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In this Stanford University lecture on Relativity, it is stated: Likewise, objects in spacetime all move at constant speed c in spacetime but if you change its direction, say by moving at speed v in the x direction, then spatial speed will change and so will the speed along the ct direction. Again, its total speed will still be c through spacetime.

http://web.stanford.edu/~oas/SI/SRGR/notes/SRGRLect6_2007.pdf

Brian Green of Columbia University also says the same thing, as do many others. Are they right?​

Qmechanic
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Astrophysics Math
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    Be aware that 4-velocity, which has a constant magnitude (4-speed?) of $c$ isn't remotely the same as ordinary velocity (3-velocity). Thinking of objects as moving through spacetime is problematic. While a particle is a point in 3-space that, in general, changes position with time, a particle is a line (a world line) in spacetime. – Alfred Centauri Apr 18 '16 at 02:41
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    @AstrophysicsMath many popular one-line quips about relativity can be mapped to mathematical statements that are true in the theory of relativity, but only by someone who is already familiar with those mathematical statements. That is, while there exists an interpretation that makes the quip true, that is not the interpretation the layperson listener will make. Some of these quips are extremely popular, so I think sometimes physicists just get lazy and repeat them. The most egregious offender is "when you are moving at close to $c$, time slows down" which is of the form "when [statement – Robin Ekman Apr 18 '16 at 03:56
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    that is meaningless in relativity] then [conclusion that is exactly opposite to what relativity says]" when read by a layperson. – Robin Ekman Apr 18 '16 at 04:00
  • If you use your mind and simply analyze "motion", it does not take long to discover that motion is constant, and that it is finite. In turn, only a change in direction of travel is left as a variable. This change in direction refers to change of direction of travel within a 4D space-time environment. Further analysis of this phenomena leads to the creation of SR equations. See http://goo.gl/fz4R0I – Sean Apr 22 '16 at 15:03

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Yes, this statement is true, in the sense that the four-velocity $u^\mu = (\gamma, \gamma \vec{v})$ always satisfies $$u^\mu u_\mu = 1$$ as you can check using the definition of $\gamma$. (I'm setting $c=1$.) Therefore the magnitude of the four-velocity is always equal to the speed of light.

However, this statement can be really misleading. It's true that a body at rest has "all its four-velocity pointing along the time direction", i.e. $$u^t = 1 \text{ and } u^x = u^y = u^z = 0.$$ But when sources are sloppy, like the lecture notes you linked, they imply that you "trade off" some of your "velocity through time" to get "velocity through space". Actually, when you acquire velocity through space, $u^x > 0$, we have $u^t > 1$, so spatially moving objects go faster through time. A longer amount of time passes per tick on their stopwatch.

knzhou
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Yes, but by definition. Not by any meaningful physics.

Imagine a path through 3-space. You can define the path by a function of time that returns a position. ${\bf f}(t)=(x(t),y(t),z(t))$. Then the velocity as a function of time is ${\bf v}(t)={\bf f}'(t)$. Easy.

You could do the same thing through 4-space, by describing a path parameterized by some other variable. ${\bf f}(k)=(t(k),x(k),y(k),z(k))$. This is useful if you want to do something as describe a constant acceleration path. ${\bf f}(k)=(\sinh(k),\cosh(k),0,0)$ is a path with constant proper acceleration! (c=1) So this isn't a dumb or useless thing to do.

But obviously, ${\bf f}(k)=(2k,0,0,0)$ and ${\bf f}(k)=(k,0,0,0)$ should describe the same path. They're different parameterizations of the same path through spacetime. We don't really care about the magnitude of ${\bf f}'(k)$. It contains no physics.

So, to discard the useless information, we normalize it. We multiply ${\bf f}'(k)$ by a number so that $t'(k)^2-x'(k)^2-y'(k)^2-z'(k)^2=1$. (this + - - - deal is the heart of special relativity). And that is the expression that "the four-velocity of a particle moving through spacetime is always the speed of light". This "multiplying by a number" is how you get the "${\bf u}=(\gamma, \gamma \vec{v})$" formula in knzhou's post.

It's a choice. A very natural choice, but a choice. Physicists like Brian Greene know that there are many choices of the phrasing you use to describe reality, and sometimes, in the interest of good popular science writing, they don't make it clear when they've described something as a valid description of the universe or the valid description of the universe. You could make the silly choice to normalize the four-velocity to twice the speed of light, and then the textbook you could write, "objects in spacetime all move at constant speed $2c$", where by "speed" I mean "magnitude of four-velocity". If you did this, all your calculations would give you the same physics.

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There's some debate in the comments, so let me come up with a precise version of my statement. What I'm really trying to say is:

No experiment can find the magnitude of a tangent vector to a curve in four-space. Only direction matters. Therefore, normalization is a choice.

ie, you're in four-space. (time and three spatial dimensions). You have a curve, a set of points, in that four-space. A tangent line to that curve should be velocity, but the magnitude of any given tangent line has no physical meaning. (The property of being timelike, spacelike, or lightlike is invariant under scalar multiplication. The three-velocity you calculate from the four velocity will be invariant under scalar multiplication of the four velocity). It's a convenient choice to make it of unit length.

  • You write, "Yes, but by definition. Not by any meaningful physics." So are you saying that "definitions of physics" and actual physics are polar opposites? Is Brian Green wrong? Is the Stanford lecture wrong? – Astrophysics Math Apr 18 '16 at 03:00
  • Actually, the normalization is physically meaningful. It's $dx^\mu/d\tau$, we are counting displacement per proper time. – knzhou Apr 18 '16 at 03:09
  • @AstrophysicsMath "Wrong" is a strong word. I'd say that Brian Greene is conveying a truth that is part of the model of special relativity that he is presenting. But other phrasings exist. –  Apr 18 '16 at 03:09
  • @AstrophysicsMath They're not wrong. It's just misleading, a lie-to-children. – knzhou Apr 18 '16 at 03:10
  • @knzhou -shrug- I think we're saying the same thing but disagreeing on word choice. If I normalized it to $u^\mu u_\mu=4c$, then in whatever calculations I did I plugged that in, I could get exactly the same results as you would with $u^\mu u_\mu=c$. I guess the most precise version of my statement: no experiment can find the magnitude of a tangent vector to a curve in four-space. Only direction matters. Therefore, normalization is a choice. –  Apr 18 '16 at 03:13
  • @knzhou awesome, glad that's settled. (Also, for any people doing comment archaeology, it should definitely be $u^\mu u_\mu=c^2$ :) –  Apr 18 '16 at 03:22
  • I've deleted some comments that seemed to be trying to pick a fight. That's not what we do here. But allow me to expand a little: physicist can and do disagree on theory, interpretation, and pedagogy and that is not a problem. And sometimes they air those disagreements in public and that too is not a problem. – dmckee --- ex-moderator kitten Apr 18 '16 at 03:24