I was trying to prove that we observe time dilation in fast moving objects, however I got something like Doppler effect: if an object is going closer to us, shouldn't we see that time in that object goes faster instead of slower? Imagine a spaceship going to Earth. The spaceship contains a clock that sends a light impulse every 1 second. It goes to us with c speed (of course it doesn't sum up with spaceship's speed - it's the same for every observer). If the ship didn't move, we would see the impulses every second (delayed a constant time), but it moves, so each impulse's delay is shorter, so time interval between 2 impulses gets shorter. It's similar to Doppler's effect - instead of pulses each second there could be times when the lightwave has the same phase. Please tell me what's wrong with my reasoning.
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3Time dilation does not mean the change in the time the light takes to reach the observer. It means that even after you've corrected for the travel time of the light clocks on the moving object will be slower. – John Rennie Jan 19 '22 at 12:03
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See my answer here: https://physics.stackexchange.com/a/307628/4993 – WillO Jan 19 '22 at 14:21
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Related : Doppler effect equation derivation. – Frobenius Jan 19 '22 at 19:05
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2The fact that two effects work in opposite directions is not a contradiction. – WillO Jan 20 '22 at 03:20
2 Answers
There is nothing wrong with your reasoning. What you have described is indeed the Doppler effect. You must not confuse the Doppler effect with time dilation, as the two have quite seperate causes.
The Doppler effect causes our view of distant events either to be speeded up if we are moving towards them, or slowed down if we are moving away from them. It is caused by the changing distance between us and the events, which causes light to have to spend more or less time travelling from them to reach us.
In SR, time dilation refers to the fact that the elapsed time between two events that occur in the same place in one frame is always less than the elapsed time between them in any moving frame (in which, as a consequence of the motion, they do not occur in the same place). Time dilation arises from the relativity of simultaneity, which means that planes of constant time in one frame a sloping slices through time in any other frame moving relative to the first.
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It's useful to think of "time dilation" and "Doppler effect" in terms of particular types of triangles on a spacetime diagram. (After all, many textbook problems in special relativity are really trigonometry problems using hyperbolic-trigonometry instead of the more familiar circular-trigonometry.)
Using graphics based on Bondi's from my article https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/ on the Bondi k-calculus (which uses the Doppler effect)...
- In the simplest case, the Doppler effect involves a triangle with two future-timelike sides from the same event, with their tips joined by a light-signal--a TLT-triangle. So, in the diagram below, OCE involves the Doppler effect, where OC is like the "period between transmissions at O and C", CE is a light-signal (OO was a degenerate light-signal) and OE is like the "period between receptions at O and E".
The ratio of these timelike legs $k=OE/OC$ is the Doppler factor.
OEN is also a Doppler effect. In fact, OEN and OCE are similar triangles in Minkowski spacetime geometry.
[Bondi didn't draw the light signals CE and EN at 45-degrees.]
More generally, in $(1+1)$-Minkowski spacetime,
the Doppler factor would involve a trapezoid with parallel lightlike edges,
which cuts two timelike rays from a common event. (For instance, use a light-signal parallel to CE and use the intersections of CE with the observer worldlines instead of the degenerate signal OO.)
The Doppler factor $k=\exp\theta$, where $\theta$ is the rapidity (Minkwoski-angle) between inertial worldlines [drawn at event O].
- The time-dilation effect involves a a Minkowski right-triangle with two timelike sides from the same event, with their tips joined by a spacelike edge which is orthogonal to one of the timelike sides--a TST right-triangle. So, in the diagram below, OPE is a Minkowski right-triangle where OP and PE are orthogonal, with Minkowski-right-angle at P, so that OE is the hypotenuse. According to the observer along OP, events P and E are simultaneous. The OP-observer says that the "elapsed time between O and E [in OP's frame]" is equal to the "elapsed time between O and P [in OP's frame]". Trigonometrically, OP is the projection of the hypotenuse OE onto OP's worldline. In terms of the rapidity $\theta$ between the timelike sides [drawn at O], OP is the adjacent side. So, $OP=OE\cosh\theta$.
The ratio of these timelike-legs $\gamma=OP/OE$ is the time-dilation factor.
Thus, $\gamma=\cosh\theta$.
The ratio of these orthogonal legs $PE/OP$ is the spatial velocity of OE according to OP.
Thus, $v=\tanh\theta$.
Since Bondi didn't draw the light rays CE and EN along 45-degree angles, it's not obvious that $CP=PN$ (that is, P is the mid-point on CN) and $|CP|=|PN|=|PE|$, where CEN is a radar measurement.
From the Bondi k-calculus viewpoint,
since CEN is a radar measurement
where inertial segment CN intersects the light cone of E,
the midpoint of CN (named P) is simultaneous with E, according to the observer along CN.
Thus, PE is Minkowski-orthogonal to CN (and to CP and to PN).
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