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The relativistic Doppler effect is based on relativity, so yes, it would be perfectly consistent.

You need to differentiate what happens at a given time ##t## from what an observer

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No, you will see it blue shifted with the pulse rate increased. The time dilation is offset by the fact that the pulses emitted later do not need to travel as far as those emitted earlier. Again, time dilation is not what you actually see, it is about what actually goes on at that specific time.

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No. The observer on Earth can correct for the changing travel time due to changing distance. What is left over is time dilation, which is the same in either direction.Wouldn't it appear to the observer on earth as though time had sped up, when it should be slower?

The section on longitudinal Doppler at wikipedia covers it. https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effectCan you reference the math for this?

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Sorry, but I feel that you are really not listening when I say

This is a very common misconception that you must get rid of if you are to have any chance of understanding relativity.You need to differentiate what happens at a given time t from what an observeractually seesat that time.

Compare to the standard Doppler effect for sound waves. When something moves towards you that something emits sound at some frequency. However, because that source is closer to you when emitting later wave crests, it will appear to you as if the frequency is higher - since the time between wave crests is smaller for you. The very same effect is at play in the relativistic Doppler effect. However, you now have the additional complication that the emitter is time dilated and therefore will take longer between the emission of wave crests so you have two effects:

- The effect of the wave crests getting closer together because the source is moving towards you and
- the effect of the wave crests being further apart because they are emitted with longer time in between.

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I am listening but not comprehending. Thanks anyway. I have more questions if you’re willing.

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RPinPA

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I am listening but not comprehending. Thanks anyway. I have more questions if you’re willing.

Let me try.

There are two different effects: One is pure geometry, the same as in analyzing the Doppler effect for sound. When something is moving toward you, each pulse is emitted from a closer location. So it takes less time to get to you. So the time between pulses is less than if the object were standing still.

Some numbers: I emit a pulse at time t = 0 that takes 2 seconds to get to you, so you get it at time t = 2.0

I emit another pulse at time t = 1, but now I'm closer and it only takes 1.5 seconds to get to you. So you get it at time t = 1 + 1.5 = 2.5. I emitted the pulses 1 second apart, but you got them 0.5 seconds apart.

This is the part that depends on whether the object is moving toward you or away from you. As I said, it's pure geometry. And this effect doesn't happen if the object is moving perpendicular to the line between you and it, because the travel time is the same for every pulse.

The relativistic part is that the time between those pulses in the object's frame are not the same as the time between pulses in your frame. That's the time dilation. And that happens even when the object is moving perpendicular to the line of sight. This is the so-called transverse doppler effect.

The full formula for relativistic doppler contains a geometric term that depends on the angle between line of sight and direction of relative motion, and a time dilation term that doesn't.

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kimbyd

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Time dilation and redshift are distinct but closely-related concepts.

There are two effects which contribute to the relativistic Doppler effect:

1) Each pulse is emitted when the moving object is in a different location.

2) Time dilation effects how quickly the clock on the moving object ticks relative to the observer's clock.

You can geometrically estimate both of these effects pretty easily. Consider the situation of an object moving directly towards you at ##\beta = v/c##. Let's say that in a span of time ##\Delta t'##, the emitted wave oscillates once, such that ##\lambda = c \Delta t'##. But in that time, the emitter has gotten ##\Delta d = \beta * \Delta t'## closer to us. Thus, the observed wavelength will be ##\lambda' = (1 - \beta) c \Delta t'##.

You may note that I consistently used ##t'## for the oscillation time. That's because this is the clock as observed by whoever is seeing the incoming signal. That time is effected by time dilation, which means ##\Delta t' = \Delta t / \sqrt{1 - \beta^2}##. Finally, the rest wavelength of the source is ##\lambda = c \Delta t##, so we have:

$$\lambda' = {1 - \beta \over \sqrt{1 - \beta^2}}\lambda$$

You can simplify the above equation by realizing that ##1 - \beta^2 = (1 + \beta)(1 - \beta)##, giving:

$$\lambda' = \lambda\sqrt{1 - \beta \over 1 + \beta}$$

Which is the formula for the relativistic blushift of an object moving towards the observer at ##\beta## (given as a fraction of the speed of light). It also works for objects moving away with ##\beta < 0##. Usually the above formula is derived for objects moving away, so the sign in front of the ##\beta## terms is reversed.

Edit: You can use the above logic to calculate the relativistic Doppler effect for an object moving in any direction. Just follow the logic in the first bit to see how much closer or further away the object gets during one oscillation period, given the angle of motion.

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The relativity of simultaneity. Your clock is moving, so its zeroing needs continual updates because "time zero" in such a construction means different things in different places. Accounting for the changing zeroing and the slow ticking will lead to your clock ticking at the rate you see.So, my question is if you assume you are the ship that is moving and the other one you pass by is the stationary ship, how/why do you still see their clock as running slower due to time dilation if you are the one moving in this perspective?

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kimbyd

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Length contraction is what makes this consistent. It's probably easiest to understand if we take a different situation: muons.

When a high-energy cosmic ray collides with an atomic nucleus in our upper atmosphere, the nucleus explodes in a shower of particles. Nearly all of those particles decay incredibly rapidly, but the muons last long enough to be detected on the ground. The reason why they last long enough is because of time dilation: their half-life is 2.2 microseconds, which isn't enough time to actually reach the ground. But because they are moving so close to the speed of light, they experience enough time dilation that large numbers of them do make it.

From the muon's perspective, clocks on the Earth are slowed way down, but there is also length contraction: the Earth itself looks, to the muon, like a pancake, rather than a sphere. Because from the muon's perspective it doesn't have as far to travel, it makes it to the ground just fine.

Note that to an observer on Earth, the incoming muon will also experience length contraction, but that isn't really observable.

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