## Dx Sentence Examples

- If there be n similar particles per unit volume, the energy emitted from a stratum of thickness
**dx**and of unit area is found from (9) by the introduction of the factor ndx. - Accordingly, if E be the energy of the primary wave, dE 87-2n (D' - D) 2 T2 E
**dx**3 D2%4 ' whence E = Eoe-hx (II) where h = 8?r 2 n (D' - D)2T2 3 D2 x 4, (12) If we had a sufficiently complete expression for the scattered light, we might investigate (12) somewhat more directly by considering the resultant of the primary vibration and of the secondary vibrations which travel in the same direction. - Consider the particles which occupy a thin stratum
**dx**perpendicular to the primary ray x. - Moreover, if OP = r, and AO=x, then r 2 =x 2 + p2, and pdp=rdr. The resultant at 0 of all the secondary vibrations which issue from the stratum
**dx**is by (3), with sin ¢ equal to unity, ndx f ? - It appears, therefore, that to the order of approximation afforded by (3), the effect of the particles in
**dx**is to modify the phase, but not the intensity, of the light which passes them. - (15) Ifµ be the refractive index of the medium as modified by the particles, that of the original medium being taken as unity, then 6=(µ - I)
**dx**, and µ - I =nT(D' - D)/2D.. - The area of the complete curve is 2a 2, and the length of any arc may be expressed in the form f(1 - x 4) - i
**dx**, an elliptic integral sometimes termed the lemniscatic integral. - Aj Hence the system of values also causes to vanish in this case;
**dx**and by symmetry aj and Fz also vanish. - In both cases ddl and dal are cogredient with xl and x 2; for, in the case of direct substitution,
**dxi**= cost**dX**i - sin 00-(2, ad2 =sin B**dX**i +cos O**dX**2, and for skew substitution dai = cos B**dX**i +sin 0d2, c-&-- 2 n d =sin -coseax2. - DW F - d -v 1+ a 7rK
**dx**dH (38) (34) [[[Magnetic Measurements]] If Ho is constant, the force will be zero; if Ho is variable, the sphere will tend to move in the direction in which Ho varies most rapidly. - If V is the volume of a ball, H the strength of the field at its centre, and re its apparent susceptibility, the force in the direction x is f= K'VH X dH/
**dx**; and if K',, and are the apparent susceptibilities of the same ball in air and in liquid oxygen, K' Q -K'o is equal to the difference between the susceptibilities of the two media. - For instance, by equating coefficients of or in the expansions of (I +x) m+n and of (I
**dx**) m . - (I
**dx**) n we obtain (22) of ï¿½ 44 (ii.). - At once by the expansion of (I
**dx**)m. - If dS =27rxdx, we have for the whole effect 27r œ sin k(at - p)x
**dx**, f P ' or, since xdx = pdp, k = 27r/A, - k fr' sin k(at - p)dp= [- cos k(at - p)]°° r. - If 2r =1000 cm., 2x = I cm., X = 6 X 105 cm., then
**dx**= 0015 cm. - (2), where S = ff sin(px+gy)
**dx**dy,. - (3), C = ffcos(px--gy)
**dx**dy,. - When, as in the application to rectangular or circular apertures, the form is symmetrical with respect to the axes both of x and y, S = o, and C reduces to C = ff cos px cos gy
**dx**dy,. - It is thus sufficient to determine the intensity along the axis of p. Putting q = o, we get C = ffcos pxdxdy=2f+Rcos 'px 1/ (R2 - x2)
**dx**, R being the radius of the aperture. - Thus, if x = p cos 4), y= p sin 0, C =11 cos px
**dx**dy =f o rt 2 ' T cos (pp cos 0) pdp do. - Nor diminishes the If we define as the " dispersion " in a particular part of the spectrum the ratio of the angular interval dB to the corresponding increment of wave-length
**dX**, we may express it by a very simple formula. - Hence, if a be the width of the diffracted beam, and do the angle through which the wave-front is turned, ado =
**dX**, or dispersion = /a .. **Dx**ru i-x) C-Fi S= „ 7 o?I x o 1 fGO**dx**eu(ti-x) - 1 2Jo x i - x Thus, if we take _ 1 `°el 1 ('°° e uxdx G 7r12 Jo 1+ x 2 ' H 7r-N/2Jo -Vx.(1-i-x2)' C = 2-G cos u+ H sin u, S =1---G sin u-H cos u.- For
**dx**=cos airv 2 .dv, dy= sin 271-v2.dv; so that s = f (**dx**2 +dy 2) =v, (30), 0= tan1 (dyldx) =171-v 2 (31). - = b 2 (
**dx**+ dy + de l (a 2 - b2)**dx**(**dx**+dy+dz) ness where a 2 and b 2 denote the two arbitrary constants. - The third equation gives 2 d dt2 = b
**dx**(3), of which the solution is = f (bt (4), where f is an arbitrary function. - If we suppose that the force impressed upon the element of mass D
**dx**dy dz is DZ**dx**dy dz, being everywhere parallel to the axis of Z, the only change required in our equations (I), (2) is the addition of the term Z to the second member of the third equation (2). - (b2V2 + n2) (a2 - b 2) = - z It will now be convenient to introduce the quantities a l, a 2', 7731 which express the rotations of the elements of the medium round axes parallel to those of co-ordinates, in accordance with the equations Ty - 1 = dz ' 3= - dy 2 =
**dx**- In terms of these we obtain from (7), by differentiation and subtraction, (b 2 v 2 + n 2) 7,3 = 0 (b 2 0 2 +n 2) .r i = dZ/dy (b 2 v 2 +n 2)', , 2 = - dZ/**dx**The first of equations (9) gives 3 = 0 (10) For al we have ?1= 47rb2, f dy e Y tkr**dx**dy dz - (11), where r is the distance between the element
**dx**dy dz and the point where a l is estimated, and k = n/b = 27r/X (12), X being the wave-length. - Thus f (= 4-rb 2;JJ Z dY (e r)
**dx**dy dz. - Phil.): - Let x, y, z be the coordinates of P in the orbit,, r t, those of the corresponding point T in the hodograph, then
**dx**dy _ dz c= ' 71 - a' - at therefore Also, if s be the arc of the hodograph, ds = v = V V1 1) j dt + (dt2) dt Equation (1) shows that the tangent to the hodograph is parallel to the line of resultant acceleration, and (2) that the velocity in the hodograph is equal to the acceleration. - On another physical assumption of constant cubical elasticity A, dp = Ad p /P, (p - po)IA= lo g (P/Po), (18) dp _ A dp (I 1 zd p dz - P ' A Po-p -z, I - p -k, A kPo ' (19) (3) P
**dx**Pdy Pdz -., (I) When the density p is f un dp/ iform, this becomes, as before in (2) § 9 P pp ==Pzz++pao constant. - The integral equation of continuity (I) may now be written l f fdxdydz+ff (lpu+mpv+npdso, (4) which becomes by Green's transformation (dt +d dz dy
**dx**(p u) + d (p v) + d (p w) l I**dxdydz**- o, dp leading to the differential equation of continuity when the integration is removed. - FlpdS= _
**dx****dxdydz**, (4) by Green's transformation. - The time rate of increase of momentum of the fluid inside S is )
**dxdydz**; (5) and (5) is the sum of (I), (2), (3), (4), so that /if (dpu+dpu2+dpuv +dpuw_ +d p j d xdyd z = o, (b)` dt**dx**dy dz**dx**/ leading to the differential equation of motion dpu dpu 2 dpuv dpuv _ X_ (7) dt +**dx**+ dy + dz with two similar equations. - These equations may be simplified slightly, using the equation of continuity (5) § for dpu dpu 2 dpuv dpuw dt
**dx**+ dy + dz =p Cat +uax+vay+waz? - Dp dpu dpv dpw -z)' reducing to the first line, the second line vanishing in consequence of the equation of continuity; and so the equation of motion may be written in the more usual form du du du du d dt +udx+vdy +wdz =X -n
**dx**' with the two others dv dv dv dv i dp dt +u**dx**+v dy +w dz - Y -P d y' dw dw dw Z w dw i d p dt +u**dx**+v dy +wd - -P dz. - To determine the component acceleration of a particle, suppose F to denote any function of x, y, z, t, and investigate the time rate of F for a moving particle; denoting the change by DF/dt, DF = 1t F(x+uSt, y+vIt, z+wSt, t+St) - F(x, y, z, t) dt at = d + u
**dx**+v dy+ w dz and D/dt is called particle differentiation, because it follows the rate of change of a particle as it leaves the point x, y, z; but dF/dt, dF/**dx**, dF/dy, dF/dz (2) represent the rate of change of F at the time t, at the point, x, y, z, fixed in space. - (5) (8) (I) The components of acceleration of a particle of fluid are consequently Du dudu du du dt = dt +u
**dx**+v dy + wdz' Dr dv dv dv dv dt -dt+udx+vdy+wdz' dt v = dtJ+udx+vdy +w**dx**' leading to the equations of motion above. - If F (x, y, z, t) =o represents the equation of a always the same particles of fluid, DF =o, or dF {-u
**dx-rzd**{ w d d = o, Trt y _ which is called the differential equation of the bounding surface. - =
**dx**dy dz the equations of motion may be Written du - 2v? - Eliminating H between (5) and (6) DS du dv dw (du dv d1zv dt u
**dx**n**dx**udx' 5 -, dzi =°' and combining this with the equation of continuity Dp du dv dw p iit**dx**+dy+ dz = °' (10) D i du n dv dw_ dt (p p**dx**p**dx**p**dx**- o, with two similar equations. - D o,
**dx**dy dz**dx**dy dz so that, at any instant, the surfaces over which tk and m are constant intersect in the vortex lines. - Equation (5) becomes, by a rearrangement, dK dmdm dm din
**dx**dt +u**dx**+ dy +Zee dz +**dx**(dt +u**dx**+v dy +w d) = o,. - Dm dm Dl (8),
**dx**-**dx**dt +**dx**dt = °' ...' - When the motion is steady, that is, when the velocity at any point of space does not change with the time, dK
**dx-**2v{ +2wn = o, .. - D - K dK dK _ dK dK dK ?
**dx**n dyd °, udx dz - ° and K=fdp/o+V+2q 2 =H (3) is constant along a vortex line, and a stream line, the path of a fluid particle, so that the fluid is traversed by a series of H surfaces, each covered by a network of stream lines and vortex lines; and if the motion is irrotational H is a constant throughout the fluid. - The osculating plane of a stream line in steady motion contains the resultant acceleration, the direction ratios of which are du du, du d i g d g 2 _ dH
**dx**+v dy + dz - 2v? - =
**dx****dx**' ...