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I am reading Landau's The Classical Theory of Fields. On page 30, section 10, the last paragraph reads:

To solve the problem, we first determine the properties of the "volume element" $dp_xdp_ydp_z$ with respect to Lorentz transformations. If we introduce a four-dimensional coordinate system, on whose axes are marked the components of the four-momentum of a particle, then $dp_xdp_ydp_z$ can be considered as the zeroth component of an element of the hypersurface defined by the equation $p^ip_i=m^2c^2$. The element of hypersurface is a four-vector directed along the normal to the hypersurface; in our case the direction of the normal obviously coincides with the direction of the four-vector $p_i$. From this it follows that the ratio $dp_xdp_ydp_z/E$ is an invariant quantity, since it is the ratio of corresponding components of two parallel four-vectors.

I don't understand the argument. Can someone explain it in simple terms without too many calculations?

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  • $\begingroup$ @Felicia: $p^0=E/c$ $\endgroup$
    – rioiong
    Feb 10, 2022 at 9:53
  • $\begingroup$ Yes! I realized. Sorry. I don't really get why the volume element is part of a four vector. Is there a relation here with differential forms? $\endgroup$ Feb 10, 2022 at 10:03
  • $\begingroup$ I think $dp_xdp_ydp_z$ is the zeroth component like $E$ is. Both fourvectors they belong to are parallel. $\endgroup$ Feb 10, 2022 at 10:13
  • $\begingroup$ @Felicia: I don't have any intuition about that. Is that related to $p^idp_i=0$? $\endgroup$
    – rioiong
    Feb 10, 2022 at 10:31
  • $\begingroup$ There is a four dimensional differential three-form associated with the differentials $dp_i$: $(dp_xdp_ydp_z)i + (dp_ydp_zdp_t)j +(dp_zdp_tdp_x)k + (dp_tdp_xdp_y)l$. This four-vector is parallel to $(E/c, p_x, p_y, p_z)$, according to the text. $\endgroup$ Feb 10, 2022 at 10:48

2 Answers 2

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So this is an interesting challenge, the easiest ways to restate this point actually dive much further into the mathematical notation, and you are asking for something which dives much less into the mathematics. Which is fine, this is physics. But yeah, the easiest way is to use either wedge products, or an orientation tensor.

Deferring the extrusion of a volume

If you looked at the 4-volume $\mathrm dw~\mathrm dx~\mathrm dy~\mathrm dz$ (where $w=ct$) this is preserved by the Lorentz boost. You can view this as part of how we define the Lorentz boost, like we wouldn't feel good if it weren't 1... or if you want to be really physically minded you would say the time dilation factor perfectly cancels the length contraction factor.

But if we want to remove, say, the $\mathrm dw$ then we won't get an invariant quantity anymore. If we remove the time dilation then we only have the contraction, it won't balance out. The problem is we have to remove some component of this because the actual 4-volume is zero, the 4-momentum has a fixed length $mc$ held rigid by the rest mass of the particle, so it lives on a “sphere.” (Scare quotes just to remind you that this is a sphere according to the Lorentzian metric... For example in Euclidean 3D space you are used to a sphere being bounded, there is a maximum that each component can attain. But there's no maximum $p_z$, it's just that as you increase that component you must increase the $p_w=E/c$ component to match.)

This is where Landau wants to get tricky and use a sort of “cross product” (you'll recall in 3D that a cross product gives a volume of a parallelepiped, Landau wants a volume). Again, we won't go into the direct construction via wedges or orientations. Instead, just suppose you had this problem in 3D, where you can calculate invariant 3D volumes but you are looking at an area on the unit sphere. Landau wants to say that the area is “almost” a 3D volume. In fact it becomes a Euclidean scalar $\mathrm dV$ when you extrude it in the radial direction $\mathrm dr$. And the area defines this normal direction $\mathbf r$ in addition to defining its own magnitude $\mathrm dA$. (Technically we would say it is a covector, $\mathrm dA_\mu = \mathrm dA~(\hat r\cdot)$, it is a quantity just waiting for you to come in with some other displacement vector $\mathrm d\mathbf s$ so that it can calculate a scalar $\mathrm dV=\mathrm dA~\hat r\cdot\mathrm d\mathbf s.$ But by now you have presumably already seen the raising of indices, every covector can be brought back to being a vector in these sorts of spaces.)

Building some intuition in the sphere

At the risk of making this very wordy, let's hop back to this simple case we know and love, the 2-sphere in Euclidean 3D. Landau wants us to see immediately that there is a vector here $\mathrm d\mathbf A=(?,?,\mathrm dx~\mathrm dy)$. Well, this is not so immediate when we start. The thing on the right after all is some sort of projection of the area on the left onto the $xy$-plane, so it is constrained by the sphere, etc.

So let me make the case this way, in spherical coordinates we would find that say the area is $\mathrm d\mathbf A = \hat r~r^2\sin\phi~\mathrm d\phi~\mathrm d\theta,$ Landau asks us to just look at the $\hat z$-component of this, $$\mathrm d\mathbf A \cdot\hat z = r^2\sin\phi\cos\phi~\mathrm d\phi~\mathrm d\theta. $$ If asked what this means, he is two steps ahead of us and says it is $\mathrm dx~\mathrm dy$, just project the area from the sphere down onto the $xy$-plane. Presumably he's thinking: well, if we multiply by $\mathrm dz$ it is a volume $\mathrm dx~\mathrm dy~\mathrm dz$, where $\mathrm dx~\mathrm dy$ is “obviously” the 2D projection of the area.... so he is seeing that this is somehow the $xy$-projection of $\mathrm dA$ being extruded into a volume in the $z$-direction. But if you want to see it directly you probably want to switch to cylindrical coordinates, $\rho = r\sin\phi,\mathrm d\rho=r\cos\phi~\mathrm d\phi,$ to rewrite as $$\mathrm d\mathbf A \cdot\hat z =\mathrm d\rho~\rho~\mathrm d\theta.$$ And that is indeed the 2D area in polar coordinates, the 3D-oriented area has indeed been projected onto the plane for this component.

So if you accept that we have the general idea then yes, we have some vector $\mathrm d\mathbf A $ with a sort of $\mathrm dx~\mathrm dy~\hat z$ component in 3D, and in 4D we'll have a $\hat w~\mathrm dp_x~\mathrm dp_y~\mathrm dp_z$ component.

Scale factors are scalars too!

Let's quickly round this up.

The final part of this argument is, Landau says, “if only we had a parallel vector!” If you have two parallel vectors they transform the same, the ratio of their magnitudes is a scalar, and it's a scalar you can get from any matching pair of components in any coordinate frame—any pair can give you the scale factor. “Oh wait, we do!!” Just as the 2-sphere was constrained to move in an area normal to $\mathbf r,$ which became the preferred direction to extrude a volume, here the momentum is constrained to move in a 3-surface normal to $p^\mu$. The $\hat w$-component of that vector is $E/c$ and so our Lorentz-invariant scale factor between these two parallel vectors is $\mathrm dp_x~\mathrm dp_y~\mathrm dp_z/E$.

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The physical meaning of the referenced text is that a particle's maximum translational speed through momentum space and in extension also spacetime is determined by the volume it occupies in space. The limits are for volume V>0 → $v<c$, thus maximum speed $v$ this particle can reach is less than c the speed of light and for V=0 → $v=c$. In layman terms, for momentum space if the particle does not occupy a 3D volume it has no mass and therefore travels at c the speed of light like the photon.

Also the phase space volume of the particle is a Lorentz invariant. Phase space volume and relativity

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