# Faster or Slower?

Discussion in 'Physics & Math' started by Motor Daddy, Apr 25, 2022.

1. ### billvonValued Senior Member

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That was your statement, not mine. "Faster simply means more distance per time. There is NO DIRECTION to the term faster."

Thus if there is no direction to the term "faster," then it cannot be going slower than the belt, since there is no such thing as a negative speed. It can only have a speed of zero or be going faster than the belt relative to the belt.

You are now argung against yourself and calling your own statements BS.

And on THAT I will agree.

3. ### Motor DaddyValued Senior Member

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There is no direction to the term "faster', it is simply a greater distance per time than some other distance per time. There is no direction to the two speeds of 10 ft/sec and 2 ft/sec, they are both speeds, and 10 ft/sec is FASTER than 2 ft/sec.

Say a cop is on the side of a country road. Two cars are approaching him from each direction of travel. The cars pass his position at the same time, going opposite directions.

According to the two cars the distance between them is increasing at the rate of 150 Miles Per Hour. Which car is faster? There is no FASTER or SLOWER when there is only 1 closing speed of 150 Miles Per Hour. There can't be a FASTER and SLOWER car when there is only 1 speed.

BUT, the cop has radar, and his radar showed the red car to be traveling 50 MPH in one direction, and the speed limit is 50 MPH, so the red car is good. The blue car, however, was traveling 100 MPH. The blue car was traveling 50 MPH FASTER than the red car, AND 50 MPH faster than the posted speed limit. He gets the ticket for traveling 100 MPH in a 50 MPH zone.

Point being, ask the two cars which is faster and all they can say is that the distance between them is increasing at the rate of 150 MPH. They CAN NOT claim that because the closing speed is 150 MPH that one car is faster than the other. The ONLY WAY the cars can determine which is faster is by comparing each of their speeds to the road (100 MPH and 50 MPH) and finding out which is faster and by how much!

You have been proven wrong, admit defeat!

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5. ### billvonValued Senior Member

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OK then. So any speed relative to the belt greater than zero is faster than the belt, per your own definition, and there can be no such thing as "slower than the belt" as you claimed.

Again, you are arguing with yourself. You can't even keep your own story straight. I'd suggest you take high school physics; such a course covers relative speeds.

7. ### Motor DaddyValued Senior Member

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That is like saying any speed greater than zero is faster than the red car. That is a NONSENSICAL statement. I never claimed "any speed greater than zero is faster than the belt." I claimed just the opposite, that the belt's speed is 10 ft/sec, and the craft's speed is 2 ft/sec. They are BOTH greater than zero, but the craft is 8 ft/sec slower than the belt in the treadmill frame.

"The belt" is equivalent to saying "the blue car" and "the craft" is equivalent to saying "the red car." The closing speed is 150 MPH between the two cars. The only way the speeds of 100 MPH and 50 MPH come up is when measured against the road (the treadmill frame). In the treadmill frame the two speeds can be compared to determine which is faster. In the road frame the speed of the blue car can be compared to the speed of the red car and determine which car is faster. They are traveling OPPOSITE directions with a closing speed of 150 MPH. They are NOT each traveling 75 MPH, nor is 1 car traveling 0 MPH and the other traveling 150 MPH!

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8. ### SsssssssRegistered Senior Member

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There are three objects here the belt and the car and the air. In the belt frame the belt is stationary and the air is moving at the speed they set on the treadmill and the car is moving faster than that so if they say the car is moving faster than the belt they're sloppily saying the car is moving faster than the air when measured in the belt frame where the air is moving at the speed the treadmill's controls show as belt speed. They're probably being sloppy because the full version is a mouthful and a half and they assume apparently incorrectly that their viewers are smart enough to disentangle one bit of shorthand.

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9. ### Motor DaddyValued Senior Member

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So your defense for the blue car doing 100 MPH is to tell the Judge that the blue car wasn't moving, it was the air that was moving 100 MPH compared to the blue car that was stationary. Let me know how that goes over.

Again, the blue car and the red car have a closing speed of 150 MPH. To claim either of those cars is "stationary" and that all of the motion is the other car is a non-starter. It turns out that one car was traveling 50 MPH and the other car was traveling 100 MPH. NEITHER car was traveling 0 MPH, and neither car was traveling 150 MPH.

To claim the belt is stationary is claiming one of the cars to be traveling 0 MPH and the other car traveling 150 MPH. That is proven FALSE by the radar gun!

Until you can find an absolute zero velocity rest frame that has a speed of 0 MPH then you have no ground to stand on claiming that the belt was in the zero velocity frame and all the motion was all the other objects.

I have a preferred rest frame that is the frame considered to be "at rest", but you do NOT!

The best you can do is measure a closing speed between the belt and the craft, just like measuring the closing speed between the red and blue car. You have NO BUSINESS claiming one of the cars was "at rest" and the other had all the motion. You are claiming one of the cars was at rest in the preferred frame, and you can't define that frame. You are claiming one of the cars had a 0 MPH speed and the other had a speed of 150 MPH. That is FALSE! To make matters worse, you claim the other car can switch the speeds, so that it can claim to be zero and the OTHER car have the 150 MPH speed. You can't even decide which is zero and which is 150! Here's a clue: NEITHER car was zero and neither car was 150!

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10. ### billvonValued Senior Member

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At this point you are arguing with yourself. Let us know who wins!

11. ### Motor DaddyValued Senior Member

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If you had the intestinal fortitude to answer my questions honestly then you would be forced to admit you are wrong, like you claimed you would be happy to do.

But oh well, I never expected you to have the balls to have an honest conversation, and I certainly didn't believe you when you claimed you would be happy to be proven wrong. Just a bunch of HOT AIR spewing from between your lips.

...and hey, let me know how the defense of the blue car goes in court when you tell the judge the car was stationary and the air was the thing moving 100 MPH. He'd probably order a mental evaluation for you.

Maybe the old switcharoo claiming the blue car was at rest and it was the red car doing 150 MPH??? It's worth a shot, right? LOL

12. ### SsssssssRegistered Senior Member

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Ah I see you're over four hundred and fifty years old and haven't updated your physics training since your childhood. Well grandad there was this young whippersnapper called Galileo who figured out that there's no absolute rest state and you can pick any state of motion and define it to be the state of rest and these young tyros on that newfangled YouTube whachamacallit are using that to simulate a cart moving across the ground in a wind by having still air and a moving ground because it's cheaper than building one o' them wind tunnel doodads.
Radar guns measure a target's speed relative to themselves so you will get different speed readings for a car from a radar gun on the road and another on a car so all you've proven FALSE which I guess is like proving something false but even more so is that neither car is travelling at 150mph with respect to the road. Duh. Try bolting the radar gun to one of the cars and see what reading you get.

13. ### Motor DaddyValued Senior Member

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I see you haven't updated your understanding of physics since I proved Einstein's second postulate to be false. Here's the proof:

https://imgur.com/fpnoaxg

Go ahead and be like billvon and talk a big talk until I show you the proof, at which point you bail on the conversation because you aren't interested in hearing the truth, only defending Einstein's BS!

I could work up the entire scenario in the preferred frame and show you, but that would be a WASTE OF TIME, since you would just bail and claim I am having a conversation with myself.

Here's some equations of motion for you, in case you're a little rusty on distance, time, acceleration...

Maybe you can find a mistake, either mathematically or conceptually??? I DOUBT IT!

https://imgur.com/QbSotVm

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14. ### SsssssssRegistered Senior Member

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Ah ha ha ha ha ha haaaaa! Thanks mate. I needed that.
Oh wait you're serious?
That's a hot mess where you can't even be bothered to state a conclusion so it'd get a failing grade in a high school science class. I reckon you're trying to point out that in the frame where the box is moving the light arrives at the x and z receivers at different times and you think that's a problem which means you forgot the relativity of simultaneity so you should probably look that up. Oh and I reckon you forgot about length contraction too unless you meant that the cube isn't a cube in its rest frame but only in the frame where it's going at 0.638971c.

15. ### Motor DaddyValued Senior Member

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The time of light travel in the box from the center to the receivers is measured in the box, which has clocks reading all the same time. There is 1 meter stick in the box which measures the distance from center to each receiver, and the distance is THE SAME from center to z and center to x. So the distance is the same in the box, and ALL clocks tick as one in the box. So THE SPEED OF LIGHT is different from center to z and center to x. There goes Einstein's BS about the speed of light being the same in a frame of reference.

...and, the box is in motion in space, which is the preferred frame. The numbers add up. When you can show otherwise then we can talk, but flapping your lips blowing hot air isn't going to save Einstein's second postulate. It's gonna take you showing a mistake in MD's Box to save his theory!
You can't do that because MD's Box is correct!

16. ### SsssssssRegistered Senior Member

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The same time in which frame?
The same distance in which frame?
Nah mate. You forgot about length contraction and the relativity of simultaneity so you're not working with Einstein's theory but instead you're using the incomplete misunderstandings that got stuck in your head when you failed to learn it and it's your own broken nonsense that doesn't make sense.

Last edited: May 5, 2022
17. ### Motor DaddyValued Senior Member

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The box is a single frame of reference. It is no different than you sitting in your living room. You are in your living room, you place a light bulb at the center of the room, equal distance to each of the four walls. You time the light from the bulb to the center of each wall.

In order for Einstein's second postulate to be correct the light MUST hit all four walls at THE SAME TIME. But that is shown to be IMPOSSIBLE in MD's Box! IMPOSSIBLE for the light sphere to contact the receivers simultaneously. IMPOSSIBLE!

Since the light sphere CAN'T contact the receivers simultaneously then Einstein's second postulate is proven FALSE! The measured speed of light is different depending on which direction you measure it.

Oh well, just goes to show he was a charlatan! More Fudge Factor at work from good old Einstein, the KING OF FUDGE!

https://en.wikipedia.org/wiki/Fudge_factor

Einstein's name is all over that wiki link! Bwahahahahahahahaaa

18. ### SsssssssRegistered Senior Member

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Nah mate. They only reach the walls at the same time in the box's rest frame because in any other frame the walls are moving so the light travels different distances in different directions and arrives at different times. This isn't a contradiction it's basic stuff that's taught in every introductory relativity class usually with the Einstein's Train thought experiment which is pretty much the thought experiment your diagram would be showing if you hadn't screwed it up. You aren't uncovering any contradictions in relativity here you are just showing us that you can't reproduce standard relativity bookwork accurately.

So far we've established that you don't understand Galileo's relativity which is why you can't understand simple explanations about the prop powered car and you don't understand Einstein's relativity which is why you can't understand how a light in a box works. I'm going to sleep now so if you feel like showing any more of your incompetence I'll try to help you in the morning but what you really need is a high school level general physics textbook and to go through it and (the really important bit) do the exercises.

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19. ### Motor DaddyValued Senior Member

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IN THE BOX the light is sent from the center. The receivers record the time the light hits the receiver. CLEARLY the light took .65 seconds to reach the z receiver, as the radius of the light sphere is 194,865,098 meters. That means light traveled for .65 seconds. 299,792,458 x .65 = 194,865,098, so the radius of the light sphere PROVES that light traveled for .65 seconds at a rate of 299,792,458 meters per second, which is the speed of light. Clearly you can see that light DID NOT reach the x receiver in .65 seconds.

The ONLY way the light sphere can contact all the receivers in .5 seconds is IF the box has an absolute zero velocity in space. But the box has motion in space, so the light sphere can not contact the receivers in .5 seconds.

20. ### SsssssssRegistered Senior Member

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Not in this frame no.
No its that only in the frame where the box is at rest does the light reach the edges of the box at the same time but in any frame where it is moving they don't because "same time" is not frame invariant not even if you write it in all caps.

The correct way to understand this is to use the Lorentz transforms but I'm gonna need the inverse transforms so I'll state those
$\begin{eqnarray*} t&=&\gamma\left(t'+\frac{v}{c^2}x'\right)\\ x&=&\gamma\left(x'+vt'\right)\\ y&=&y'\\ z&=&z' \end{eqnarray*}$
And $\gamma=\left(1-\frac{v^2}{c^2}\right)^{-1/2}$. Now I can work in the primed frame where the box is at rest and if I choose the origin to be the middle of the box when the light is turned on then it's easy to write the coordinates of the receiver events so your Z receiver receives light at $x'=0,\ z'=0.5,\ t'=0.5$ if I measure distance in light seconds and time in seconds and I ignore $y'$ because it's always zero for everything and your X receiver receives light at $x'=0.5,\ z'=0,\ t'=0.5$. You chose $v=0.638971c$ which makes $\gamma=1.3$ and it's easy to feed that through the inverse transforms to get that in the frame where the box is moving your Z receiver receives light at $x=0.4153,\ z=0.5,\ t=0.65$ and your X receiver receives light at $x=1.0653,\ z=0,\ t=1.0653$ and you can see the times aren't the same and relativity has no problem with this it's only you who gets all confused by it. You can easily see that light moved 0.5 light seconds in 0.5 seconds in the box frame and if you calculate $\sqrt{0.4153^2+0.5^2}=0.65$ you can see that light moved 0.65 light seconds in 0.65 seconds and 1.0653 light seconds in 1.0653 seconds in the unprimed frame where the box is moving so it's always travelling at one light second per second in both frames.

If you go and compare my figures to the ones in your picture you'll see that I've got the same answer as you did for when the light reaches your Z receiver but a different answer for when it reaches the X receiver and that's because you forgot about length contraction so you drew a box that's square in a frame of reference where it's contracted in its direction of motion by 1/1.3 so would only be 0.769 light seconds long if it was a cube in its rest frame. If you want to work with a box that's 1 light second long in the frame where it's moving at 0.638971c then you need to make it 1.3 light seconds long in the x' direction and 1 light second in the z' direction in its rest frame so it isn't a cube there and in that case the X receiver receives light at $x'=0.65,\ z'=0,\ t'=0.65$ which inverse transforms to $x=1.1385,\ z=0,\ t=1.1385$ which is what you got.

So yeah relativity has no problem with light arriving simultaneously in some frames and not others. I realise that you've probably missed that because it's only stated in every single work on relativity ever even Einstein's first paper where he says "the observers moving with the moving rod, thus would not find the clocks synchronous, though the observers in the stationary system would declare the clocks to be synchronous" in the last but one paragraph of §2 so you might have missed it if you'd never even looked at a serious source. And I know you have trouble parsing simple sentences because that's what got this conversation started so before you get your knickers in a twist over Einstein talking about the stationary system he explicitly states that it's an arbitrarily chosen system that he's decided to call stationary for convenience at the top of §1.

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21. ### exchemistValued Senior Member

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Nice, clear explanation.

22. ### Motor DaddyValued Senior Member

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Not in ANY frame. The point in time that the light sphere contacts the x receiver is the same point in time in EVERY frame. Just like the Earth completes a revolution at the same point in time in EVERY frame.

What you are proclaiming is that the Earth completes a revolution infinite times because there are infinite frames. That is simply not the case. The Earth completes a revolution at a single point in time for every frame!

What you are claiming is that the light sphere has a different radius in the x direction than it does in the z direction. So you're saying light travels a different distance in different directions in the same rime.

If light travels a different distance in the same time (due to your length contraction) then the speed of light is different in that direction. If light travels the same distance in a different amount of time then the speed of light is different.

In order for the speed of light to be the same in both directions it MUST travel the SAME distance in the SAME time! A sphere is a sphere in EVERY FRAME! What you are proclaiming is that a ball's shape changes depending on the speed you travel past it. That is ABSURD! What you are proclaiming is that 5 apples turn into 2.5 apples if you travel at a velocity of .866c past them. That is ABSURD! 5 apples are 5 apples, regardless of how fast you travel past them, and a ball does NOT change shape just because you travel real fast past it.

Length contraction and time dilation are FUDGE FACTORS in order to keep the speed of light the same in all frames.

23. ### SsssssssRegistered Senior Member

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Well since you were actually talking about something that isn't a cube in its rest frame you actually only get simultaneity in the frame where the box is doing 0.23c to the left but there definitely is such a frame.
I just finished showing you that this is not the case in relativity so if you didn't understand that it would be better if you asked questions about what specific parts you didn't understand instead of repeating your errors.
What you are calling a point in time is actually a three dimensional hypersurface in spacetime and different frames choose different hypersurfaces to call a point in time so you are incorrect.
No I am saying that different frames measure different times for the same process. Once again simply repeating your errors as if they were true isn't going to help you learn.
Once again I just showed you that the light sphere is a sphere in every frame and if you didn't understand that you should ask questions rather than repeatedly state your errors.
Once again I just showed you that light travels at the same speed in every frame and if you didn't understand that you should ask questions rather than repeatedly state your errors.
Length contraction is a logical consequence of the invariance of light speed so it cannot lead to light travelling different distances in the same time.
Once again I just showed you that the speed of light was the same in different frames and if you didn't understand that you should ask questions rather than repeatedly state your errors.
No it must travel some distance in some time such that the ratio of the two is $c$ because travelling one light second in one second is the same speed as travelling two light seconds in two seconds. This should be taught in high school physics so as I said before you desperately need to study a textbook at that level as a prerequisite for understanding things like relativity.
Simply stating that something is absurd does not make it absurd even if you write it in all caps and if you don't understand how this can make sense you should ask specific questions about that rather than repeatedly state your errors.
Relativity makes no such claim since the number of things is an invariant and you will count five apples at any speed.
No they are logical corollaries of assuming the principle of relativity and that there is a finite speed that is invariant under frame change. They are frame dependant quantities so it's arguable whether they can strictly speaking be measured but to the extent that they can be measured time dilation has been measured and length contraction has been indirectly measured via Purcell's interpretation of the electromagnetic field around a wire and effects that are invariant such as the different ages of twins in the twin paradox have been directly measured.

In summary your post is saying that there are quantities that you want to be invariant that aren't and it's not fair and you're going to tell your mommy. The absolute space viewpoint you want to believe was abandoned more than four centuries ago because it was obvious even then that it did not describe reality accurately and your close-minded EVERYTHING IS STUPID EXCEPT WHAT I WANT TO BELIEVE attitude is stopping you from understanding why. If you have substantive questions that show you've actually engaged with some learning then I will answer them but until then I think I've wasted enough time here.