The direction of the big bang

Seedling said:

Two cars are traveling towards each other on a road. One car has a speed of 30 MPH along the road, the other car has a speed of 70 MPH along the road. The distance between them is closing at the rate of 100 MPH. Each driver observes the distance between them as closing at that rate. According to SR, does each driver see the same number of meter sticks between them at every point in time? What I'm getting at is, if the distance between the cars in the road frame is 100 miles, and each driver sees that distance closing at the rate of 100 MPH, then all the frames (road, driver a, and driver b) see that distance as 100 miles, and they also see the time elapsing at the same rate, correct?

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Your first question:

Yes, both drivers see the same number of meter sticks between them. A person standing at the burger king DOES NOT. At relative speeds of 100mph the lorentz factor approaches 1. "sqrt(1-(v/c)^2)" = 1 when v <<< c

Say that instead of 30mph and and 70mph they are going .31c and .71c (as measured from the ground). The person at the burger king (in the same frame as the ground) measures them at 0.31c and 0.71c. He calculates their closing speed to be 1.02c, FASTER than the speed of light! thats insane! But what does the driver see? When they measure the other drivers speed he measures some value slightly less than the speed of light.

Say that the burger king measures each of the cars speeds to be .99c. He then measures their closing speed to be 1.98c. Each driver still perceives the other driving to be driving at LESS than the speed of light. This is the reason your box diagram does not make sense. I can measure a closing speed between the light and the sensors from MY point of view and calculated the speed you are moving RELATIVE to me. Inside the cube though you will measure the speed of light to be exactly c REGARDLESS of your direction or speed or the light sources direction or speed.

I know it seems crazy and counter intuitive, but that is simply how it works. No matter where or when you measure the speed of light, even if you are moving, or if the light source is moving, you WILL measure it to be exactly c, and that has been proven by experiment after experiment.

Remember my example about the baseball and the train? Throwing a baseballs 50mph relative to the train, and the train moving 50mph relative to the ground? The entire reason I brought that example up was to illustrate that it SEEMS like the ball is going 100mph relative to the ground, but it is NOT. It is going slightly less than 100mph. The difference between the actual speed and 100mph is going to be imperceivable at those speeds. You will not be able to even measure it the difference will be so small. But if you replace those values with 0.6c you should see the error in your ways.

Seedling said:

Another question:

At what rate does the center point of the cars change along the road as the distance between the cars decreases over time? For instance, there's a distance of 100 miles between the cars at t=0. Their center point is a Burger King that is 50 miles away from each of them. In one hour when the cars crash into each other, the car that was traveling 70 MPH is 20 miles past the Burger King, and the car that was traveling 30 MPH is 20 miles short of the Burger King. So the center point was the Burger King at t=0 seconds, and the final center point was 20 miles away from that point at t=3600 seconds. The center point changed at a rate of 20 MPH. Put that into your pipe and smoke it!!!!

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I don't understand your point. They were both moving at different speeds relative to the burger king. That was the premise you set up.

guy incognito said:

Remember my example about the baseball and the train? Throwing a baseballs 50mph relative to the train, and the train moving 50mph relative to the ground? The entire reason I brought that example up was to illustrate that it SEEMS like the ball is going 100mph relative to the ground, but it is NOT. It is going slightly less than 100mph. The difference between the actual speed and 100mph is going to be imperceivable at those speeds. You will not be able to even measure it the difference will be so small. But if you replace those values with 0.6c you should see the error in your ways.

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So which is it, did you throw the ball at 50 MPH relative to you, or did you throw the ball slightly less than 50 MPH relative to you? Or maybe you're saying that the train is traveling slightly less than 50 MPH and you actually threw the ball at 50 MPH relative to you. Which is it??? Do you see how you bait and switch?

Seedling said:

So which is it, did you throw the ball at 50 MPH relative to you, or did you throw the ball slightly less than 50 MPH relative to you? Or maybe you're saying that the train is traveling slightly less than 50 MPH and you actually threw the ball at 50 MPH relative to you. Which is it??? Do you see how you bait and switch?

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Seedling said:

So which is it, did you throw the ball at 50 MPH relative to you, or did you throw the ball slightly less than 50 MPH relative to you? Or maybe you're saying that the train is traveling slightly less than 50 MPH and you actually threw the ball at 50 MPH relative to you. Which is it??? Do you see how you bait and switch?

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No bait and switch. I'm saying I throw the ball RELATIVE TO ME at 50mph. The train is moving RELATIVE TO YOU at 50 mph. YOU (in YOUR frame of reference on the ground) cannot simply assume that the ball will be traveling at 50mph +50mph = 100mph. Those 50mph measurements were taken from different reference frames (one from the ground [speed of the train] and one from the train [speed of the ball]).

Say for example I am standing on a platform, and a train goes by with the simultaneous lighting strike from chapter 9. The man measures the distance between point A and M and B and M, and concludes the strikes happened simultaneously, therefore he is at absolute rest. Agreed?

(we don't actually agree...this is wrong and i'll point out why shortly..)

So now the man is standing there and sees another train coming, this time at 0.99c. The front of the train has a gun that can fire a bullet at 0.99c relative to the train. The train engineer fires a bullet and he measures it speed ahead of him at 0.99c and concludes his gun is working. What speed does the man measure the bullet?

The train fires a second bullet after passing the man. Does the man ever see the 2nd bullet?

And another question...

The man sees yet a 3rd train, this one is coming from the opposite direction. He measures it 0.99c. Now there is train 2 going away from him in one direction at 0.99c, and train 3 going away from him in the opposite direction at 0.99c. The man is at absolute rest (according to your definitions and "logic" that you laid out previously in this thread). The trains point lasers at each other. Are they able to see the light from the lasers?

guy incognito said:

No bait and switch. I'm saying I throw the ball RELATIVE TO ME at 50mph. The train is moving RELATIVE TO YOU at 50 mph. YOU (in YOUR frame of reference on the ground) cannot simply assume that the ball will be traveling at 50mph +50mph = 100mph. Those 50mph measurements were taken from different reference frames (one from the ground [speed of the train] and one from the train [speed of the ball]).

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You are basically a point on a train that is traveling 50 MPH in the track frame. Your frame overlaps distance in the track frame at every point in time. At 12:00 you are standing on a 1 mile long train at the midpoint. The front of the train is .5 miles from the station. You are 1.0 mile from the station and the rear of the train is 1.5 miles from the station. At 12:01 the front of the train is .25 miles from the station, you are .75 miles from the station, and the rear of the train is 1.25 miles from the station. AT EVERY POINT IN TIME your frame distance is equal to the track frame distance. You do not have separate length in your respective frame, you have identical lengths in each frame and there is no question about it. At EVERY POINT IN TIME, you are in the frame of the tracks. You are also in the preferred frame, as is the train, the tracks, the ice cream man, and big bird!!! They are all in the preferred frame!

Seedling said:

You are basically a point on a train that is traveling 50 MPH in the track frame. Your frame overlaps distance in the track frame at every point in time. At 12:00 you are standing on a 1 mile long train at the midpoint. The front of the train is .5 miles from the station. You are 1.0 mile from the station and the rear of the train is 1.5 miles from the station. At 12:01 the front of the train is .25 miles from the station, you are .75 miles from the station, and the rear of the train is 1.25 miles from the station. AT EVERY POINT IN TIME your frame distance is equal to the track frame distance. You do not have separate length in your respective frame, you have identical lengths in each frame and there is no question about it. At EVERY POINT IN TIME, you are in the frame of the tracks. You are also in the preferred frame, as is the train, the tracks, the ice cream man, and big bird!!! They are all in the preferred frame!

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You just don't get it.

Seedling said:

You are basically a point on a train that is traveling 50 MPH in the track frame.

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Again, this depends on your point of view. YOU see me as a point on a train traveling 50mph. I do not detect any motion at all. I see YOU as a point on the ground that is traveling 50mph.

Seedling said:

Your frame overlaps distance in the track frame at every point in time.

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No it doesn't, no it doesn't, no it doesn't! For the umpteenth time NO IT DOESN'T. My measurement of distance and time is DIFFERENT from yours if I am moving relative to you.

Seedling said:

At 12:00 you are standing on a 1 mile long train at the midpoint. The front of the train is .5 miles from the station. You are 1.0 mile from the station and the rear of the train is 1.5 miles from the station. At 12:01 the front of the train is .25 miles from the station, you are .75 miles from the station, and the rear of the train is 1.25 miles from the station. AT EVERY POINT IN TIME your frame distance is equal to the track frame distance. You do not have separate length in your respective frame, you have identical lengths in each frame and there is no question about it. At EVERY POINT IN TIME, you are in the frame of the tracks. You are also in the preferred frame, as is the train, the tracks, the ice cream man, and big bird!!! They are all in the preferred frame!

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Ok, back to the baseball example again. Remember when I said like 25 times that your notion of classical physics only works at extremely low velocities? by low I mean in relation to c. so even at speeds of 50,000 mph (which is practically 0 when compared to c) you cannot notice the effects of relativity. Your line of thinking APPEARS to be correct at these low speeds because v <<<<<< c, even at speeds of 50,000mph. If you take measurements of baseballs, bullets, trains, and cars everything will APPEAR to add up exactly like you think it will. However as v approaches c the effects become significant.

Let me try to explain this another way. I say gravity exists. I give you the equations to calculate the force of gravity. You seem skeptical so you set up this experiment:

Get a high powered rifle. Set up a target 1 mile away from you. Nothing is between you and the target. You fire the bullet at the target with deadly accuracy hitting the bullseye. Now you place a large mass, lets say your house at the midway point between you and the target, but slightly to the right. You fire again without moving the gun and get a bullseye again.

"AHA!" you exclaim. "That guy_incognito is such a bullshitter! The house didn't affect the bullet with it's so called "gravity" at all!"

The reality of of the situation is that you just could not measure the difference because the gravitational force generated between the bullet and the house was so small compared to the force that the bullet was fired with, that it was literally negligible to you.

However, if you took the moon, and compressed it to fit in your house, and you repeated the expiriment you would find the gravitational force would significantly alter the path of the bullet. To the point that the bullet may actually start orbiting the house like a satellite. Hell it might even be so much gravitational force that the bullet plummets INTO the house.

Your argument is basically this:

I can fire my bullet at the target, and my house does not effect the path of travel because of gravitational force, therefore gravitational force must not exist. So when you place the moon next to my line of fire it also will not effect the path of my bullet.

guy incognito said:

Again, this depends on your point of view. YOU see me as a point on a train traveling 50mph. I do not detect any motion at all. I see YOU as a point on the ground that is traveling 50mph.

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We both detect a 50 MPH closing speed. We each see the same distance close in the same amount of time. Do you agree?

Seedling said:

We both detect a 50 MPH closing speed. We each see the same distance close in the same amount of time. Do you agree?

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Yes, but that is not what you said.

guy incognito said:

Yes, but that is not what you said.

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So you agree that what I say is 50 miles between us at 12:00 is what you say is 50 miles between us at 12:00? Do you also agree that at 12:30 if I say there is 25 miles between us that you also say there is 25 miles between us at 12:30?

Beefbisquit said:

You just don't get it.

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Until he answers the charges that real-world experiments have falsified his particular hypotheses, I don't see any reason to take him seriously.

mindphuk said:

Until he answers the charges that real-world experiments have falsified his particular hypotheses, I don't see any reason to take him seriously.

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Maybe you care to help your friend out in time of need? I have him by the balls!

Closing speed is always agreed on, and in order for that to happen the distance and time must be agreed on, by every observer. There is no hanky panky smoke and mirrors with closing speeds, because the closing speed is always in the preferred frame, just like all of Moma's kiddies!

Seedling said:

Maybe you care to help your friend out in time of need? I have him by the balls!

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Are you interested in demonstrating your theory or grabbing other peoples balls? The goal of debate is to find common ground.

Heisenberg said:

Are you interested in demonstrating your theory or grabbing other peoples balls? The goal of debate is to find common ground.

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It's a figure of speech, you know, like got him backed into a corner. (rolls eyes) Common ground is the preferred frame!

Seedling said:

It's a figure of speech, you know, like got him backed into a corner. (rolls eyes) Common ground is the preferred frame!

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Sorry for expecting you to extract a point from my words. Let me be more clear.

I continued your premise. I don't think you want to actually grab balls, which would be hard to do through a computer screen. If your position hinges on backing someone into a corner instead of it's legitimacy, then there is no common ground to be had. You are a denialist who seeks to confine the debate to the parts you have thought through, and to avoid or discount the parts you haven't. You want to perform the song and dance you have practiced rather than actually counter the other person's moves. You are too busy rolling your eyes to have a look at what you're actually talking about. Anytime the conversation steers outside your comfort zone, you seek to steer it back. You want other's to be confined to the same low-information perspective as you.

Heisenberg said:

Sorry for expecting you to extract a point from my words. Let me be more clear.

If your position hinges on backing someone into a corner instead of it's legitimacy, then there is no common ground to be had.

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This isn't a compromise, it's my way or the highway! You don't seem to understand, I am telling you that there is no other valid choice other than the way I have it laid out in my diagram. That is the geometry of distance and time, as defined. This is not smoke and mirrors using a box of band-aids to keep alive a pet theory, these are the facts of distance and time.

Seedling said:

This isn't a compromise, it's my way or the highway! You don't seem to understand, I am telling you that there is no other valid choice other than the way I have it laid out in my diagram. That is the geometry of distance and time, as defined. This is not smoke and mirrors using a box of band-aids to keep alive a pet theory, these are the facts of distance and time.

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Finding common ground does not need to involve compromise. The point is to find where you agree so that you can identify the place where theories diverge, which is what everyone but you is attempting to do. Each time they do so and attempt to explain their reason for divergence, you double down and talk about how you must be right, but fail to actually demonstrate anything. You are unable to answer questions which should have easy answers. You resist when the conversation leads to real world data, and fall back again on your flawed hypotheticals. Your theory requires us to devalue the scientific method to accept it, but you are unable to explain why it deserves special consideration, making you a denialist or a moron. Since morons don't tend to understand math, I suspect the former. You are not interested in progress or sharing knowledge, you are interested in contradiction and conflict.