Anybody fortunate enough to have reached the levels of physics in which one is taught the special theory of relativity will most likely encounter two components in the narrative: A light pulse bouncing between two mirrors on a train and two lightning bolts hitting said train on both ends. They are helpful analogies, they help you to visualise abstract concepts like inertial frames and length contraction. A certain group of students however doesn't need or even like analogies, they prefer the purity of mathematical equations and derivations. It is for this group of students that this is written.
First of all that you enjoy yourself. Working though the mathematical puzzles and quandaries that are associated with theoretical physics can be hard but also very rewarding. We also hope to instil a newfound appreciation for the mathematical art behind physics. Theoretical physics might not physically launch a billion dollar space telescope, it was theory and mathematics that made the workings and launching of said telescope possible.
The main goal of the calculations is to work towards four well-known equations:
$$v''=\frac{v'+v}{1+\frac{v'v}{c^2}}$$ | $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ |
which is the equation to add two velocities and the Lorentz factor. Also:
$$\Delta t'=\frac{\Delta t_0}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$ | $$l=l_0\sqrt{1-\frac{v^{2}}{c^{2}}}$$ |
which are the equations for time dilation and length contraction.
To get these equations, we're going to transform time-coordinates and place-coordinates from one inertial frame into another inertial frame.
To make full use of this document, you should have a notebook ready so you can do all the mathematics yourself. To make this manageable, the mathematical process is divided up into step-by-step questions and explanations. With most questions, you will find hints to help you on your way and of course, the full workings out. Perhaps the most useful are the tooltips that might appear at a question. These mathematical tricks might come in handy when you're working on other assignments and are a good addition to your mathematical toolbox.
Depending on your country or region, there might be differences in notations and these (even in the same region) change over the course of your academic career. At a certain moment \(v_x\) would just have meant the component of \(v\) in the x-direction. When different kinds of vectors are introduced, \(v_x\) and \(v^x\) are vastly different entities that require a different form of calculations but they still both mean the component of \(v\) in the x-direction. In this work, the correct notation is maintained as a matter of principle.
In some countries, it is common to use the prime (apostrophe or accent) to denote a derivative: \(f'_{(x)}\) is the derivative of \(f_{(x)}\). In this work, the prime will denote in which inertial frame a certain coordinate is seen. When a point has an x-coordinate \(x\) in inertial frame \(I\), seen from frame \(I'\) it will be \(x'\) (and a different number of course) and \(x''\) in frame \(I''\).
At a certain moment, we'll use \(v\) as the velocity of a second frame as seen from the first and we'll need the opposite, the velocity of the first frame as seen from the second. Since we'll have to use \(v'\) later as the velocity between the second and third frame, we'll use \(v^*\). This won't mean the complex conjugate.
Einstein based his thought experiments on two assumptions; In an inertial frame, all laws of physics behave like in all other inertial frames, regardless of the time, location or velocity and; Light is observed to be traveling at the same velocity from any reference point. The former can be seen in Newton's first law and the latter follows from Maxwell's equations. They're both valid assumptions but the latter is not necessary. Indeed, it might even cause misconceptions.
Inertial frames are basically coordinate systems that you use in mathematics. Contrary to the static coordinate systems that mathematicians are more comfortable with, an inertial frame can have a constant motion as seen from another inertial frame. In physics, the frame is usually centred on a physical object like a particle or vehicle but is equally valid for any mathematical point without a physical body.
Even though this is a mathematical exercise, we can't escape the fact that it is describing moving points. As such, you need to know about velocity and position. Since the velocity is the rate and direction in which a point changes its position, when the derivative of an equation that describes the position over time is taken (over time), you will get an equation that describes velocity over time.
To start our derivation, we'll need two inertial frames, \(I\) and \(I'\), traveling past each other in the x-direction at velocity \(v\). At a certain point, the "zero's" of both the space and time coordinates are equally zero.
The first assumption means that we'll deal with just one spatial and one temporal dimension. Since in any motion, you are allowed to chose which time is \(t=0\) and what position constitutes \(x=0\), the second assumption just means we don't have to add a constant to every coordinate.
In frame \(I\), any point can be described by the space-time coordinates \((t,x;y;z)\) and velocity \((u^x,u^y,u^z)\).
To describe the position of the point from the perspective of \(I'\), we have this set of equations with which we can transform the coordinates from one perspective to the other:
In which \(\gamma\), \(\kappa\), \(\beta\) and \(\eta\) are either constants or functions. Note that \(\gamma\) should be positive otherwise a series of points at successive time-coordinates in \(I\) would have the points in a reverse order through time in \(I'\). In other words, time would appear to run backwards. Mathematically it would be allowed but we haven't seen anything yet in physics to suggest the universe allows it too.
The frames are only moving in respect to each other in the x-direction. This means that the y and z coordinates will appear the same as seen from the other frame. The reason that we stated that the "zero's" had to be at the same place at some point is because of this, otherwise we'd have to add a constant value to these coordinates all the time.
In this derivation, we'll mainly deal with "empty" coordinates but the equations should also be applicable to coordinates of objects in an inertial frame. These objects should be allowed to move and for that you need a change in position \(x\). Without the \(x\) in the equation, you wouldn't be able to account for motion. Also, when an object is stationary in frame \(I\) and frame \(I'\) has a velocity in respect to \(I\), the objects has a velocity when seen from frame \(I'\).
In a broader scientific view, you may not just assume the position has no effect on the time. If the mathematics show that it doesn't, \(x\) can be eliminated from the equations through mathematical operations. You would be remiss if you eliminate \(x\) beforehand based on an assumption.
To achieve our main goal, we should now transform the equations from set 1 into equations that are dependant on \(v\), the velocity between the two frames. This means we must either get rid of \(\gamma\), \(\kappa\), \(\beta\) and \(\eta\) or define them so they are dependant on \(v\). At a certain moment, a new quantity may be introduced which will contain \(c\).
To get rid of \(\beta\) and \(\eta\) we need to have a look at the velocity of the point within frame \(I\), velocity \(u\) that we mentioned earlier. If we can transform positions in space-time, we can also transform velocities like \(u\) into \(u'\).
$$u'^{x}=\frac{dx'}{dt'}=\frac{\beta+\eta u^{x}}{\gamma+\kappa u^{x}}$$ |
And...
$$u'^{y}=\frac{dy'}{dt'}=\frac{u^{y}}{\gamma+\kappa u^{x}}$$ | $$u'^{z}=\frac{dz'}{dt'}=\frac{u^{z}}{\gamma+\kappa u^{x}}$$ |
You can easily derive \(x'\) to \(t\) but not so easily to \(t'\). Use this:
\(=(dx')^{1}(dt')^{-1}(dt)^{1}(dt)^{-1}=(dx)^{1}(dt)^{-1}(dt)^{1}(dt')^{-1}=\)
And remember that:
Since \(x\) is dependant on \(t\) you can't just assume it's a constant and claim that \(\frac{dx'}{dt}=\beta\). Remember that the derivation of place with respect to time is velocity so the derivation of \(x\) with respect to \(t\) becomes...
$$u'^x=\frac{dx'}{dt'}=\frac{dx'}{dt}\cdot\frac{dt}{dt'}$$ $$\frac{dx'}{dt}=\beta+\eta u^{x}$$ $$\frac{dt'}{dt}=\gamma+\kappa u^{x}$$ $$u'^x=\frac{dx'}{dt}\cdot\frac{1}{\frac{dt'}{dt}}=\frac{\beta+\eta u^x}{\gamma+\kappa u^x}$$
$$u'^{y}=\frac{dy'}{dt'}=\frac{dy'}{dt}\cdot\frac{dt}{dt'}$$ $$\frac{dy'}{dt}=u^{y}$$ $$\frac{dt'}{dt}=\gamma+\kappa u^{x}$$ $$u'^{y}=\frac{dy'}{dt}\cdot\frac{1}{\frac{dt'}{dt}}=\frac{u^{y}}{\gamma+\kappa u^{x}}$$
Let's take a point that is at rest in reference frame \(I'\), this would mean that \(u'^{x}=0\). When the frame \(I'\) is moving past \(I\) with velocity \(v\), the point as seen from \(I\) would have velocity \(v\) so \(u^{x}=v\). Reversely, a point at rest in \(I\) so that \(u^{x}=0\) will appear to have \(u'^{x}=v^{*}\) in frame \(I'\). Obviously, \(v^{*}=-v\).
This leads to:
$$v=-\frac{\beta}{\eta}$$ | |
$$v^{*}=\frac{\beta}{\gamma}$$ |
When a fraction like \(\frac{\beta+\eta u^{x}}{\gamma+\kappa u^{x}}\) has to produce zero, you only need to state that the numerator (term on top) equals zero. The denominator (the term on the bottom) isn't allowed to be zero at the same time but otherwise, doesn't matter.
Applying a relation to an equation often just means that you substitute (replace) quantities from the equation just as the relation states. In the case of equation 5, just take equation 2 and replace all instances of \(u'^{x}\) with \(v^{*}\) and all instances of \(u^{x}\) with zero.
$$u'^{x}=\frac{\beta+\eta u^{x}}{\gamma+\kappa u^{x}}$$ $$0=\frac{\beta+\eta v}{\gamma+\kappa v}$$ $$0=\beta+\eta v$$ $$-\beta=\eta v$$ $$v=-\frac{\beta}{\eta}$$
$$u'^{x}=\frac{\beta+\eta u^{x}}{\gamma+\kappa u^{x}}$$ $$v^{*}=\frac{\beta+\eta\cdot 0}{\gamma+\kappa\cdot 0}=\frac{\beta}{\gamma}$$
In the set of equations 1, we transform coordinates from \(I\) to \(I'\), this of course should also be possible in reverse, from \(I'\) to \(I\). This gives us:
$$t=\frac{\eta t'-\kappa x'}{\gamma\eta-\kappa\beta}$$ | $$x=\frac{\gamma x'-\beta t'}{\gamma\eta-\beta\kappa}$$ | $$y=y'$$ | $$z=z'$$ |
Rewrite the equations for \(t'\) and \(x'\) from set 1 into equations for \(x\) and \(t\). This should give you a total of four equations.
When you have the four equations, two are expressions of \(t\) so they can be equated, the same goes for the two expressions of \(x\).
\(t'=\gamma t+\kappa x\) |
| \(x'=\beta t+\eta x\) |
|
\(t=\frac{t'-\kappa x}{\gamma}\) | \(x=\frac{t'-\gamma t}{\kappa}\) | \(x=\frac{x'-\beta t}{\eta}\) | \(t=\frac{x'-\eta x}{\beta}\) |
\(\frac{t'-\kappa x}{\gamma}=\frac{x'-\eta t}{\beta}\) |
| \(\frac{t'-\gamma t}{\kappa}=\frac{x'-\beta t}{\eta}\) |
|
\(\beta t'-\beta\kappa x=\gamma x'-\gamma\eta x\) |
| \(\eta t'-\eta\gamma t=\kappa x'-\kappa x'-\kappa\beta t\) |
|
\(-\beta\kappa x+\gamma\eta x=\gamma x'-\beta t'\) |
| \(-\eta\gamma t+\kappa\beta t=\kappa x'-\eta t'\) |
|
\(x(\gamma\eta-\beta\kappa)=\gamma x'-\beta t'\) |
| \(t(\kappa\beta-\eta\gamma)=\kappa x'-\eta t'\) |
|
\(x=\frac{\gamma x'-\beta t'}{\gamma\eta-\beta\kappa}\) |
| \(t=\frac{\kappa x'-\eta t'}{\kappa\beta-\eta\gamma}\) |
|
|
| \(t=\frac{\eta t'-\kappa x'}{\eta\gamma-\kappa\beta}\) |
|
In these equations, all constants we introduced are still present: \(\gamma\), \(\kappa\), \(\beta\) and \(\eta\). We can however already elinimate \(\eta\).
substitute both equation 4 and 5 into \(v^{*}=-v\)
$$v^{*}=-v$$ $$\frac{\beta}{\gamma}=--\frac{\beta}{\eta}$$ $$\gamma=\eta$$
Succes! From now on, we can replace \(\eta\) with \(\gamma\), we have eliminated \(\eta\)!
This victory does pose a problem when combined with the equations from set 1 and 6.
When \(t\) is transformed into \(t'\) it is done with the equation from set 1: \(t'=\gamma t+\kappa x\). When it is transformed back, it is done with the equation from set 6: \(t=\frac{\eta t'-\kappa x'}{\gamma\eta -\kappa\beta}\). Now that we now that \(\gamma =\eta\) we could say that \(t=\frac{\gamma t'-\kappa x'}{\gamma\eta -\kappa\beta}\). The numerator looks a lot like the equation from set 1 but in reverse, as you would expect. The problem is in de nominator. When \(t'\) is transformed back into \(t\), the value is reduced by a factor of \(\gamma\eta -\kappa\beta\). This would mean that by transforming \(t\) back and forth between \(I\) and \(I'\), its value will change and that shouldn't be. This means that \(\gamma\eta -\kappa\beta\) must be equal to 1.
$$\gamma=\eta$$ | $$\gamma\eta -\kappa\beta =1$$ |
This allows us to write the transformations of \(x\) and \(x'\) more easily.
$$x'=\gamma (x-vt)$$ | $$x=\gamma (x'+vt')$$ |
You can introduce \(v\) into the equation by substituting \(\beta\) using equation 4.
Use the second equation from set 7 to get rid of the nominator first.
$$\left\{\begin{array}{ll}x'=\beta t+\eta x\\ \beta=-v\eta\end{array}\right.$$ $$x'=-v\eta t+\eta x$$ $$x'=-v\gamma t+\gamma x$$ $$x'=\gamma(x-vt)$$
$$\left\{\begin{array}{ll}x=\frac{\gamma x'-\beta t'}{\gamma\eta -\beta\kappa} \\ \gamma\eta -\beta\kappa =1\end{array}\right.$$ $$x=\frac{\gamma x'-\beta t'}{1}$$ $$x=\gamma x'-\beta t'$$ $$x=\gamma x'--v\eta t'$$ $$x=\gamma x'=v\eta t'$$ $$x=\gamma x'=v\gamma t'$$ $$x=\gamma (x'+vt')$$
Succes! From now on, we'll only use these equations for \(x\) and \(x'\) because we have eliminated \(\beta\) and \(\eta\) from them. We need to do the same for the equations for \(t\) and \(t'\) so \(\beta\) and \(\eta\) will have been eliminated from all transformations.
They're in set 1 and set 6.
\(t'=\gamma t+\kappa x\) doesn't contain \(\beta\) or \(\eta\) so it's already done. $$t=\frac{\eta t'-\kappa x'}{\gamma\eta -\kappa\beta}$$ $$t=\frac{\gamma t'-\kappa x'}{1}$$ $$t=\gamma t'-\kappa x'$$
We can now also express \(\kappa\) as a function of \(\gamma\) and v. We'll need to do some more calculations so for convenience sake, we'll keep using \(\kappa\) but later on, when it's time to eliminate \(\kappa\), we'll have the means to do so.
$$\kappa =\frac{1-\gamma^{2}}{v\gamma}$$ |
Start with the second equation from set 7 and use equation 4 to elinimate \(\beta\).
Replace all \(\eta\) with \(\gamma\) and solve.
$$\left\{\begin{array}{ll}\gamma\eta-\kappa\beta=1 \\ -\beta =v\eta\end{array}\right.$$ $$\gamma\eta +\kappa v\eta =1$$ $$\gamma^{2} +\kappa v\gamma =1$$ $$\kappa v\gamma =1-\gamma^{2}$$ $$\kappa =\frac{1-\gamma^{2}}{v\gamma}$$
Let's review for a moment, we now have the following equations for the transformation of a point in \(I\) to \(I'\) and back to \(I\).
$$t'=\gamma t+\kappa x$$ | $$x'=\gamma (x-vt)$$ | $$y'=y$$ | $$z'=z$$ | |
$$t=\gamma t'+\kappa x'$$ | $$x=\gamma (x'+vt)$$ | $$y=y'$$ | $$z=z'$$ |
All of the equations are dependant on velocity \(v\) between the two inertial frames and one remaining constant/function \(\gamma\). \(\kappa\) is still there too but can be replaced by equation 9 so actually only \(\gamma\) and \(v\) remain.
To investigate what \(\gamma\) is, we need to remove \(t\) and \(x\) from the equations. This seems odd since the equations are specifically meant to transform the coordinates \(t\) and \(x\) from one frame into another but that transformation should not be dependant on \(t\) or \(x\). A different value of \(t\) will produce a different value of \(t'\) but the proces itself should remain the same.
To start this investigation, we need a few more equations to work with and to achieve that, we will introduce a third inertial frame \(I''\). The three frames are all moving in respect to each other. We already established that frame \(I'\) travels at velocity \(v\) in respect to \(I\) and now we introduce \(I''\) that travels at \(v'\) in respect to \(I'\) and at \(v''\) in respect to \(I\). This logic follows that every time you advance to the "next" frame, you add a prime. Using this logic, the transformation of a time-coordinate from \(I'\) to \(I''\) would be \(t''=\gamma't'+\kappa'x'\).
\(I\), \(I'\) and \(I''\) | The three inertial frames |
\(t\) and \(x\) | A time and space coordinate in frame \(I\) |
\(t'\) and \(x'\) | The same time and space coordinate as seen from frame \(I'\) |
\(t''\) and \(x''\) | The same time and space coordinate as seen from frame \(I''\) |
\(v\) | The velocity of \(I'\) as seen from \(I\) |
\(v'\) | The velocity of \(I''\) as seen from \(I'\) |
\(v''\) | The velocity of \(I''\) as seen from \(I\) |
\(\gamma\) and \(\kappa\) | Constants/functions to transform coordinates from \(I\) to \(I'\) |
\(\gamma'\) and \(\kappa'\) | Constants/functions to transform coordinates from \(I'\) to \(I''\) |
\(\gamma''\) and \(\kappa''\) | Constants/functions to transform coordinates from \(I\) to \(I''\) |
$$t''=\gamma'(\gamma t+\kappa x)+\kappa'\gamma(x-vt)$$ | ||||
$$x''=\gamma'(\gamma(x-vt)-(v'(\gamma t+\kappa x))$$ |
\(t''=\gamma't'+\kappa'x'\) and \(x''=\gamma'(x'-v't')\) basically, add a prime to each symbol or a second prime if there already was one.
Substitue all instances of \(t'\) and \(x'\) for their equations from set 10
$$t''=\gamma't'+\kappa'x'$$ $$t''=\gamma'(\gamma t+\kappa x)+\kappa'\gamma (x-vt)$$
$$x''=\gamma'(x'-v't')$$ $$x''=\gamma'(\gamma (x-vt)-v'(\gamma t+\kappa x))$$
This third frame and the transformations to get into that frame allows for some more equations because if we transform coordinates from \(I\) to \(I'\) and then from \(I'\) to \(I''\), we should get the same result as a transformation straight from \(I\) to \(I''\). This means that the transformations from set 11 should produce the same result as the following equations that transform \(t\) and \(x\) into \(t''\) and \(x''\):
$$t''=\gamma''t+\kappa''x$$ | ||||
$$x''=\gamma''(x-v''t)$$ |
We're not going to equate the first \(t''\) with the second \(t''\) straight away, first we're going to make sure all the equations have the same shape.
In front of \(t\), gather everything from the original equation that gets multiplied by \(t\).
All the red quantities are mutiplied by \(t\): \(t''=\)
This means that the first part of the equation is going to be: \(t''=(\gamma'\gamma-\kappa'\gamma v)t+\)...
From the first of set 11: | \(t''=(\gamma'\gamma-\kappa'\gamma v)t+(\gamma'\kappa+\kappa'\gamma)x\) | |||
From the second of set 11: | \(x''=-(\gamma'\gamma v+v'\gamma'\gamma)t+(\gamma'\gamma-\gamma'\kappa v')x\) | |||
The first of set 12: | \(t''=\gamma''t+\kappa''x\) | |||
From the second of set 12: | \(x''=-(\gamma''v'')t+(\gamma'')x\) |
Now that the equations have the same shape, we're going to equate them in parts.
If you know that \(t''=A\cdot t+B\cdot x\) and also \(t''=C\cdot t+D\cdot x\), you know that \(A=C\) and \(B=D\)
\(\gamma''=\gamma\gamma'-\gamma\kappa'v\) | ||||
\(\kappa''=\gamma'\kappa+\kappa'\gamma\) | ||||
\(\gamma''v''=\gamma\gamma'(v+v')\) | ||||
\(\gamma''=\gamma\gamma'-\gamma'\kappa v'\) |
The first equation from set 13 is derived by taking the factor in front of \(t\) from \(t''=\gamma''t+\kappa''x\) and equating it to the factor in front of \(t\) from \(t''=(\gamma'\gamma-\kappa'\gamma v)t+(\gamma'\kappa+\kappa'\gamma)x\)
This produces \(\gamma''=\gamma'\gamma-\kappa'\gamma v\)
From the equations starting with \(t''\) take the two factors in front of the \(x\)
\((\gamma'\kappa + \kappa'\gamma )=\kappa''\)
\(\kappa''=\gamma'\kappa + \kappa'\gamma\)
From the equations starting with \(x''\) take the two factors in front of the \(t\)
\(-(\gamma'\gamma v+v'\gamma'\gamma)=-\gamma''v'' \)
\( \gamma''v''=(\gamma'\gamma v+v'\gamma'\gamma) \)
\( \gamma''v''=\gamma\gamma'(v+v')\)
From the equations starting with \(x''\) take the two factors in front of the \(x\)
\((\gamma'\gamma-\gamma'\kappa v')=\gamma''\)
\(\gamma''=\gamma\gamma'-\gamma'\kappa v'\)
The equations for \(\kappa''\) and \(\gamma'' v''\) are not that exciting but the two equations for \(\gamma''\) open up a new possibility:
$$\frac{\kappa'}{\gamma'v'}=\frac{\kappa}{\gamma v}$$ |
\(\gamma\gamma'-\gamma\kappa'v=\gamma\gamma'-\gamma'\kappa v'\)
\(-\gamma\kappa'v=-\gamma'\kappa v'\)
\(\frac{\kappa'}{\gamma'v'}=\frac{\kappa}{\gamma v}\)
This is something remarkable. Remember that the regular factors and velocity are used to transform coordinates from frame \(I\) to frame \(I'\) and the primed factors and velocity are used to transform from frame \(I'\) to \(I''\). Now we have proven that they are the same. In hindsight, it makes perfect sense because it shouldn't matter which frame you call \(I\) and which one you call \(I'\) but it also shows something more profound; There is a factor that is the same for each transformation. This factor then only needs to be combined with the velocity to get the transformation specific for a certain pair of frames but the factor itself doesn't change, it's an invariant.
Let's call this invariant \(a\) and define it like this:
$$\frac{\kappa}{\gamma v}=-a$$ |
It doesn't seem to make sense that the invariant \(a\) contains the velocity but even though we can't see it yet, so do the definitions of \(\kappa\) and \(\gamma\). Later on, we'll be able to see that \(a\) is a truly independent invariant.
To set up the equation to add velocities, consider the following: Frame \(I'\) moves at velocity \(v\) in respect for frame \(I\) and frame \(I''\) move at velocity \(v'\) with respect to frame \(I'\). This means that the velocity at which frame \(I''\) moves in respect to frame \(I\) should be the addition of \(v\) and \(v'\). Common, everyday sense would claim that \(v''=v+v'\) but Einstein has shown us that this is not the case. With the definition of \(a\), the third equation of set 13 and the fourth equation of set 13, we can begin to set up the equation to add velocities.
$$v''=\frac{v+v'}{1+avv'}$$ |
Since we want an equation that starts with \(v''=\), start with the third equation of set 13
Substitute the fourth equation to get rid of \(\gamma''\) and equation 15 to get rid of \(\kappa\).
In the nominator, you have two instances of \(\gamma\gamma'\), establish parenthesis with \(\gamma\gamma'\) outside.
\(\gamma'' v''=\gamma\gamma'(v+v')\)
\(v''=\frac{\gamma\gamma'(v+v')}{\gamma''}\)
\(v''=\frac{\gamma\gamma'(v+v')}{\gamma\gamma'-\gamma'\kappa v'}\)
\(v''=\frac{\gamma\gamma'(v+v')}{\gamma\gamma'+\gamma'a\gamma vv'}\)
\(v''=\frac{\gamma\gamma'(v+v')}{\gamma\gamma'(1+avv')}\)
\(v''=\frac{v+v'}{1+avv'}\)
This gets us really close to one of our main goals, the known equation to add velocities but \(a\) contains \(\kappa\) and we seem to have lost \(\gamma\). We can restore this by eliminating \(\kappa\) from the definition of \(a\).
$$\gamma=\frac{1}{\sqrt{1-av^2}}$$ |
Substitute equation 9 into 15
When you're confronted by something like \(A+x=B\cdot x\), you can still collate the x's by dividing left and right by x. You'll get \(\frac{A}{x}+1=B\)
$$\left\{\begin{array}{ll}\frac{\kappa}{\gamma v}=-a \\ \kappa=\frac{1-\gamma^2}{\gamma v}\end{array}\right.$$
$$\frac{1-\gamma^2}{\gamma^2 v^2}=-a$$
$$1-\gamma^2=-a\gamma^2 v^2$$
$$\frac{1}{\gamma^2}-1=-av^2$$
$$\frac{1}{\gamma^2}=1-av^2$$
$$\gamma^2=\frac{1}{1-av^2}$$
$$\gamma=\sqrt{\frac{1}{1-av^2}}$$
$$\gamma=\frac{1}{\sqrt{1-av^2}}$$
Let's appreciate that we've achieved one of our main goals: we've derived the Lorentz factor! The only difference between equation 17 and the Lorentz factor as we know it is the fact that apparently:
$$a=\frac{1}{c^2}$$ |
Once you know this, it's also easy to eliminate \(a\) from equation 16 and with that, we've achieved another main goal: the derivation of the known equation to add velocities!
This leaves us with the equations for time dilation and length contraction and for that, we have to back all the way to set 1. let's start with time dilation:
$$t'=\gamma t+\kappa x$$ |
We should realise that time dilation by its very definition has to do with a length of time, not a point in time. The equation transforms a point in time so we need to change it slightly:
$$dt'=\gamma dt+\kappa dx$$ |
With this, we have taken a little step towards physics and away from the very pure mathematics of just points in coordinate systems.
We can now get rid of the \(\gamma\) and the \(\kappa\).
$$dt'=\gamma(dt-\frac{v}{c^2}dx)$$ |
\(\left\{\begin{array}{ll}dt'=\gamma dt+\kappa dx \\ \kappa=-a\gamma v\end{array}\right.\)
\(dt'=\gamma dt-a\gamma vdx\)
\(dt'=\gamma(dt-avdx)\)
\(\left\{\begin{array}{ll}dt'=\gamma(dt-avdx) \\ -a=\frac{1}{c^2}\end{array}\right.\)
\(dt'=\gamma(dt-\frac{v}{c^2}dx)\)
This is the normal form of the Lorentz transformation but you'll notice that \(dx\) has not been eliminated. This is because the known equation for time dilation deals with a special situation: the start and end of the time period \(dt\) must occur on the same position in frame \(I\). Before we try and get to that know equation, let's derive the Lorentz transformation for \(dx'\) first.
$$x'=\beta t+\eta x$$ |
$$dx'=\gamma(dx-vdt)$$ |
\(\left\{\begin{array}{ll}x'=\beta t+\eta x \\ \beta=-v\eta\end{array}\right.\)
\(dx'=-v\eta dt+\eta dx\)
\(\left\{\begin{array}{ll}dx'=-v\eta dt+\eta dx) \\ \eta=\gamma\end{array}\right.\)
\(dx'=-v\gamma dt+\gamma dx\)
\(dx'=\gamma(dx-vdt)\)
Have a look at equations 20 and 21. If we were to assume that \(dx=0\) in equation 20 and \(dt=0\) in equation 21, we would get similar equations: the primed quantity is the unprimed quantity multiplied by \(\gamma\). When we observe the known equations for time dilation and length contraction however, we notice that one of them contains a multiplication with \(\gamma\) and the other one contains a division. The question then: Why? Why is \(dt'=dt\gamma\) and \(dx'=\frac{dx}{\gamma}\)? The answer can't be given by just mathematics.
We already made a slight step towards physics before when we had to use \(dt\) and \(dx\) but now we need to take another step. Since the equations for time dilation and length contraction describe what happens to an actual measured amount of time and distance, we need to set up a situation in which that time and distance might actually be measured. That situation doesn't have to be practical because the velocities involved in relativity are far from what we can now achieve in real life. We don't even have to use numbers or analogies, but we must describe a situation in which it is clear what the duration in a certain frame is and from which frame the observer is looking at it. This is done by introducing events.
Events can be anything, the explosion of a piece of fireworks, the turning on of a lamp, the turning off of a lamp, passing a point, it really doesn't matter. The only requirements are that the event itself has no timespan and the event takes place at one position (the former of course leads to the latter).
Situation 1: in frame \(I\), two events take place, \(E_{1}\) and \(E_{2}\). As seen from frame \(I\), \(E_{1}\) takes place before \(E_{2}\) but they both take place at the same position. If you want, a lamp turns on and after a while, the lamp turns off. The observer is located in frame \(I'\) which is still traveling at velocity \(v\) along the \(x\)-direction with respect to frame \(I\). This means that:
\(\left\{\begin{array}{ll} dx=0 \\ dt\neq0 \\ dx'\neq0 \\ dt'= ? \end{array}\right.\)
\(dx=0\) because the events take place at the same position as seen from frame \(I\).
\(dt\neq0\) because event \(E_{2}\) takes places after \(E_{1}\). We don't now how long so we can't specify what \(dt\) exactly is but it isn't zero.
\(dx'\neq0\) because frame \(I'\) moves past the position at which the events take place and it takes its axes with it. The distance between the event and position zero of frame \(I'\) will have changed in the time between the two events.
\(dt'= ?\) because we don't know enough yet about this and can't rule out anything.
\(dt'=\gamma(dt-\frac{v}{c^2}dx)\)
\(dt'=\gamma(dt-\frac{v}{c^2}\cdot 0)\)
\(dt'=\gamma dt\)
\(dt'=\frac{dt}{\sqrt{1-\frac{v^2}{c^2}}}\)
Since \(dt'\) is the time that the observer percieves to be passing in frame \(I'\) relative to his own time \(dt\), a more common notation is:
$$\Delta t=\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
And with that, we have succeeded in our third goal, the well known equation of time dilation.
Now for the last goal, the known equation of length contraction. We'll need another situation.
Situation 2: The observer is again located in frame \(I\) and frame \(I'\) is still passing him at velocity \(v\). The two events occur at a certain distance from each other in frame \(I'\) and for the observer in frame \(I\), they occur at the same time. We now want to know how the distance in frame \(I'\) is perceived by the observer in frame \(I\). We'll also introduce the proper symbols for that distance right away. The length \(l_0\) is the length all observers would perceive at \(v=0\) and \(l\) is the length that the observer in \(I\) perceives when \(v\neq 0\).
If you want: A train is moving past the observer and as seen by the observer, the head and tail lights are turned on at the same time, what is the length of the train that he perceives?
The following statements are now known:
\(\left\{\begin{array}{ll} dx=l \\ dt=0 \\ dx'=l_0 \\ dt'= ? \end{array}\right.\)
\(dx=l\) This is the length that the observer perceives in his frame \(I\).
\(dt=0\) because the observer sees the events at the same time in his frame
\(dx'=l_0\) because if the velocity would have been zero, this is what everybody would perceive. In the analogue of the train, this is the length that an observer in the train would perceive. It is also called the true length.
\(dt'= ?\) because we don't know enough yet about this and can't rule out anything.
\(dx'=\gamma(dx-vdt)\)
\(l_0=\gamma(l-v\cdot 0)\)
\(l=\frac{l_0}{\gamma}\)
$$l=l_0\sqrt{1-\frac{v^2}{c^2}}$$
And with that, we have succeeded in the last of our four goals, the well known equation of length contraction.
The two situations seem to be constructed to produce a certain outcome but they are actually chosen so the outcome can be reached by just one of the equations 20 and 21. To reach the same conclusions in different situations we'll need both equations and even more physics in the form of \(distance=velocity \cdot time\).
Situation 3: The observer is located in frame \(I\) which is passing frame \(I'\) at velocity \(v\). A certain length is at rest in frame \(I'\) and the observer needs to determine the length that he will perceive. If you want, a train driver is moving past a platform and he wants to determine the length of the platform using his velocity and the time he believes it takes him to travel from one end of the platform to the other end.
The following statements are now known:
\(\left\{\begin{array}{ll} dx=0 \\ dt = \frac{l_{perceived}}{v} \\ dx'= l_0 \\ dt'= ? \end{array}\right.\)
In this, \(l_{perceived}\) is the perceived length of the platform.
\(dx=0\) The observer is at rest in his frame \(I\), the cabin of the train.
\(dt = \frac{l_{perceived}}{v} \) This is what the observer measures and because of \(distance=velocity \cdot time\), it is equal to \(\frac{length_{perceived}}{velocity}\)
\(dx'= l_0 \) As before, this is the length that everybody would agree on when not moving relative to the frame it's in.
\(dt'= ?\) This is the time the observer believes has elapsed on the platform (which according to him is in motion).
\(dx'=\gamma(dx-vdt)\)
\(l_0=\gamma(0-v\cdot\frac{l_{perceived}}{v} \)
\(l_0=\gamma(-l_{perceived})\)
\(-l_{percieved}=\frac{l_0}{\gamma}\)
$$l=l_0\sqrt{1-\frac{v^2}{c^2}}$$
The minus-sign just tells you the direction of the distance
\(dt'=\gamma(dt-\frac{v}{c^2}dx)\)
\(dt'=\gamma(dt-\frac{v}{c^2}\cdot 0)\)
\(dt'=\gamma(dt)\)
$$\Delta t=\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
And with that, we have achieved all of our goals. Using nothing but mathematics and a only a little bit of physics, we've derived the relativistic equation to add velocities, the Lorentz-factor and the well-known equations of time dilation and length contraction. We hope you've enjoyed yourself and in the process, learnt a bit of mathematics and its importance to physics.