I'm reading about relativity and I thought I understood things clearly about proper time / coordinates until I read the analysis of the Doppler effect.

A and B are 2 fixed points, there is a clock in A and one in B. The clock in A delivers a periodic signal of period $T_A$, the gravitational potential in A is $U_A$ and the signal is received in B.

In the book: $d\tau_A = \sqrt{1-\frac{2 U_A}{c^2}} dt$ so $\Delta t_A = \frac{T_A}{\sqrt{1-\frac{2 U_A}{c^2}}}$, same for B and equating $\Delta t_A = \Delta t_B$ leads to the Doppler frequency shift formula.


I don't understand why $d\tau_A$ differs from $dt$ in the first place as the points are not moving and why $\Delta t_A = \Delta t_B$. It makes me feel that $dt$ is a 'universal' clock. What is a good mental representation of $dt$ vs $d\tau$?

I looked at many questions including this one but it didn't help. Thanks in advance

  • $\begingroup$ The points aren't moving relative to each other? I may be wrong but that looks to me like gravitational Doppler effect in general relativity and not the classical Doppler effect of special relativity due to the motion of the sources relative to each other $\endgroup$ Dec 29, 2018 at 12:12
  • $\begingroup$ @Runlikehell no they are not moving, it is indeed the GR Doppler effect $\endgroup$
    – Thomas
    Dec 29, 2018 at 12:41
  • $\begingroup$ So what's that you don't understand? $\endgroup$ Dec 29, 2018 at 13:22
  • $\begingroup$ @Runlikehell it might sound stupid formulated like this but... what is dt? I would like to say that it is the time in some referential but it doesn't work out $\endgroup$
    – Thomas
    Dec 29, 2018 at 15:07
  • $\begingroup$ you may find your answer here, in the section called "outside a non-rotating sphere" en.m.wikipedia.org/wiki/Gravitational_time_dilation tell me if it's clear after you read that $\endgroup$ Dec 29, 2018 at 15:19

1 Answer 1


What is a good mental representation of $dt$ vs $d\tau$?

I'm not quite sure I understand the question. Here's how I "mentally" interpret $t$.

$t$ is a coordinate. I wouldn't actually picture it as "time flowing", as I wouldn't picture the coordinate $r$ as "distance from the center". I would picture it simply as a coordinate, used to describe your world. So, in a sense, we may say it's common to all the observers. For example, instead of $(t,r,\theta,\phi)$ you may decide to use the Kruskal–Szekeres coordinates $(T,X, \theta, \phi)$, where here the time coordinate is $T=T(t,r)$. So, should we use $t$ or $T$, what is our "real" time, which is the best one? None of them. Depending on what physical situation you are analyzing, a set of coordinates may be more useful than another one, describing the physical context in a clearer way. Coordinates are a tool you are using to describe your world.

Let's focus on the Schwartzchild metric, as it's the most simple one: \begin{equation} ds^2=\Big(1-\frac{2m}{r}\Big)dt^2-\Big(1-\frac{2m}{r}\Big)^{-1}dr^2-r^2d\Omega^2 \end{equation} which is asymptotically flat (i.e. we recover the Minkowski metric far away from the origin). In this case, $dt$ has a physical meaning asymptotically, as it represents the proper time measured by an observer at rest at $r\to\infty$. So we may say that $t$ represents a "real time" only for an observer at rest at infinity.

Let's make an example to clarify the difference between $dt$ and $d\tau$ and how they are related. Let's consider a photon in a circular orbit ($r=3m$) in the plane $\theta=\pi/2$. This is a light-like event, so we have $ds^2=0$, thus computing from the Schwartzschild metric, we get \begin{equation} 0=\frac{1}{3}dt^2-9m^2d\phi^2\implies dt^2=27m^2d\phi^2 \end{equation} and integrating from $0$ to $2\pi$ we obtain $\Delta t=6\pi m\sqrt{3}$.

Now, say we want to know how much time passes for an observer who is at rest at $r=3m$. You can picture it this way: an observer at $r=3m$ "shoots" a photon with a laser and he starts a chronometer, and he stops it as he sees "his" photon again, once the photon has successfully completed the orbit. What we need is the relation \begin{equation} d\tau^2_{r=3m}=\frac{1}{3}dt^2 \implies \Delta\tau_{r=3m}=\frac{1}{\sqrt{3}}\Delta t \end{equation} obtained by setting $dr=d\phi=d\theta=0$ (as the observer is at rest) and $r=3m$. Substituting $\Delta t=6\pi m\sqrt{3}$ in the equation we get $\Delta\tau_{r=3m}=6\pi m$. What would a distant observer at rest say about an event "taking" $\Delta t=6\pi m\sqrt{3}$ coordinate time? Again, from the Schwartzschild metric \begin{equation} d\tau^2_{r=\infty}=dt^2\implies \Delta\tau_{r=\infty}=\Delta t=6\pi m\sqrt{3} \end{equation}

I hope I got the point of your question, at least partially.


Previously I answered to the general question regarding the difference between $dt$ and $d\tau$, without focusing on the specific problem, because it seemed to me that the main source of the problem was the "nature" of these two quantities. Now I'll try to focus on the specific problem you posed, which may seem easy but it's a bit subtle.

Let's ask ourselves what's happening here. We are measuring the period of a light wave, that is to say the time difference between two wave crests, at the point $P_1$. Then the wave travels, and it reaches the point $P_2$, where we measure the period as well. You can see that, even if it seems that we are analyzing the same situation, we are no longer dealing with the same space-time interval between the same events. In the first case we are measuring something specific happening at the point $P_1$, while in the second case we are studying something else (even if it may seem identical) happening at the point $P_2$. But what happens in between? The more specific question should be: if I have a wave the coordinate-time difference between two wave crests stays the same no matter "where/when" I measure it, no matter the point $(t,r,\theta, \phi)$ my observer at rest is in space-time?

The answer is: yes, if your space-time is stationary.

Indeed, let's focus on the following space-time interval: the difference between the departure of a wave-crest $A$ from point $P_1$ and the departure of a wave-crest $B$ from the same point. In terms of coordinates (we're not interested in physical measurements now, we want to see what's happening in our geometrical manifold), this infinitesimal interval will be characterized by $dt_1, dr_1, d\theta_1, d\phi_1$, the last three being zero as we are dealing with two events (departure of crest $A$, departure of crest $B$) happening at the same point $P_1$.

When the crest $A$ reaches the point $P_2$, followed by the crest $B$, it's this interval we are analyzing: the arrival of the crest $A$ in $P_2$, followed by crest $B$ at the same point. This infinitesimal interval will be characterized by $dt_2, dr_2, d\theta_2, d\phi_2$, the last three being zero. How can we be sure that $dt_2=dt_1$? Well, if the coordinate-time $\Delta t_{A,P_1\to P_2}$ taken by the crest $A$ to travel from point $P_1$ to point $P_2$ is equal to $\Delta t_{B,P_1\to P_2}$, coordinate-time taken by the crest $B$ to travel from point $P_1$ to point $P_2$, then obviously $dt_1=dt_2$.

It happens that $\Delta t_{A,P_1\to P_2}=\Delta t_{B,P_1\to P_2}$ if we are dealing with a stationary space-time. To show this it suffices to see that the integral \begin{equation} \Delta t_{P_1 \to P_2}= \int dt \end{equation} can be recast as an integral over the path going from $P_1$ to $P_2$ if we re-express $dt$ using the relation \begin{equation} g_{\mu\nu}\frac{dx^\mu}{d \lambda}\frac{dx^\nu}{d \lambda}=0 \end{equation} as \begin{equation} dt=\frac{1}{g_{00}}\sqrt{-g_{i0}dx^i-[g_{i0}g_{j0}-g_{ij}g_{00}]dx^idx^j} \end{equation} If the spacetime is stationary, there is no dependence on the time-coordinate, but only on the spatial ones. We thus get $\Delta t_{A,P_1\to P_2}=\Delta t_{B,P_1\to P_2}$ and we can say that $dt_1=dt_2=dt$.

I want to stress out that this is really a subtle issue. In general, in GR problem you encounter at the beginning, you always deal with analyzing the same space-time interval, as the photon orbiting, or the infalling particle, etc. In other cases, such as this one, you deal with light waves and you always have a stationary metric, so you can safely set $dt_1=dt_2=dt$.

I didn't want to add confusion but I remember that when I was studying GR for the first time I completely missed it and it puzzled me seeing, in the case of the redshift you just mentioned, sometimes some author say "so let's set $dt_1=dt_2$" because to me it didn't make any sense.

  • $\begingroup$ Thanks! Very useful. So in the Doppler case, $ds^2 = (1- \frac{2 U}{c^2}) dt^2$ so this give $\Delta t$, which is the quantity that we can equate $\endgroup$
    – Thomas
    Dec 29, 2018 at 17:14
  • $\begingroup$ I edited the answer to be more specific on your problem. $\endgroup$
    – Luthien
    Dec 29, 2018 at 21:22
  • $\begingroup$ I would say you could as well skip it for now because I don't want to add confusion. Just stick to the first part if your issue is just the difference between $dt$ and $d\tau$ for what concerns a given space-time interval. If you want to know what happens specifically on your case read the edit: it gives you an insight on what happens in that particular case but, as it is really specific, it may also add a bit of confusion. I tried to explain myself in the clearest way possible $\endgroup$
    – Luthien
    Dec 29, 2018 at 21:25
  • $\begingroup$ Happy I could help! :) $\endgroup$
    – Luthien
    Dec 30, 2018 at 16:28
  • $\begingroup$ Why is the coordinate time not the proper time when r doesn’t tend to infinity in schwarzschild metric. The definition of proper time is the time interval for an an observer at rest. In Minkowski space time $ds^2 = dt^2 - dx^2$, for dx=0, dt= $d\tau$. Similarly in Schwarzschild Metric if we have an observer at rest then dr=0...etc and then proper time should be dt=d$\tau$ ( like in the SR case giving $ds= (1-2M/r) dt= (1-2M/r)d\tau$. Why is tau not equal to t here. $\endgroup$
    – Shashaank
    Dec 23, 2020 at 17:02

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