Ok, I will, though I do apologize if my last derivation may have been difficult to follow. My time is limited and working with LaTeX in Blogger is about as fun as grinding with a porcupine. I am still getting used to it, so please, bear with me.
Let's back track somewhat and talk a little about an aspect of Special Relativity, that frankly, I find much cooler than Lorentz Contraction and Time Dilation: Relativistic Momentum and Energy. In Newtonian Mechanics, for those who don't know, we have momentum, p:
\begin{equation}
p = \int \! F \ \mathrm{d}t = mv
\end{equation}
where F is the net force, m is mass, and v is velocity.
And kinetic energy, T:
\begin{equation}
T = \int \! F \cdot \ \mathrm{d}s = \frac{mv^2}{2}
\end{equation}
where s is the path.
In SR, (1) is replaced by
\begin{equation}
p = \mathrm{MAAAGIC} = mv\gamma
\end{equation}
and (2) is replaced by
\begin{equation}
T^2 = \mathrm{ME \> LUCKY \> CHARMS!} = m^2c^4 + p^2c^2
\end{equation}
where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is the Lorentz factor.
You'll notice that the intermediate steps involve mystical invocations and supplications to supernatural beings (the kind you find on cereal boxes). For all intents and purposes, this is how derivation of relativistic momentum and energy is presented in all explanations I have encountered.
An extremely common method used to describe this, which I confess, I have never taken the time to truly appreciate, is one involving a glancing, elastic, collision between two particles. First, they show that if we assume $p=mv$ momentum, momentum is not conserved. Then, they skip the most important step and guess that p satisfies (3) and then show that under that arbitrary assumption, that momentum will be conserved. That's right! In order to show people how to derive relativistic momentum, they don't actually derive it, but magically guess that it must take that form.
Please permit me to take a moment to tell you how awful, and completely unhelpful this type of pedagogical strategy is. As physicists, it is in our best interest to teach upcoming physicists not just how to remember equations, but how to apply sound physical reasoning to arrive at those results, in hopes that some day, they will be able to apply those same vital skills to arrive at novel results. How is this going to help them do that?
The sad thing is, this kind of strategy is rampant in mathematically oriented subjects. Too often, especially in upper level courses, new concepts are presented without precedent or even the barest motivation--as if they sprang, fully formed, from the foreheads of their originators. There is no hint of the hours of reasoning that lead to these results. Even worse, as an artifact of progress, the original derivation, though correct, is often replaced with a sleeker, more efficient, derivation that turns out to be less instructive than its predecessor. I want to talk more about this phenomenon--perhaps in the next blog post, but for now, I just want to point out how misguided this particular method is. If you are going to derive something, then for heaven's sake, derive it! Don't just invoke Baphomet, and then act as if you have actually explained anything worthwhile!
Anyway, it turns out that there is a really illuminating way to look at this situation and obtain the necessary physical insight for deriving this result. Let's start with relativistic acceleration. Again, for pedagogical purposes, it suffices to restrict ourselves to 1+1 dimensions. Let's assume the canonical set up where B passes A moving at a relative velocity of v. Suppose that B sees some object, say, a fly, moving with a velocity, $v'_f$, in some event, E. Then to A, the fly will be moving with velocity,
\begin{equation}
v_f = \frac{v'_f + v}{1+\frac{v'_fv}{c^2}}
\end{equation}
in that same event.
I will not explain why $v_f$ takes this value in this post, though it is straightforward to derive from the Lorentz Transformation.
Now, let's change things up a little. Let's assume that at time t' on B's clock, B is instantaneously at rest with the fly. Now, suppose that B measures a force, F', to accelerate the fly at time t' on his clock. Since B is at rest with the fly in this instant, then B does not measure the fly to be moving at relativistic speeds relative to himself, at least not at first. In other words, during a small amount of time $\Delta t'$ after $t'$, B would measure the fly to be moving, but not at relativistic speeds. In fact, if $\Delta t'$ is small enough the fly's speed will be nearly 0 m/s. Now, let's think about the principle of relativity:
Keeping this in mind, let's ask the following question: What if, instead, A were seeing this fly accelerate from rest, instead of B? Then, we would reasonably assume that in this small interval of time, the fly should obey regular old Newtonian mechanics--more formally, as $\Delta t'$ approached 0, the laws of physics would approach Newtonian physics. Namely, we would assume the fly had a mass, m, and that this force F', accelerated this fly in the time interval between $t'$ and $t'+\Delta t'$ by approximately $\frac{F'}{m}\Delta t'$
Well, by the First Principal of Relativity, B would have to measure this exact same scenario. Now, we're in business! So, let's see what this implies. At time $t'$, B sees the fly at some position, $x'$, at rest. Then, at time $t'+\Delta t'$, B sees the fly at position, $x'$, but moving with velocity, $\frac{F'}{m}\Delta t'$. The following is a table relating how A and B measure things:
Boy, that looks beautiful, doesn't it? You just want to smear chocolate all over that table and rub it on your chest. Oh yeah...
Oh, where was I? Oh yes, this horrific table. Now, as much as I enjoy writing in LaTeX (having spent the last hour fighting with that stupid table), I am going to skip a few steps. They are straightforward, and you should be able to do them yourself. But, basically, we would find that where B sees the fly accelerate by $\frac{dv'}{dt'}=\frac{F'}{m}$, A sees the fly accelerate by $\frac{dv}{dt} = \frac{F'}{m}\gamma^{-3}$
We can rewrite this as
\begin{equation}
F'=m\frac{dv}{dt}\gamma^3
\end{equation}
Now, what are we to make of this? Well, intuitively, if B sees the fly accelerate more from his frame, A will see it accelerate less, so we should expect that should somehow translate into the fly being "harder to push" when it is moving. So, even though we have an expression for acceleration, we need some way to connect this to the actual force that A measures. Personally, I don't think it is immediately clear how to do this. The answer, however, lies, in finding a force that both A and B can agree on.
There are a number of ways to find such a force. Perhaps the most obvious one (at least to me) is a uniform gravitational field, but this is immediately problematic--at least conceptually, because we know that, in a gravitational field, everything accelerates at the same rate. In this example, we have found that the acceleration should differ. Of course, we haven't actually found a contradiction. Rather, all we have found is that, at least algebraically, the expressions for accelerations differ, but have not actually shown that the numerical values should be different. This, of course, indicates that F', m, and $\gamma$ must take on appropriate values to preserve the acceleration, but, as you can see, things are already getting hairy, so it might help to clear things up by using a different force. [UPDATE: I can't believe that I just realized the error I made in this paragraph. Everyone in a uniform gravitational field experiences the same acceleration, not the same force. FAILURE! Sorry about that. You know, I am told that a while ago, an actual physics professor visited my blog and concluded that I don't know what I am talking about. The above paragraph could be what induced him to arrive at this conclusion. Oh well.]
And for that, we'll choose the good ol' Coloumb force. Let's imagine for a moment that the fly is a dipteran Piotr Nikolaievitch Rasputin, and has used his mutant powers to turn completely metal, and that dipteran storm charged him with lightning. Furthermore, suppose that everybody is situated between two enormous, charged, parallel, metal plates, orthogonal to $v$, which generate a uniform electric field, which serves as the force. At the limit that the plates become infinitely large, not only is this field uniform, but it only depends on their charge and area. As both A and B agree on these quantities, they would agree on the force between the plates. Now, invoking the First Postulate of Relativity again, we immediately see that both A and B must agree that the force on the fly is the same.
Hence,
\begin{equation}
F = F' = m\frac{dv}{dt}\gamma^3
\end{equation}
So, what exactly is this telling us? Well, basically, any time you push on a moving object with a force, F' in the moving frame, an equivalent push in the rest frame will result in a lower accelerations. Consequently, it should take a stronger force to achieve that same acceleration in the rest frame. So, it seems that we have arrived at the expression for force, as A measures it, on an accelerating object.
From here on in, things get relatively easy, so to speak (depending on how you look at things). We have done most of the conceptual work with this thought experiment. Now, let's invoke the handy First Postulate again to find momentum and energy.
In this case, since B would use $\frac{dp'}{dt'} = F'$, then A should use $\frac{dp}{dt}=F$. Thus, integrating F, we find that $p=mv\gamma$.
Let's write v in terms of p because that will become useful in a moment:
\begin{equation}
v=\frac{p}{\sqrt{m^2 + \frac{p^2}{c^2}}}
\end{equation}
Likewise, we can calculate the kinetic energy
\begin{equation}
T=\int{F\mathrm{d}x}
\end{equation}
Noting that $F=\frac{dp}{dt}$, let's make this easier by changing the variables:
\begin{equation}
T=\int{\frac{dp}{dt}dx} = \int{\frac{dx}{dt}dp} = \int{v \ dp} = \int{\frac{p}{\sqrt{m^2 + \frac{p^2}{c^2}}} \ dp}
\end{equation}
Therefore,
\begin{equation}
T=c^2\sqrt{m^2 + \frac{p^2}{c^2}} = \sqrt{m^2c^4 + p^2c^2}
\end{equation}
And so, we have just derived equations (3) and (4), as we set out to do.
p = \int \! F \ \mathrm{d}t = mv
\end{equation}
where F is the net force, m is mass, and v is velocity.
And kinetic energy, T:
\begin{equation}
T = \int \! F \cdot \ \mathrm{d}s = \frac{mv^2}{2}
\end{equation}
where s is the path.
In SR, (1) is replaced by
\begin{equation}
p = \mathrm{MAAAGIC} = mv\gamma
\end{equation}
and (2) is replaced by
\begin{equation}
T^2 = \mathrm{ME \> LUCKY \> CHARMS!} = m^2c^4 + p^2c^2
\end{equation}
where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is the Lorentz factor.
You'll notice that the intermediate steps involve mystical invocations and supplications to supernatural beings (the kind you find on cereal boxes). For all intents and purposes, this is how derivation of relativistic momentum and energy is presented in all explanations I have encountered.
An extremely common method used to describe this, which I confess, I have never taken the time to truly appreciate, is one involving a glancing, elastic, collision between two particles. First, they show that if we assume $p=mv$ momentum, momentum is not conserved. Then, they skip the most important step and guess that p satisfies (3) and then show that under that arbitrary assumption, that momentum will be conserved. That's right! In order to show people how to derive relativistic momentum, they don't actually derive it, but magically guess that it must take that form.
Please permit me to take a moment to tell you how awful, and completely unhelpful this type of pedagogical strategy is. As physicists, it is in our best interest to teach upcoming physicists not just how to remember equations, but how to apply sound physical reasoning to arrive at those results, in hopes that some day, they will be able to apply those same vital skills to arrive at novel results. How is this going to help them do that?
The sad thing is, this kind of strategy is rampant in mathematically oriented subjects. Too often, especially in upper level courses, new concepts are presented without precedent or even the barest motivation--as if they sprang, fully formed, from the foreheads of their originators. There is no hint of the hours of reasoning that lead to these results. Even worse, as an artifact of progress, the original derivation, though correct, is often replaced with a sleeker, more efficient, derivation that turns out to be less instructive than its predecessor. I want to talk more about this phenomenon--perhaps in the next blog post, but for now, I just want to point out how misguided this particular method is. If you are going to derive something, then for heaven's sake, derive it! Don't just invoke Baphomet, and then act as if you have actually explained anything worthwhile!
Anyway, it turns out that there is a really illuminating way to look at this situation and obtain the necessary physical insight for deriving this result. Let's start with relativistic acceleration. Again, for pedagogical purposes, it suffices to restrict ourselves to 1+1 dimensions. Let's assume the canonical set up where B passes A moving at a relative velocity of v. Suppose that B sees some object, say, a fly, moving with a velocity, $v'_f$, in some event, E. Then to A, the fly will be moving with velocity,
\begin{equation}
v_f = \frac{v'_f + v}{1+\frac{v'_fv}{c^2}}
\end{equation}
in that same event.
I will not explain why $v_f$ takes this value in this post, though it is straightforward to derive from the Lorentz Transformation.
Now, let's change things up a little. Let's assume that at time t' on B's clock, B is instantaneously at rest with the fly. Now, suppose that B measures a force, F', to accelerate the fly at time t' on his clock. Since B is at rest with the fly in this instant, then B does not measure the fly to be moving at relativistic speeds relative to himself, at least not at first. In other words, during a small amount of time $\Delta t'$ after $t'$, B would measure the fly to be moving, but not at relativistic speeds. In fact, if $\Delta t'$ is small enough the fly's speed will be nearly 0 m/s. Now, let's think about the principle of relativity:
The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of coordinates in uniform translatory motion. OR: The laws of physics are the same in all inertial frames of reference.
Keeping this in mind, let's ask the following question: What if, instead, A were seeing this fly accelerate from rest, instead of B? Then, we would reasonably assume that in this small interval of time, the fly should obey regular old Newtonian mechanics--more formally, as $\Delta t'$ approached 0, the laws of physics would approach Newtonian physics. Namely, we would assume the fly had a mass, m, and that this force F', accelerated this fly in the time interval between $t'$ and $t'+\Delta t'$ by approximately $\frac{F'}{m}\Delta t'$
Well, by the First Principal of Relativity, B would have to measure this exact same scenario. Now, we're in business! So, let's see what this implies. At time $t'$, B sees the fly at some position, $x'$, at rest. Then, at time $t'+\Delta t'$, B sees the fly at position, $x'$, but moving with velocity, $\frac{F'}{m}\Delta t'$. The following is a table relating how A and B measure things:
| $\left(x',\> t'\right)$ | $v'_f$ | $(x,t)$ | $v_f$ |
| $\left(x',\> t'\right)$ | $0 \mathrm{ m/s}$ | $\left((x'+vt')\gamma,\>(t'+\frac{x'v}{c^2})\gamma\right)$ | $v$ |
| $\left(x',\> t'+\Delta t'\right)$ | $\frac{F'}{m}\Delta t'$ | $\left(\left(x'+v\left(t'+\Delta t'\right)\right)\gamma,\>\left(\left(t'+\Delta t'\right)+\frac{x'v}{c^2}\right)\gamma\right)$ | $\frac{\frac{F'}{m}\Delta t' + v}{1+\frac{\left(\frac{F'}{m}\Delta t'\right) v}{c^2}}$ |
Boy, that looks beautiful, doesn't it? You just want to smear chocolate all over that table and rub it on your chest. Oh yeah...
Oh, where was I? Oh yes, this horrific table. Now, as much as I enjoy writing in LaTeX (having spent the last hour fighting with that stupid table), I am going to skip a few steps. They are straightforward, and you should be able to do them yourself. But, basically, we would find that where B sees the fly accelerate by $\frac{dv'}{dt'}=\frac{F'}{m}$, A sees the fly accelerate by $\frac{dv}{dt} = \frac{F'}{m}\gamma^{-3}$
We can rewrite this as
\begin{equation}
F'=m\frac{dv}{dt}\gamma^3
\end{equation}
Now, what are we to make of this? Well, intuitively, if B sees the fly accelerate more from his frame, A will see it accelerate less, so we should expect that should somehow translate into the fly being "harder to push" when it is moving. So, even though we have an expression for acceleration, we need some way to connect this to the actual force that A measures. Personally, I don't think it is immediately clear how to do this. The answer, however, lies, in finding a force that both A and B can agree on.
There are a number of ways to find such a force.
And for that, we'll choose the good ol' Coloumb force. Let's imagine for a moment that the fly is a dipteran Piotr Nikolaievitch Rasputin, and has used his mutant powers to turn completely metal, and that dipteran storm charged him with lightning. Furthermore, suppose that everybody is situated between two enormous, charged, parallel, metal plates, orthogonal to $v$, which generate a uniform electric field, which serves as the force. At the limit that the plates become infinitely large, not only is this field uniform, but it only depends on their charge and area. As both A and B agree on these quantities, they would agree on the force between the plates. Now, invoking the First Postulate of Relativity again, we immediately see that both A and B must agree that the force on the fly is the same.
Hence,
\begin{equation}
F = F' = m\frac{dv}{dt}\gamma^3
\end{equation}
So, what exactly is this telling us? Well, basically, any time you push on a moving object with a force, F' in the moving frame, an equivalent push in the rest frame will result in a lower accelerations. Consequently, it should take a stronger force to achieve that same acceleration in the rest frame. So, it seems that we have arrived at the expression for force, as A measures it, on an accelerating object.
From here on in, things get relatively easy, so to speak (depending on how you look at things). We have done most of the conceptual work with this thought experiment. Now, let's invoke the handy First Postulate again to find momentum and energy.
In this case, since B would use $\frac{dp'}{dt'} = F'$, then A should use $\frac{dp}{dt}=F$. Thus, integrating F, we find that $p=mv\gamma$.
Let's write v in terms of p because that will become useful in a moment:
\begin{equation}
v=\frac{p}{\sqrt{m^2 + \frac{p^2}{c^2}}}
\end{equation}
Likewise, we can calculate the kinetic energy
\begin{equation}
T=\int{F\mathrm{d}x}
\end{equation}
Noting that $F=\frac{dp}{dt}$, let's make this easier by changing the variables:
\begin{equation}
T=\int{\frac{dp}{dt}dx} = \int{\frac{dx}{dt}dp} = \int{v \ dp} = \int{\frac{p}{\sqrt{m^2 + \frac{p^2}{c^2}}} \ dp}
\end{equation}
Therefore,
\begin{equation}
T=c^2\sqrt{m^2 + \frac{p^2}{c^2}} = \sqrt{m^2c^4 + p^2c^2}
\end{equation}
And so, we have just derived equations (3) and (4), as we set out to do.
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