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            <span id="subtitle">【迁移】时空图：以天狼星单程旅行为例</span>
          
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            <h1 id="seo-header">【迁移】时空图：以天狼星单程旅行为例</h1>
            
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                <p>最近看《从一到无穷大》，对里面的闵氏时空图很感兴趣，恰好里面有去天狼星的例子，因此想用时空图画画看。为简化问题，这里取天狼星到地球的距离为9光年，实际约为8.6光年。同时为了便于画图，纵轴时间轴以年×光速为单位，横轴空间轴以光年为单位。使用 <a target="_blank" rel="noopener" href="https://www.geogebra.org/calculator">geogebra</a> 进行绘图。</p>
<h2 id="出发前的时空图">出发前的时空图</h2>
<p><img src="https://img.limour.top/2023/08/30/64ef0e979b29f.webp" srcset="https://jscdn.limour.top/gh/Limour-dev/Sakurairo_Vision/load_svg/inload.svg" lazyload alt="图1 出发前的时空图"><br>
如图1所示，线段g（CAD）表示飞船，A点为质心。直线f表示天狼星B的世界线。射线h（AE）表示A的光锥。射线i（AF）表示飞船的将来的世界线，其速度为 tanα=0.9 倍光速。</p>
<h2 id="出发后的时空图">出发后的时空图</h2>
<p><img src="https://img.limour.top/2023/08/30/64ef0e5f5bd5a.webp" srcset="https://jscdn.limour.top/gh/Limour-dev/Sakurairo_Vision/load_svg/inload.svg" lazyload alt="图2 出发后的时空图"><br>
如图2所示，为了便于绘图，减小了α。直线AF与直线AL构成了以飞船为参考系的新时空坐标系。</p>
<ul>
<li>x’为0的直线AF为新的纵轴，t’为0的直线AL为新的横轴，两者关于AE轴对称</li>
<li>AB在新坐标系的几何位置为（AL，AK），物理位置为（ALβ，AKβ）</li>
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<p><img src="https://img.limour.top/2023/08/30/64ef0eab0820b.webp" srcset="https://jscdn.limour.top/gh/Limour-dev/Sakurairo_Vision/load_svg/inload.svg" lazyload alt="图3 一些计算过程"></p>
<h2 id="一些有趣的发现">一些有趣的发现</h2>
<ol>
<li>AB的时空距离原来为 9，现在AB’的时空距离依然为 9，即 $9^2 + (0i)^2 = 20.65^2 + (-18.58i)^2$</li>
<li>AF、AL上的单位对应的几何长度均变为原来的 1/β，即 $\lvert\overrightarrow{AF}\rvert\beta = t’<em>{ \overrightarrow{AF’}} \quad \lvert\overrightarrow{AL}\rvert\beta = x’</em>{ \overrightarrow{AB’}}$</li>
<li>对于出发的飞船上的人来说，旅行只花了 4.36 年，小于 10 年，即 $t’<em>{ \overrightarrow{AF’}} = 4.36 \lt 10 = t</em>{ \overrightarrow{AF}}$</li>
<li>对于出发的飞船上的人来说，天狼星距离为 3.92 光年，小于9光年，B’点的天狼星是 18.58 年前的幻影，现在的天狼星是M’点的天狼星，即 $x’<em>{ \overrightarrow{AM’}} = (\frac{ \lvert\overrightarrow{AB}\rvert}{cos\alpha})\beta = 3.92 \lt 9 = x</em>{ \overrightarrow{AB}}$</li>
<li>对于出发的飞船上的人来说，天狼星靠近的速度 v = 3.92/4.36 = 0.9 倍光速，与外界观察飞船飞行的速度一致</li>
<li>飞船感知外界的空间长度 $x_{\overrightarrow{AB}} = 9$ 变成了空间长度 $x’_{\overrightarrow{AB’}} = \lvert\overrightarrow{AL}\rvert\beta = 20.65$，变长了γ倍，外界尺子变长了</li>
<li>飞船感知外界的时间长度 $t_{\overrightarrow{AF}} = 10$ 变成了时间长度 $t’_{ \overrightarrow{AF’}} = \lvert\overrightarrow{AF}\rvert\beta = 4.36$，缩短为1/γ，外界时间变快了</li>
<li>飞船认为外界“尺长钟快”，因此外界认为飞船上“尺缩钟慢”</li>
<li>根据四维时空距离不变性，可以轻松计算飞船感知的旅程时间耗费，飞船参考系下的AF’的时空长度与外界参考系下AF的时空长度一致，即 $(i\lvert\overrightarrow{AF’}\rvert)^2 = (i\lvert\overrightarrow{AF}\rvert\beta)^2=AB^2+(iBF)^2$</li>
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