Hi, you have wandered into a morass of contradictions and ignorance. Welcome to Motor Daddy's universe.
Light at Light Speed
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Hi, you have wandered into a morass of contradictions and ignorance. Welcome to Motor Daddy's universe.
Motor Daddy
Valued Senior Member
The speed of light is 299,792,458 m/s, and my calculations reflect as much. An observer on the train must take into account the train's velocity in any calculations. Surely he understands that he can have a velocity, correct? Relative to what, you ask? Relative to the speed of light in a vacuum. Does it take light more time to reach you if it has to travel further to reach you? Surely you understand that simple concept?
No, I do not understand that simple concept, because it is in not supported by any evidence. Please present your evidence that supports your conclusion that the velocity of the observer or the light source results in a change in the speed of light.
PS. Your incredulity and/or 'common sense' is not evidence.
Motor Daddy
Valued Senior Member
I didn't say the speed of light changes. Where did I state that? I maintain the speed of light is a constant.
You don't understand the concept that it takes more time for light to travel a greater distance?
And yet whenever he measure his speed, he always finds that his speed is zero.
You're making assumptions that don't match reality, MD.
When the map disagrees with the ground, it's silly to insist that the ground is wrong, wouldn't you agree?
Of course. But as far as the train observer can tell, the light from the back traveled the same distance as the light from the front.
Is that so hard to understand?
But we've been over this many times before, MD. It's getting pretty boring.
Here's a deal for you:
We're considering two mathematical worlds: Newton's world and Einstein's world.
You think that Newton's world is a better match for the real world than Einstein's.
I think that Einstein's world is a better match.
If you agree that only actual measurements of the real world (ie experiments) can decide who is right, then I'll show you the numbers in Einstein's world, one small step at a time, so you can point out any problems.
Deal?
"belive"?
Motor Daddy
Valued Senior Member
That's simply false, Pete. The speed of light is a constant. If light has to travel a greater distance it takes more time. The speed of light never changes, but the amount of light travel time does.
If I perform two separate experiments timing light from the rear of the train to the front of the train, one while the train is moving down the tracks, and the other while the train is stationary compared to the tracks, the two times will be different. It's not even debatable.
If I do the same thing with a lamp post, standing 25 meters away from the lamp and time how much time it takes for the light to reach me, and then perform the experiment again, but this time as soon as the light starts to leave the lamp post I run away from the lamp post, it simply takes more time for the light to reach me because the light had to travel a greater distance to get to me, which was more than 25 meters. It's simply irrefutable, and yet you say that's not the real world. Prove it! That's the real world and the basic facts of distance and time.
I can certainly see why the train observer would think the light traveled the same distance, in his limited narrow minded vision, disregarding that he could posses a velocity. His mistake is that he assumes the speed of light is dependent on the train, which it is not. If he would have tested for a train velocity first he wouldn't have made such a stupid mistake. His calculations would be correct, but he didn't do that, did he?
Yup, and you still don't get it, and I still don't have numbers from you. That is a red flag that says you can't make your method work. Prove it, show me the numbers!!!
How about this. We have a deal, but first you prove to me a relativity of simultaneity exists before you start using it in your method of calculations. Show me your numbers of Chapter 9 and prove to me that a relativity of simultaneity actually exists as Einstein claims it does. Prove it! Show me the numbers!
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Special relativity has three equations; one each for time, distance, and mass. The motion of the light on the moving train will result in an energy change for the light; red shift where the light loses energy, or blue shift where it gains energy. There is a conservation of energy. However, this is not reflected in C, but rather the energy conservation is reflected in the t,d,m where t=frequency, d=wavelength, m=mass/energy which we see as the doppler shift.
The ball on the train will gain an additive velocity with the train to maintain energy conservation; kinetic energy. But light remains at C, with energy conservation reflected by changes in the wavelength, frequency and energy value of the photons. This allows energy to be conserved in relative motion while not requiring C to ever change.
An analogy would be instead of the ball moving with linear motion that adds its velocity with the train, we could get the same energy balance for the bal if the ball maintained train velocity, but gained all its energy through angular momentum changes. Relative to light this gives frequency changes which amounts also to energy and wavelength changes since these three are all connected by the math.
The ball on the train will gain an additive velocity with the train to maintain energy conservation; kinetic energy. But light remains at C, with energy conservation reflected by changes in the wavelength, frequency and energy value of the photons. This allows energy to be conserved in relative motion while not requiring C to ever change.
An analogy would be instead of the ball moving with linear motion that adds its velocity with the train, we could get the same energy balance for the bal if the ball maintained train velocity, but gained all its energy through angular momentum changes. Relative to light this gives frequency changes which amounts also to energy and wavelength changes since these three are all connected by the math.
I see the problem is that you do not understand the most basic part of relativity - namely that the speed of light is a constant.
So lets just assume that your idea that the speed of light and the motion of the source is additive. If the person on the train was in a closed car and did not know that he was moving and measured the time for the light to reach front of the car (using your conjecture) then he would divide the distance/the time and find that the apparent velocity is less than c. He would deduce that he is moving because the speed of light is constant (your definition of constant).
The problem with this conjecture is that all experimental evidence shows this to be wrong.
So again I ask you for ANY evidence that shows that the speed of light and the motion of the source or observer is additive.
You must have some compelling reaon to disregard 100 years of experimental evidence that shows that your conjecture is wrong.
Please supply some evidence, ANY evidence that supports what you say.
Motor Daddy
Valued Senior Member
I never said the speed of light and the motion of the source is additive. Where did you get that from? Where does it assume that in my calculations? The speed of light is a constant in a vacuum.
If the train has a .5c velocity going down the tracks, and the train turns on a headlight that's located at the front of the train, 1 second later the light will be 149,896,229 meters in front of the train. The light traveled at c (299,792,458 m/s) for 1 second, and the train traveled at .5c (149,896,229 m/s) for 1 second, so the light is 149,896,229 meters in front of the train after 1 second.
How is that additive?
Show me your numbers of Chapter 9 if you think you are right. Can't do it, can you?
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Lets see if I understand what Motor Daddy is saying.
A = Source
B = Target
C = Distance between A and B
D = Distance light travels from A to B
If A and B are stationary light travels C in a given time.
If both A and B are moving together in the direction of B, then light travels D which has to take more time than C, because it's traveling at the constant speed of light for a greater distance.
I fail to see any problem with that reasoning.
And I agree.
Here is the problem, which is at the heart of MD's misconception - it depends on the reference frame of the observer.
Lets say that a light is in the center of the train car and the train is moving at a high rate of speed. The light flashes and a beam is sent to the front and the back of the car. MD states that since the light and the train are moving then the light will strike the back of the car before the front of the car.
Here is the important (vital) part. For someone sitting outside of the car this will be completely true - because the speed of light is constant, so the beams will move out from the source at a fixed rate and the back of the car will encounter the beam before the front of the car. However, for someone inside the car the light will strike the front of the car and the back of the car at the same time. How is that possible? Because the speed of light is constant for any observer. If the beams were to hit the back of the car before the front of the car then the observer would be detecting difference speeds depending on the direction of travel.
These conceptss are not a mathematical construct. They are not a thought experiment. These results have been shown consistently for over 100 years.
This is only true for an outside observer. For someone on the train the light will be 299,792,458 meters in front of the train after 1 sec. The speed of light is constant so it will be traveling at 299,792,458 m/s away from the train. You dissagree with this because you think that speed of the source affects the speed of the light that is how it is addditive.
Get it. Didn't think so...
So once again show me any evidence that what you think is correct. Come on it should be easy if it is true!
Okay I don't have a problem with that explanation, very good.
However, I would make a correction.
The light will strike the back of the car before the front of the car. This is what happens in reality.
This reality can be perceived erroneously.
This is the case for someone inside a car. He will perceive the light will strike the front of the car and the back of the car at the same time.
But if he is a scientist or a smart man, he will make the necessary corrections.
You really think when we see that something happens with a star is exactly when that it happens? Or you make a correction?
(In reference to post #72 by KilljoyKlown) Except that both the distance C and the time light needs to cover D are functions of the state of motion of the person with the rulers and clock. Hidden in KilljoyKlown's synopsis is an assumption of a model of space and time when the words "stationary" and "moving together" ae used. If A and B are "moving together" how is that different from "stationary" as seen from A's viewpoint?
So let's say A and B are in the same state of uniform motion as described by one observers's clocks and rulers. Then there is a linear relationship between position at time for each observation of A and B.
So if we observe A at two points of space and time: $$(x,t) \quad \textrm{and} (\chi,\tau)$$ we can construct a general rule for all other observations. And likewise for B.
$$ \frac{ x_A - \chi_A }{ t_A - \tau_A} = u = \frac{ x_B - \chi_B }{ t_B - \tau_B} $$
which we can simplify by introducing a constant of the motion-- the position at time t=0.
$$ x_A - u t_A = \chi_A - u \tau_A = p_A$$
$$ x_B - u t_B = \chi_B - u \tau_B = p_B$$
So now we don't need $$\chi$$ and $$\tau$$ anymore. What is left is the fundamental Cartesian description of two straight lines in space and time. So there is a geometric as well as algebraic description of the situation.
$$ x_A - u t_A = p_A$$
$$ x_B - u t_B = p_B$$
In a similar way, we describe a ray of light as moving with a different uniform motion. Here we use the conventional c for a description of its state of motion.
$$ x_R - c t_R = p_R$$
Thus geometry or algebra gives us the two important points of intersection between R and the parallel lines of A and B (if $$u \ne c$$)
$$(x_0, t_0) = \left( \frac{c p_A - u p_R}{c - u} \quad , \quad \frac{p_A - p_R}{c - u} \right)$$
$$(x_1, t_1) = \left( \frac{c p_B - u p_R}{c - u} \quad , \quad \frac{p_B - p_R}{c - u} \right)$$
If we assume $$t_0 < t_1$$ or $$0 < \frac{p_B - p_A}{c - u}$$ then we can fairly say A is the source of the light and B is the target.
What is $$p_B - p_A$$? It is the difference between the positions of A and B at time t=0 (and any other fixed time) as measured by the observer's clocks and rulers. $$C = \left| p_B - p_A \right|$$.
Further, we can say the distance the light travels as measured by the observer's rulers is $$D = \left| x_1 - x_0 \right| = \left| \frac{c p_B - c p_A}{c - u} \right| = \left| \frac{c}{c - u} \left( p_B - p_A \right) \right| $$.
Clearly if u = 0 then D = C, and if 0 < u < c, then D > C.
So what is the same physical situation seen from the perspective of A's clocks and rulers? Here I borrow the "test theory" of space-time which encompasses both Galilean and Lorentzian transforms:
$$x' = X' + \frac{1}{\sqrt{1 - K v^2}} x + \frac{v}{\sqrt{1 - K v^2}} t$$
$$t' = T' + \frac{1}{\sqrt{1 - K v^2}} t + \frac{K v}{\sqrt{1 - K v^2}} x$$
and the velocity addition law of:
$$u' = \frac{u + v}{1 + K u v }$$.
http://www.sciforums.com/showthread.php?p=2039656#post2039656
Clearly, in A's perspective A does not move relative to itself, so u' = 0 which implies v = -u.
Starting with $$ x_A - u t_A = p_A$$ we solve for t in terms of t'
$$t_A = \frac{1}{\sqrt{1 - K u^2}} (t'_A - T') + \frac{K u p_A}{1 - K u^2}$$
And so:
$$x'_A = X' + \frac{1}{\sqrt{1 - K u^2}} (p_A + u t_A) - \frac{u}{\sqrt{1 - K u^2}} t_A \\ = X' + \frac{p_A \sqrt{1 - K u^2} }{1 - K u^2} $$
or $$x'_A + 0 t'_A = X' + \frac{p_A}{\sqrt{1-K u^2}}$$ .
Likewise $$x'_B + 0 t'_B = X' + \frac{p_B}{\sqrt{1-K u^2}}$$ .
Less cancellations happen for the ray of light:
$$x'_R = X' + \frac{p_R \sqrt{1 - K u^2} - (c-u) T'}{1 - K c u} + \frac{c-u}{1 - K c u} t'_R$$.
Thus, the transformations (Galilean or Lorentian) transform uniform motion into uniform motion.
$$ x'_A - u' t'_A = p'_A$$
$$ x'_B - u' t'_B = p'_B$$
$$ x'_R - c' t'_R = p'_R$$
And the constants of this motion are:
$$u' = 0 \quad , \quad c' = \frac{c-u}{1 - K c u} \quad , \quad p'_A = X' + \frac{p_A}{\sqrt{1-K u^2}} \quad , \quad p'_B = X' + \frac{p_B}{\sqrt{1-K u^2}} \quad , \quad p'_R = X' + \frac{p_R \sqrt{1 - K u^2} - (c-u) T'}{1 - K c u}$$
So from A's clocks and rulers, we have (for the same physical situation described above by C and D):
$$C' = \left| p'_B - p'_A \right| = C \frac{1}{\sqrt{1-K u^2}}$$
and
$$D' = \left| \frac{c'}{c' - u'} \left( p'_B - p'_A \right) \right| = C'$$.
Now, if we assume c' = c - u, then K = 0 and we recover the Galilean predictions.
$$c' = c-u \quad , \quad p'_A = X' + p_A \quad , \quad p'_B = X' + p_B \quad , \quad p'_R = X' + p_R - (c-u) T' \quad , \quad C' = C$$
These do not match our universe. Logically, you can work with these assumptions and not reach a contradiction, but you won't be talking about the physics of our universe and will be wrong when you address topics in electromagnetism, etc.
But if we assume c' = c, then $$K = \frac{1}{c^2}$$ and we recover the Lorentzian predictions, which do match our universe.
$$c' = c \quad , \quad p'_A = X' + \frac{p_A}{\sqrt{1- \frac{u^2}{c^2}}} = X' + \gamma (u) p_A \quad , \quad p'_B = X' + \gamma (u) p_B \quad , \quad p'_R = X' - c T' + (1 + \frac{u}{c}) \gamma (u) p_R \quad , \quad C' = \gamma (u) C$$
For Motor Daddy to assert both $$c' = c$$ and that there is no relativity of simultaneity means that Motor Daddy's description of the universe is not even internally consistent and therefore he is wrong.
So let's say A and B are in the same state of uniform motion as described by one observers's clocks and rulers. Then there is a linear relationship between position at time for each observation of A and B.
So if we observe A at two points of space and time: $$(x,t) \quad \textrm{and} (\chi,\tau)$$ we can construct a general rule for all other observations. And likewise for B.
$$ \frac{ x_A - \chi_A }{ t_A - \tau_A} = u = \frac{ x_B - \chi_B }{ t_B - \tau_B} $$
which we can simplify by introducing a constant of the motion-- the position at time t=0.
$$ x_A - u t_A = \chi_A - u \tau_A = p_A$$
$$ x_B - u t_B = \chi_B - u \tau_B = p_B$$
So now we don't need $$\chi$$ and $$\tau$$ anymore. What is left is the fundamental Cartesian description of two straight lines in space and time. So there is a geometric as well as algebraic description of the situation.
$$ x_A - u t_A = p_A$$
$$ x_B - u t_B = p_B$$
In a similar way, we describe a ray of light as moving with a different uniform motion. Here we use the conventional c for a description of its state of motion.
$$ x_R - c t_R = p_R$$
Thus geometry or algebra gives us the two important points of intersection between R and the parallel lines of A and B (if $$u \ne c$$)
$$(x_0, t_0) = \left( \frac{c p_A - u p_R}{c - u} \quad , \quad \frac{p_A - p_R}{c - u} \right)$$
$$(x_1, t_1) = \left( \frac{c p_B - u p_R}{c - u} \quad , \quad \frac{p_B - p_R}{c - u} \right)$$
If we assume $$t_0 < t_1$$ or $$0 < \frac{p_B - p_A}{c - u}$$ then we can fairly say A is the source of the light and B is the target.
What is $$p_B - p_A$$? It is the difference between the positions of A and B at time t=0 (and any other fixed time) as measured by the observer's clocks and rulers. $$C = \left| p_B - p_A \right|$$.
Further, we can say the distance the light travels as measured by the observer's rulers is $$D = \left| x_1 - x_0 \right| = \left| \frac{c p_B - c p_A}{c - u} \right| = \left| \frac{c}{c - u} \left( p_B - p_A \right) \right| $$.
Clearly if u = 0 then D = C, and if 0 < u < c, then D > C.
So what is the same physical situation seen from the perspective of A's clocks and rulers? Here I borrow the "test theory" of space-time which encompasses both Galilean and Lorentzian transforms:
$$x' = X' + \frac{1}{\sqrt{1 - K v^2}} x + \frac{v}{\sqrt{1 - K v^2}} t$$
$$t' = T' + \frac{1}{\sqrt{1 - K v^2}} t + \frac{K v}{\sqrt{1 - K v^2}} x$$
and the velocity addition law of:
$$u' = \frac{u + v}{1 + K u v }$$.
http://www.sciforums.com/showthread.php?p=2039656#post2039656
Clearly, in A's perspective A does not move relative to itself, so u' = 0 which implies v = -u.
Starting with $$ x_A - u t_A = p_A$$ we solve for t in terms of t'
$$t_A = \frac{1}{\sqrt{1 - K u^2}} (t'_A - T') + \frac{K u p_A}{1 - K u^2}$$
And so:
$$x'_A = X' + \frac{1}{\sqrt{1 - K u^2}} (p_A + u t_A) - \frac{u}{\sqrt{1 - K u^2}} t_A \\ = X' + \frac{p_A \sqrt{1 - K u^2} }{1 - K u^2} $$
or $$x'_A + 0 t'_A = X' + \frac{p_A}{\sqrt{1-K u^2}}$$ .
Likewise $$x'_B + 0 t'_B = X' + \frac{p_B}{\sqrt{1-K u^2}}$$ .
Less cancellations happen for the ray of light:
$$x'_R = X' + \frac{p_R \sqrt{1 - K u^2} - (c-u) T'}{1 - K c u} + \frac{c-u}{1 - K c u} t'_R$$.
Thus, the transformations (Galilean or Lorentian) transform uniform motion into uniform motion.
$$ x'_A - u' t'_A = p'_A$$
$$ x'_B - u' t'_B = p'_B$$
$$ x'_R - c' t'_R = p'_R$$
And the constants of this motion are:
$$u' = 0 \quad , \quad c' = \frac{c-u}{1 - K c u} \quad , \quad p'_A = X' + \frac{p_A}{\sqrt{1-K u^2}} \quad , \quad p'_B = X' + \frac{p_B}{\sqrt{1-K u^2}} \quad , \quad p'_R = X' + \frac{p_R \sqrt{1 - K u^2} - (c-u) T'}{1 - K c u}$$
So from A's clocks and rulers, we have (for the same physical situation described above by C and D):
$$C' = \left| p'_B - p'_A \right| = C \frac{1}{\sqrt{1-K u^2}}$$
and
$$D' = \left| \frac{c'}{c' - u'} \left( p'_B - p'_A \right) \right| = C'$$.
Now, if we assume c' = c - u, then K = 0 and we recover the Galilean predictions.
$$c' = c-u \quad , \quad p'_A = X' + p_A \quad , \quad p'_B = X' + p_B \quad , \quad p'_R = X' + p_R - (c-u) T' \quad , \quad C' = C$$
These do not match our universe. Logically, you can work with these assumptions and not reach a contradiction, but you won't be talking about the physics of our universe and will be wrong when you address topics in electromagnetism, etc.
But if we assume c' = c, then $$K = \frac{1}{c^2}$$ and we recover the Lorentzian predictions, which do match our universe.
$$c' = c \quad , \quad p'_A = X' + \frac{p_A}{\sqrt{1- \frac{u^2}{c^2}}} = X' + \gamma (u) p_A \quad , \quad p'_B = X' + \gamma (u) p_B \quad , \quad p'_R = X' - c T' + (1 + \frac{u}{c}) \gamma (u) p_R \quad , \quad C' = \gamma (u) C$$
For Motor Daddy to assert both $$c' = c$$ and that there is no relativity of simultaneity means that Motor Daddy's description of the universe is not even internally consistent and therefore he is wrong.
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Really? So when we are stationary in a room on earth and we shoot a beam of light from the center of the room and it hits both sides of the room at exactly the same time this is not reality, because the earth is moving? So how would a 'smart man' make a correction. What is the speed of earth relative to the universe? We need to know this speed to make the 'correction' you allude to.
The truth is there is no correction to make because as I have said multiple times the speed of light is always measured at c, regardless of the speed of the observer or the source. This is experimental fact.
That has nothing to do the discussion. No one gave any hint of a belief that light traveled from point to point instantaneously.