caledonia Posted April 10, 2014 Posted April 10, 2014 With my telescope I can see a distant clock, keeping 'normal' time. (taking both time dilation and doppler into account), What do I SEE if I move away from the clock at high speed, say 3/4 speed of light ? What do I see if my speed approaches c ? On the other hand, what do I SEE if I move quickly, say 3/4 c, towards the clock ? What do I see if my speed approaches the speed of light ? Four questions . . . Thank you.
xyzt Posted April 10, 2014 Posted April 10, 2014 (edited) With my telescope I can see a distant clock, keeping 'normal' time. (taking both time dilation and doppler into account), What do I SEE if I move away from the clock at high speed, say 3/4 speed of light ? What do I see if my speed approaches c ? On the other hand, what do I SEE if I move quickly, say 3/4 c, towards the clock ? What do I see if my speed approaches the speed of light ? Four questions . . . Thank you. Let's say that in the absence of relative motion you observe the distant clock to tick at the frequency [math]f_s[/math]. When you move towards the distant clock at speed [math]v[/math] you will measure the frequency to increase to: [math]f_o=f_s \sqrt{\frac{1+\beta}{1-\beta}}[/math] where [math]\beta=\frac{v}{c}[/math] When you move away from the distant clock at speed [math]v[/math] you will measure the frequency to decrease to: [math]\displaystyle{f_o=f_s \sqrt{\frac{1-\beta}{1+\beta}}}[/math] Edited April 10, 2014 by xyzt 3
caledonia Posted April 10, 2014 Author Posted April 10, 2014 Thank you. Can you explain these formulae in terms of the basic postulate viz. the constancy of the speed of light ?
xyzt Posted April 10, 2014 Posted April 10, 2014 Thank you. Can you explain these formulae in terms of the basic postulate viz. the constancy of the speed of light ? These are the basic relativistic Doppler formulas. Do you want a derivation from base principles? 1
swansont Posted April 10, 2014 Posted April 10, 2014 Can you explain these formulae in terms of the basic postulate viz. the constancy of the speed of light ? The speed of light will remain the same. What the formulae show is that for motion toward the source (or it toward you) then the frequency of the light increases (a "blue" shift), and for motion away means the frequency decreases (a "red" shift). It does not matter if the source or you is moving. 1
caledonia Posted April 11, 2014 Author Posted April 11, 2014 These are the basic relativistic Doppler formulas. Do you want a derivation from base principles? Er, yes please. But let me explain further my problem. I read that Einstein in his tramcar, heading away from the town clock in Berne, considered what would happen if his speed approached that of light. He deduced that the clock would appear to stop. I can 'understand' this, by assuming that he would be 'riding', i.e. keeping pace with, the wavefront. But if I reverse this situation and try to apply similar reasoning, I find that the clock is then seen to be moving at double speed. Whereas SR says infinite speed. How does the model change when taking into account the basic postulate (for light) of SR ?
xyzt Posted April 11, 2014 Posted April 11, 2014 Let's say that in the absence of relative motion you observe the distant clock to tick at the frequency [math]f_s[/math]. When you move towards the distant clock at speed [math]v[/math] you will measure the frequency to increase to: [math]f_o=f_s \sqrt{\frac{1+\beta}{1-\beta}}[/math] where [math]\beta=\frac{v}{c}[/math] When you move away from the distant clock at speed [math]v[/math] you will measure the frequency to decrease to: [math]\displaystyle{f_o=f_s \sqrt{\frac{1-\beta}{1+\beta}}}[/math] In answer to "caledonia" question(s): -if [math]v->c[/math] then, when you move away from the distant clock at speed [math]v[/math] you will measure the frequency to decrease to: [math]\displaystyle{f_o=f_s \sqrt{\frac{1-\beta}{1+\beta}}}->0[/math] because [math]\beta->1[/math]. In other words, it would appear that the frequency of the clock under observation goes to zero when it "separates" from the observer at a speed approaching the speed of light. On the other hand, when the observer moves towards the distant clock at speed [math]v->c[/math] then the measured frequency goes to infinity: [math]f_o=f_s \sqrt{\frac{1+\beta}{1-\beta}}->\infty[/math] because [math]\beta->1[/math] Makes sense? But if I reverse this situation and try to apply similar reasoning, I find that the clock is then seen to be moving at double speed. Whereas SR says infinite speed. How does the model change when taking into account the basic postulate (for light) of SR ? Not "double" , nor "infinite" speed. The clock and the observer are moving towards each other at [math]v->c[/math]
Delta1212 Posted April 11, 2014 Posted April 11, 2014 In answer to "caledonia" question(s): -if [math]v->c[/math] then, when you move away from the distant clock at speed [math]v[/math] you will measure the frequency to decrease to: [math]\displaystyle{f_o=f_s \sqrt{\frac{1-\beta}{1+\beta}}}->0[/math] because [math]\beta->1[/math]. In other words, it would appear that the frequency of the clock under observation goes to zero when it "separates" from the observer at a speed approaching the speed of light. On the other hand, when the observer moves towards the distant clock at speed [math]v->c[/math] then the measured frequency goes to infinity: [math]f_o=f_s \sqrt{\frac{1+\beta}{1-\beta}}->\infty[/math] because [math]\beta->1[/math] Makes sense? Not "double" , nor "infinite" speed. The clock and the observer are moving towards each other at [math]v->c[/math] By "moving" I think he meant the rate of time passing on the clock, rather than approach speed.
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