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1 Uniform Acceleration Physics 139 Relativity Relativity Notes 2003 G. F. SMOOT Oce 398 Le Conte DepartmentofPhysics, University of California, Berkeley, USA 94720 Notes to b e found at: http://aether.lbl.gov/www/classes/p139/homework/homework.html 1 Uniform Acceleration This material is to prepare a transition towards General Relativity via the Equivalence Principle by rst understanding uniform acceleration. The Equivalence Principle stated in a simple form: Equivalence Principle: A uniform gravitational eld is equivalent to a uniform acceleration. This is not very precise statement and one lesson wehave learned in Sp ecial Relativity is the need to b e precise in our statements, de nitions, and use of co ordinates. We will come to a more precise statement of the Equivalence Principle in terms like at a space-time p oint with gravitational acceleration ~g there is a tangent reference frame undergoing uniform acceleration that is equivalent. This is similar to the instantaneous rest frame of Sp ecial Relativity in the case of an ob ject undergoing acceleration. We rst need to understand carefully what is a uniform acceleration reference frame, whichwe will do in steps. First imagine a reference frame { a rigid framework of rulers and clo cks, our standard reference frame { undergoing uniform acceleration. In classical nonrelativistic physics we can imagine a rigid framework to whichwe can apply a force which will cause it to move with constant acceleration. However, in Sp ecial Relativity no causal impulse can travel faster than the sp eed of light, thus the frame work cannot b e in nitely rigid. When the force causing the acceleration is rst applied, the p oint where the force is rst applied b egins to accelerate rst and as the casual impulse moves out, the other p ortions join in the acceleration. Consider a simple long ro d as an example: If one pulls on a long ro d, it will lengthen at rst as the end b eing pulled starts moving b efore the other end even knows it is. Then as it gains sp eed, Lorentz-FitzGerald contraction will cause it to shorten. If one pushes on the long ro d from b ehind, it will rst shorten as the end with the force moves toward the other end which sits there unaware of the some to arrive acceleration. All ob jects, however rigid, evidently display some degree of elasticity during acceleration. It is clear that in Sp ecial Relativity no ro d can b e in nitely rigid but must b e elastic at some level. Home work problem: prove that since the sp eed 1 of sound is less than or equal to the sp eed of light, that the rigidityofany material is less than xxx? As a b o dy accelerates, it moves in a continuous fashion from one inertial system to another. If it is to retain its same rest length in its instantaneous rest system, then it length relative to its original inertial system will have to decrease continuously b ecause of Lorentz-FitzGerald length contraction. If, on the other hand, it retained the same length relative to the original inertial system, then the Lorentz-FitzGerald contraction would require its rest length to increase as its gains sp eed. This is not very satisfactory. Either way, the metric will dep end up on time. If wewant a direct comparison to gravity,we need to require an accelerated co ordinate system to have a time indep endent form. 1.1 Accelerating a Point Mass A uniformly accelerating p oint mass is one that is sub ject to the same force in each and every one of its instantaneous rest systems. I.e. a uniformly accelerating p oint ~ mass is sub ject to a constant force F = m ~g along the +x-axis in a co ordinate system o which is the inertial frame where its velo city is zero instantaneous rest frame. Instantaneous 0 Lab oratory Frame S Rest Frame S 0 ct = ct ct L 6 6 @ @ 6 @ @ @ - @ 0 - x x = x L @ x 0 @ @ @ Light Cone Acceleration transforms as 2 0 d x a x a = = 1 x 3 0 2 vu dt 3 x 1+ 2 c 3 0 so that in the instantaneous rest frame a = a . In the instantaneous rest frame 0 F = F .Nowwe can solve the equation of motion in either of twoways: from the x x acceleration or from the force. In Problem Set 2 we solved the problem for a uniformly 1 accelerating ro cket using the acceleration transformation. 1 0 0 In ro cket frame the acceleration was a = g Note reversal of S and S compared to discussion x 2 Here we use force transformation. 0 dp x 0 F = = m g = F o x x 0 dt 2 d m c o = m g o 0 dct g 0 d = dct 2 c 0 g gt 0 = ct = 2 2 c c where the constantofintegration is set equal to zero b ecause we de ne the time zero to b e when = 0. This can b e turned into an equation for alone: 0 g gt 0 p = ct = = 2 2 c c 1 ! 2 2 0 gt = 2 1 c 0 gt =c q = 3 2 0 1+gt =c So that, from the lab oratory, observing the test particle start from rest we rst see its velo city increasing linearly with time as we classically exp ect for a particle under uniform acceleration. Then as the velo city b egins to b e a signi cant fraction of the sp eed of light, the term in the denominator b ecomes increasing imp ortant and the velo city increases ever more slowly in time and only approaches the sp eed of light asymptotically. The shap e of the tra jectory of a particle undergoing uniform acceleration is a hyp erb ola and not the classical parab ola, but for lowvelo cities they are indistinguishable conics. Note also s 2 0 gt 1 p = 1+ 4 = 2 c 1 If we lo ok at the Lorentz factor ,we see that it is rst very nearly unity and then as the velo city b egins to saturate, increases linearly with time. This is simply conservation of energy, as the constant acceleration force in the instantaneous rest frame is constantly doing work W = cF . Now take a side step and evaluate in terms of proper time. 0 0 dt dt r d = = 2 0 gt 1+ c 3=2 2 2 in this section. Thus the acceleration in the Earth frame was a = 1 v =c g = dv =dt. x x p 3=2 2 2 2 2 Regrouping we had gdt = du = 1 v =c and integrating gives gt = v= 1 v =c or x p 2 1+g t=c . v=c = g t=c= 3 ! Z 0 0 dt c gt 1 r = = sinh 2 0 g c gt 1+ c 0 or inverting the equation to nd t in terms of prop er time g g 0 sinh 5 t = c c end aside 0 0 0 Nowwe can solve for x using the de nition of = dx =dct . 0 0 dx = dct Z Z 0 0 0 0 0 ct =ct ct =ct gt =c 0 0 0 q x = x + dct =x + dct o o 0 0 2 ct =0 ct =0 0 1+gt =c s 2 2 c g 0 0 ct =ct 0 0 = 1+ ct j + x 0 t =0 o 2 g c q 2 2 c c 2 0 0 + x 6 = 1+gt =c o g g De ne 2 c 0 0 x x 7 P o g Then our equation b ecomes s 2 2 c g 0 0 0 ct x x = 1+ P 2 g c 2 2 g g 0 0 0 x x = 1+ ct P 2 2 c c 2 2 g g 0 0 0 = 1 8 x x ct P 2 2 c c This last equation describ es a hyp erb ola. Note that we can nd a formula for x in 0 0 terms of given the last equation for x in terms of t . 2 0 2 2 2 0 0 gt =c =1 g=c x x P 2 0 2 2 2 2 2 0 0 =1+gt =c =1+sinh g =c=cosh g =c g=c x x P or 2 2 2 c c g c g 0 0 0 x = x + = x cosh + cosh P 0 g c g g c 0 Note that x is directly related to the x in the accelerating frame. 0 0 Because the world line is a hyp erb ola in Minkowski space, the world line of the p oint mass approaches the light line asymptotically. This means all events on the 4 world line will have a space like relationship to all events to the left of the fo cal p oint 0 P 0;x . P 2 c 0 0 x = x 9 P o g So that the distance b etween the rest p oint and fo cal p oint is prop ortional to the inverse of the acceleration. insert gure here showing frames with small acceleration and with large accelerations. s 2 0 0 gt =c g gt 0 0 r = x x = 1+ 10 p 2 2 0 c c gt 1+ c Therefore 0 ct = = tan 11 0 0 x x P where is the horizontal angle.
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