This isn’t stuff I just memorized. I know why it’s true. I know how to derive it from first principles. In fact, I have.
I’ve seen multiple sources talking about it. I’ve read about the experiments proving it from multiple sources.
Physics is the same in all reference frames. Light travels at a constant speed relative to any reference frame. Everyone knows this. I’ve never seen it disputed in any good source.
My confidence level is well over 99.9%, at least. Being wrong about this means either I have false memories, or consistent fiction is being spread on a massive scale.
Ok if you are interested I will walk through your calculation to prove:
“If you’re moving away from Earth at 87% of the speed of light, time dilation would make it look like time on Earth is passing half as fast. From your point of reference, everyone will live twice as long.”
Hopefully by the end one of us sees a flaw in our reasoning and/or wording.
Imagine you’re moving directly away from the Earth at sqrt(3)/2 c. From your point of reference, it’s moving away from you. Someone on Earth has a set of mirrors with a photon bouncing between them. The mirrors are perpendicular to your path, so there’s no length contraction. (That part can be proven separately, if you wish.) For simplicity, let’s assume the mirrors are a meter apart.
From your point of reference, the x component of the velocity of the photon is sqrt(3)/2c. The total speed is c.
v_total^2 = v_x^2 + v_y^2
c^2 = [sqrt(3)/2 c]^2 + v_y^2
v_y^2 = c^2 − 3⁄4 c^2
v_y^2 = 1⁄4 c^2
v_y = 1⁄2 c
Since the y component of the velocity is half the speed of light, the clock will only run at half speed from your point of reference. It will run at full speed from the point of reference of someone standing right next to it, so from your point of reference, they’re moving at half speed.
Separate thread. Wikipedia often does not do a great job on this type of topic for some reason. Example:
The standard textbook approach treats the twin paradox as a straightforward application of special relativity. Here the Earth and the ship are not in a symmetrical relationship: the ship has a turnaround in which it undergoes non-inertial motion, while the Earth has no such turnaround. Since there is no symmetry, it is not paradoxical if one twin is younger than the other. Nevertheless it is still useful to show that special relativity is self-consistent, and how the calculation is done from the standpoint of the traveling twin.
The argument completely neglect the initial acceleration of the ship and only considers the U-turn acceleration. The symmetry is broken with initial acceleration.
It doesn’t really start until after the initial acceleration.
Imagine that the ship just passed the Earth, rather than taking off from it. Everything would work out the same, but turning around is clearly the only break in symmetry.
No the symmetry breaking event would just be further back in time. Also you could not call it the twin paradox, because you would not be able to explain how one of the twins got into the moving reference frame with out accelerating.
It doesn’t matter how they accelerated before the experiment began, so long as their clocks are synchronized. If they’re in the same place, they can synchronize their clocks.
You seem to missing some key information like why the photon has an x and y velocity rather then just an x or a y.
I suggest that you start with the space time interval, since it is conserved over Lorentz transformation and solve for t1 where t1 is the time rest frame. (t1>t0) where t0 is the start of the experiment and t1 marks the end of the experiment.
It has to have an x and y velocity. It’s just that one of them can be zero. I calculated what they were, and neither was zero.
What do you mean by the start and end of the experiment? The moment the photon leaves one mirror, and the moment it gets to the other?
I’ll let t = t1 - t0, x = x1 - x0, and y = y1 - y0, where (x0, y0, t0) marks the start of the experiment and (x1, y1, t1) marks then end. You can’t give just the time due to the relativity of simultaneity.
The spacetime interval is zero (it always is with light). So that’s sqrt(x^2 + y^2 - t^2) = 0. Setting y to one and solving for x, you get x = sqrt(t^2 − 1). Solving for t = 2, you get x = sqrt(3). This means that, from the point of reference where the light is moving up one meter and horizontally sqrt(3), it takes two meters per c, as opposed to one in the planet’s frame. The photon is moving horizontally at the same rate as the planet. Thus, if the planet moves sqrt(3) meters horizontally between the light moving from one mirror to the other, everything takes twice as long as if it doesn’t move at all.
This isn’t stuff I just memorized. I know why it’s true. I know how to derive it from first principles. In fact, I have.
I’ve seen multiple sources talking about it. I’ve read about the experiments proving it from multiple sources.
Physics is the same in all reference frames. Light travels at a constant speed relative to any reference frame. Everyone knows this. I’ve never seen it disputed in any good source.
My confidence level is well over 99.9%, at least. Being wrong about this means either I have false memories, or consistent fiction is being spread on a massive scale.
Ok if you are interested I will walk through your calculation to prove:
“If you’re moving away from Earth at 87% of the speed of light, time dilation would make it look like time on Earth is passing half as fast. From your point of reference, everyone will live twice as long.”
Hopefully by the end one of us sees a flaw in our reasoning and/or wording.
You may be interested in the twin paradox.
Imagine you’re moving directly away from the Earth at sqrt(3)/2 c. From your point of reference, it’s moving away from you. Someone on Earth has a set of mirrors with a photon bouncing between them. The mirrors are perpendicular to your path, so there’s no length contraction. (That part can be proven separately, if you wish.) For simplicity, let’s assume the mirrors are a meter apart.
From your point of reference, the x component of the velocity of the photon is sqrt(3)/2c. The total speed is c.
v_total^2 = v_x^2 + v_y^2
c^2 = [sqrt(3)/2 c]^2 + v_y^2
v_y^2 = c^2 − 3⁄4 c^2
v_y^2 = 1⁄4 c^2
v_y = 1⁄2 c
Since the y component of the velocity is half the speed of light, the clock will only run at half speed from your point of reference. It will run at full speed from the point of reference of someone standing right next to it, so from your point of reference, they’re moving at half speed.
Separate thread. Wikipedia often does not do a great job on this type of topic for some reason. Example:
The argument completely neglect the initial acceleration of the ship and only considers the U-turn acceleration. The symmetry is broken with initial acceleration.
It doesn’t really start until after the initial acceleration.
Imagine that the ship just passed the Earth, rather than taking off from it. Everything would work out the same, but turning around is clearly the only break in symmetry.
No the symmetry breaking event would just be further back in time. Also you could not call it the twin paradox, because you would not be able to explain how one of the twins got into the moving reference frame with out accelerating.
It doesn’t matter how they accelerated before the experiment began, so long as their clocks are synchronized. If they’re in the same place, they can synchronize their clocks.
You seem to missing some key information like why the photon has an x and y velocity rather then just an x or a y.
I suggest that you start with the space time interval, since it is conserved over Lorentz transformation and solve for t1 where t1 is the time rest frame. (t1>t0) where t0 is the start of the experiment and t1 marks the end of the experiment.
It has to have an x and y velocity. It’s just that one of them can be zero. I calculated what they were, and neither was zero.
What do you mean by the start and end of the experiment? The moment the photon leaves one mirror, and the moment it gets to the other?
I’ll let t = t1 - t0, x = x1 - x0, and y = y1 - y0, where (x0, y0, t0) marks the start of the experiment and (x1, y1, t1) marks then end. You can’t give just the time due to the relativity of simultaneity.
The spacetime interval is zero (it always is with light). So that’s sqrt(x^2 + y^2 - t^2) = 0. Setting y to one and solving for x, you get x = sqrt(t^2 − 1). Solving for t = 2, you get x = sqrt(3). This means that, from the point of reference where the light is moving up one meter and horizontally sqrt(3), it takes two meters per c, as opposed to one in the planet’s frame. The photon is moving horizontally at the same rate as the planet. Thus, if the planet moves sqrt(3) meters horizontally between the light moving from one mirror to the other, everything takes twice as long as if it doesn’t move at all.
You are going to have to set up your problem more rigorously and clear if we are to continue this conversation.
edit: typo
The light is staying between the mirrors, which means it’s moving diagonally, which means it will take longer.
Where in that are you getting lost?