Feynman's Four Gradient in Special Relativity

I am having trouble understanding Feynman's explanation of four gradient.

In section 25-3 of Vol. 2 of the Feynman lectures, he explains why the four gradient is not $(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$ by showing that derivatives for $\phi$ become (we let c = 1):

\begin{align} \frac{\partial \phi}{\partial t} &= \gamma \left(\frac{\partial \phi}{\partial t'} - v \frac{\partial \phi}{\partial x'}\right)\tag{1}\\ \frac{\partial \phi}{\partial x} &= \gamma \left(\frac{\partial \phi}{\partial x'} - v \frac{\partial \phi}{\partial t'}\right)\tag{2} \end{align}

Then he states that since the correct Lorentz transformations are:

\begin{align} t &= \gamma \left(t' + v x'\right)\tag{3}\\ x &= \gamma \left(x' + v t'\right)\tag{4} \end{align}

And concluded that since the signs don't match, (1) and (2) must be wrong. A full explanation can be found here: http://www.feynmanlectures.caltech.edu/II_25.html

I did not understand Feynman's logic, so I decided to check the correct formulas for the four gradient using (3) and (4).

I looked at the partial derivative of $\phi$ with respect to $t$. The correct four gradient states that it should be:

$$\frac{\partial \phi}{\partial t} = \gamma \left(\frac{\partial \phi}{\partial t'} + v \frac{\partial \phi}{\partial x'}\right)\tag{*}$$

We differentiate (3), (4):

\begin{align} \frac{\partial t}{\partial t'} &= \gamma \left(1 + v \frac{\partial x\prime}{\partial t'}\right)\tag{5}\\ \frac{\partial t}{\partial x'} &= \gamma \left(\frac{\partial t'}{\partial x'} + v\right)\tag{6} \end{align}

For simplicity, let $u\equiv \partial x'/\partial t'$. We put (5), (6) into (*) to get:

\begin{align} \frac{\partial \phi}{\partial t} &= \gamma \left(\frac{\partial \phi}{\partial t} \frac{\partial t}{\partial t'} + v \frac{\partial \phi}{\partial t} \frac{\partial t}{\partial x'}\right) \\ &= \gamma \left(\frac{\partial \phi}{\partial t}\right) \left( \frac{\partial t}{\partial t'} + v \frac{\partial t}{\partial x'}\right) \end{align}

\begin{align}1 &= \gamma \left(\gamma(1 + vu) + v \gamma\left(\frac{1}{u} + v\right)\right) \\ 1 - v^2 &= 1 + vu + \frac{v}{u} + v^2\end{align}

And this last equality is clearly false. So where did I make the mistake? And how can I correctly prove (*)?

Also, another question: when doing implicit differentiation on the x, t values, we can see that:

$$\frac{\partial t}{\partial t'} = \gamma \left(1 + v \frac{\partial x'}{\partial t'}\right),$$ but, $$\frac{\partial t'}{\partial t} = \gamma \left(1 - v \frac{\partial x}{\partial t}\right).$$ So then, $$1 = \frac{\partial t}{\partial t'} \frac{\partial t'}{\partial t}$$ is false. How can this be so!?

• The variables $\:x',t'\:$ are independent as space-time coordinates (like $\:x,y,z \in \mathbb{R}^3)\:$. So $$\dfrac{\partial x'}{\partial t'}=0 \tag{01}$$ The variable $\:t\:$ is a function of $\:x',t'\:$ : $\:t(x',t')\:$ and $$\dfrac{\partial t}{\partial t'}=\gamma \tag{02}$$ Jul 11, 2018 at 15:14

The wording is a bit unfortunate. Equations (1) and (2) are definitely not wrong: they are how derivatives transform, nothing we can do about that. What Feynman is saying is that they look wrong if we want to interpret $(\partial_t \phi, \nabla \phi)$ as a four-vector: the signs are different from the ones in the standard Lorentz transformation of a four-vector, equations (3) and (4). If we define $\nabla_\mu \phi = (\partial_t \phi, -\nabla \phi)$, we will find that it transforms as equations (3) and (4).

Regarding your second question, the answer lies in the multivariable chain rule: the correct equation is

$$\frac{\partial t}{\partial t'} \frac{\partial t'}{\partial t} + \frac{\partial t}{\partial x'} \frac{\partial x'}{\partial t} + \frac{\partial t}{\partial y'} \frac{\partial y'}{\partial t} + \frac{\partial t}{\partial z'} \frac{\partial z'}{\partial t} = 1$$

since the left hand side is the correct expansion of $\partial t/\partial t$. Remember, partial derivatives are tricky, we can't just cancel differentials.

Lastly, a warning for the future: Feynman is using a slightly unconventional notation for four-vectors, probably for pedagogical reasons. You will eventually learn that there is a difference between $a^\mu$ and $a_\mu$, and some of Feynman's equations will seem wrong in the common convention. Don't worry about it right now, just remember in case you ever come back to Feynman after learning more about four-vectors.

• thanks for the clarification! I am still confused about what you said: "What Feynman is saying is that they look wrong if we want to interpret (∂tϕ,∇ϕ) as a four-vector: the signs are different from the ones in the standard Lorentz transformation of a four-vector, equations (3) and (4)." How can we really compare the signs between the first 4 equations, if (3), (4) are normal equations, but (1) and (2) are differentials, so the essentially mean something different? Jul 11, 2018 at 16:56
• @A.Remorov $(\partial_t \phi, \nabla \phi)$ is a set of four numbers that you can calculate in each reference frame. But these four numbers only form a four-vector if, when switching frames, they transform as in equations (3)-(4). Equations (1)-(2) show that they do not transform as a four-vector does. Jul 11, 2018 at 17:03

So this is a hazard that you have to deal with whenever you are using partial derivatives: in the mathematically pure world, you would have some multivariable function $f(x, y)$ and you would describe partial derivatives "with respect to its first argument holding its other arguments constant," thus coming to $$f_{[1]}(x, y) = \lim_{h\to 0} \frac{f(x + h, y) - f(x, y)}h.$$ The issue is that in terms of pure mathematics the function doesn't "know" what its arguments are, except for what order they are in. Similarly you wish to define ($w = ct$), $$\phi(w, x) = \bar\phi(\bar w(w, x), \bar x(w, x))$$where $\bar w(w, x) = \gamma~(w - \beta x),~~ \bar x(w, x) = \gamma~ (x - \beta w).$ The derivatives of this are straightforwardly:\begin{align}\phi_{[1]}(w, x) &= \bar\phi_{[1]}~\bar w_{[1]}(w, x) + \bar\phi_{[2]}~\bar x_{[1]}(w, x)\\ &=\bar\phi_{[1]}~\gamma - \bar\phi_{[2]}~\gamma~\beta,\end{align}Which is Feynman's expression.

Now in our physicist ways we would like to identify these two fundamentally different mathematical functions, $\phi$ with $\bar \phi$, and say that they are both "the same $\phi$" but expressed "over different coordinates." (You could create a suitable mathematical definition, something like defining a set of smooth invertible coordinate-change maps $\mathbb U_N \subset C^\infty(\mathbb R^N, \mathbb R^N)$ and then defining some sort of equivalence class of $(u, \phi\circ u): u \in \mathbb U_2$ or something. We do something slightly different but in the same vein when we get to general relativity where we talk about manifolds; this just says that there is an abstract space of points and $\phi$ is a smooth scalar map from those points to real numbers; your coordinates $x,y,z$ are also some other smooth scalar maps from those points to real numbers, and you have to be careful how you combine these two to "coordinatize" your "tensor" expressions.)

The long-and-short of this is that if you're going to erase the difference between these functions $\phi$ and $\bar \phi$ which are fundamentally different mathematical machines, and say that they are "the same scalar field", you have to constantly specify both (1) what the variables are that we are differentiating with respect to, and (2) what the other variables are that we are holding constant. You have to unambiguously specify the line that you are following and the relative size of the steps you are taking along that line, to take this derivative!

With that we can re-express the above as:\begin{align}\left({\partial\phi\over\partial w}\right)_x &= \left({\partial\phi\over\partial \bar w}\right)_{\bar x} \cdot\left({\partial \bar w\over\partial w}\right)_{x} + \left({\partial\phi\over\partial \bar x}\right)_{\bar w}\cdot\left({\partial \bar x\over\partial w}\right)_{x}\\ &=\left({\partial\phi\over\partial \bar w}\right)_{\bar x}~\gamma - \left({\partial\phi\over\partial \bar x}\right)_{\bar w}~\gamma~\beta. \end{align}We can sometimes drop the outward statements that we are taking a partial derivative with respect to $\bar w$ holding $\bar x, \bar y, \bar z$ constant by understanding that $\bar w$ is part of the $(\bar w, \bar x, \bar y, \bar z)$ "barred" coordinate system and whenever I take a partial derivative with respect to the one, I understand implicitly that the others of these variables are expected to be held constant.

If we understand this implicitly then we understand that your tricky terms,$\frac{\partial \bar x}{\partial \bar w} = \frac{\partial x}{\partial w} = 0,$ because $x$ is implicitly being held constant while taking this partial derivative.

Well, that still gives the strange, $$\left(\frac{\partial \bar w}{\partial w}\right)_x\cdot \left(\frac{\partial w}{\partial \bar w}\right)_{\bar x} = \gamma^2,$$ but is this really so strange when we're holding different expressions constant for these two formulas? Consider if we tried to take instead $\left(\frac{\partial w}{\partial \bar w}\right)_{x}$, we would first have to remove $\bar x$ from the expression, \begin{align}w &= \gamma~(\bar w + \beta \bar x)\\ &= \gamma~\big(\bar w + \beta~\gamma~(x - \beta w)\big). \end{align}To handle the fact that we just regenerated $w$ in the last expression we would need to pull both $w$ terms to the left hand side: $$(1 + \beta^2\gamma^2)~w = \gamma~\bar w + \beta \gamma^2~ x.$$Now that's a strange coefficient on the left so let's dig into it, we have $\gamma=1/\sqrt{1-\beta^2}$ so this is $$1 + \frac{\beta^2}{1-\beta^2} = \frac{1-\beta^2}{1-\beta^2} + \frac{\beta^2}{1-\beta^2} = \frac{1}{1-\beta^2} = \gamma^2.$$ So now we know that we can divide through by it; $\gamma$ is never zero. And when we do we find: $$w = \frac1\gamma~\bar w + \beta~x,$$so that as you might have expected, $$\left(\frac{\partial \bar w}{\partial w}\right)_x\cdot \left(\frac{\partial w}{\partial \bar w}\right)_{x} = \gamma \cdot \frac1\gamma = 1.$$They only cancel out this way if you are very careful to make sure that you are holding the same other things constant. If you don't do this then all bets are off!

If we have a reference frame $S$ with space and time coordinates $x$ and $t$ and another reference frame $S'$ with space and time coordinates $x'$ and $t'$, where $S'$ is moving in the positive $x$ direction with a velocity $v$ when observed from $S$. The Lorentz transformations relating the two reference frames are;$$x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}},t'=\frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$$and$$x=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}},t=\frac{t'+\frac{vx'}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$$ Let us have a scalar function $\phi(x,t)$ that is invariant under the Lorentz transformations. Using our knowledge of multivariable calculus we can write; $$\Delta\phi=\frac{\partial\phi}{\partial{x}}\Delta{x}+\frac{\partial\phi}{\partial{t}}\Delta{t}$$and$$\Delta\phi=\frac{\partial\phi}{\partial{x'}}\Delta{x'}+\frac{\partial\phi}{\partial{t'}}\Delta{t'}$$ By holding $x$ constant and applying the Lorentz transformation to $\Delta{t'}$ and $\Delta{x'}$, we see that:$$\frac{\partial\phi}{\partial{t}}=\gamma(\frac{\partial\phi}{\partial{t'}}-v\frac{\partial\phi}{\partial{x'}})$$Now if we hold $t$ constant and apply the same procedures as above, we see that:$$\frac{\partial\phi}{\partial{x}}=\gamma(\frac{\partial\phi}{\partial{x'}}-\frac{v}{c^2}\frac{\partial\phi}{\partial{t'}})$$Now let our four-gradient be(let $c=1$) $$\nabla_\mu=(\frac{\partial}{\partial{t}},\nabla)$$Remember that the scalar product of two four-vectors must be invariant under the Lorentz transformation. The issue arises when we try to take the scalar product of $\nabla_\mu$ and $\nabla_\mu\phi$,we find that the scalar product isn't invariant under the Lorentz transformation. When we change the four-gradient to the form;$$\nabla_\mu=(\frac{\partial}{\partial{t}},-\nabla)$$we see that$$\nabla_\mu(\nabla_\mu\phi)=\frac{\partial^2\phi}{\partial{t}^2}-\frac{\partial^2\phi}{\partial{x^2}}$$which is invariant under the Lorentz transformation. So that is why the space components of the four-gradient must have a negative sign, so the d'Alembertian can remain Lorentz invariant.