Seeing things in their past? You are full of beans!

[Polymath257]

Sorry, that ship build have been

t^2 - x^2 /c^2

 

[TrueBeliever37]

Use the formula I gave. For light, x=ct.



I'll be back home tonight.

That looks like c = distance/time = 186,000 miles/sec

Are you out on vacation? We just got back from a trip to South Dakota. Saw the Badlands, and the Black Hills, and Mt Rushmore, etc. It's a beautiful state.

Why did you get away from the equation you used before? observer time = proper time/ square root of 1-(v/c)^2
That was the one you used before and said it showed how t=0 for light. I was ready to dispute your use of it, but now you have changed equations on me.

 

[Polymath257]

That looks like c = distance/time = 186,000 miles/sec

Are you out on vacation? We just got back from a trip to South Dakota. Saw the Badlands, and the Black Hills, and Mt Rushmore, etc. It's a beautiful state.

Why did you get away from the equation you used before? observer time = proper time/ square root of 1-(v/c)^2
That was the one you used before and said it showed how t=0 for light. I was ready to dispute your use of it, but now you have changed equations on me.

Yes, on vacation in New Orleans.

The two equations are equivalent. Let x=vt.

 

[TrueBeliever37]

Yes, on vacation in New Orleans.

The two equations are equivalent. Let x=vt.

Well have a great vacation. Definitely don't hurry back on account of these debates.

 

[Polymath257]

Well have a great vacation. Definitely don't hurry back on account of these debates.

Heading back tonight. Will have more details tomorrow. Until then: for two events a distance x apart and a time t apart, the proper time between the events is the square root of
t^2 - x^2/c^2.

Here, x and t are measured in the same frame. But all frames give the same result when this is calculated. In particular, in the rest frame, x=0, so the proper time and coordinate time are the same: the experienced time.

If x=vt, the proper time is t times the square root of 1 -v^2/c^2. This is the other formula you have seen hete.

 

[Polymath257]

[QUOTE="TrueBeliever37, post: 5685301]
Are you out on vacation? We just got back from a trip to South Dakota. Saw the Badlands, and the Black Hills, and Mt Rushmore, etc. It's a beautiful state.[/QUOTE]

I once camped in the badlands and had a bison walk by the tent in the middle if the night. That was exciting!

 

[Subduction Zone]

[QUOTE="TrueBeliever37, post: 5685301]
Are you out on vacation? We just got back from a trip to South Dakota. Saw the Badlands, and the Black Hills, and Mt Rushmore, etc. It's a beautiful state.

I once camped in the badlands and had a bison walk by the tent in the middle if the night. That was exciting![/QUOTE]

That might get my attention. I am not sure if it would have rated as high as when a marmot (basically a mountain woodchuck) cruised right between me and another student during a field geology class. Though it was more exciting for the other guy. We were sitting with our butts on the ground and our feet downhill with a strong bend at the knees. Before it passed between the two of us it scurried from our left under his knees first.

 

[Polymath257]

Polymath,

I always felt like you guys were applying the math wrong in some kind of way. But I just couldn't put my finger on it. When I would look at the equation you were using, I thought well it does appear to let what you call the proper time approach 0 as the velocity approached the speed of light. But I just knew it couldn't be the way you guys were applying it.

So I got to thinking about it more, wondering what does the proper time really represent in that equation. I started wondering why would the speed and distance a photon experiences be effected by the velocity of a ship? Because the photon itself is always traveling at c.

The proper time is NOT affected by the motion of the ship. The *coordinate* time is, however. But all ships would compute the same proper time even though they measure different coordinate times (on their clocks).

Then I realized the proper time approaching 0 in that equation is not what the photon is experiencing. The proper time is what an observer B on the ship would record for the photon from his perspective or relative to himself.

No, that is the coordinate time.

And it makes sense then, because as the ship velocity approaches c, observer B records smaller and smaller times and distances for the photon relative to himself. Once the ship reaches the speed of light, observer B can record no time or distance for the photon relative to himself. But the photon itself is still experiencing 186,000 miles/sec. It's just that so would observer B at that point. Relative to each other there is no time or distance experienced.

Stop a bit here. How does the ship in this case measure the speed of light? Remember, in *all* reference frames (moving or not), the speed of light is determined to be that same 186,282 mps.

Much like observer B with a softball in his lap traveling on a plane moving at 500 mph could record no time or distance for the softball. Although observer A on the earth would record it as moving at 500 mph.

No, the observer on the plane would measure not distance for the motion of the softball on the plane, but *would* measure a time. In the reference frame of the plane, the softball is at rest...distance=0, time not zero, so dist/time=0.

 

[TrueBeliever37]

The proper time is NOT affected by the motion of the ship. The *coordinate* time is, however. But all ships would compute the same proper time even though they measure different coordinate times (on their clocks).



No, that is the coordinate time.



Stop a bit here. How does the ship in this case measure the speed of light? Remember, in *all* reference frames (moving or not), the speed of light is determined to be that same 186,282 mps.



No, the observer on the plane would measure not distance for the motion of the softball on the plane, but *would* measure a time. In the reference frame of the plane, the softball is at rest...distance=0, time not zero, so dist/time=0.

I hear your words but the responses just don't make sense. The proper time was what you told me approached 0 , as the v of the ship approached c. Remember the equation you used was something like: observer time = proper time / square root of 1- (v/c)^2 Now you are saying proper time is not affected by the v. That's the only thing besides time that can change in the equation.

We were discussing limits, you said the limit for the proper time was 0, as v approached c. So how can the velocity not be affecting the proper time?

What is the proper time when the v = 0.5c?

The observer on the plane would know the ball wasn't moving any distance, because it would be sitting on his lap, still relative to him. He could also measure time.

 

[TrueBeliever37]

Polymath, Here is the equation we were discussing. You took proper time approaching 0 as v approached c, and tried to use it to show that for a photon t=0

True or am I mistaken?

Δt = the observer time, or two-position time (s)

Δt0 = the proper time, or one-position time (s)

v = velocity (m/s)

c = speed of light (3.0 x 10^8 m/s)

All you need to do is plug 3.0 x 10^8 m/s into v and solve the equation. You will see that you get a 0 in the denominator. Dividing by zero is what others have referred to as a limit, and at that limit it doesn't make sense to talk about there even being time for something moving at the speed of light. The very same equation is used for length contraction.

 

[TrueBeliever37]

[QUOTE="TrueBeliever37, post: 5685301]
Are you out on vacation? We just got back from a trip to South Dakota. Saw the Badlands, and the Black Hills, and Mt Rushmore, etc. It's a beautiful state.

I once camped in the badlands and had a bison walk by the tent in the middle if the night. That was exciting![/QUOTE]

I bet that did get your attention. Some bison crossed in front of the car in Custer State Park.

 

[Polymath257]

I hear your words but the responses just don't make sense. The proper time was what you told me approached 0 , as the v of the ship approached c. Remember the equation you used was something like: observer time = proper time / square root of 1- (v/c)^2 Now you are saying proper time is not affected by the v. That's the only thing besides time that can change in the equation.

OK, let's get straight what this equation is saying.

In this, there are two observers. Observer A is measuring the 'observer time'. Observer B is moving at speed v with respect to observer A and is the one measuring the proper time (we want the proper time between events on observer B's path).

So, to use this equation, you need to know the relative speed, v, and *either* the time it takes for observer A or the time it takes for observer B. Notice that the distance traveled by B as measured by A is not mentioned, but if you call it x, then x=vt.

We were discussing limits, you said the limit for the proper time was 0, as v approached c. So how can the velocity not be affecting the proper time?

OK, suppose there is *another* observer, C, that is watching B go from the Earth to the sun (or earth to a star--it doesn't matter). Observer C will measure a different value for v (the relative speed) and a different value for the *observer time* for the journey of B. BUT, when C puts *their* values for v and observer time into the same equation, they get the same value of proper time out. So, any two observers watching B will compute the same value of proper time no matter how A and C are moving.

Now, let B travel between the same two points in space, but at a different speed v. The *new* proper time for the trip is calculated with the *new* v and the *new* value(s) for the observer time. In each case, the calculated proper time is the time that B actually experiences as it travels.

Now, do several different trips for B, each with a different speed v and see what happens as v gets closer and closer to c. So, both the measured time for the trip (observer time) and the speed (v) are changing. We calculate the limit for the proper time as v gets closer and closer to c. That limit is 0.

What is the proper time when the v = 0.5c?

You need more information than that. You need to know the observer time (the time measured by A or C) also. Remember that v=.5c only makes sense once you have determined a reference frame.

The observer on the plane would know the ball wasn't moving any distance, because it would be sitting on his lap, still relative to him. He could also measure time.

Exactly. The ball is *at rest* in the frame of the plane. So the proper time uses v=0 and is then the same as the observer time (for an observer in the plane).

 

[TrueBeliever37]

OK, let's get straight what this equation is saying.

In this, there are two observers. Observer A is measuring the 'observer time'. Observer B is moving at speed v with respect to observer A and is the one measuring the proper time (we want the proper time between events on observer B's path).

So, to use this equation, you need to know the relative speed, v, and *either* the time it takes for observer A or the time it takes for observer B. Notice that the distance traveled by B as measured by A is not mentioned, but if you call it x, then x=vt.



OK, suppose there is *another* observer, C, that is watching B go from the Earth to the sun (or earth to a star--it doesn't matter). Observer C will measure a different value for v (the relative speed) and a different value for the *observer time* for the journey of B. BUT, when C puts *their* values for v and observer time into the same equation, they get the same value of proper time out. So, any two observers watching B will compute the same value of proper time no matter how A and C are moving.

Now, let B travel between the same two points in space, but at a different speed v. The *new* proper time for the trip is calculated with the *new* v and the *new* value(s) for the observer time. In each case, the calculated proper time is the time that B actually experiences as it travels.

Now, do several different trips for B, each with a different speed v and see what happens as v gets closer and closer to c. So, both the measured time for the trip (observer time) and the speed (v) are changing. We calculate the limit for the proper time as v gets closer and closer to c. That limit is 0.




You need more information than that. You need to know the observer time (the time measured by A or C) also. Remember that v=.5c only makes sense once you have determined a reference frame.



Exactly. The ball is *at rest* in the frame of the plane. So the proper time uses v=0 and is then the same as the observer time (for an observer in the plane).

Ok, now having said all that, please explain again how you came up with t=0 for a photon.

 

[Polymath257]

Ok, now having said all that, please explain again how you came up with t=0 for a photon.

OK, observer A watches the photon (B) go from the Earth to the sun. There is a non-zero observer time, t. But, we can solve for the proper time in the equation to get

proper time = (observer time)*sqrt( 1 -v^2 /c^2 )

Now, in this, the photon is going at v=c with respect to A. So, the square root is 0. This gets multiplied by the observer time (for an observer on the Earth, this would be 500 seconds). And 500*0=0.

So the proper time for the photon is 0.

Note: observer A can also determine the observer time by looking at the distance traveled by the photon (93 million miles) and dividing by the speed (in this case 186000 mps).

 

[TrueBeliever37]

OK, observer A watches the photon (B) go from the Earth to the sun. There is a non-zero observer time, t. But, we can solve for the proper time in the equation to get

proper time = (observer time)*sqrt( 1 -v^2 /c^2 )

Now, in this, the photon is going at v=c with respect to A. So, the square root is 0. This gets multiplied by the observer time (for an observer on the Earth, this would be 500 seconds). And 500*0=0.

So the proper time for the photon is 0.

Note: observer A can also determine the observer time by looking at the distance traveled by the photon (93 million miles) and dividing by the speed (in this case 186000 mps).

See you guys go back and forth. In your Note, you say the photon traveled 93 million miles. I agree with that.

I could modify your equation to proper time = (observer time)*sqrt( 1 - v^3/ c^3 ) and get your same result , since v=c with respect to A, So, the square root is 0.

Or I could let v,c both be to 4th or 5th or whatever power, and still get proper time = 0

One of you guys said this is not a universal equation and only works for light. So how can we know it is accurate? It could be an equation just developed to give the answers wanted to fit the theory.

 

[Polymath257]

See you guys go back and forth. In your Note, you say the photon traveled 93 million miles. I agree with that.

In the frame of the Earth. In a different frame, it would be a different distance.

I could modify your equation to proper time = (observer time)*sqrt( 1 - v^3/ c^3 ) and get your same result , since v=c with respect to A, So, the square root is 0.

Or I could let v,c both be to 4th or 5th or whatever power, and still get proper time = 0

One of you guys said this is not a universal equation and only works for light. So how can we know it is accurate? It could be an equation just developed to give the answers wanted to fit the theory.

The formula I gave works for any speed v and has been tested. We apply it to light with v=c.

 

[TrueBeliever37]

In the frame of the Earth. In a different frame, it would be a different distance.


The formula I gave works for any speed v and has been tested. We apply it to light with v=c.

So are you saying it will work for things besides light? Or it only works for light, when v=c?

 

[Polymath257]

So are you saying it will work for things besides light? Or it only works for light, when v=c?

The formula works for any v. In particular, it works with v=c.

More generally, if there are two events which are a distance x apart and a time t apart, as measured in some reference frame (coordinate distance and time), then the proper time between the events is

proper time = sqrt( t^2 - x^2 /c^2 ).

For the two events in question, this proper time will be the same value in *all* reference frames.

In this case, x=vt because B is moving at a speed of v in the reference frame of A. Plugging this in gives

proper time = sqrt ( t^2 - v^2 t^2 /c^2 ) = sqrt( t^2 (1 - v^2 /c^2 ) ) = t * sqrt( 1 - v^2 /c^2 ). This is the other equation we have been using.

So, yes, this equation works for *any* two events, whether on two ends of a light path or not. In the particular case of light though, v=c, or alternatively, x=ct. This gives a proper time for the path of the light of 0.

 

[TrueBeliever37]

The formula works for any v. In particular, it works with v=c.

More generally, if there are two events which are a distance x apart and a time t apart, as measured in some reference frame (coordinate distance and time), then the proper time between the events is

proper time = sqrt( t^2 - x^2 /c^2 ).

For the two events in question, this proper time will be the same value in *all* reference frames.

In this case, x=vt because B is moving at a speed of v in the reference frame of A. Plugging this in gives

proper time = sqrt ( t^2 - v^2 t^2 /c^2 ) = sqrt( t^2 (1 - v^2 /c^2 ) ) = t * sqrt( 1 - v^2 /c^2 ). This is the other equation we have been using.

So, yes, this equation works for *any* two events, whether on two ends of a light path or not. In the particular case of light though, v=c, or alternatively, x=ct. This gives a proper time for the path of the light of 0.

Would an example of two ends of a light path, be from star to earth? Please explain how you have two events(involving light) on that light path, that are both in the same reference frame.

 

[TrueBeliever37]

OK, let's get straight what this equation is saying.

In this, there are two observers. Observer A is measuring the 'observer time'. Observer B is moving at speed v with respect to observer A and is the one measuring the proper time (we want the proper time between events on observer B's path).

So, to use this equation, you need to know the relative speed, v, and *either* the time it takes for observer A or the time it takes for observer B. Notice that the distance traveled by B as measured by A is not mentioned, but if you call it x, then x=vt.

Isn't that what I did in my example? Observer A, measures the time it takes for light to get from star to earth. Observer B is in a ship moving at speed v with respect to observer A and is the one measuring the proper time. So when his ship reaches v=c, he measures proper time=0

What are you calling two events in this example, are you meaning the two different observations?