james blunt
Well-Known Member
Energy takes time to travel from a star to a planet, the stars field is always at the planet.
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Seeing things in their past? You are full of beans!
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Energy takes time to travel from a star to a planet, the stars field is always at the planet.
TrueBeliever37
Well-Known Member
I do acknowledge you are trying to use two reference frames, but one is actually nonexistent.
In what way is the photon traveling at C, if the time and distance are both 0?
Subduction Zone
Veteran Member
You do not seem to understand what an inertial frame of reference is. An observer in an inertial frame of reference is not "traveling". The observer feels that he is stationary. When you are traveling in a car both you and a passenger in that car are at rest relative to the world. It looks like the world is "traveling". Let's scale up a bit. If you are standing on the ground it always feels as if you were not "traveling" Yet someone outside the Solar System could see both the Earth traveling around the Sun and perhaps the Sun traveling in the galaxy. As one pulls further and further back what is a "rest" keeps changing. Zooming back in if one is moving at just below the speed of light in a space ship relative to the Earth one still feels that one is stationary and it is the Earth zooming past, or more worrisome zooming towards you.
james blunt
Well-Known Member
An inertia observer feels stationary but is travelling at ~ 1000 mph and they also are experiencing an acceleration of 9.82 m/s . Feelings are nothing.
TrueBeliever37
Well-Known Member
You replied but didn't answer my question.
when I am in a car I am traveling whether I feel like I am stationary or not.
Subduction Zone
Veteran Member
Actually you aren't. The problem is that you can't come up with a consistent definition of "traveling" that is not refuted by reality at some point.
What is "traveling"?
Subduction Zone
Veteran Member
Relativistically if one feels an acceleration I do believe that one is not in an inertial frame of reference. You might want to check with @Polymath257 for that one.
Both proper time and observer time appear in the formula. The limit is not as time goes to 0, but as v goes to c. When that happens, the proper time (but not the observer time) goes to 0.
james blunt
Well-Known Member
You do feel the acceleration, its called your weight.
james blunt
Well-Known Member
No it doesn't , or there would be no velocity measured. You can't have velocity without t/d.
But different answers won't be obtained from different locations as long as the two observers are at rest with respect to each other again, we are ignoring gravity--where position *is* important because of curvature).
It isn't the different location that gives different observer times, it is relative motion.
Subduction Zone
Veteran Member
And that means that you are not in a relativistic inertial frame of reference. Once again, check with @Polymath257 , I could be wrong on this one.
For every reference frame, there is an elapsed time for the motion of the photon. The *proper time* for the motion of the photon is 0. Choose whether you want the proper time of the clock time in a reference frame.
The 8 minutes, 20 seconds is the *observer time* for the reference frame of the earth. The *proper time* is not the same, in this case, as any observer time.
james blunt
Well-Known Member
The electrical n-field of space is the inertia frame.
A single object in this inertia frame has no velocity. But never mind....this sort of stuff is for proper scientists.
This is correct. If you feel an acceleration, then you are not in a Lorentzian reference frame. Inertial frames, by definition, are those for constant velocity motion--no acceleration.
There is no frame in which the distance and time are 0. However, we can still talk about the proper time and for a photon, that proper time is 0.
BTW, proper time is ONLY defined for events in the light cones of each other. For events that would take 'faster than light' to get between, the proper time is not defined.
if you feel an acceleration, you are NOT in an inertial frame. Now, an acceleration of 9.8m/s^2 is small enough that it doesn't make much of a different locally. But it does if you don't stay local.
Thermos aquaticus
Well-Known Member
Thank you for the clarification! Science is one of those fields that requires very precise language so there can be some confusion at times.
Thermos aquaticus
Well-Known Member
For any observer whose velocity is below the speed of light it will take time for light to travel a distance.
Thermos aquaticus
Well-Known Member
You don't feel velocity. You feel acceleration. That's the point we are trying to make.
If we had a perfectly smooth and straight track and a car with really soft suspension and blacked out windows you wouldn't be able to tell if you were sitting still or travelling at a constant 60 mph. The only forces you would detect is when we accelerate the car to 60 mph or decelerate (i.e. hit the brakes) to slow down from 60 mph. Any frame with constant velocity is the same in physics. All of the physical constants and laws are the same for all frames of reference with constant velocity.