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

[Thermos aquaticus]

Is this a universal type equation that could be used to determine times for other things if speed of light was replaced with say speed of a boat crossing a lake?

As others have stated, you can't replace a physical constant in a physics equation with any old number you want and expect the equation to still work. c is a physical constant, the speed of light. If you want to measure the gravitational attraction between two bodies using Newton's formulas you can't replace the gravitational constant with whatever number you want. If you want to calculate how many molecules there are in a mole you can't replace Avogadro's number with any value you want. Scientists use specific values for physical constants for a reason.

 

[Polymath257]

You replied but didn't answer my question.

when I am in a car I am traveling whether I feel like I am stationary or not.

And this is the fundamental issue. There is no way to tell whether *you* are traveling along the road or the *earth* is traveling backwards with the road going under you. The laws of physics are exactly the same as long as you move with uniform velocity. Now, in this case, the gravity of the earth is relevant to some extent.

So, suppose you are in a huge spaceship (large enough to do any experiment you want) that is otherwise in a vacuum. There is *no* way to tell whether you are traveling or not. Any experiment you do would turn out the same whether you are 'at rest' or moving with constant speed.

You can measure acceleration (in special relativity), but you cannot measure uniform motion. This means there is no *absolute* meaning to 'traveling'. You can *always* just assume you are at rest and everything else is moving past you. The laws of physics can't tell the difference.

 

[TrueBeliever37]

You don't feel velocity. You feel acceleration. That's the point we are trying to make.

If we had a perfectly smooth and straight track and a car with really soft suspension and blacked out windows you wouldn't be able to tell if you were sitting still or travelling at a constant 60 mph. The only forces you would detect is when we accelerate the car to 60 mph or decelerate (i.e. hit the brakes) to slow down from 60 mph. Any frame with constant velocity is the same in physics. All of the physical constants and laws are the same for all frames of reference with constant velocity.

Right, I agree with that. I just think if you are traveling at a constant speed you still experience time and distance. You may not realize it , but you would be traveling even if blindfolded.

 

[TrueBeliever37]

The equations are found in one of my earlier posts. Here it is again:

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.

Isn't the limit = 0 you are referring to as v approaches c, actually the limit of time for the observer? and not for what you are calling proper time?

 

[Thermos aquaticus]

Right, I agree with that. I just think if you are traveling at a constant speed you still experience time and distance.

As long as you aren't going the speed of light, you will experience time and distance. Since you are made of matter that has mass you can never travel the speed of light, so that's kind of a moot point anyway.

You may not realize it , but you would be traveling even if blindfolded.

And you are travelling right now in many other reference frames, even if you are sitting in front of your computer. The Earth is rotating at a certain speed. The Earth is moving about the Sun at a certain speed. The whole solar system is moving around the center of the Milky Way at a certain speed. The Milky Way is moving at a certain speed around the barycenter of the Virgo Supercluster that we are a part of.

 

[Thermos aquaticus]

Isn't the limit = 0 you are referring to as v approaches c, actually the limit of time for the observer? and not what you are calling proper time?

That would be a question better answered by @Polymath257. My calculus-fu isn't as strong as it used to be. From my very limited understanding (pun intended), you reach limits when you have infinities and zero's in the denominator. If you put in c for the velocity in that equation you get a zero in the denominator, and I believe that defines the limit, although I could (and probably do) have this wrong.

 

[TrueBeliever37]

As long as you aren't going the speed of light, you will experience time and distance. Since you are made of matter that has mass you can never travel the speed of light, so that's kind of a moot point anyway.



And you are travelling right now in many other reference frames, even if you are sitting in front of your computer. The Earth is rotating at a certain speed. The Earth is moving about the Sun at a certain speed. The whole solar system is moving around the center of the Milky Way at a certain speed. The Milky Way is moving at a certain speed around the barycenter of the Virgo Supercluster that we are a part of.

Right, I agree about speeds relative to other things. I think it is just that if I were approaching the speed of light in a ship, the time that is approaching zero would have to be for me as an observer, relative to the light/photon. But the photon itself would still be traveling at c, and experiencing time and distance.

 

[TrueBeliever37]

That would be a question better answered by @Polymath257. My calculus-fu isn't as strong as it used to be. From my very limited understanding (pun intended), you reach limits when you have infinities and zero's in the denominator. If you put in c for the velocity in that equation you get a zero in the denominator, and I believe that defines the limit, although I could (and probably do) have this wrong.

No I think you are right, that is what defines the limit. But it should be the limit of time for the observer, and not the proper time as experienced by the photon.

 

[Polymath257]

Right, I agree about speeds relative to other things. I think it is just that if I were approaching the speed of light in a ship, the time that is approaching zero would have to be for me as an observer, relative to the light/photon. But the photon itself would still be traveling at c, and experiencing time and distance.

THE PHOTON DOES NOT EXPERIENCE TIME OR DISTANCE. There is no reference frame for the photon!!!

 

[Polymath257]

No I think you are right, that is what defines the limit. But it should be the limit of time for the observer, and not the proper time as experienced by the photon.

You have now *two* observers. the one on earth and the one in the spaceship. The v is associated with the relative motion of these two.

So, as v-->c, the proper time for the spaceship goes to 0. But so does the denominator. The *ratio* of these two things going to 0 is the (earth) observer time.

 

[Thermos aquaticus]

Right, I agree about speeds relative to other things. I think it is just that if I were approaching the speed of light in a ship, the time that is approaching zero would have to be for me as an observer, relative to the light/photon. But the photon itself would still be traveling at c, and experiencing time and distance.

You wouldn't notice anything different about the laws of physics if you were on a spaceship travelling near the speed of light. You would observe that the speed of light is ~3E8 m/s.

The observer that would notice something wonky going on in the spaceship would be an observer that sees the spaceship going by at near the speed of light. If that observer had a telescope and could see what was going on in the spaceship they would see that the spaceship is squished in the direction of its velocity. On top of that, the observer would see that the clocks on the spaceship are moving really slow.

 

[Thermos aquaticus]

No I think you are right, that is what defines the limit. But it should be the limit of time for the observer, and not the proper time as experienced by the photon.

Are we bouncing between the observer and the photon again?

 

[TrueBeliever37]

THE PHOTON DOES NOT EXPERIENCE TIME OR DISTANCE. There is no reference frame for the photon!!!

I know you keep saying that, but isn't the limit for time being evaluated, actually the limit of time for the observer?

What formula are you using to come up with time=0 for the proper time?

 

[Subduction Zone]

Here is a helpful suggestion so that you might understand the nature of the limit. You could put real numbers into the equations given to you and see which ways that they go. To make it easier you should express velocity as a fraction of c, that way the speed of light cancels out in the answer.

 

[Polymath257]

I know you keep saying that, but isn't the limit for time being evaluated, actually the limit of time for the observer?

Which observer? The one on the earth? Or the one is a ship going at some v, where we take the limit as v goes to c?

It is the observer *on the ship* that is the one we take the limit for. And the proper time *for the ship* goes to 0 as v-->c. The ratio of this proper time and sqrt(1-(v/c)^2 ) is always the earth time.

 

[TrueBeliever37]

Are we bouncing between the observer and the photon again?

No I am looking at the formula you gave me and asking which time the limit is applying to?

 

[TrueBeliever37]

Here is a helpful suggestion so that you might understand the nature of the limit. You could put real numbers into the equations given to you and see which ways that they go. To make it easier you should express velocity as a fraction of c, that way the speed of light cancels out in the answer.

I understand that, but the limit is for the observers time.

 

[Thermos aquaticus]

No I am looking at the formula you gave me and asking which time the limit is applying to?

The limit would be c. When you plug that number into the formula you get a zero in the denominator.

 

[Thermos aquaticus]

I understand that, but the limit is for the observers time.

The observer isn't at the limit, so why does it matter?

 

[TrueBeliever37]

Which observer? The one on the earth? Or the one is a ship going at some v, where we take the limit as v goes to c?

It is the observer *on the ship* that is the one we take the limit for. And the proper time *for the ship* goes to 0 as v-->c. The ratio of this proper time and sqrt(1-(v/c)^2 ) is always the earth time.

I agree it's the observer in the ship. As I said, if I was in a ship as I approached the speed of light, time would approach 0 for me relative to the light. But once again the light/photon would still be traveling at c and experiencing time and distance.