# Standard SRT problem need a standard SRT solution.

Discussion in 'Physics & Math' started by Quantum Quack, Mar 26, 2005.

1. ### PeteIt's not rocket surgeryRegistered Senior Member

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I'm saying exactly what I said.
Pulses from B arrive at A three times faster than A's pulses leave.

Or, if you prefer, three pulses from B strike A's reflector in the time it takes A's clock to tick over by one second.

3. ### Quantum QuackLife's a tease...Valued Senior Member

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Is that result a combination of dilation and superluminal effects?

5. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,166
I'm not sure exactly what you're asking, but I'll give an answer to the question I think you might be asking.

If C monitors all pulses, C will receive B's pulses reflected from A three times faster than A's pulses received directly.

You might like to review [post=798202]this post[/post], which includes the workings of the three to one ratio from both A's and C's reference frames.

7. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,166
Following on...

The actual timing of the pulses received by C will vary as ships A and B approach and pass, but if C collects distance data and subtracts the light travel time for each post, she can determine that:
A's pulses are transmitted every 1.155 seconds.
A's pulses arrive at B every 0.385 seconds.
B's pulses are transmitted every 1.155 seconds.
B's pulses arrive at A every 0.385 seconds.

8. ### PeteIt's not rocket surgeryRegistered Senior Member

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It's the result of a correct application of SRT. I don't consider SRT as separate effects, it's a complete model. I'm not even sure what you mean by "superluminal effects".

I can calculate the arrival time of B's pulses either by using the relativistic doppler formula, or by determining the times and positions at which B fires its pulses, and adding the time taken for the pulses to travel from A to B, if that's what you mean.

But regardless, this is the SRT prediction of what A actually observes (note - not what A calculates they observe. This is What A actually observes.)

9. ### Quantum QuackLife's a tease...Valued Senior Member

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18,220
yep I think you are doing what I think you are doing......hmmmmmmm......
from what I see there are two factors involved:
1] the dilated time at the source of the pulses and
2] the increased rate of reciept due to the velocity of the source. [the doppler effect. ]

Ok...let me think on it...some more .......thanks BTW

Can you post the formula ...the relativistic doppler formula?

10. ### PeteIt's not rocket surgeryRegistered Senior Member

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(observed frequency) = (source frequency).&radic;( (1 + v/c)/(1-v/c) )

Where v is positive for an approaching source, and negative for a departing source.

Here it is at Hyperphysics

I think it's more informative to work it from more fundamental principles. Formulas can be dangerous and easy to misapply if you're not sure where they come from.