Reference - 19 - Space-Time.ipynb

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    "# Space-Time"
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    "\n",
    "### Peter Onyisi\n",
    "<img src=\"images/texas_logo.png\" width=\"400\" align=\"left\"/>"
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    "## The End of Simultaneity\n",
    "\n",
    "The phenomena of time dilation and length contraction are special cases of a broader phenomenon: special relativity says that space and time are not distinct entities, but can be interchanged with each other by a change of reference frame.\n",
    "\n",
    "As an example of this sort of thing, consider system of a flashing light bulb at $x'=0$ and two light detectors at $x'=\\pm L$. If the light flashes at $t'=0$, we expect the two light detectors to detect the light at the same time, $t'_\\textrm{left}=t'_\\textrm{right}=L/c$. So this is determined to be simultaneous. (Why are we using primes on the variables? In order to make things consistent with later notation.)\n",
    "\n",
    "In a reference frame where the bulb/detector system moves towards $+x$ at speed $v$, This is no longer the case.  Let's denote the coordinates in the new frame by $x$ and $t$, and let's synchronize the origins of the two reference frames so that $x=t=0$ is also $x'=t'=0$ (i.e. in both frames, when the light flashes, the light is at the origin).  Then in the \"unprimed\" frame, the left detector is moving with speed $v$ towards the left-traveling light flash (which is itself moving with speed $c$), while the right detector is moving away from the right-traveling light flash.  So the left detector will observe the light before the right one does: the detection is no longer simultaneous.  The position of the detectors partially determines the observed time in the unprimed frame.\n",
    "\n",
    "To find the exact times $t$ of detection, we need to account for the length contraction of the system from the frame in which it is stationary: in the unprimed frame the detectors are $L/\\gamma$ away from the light. So the times of detection are\n",
    "\n",
    "$$\\begin{align*}\n",
    "t_\\textrm{left} &= \\frac{L\\sqrt{1-(v/c)^2}}{c+v}\\\\\n",
    "&= \\frac{L}{c}\\frac{1-(v/c)^2}{\\sqrt{1-(v/c)^2}}\\frac{1}{1+v/c}\\\\\n",
    "&= \\frac{L}{c}\\frac{1}{\\sqrt{1-(v/c)^2}}\\frac{(1-v/c)(1+v/c)}{1+v/c}\\\\\n",
    "&= \\frac{1}{\\sqrt{1-(v/c)^2}}\\left(\\frac{L}{c}-\\frac{L}{c}\\frac{v}{c}\\right)\\\\\n",
    "&= \\frac{1}{\\sqrt{1-(v/c)^2}}\\left(\\frac{L}{c}-vL\\frac{1}{c^2}\\right)\\\\\n",
    "&= \\gamma\\left(t'_\\textrm{left}+vx'_\\textrm{left}/c^2\\right)\\\\\n",
    "t_\\textrm{right} &=  \\frac{L\\sqrt{1-(v/c)^2}}{c-v}\\\\\n",
    "&= \\frac{1}{\\sqrt{1-(v/c)^2}}\\left(\\frac{L}{c}-v(-L)\\frac{1}{c^2}\\right)\\\\\n",
    "&= \\gamma\\left(t'_\\textrm{right}+vx'_\\textrm{right}/c^2\\right)\\\\\n",
    "\\end{align*}\n",
    "$$\n",
    "\n",
    "This fact (events simultaneous in one frame need not be simultaneous in other frames) is the time equivalent of the statement that events that occur at the same coordinate in one frame need not occur at the same coordinate in other frames. This latter statement is familiar from our daily lives: if we put ourselves at the origin of our coordinate system, then everything that we do happens at $x=y=z=0$, but that is obviously not how it would be determined in a reference frame fixed to the Earth.\n",
    "\n",
    "If we were to solve for the times as seen in the _primed_ frame (the one moving with the bulb and light detectors), we would get\n",
    "\n",
    "$$\n",
    "\\begin{align*}\n",
    "t'_\\textrm{left} &= \\gamma\\left(t_\\textrm{left}-vx_\\textrm{left}/c^2\\right)\\\\\n",
    "t'_\\textrm{right} &= \\gamma\\left(t_\\textrm{right}-vx_\\textrm{right}/c^2\\right)\\\\\n",
    "\\end{align*}\n",
    "$$\n",
    "where we bring in the observed _positions_ of detections in the unprimed frame $x_\\textrm{left}$ and $x_\\textrm{right}$."
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    "## The Geometry of Space-Time\n",
    "\n",
    "In ordinary 3D space, we are used the idea that coordinates are, fundamentally, arbitrary: we can move the origin around and we can rotate the coordinates, while not fundamentally changing the physics (we merely change the numbers that go into the same equations).\n",
    "\n",
    "Let's look specifically at rotations in 3D space. These mix (e.g.) the $x$ and $y$ coordinates together, but in such a way as to keep the distances between all points constant - i.e. under any rotation that changes $(x, y, z)$ to $(x', y', z')$, the separation (or, here, the distance squared) between two points is preserved:\n",
    "\n",
    "$$ \\Delta x^2 + \\Delta y^2 + \\Delta z^2 = \\Delta x'^2 + \\Delta y'^2 + \\Delta z'^2 $$\n",
    "\n",
    "It turns out that this requirement (along with an additional technical requirement) is enough to tell us that the coordinate changes that keep the origin at the same location are just rotations. For example, the transformations that keep the $z$ coordinate the same are just 2D rotations through an angle $\\phi$\n",
    "\n",
    "$$ \\left(\\begin{array}{c}x'\\\\y'\\end{array}\\right) = \\left(\\begin{array}{cc}\\cos\\phi & -\\sin\\phi\\\\ \\sin\\phi & \\cos\\phi\\end{array}\\right)\\left(\\begin{array}{c}x\\\\y\\end{array}\\right)$$\n",
    "\n",
    "A similar, though not identical, kind of thing can be used to describe changes of coordinate in four-dimensional space-time. Basically, we would like to consider the changes in $x$ and $t$ cause by motion along $x$ as a change of coordinates and find the relationship between the two.  \n",
    "\n",
    "### Units\n",
    "\n",
    "First, we would like time and space to be measured in the same units. The ratio of the units is a velocity, and since the one universal velocity seems to be the speed of light, we'll measure $t$ in terms of the distance that light goes in the corresponding number of seconds. In these units, a second of time is $3 \\times 10^8$ meters of time.  (Alternatively we could have measured space in seconds: a light-second, $3 \\times 10^8$ m, is a second of space.)  In these units, velocities are a ratio of one distance to another, and therefore are unitless (the same way that a slope in 2D coordinates has no dimensions). The speed $c$ is one meter per meter, and hence 1.\n",
    "\n",
    "### Coordinates\n",
    "\n",
    "Two frames that are in motion relative to each other can share a 4D origin (i.e., the points $(t, x, y, z) = (0,0,0,0)$ and $(t', x', y', z') = (0,0,0,0)$ correspond to the same event). The idea of \"rotations\" in space-time runs into a few issues that show that there are differences from the \"usual\" idea of space rotations:\n",
    "* With regular rotations, if I rotate by 180 degrees I reverse the sign of both coordinates. However there is no evidence that you can reverse the sign of time by just going fast enough.  It's important that we cannot do this because then we could disrupt causality: we think that the order of events that cause each other is important.\n",
    "* If we plot $t$ versus $x$, the position of a flash of light describes a line $t = x$.  The condition that the speed of light is the same for all reference frames means that the slope of this line must remain constant in all reference frames. But ordinary rotations change the slopes of lines.\n",
    "\n",
    "The latter point, in particular, means that the thing that is preserved cannot be, say. $\\Delta t^2 + \\Delta x^2$; applied to the propagation of a flash of light, this would imply that in some frame we could stop the light (the separation would be all $\\Delta t$) or to have it move infinitely fast (the separation would be all $\\Delta x$).  However, we can see the outlines of the solution: we could instead demand that $\\Delta t^2 - \\Delta x^2$ remain constant. That would at least preserve the speed of light in all frames, as the trajectory of a flash of light will always satisfy $\\Delta t = \\Delta x$.\n",
    "\n",
    "### Space-Time Invariant\n",
    "\n",
    "In the full four dimensions, we therefore require that all coordinates (in inertial reference frames) keep the following unchanged:\n",
    "\n",
    "$$ \\Delta t^2 - (\\Delta x^2 + \\Delta y^2 + \\Delta z^2) = \\Delta t'^2 - (\\Delta x'^2 + \\Delta y'^2 + \\Delta z'^2) $$\n",
    "\n",
    "This is called the _space-time invariant_ and, unlike a usual distance, can be positive or negative. Normal 3D rotations are a subset of acceptable coordinate transformations since they leave the time part untouched and only touch the space coordinates, and by definition preserve $\\Delta x^2 + \\Delta y^2 + \\Delta z^2$.  However we are permitted a new freedom, to mix $t$ and the space coordinates.  \n",
    "\n",
    "The invariant can be positive or negative, and this has implications. If the invariant between two events is *positive*, this implies that the *time* separation $\\Delta t$ between the two is greater than the *space* separation $\\Delta r = \\sqrt{\\Delta x^2 + \\Delta y^2 + \\Delta z^2}$.  This means that an object traveling with uniform velocity between the two events would have speed $\\Delta r/\\Delta t < 1$, i.e. it would be moving slower than the speed of light.  From the reference frame of the object, a time $\\Delta t'$ elapses between the two events, and the two events happen in the same place ($\\Delta r' = 0$). Since this is the smallest $\\Delta r'$ possible, it must be the smallest $\\Delta t'$ possible - this is the *proper time* between the two events. This confirms our idea that the proper time between events is the time measured in an inertial reference frame in which the events happen in the same location. Events with positive space-time invariant between them are called *timelike separated*.\n",
    "\n",
    "If the invariant is *negative*, things change. If it is possible to send an object between the two events, then in the object's reference frame $\\Delta t' > 0$ and $\\Delta r' = 0$, which would give a *positive* invariant and contradict the universality of the invariant. So we conclude *it is not possible for an object to travel between two events separated by a negative space-time invariant*. Such an object would need to be moving with speed $\\Delta r/\\Delta t > 1$, which would be faster than light. Therefore we conclude that *it is not possible for objects to travel faster than the speed of light.*  Events with negative space-time invariant between them are called *spacelike separated*.  For such events, it is possible to find a reference frame in which the events occur at the same time ($\\Delta t' = 0$).  (The $\\Delta r'$ in this frame is _not_ the proper length that gets considered for length contraction!)\n",
    "\n",
    "Events that are separated by an invariant of zero are called *lightlike separated*, which means that something moving at the speed of light could connect them.  For a given event, the set of all other events that are lightlike separated from it is called the _light cone_ of that event. The timelike-separated events are _inside_ the light cone and the spacelike-separated ones are _outside_ it.\n",
    "\n",
    "![Light cones](images/lightcones.png)\n",
    "\n",
    "Because nothing can move faster than light, if an event is going to send a signal to another event, they need to be timelike or lightlike separated. Therefore causal relationships can only exist between events that lie within (or on) each other's light cones. This is a standard problem in space missions: control centers on Earth can only influence what happens on Mars at some point in the future, not what is happening \"now\".\n",
    "\n",
    "### Lorentz Transformations\n",
    "\n",
    "Let's consider the case of mixing $t$ and $x$.  The linear transformations that preserve the space-time invariant $\\Delta t^2 -\\Delta x^2$ in this case have the form \n",
    "\n",
    "$$ \\left(\\begin{array}{c}t'\\\\x'\\end{array}\\right) = \\left(\\begin{array}{cc}\\cosh\\phi & -\\sinh\\phi\\\\ -\\sinh\\phi & \\cosh\\phi\\end{array}\\right)\\left(\\begin{array}{c}t\\\\x\\end{array}\\right) $$\n",
    "\n",
    "where $\\phi$ is a number which could vary from $-\\infty$ to $\\infty$ (called the *rapidity*). This looks like a rotation matrix, but now with _hyperbolic_ trig functions instead of the usual ones. To match what we know from time dilation and length contraction, we need to identify\n",
    "\n",
    "$$\n",
    "\\begin{align*}\n",
    "v &= \\tanh \\phi\\\\\n",
    "\\gamma &= \\cosh \\phi\\\\\n",
    "\\gamma v &= \\sinh \\phi\n",
    "\\end{align*}\n",
    "$$\n",
    "\n",
    "(remember that we are using units where $v$ is dimensionless and $c=1$).  Therefore\n",
    "\n",
    "$$\n",
    "\\left(\\begin{array}{c}t'\\\\x'\\end{array}\\right) = \\left(\\begin{array}{cc}\\gamma & -\\gamma v\\\\ -\\gamma v & \\gamma\\end{array}\\right)\\left(\\begin{array}{c}t\\\\x\\end{array}\\right)\n",
    "$$\n",
    "\n",
    "or\n",
    "$$\n",
    "\\begin{align*}\n",
    "t' &= \\gamma t - \\gamma v x = \\frac{1}{\\sqrt{1-v^2}}(t - vx)\\\\\n",
    "x' &= \\gamma x - \\gamma v t = \\frac{1}{\\sqrt{1-v^2}}(x - vt)\\\\\n",
    "\\end{align*}\n",
    "$$\n",
    "\n",
    "![Space-time coordinates with Lorentz transformed axes](images/space_time_diagram_lorentz.png)\n",
    "\n",
    "This is called the *Lorentz transformation* and replaces the Galilean transformation. The definition of the frames here is that the origin of the _primed_ frame is moving towards $+x$ in the unprimed frame, with speed $v$. (This is consistent with our notation when discussing the flashing light bulb and detectors; the primed frame, where the bulb and detectors are at rest, is moving to $+x$ at speed $v$ relative to the unprimed frame.)  We can invert this to get the *inverse Lorentz transformation* from primed to unprimed frame:\n",
    "\n",
    "$$\n",
    "\\begin{align*}\n",
    "t &= \\gamma t' + \\gamma v x' = \\frac{1}{\\sqrt{1-v^2}}(t' + vx')\\\\\n",
    "x &= \\gamma x' + \\gamma v t' = \\frac{1}{\\sqrt{1-v^2}}(x' + vt')\\\\\n",
    "\\end{align*}\n",
    "$$\n",
    "\n",
    "This is in fact just the Lorentz transformation with velocity $-v$ so you don't really need to memorize two sets of transformations. (If you look at the bulb/detector system, the first equations we determined were effectively the inverse Lorentz transformation.)  Just remember the Lorentz transformation takes the coordinates that you measure in some (unprimed) frame, then uses those to determine the coordinates as measured in another (primed) frame, whose origin is moving with velocity $v$ along the $x$-axis.\n",
    "\n",
    "Note that to use the Lorentz transformation you need to have fully defined 4D points (events) in space-time. You cannot Lorentz transform just a space distance, or just a time!\n",
    "\n",
    "Let's use the Lorentz transformation for the cosmic ray decay experiment to relate the frames of the Earth and the cosmic ray muon.  We will set the origin of the four-dimensional coordinate systems for the Earth and the muon to be the top of Mt. Washington when the muon passes that altitude. Let's also set $x$ to increase going down. That means that in the Earth frame, $x_\\textrm{start}=0$ is the top of Mt. Washington, $x_\\textrm{end}=1907$ m is the altitude of Cambridge (both are at rest in this frame), $t_\\textrm{start}=0$ is when the muon passes the top of Mt. Washington, and $t_\\textrm{end} = 6.39\\ \\mu\\textrm{s} = 1917$ m is the time when the muon reaches Cambridge. (We need to use the same units for space and time to use the Lorentz transformation as given above. It's not a coincidence that $x \\approx t$; the speed of the muon is close to $c$.) Then the primed frame is the frame of the cosmic ray muon, with $v = 0.995$ (the muon is moving towards $+x$ in the Earth frame) and $\\gamma = 10.0$.  Then $x'_\\textrm{start}=0$, $t'_\\textrm{start}=0$, and\n",
    "\n",
    "$$\\begin{align*}\n",
    "t'_\\textrm{end} &= \\gamma(t_\\textrm{end} - vx_\\textrm{end}) = 195\\textrm{ m} = 0.65\\ \\mu \\textrm{s}\\\\\n",
    "x'_\\textrm{end} &= \\gamma(x_\\textrm{end} - vt_\\textrm{end}) = 0\n",
    "\\end{align*}\n",
    "$$\n",
    "\n",
    "Nothing we have written here depends on one of the events being at the origin; the only requirement is that the two reference frames share a common coordinate origin and direction of the $x$ axis, and that the motion is along $x$. Let's illustrate this by shifting the coordinates so that the origin of the two frames (as measured in the Earth frame) is 1 km above the top of Mt. Washington and 10 microseconds earlier.  (In this situation the muon is no longer at the origin of the primed frame, although it is still at rest in the primed frame.) We then have\n",
    "$$\\begin{align*}\n",
    "x_\\textrm{start} &= 1000\\textrm{ m}\\\\\n",
    "x_\\textrm{end} &= (1907+1000)\\textrm{ m} = 2907\\textrm{ m}\\\\\n",
    "t_\\textrm{start} &= 10\\ \\mu\\textrm{s} = 3000\\textrm{ m}\\\\\n",
    "t_\\textrm{end} &= (10 + 6.39)\\ \\mu\\textrm{s} = 16.39\\ \\mu\\textrm{s} = 4917\\textrm{ m}\\\\\n",
    "x'_\\textrm{start} &= \\gamma(x_\\textrm{start} - vt_\\textrm{start}) = -19850\\textrm{ m}\\\\\n",
    "x'_\\textrm{end} &= \\gamma(x_\\textrm{end} - vt_\\textrm{end}) = -19850\\textrm{ m}\\\\\n",
    "t'_\\textrm{start} &= \\gamma(t_\\textrm{start} - vx_\\textrm{start}) = 20050\\textrm{ m} = 66.83\\ \\mu \\textrm{s}\\\\\n",
    "t'_\\textrm{end} &= \\gamma(t_\\textrm{end} - vx_\\textrm{end}) = 20245\\textrm{ m} = 67.48\\ \\mu \\textrm{s}\\\\\n",
    "\\end{align*}\n",
    "$$\n",
    "\n",
    "After all this, we find again that $t'_\\textrm{end}-t'_\\textrm{start} = 0.65\\ \\mu\\textrm{s}$. The _coordinate values_ in the new primed frame are different than the ones we had before, but the physics conclusions are the same."
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