transformation<\/a>,<\/p>\n\n\n\n\\begin{equation}\n \\vb \\Lambda({\\vb v}) \\left(\\ket{\\vb{x},0} +c t \\ket{\\vb{0},1}\\right) = \\ket{\\vb{x’},0} +c t’ \\ket{\\vb{0},1} \\label{lk}\n\\end{equation}\n\n\n\n\\begin{equation}\n\\vb \\Lambda({\\vb v})=\n\\beta \\ket{\\vb{e}_1,0}\\bra{\\vb{e}_1,0} -\\ds\\beta\\frac{v}{c}\\ket{\\vb{e}_1,0}\\bra{\\vb{0},1}+\n(\\vb{I’}-\\ket{\\vb{e}_1,0}\\bra{\\vb{e}_1,0})\n-\\beta\\frac{ v} {c} \\ket{\\vb{0},1}\\bra{\\vb{e}_1,0} \n+\\beta \\ket{\\vb{0},1}\\bra{\\vb{0},1}\\\\\n\\label{lambda_1}\n\\end{equation}\n\n\n\nand the velocity of the body in $F$ is computed from the time derivative of equation (\\ref{lk}) with respect to the time $t$,<\/p>\n\n\n\n\\begin{equation}\n\\vb \\Lambda({\\vb v}) (\\ket{\\vb{u},0} +c \\ket{\\vb{0},1}) = \\beta\\left(-\\frac{ v} {c^2}\\bra{{\\vb e}_1}\\ket{\\vb u}+1 \\right) (\\ket{\\vb{u’},0} +c \\ket{\\vb{0},1}) \\label{nm}\n\\end{equation}\n\n\n\n
where the time $t’$ is a function of the time $t$ given by the time component of the Lorentz transformation<\/p>\n\n\n\n\\begin{equation} \nt’=\\beta\\left(-\\frac{ v} {c^2}\\bra{{\\vb e}_1}\\ket{\\vb x}+t \\right)\n\\end{equation}\n\n\n\n
from which we easily compute the derivative with respect to $t$<\/p>\n\n\n\n\\begin{equation} \n\\frac{dt’}{dt}=\\beta\\left(-\\frac{ v} {c^2}\\bra{{\\vb e}_1}\\ket{\\vb u}+1 \\right) \\label{pkl}\n\\end{equation}\n\n\n\n
We write equation (\\ref{nm}) explicitly, <\/p>\n\n\n\n\\begin{gather*}\n(\\beta \\ket{\\vb{e}_1,0}\\bra{\\vb{e}_1,0} -\\ds\\frac{v}{c}\\beta \\ket{\\vb{e}_1,0}\\bra{\\vb{0},1}\n+ \\ket{\\vb{e}_2,0}\\bra{\\vb{e}_2,0} \n+\n\\ket{\\vb{e}_3,0}\\bra{\\vb{e}_3,0} +\n-\\beta\\ds\\frac{v}{c} \\ket{\\vb{0},1}\\bra{\\vb{e}_1,0}+\n\\beta\\ket{\\vb{0},1}\\bra{\\vb{0},1}\n)\\cdot (\\ket{\\vb{u},0} +c \\ket{\\vb{0},1} ) \\\\ \n= \\\\ \n\\beta\\left(-\\frac{ v} {c^2}\\bra{{\\vb e}_1}\\ket{\\vb u}+1 \\right) (\\ket{\\vb{u’},0} +c \\ket{\\vb{0},1})\n\\end{gather*}\n\n\n\n
which simplifies to,<\/p>\n\n\n\n\\begin{gather*} \n\\beta \\ket{\\vb{e}_1,0}\\bra{\\vb{e}_1,0} \\ket{\\vb{u},0}\n-{v}\\beta \\ket{\\vb{e}_1,0} \n+\\ket{\\vb{e}_2,0}\\bra{\\vb{e}_2,0} \\ket{\\vb{u},0} \n+\\ket{\\vb{e}_3,0}\\bra{\\vb{e}_3,0} \\ket{\\vb{u},0}\n-\\beta\\ds\\frac{v}{c} \\ket{\\vb{0},1}\\bra{\\vb{e}_1,0} \\ket{\\vb{u},0}\n+c\\beta\\ket{\\vb{0},1} \\\\\n = \\\\\n-\\beta\\frac{ v} {c^2}\\ket{\\vb{u’},0}\\bra{{\\vb e}_1}\\ket{\\vb u}\n-\\beta\\frac{ v} {c}\\ket{\\vb{0},1}\\bra{{\\vb e}_1}\\ket{\\vb u} +\\beta\\ket{\\vb{u’},0} \n+\\beta c\\ket{\\vb{0},1} \n\\end{gather*}\n\n\n\n
and finally,<\/p>\n\n\n\n\\begin{equation} \n\\beta \\ket{\\vb{e}_1 +\\frac{ v} {c^2}\\vb{u’},0}\\bra{{\\vb e}_1,0}\\ket{\\vb u,0}\n+\\ket{\\vb{e}_2,0}\\bra{\\vb{e}_2,0} \\ket{\\vb{u},0} \n+\\ket{\\vb{e}_3,0}\\bra{\\vb{e}_3,0} \\ket{\\vb{u},0} =\n{v}\\beta \\ket{\\vb{e}_1,0} \n+\\beta\\ket{\\vb{u’},0} \\label{df}\n\\end{equation}\n\n\n\n
The problem with the above equation is that $\\vb u’$ appears both on the left and right sides, and, in order to separate the two velocities, we have to write the above expression as the product of an operator depending only on $\\vb u’$, times $\\vb u$. <\/p>\n\n\n\n
The first step is to decompose $\\vb u’$ along the versors, <\/p>\n\n\n\n\\begin{equation}\n{\\vb u’} = u’_1\\ket{{\\vb e}_1,0} +\nu’_2\\ket{{\\vb e}_2,0} +u’_3\\ket{{\\vb e}_3,0} \\label{hj}\n\\end{equation}\n\n\n\n
so that equation (\\ref{df}) assumes the form,<\/p>\n\n\n\n\\begin{gather*} \n\\left(\\beta \\left(1 +\\ds\\frac{ v} {c^2}u’_1\\right)\\,\\,\\ket{{\\vb e}_1,0}\\bra{{\\vb e}_1,0} + \n\\beta\\frac{ v} {c^2}u’_2\\,\\,\\ket{{\\vb e}_2,0}\\bra{{\\vb e}_1,0}+\n\\beta\\ds\\frac{ v} {c^2}u’_3\\,\\,\\ket{{\\vb e}_3,0}\\bra{{\\vb e}_1,0} +\\ket{\\vb{e}_2,0}\\bra{\\vb{e}_2,0} + \\ket{\\vb{e}_3,0}\\bra{\\vb{e}_3,0}\n \\right)\\cdot\\ket{\\vb{u},0}\n\\\\ = \\\\\n{v}\\beta \\ket{\\vb{e}_1,0} \n+\\beta\\ket{\\vb{u’}}\\label{yu1}\n\\end{gather*}\n\n\n\n
We have on the right side of the equation an operator that only depends on $\\vb u’$,<\/p>\n\n\n\n\\begin{equation}\n\\beta \\left(1 +\\ds\\frac{ v} {c^2}u’_1,0\\right)\\,\\,\\ket{{\\vb e}_1,0}\\bra{{\\vb e}_1,0} +\n\\beta\\ds\\frac{ v} {c^2 }u’_2\\,\\,\\ket{{\\vb e}_2,0}\\bra{{\\vb e}_1,0}+\\beta\\ds\\frac{ v} {c^2}u’_3\\,\\,\\ket{{\\vb e}_3,0}\\bra{{\\vb e}_1,0} \n+ \\ket{\\vb{e}_2,0}\\bra{\\vb{e}_2,0} +\\ket{\\vb{e}_3,0}\\bra{\\vb{e}_3,0}\n \\end{equation}\n\n\n\n
which can be easily inverted as,<\/p>\n\n\n\n\\begin{equation}\n\\ds\\frac{1}{\\beta \\left( 1 +\\ds\\frac{ v} {c^2}u’_1\\right)}\\,\\,\\ket{{\\vb e}_1,0 }\\bra{{\\vb e}_1,0} \n%\n\\ds-\\frac{ v} {c^2}\\frac{u’_2}{(1 +\\ds\\frac{ v} {c^2}u’_1)}\\,\\,\\ket{{\\vb e}_2,0}\\bra{{\\vb e}_1,0} \n%\n\\ds-\\frac{ v} {c^2}\\frac{u’_3}{(1 +\\ds\\frac{ v} {c^2}u’_1)}\\,\\,\\ket{{\\vb e}_3,0}\\bra{{\\vb e}_1,0}+ \\ket{\\vb{e}_2,0}\\bra{\\vb{e}_2,0} +\\ket{\\vb{e}_3,0}\\bra{\\vb{e}_3,0}\n \\end{equation}\n\n\n\n
We can express the velocity measured in $F$ as,<\/p>\n\n\n\n\\begin{equation}\n\\ket{\\vb{u},0}=\n\\left(\n\\ds\\frac{1}{\\beta (1 +\\ds\\frac{ v} {c^2} u’_1)}\\,\\,\\ket{{\\vb e}_1,0}\\bra{{\\vb e}_1,0} \n%\n\\ds-\\frac{ v} {c^2}\\frac{u’_2}{(1 +\\ds\\frac{ v} {c^2}u’_1)}\\,\\,\\ket{{\\vb e}_2,0}\\bra{{\\vb e}_1,0} \n%\n\\ds-\\frac{ v} {c^2}\\frac{u’_3}{(1 +\\ds\\frac{ v} {c^2}u’_1)}\\,\\,\\ket{{\\vb e}_3,0}\\bra{{\\vb e}_1,0}+ \\ket{\\vb{e}_2,0}\\bra{\\vb{e}_2,0} + \\ket{\\vb{e}_3,0}\\bra{\\vb{e}_3,0}\n\\right)\\cdot \\left({v}\\beta \\ket{\\vb{e}_1,0} \n+\\beta\\ket{\\vb{u’}}\\right)\n\\end{equation}\n\n\n\n
which finally simplifies to,<\/p>\n\n\n\n\\begin{equation}\n\\ket{\\vb{u},0}=\n\\ds \\frac{1}{(1 +\\ds\\frac{ v} {c^2}u’_1)} \\left(\n\\ds (v+u’_1)\n\\,\\,\\ket{{\\vb e}_1,0} –\n%\n\\ds\\sqrt{1-\\ds\\frac{v^2}{c^2}}\\; u’_2 \\;\n\\ket{\\vb{e}_2,0} –\n\\sqrt{1-\\ds\\frac{v^2}{c^2}} \\; u’_3 \\;\n\\ket{\\vb{e}_3,0} \\right)\n \\label{p12}\n\\end{equation}\n\n\n\n
Equation (\\ref{p12}) assumes a more elegant form when we drop the explicit dependence on the time component <\/p>\n\n\n\n\\begin{equation}\n\\ket{\\vb{u}}= \\frac{1}{ 1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2}\n }\n\\left(\n\\ket{\\vb v} +\n%\n\\alpha_v \\ket{\\vb u’}\n+\\left(1-\\alpha_v\\right)\\frac{ \\bra{\\vb u’}\\ket{\\vb v}}{v^2}\\ket{\\vb v}\n\\right)\n \\label{nice} \n\\end{equation}\n\n\n\n
where,<\/p>\n\n\n\n\\begin{align}\n\\ket{\\vb v}&=v\\,\\, \\ket{\\vb{e}_1,0} \\nonumber\\\\\n\\alpha_v&=\\frac{1}{\\beta}=\\ds\\sqrt{1-\\ds\\frac{v^2}{c^2}}\\nonumber\n\\end{align}\n\n\n\n
Considering the components of $u$ along the vectors $\\ket{\\vb{e}_1,0}, \\ket{\\vb{e}_2,0}$ and $\\ket{\\vb{e}_3,0}$<\/p>\n\n\n\n\\begin{equation}\n \n{\\vb u} = u_1\\ket{\\vb{e}_1,0} +u_2\\ket{\\vb{e}_2,0} +u_3\\ket{\\vb{e}_3,0} \n\\end{equation}\n\n\n\n
equation ($\\ref{p12}$) assumes a lighter and more familiar form<\/p>\n\n\n\n\\begin{align}\nu_1 &= \\frac{u’_1+v}{1+\\ds\\frac{vu’}{c^2}}\\nonumber\\\\\nu_2 & = \\frac{\\sqrt{1-\\ds\\frac{v^2}{c^2}}}{1+\\ds\\frac{vu’_1}{c^2}}u’_2\\nonumber\\\\\nu_3 &= \\frac{\\sqrt{1-\\ds\\frac{v^2}{c^2}}}{1+\\ds\\frac{vu’_1}{c^2}}u’_3\\label{velComp}\\\\\n\\end{align}\n\n\n\n
Composition of Accelerations<\/h1>\n\n\n\n
The composition of accelerations is computed by performing the time derivative of the velocity ($\\ref{nice}$)<\/p>\n\n\n\n\\begin{align*}\n\\ket{\\vb{a}}&= -\\frac{1}{\\left( 1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2}\\right)^2}\n\\left(\n\\ket{\\vb v} +\n%\n\\alpha_v \\ket{\\vb u’}\n+\\left(1-\\alpha_v\\right)\\frac{ \\bra{\\vb u’}\\ket{\\vb v}}{v^2}\\ket{\\vb v}\n\\right)\\frac{ \\bra{\\vb a’}\\ket{\\vb{v} }} {c^2}\\frac{dt’}{dt} \\\\\n&+\\frac{1}{ 1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2}\n }\n\\left(\n\n%\n\\alpha_v \\ket{\\vb a’}\n+\\left(1-\\alpha_v\\right)\\frac{ \\bra{\\vb a’}\\ket{\\vb v}}{v^2}\\ket{\\vb v}\n\\right)\\frac{dt’}{dt}\n\\end{align*}\n\n\n\n
The derivative $dt’\/dt$ reads, using equation (\\ref{pkl}) and (\\ref{nice}),<\/p>\n\n\n\n\\begin{equation}\n\\frac{dt’}{dt} = \\frac{\\alpha}{1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2}}\n\\end{equation}\n\n\n\n
substituted in the acceleration expressions $\\ket{\\vb a}$ yields<\/p>\n\n\n\n\\begin{align*}\n\\ket{\\vb{a}}&= \\frac{\\alpha}{\\left( 1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2}\\right)^3}\n\\left[\n-\\left(\n\\ket{\\vb v} +\n%\n\\alpha_v \\ket{\\vb u’}\n+\\left(1-\\alpha_v\\right)\\frac{ \\bra{\\vb u’}\\ket{\\vb v}}{v^2}\\ket{\\vb v}\n\\right)\\frac{ \\bra{\\vb a’}\\ket{\\vb{v} }} {c^2} \\right. \\\\\n&\\left.+( 1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2})\n \n\\left(\n\n%\n\\alpha_v \\ket{\\vb a’}\n+\\left(1-\\alpha_v\\right)\\frac{ \\bra{\\vb a’}\\ket{\\vb v}}{v^2}\\ket{\\vb v}\n\\right) \\right]\n\n\\end{align*}\n\n\n\n
which simplifies to the final expression,<\/p>\n\n\n\n\\begin{equation}\n\\ket{\\vb{a}}= \\frac{\\alpha^2}{\\left( 1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2}\\right)^2} \\ket{\\vb a’} + \\;\n\n \\frac{\\alpha^2(\\alpha-1)\\bra{\\vb a’}\\ket{\\vb v}}{v^2\\left( 1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2}\\right)^3} \\ket{\\vb v}\\:\n\n– \\frac{\\alpha^2 \\bra{\\vb a’}\\ket{\\vb v}}{c^2\\left( 1 +\\ds\\frac{ \\bra{\\vb v}\\ket{\\vb{u’} }} {c^2}\\right)^3}\\ket{\\vb u’}\\label{acc}\n\\end{equation}\n","protected":false},"excerpt":{"rendered":"
\\( \\def\\UP{(\\tilde{c}-\\tilde{v}){0\\rightarrow1}} \\def\\DO{(\\tilde{c}-\\tilde{v}){1\\rightarrow2}} \\def\\UPM{|\\tilde{c}-\\tilde{v}|_{0\\rightarrow1}} \\def\\DOM{|\\tilde{c}-\\tilde{v}|_{1\\rightarrow2}} \\def\\ds{\\displaystyle} \\def\\mm{ \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & -1 \\end{pmatrix}} \\) Composition of Velocities In classical mechanics, a velocity ${\\vb u’}$ in the moving frame […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"nf_dc_page":"","ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"0","ocean_second_sidebar":"0","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"0","ocean_custom_header_template":"0","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"0","ocean_menu_typo_font_family":"0","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"default","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"default","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":"0","footnotes":""},"_links":{"self":[{"href":"https:\/\/physicsandmusic.com\/wp-json\/wp\/v2\/pages\/8659"}],"collection":[{"href":"https:\/\/physicsandmusic.com\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/physicsandmusic.com\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/physicsandmusic.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsandmusic.com\/wp-json\/wp\/v2\/comments?post=8659"}],"version-history":[{"count":5,"href":"https:\/\/physicsandmusic.com\/wp-json\/wp\/v2\/pages\/8659\/revisions"}],"predecessor-version":[{"id":11095,"href":"https:\/\/physicsandmusic.com\/wp-json\/wp\/v2\/pages\/8659\/revisions\/11095"}],"wp:attachment":[{"href":"https:\/\/physicsandmusic.com\/wp-json\/wp\/v2\/media?parent=8659"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}