# In which direction is the relation between the time-component of celerity and the Lorentz factor defined?

Celerity (a.k.a. proper velocity) is defined as $$w^\alpha=\frac{\mathrm{d}X^\alpha}{\mathrm{d}\tau}$$, where $$\mathrm{d}X^\alpha=(\mathrm{d}t,\mathrm{d}x,\mathrm{d}y,\mathrm{d}z)$$ and $$\mathrm{d}\tau$$ is the proper time defined by the metric, $$\mathrm{d}s^2=c^2\mathrm{d}\tau^2=g_{\mu\nu}\mathrm{d}X^\mu\mathrm{d}X^\nu$$. Sometimes it's said that you can define it in terms of the regular velocity and the Lorentz factor $$\gamma=\frac{\mathrm{d}t}{\mathrm{d}\tau}$$, which is notably the time component of celerity, because of the chain rule:

$$w^i=\gamma v^i=\frac{\mathrm{d}X^i}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}\tau}.$$

The Lorentz factor is also directly defined in many special relativity texts as

$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}},$$

for the magnitude of velocity $$v$$ and the speed of light $$c$$.

My question is in which direction the Lorentz factor is related to the time component of proper velocity. What I mean is, is $$\gamma$$ always equal to exactly that value above, and all metrics that don't lead to that result are somehow in error, or does the metric define the time component of proper velocity and thus the Lorentz factor?

I think this question is motivated because from general relativity you can prove that that definition of the Lorentz factor is valid at least in Minkowski space:

$$\mathrm{d}s^2=c^2\mathrm{d}\tau^2=c^2\mathrm{d}t^2-\mathrm{d}x^2$$

(in one dimension for simplicity) leads to

$$c^2\mathrm{d}\tau^2=(c^2-v^2)\mathrm{d}t^2,$$

which then gives

$$\mathrm{d}\tau^2=\left(1-\frac{v^2}{c^2}\right)\mathrm{d}t^2,$$ $$\mathrm{d}\tau=\mathrm{d}t\sqrt{1-\frac{v^2}{c^2}},$$ $$\frac{\mathrm{d}\tau}{\mathrm{d}t}=\sqrt{1-\frac{v^2}{c^2}},$$ $$\frac{\mathrm{d}t}{\mathrm{d}\tau}=\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.$$

So I guess another way of saying my question is: in that last equality, is it the former two terms that are necessarily always equal by definition (even in curved spacetime and for nontrivial metrics), or the latter two? It's trivial to come up with a metric where the time component of celerity $$\frac{\mathrm{d}t}{\mathrm{d}\tau}$$ and the classic Lorentz factor $$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ are not equal (even the Schwarzschild, Kerr, Reissner-Nordstrom, Kerr-Newman, Kerr-Newman-de Sitter black holes have this property). So in one of those scenarios, is the Lorentz factor still the same as in Minkowski space, or does it change in curved spacetime?

• The term "Celerity (a.k.a. four-velocity)" does not agree with the definition of "celerity" given in en.wikipedia.org/wiki/Proper_velocity . Commented Jul 21 at 20:48
• @robphy That's actually where I took the definition from originally to write this question. Where did I make a mistake? Commented Jul 21 at 21:57
• The 4-velocity is an observer-independent timelike 4-vector. Celerity is an observer-dependent spacelike component of a 4-velocity, akin to $\sin\theta$ of a Euclidean unit-vector. (I don't like the term "proper velocity" because it is observer-dependent, unlike "proper time" and "proper acceleration" which are observer-independent.) In my opinion, focus on 4-velocity. The time-dilation factor is an observer-dependent timelike-component akin to $\cos\theta$ and the spatial velocity is akin to "slope" as $\tan\theta$. Since celerity and proper-velocity are not as useful, I avoid them. Commented Jul 22 at 1:36
• Sorry @robphy, your answer points out my error - I meant proper velocity not four velocity. My bad Commented Jul 22 at 5:38
• Since an inertial observer with 4-velocity $\hat t$ decomposes a 4-velocity vector $\hat\tau$ into a temporal-part $\gamma \hat t$ and a spatial-part $\beta\gamma \hat t_{\perp}$ (called the celerity or proper-velocity), the proper-velocity has no time-component. So, in your question, "the time component of proper velocity" is zero. Commented Jul 22 at 6:41

My question is in which direction the Lorentz factor is related to the time component of proper velocity.

As I mentioned in my comment, the term "celerity" (sometimes called "proper velocity") is not the same as "4-velocity". So, your question doesn't make sense to me.

Hopefully, the following diagram and explanation clarifies some of these terms.

Using the spacetime diagram from my answer to Physical interpretation of relativistic velocity (distance by proper time)

For simplicity, suppose we have an inertial observer along OP and an inertial particle along OV.

• $$\hat t= \hat{OP}=\frac{\vec{OP}}{\sqrt{\vec{OP}\cdot\vec{OP}}}$$ is the 4-velocity of the inertial observer drawing this spacetime diagram, where the dot-product will be more fully defined below.

• $$\hat \tau= \vec{OV}$$ is the 4-velocity of a particle at event O, where $$\tau$$ is the proper-time wristwatch time of the particle

• $$\theta$$ is the rapidity (hyperbolic-angle) between the 4-velocities (which are both future-timelike unit-4-vectors)

• $$\gamma=\hat\tau\cdot\hat t=\cosh\theta=\frac{ADJ}{HYP}$$ is the time-dilation factor, so that $$\gamma \hat t$$ is the time-component of $$\hat\tau$$ according to $$OP$$. [The dot-product uses the spacetime metric: $$\gamma=(\hat\tau\cdot\hat t)=g_{ab}\tau^a t^b$$. That $$\hat t$$ and $$\hat\tau$$ are unit-timelike can be written as $$\hat t\cdot \hat t=1$$ and $$\hat \tau\cdot \hat \tau=1$$.]

• $$\beta\gamma=\sinh\theta=\frac{OPP}{HYP}$$ is the celerity (or proper-velocity https://en.wikipedia.org/wiki/Proper_velocity ) of $$\hat\tau$$ according to $$OP$$, so that $$(\hat\tau-\gamma \hat t)=\beta\gamma \hat t_{\perp}$$ is the space-component of $$\hat\tau$$ according to $$OP$$. That is, $$OP$$ decomposes $$\hat\tau$$ as the sum of a part parallel to $$\hat t$$ and the rest perpendicular to $$\hat t$$: $$\hat\tau= \gamma \hat t + (\beta\gamma)\hat t_{\perp}= (\cosh\theta) \hat t + (\sinh\theta)\hat t_{\perp}.$$

• $$\beta=\frac{v}{c}=\tanh\theta=\frac{OPP}{ADJ}$$ is the dimensionless spatial velocity of $$\tau$$ according to $$OP$$.
$$\displaystyle \beta \hat t_{\perp} =\frac{\beta\gamma \hat t_{\perp}}{\gamma}$$ is a spacelike-vector that is related to the "spatial 3-velocity of $$\hat\tau$$ according to $$OP$$".

• Using hyperbolic trig identities, one can show $$\gamma=\frac{1}{\sqrt{1-\beta^2}}$$.

[More technically, these operations are being done in the vector space at event $$O$$.]

• The trig identities are natural from special relativity; the question is do these also hold in curved spacetime with a nontrivial (non-Minkowski) metric? It would seem not to be the case to me. Commented Jul 22 at 5:37
• At each point in a riemannian space, there is a tangent vector space, which has a vector space structure with a Euclidean metric. Similarly, at each event in a general spacetime, there is a tangent vector space, which has a vector space structure with a Minkowski metric. Commented Jul 22 at 6:35
• That still is difficult to understand. It seems (to me at least) that the relation between $\mathrm{d}\tau$ and $\mathrm{d}t$ as given in flat spacetime (or equivalently locally-flat spacetime) shouldn't hold up in curved spacetime if the metric is to be believed. Commented Jul 22 at 16:45