5 To preserve the parallel orientation of the x and x'axes Squaring and rearranging give we have to choose the plus sign(as can be seen by taking thev→0 limit),so that ("2= (68】 1 品） Y= (61) V1-v2f or. Thus, "=±+心 1+vv'f (69) 同-同 62) Choosing the plus sign(to make sure that u"=v when v'=0),we get and 月-，间 U+v (63) U”=1+w时 (70) where,we recall,f=f(v2). Step 12:The universal constant f cannot be negative .. Step 10:Two Lorentz transformations performed in succession is a Lorentz transformation ..because in that case the conclusions of relativistic dynamics would violate experimental observations!For This step is crucial for everything that we've been do- example,the force law. ing so far,for it shows that f is a constant,which will be identified with 1/c2,where c is Nature's limiting speed. d mvp=F, (71) We have a sequence of two transformations:from(,t) to (x',t),and then from ('t')to (x",t"), 1-哈别 同=[r where vp is the velocity of a particle,would get messed up.In particular,such law would violate the observed fact that it requires an infinite amount of work (and. thus,energy)to accelerate a material particle from rest to speeds approaching 3x 108 m/s](this argument is due [1+v'f-(v+)「x to Terletskii [4]).In fact,it would become "easier"to ac- [-(f+v'f)1+wv'f t (64) celerate the particle,the faster it is moving.Incidentally, this erperimental fact is what "replaces"Einstein's Sec- where v is the velocity of K'relative to K(as measured ond Postulate in the present derivation!Thus,Eq.(70)is in K using the (z,t)coordinates),and v'is the velocity the limit to which our (actually,Ignatowski's)derivation of K"relative to K(as measured in K'using the (x',t') can be pushed. coordinates).But this could also be written as a single REMARK:Relativistic dynamics has to be discussed transformation from (r,t)to (r",t") separately.We won't do that here,but maybe you can suggest a different reason for f not to be negative?See (65) [3]and [5]for possible approaches. with v"being the velocity of K"relative to K(as mea- sured in K using the (z,t)coordinates).This shows that Step 13:Existence of the limiting speed the(1,1)and(2,2)elements of the transformation ma- trix (64)must be equal to each other and,thus, Denoting f=f', (66) which means that f is a constant that has units of inverse f三2 (72) speed squared,[s2/m2].Wow!! we get the velocity addition formula, v+v Step 11:Velocity addition formula (for reference v”= 1+g (73) frames) If we begin with v'<c and attempt to take the limit To derive the velocity addition formula (along the x- v'→c,we'll get axis)we use Eqs.(64)and (65)to get y'y(v+v')=y"". (67) v" v+c 1+警s6 (74)5 To preserve the parallel orientation of the x and x 0 axes we have to choose the plus sign (as can be seen by taking the v → 0 limit), so that γ = 1 p 1 − v 2f . (61) Thus,  x 0 t 0  = 1 p 1 − v 2f  1 −v −vf 1  x t  , (62) and  x t  = 1 p 1 − v 2f  1 v vf 1  x 0 t 0  , (63) where, we recall, f = f(v 2 ). Step 10: Two Lorentz transformations performed in succession is a Lorentz transformation This step is crucial for everything that we’ve been do￾ing so far, for it shows that f is a constant, which will be identified with 1/c2 , where c is Nature’s limiting speed. We have a sequence of two transformations: from (x, t) to (x 0 , t0 ), and then from (x 0 , t0 ) to (x 00, t00),  x 00 t 00  = γ 0  1 −v 0 −v 0f 0 1  x 0 t 0  = γ 0  1 −v 0 −v 0f 0 1  γ  1 −v −vf 1  x t  = γ 0 γ  1 + vv0f −(v + v 0 ) −(vf + v 0f 0 ) 1 + vv0f 0  x t  , (64) where v is the velocity of K0 relative to K (as measured in K using the (x, t) coordinates), and v 0 is the velocity of K00 relative to K0 (as measured in K0 using the (x 0 , t0 ) coordinates). But this could also be written as a single transformation from (x, t) to (x 00, t00),  x 00 t 00  = γ 00  1 −v 00 −v 00f 00 1  x t  , (65) with v 00 being the velocity of K00 relative to K (as mea￾sured in K using the (x, t) coordinates). This shows that the (1, 1) and (2, 2) elements of the transformation ma￾trix (64) must be equal to each other and, thus, f = f 0 , (66) which means that f is a constant that has units of inverse speed squared, [s2/m2 ]. Wow!! Step 11: Velocity addition formula (for reference frames) To derive the velocity addition formula (along the x￾axis) we use Eqs. (64) and (65) to get γ 0 γ(v + v 0 ) = γ 00v 00 . (67) Squaring and rearranging give (v 00) 2 =  v + v 0 1 + vv0f 2 , (68) or, v 00 = ± v + v 0 1 + vv0f . (69) Choosing the plus sign (to make sure that v 00 = v when v 0 = 0), we get v 00 = v + v 0 1 + vv0f . (70) Step 12: The universal constant f cannot be negative . . . . . . because in that case the conclusions of relativistic dynamics would violate experimental observations! For example, the force law, d dt q mvp 1 − v 2 p f = F, (71) where vp is the velocity of a particle, would get messed up. In particular, such law would violate the observed fact that it requires an infinite amount of work (and, thus, energy) to accelerate a material particle from rest to speeds approaching 3×108 [m/s] (this argument is due to Terletskii [4]). In fact, it would become “easier” to ac￾celerate the particle, the faster it is moving. Incidentally, this experimental fact is what “replaces” Einstein’s Sec￾ond Postulate in the present derivation! Thus, Eq. (70) is the limit to which our (actually, Ignatowski’s) derivation can be pushed. REMARK: Relativistic dynamics has to be discussed separately. We won’t do that here, but maybe you can suggest a different reason for f not to be negative? See [3] and [5] for possible approaches. Step 13: Existence of the limiting speed Denoting f ≡ 1 c 2 , (72) we get the velocity addition formula, v 00 = v + v 0 1 + vv0 c 2 . (73) If we begin with v 0 < c and attempt to take the limit v 0 → c, we’ll get v 00 → v + c 1 + vc c 2 = c, (74)