = (1/2)mv2, but if v/c is small, then using the approximation: 1/[1-(v/c)2]1/2= 1 + (1/2)v2/c2 + ... we do get: KE = (1/2)mov2 + negligible terms

  26. Relativistic Energy kinetic energy: KE(m) = mc2 - moc2, and KE(v) = moc2 ({1/[1-(v/c)2]1/2 } - 1). Note that as v approaches c, that KE approaches infinity! This means that we would need infinite energy to reach the speed of light - an impossibility! Thus we cannot go faster than light!

  27. Experimental Evidence • Michelson-Morley experiment • Pair Production and Pair Annihilation Note: E = mc2 and pair annihilation says that if we had 1 lb of anti-matter, we could combine it with 1 lb of regular matter and have about 1 kg for m. The the energy released would be: 1 kg x (3x108 m/s)2 = 9x1016 J, or 1000 MW*9x107 sec = 1,000 MW for 3 years!

  28. Experimental Evidence In the cyclotron, accelerate a charge particle with voltage difference, but make particle go in a circle with magnetic field: F = qv x B and F = ma, so qvB=m(v2/r), or qB = mv/r (but v = r), so  = qB/m . Normally,  is a constant, independent of r and v. But as v approaches c, we need to slow  of Voltage source down to keep giving particle more energy: SYNCHROTRON

  29. Experimental Evidence In radioactive decay, the mass of the radioactive atom is larger than the masses of the resulting particles. In addition, the mass of carbon-12 (which consists of 6 protons and 6 neutrons) is less than the mass of 6 protons and 6 neutrons! How can these be? Where is the “missing mass” in each case?

  30. Experimental Evidence In the labs, when muons are relatively slow moving, they have a half life of about 2.2s But when they are created in the upper atmosphere and have a very high speed, they seem to go further than the x=vt = 3x108 m/s * 2.2x10-6 s = 660 m . This can be explained by time dilation and length contraction.

  31. Consequences If we can not go faster than c, and if we cannot live longer than 100 years, are we confined to a radius of 100 lt-yrs in our universe? Consider which time (proper or non-proper) the 100 years is. Consider which distance (proper or non-proper) the 100 lt-yrs is.

  32. Relativity and Light • Does light have mass? But m = mo /[1-(v/c)2]1/2 , but with v=c, m=mo/0, or mo = m * 0 ! This means that mo=0! But light never stops (if it remains light), so can’t directly measure mo! • Does light have energy? E = hf, E = mc2 therefore, light does have mass: m = hf/c2 , just no rest mass • If light has mass and is moving, does light have momentum? p = mv = mc = (E/c2)c = E/c = hf/c, but f=c, or =c/f, so p = h/(The DeBroglie relation!)

  33. Relativity and Light If light has mass, does gravity act on it (bend it)? Since light moves so fast, gravity will not have much time to deflect it. If we have starlight go very close to the sun, we do indeed see a slight deflection! We even have BLACK HOLES in which the gravity attracts the light so strongly that it cannot escape!

  34. Relativity Example #5 How much energy would it take to accelerate a 1 kg object from rest to 0.5*c a) classically? b) relativistically? a) KE = (1/2)mv2 - 0 = (1/2)*(1 kg)* (1.5x108 m/s)2 - 0 = 1.125x1016 J

  35. Relativity Example #5 b) KE = moc2(1/[1-(v/c)2]1/2) - moc2 = (1 kg)*(3x108 m/s)2 *{1/ [1-(0.5)2]1/2 - 1} = 9x1016 J *{1.155 - 1} = 1.392x1016 J . (Recall from part a, that classically, KE = 1.125x1016 J.) The relativistic amount is larger than the classical amount, as we should have expected.

  36. Relativity Example #6 How fast would an electron be going and what would be the mass of this electron if we gave it 1.2 MeV of kinetic energy? [The rest mass energy of an electron is: moc2 = (9.1x10-31 kg)*(3x108 m/s)2= 8.19x10-14 J x (1 eV / 1.6x10-19 J) = 511,000 eV = .511 MeV. KE > moc2 makes the problem relativistic.]

  37. Relativity Example #6 KE = moc2 ({1/[1-(v/c)2]1/2 } - 1) , so 1.2 MeV = 0.511 MeV * ({1/[1-(v/c)2]1/2 } - 1) (1.2 + 0.511)MeV = 0.511 MeV/[1-(v/c)2]1/2 } or 1 - (v/c)2 = (0.511 / 1.711)2 , or (v/c) = [1 - (0.511/1.711)2]1/2 , or v = .954*c ; and m = mo / [1-(v/c)2]1/2 m = mo/2.99 = 3.348*mo = 3.05x10-30 kg .

  38. Relativity Example #7 Let’s contrast this with a photon with the same 1.2 MeV of energy. What is the mass of an x-ray of KE=1.2 MeV? KE = E (since mo for photon=0) = mc2 = hf = hc/ = 1.2 MeV*1.6x10-13J/MeV = 1.92x10-13 J. m = E/c2 = 1.92x10-13 J / (3x108 m/s)2 = 2.13x10-30 kg. (Note that this is a little less than the mass of an electron of KE=1.2 MeV.) = hc/E = (6.63x10-34Js)*(3x108m/s)/1.92x10-13J =1.036x10-12 m(in the x-ray range).

More Related