- #36
StarThrower
- 220
- 1
David,
There is nothing for Einstein to recant, because the assumption that the speed of light is c in any inertial reference frame, leads to the conclusion that the length of a ruler which is moving in someone's inertial reference frame, has a length in that frame which is less than the ruler's proper length. You really aren't understanding relativity at all. There is a simple derivation of the time dilation formula, and length contraction formula which will prove that the contraction occurs in absence of any force, and is solely due to relative motion alone. Of course these conclusions are contingent on the postulate that the speed of light is c in any inertial reference frame, but they are the conclusions. Here is their derivation.
Consider an experiment designed to measure the speed of light. A photon is going to be fired from a photon gun at a mirror, it will then bounce back, and the time of travel will be measured by a single clock at rest with respect to the photon gun, and the mirror. Let the distance from the photon gun to the mirror be denoted by d, and let the time it takes for the photon to move from the gun back to the gun be denoted by [tex] \Delta t [/tex].
Now, the distance the photon travels in time delta t, is 2d. Let us denote the speed of the photon in this frame be denoted by c. Thus, in this frame the speed of light is given by:
[tex] c = 2d/ \Delta t [/tex]
Now, consider things from another inertial reference frame which is moving at a constant speed v relative to the photon gun system. In this inertial frame the photon travels an isosceles triangular path. Let a clock in this frame measure the time of the event as:
[tex] \Delta t' [/tex]
Now, the altitude of the triangle is d, the base of the triangle is
[tex] v \Delta t' [/tex]
Now, the sides of the triangle can be found from the Pythagorean theorem. Now each side of the triangle has a length which is equal to the speed of light in this frame (which I will denote by c') times half the time of the event in this frame which is:
[tex] \Delta t'/2 [/tex]
Thus, the side length of the triangular path of the photon is:
[tex] (c' \Delta t')/2 [/tex]
And the height of this triangle is d.
And the base is given by:
[tex] v \Delta t')/2 [/tex]
Hence by the Pythagorean theorem the following relationship holds:
[tex] d^2 + [v \Delta t')/2 ]^2 = [c' \Delta t'/2]^2 [/tex]
From which it follows that:
[tex](2d)^2 + [v \Delta t')]^2 = [c' \Delta t']^2 [/tex]
[tex](2d)^2 = [c' \Delta t']^2 - [v \Delta t')]^2 [/tex]
[tex](2d)^2 = \frac{[(c')^2 - v^2]}{(\Delta t')^2}[/tex]
And we already know that 2d = c \Delta t
From which it follows that:
[tex](c \Delta t)^2 = \frac{[(c')^2 - v^2]}{(\Delta t')^2}[/tex]
From which it follows that:
[tex](c \Delta t')^2 = [(c')^2 - v^2](\Delta t)^2[/tex]
From which it follows that:
[tex](\Delta t')^2 = [(c'/c)^2 - v^2/c^2](\Delta t)^2[/tex]
Now, the previous formula is true, regardless of whether or not the special theory is true. The fundamental postulate of SR is that c=c', let us make that assumption at this point in the mathematical analysis. Hence we have:
[tex](\Delta t')^2 = [1 - v^2/c^2](\Delta t)^2[/tex]
Now, take the square root of both sides of the above equation:
[tex]\Delta t' = \sqrt{[1 - v^2/c^2]}\Delta t [/tex]
And so finally we get the time dilation formula:
[tex]\frac{\Delta t'}{\sqrt{[1 - v^2/c^2]}} = \Delta t [/tex]
Half of the work is done. The other half of the work consists of deriving the formula for Length contraction, from the above formula for time dilation. I will let you do the work. The point is, there will be length contraction of rulers, and of distances traveled. There are multiple ways to draw the conclusion, and the only assumption that was made, was that c=c', which is the Postulate of relativity. Thus, length contraction is a consequence of the postulate of relativity, and has absolutely nothing whatsoever to do with applied forces. It is a relativistic effect, that only has to do with relative motion, as a derivation of the formulas fully reveals.
A classical approach would have said that not (c=c'), but the times of the event in the frames was equal.
There is nothing for Einstein to recant, because the assumption that the speed of light is c in any inertial reference frame, leads to the conclusion that the length of a ruler which is moving in someone's inertial reference frame, has a length in that frame which is less than the ruler's proper length. You really aren't understanding relativity at all. There is a simple derivation of the time dilation formula, and length contraction formula which will prove that the contraction occurs in absence of any force, and is solely due to relative motion alone. Of course these conclusions are contingent on the postulate that the speed of light is c in any inertial reference frame, but they are the conclusions. Here is their derivation.
Consider an experiment designed to measure the speed of light. A photon is going to be fired from a photon gun at a mirror, it will then bounce back, and the time of travel will be measured by a single clock at rest with respect to the photon gun, and the mirror. Let the distance from the photon gun to the mirror be denoted by d, and let the time it takes for the photon to move from the gun back to the gun be denoted by [tex] \Delta t [/tex].
Now, the distance the photon travels in time delta t, is 2d. Let us denote the speed of the photon in this frame be denoted by c. Thus, in this frame the speed of light is given by:
[tex] c = 2d/ \Delta t [/tex]
Now, consider things from another inertial reference frame which is moving at a constant speed v relative to the photon gun system. In this inertial frame the photon travels an isosceles triangular path. Let a clock in this frame measure the time of the event as:
[tex] \Delta t' [/tex]
Now, the altitude of the triangle is d, the base of the triangle is
[tex] v \Delta t' [/tex]
Now, the sides of the triangle can be found from the Pythagorean theorem. Now each side of the triangle has a length which is equal to the speed of light in this frame (which I will denote by c') times half the time of the event in this frame which is:
[tex] \Delta t'/2 [/tex]
Thus, the side length of the triangular path of the photon is:
[tex] (c' \Delta t')/2 [/tex]
And the height of this triangle is d.
And the base is given by:
[tex] v \Delta t')/2 [/tex]
Hence by the Pythagorean theorem the following relationship holds:
[tex] d^2 + [v \Delta t')/2 ]^2 = [c' \Delta t'/2]^2 [/tex]
From which it follows that:
[tex](2d)^2 + [v \Delta t')]^2 = [c' \Delta t']^2 [/tex]
[tex](2d)^2 = [c' \Delta t']^2 - [v \Delta t')]^2 [/tex]
[tex](2d)^2 = \frac{[(c')^2 - v^2]}{(\Delta t')^2}[/tex]
And we already know that 2d = c \Delta t
From which it follows that:
[tex](c \Delta t)^2 = \frac{[(c')^2 - v^2]}{(\Delta t')^2}[/tex]
From which it follows that:
[tex](c \Delta t')^2 = [(c')^2 - v^2](\Delta t)^2[/tex]
From which it follows that:
[tex](\Delta t')^2 = [(c'/c)^2 - v^2/c^2](\Delta t)^2[/tex]
Now, the previous formula is true, regardless of whether or not the special theory is true. The fundamental postulate of SR is that c=c', let us make that assumption at this point in the mathematical analysis. Hence we have:
[tex](\Delta t')^2 = [1 - v^2/c^2](\Delta t)^2[/tex]
Now, take the square root of both sides of the above equation:
[tex]\Delta t' = \sqrt{[1 - v^2/c^2]}\Delta t [/tex]
And so finally we get the time dilation formula:
[tex]\frac{\Delta t'}{\sqrt{[1 - v^2/c^2]}} = \Delta t [/tex]
Half of the work is done. The other half of the work consists of deriving the formula for Length contraction, from the above formula for time dilation. I will let you do the work. The point is, there will be length contraction of rulers, and of distances traveled. There are multiple ways to draw the conclusion, and the only assumption that was made, was that c=c', which is the Postulate of relativity. Thus, length contraction is a consequence of the postulate of relativity, and has absolutely nothing whatsoever to do with applied forces. It is a relativistic effect, that only has to do with relative motion, as a derivation of the formulas fully reveals.
A classical approach would have said that not (c=c'), but the times of the event in the frames was equal.
Last edited: