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BiGyElLoWhAt
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I've read a lot of places about length and time dilation or length contraction of an object relative to an observer, but what I haven't found is whether the space that the object occupies is actually being altered. It makes more sense to me that it would only appear as though the length changed.
I suppose my explanation for this phenomena would be as follows:
Imagine an object of 'rest length' L_r and an observer a distance R_b from the base of our object, distance R_c from the center of our object, and distance R_t from the top of our object at time t_observer = t_object = 0. (I'm assuming for simplicity that the times are sychronized in the rest frame).
If we accelerate both our observer and our object at a rate of a=a_0, then (since we base size off of the received light), we can say that our inertial length (sorry if I'm botching these terms, I'm pretty new to these concepts, hence this thread) is going to be the same as it was in the rest frame, assuming that L_r is negligibly small compared to R_c, as our object and observer are moving at the same velocity and can thus be considered at rest with each other. This length should be
...
I have to go now, I'll finish my post when i get home from work. #No save as draft function?
[Edit] Part 2:
Ok, firstly, I want to amend my situation slightly. I want to put my observer on top of our light source, giving him the ability to switch it on and off, and also a way to measure time immensely accurately.
Call ##P(t)_{top}## the position of the top of our mobile object and let ##P(t)_{top} = P(0)_{top} + \dot(P)t## (I'm assuming that the whole object moves as one, with the same velocity, and it is thus unnecessary to label the velocities Ptop and Pbottom)
Call ##P(t)_{bot}## the position of the bottom of our mobile object and let ##P(t)_{bot} = P(0)_{bot} + \dot{P}t##
Call ##X(t)## the position of our mobile observer and let ##X(t) = X_{0} + \dot{X}t##
looking at the case of a stationary object and observer, we can see that the time it takes for a "ray of light" to leave our observer and hit the top of the object is
##\Delta t_{top} = \frac{P(0)_{top} - X(0)}{c}##
and bottom is
##\Delta t_{bot} = \frac{P(0)_{bot} - X(0)}{c}##
(remember our observer is starting and stopping the clocks)
Thus the "rest length" of our object would be ##c \Delta t_{top} -c \Delta t_{bot}##
Now let us give our objects velocities.
If we focus in on the top of our object, we can see that the time it takes light to leave our observer, hit the top of our object, and once again arrive at our observer will be equal to
##\Delta t_{top} = ## ...
OK this ended up being a lot more tedious than I thought.
I have solved for the distance 1 way (light leaves the observer and reaches the object)
##\Delta t_{top_{on\ the\ way\ there}} = \sqrt{\frac{P(0)_{top} - X(0)}{c^2 - \dot{P}^2}}##
It might end up taking me a while to solve for the time it takes to get back and hit the observer, but once I get that solving for the bottom will merely be a matter of substitution.
taking ##P(t)_{top} = \dot{P}(\Delta t_{top_{on\ the\ way\ there}} + \Delta t_{top_{on\ the\ way\ back}})## and subtracting ##P(t)_{bot}## which will be a very similar equation, we will get the illusion of length dilation, (I presume)
I solved this using pythagorean's theorem with the base being equal to ##P_{top_{0}} - X_{0}##, the height being equal to ##\dot{P}\Delta t## and ##\Delta t ## being equal to the hypotenuse (distance traveled by the light to intersect with an object moving at ##\dot{P}## ) divided by c. Solve for ##\Delta t##
The way back won't be as simple because I have to account for the velocity of my observer, which adds another unknown.
I think it is apparent what I am going for, and through this method I will arrive at (at least conceptually, not necessarily mathematically) the same conclusion of length dilation as relativity (I'm not sure if it's GR or SR).
Can I get some input?
Please note that I'm not trying to discredit the relativities, merely understand what they are saying by providing how I'm interpreting the phenomena, and compare and contrast the two.
I suppose my explanation for this phenomena would be as follows:
Imagine an object of 'rest length' L_r and an observer a distance R_b from the base of our object, distance R_c from the center of our object, and distance R_t from the top of our object at time t_observer = t_object = 0. (I'm assuming for simplicity that the times are sychronized in the rest frame).
If we accelerate both our observer and our object at a rate of a=a_0, then (since we base size off of the received light), we can say that our inertial length (sorry if I'm botching these terms, I'm pretty new to these concepts, hence this thread) is going to be the same as it was in the rest frame, assuming that L_r is negligibly small compared to R_c, as our object and observer are moving at the same velocity and can thus be considered at rest with each other. This length should be
...
I have to go now, I'll finish my post when i get home from work. #No save as draft function?
[Edit] Part 2:
Ok, firstly, I want to amend my situation slightly. I want to put my observer on top of our light source, giving him the ability to switch it on and off, and also a way to measure time immensely accurately.
Call ##P(t)_{top}## the position of the top of our mobile object and let ##P(t)_{top} = P(0)_{top} + \dot(P)t## (I'm assuming that the whole object moves as one, with the same velocity, and it is thus unnecessary to label the velocities Ptop and Pbottom)
Call ##P(t)_{bot}## the position of the bottom of our mobile object and let ##P(t)_{bot} = P(0)_{bot} + \dot{P}t##
Call ##X(t)## the position of our mobile observer and let ##X(t) = X_{0} + \dot{X}t##
looking at the case of a stationary object and observer, we can see that the time it takes for a "ray of light" to leave our observer and hit the top of the object is
##\Delta t_{top} = \frac{P(0)_{top} - X(0)}{c}##
and bottom is
##\Delta t_{bot} = \frac{P(0)_{bot} - X(0)}{c}##
(remember our observer is starting and stopping the clocks)
Thus the "rest length" of our object would be ##c \Delta t_{top} -c \Delta t_{bot}##
Now let us give our objects velocities.
If we focus in on the top of our object, we can see that the time it takes light to leave our observer, hit the top of our object, and once again arrive at our observer will be equal to
##\Delta t_{top} = ## ...
OK this ended up being a lot more tedious than I thought.
I have solved for the distance 1 way (light leaves the observer and reaches the object)
##\Delta t_{top_{on\ the\ way\ there}} = \sqrt{\frac{P(0)_{top} - X(0)}{c^2 - \dot{P}^2}}##
It might end up taking me a while to solve for the time it takes to get back and hit the observer, but once I get that solving for the bottom will merely be a matter of substitution.
taking ##P(t)_{top} = \dot{P}(\Delta t_{top_{on\ the\ way\ there}} + \Delta t_{top_{on\ the\ way\ back}})## and subtracting ##P(t)_{bot}## which will be a very similar equation, we will get the illusion of length dilation, (I presume)
I solved this using pythagorean's theorem with the base being equal to ##P_{top_{0}} - X_{0}##, the height being equal to ##\dot{P}\Delta t## and ##\Delta t ## being equal to the hypotenuse (distance traveled by the light to intersect with an object moving at ##\dot{P}## ) divided by c. Solve for ##\Delta t##
The way back won't be as simple because I have to account for the velocity of my observer, which adds another unknown.
I think it is apparent what I am going for, and through this method I will arrive at (at least conceptually, not necessarily mathematically) the same conclusion of length dilation as relativity (I'm not sure if it's GR or SR).
Can I get some input?
Please note that I'm not trying to discredit the relativities, merely understand what they are saying by providing how I'm interpreting the phenomena, and compare and contrast the two.
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