You may be thinking of the light clock, where in the clock's rest frame the light travels vertically up and down, but in another frame where the clock is moving, it travels on a slanted path that's the hypotenuse of a right triangle whose vertical side is the same height H as in the previous frame, and whose horizontal side has a length equal to the clock's velocity v times the time T it takes for the light to get from the bottom to top in this frame. Since the light must travel at c in all frames, the length of the hypotenuse must be equal to c*T, so by the Pythagorean theorem we have H2 + (vT)2 = (cT)2, from this you can find the time T for the light to get from in this frame, which will be greater than in the clock's rest frame, where it the light just has to travel vertically a distance H so the time is H/c.frankinstein said:Has anyone heard of the speed of light being derived from the Pythagorean Theorem? Obviously I'm referring to using the time dilation effect of motion.
Frank
JesseM said:You may be thinking of the light clock, where in the clock's rest frame the light travels vertically up and down, but in another frame where the clock is moving, it travels on a slanted path that's the hypotenuse of a right triangle whose vertical side is the same height H as in the previous frame, and whose horizontal side has a length equal to the clock's velocity v times the time T it takes for the light to get from the bottom to top in this frame. Since the light must travel at c in all frames, the length of the hypotenuse must be equal to c*T, so by the Pythagorean theorem we have H2 + (vT)2 = (cT)2, from this you can find the time T for the light to get from in this frame, which will be greater than in the clock's rest frame, where it the light just has to travel vertically a distance H so the time is H/c.
The derivation depends on the assumption that light has a speed of c in both frames, which is one of the basic postulates of relativity--for slower-than-light objects like a ping pong ball there's no postulate demanding that the speed be the same in both frames. I don't understand what you mean by the phrase "to reach zero time"...frankinstein said:Yes I was thinking of the light clock, but wouldn't the path of anything in motion be lengthen, its doesn't have to be a photon, it could be a ping pong ball, or a bullet in a vaccum, the path distortion is dependent on the motion of the frame. If that's the case then to reach zero time we should derive the speed of light or something very close?
JesseM said:The derivation depends on the assumption that light has a speed of c in both frames, which is one of the basic postulates of relativity--for slower-than-light objects like a ping pong ball there's no postulate demanding that the speed be the same in both frames. I don't understand what you mean by the phrase "to reach zero time"...
frankinstein said:Has anyone heard of the speed of light being derived from the Pythagorean Theorem? Obviously I'm referring to using the time dilation effect of motion.
Right, that's why the argument concludes that time dilation must exist in relativity, but my point is you need light for the argument to work, not ping pong balls or other slower-than-light objects.frankinstein said:Understood but if the speed is constant in both frames the time dilation remains because of the longer path formed by moving the frame.
It can't really stop the clock because it's impossible to accelerate an object with mass to exactly the speed of light, although you can get it moving at a speed arbitrarily close to c in your frame.frankinstein said:What I mean't by "to reach zero time" is the lengthening of the path that ultimately stops the tick of the clock.
Did you get that just by solving the time dilation equation t = tau/sqrt(1 - v2/c2) for c, or did you use some other argument like the light clock? Just solving for c in the time dilation equation wouldn't require use of the Pythagorean theorem...GrayGhost said:By Pathagorus' theorem ...
c = vt/(t2-tau2)1/2
where vt = x
JesseM said:Right, that's why the argument concludes that time dilation must exist in relativity, but my point is you need light for the argument to work, not ping pong balls or other slower-than-light objects.
JesseM said:Did you get that just by solving the time dilation equation t = tau/sqrt(1 - v2/c2) for c, or did you use some other argument like the light clock? Just solving for c in the time dilation equation wouldn't require use of the Pythagorean theorem...
Ah, gotcha, that was the step I didn't think of, that since the heights are equal you can substitute an expression involving proper time into the equation for the three sides of the triangle in the frame where the clock is moving. My method was just to solve for T in the equation H2 + (vT)2 = (cT)2 and compare it to the proper time of H/c in the clock's rest frame, but yours works equally well.GrayGhost said:Add, per system k, Y=cTau, so y=Y=cTau.
JesseM said:By the way, the fellow's name was spelled "Pythagoras" ;)
GrayGhost said:JesseM,
By Pythagoras' theorem ...
It's just from Einstein's kinematic model ... in a 2-space diagram of system K (ie. x,y,z) with time (t) implied, imagine the system k (ie X,Y,Z, time Tau implied) moving at inertial v along +x with x & X colinear and y & Y always parallel. The origin is marked by the momentary colocation of both system origins at t=Tau=0. Imagine a spherical EM pulse emitted from said origin, and a subsequent consideration of the intersection of said expanding EM sphere with the moving Y-axis in quadrant 1 (ie +x,+y). This interval represents the path of a single photon from origin to the moving Y-axis at some arbitrary time t. This lightpath is defined in system K by Pathagorus' theorem ...
(ct)2 = (vt)2+y2
where x=vt is the location of the system k origin in system K at time t. Since no length contractions exist wrt axes orthogonal to the direction of motion, then y=Y. Add, per system k, Y=cTau, so y=Y=cTau. So ...
(ct)2 = (vt)2+y2
(ct)2 = (vt)2+(cTau)2
then solving for c, this becomes ...
c= vt/(t2-tau2)1/2
The geometry of flat spacetime is the same geometry regardless of what frame of reference you use. In fact, you can even use arbitrary coordinates, and the geometry is still the same. GR defines curvature in a manner that is coordinate-independent.frankinstein said:But the effect still happens without having a light clock. So could a constant c actually be caused by the geometry change or manifold expansion due to motion?
The distortion grows without bound as the velocity approaches c, so there is no maximum. Material objects can't move at c.frankinstein said:When objects are moving much slower to c then the effects to the geometry are negligible but at c we reach the maximum distortion?
bcrowell said:The distortion grows without bound as the velocity approaches c, so there is no maximum. Material objects can't move at c.
frankinstein said:For some reason Latex is not working on my browser so please bare with the awkward notation, thanks.
From Pythagoras:
A frame in motion has the following distortion, irrelevant of relativistic terms:
D = 2*sqrt(L^2 + (1/2*vt)^2)
v = sqrt(D^2 - 2*L^2)/t
So evaluating the limit of sqrt(D^2 - 2*L^2)/t, as D and t approach infinity the function should limit toward c?
If this wrong please explain, and if not how do I find the solution of the limits?
Thx in advance.
Frank
frankinstein said:My real question however is; Has anyone derived c from the limit of the function for distortions caused by motion of a frame without introducing c?
frankinstein said:For some reason Latex is not working on my browser so please bare with the awkward notation, thanks.
Can you explain what the terms are supposed to represent physically? Are you still dealing with a light-clock-like scenario where a signal travels vertically in the clock's rest frame, but along a diagonal in the frame where the clock is moving? If so, is v the velocity of the clock, rather than the velocity of the signal, while D/2 is the length of the diagonal path (just from bottom to top, while D is the length of two diagonals, from bottom to top and back to the bottom) and L is the vertical height? And t is the time for the signal to go from bottom to top and back to the bottom, so t/2 is the time for it to just go from bottom to top? If so that would make sense of D = 2*sqrt(L^2 + (1/2*vt)^2). Then for a single diagonal from bottom to top we'd have the horizontal side as v*(t/2), the vertical side as L, and the diagonal side as D/2. But that doesn't seem equivalent to your second equation, since that gives us:frankinstein said:From Pythagoras:
A frame in motion has the following distortion, irrelevant of relativistic terms:
D = 2*sqrt(L^2 + (1/2*vt)^2)
v = sqrt(D^2 - 2*L^2)/t
JesseM said:that gives us:
v^2*t^2/4 + L^2 = D^2/4
which simplifies to
v^2 = 4/t^2 * (D^2/4 - L^2)
v = sqrt(D^2 - 4*L^2)/t
...which isn't quite what you had.
Anyway, with the modified equation for v above, both equations would be as true in relativity as they are in Newtonian mechanics. But I don't see why you think taking the limit as D approaches infinity should limit towards c, that just seems like a complete non sequitur since D does not approach infinity in relativity, ...
Yes, I would say D=ct...I assume by "never departing it" you mean that the photon always remains directly above the current position of the emitter, even as the emitter moves horizontally with velocity v, right?GrayGhost said:Ahhh, I see.
However, I'm trying to figure out what you mean by the highlight here. I'm supposing that you'll say D = ct, because the photon travels with the emitter never departing it, yes?
Why would it take an infinite amount of time? Perhaps you think that because of the next part...GrayGhost said:So D depends on the duration t considered, and the specific value of invariant c assumed. Yet, if D/2 is defined by the reflection event, then it must take an infinite amount of time t
v is the horizontal velocity of the clock, not the velocity of the signal, so it wouldn't make sense to me to set v=c in relativity.GrayGhost said:and so when v=c
JesseM said:Yes, I would say D=ct...I assume by "never departing it" you mean that the photon always remains directly above the current position of the emitter, even as the emitter moves horizontally with velocity v, right?
JesseM said:Why would it take an infinite amount of time? Perhaps you think that because of the next part...
v is the horizontal velocity of the clock, not the velocity of the signal, so it wouldn't make sense to me to set v=c in relativity.
OK, I thought that by considering the limit as D and t approached infinity, frankinstein wanted to find a value for some quantity that would match what relativity says when you consider the case where the signal's velocity is c, not the clock's. The equations frankinstein wrote (with the small correction I noted) are very general, and would apply equally well to either a light clock or a clock where the signal bouncing between top and bottom had some velocity other than c (and they'd work in either Newtonian physics or relativity). And c does have a role in relativity similar to infinite velocity in Newtonian physics, which was why I was thinking along these lines.GrayGhost said:Indeed, v cannot reach c. Yet, we are imagining what the math requires if v did magically reach c.
Yes, your interpretation does seem to make more sense of frankinstein's other comments, I had mainly been looking at the post where he actually wrote the equations, but elsewhere I see now he did talk about the idea of a frame having a speed of c. In any case I agree with you that there's not going to be any way of deriving c from this, since the signal will just move along with the emitter forever and not rise at all in this limit. And the second part of my original comment about this still applies:GrayGhost said:When frankinstein says an infinite length is required for v=c, he's talking the limit as v approaches c.
you can't derive the precise value of c from any classical limit, that's an empirical matter (after all we could imagine a variety of theories that had the same equations as relativity but different values of c).
Here's were I'm ultimately heading at:JesseM said:Yes, your interpretation does seem to make more sense of frankinstein's other comments, I had mainly been looking at the post where he actually wrote the equations, but elsewhere I see now he did talk about the idea of a frame having a speed of c. In any case I agree with you that there's not going to be any way of deriving c from this, since the signal will just move along with the emitter forever and not rise at all in this limit. And the second part of my original comment about this still applies:
D = 2*sqrt(L^2 + (1/2*vt)^2)frankinstein said:Here's were I'm ultimately heading at:
a: acceleration
v: frame final velocity
v': frame initial velocity
D: frame distortion
t: time
fd: frame's traveled distance
a = (v-v')/t
v = v' + at
a = ((fd + D)/t - (fd' + D')/t')/t
v = v' + ((fd + D)/t - (fd' + D')/t')
frankinstein said:D = 2*sqrt(L^2 + (1/2*vt)^2)
a = ((fd - D)/t - v')/t
fd = vt
a = ((vt-D)/t - v')/t
v = v' + at-D
I listed the equation for distortion as a quick reference and simplified the other equations, also made a sign change to the distortion. And now its pretty obvious the distortion counters the velocity as an object accelerates.
The Pythagorean Theorem can be used to calculate the speed of light by using the formula c = √(a^2 + b^2), where c is the speed of light, a is the distance between the light source and the observer, and b is the distance between the light source and the point where the light is reflected.
The Pythagorean Theorem is used to derive the speed of light because it relates the distance traveled by light to the time it takes to travel that distance. This allows us to calculate the speed of light by measuring the distance and time accurately.
Yes, the Pythagorean Theorem can be used to calculate the speed of light accurately as long as the distance and time measurements are precise. However, there may be other factors that can affect the accuracy of the calculated speed, such as the medium through which the light is traveling.
Yes, there are some limitations to using the Pythagorean Theorem to derive the speed of light. This method assumes that the light is traveling in a straight line and that there are no other factors affecting its speed, such as refraction or interference.
The Pythagorean Theorem does not directly relate to the speed of light in different mediums. However, it can be used to calculate the speed of light in a vacuum, which is known as the speed of light in a vacuum (c). This value can then be compared to the speed of light in different mediums, which is typically slower due to factors such as refraction and absorption.