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md65536 last won the day on September 3

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  1. Ah I made a mistake, I get several answers that fit. I get one solution if I assume that Mr. Johnson knows the answer in the end.
  2. I have two possible answers that seem to fit, one for whether Mrs. Albert replies "no" and one for "yes." Is there supposed to be a single solution? I'm assuming 1) "smallest" means smaller than the number of Charles family children, though technically the statements would be true if they were the same, 2) Mr. Johnson only asks questions he can't already know the answer to.
  3. You left out that they measure the distance in their respective frames, which are different.
  4. Sorry to have to confuse you, but the light that makes up an image, and the subject represented by the light, are two different things. Suppose you print a photograph of yourself. The photo will age: discolor, disintegrate, etc. but that doesn't mean that it will show you aging in the image. In the case of beaming light across some great distance, neither would you age in the image (same as with a photo), nor would the light itself change with respect to time (as measured by an observer).
  5. Just to be clear, this is entirely due to delay time of light. Your reflection travels at a speed of c and doesn't age. Time dilation applies to a moving mirror's age, but that doesn't affect the timing (or speed) of the reflection. As for the equations, sqrt((1-v/c)/(1+v/c)) = sqrt((c-v)/(c+v)) is the inverse Doppler factor, which is how many times the source frequency is multiplied to get what you see. (sqrt((c+v)/(c-v)) - 1) I think is the redshift, which is what fraction of the source wavelength is added to get what you see. Maybe someone can correct this or state it better. The source moving away from you has a positive velocity.
  6. The "perhaps" is right, the mirror doesn't have to be moving. If it's stationary relative to you it will only make you appear half your age at one specific age. If you set up a series of mirrors each 1/4 light years further away from you, then once every year, one of the mirrors would show you at half your current age. (1/4 LY away takes half a year for light to make a 2-way trip, so at age 1, light that left you half a year ago returns. At age 2 it takes 1 year return trip to a mirror 1/2 LY away, etc.) If a mirror is moving away from you at any speed, your reflection is actually doubly Doppler shifted. If you have any object moving away from you, and it is emitting light (eg. if it is self-lit or lit by a source in its own inertial frame) then it is red shifted according to the relativistic Doppler shift (even at non-relativistic speeds). Symmetrically, an observer in the object's frame sees you red-shifted by the same amount. As for the reflection, it is equivalent to an observer receiving a red-shifted image of you, then projecting that image back to you, which you see as red-shifted again. So say it's a mirror with a wooden border and a light on it. The wood appears at an intensity and frequency that is sqrt((1-v/c)/(1+v/c)) times its emitted intensity and frequency. Your reflection appears at intensity and frequency of (1-v/c)/(1+v/c) times what a relatively stationary mirror shows. The formulae will differ depending on whose frame you're talking about (and who has +v or -v in your setup), whether you're talking about source or observer, whether you're talking wavelength or frequency, etc (all these cases means is an inverse in the formula).
  7. It you had a mirror that left you at birth and traveled at a speed of c/3 away from you, you would always look half your age in the mirror. Your reflected image would appear to be n/2 light years away when you see it at age n. Pop quiz! This mirror's moving at a relativistic speed. Why am I not using gamma or relativistic composition of velocities? What's the Doppler shift of your image, and why isn't it the relativistic Doppler shift?
  8. Choosing a reference frame doesn't determine what happens. His trip happens in all reference frames. You choose a reference frame to measure/describe what happens, and the measurements from both reference frames implied here are valid. Only relative to various observers. For the traveling twin, the biological processes occurred at the usual rate (1 s/s). The traveling twin *is* aware of the difference in aging. She can calculate the aging of Earth, and she can also see it happening. No. This is just justifying misunderstanding SR.
  9. In case anyone reading is confused about whether the effects are "real", there's no argument here from anyone that they're not. They're real. "It can't be both" (a ruler can't be .5% shorter and 3% shorter) can be understood to mean, "It can't be both in any single frame of reference." Because length and time are relative, it is perfectly reasonable for example for one observer to say "Clock A is slowed but clock B isn't; ruler X is .5% shorter" and another to say "Both clocks A and B are slowed; ruler X is 3% shorter", and another to say "Clock B is slowed but clock A isn't; ruler X is not length-contracted." All of those are real physically measurable quantities made by individual observers, none of them need to be interpreted by another observer. It is not reasonable to say a rocket is length-contracted to 90% in all frames but the other frames measure it differently because their rulers are also length-contracted differently.
  10. Because time dilation is real. And yet, the atomic clock recorded time at one second per second, never slowing down or "being affected" by time dilation in its own frame. This is all consistent with relativity. Human interpretation??? What do you mean by that? The length contraction is just as real as time dilation. What does it have to do with humans? Do you need the humans' frame to describe the proper time measured by the muon? Does the humans' frame have some special status?
  11. In the scientists frame, the process is really slowing down. In the muon's frame, the process is really NOT slowing down, but the height of the atmosphere is really length contracted, meaning it can pass through it in a shorter proper time. The unhelpful part is trying to describe what is "really" happening in terms of only one frame of reference (like the scientists'), because many people think that different frames' measurements include one "real" measurement and just how it "falsely appears" in other frames. And once again, what the muon observes, it measures in its own frame, where the process really does not slow down. The muon doesn't have to transform its measurements to the Earth frame (or any other), to say "the process really slowed down but my clock is also slowed down." It can just say, "My clock did not slow down."
  12. Perfect! You've also answered the bonus question. d=v*t' (or v*tau) measured in the traveler's frame. Half the proper time during the trip means double the speed, to keep the distances the same. Half the proper time also means double gamma. So 2v_a = v_b, and 2/sqrt(1-v_a^2) = 1/sqrt(1-v_b^2), solve and you get 1/sqrt(5), 2/sqrt(5), which is what you got. The red curve shows something I'd hoped might be intuitive.
  13. You could set it up so that in the "lab" frame, the light is emitted from one point along the circumference, passes through the center, and hits a detector on the opposite side, all in a straight line. This is true no matter what frame the light source is in (it would just be pointed differently in different frames, due to "aberration of light"). If the detector was rotating around the circumference, it wouldn't be hit directly opposite the emitter, in the rotating frame. The observer moving around the circumference would see the same beam of light aimed "backward" of center, curve "forward" (in the direction the observer is moving) so that it hits the center, and continue curving so that it hits the circumference a bit "closer" along the circumference rather than directly across from it. All observers would agree on this: if the light is emitted as the observer passes by, the rotating frame turns during the time it takes the light to travel across the diameter, and "directly opposite the observer in the rotating frame" has moved on by the time the light reaches the other side. There's probably a better way to describe it. You can figure out a lot by describing things in terms of events (happening at one time and place, so all observers agree that it happens (like if a particular beam passes through the center, it does so in all frames)), adding extra measuring tools (like a marked disk that turns with the centrifuge), replacing the accelerating observer with an inertial one that shares a momentary inertial frame at an event, etc.
  14. Sounds like "begging the question" fallacy. You're assuming that the maths are based on physical processes, and concluding that the paradox has a physical resolution.
  15. Local would have to be a lot smaller than that, eg. "measured at the center of the centrifuge" might work. If you're moving fast enough for it to not be negligible, and you send light in a path around the edge of the centrifuge, then an observer moving around the centrifuge will measure that light will take more time to make a round trip (from observer back to observer) in the direction the observer is going, and less than usual in the opposite direction. Both path lengths are measured to be the same in either direction. You can confirm the timing in the "outside" inertial frame: by the time the light has made a full circle in the outside frame, the observer has moved on from its original position, and light must make more than a full circle when sent in the same direction the observer is moving, and less in the opposite direction. However, this is not really a valid measure of the speed of light. You could call it the "coordinate speed" of light, and it's been argued on these forums that it's a meaningless measure. I'd say it'd be like making individual measurements in different momentary inertial frames of the revolving observer; you can get similar invalid measures of the speed of light if you switch between inertial frames in SR without properly accounting for the switch.
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