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Chris Miller
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Given a perfect sphere with the ability to measure any of its diameters at any time, if traveling at relativistic velocities, could I use length-foreshortening to determine my speed and direction?
If they knew it was a sphere to begin with and account properly for the travel time of light. Frankly, there would be easier ways of measuring the velocity.Chris Miller said:Thanks so much for the explanation. So, if someone on Earth could (somehow) take measurements of my sphere, they could determine my direction and velocity relative to them?
This might be of interest to you:Chris Miller said:This all stems from an obsession I've developed in knowing how the universe would appear to me as I near v=c.
Chris Miller said:Thank you, Orodurin. Of course. It's more a thought experiment than technological suggestion. This all stems from an obsession I've developed in knowing how the universe would appear to me as I near v=c. Like, if I'm approaching a 100 light year distant (by Earth measurement) star at v sufficient to bring it to within a few miles, how does it present? How wide? How bright/hot. How do I experience its gravitational forces? And how fast is the Earth spinning around the sun?
Chris Miller said:Given a perfect sphere with the ability to measure any of its diameters at any time, if traveling at relativistic velocities, could I use length-foreshortening to determine my speed and direction?
There is no such thing as "entering a frame of reference" - everything is always in all frames at all times. The frame is a convention that you as an observer use to assign times and positions to events.Chris Miller said:At that relative velocity, even radio waves from the star would be shifted into gamma ray part of the spectrum."
Very interesting. Is this because of length foreshortening? If so, wouldn't they revert to their original wavelengths after entering my frame of reference?
Frames of reference are not things that objects enter into or exit from. All objects at all times are "in" all frames of reference. Frames of reference should be thought of as different ways of assigning coordinates to events. The light does not change wavelength. It is measured to have different wavelengths depending on exactly how one does the measurement.Chris Miller said:If so, wouldn't they revert to their original wavelengths after entering my frame of reference?
In a year of proper time, you'll still be at rest in your instantaneous tangent inertial rest frame. And you won't be going at c in anyone's inertial frame. Relativistic velocity addition prevents you from ever accelerating to c.Chris Miller said:All I wanted to do was set out at 1g acceleration for until I hit c (a bit < year, my time)
Change reference frame and you change wavelength. But the light didn't change. Only the way you labelled it changed.Chris Miller said:so length foreshortening doesn't impact EM wavelengths?
Chris Miller said:pervcet: If I understand russ_watters correctly, a perfect sphere created in any frame of reference will be perfect in all frames of reference (and imperfect examined from any other)?
As far as you are concerned, you are not time dilated, everything else is moving at close to c with respect to you and it time dilated.Chris Miller said:Great forum! Your answers have really helped clear up some of my thinking.
Janus: Your thoughtful remarks have raised further questions.
I understand that, "As you are passing the Earth [i.e., on my way toward the star], you and the Earth are both seeing light that left that star at the same time." But by my dilated time, wouldn't it be only about 8 minutes old?
You misunderstood. I'm not talking about light being reflected back to the star, I'm asking you to start at the moment you pass the Earth and mentally run time backwards from that moment. You are in essence back-tracking the light and the star to the moment the light you see at the moment you pass the Earth left the star. It's like this. You have two friends that left the same gas station at the same time on their way to you. One is driving at 60 mph and the other at 50. At the moment the first friend arrives, you get a call from the second one saying he is five miles away. How far away is the gas station that they left? The answer (30 miles), is he same whether you run events forward from when they left the gas station or if you run them backwards from the moment one friend arrives and you get the call."The light you saw from the star is now traveling away from you and back to the star at c."
How is the light from the star now traveling (reflected?) back to the star?
It's best not to bring acceleration into a discussion of Relativity until you are really comfortable with it under uniform velocity conditions, as it opens a whole new can of worms. But you can think of it like this. Imagine sitting in a parked car during a snowfall. The snow is falling straight down, so the snow falling on the car is coming from directly over head. Now start driving, The same snow now appears to be coming from somewhere ahead of the car at an angle, the faster you drive the closer this angle becomes to being horizontal. You get the same effect with light. Light that would appear to come from a point 45 degrees from directly in front of you if you are at rest with respect to the source, will come at you from a narrower angle if you are moving towards the source, the greater your speed, the narrower the angle. It is caller the aberration of light."The star will actually visually appear much much smaller to you than it does to the Earth."
So, if I could instantaneously accelerate to 0.999999999875c from Earth toward the sun, the sun would appear smaller than if it were 100 light years away from Earth's vantage? At what v would this hold true when I was only 1 km from the sun? Blows my mind a little.
It is Doppler shift. And while there are Relativistic effects that contribute to the exact degree of the shift, it can be basically explained as due to a decrease in propagation delay as the distance between you and the source decreases. Imagine a peak of an electromagnetic wave emitted by a source 10 light sec away and moving at .5c towards you. The peak will take 10 sec to reach you. The source is emitting a 1 hz wave, so 1 sec later it emits the next peak. By this time its has moved 0.5 light seconds closer to you, and takes 9.5 sec to reach you. Since it was emitted 1 sec after the first peak was, it reaches you 1+9.5 =10.5 secs after the first peak was emitted, this means you will only see a 10.5-10 = 0.5 sec time difference between the arrival of the peaks of the successive waves. The waves arrive at a higher frequency for you than emitted at the source."At that relative velocity, even radio waves from the star would be shifted into gamma ray part of the spectrum."
Very interesting. Is this because of length foreshortening? If so, wouldn't they revert to their original wavelengths after entering my frame of reference?
Mister T said:An object that's measured to be a sphere by Reference Frame A will not be a perfect sphere when measured by any other reference frame that's in motion relative to Reference Frame A.
Constant in all inertial frames of reference using the simultaneity convention that is natural in such frames (Einstein clock synch). If you base your frame of reference on a constantly accelerating rocket, you're no longer using an inertial frame and there is no natural simultaneity convention to use.Chris Miller said:jbriggs444: "groping" is definitely the right word here. Interesting link, though mostly way over my head. The idea that light would not catch up to me from behind or the sides seems to clash with c being a constant in all frames of reference?
The light would hit with a higher frequency. The difference between the snow and light is that as you drive faster and faster, the snow hits your car at a higher and higher speed. Light always travels at c with respect to you, so it show this as an increase in its frequency. you measure light emitted at one frequency from the star as having a higher frequency.Chris Miller said:Janus: Thanks for your clarifications, and explanation of light aberration and EM Doppler effect. I didn't intend to bring acceleration into the mix. Assume I'm passing Earth at ~c. In your snow/car analogy, the faster I drive, the more snow hits my windshield. Wouldn't this be true of light from that encroaching star (and which would be more analogous to my driving straight up into the snow)?
As far as you are concerned, it is the star that is moving towards you, and thus its clock runs slow compared to yours. So even after you account for Doppler shift, it is, according to you, emitting light at a lower frequency than what be measured by the star itself."Of course in this example I didn't account for Relativity. When you do, the answer comes out to 1.732 times the source frequency and not 2 times." So relativistic effects partially counteract the Doppler effect? I'd have thought the opposite.
Thanks again, Janis. Very helpful. I'm still trying to get my head around each seeing the other's clock as running slower. I always thought that @ near c, I'd see Earth's clock run out of time, but now I understand we each see the other's as frozen. How does my v=~c affect my measurement of H?Janus said:The light would hit with a higher frequency. The difference between the snow and light is that as you drive faster and faster, the snow hits your car at a higher and higher speed. Light always travels at c with respect to you, so it show this as an increase in its frequency. you measure light emitted at one frequency from the star as having a higher frequency. As far as you are concerned, it is the star that is moving towards you, and thus its clock runs slow compared to yours. So even after you account for Doppler shift, it is, according to you, emitting light at a lower frequency than what be measured by the star itself.
A universal compass/speedometer is a device that combines the functions of a traditional compass and a speedometer into one device. It can be used to determine direction and measure speed at the same time.
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